How old is the oldest light visible from Earth? Because light can only travel so fast, all of the light we see in the sky was emitted at a previous moment in time.  So if for example we see a supernova or some other great stellar event, by the time we see it, it maybe long over.  That made me kind of curious, what is the most ancient light we can see from earth?

The universe is supposedly ~13+ billion years old, but we are probably not at the very edge of the known universe so all the light we see is probably less than 13 billion years old.  So what is the oldest light we can see? and as an optional follow-up question how do we know the age of that light?

I guess the light itself may not actually be literally 'old', but its probably obvious what I'm asking here, put another way: what's the longest distance that now earth visible light emitted has traveled to reach the earth?  Though that reformation of the question gets kind of tangled with lensing effects.

The oldest light in the universe is the cosmic microwave background. Roughly 380,000 years after the Big Bang, protons and electrons "recombined"1 into hydrogen atoms. Before this, any photons scattered off the free electrons in the plasma filling space, and the universe was essentially opaque to light. Once recombination occurred, however, photons were able to "decouple" from the electrons and move through space unimpeded. This relic radiation is still observable today; it has been redshifted and cooled.

We can detect light from very distant objects, and we have. It makes more sense to talk about distance in terms of redshift; the larger the redshift, the farther away an object is. There are a number of extremely high-redshift objects, some of which have had their measurements confirmed, and others of which have not. Candidates include

MACS0647-JD (redshift $z=10.7^{+0.6}_{-0.4}$), a very distant galaxy.
UDFj-39546284 (redshift from $z=10$ to $z=12$), another galaxy or protogalaxy (although its distance is uncertain, and its exact nature is disputed).
GN-z11 (redshift $z=11.09^{+0.08}_{-0.12}$), a very luminous distant galaxy.

All of these objects would have formed some hundreds of millions of years after the Big Bang, however, so the light we see from them is much "younger" than that of the cosmic microwave background.

1 I've never liked the usage in this context, as this was the first time they combined; the "re" is kind of misleading.


What are Kepler's laws (as he wrote them)? There are of course many, many sources that quote Kepler's laws of planetary motion. This is preventing me from finding out what I really want to know: which is - what are Kepler's laws as he wrote them? As opposed to how they may have been later interpreted.

Very specifically I am interested in whether "Kepler's First Law", as proposed by Kepler, which is usually stated in the form that orbiting objects in the solar system execute elliptical orbits with the Sun at one focus, actually recognised, or had a caveat, that in a 2-body system it is the barycentre that is at the focus of an ellipse for both bodies, and that it is even more complex in the case of 3-bodies or more.

Short Answer: Kepler expressed his laws with the sun at a foci rather than a barycenter.
Long Answer: In Astronomia Nova (pub. 1609) Kepler presents the first version of something we can recognize as Kepler’s first law:
On page 285 of Astronomia Nova:

CAPVT  LIX.   Demonftratio, quod orbita MARTIS, librati in diametro
epicycli, fiat perfecta ellipfis: Et quod area circuli metiatur fummam
diftantiarum, ellipticae circumferentiae punctorum.

The translation by William H. Donahue New Astronomy is:

Chapter 59 Demonstration that when Mars reciprocates on the diameter
of the epicycle, its orbit becomes a perfect ellipse; and that the
area of the circle measures the sum of the distances of points on the
circumference of the ellipse.

It isn’t immediately clear from Kepler’s own statement of his first law if the sun or the barycenter is the occupied focus.  Indeed, I couldn’t even find a mention of a focus in Astronomia Nova.  The closest I could find is when Kepler says, “a planet is moved less vigorously when it recedes from the point whence the eccentricity is computed.”  Here, it appears that Kepler is referring to the fastest speed of a planet at Perihelion, and the slowest speed at Aphelion.  The point from "whence the eccentricity is computed" is also clearly the focus of the ellipse.  Here is one of Kepler's original woodcuts, and it can be seen that the sun is at the focus. Kepler used the metaphor of a row boat for the planet's motion, and his drawings of oars in the water complement this cutting.

Kepler believes the sun is the center of this system from “Chapter 33 – The power that moves the planets resides in the body of the sun.”  On page 280 of Donahue’s translation, Kepler says that:

Therefore, with the sun belonging in the center of the system, the
source of the motive power, from what has now been demonstrated,
belongs in the sun, since it too has now been found to be in the
center of the world.

Summary: Kepler believed the sun was the center of the solar system, and provided the motive power for the planets.  He used the location of the sun to calculate the eccentricity of Mars's orbit and later extended his findings to the rest of the planets.  While we can't rule out the possibility that Kepler speculated about a barycenter, he certainly didn't express any such speculations in his original formulations of his laws.
Notes:

Lack of explanation of Kepler’s stated 1st law: Kepler’s 1st law as stated in Astronomia Nova isn’t very recognizable as our modern Kepler’s 1st law.  Explaining how they are equivalent gets complicated due to the ancient math and breadth of terminology.  I do not attempt to bridge that gap in this answer, though I would be happy to provide insight, information, and resources on request.  At the end of Chapter 59, in a conclusion of the book, Kepler himself acknowledges the difficulty of reading his book: “If anyone thinks that the obscurity of this presentation arises from the perplexity of my mind, I shall myself only thus acknowledge to him my guilt, that I was unwilling to leave anything untested, no matter how utterly obscure”…  “He will see that there are some matters which no mind, however gifted, can present in such a way as to be understood in a cursory reading.  There is need of meditation, and a close thinking through of what is said.” If you want to understand what Kepler is talking about, you should buy a copy of Donahue's English translation and pack a lunch. I also found this guide quite helpful in my own journey: Tutorial.

Keplers 2nd and 3rd laws: Kepler’s 2nd law is also articulated in Astronomia Nova in 1609, but I didn’t choose to treat it in this response. I think it needlessly distracts from the deep point of the original question, which is whether or not Kepler included barycenters in his laws.  The 3rd law was published in Harmonices Mundi in 1618.  Though all three laws were asked for in the question, I don’t provide them in this answer, making it somewhat incomplete.

First mention of the barycenter:  The first mention of something like a barycenter that I could find is in a translation of Newton’s Principia, published in 1687, proposition 12, theorem 12: https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/BookIII-Prop2.  This includes the quote that: “the common centre of gravity of Jupiter and the sun will fall upon a point a little without the surface of the sun” which to me demonstrates a clear understanding and articulation of the idea of a barycenter.  Does anyone know an earlier clear reference to a barycenter?

Implications:  The fact that Kepler’s laws didn’t include a barycenter is not just a historical curiosity, but also informs our modern conventions.  A Kepler orbit assumes that a less massive body orbits around a more massive one in an ellipse, with the more massive body at one of the foci of that ellipse.  Though Kepler’s laws can be generalized to replace the sun with the barycenter, they are no longer recognizable as Kepler’s laws and will no longer conform to convention. All the orbital elements would be different.  For example, the periapsis and apoapsis are always measured from the central body, rather than the barycenter.  Here is a diagram from wikipedia:


Black Holes and Worm Holes Can a Black Hole enter inside a Worm hole? If yes then what will a observer observe if it present at the another side of Worm hole?

Wormholes are a theoretical solution to Einstein's equations, but, unlike black holes, they have never been observed and probably don't exist.

If they did exist, they would almost immediately close due to the presence of matter in them. To stop them from closing you would need some "exotic matter" with remarkable properties, like "negative mass". Exotic matter has never been observed, and probably doesn't exist.

Trying to push something massive, like a black hole close to a wormhole would almost certainly cause the wormhole to close. Perhaps if the wormhole were sufficiently big with enough negative mass to stabalise it it would be possible for the black hole to traverse it. The black hole would pass through the worm hole just like anything else. At the other end you would see the black hole coming through the wormhole.

But remember, fundamentally, wormholes don't exist.


On a Hertzsprung-Russell diagram, why is the main-sequence line squiggly? When I look at a Hertzsprung-Russell diagram, I'm able to see a main-sequence line from the most luminous main-sequence stars at the top left to the dimmest main-sequence stars at the bottom right. However, this line isn't straight - it's quite squiggly. Why is the line squiggly, and not defined by a polynomial or rational function?

Here is another plot of a Hertzsprung Russell diagram (luminosity versus temperature), but this time based on theoretical models. (The plot is from D. Prialnik 2000, An Introduction to the theory of stellar structure and evolution). Note that the zero age main sequence is well behaved in this plot. Luminosity and temperature are related by smoothly changing power laws that are largely due to the varying temperature dependence of energy generation and modes of energy transport within stars of different mass.

Your diagram is actually an absolute magnitude versus colour diagram. This is not the same thing. The transformation between absolute magnitude and luminosity and especially between colour and temperature are distinctly non-linear and dependent on what particular spectral features happen to be present in the photometric bands that are used to construct the plot. For example, there is a knee for cool M-dwarfs because the B-V colour "saturates" at a value of about 1.6 because there is hardly any flux in the B and V bands and their ratio doesn't change very much with temperature because of the characteristics of the opacity sources in the atmospheres of these stars at those wavelengths.

To demonstrate that the plot is not necessarily "wiggly", the plot below shows the colour-magnitude for the Hyades and Praesepe clusters (basically at the ZAMS) in the Gaia filter system (from Babusiaux et al. 2018). Nice and smooth - though it doesn't cover as big a luminosity range as the Hipparcos plot in your question; the main wiggle at B-V~0.3 is still in the Gaia data (off the top of the plot), but the M-dwarf "knee" is smoothed out by the redder filter response.


Why the continued obsession in measuring the year as a multiple of days? Pick a any notion of year and any type of day measurement.  Outside of history and the common cycle of day an night, why are these viewed as integrally or even fractionally commensurate?

Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks?  Why not decouple the two?

Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks?

Science exists to serve mankind, not to rule it. Calendars were one of the first concepts developed by humankind, arguably predating writing. Timekeeping is even more important to modern society than it was to the ancients. You're not going to get rid of calendars.

One key purpose of the Gregorian calendar is to keep the calendar in sync with the seasons. Every solar calendar requires some intercalation mechanism. The Gregorian calendar uses leap days. The Gregorian scheme (years divisible by 4 have an extra day, except for years that are multiples of 100 that aren't divisible by 400) will fall out of sync with the seasons at at the very slow rate of one day per 3030 years.

Leap seconds are a rather different beast. They are inserted, unpredictably, to keep the clock in sync with the day/night cycle. Without leap seconds, counting 86400 atomic clock seconds = one day would already have the clock out of sync by about half a minute. Leap seconds are problematic for business. There is an ongoing effort to remove leap seconds and just live with a day of 86400 seconds going out of sync with the day, at least until it truly becomes noticeable.

  Why not decouple the two and say that day X belongs to year Y if at 12:00 noon of that day is "in" year Y?

What's this "year Y" thing you're writing about? How do you know what year it is without a calendar?


How did the Ancients understand the ecliptic using the Ptolemaic system I studied the Farnese Atlas and I wonder how did the Ancients discover the ecliptic, the equator, etc. with the Ptolemaic system namely without knowing that the Earth rotates on itself, especially for the Ecleptic.

I ask the question because if in their model the Earth does not rotate, so how do they explain the two revolutions of the sun around the earth:

The first one being the daily cycle (sunrise and sunset)
The second one being the tropical year

With a model where the Earth does not rotate, I do not understand how they found out that the Sun make a revolution of a one year around the Earth (whence the ecliptic) with a second daily revolution. How could they conjugate these two Sun's cycles?

The ecliptic was understood to be the path of the sun relative to the sky.

Ptolemy believed the sky rotated daily around the Earth. All the planets, including the Moon and the Sun, shared in this motion.

However, the planets had secondary motions, relative to the sky. And in particular relative to the stars that they thought were fixed to the sky. This motion was discovered by careful observations of the sky, and was very well known, long before Ptolemy.

The motion of the stars was very simple: therefore when Ptolemy was describing his system, he only concerned himself with the motion of the planets (including the moon and sun) relative to the stars. It was clear that the planets all approximately followed a great circle relative to the stars. This great circle is the ecliptic. The Sun's motion was the simplest. It was exactly on the great circle, and it's position relative to the stars moved almost evenly. The other planets (including the moon) had orbits that were less even, but by introducing a system of Eccentrics, Deferents, Epicycles and Equants, Ptolemy was able to account for nearly all their motions.

The precession of the equinoxes was noticed because the Greeks used a calendar based on the heliacal rising (the date that stars are just visible before sunrise) of certain stars. And it was noticed that the heliacal rising was changing relative to the seasons and the equinox. We understand the reason for this to be the precession of the equinoxes.


Observing the night sky with a 6-year old We currently possess a Celestron NexStar 130 SLT computerized reflector telescope and have an opportunity to do backyard astronomy with my 6-year old daughter from time to time.

She is quite interested in space and astronomy, knows a lot of objects in the Solar System, which have atmosphere and what don't, knows major planet satellites and thinks there is life on Titan.

But, I find it quite challenging to actually do telescope observations with her:

it takes time to set things up, assemble the equipment and align the telescope. She is not patient enough and that sometimes "kills" all the fun and excitement
it is not easy for her to get used to looking through an eyepiece
it is easy to touch the telescope which leads to a shaking image

What are the general guidelines to observe the night sky with the 6-year old? What can I do to make the process less tedious, more fun and engaging?

(we also have the 15x70 Binoculars)

Your issues seem to be rather in line with what one would expect from a 6-year old getting in touch with observational astronomy. Actually, adults can sometimes experience these difficulties too when they first get in touch with it. I'm speaking out of experience of having shown objects of the night sky through a telescope to visitors. Some people just don't understand that when they touch the telescope, it moves. I would recommend the following guidelines to "fix" your proposed problems (expanding on hartacus' answer, which appeared while I was writing this):

Someone else could keep her busy while you set up the telescope. Since it is computerized, you won't waste time looking for objects and the only downtime will be due to the moving of the telescope. You can minimise this by planning your observation ahead of time to minimise the distance between two following objects.
As for the eyepiece, just tell her how to use it. She'll learn it in time. Even adults who have never used a telescope before generally need to be told how to look through it.
Most telescopes have some mechanism to lock the primary axes of the telescope mount. You can use this to make sure the telescope always points to the same spot. If you find that it is the tripod that is moving, you could try to find a way to lock that as well. Maybe surround its legs with loose bricks?
To minimise the last two difficulties at the same time, the best position to look through a telescope is to have both hands behind your back, move one eye as close to the eyepiece as possible without touching it, and close your other eye.

As a final remark on looking through the telescope, she might not be seeing the same thing as you when she looks through the telescope, because her eyes are different from you. I don't know whether this is applicable, but most telescopes have a way to adjust their focus to make the image as sharp as possible. People with near - or farsightedness generally require the focus of a telescope to be different.

These, however, are difficulties you will definitely overcome in time. When these are fixed, the most important part of the experience will be in the objects you show her. As hartacus mentioned, show her things she can relate to (planets are great), and tell her about the things she is looking at.
Apart from planets, star clusters can be impressive too. She'll surely be astonised if you tell her how many stars there are in one cluster. Some open clusters can even be thought of as 'micro-constellations' (more specifically, this one).
There are some nebulae / galaxies that are nice to watch, but I would restrain from them in the beginning because most will appear as blurry grey spots through an average telescope, which is not that impressive.

One last thing: you mentioned a binoculars. You can already see a lot with binoculars, although most impressive objects would be too faint. The main issue with the binoculars is that you can't aim them for her (unless you have a mount for them, which exists for some). If you can teach her to aim them herself however, there are some spots in the night sky which have 'mini-constellations': sets of stars in which you can recognize images, but they are too small to see with the naked eye, and too big to see with a telescope.


Why are retrograde orbits more stable than prograde ones? Regarding solar system dynamics, i.e. planets in stellar systems and moons in planetary systems, this is often mentioned in the literature, but it is difficult to find a good analysis/explanation of this phenomenon.

In any system of orbiting bodies, the length of time two bodies spend in proximity determines the mutual gravitational perturbation of each body on the other.

If both bodies are orbiting in the same direction, the inner body (closer to the central mass [primary star, for instance]) will overtake the outer one, because it is orbiting at a faster angular velocity. Because both bodies are orbiting in the same direction, they will have a great deal of time close to one another, and more time mutually to disturb one another. This is especially pronounced when the two orbits are adjacent and close.

On the other hand, if two bodies are orbiting in opposite directions, they spend very little time in close proximity, so gravitational perturbation is minimized. Passing one another happens very quickly.

By analogy, compare the amount of time two vehicles spend close to one another on a highway when one is passing the other and going in the same direction. A passenger in one could have time to study the other vehicle in great detail. Then, compare two vehicles moving in opposite directions; any passenger could only gain only a brief glimpse of the other vehicle up close. The amount of time for a passenger to study the other vehicle is analogous to the amount of gravitational disturbance one orbiting body will have on another.

A similar problem occurs when the orbital periods of two bodies are simple ratios of one another (orbital resonance). The repeated tugging of both bodies on one another at the same place in their orbits creates an additive distortion at that location in their orbits. This problem would occur in both situations -- orbiting in the same direction and opposing directions.

Though the reasoning is simple, the exact amount of perturbation is quite complex. A "simple" 3-body problem requires the use of advanced mathematics (celestial mechanics).


Is antimatter present on Earth? As we all know, antimatter is present across all of space. Is it also present on Earth? If it is present on Earth then by the interaction of matter and antimatter, why don't we feel the energy around us?

Antimatter is present on Earth and is being naturally created all the time (by Beta decay) as well as being created as product of cosmic ray collisions and in particle accelerators.

However the universe appears to be principally made of matter and so the antimatter thus created is annihilated: and although the production of antimatter through nuclear decay is essentially continuous, it is not of a high volume so the energy created in this way is not significant in the normal course of events. The real question is why is matter so much more abundant than antimatter: which is an open question though various theories have been proposed about why we might live in such a universe.


Where can I find data for fast radio bursts like FRB 180916? Where can I find Strength over Time for FRB 180916?
Or at least day+time of remarkable events.

Is there some good data for other FRB that hit the earth for the last 2 years?

There are parameters from the data listed here There is a site with raw data for FRB 121102, another repeating FRB, but the data set is 400 TB. The first site has links and may take you to the site where the data is stored. There is a Wikipedia page with your specific FRB listed among others. The sources listed for 180916 may take you to the data site.


We know what a nova is, but how? I work with astrophysicists and require some basic knowledge of many astronomical sources, however research priorities often demand that most of human knowledge on a subject be taken for granted.

I am currently researching galactic novae, and I find it relevant to summarize their history briefly when presenting my research to certain audiences. Unfortunately, I am unable to find any source material which describes how we know one key aspect of the events: that they are an accreting white dwarf in a stellar binary. This fact appears to be so well founded that no scientific paper feels obligated to cite it when stated, but basic resources like astronomical encyclopedia also make no reference that I've seen.

How do we know that novae are binary systems?

E.g., have follow-up observations clearly identified the white dwarf and its companion? Or do other astronomical measurements strongly confirm this binary hypothesis (and make it all but obviously true)? I apologize if it's as simple as "someone looked through a telescope, and it was pretty obvious" -- in my experience no revelation in astrophysics is nearly so simple, but certainly this could be the case.

Following a reference to Darley et al., ApJ 746, 61 (2012) from your Wikipedia link gives a (very technical) discussion of nova progenitors, including distinctions between nova systems where the secondary stars are main sequence or supergiant stars, and distinctions among white dwarfs with different chemistries.
The first sentence of that paper is

  A classical nova (CN) outburst occurs in an interacting
  binary system comprising a white dwarf (WD, the primary) and
  typically a late-type main-sequence (MS) star (the secondary)
  that fills its Roche lobe (Crawford & Kraft, 1956).

That suggests the 1956 paper is the original proposal for the Roche overflow model of the classical nova.
Like many original-idea papers, it's a pretty clear read.
But for your question, Crawford and Kraft seem to hedge about whether the "blue star" in their particular pair needs to be a white dwarf:

  [T]he
  observed
  luminosity
  of
  the
  blue
  star
  is
  essentially
  due
  to
  the
  energy
  released
  by
  the
  accreted
  material.
  This
  view
  is
  strengthened
  also
  by
  the
  fact
  that
  the
  blue
  star
  occupies
  a
  peculiar
  position
  in
  the
  H-R
  diagram.
  It
  lies
  10.5
  vis.
  mag.
  below
  the
  main
  sequence
  but
  about
  4
  mag.
  above
  the
  most
  luminous
  white
  dwarfs,
  whose
  effective
  temperature
  it
  exceeds
  by
  about
  8000°
  K.
  Unless
  the
  blue
  star
  is
  essentially
  degenerate,
  it
  can
  readily
  be
  shown
  that
  the
  small
  radius
  implies
  such
  a
  high
  internal
  temperature
  that
  electron
  scattering
  is
  the
  principal
  source
  of
  opacity.
  A
  simple
  calculation
  based
  on
  the
  standard
  model
  then
  yields
  a
  luminosity
  8
  mag.
  brighter
  than
  is
  observed.

In other words, Crawford and Kraft don't come out and say "definitely a WD," but if it's a non-degenerate star, it's a very strange one.
More modern observations of novae are compared to detailed models of the dynamics of the surface dynamics, models which have been debated vigorously for decades; the current generation of comparisons to data are sensitive to details like the amount of helium accumulating on the white dwarf's surface during the nova event. It seems unlikely that such details could even come close if the underlying assumptions about the basic physics of the erupting star were wrong.

Note that a classical nova system can be thought of as a type of contact binary star.
For any reasonable estimate of the size of the giant star, a distance of 10 AU between the two members of the pair seems like an overlarge estimate.
Ten astronomical units of separation viewed from a distance of 50 parsecs is already a gap of 0.1 seconds of arc. I wouldn't expect to see visible-light photographs showing both the giant star and the white dwarf, but rather that all of the information about the binary systems comes from spectroscopy.


How did Jocelyn Bell Burnell discover the periodicity of CP 1919? According to Wikipedia,
in 1967, a radio signal was detected using the Interplanetary Scintillation Array of the Mullard Radio Astronomy Observatory in Cambridge, UK, by Jocelyn Bell Burnell. The signal had a 1.337302088331-second period and 0.04-second pulsewidth.[4] It originated at celestial coordinates  19h 19m right ascension, +21° declination. It was detected by individual observation of miles of graphical data traces. Due to its almost perfect regularity, it was at first assumed to be spurious noise, but this hypothesis was promptly discarded. The discoverers jokingly named it little green men 1 (LGM-1).
Also from Wikipedia,

My question is the signal was just measured from the chart directly?

Yes - at that time, the chart was the source of data used for analysis. Jocelyn visually reviewed hundreds of metres of traces every night looking for quasars (the reason the Interplanetary Scintillation Array was built)
The wiggle looked different from expected signals, and in fact was misidentified as interference at first, but when she noticed it again, and again, she conducted further analysis, identifying its repetition rate of 1.337 seconds.
From the New Yorker:

Bell Burnell - who analyzed the roughly seven hundred feet of paper generated each week as galactic radio waves were recorded in inked peaks - noticed a faint signal arriving from one slice of sky. Then it disappeared. In November, she saw it again. By adjusting the speed of the recording device, she determined that the signal came in every 1.34 seconds, a regular beat against the background static of the cosmos.

All from visual measurement of the trace!


Why black holes look black to us? Why black holes look black to us?
The question might sound weird but my head stuck in it. As we know objects that fall into the black holes seem to remain unchanged in its event horizon due to the nature that lights cannot escape from it. For example, in the below image (image with white astronaut) of the black hole we will see the astronaut remains in its event horizon.

Now, we also know that the black holes are constantly pulling all the interstellar objects & meteors around it, thus if these objects are grabbed by the black hole, we are supposed to see many of these objects remain unchanged on its event horizon surface hence the color of the black hole is not supposed to be black instead the distorted color by the objects (i.e. below image). But still we see its color a pure black in most of the representations and photos.

So why do we see black holes black instead of distorted colors of the space objects?

When the astronaut is far outside the black hole we can see the astronaut normally.
As the astronaut falls towards the black hole, things get strange. We don't see the astronaut pass the event horizon, instead as the astronaut falls towards the event horizon, he will seem to be time dialated.  The clock on his wrist will seem (from our perspective) to run slower (he won't notice this from his perspective, the watch is normal).  As he gets closer to the event horizon, light reflected off him will be more and more redshifted.  As he gets closer to the event horizon the light will become redshifted to infra-red (at which point he will appear invisible) then microwaves then longer and longer wavelengths carrying less and less energy.  He will turn black and invisible to us.
So you can't see the objects that have fallen to the black hole.
From the point of view of the astronaut, he falls down and in a short amount there is a singularity in the equations. It seems that his body is compacted into a single point of infinite density, it seems that general relativity can't yet describe this state, which will require a quantum theory of gravity.


Is there a consensus as to where terrrestrial planet atmospheres in our solar system came from? I tried to research this and I'm not sure if there is a consensus where the $CO_2$, $N_2$, etc. comes from on terrestrial planets like Venus, Earth and Mars. Possible sources would be accretion from the protoplanetary disk, outgassing (which I would think would require plate tectonics/ molten core) or reactions at the surface caused by stellar radiation. Also, reactions in the atmosphere can alter the composition. Would it be a combination of all of the above and, if so, do we know if one of the processes dominated over the others? Does anybody have a good link that discusses this?

Three decent answers in comments converted to a community wiki.

A quite recent study about the origin and development of terrestrial planet atmospheres is the 2015 paper "The Atmospheres of the Terrestrial Planets: Clues to the Origins and Early Evolution of Venus, Earth, and Mars" (Baines et al.) includes a fairly comprehensive analysis of chemical evidence from Earth and from Martian meteorites and observations of the atmosphere of Venus.

The general consensus is that much of the atmospheric gases were brought through comet and asteroid impacts.

The pre-planetary disk would have been centrifugally sorted. Many iron, stone, carbonacious meteorites, watery, ice-iron-rock agglomerates would have brought a very diverse amalgam of different stones, ices, and room temperature gas elements, throughout the formation. CO2 and N2 objects are a bit less stable in direct solar radiation at 1 AU. CO2 and N2 would have pounded into the world in considerable quantities throughout the accretion process, with a wide range of particulate sizes from microns to kilometers in diameter without a marked difference in element type from start to end.


What is the current accepted hypothesis of what caused Triton's retrograde orbit around Neptune? According to the NASA overview about Triton, this, the largest satellite of Neptune is unique in that it has a retrograde orbit around Neptune.  The page also suggests that Triton is a captured Kuiper Belt object.

What is the current accepted hypothesis as to what caused Triton's retrograde orbit?  Also, is this orbit sustainable - meaning, is it degrading?

According to this source from SOA/NASA Astrophysics Data System on Tidal evolution in the Neptune-Triton system (Chyba, C. F., Jankowski, D. G., & Nicholson, P. D.),

  Transcript: Many investigators have speculated on the reason for
  Triton's retrograde orbit. Lyttleton (1936) suggested that both Pluto
  and Triton originated as prograde satellites of Neptune, only to
  experience a catastrophic gravitational interaction. McCord (1966)
  showed that tidal evolution could have brought Triton inward from a
  near-parabolic orbit, lending plausibility to a capture origin for the
  satellite. Jarrington and Van Flandern (1979) suggested Triton's
  peculiar orbit and Pluto's supposed "escape" were caused by ab
  encounter with a massive "rogue" body, a conclusion contested ny
  Farinella et al. (1980). Finally, McKinnon (1984) combined momentum
  and energy constraints with an improved knowledge of Pluto's mass to
  show that all Pluto-Triton interaction scenarios for the origin of
  Triton's retrograde orbit are untenable, leaving the capture
  hypothesis as the most likely possibility.

Now, that summarizes all the theories one might have about it, at least all I could find.

As far as evolution of the system is concerned, the same document follows up to it and demonstrates how tidal effects are responsible for the evolution of the system. "Degrading" is a strange term. Let us go with evolution.

One of the possiblities stated is that Triton will reach a perfectly retrograde state and stay in it for ever! For a detailed analysis, you may read up section 6 in the link provided, "The future of Triton"


When the set of official dwarf planets is expected to increase? When the set of official dwarf planets is expected to increase or new bodies considered for inclusion?

As I noted in a previous answer, the official body for naming and classification of astronomical objects is the International Astronomical Union, or IAU. They are the ones responsible for designating bodies as "planets," "dwarf planets," etc. The scientific community recognizes them and their decisions as "official."

The IAU meets periodically for General Assemblies to discuss general issue; the next such meeting is scheduled for 2015. The topics discussed may range widely, covering all sub-fields of astronomy. Unfortunately, I don't know the exact topics for discussion, and while I imagine there will be a wide range, issues relating to dwarf planets may not be covered. However, the IAU does meet in between these meetings to discuss a variet of topics.

That might answer your question, but if you're talking about the possible discovery of dwarf-planet-like bodies, I can't help you. There are probably many on-going efforts to search for small bodies within our solar system, but I don't know of any recent discoveries of possible dwarf planet candidates.

I hope this helps.


Does the recent news of "ten times more galaxies" imply that there is correspondingly less dark matter? Nature: Universe has ten times more galaxies than researchers thought

NASA feature: Hubble Reveals Observable Universe Contains 10 Times More Galaxies Than Previously Thought

Headlines sometimes oversimplify. But if this is really true, that there seems to be about ten times more galaxies than had been generally assumed, does this impact other assumptions? Does this imply the universe is ten times more massive, or just that there is less dark matter than was previously thought.

All Conselice et al. (2016) appear to suggest is that when you look at something like the Hubble deep field, there are many faint (and presumably low mass) galaxies that are not seen. This has absolutely no effect on the need for dark matter.

The main results are: (i) as you look back in time, the overall (co-moving) density of galaxies (more massive than a million times that of the Sun) increases. (ii) But the density of more massive galaxies actually decreases. This is consistent with hierarchical merger picture where small galaxies merge to become larger galaxies.
This really doesn't have any influence on the need for dark matter.

First, the presence of dark matter is inferred from many different observations. Some of these (e.g. galaxy rotation curves) are not influenced at all if there are lots of extra galaxies.

Second, the "missing" galaxies are at high redshift, not (or not all) in the present day universe, so they cannot significantly affect a calculation of how much normal matter there is in the universe today. Presumably, many of these small galaxies then merge to become larger galaxies and the total mass is conserved.

Third, just because there are lots of them does not mean they contain much mass anyway. The "mass function" (number density as a function of mass) of galaxies goes roughly as $\phi(M) \propto M^{-1}$ at low masses. This means the mass contained in any interval is
 $$M_{\text{tot}}\propto \int^{M_2}_{M_1} M\phi \ \mathrm{d}M \ = M_2 - M_1$$
So although, low mass galaxies may be ten times more frequent, they are ten times less massive and so don't change the total mass very much. I will need to read the paper more carefully to see if the authors are suggesting that low mass galaxies are much more common in the early universe than was already supposed.

Fourth, primordial nucleosynthesis calculations tell us that only 4 per cent (as a fraction of the critical density) of the energy density of the universe is in the form of baryonic mass. Observations of gravitational lensing, cluster dynamics and the cosmic microwave background tell us that the mass density is actually around 30 per cent of the critical density. Thus most of the dark matter is non-baryonic and cannot be in the form of missing faint galaxies, or any other form of normal baryonic matter.


How much do the mascons on the moon affect surface gravity? The gravity anomalies on the moon (generally accepted to be caused by mascons) are powerful enough to affect the orbit of satellites.  I'm trying to determine how much effect they would have on a person standing on the surface.  If we experience about 17% of Earth's gravity on the lunar surface, how much would that differ within one of these anomalies?

Upon some google-ing and wiki-ing I found this image of a gravity map of the moon:

That scale up the top is measured in milli-Gal which is thousandths of a cm/s^2. For scale gravity is ~9.81m/s^2 which equals ~981000mGal. The difference between gravity at sea level and the top of mount everest is 2Gal, or 2000mGal, which is 0.2% of average gravity. On the moon, the variation seems to be about 1000mGal, with the average being 162000mGal, which is about 0.6% of average gravity.

In conclusion, on the surface, there is no noticeable difference. Not to a human anyway.


If space is negative energy and matter is positive energy then does that mean the universe is finite? I was reading a book by Stephen Hawking, it was written that universe is made up of 2 ingredients,i.e., space and energy and space was the negative energy, so they add up to 0. But since space is energy does that mean it is finite?

As everyone is saying, this is a very loose "plain English" version of some complex ideas, but if space was infinite, you could view this suggestion (that the overall total energy of the universe is zero) as applying to any sufficiently large region of space. So if you pick any 10 billion lightyear diameter sphere you would expect to find the energy of the matter in that sphere more or less cancelled out by the gravitational potential energy arising from the separation of that matter.


Calculating Field of View of a Telescope Given Time of Another Object Given the celestial coordinates of a certain star and the time it takes to pass across your telescope's eyepiece, how can you calculate the field of view of the telescope?

This technique for determining the field of view of a telescope and eyepiece is called drift timing. Since the sidereal day is about 4 minutes shorter than the usual solar day, a fixed point on the celestial equator rotates at about 15.04 arc seconds per second of time. Thus the true field of view in arc seconds is given by
$$TFOV=15.04 \hspace{2pt} t \cos(\delta)$$
where $t$ is the time in seconds for the star to drift across the field, and $\delta$ is the declination of the star. Divide by 60 to convert to arc minutes, or by 3600 to degrees.


Path of sun calculation using java if I am not using very odd programming language for Astrophysicists, here is my code for calculating path of sun. I am following this reference

I want to calculate the local time as a function of zenith (e.g. sunset, sunrise, golden hour etc)

import java.time.LocalTime;
import java.util.Calendar;
import static java.lang.Math.acos;
import static java.lang.Math.cos;
import static java.lang.Math.sin;
import static java.lang.Math.tan;

/**
 * Created by rudra on 17/02/18.
 */

public class PathOfSun {

    static public double getTime() {
        LocalTime loacltime = LocalTime.now();//.getHour();
        double tzone = 5.5; //TODO: placeholder, will be calculated
        //    int localhour = time.getHour();
        int hour = loacltime.getHour();
        int min = loacltime.getMinute();
        int sec = loacltime.getSecond();
        double lat  =  28.6;//latlang.Lat;
        double lang =  77.2;//latlang.Lang;
        Calendar calendar = Calendar.getInstance();
        int dayofyear = calendar.get(Calendar.DAY_OF_YEAR);
        int daysthisyear = calendar.getActualMaximum(Calendar.DAY_OF_YEAR);
        double pi = Math.PI;

        double gamma = (2 * pi / daysthisyear) * (dayofyear - 1 + (hour - 12) / 24);
        double eqtime = 229.18 * (0.000075 + 0.001868 * cos(gamma) - 0.032077 * sin(gamma) -
                0.014615 * cos(2 * gamma) - 0.040849 * sin(2 *gamma));
        double decl = 0.006918 - 0.399912 * cos( (gamma)) + 0.070257 * sin( (gamma)) -
                0.006758 * cos(2 *  (gamma)) + 0.000907 * sin(2 *  (gamma)) -
                0.002697 * cos(3 *  (gamma)) + 0.00148 * sin(3 *  (gamma));
        double toffset = 4 * lang*60 + 60*(tzone) + eqtime;
        double tst = 60 * hour + min + sec / 60.0 + toffset;
        double ha = (tst/(4*60)) - 180;

        // Zenith
        double phi = acos(sin(Math.toRadians(lat)) * sin(decl) +
                      cos(Math.toRadians(lat)) * cos((decl)) * cos(Math.toRadians(ha*60)));
        // Azimuth
        double theta = pi - (acos((sin(Math.toRadians(lat)) * cos(phi) - sin(decl)) / (cos(Math.toRadians(lat)) * sin(phi))));

        // Calculating the hourAngle as in second page
        // At sunrise
        double ha2 = acos(cos(Math.toRadians(90.833))/cos( Math.toRadians(lat))*cos((decl))-
                tan(Math.toRadians(lat))*tan((decl)));

        double stime = (720 - 4*(lang + Math.toDegrees(ha)) - eqtime)/60.0;

        System.out.println("DayOfYear"+ dayofyear+ "  DaysThisYear"+daysthisyear);
        System.out.println("lat "+lat+"  lang "+lang+" ltime "+ loacltime);
        System.out.println("eqtime "+eqtime+"  ha "+ha+" , "+ Math.toRadians(ha)+ "  ha2 " + ha2+
                "\ndecl "+Math.toDegrees(decl)+" azimuth "+ Math.toDegrees(theta)+" Zenith "+Math.toDegrees(phi));
        System.out.println("Time of Sun "+ stime);

        return stime;
    }
}

The problem is its giving me more or less OK value for equationoftime and declination, but azimuth etc are grossly incorrect.

ANy idea where I am going wrong? I am following this page for reference.

The NOAA formula sheet
expresses some quantities in radians, some in degrees, and some in minutes of time.
It does say which is which, but it's easy to miss. A few minor code changes can help expose bugs:

Store angles in radians. Trigonometric expressions free of Math.toRadians calls are easier to verify against a reference document.
Put that document's URL in a comment so it's easy to find.
If you need to store something in other units,
make it obvious
with a naming convention such as long_deg for longitude in degrees.

The other quirks are the $\cos(\phi)$ and $\cos(180 - \theta)$
on the left hand sides of those equations.
$\arccos$ has a restricted range of $[0, \pi]$,
which is OK for zenith angle but leads to 180-degree errors in azimuth
unless you code east/west cases or derive new formulas where you can use Math.atan2.

The solar hour angle should equal 0 at noon and increase by 1 degree every 4 minutes.
Sunset hour angle should be about 90 degrees near an equinox, smaller in winter, larger in summer.

The sunrise/sunset formulas produce times in minutes past midnight UTC, so you'll need to apply a time zone offset to get local times.


How gently could a comet/asteroid/meteorite "hit" Earth? Could an object from outer space with the right velocity and orbit come into contact with the surface of our planet in a "slowish" manner?

Yes, spacecraft do it all the time by using air resistance (and sometimes rockets) to slow down.
Meteorites enter the atmosphere at high speed, typically 10-70 km/sec, but the smaller ones are slowed by air resistance, so they typically hit the surface at just a few hundred kilometers per hour. Reference: http://csep10.phys.utk.edu/astr161/lect/meteors/impacts.html (that link is now dead).
As gerrit points out in comments, a sufficiently small micrometeroid, like a dust particle, probably could hit the surface gently.
Larger bodies are not affected nearly as much by the atmosphere. For any such body, the velocity at impact is going to be at least several kilometers per second. A body moving obliquely in the same direction as the Earth's rotation would have a slightly lower speed at impact, but the impact still could not be described as "gentle"; the Earth's rotational speed is still a small fraction of orbital or escape velocity.
Here's one way to think of it. Freefall trajectories are reversible. If you watch a movie of an incoming body in reverse, it still makes physical sense (ignoring air resistance). Any meteorite hitting the surface has to have been in deep space at some point before the impact. If there were a trajectory that allowed such a body to have a "gentle" impact speed, then it would be possible to start with the same body near the surface at the same "gentle" speed, but in the opposite direction, and have it reach deep space. Unless the meteorite has its own propulsion system, that's just not going to happen.
You can get to space with a lower starting speed by moving along with the Earth's rotation - which is why most rockets are launched to the east, to take advantage of that. Reversing such a trajectory can result in a slightly slower impact, but only slightly.


Is it possible to know if a planet is located in the Habitable Zone knowing it's properties? Is there an equation that can give me the possibile distance of the planet from his star knowing planet's temperature (or other datas)?

I found this method for calculating the Habitable Zone but it's based only on star's properties.

The habitable zone is, roughly, defined as the region around a star in which a planet could have liquid surface water. So if the surface temperature of the planet is between 0 and 100*C then the planet is in the habitable zone. The converse, however, is not true.

The reason for this definition is that the sort of chemical reactions that are needed for life work best when there are complex molecules dissolved in some kind of solvent, and the best polar solvent is H2O.

If you know the distance a planet is from its star, the temperature can be estimated from the properties of the star and the atmosphere of the planet. Not every planet in the habitable zone actually has liquid water. Venus, Earth and Mars are all in the Habitable zone of the sun, since they could all have liquid water if they had the right atmosphere. But of course only Earth actually does have liquid water in any substantial amounts.

So just being in the habitable zone, doesn't make a planet certainly suitable for life, and even if water does exist on the surface, it does not mean that humans could inhabit the planet.


Two black holes photo interpretation Please see this image:

I wonder two things about these black holes. The first is, what are the oblong, smaller black hole-looking voids outside the perimeter of the major black holes? Also, why are these black holes not entirely circular, given that their mass is concentrated in a point?  Is this just visual distortion?

The image is about a merger of two black holes, before the actual merger takes place.
The two largest of the dark regions are slightly distorted images of two the black holes. Distortion is due to light being bended asymmetrically by the combined gravity of the two black holes.

The two smaller dark areas show the respective other black hole in a strongly distorted version, since the light - or missing light - of the other black hole is bended around the respective less distorted appearing black hole by about 180 degrees, somewhat similar to an arrangement of uneven mirrors.

You may interprete the phenomenon of the smaller dark areas as an extreme version of Einstein rings.


Effects of the solar tide on planets It’s well known that (lunar) tides on Earth result in a transfer of angular momentum from Earth proper to the Earth–Moon orbital motion. That’s why the Moon resides now in a high Earth orbit, and Earth’s axial rotation is nowadays much slower than in the past (Hadean, Archean etc.).

What about effects of solar tides, indeed? The angular momentum certainly must be transferred from Earth proper to the Earth–Sun orbital motion. Does it result in a tiny increase of (Earth+Moon)’s heliocentric orbital angular momentum over millenia? What about Mars and Ceres: does the solar tide brake their axial rotation and change slightly their orbits? And what about Venus: does its solar tidal braking explain the slowness of rotation completely?

Addendum. The solar tide may not be confused with “dual” effect: planetary tide on the central star. It exists as well and has much significance for (extrasolar) planets on near orbits — but this question is not about tides on Sun. Ī̲’d accept an answer assuming gravity of the Sun to be spherically symmetric, neglecting Sun’s deviation from spherical symmetry, as well as rotation (even going to account for General Relativity). The solar tide is interaction between this gravity and the orbiting object. By Newton’s laws, gravitation affects Sun’s momentum reciprocally (and, hence, velocity), and we may ignore here any effect on Sun beyond its translational motion (momentum/velocity/position).

You're right; the effect of solar tides on the Earth would be to increase the Earth's (or Earth+Moon's) orbital angular momentum around the Sun. But it would be difficult to compute this properly, and probably impossible to measure the effect, because of all the other small alterations to the Earth's orbit going on.

A simple way to see the effect (which I gather you've already worked out) is to imagine two equal-mass objects in circular orbit about their common center of mass (midway between the two), one effectively a point mass (e.g., black hole) and the other being the Earth. If all motion is prograde (Earth rotates in same direction as orbit) and the Earth rotates faster than the orbital period, then the Earth's tidal bulge will be slightly ahead of the Earth-Sun direction. Then the near-side bulge will experience a retarding force from the point mass (slowing the Earth's rotation) while simultaneously pulling on the point mass in approximately the same direction as the latter's orbital motion. So the point mass is accelerated, adding to the angular momentum and making the orbit larger.

I believe that most attempts to estimate the solar tidal torque on the Earth end up with it begin about 5 to 10 times weaker than the lunar torque, so most of the slowdown in the Earth's rotation is still due to the Moon.

Of course, in the real Solar System the Sun also experiences a tidal bulge raised by the Earth, which (because the Sun rotates faster than the Earth's orbit) will accelerate the Earth forward in its orbit, also adding to the orbital angular momentum. But then you have remember that Venus and Mercury will raise larger tidal bulges on the Sun with different periods (relative to the Earth's orbit), and so the combined effect gets complicated.

As for Mars and Ceres: yes, in principle the same effect would apply. But it would be weaker for Mars (smaller planetary radius, larger distance from Sun) and much weaker for Ceres (same reasons, except moreso).

In addition, all of the orbits are being affected by N-body interactions with the rest of the planets (especially Jupiter), which I strongly suspect means that any potential changes in orbital angular momentum due to solar tides would be washed out by the various N-body and resonant interactions.

(My understanding is that solar tidal braking could explain the current rotations of both Mercury and Venus, though you really can't be sure because you don't know what their initial rotations were.)


How did Arecibo detect methane lakes on Titan, and image Saturn's rings? This answer to Farthest distance to a solar system object that's been measured by radar? mentions that Saturn's rings, and the Uncover Travel post Arecibo Observatory, Puerto Rico – The World’s Largest Radio Telescope For Over 50 Years mentions:

Among other accomplishments of the Arecibo Observatory are:

Direct imaging of an asteroid for the first time in history.
Discovery of water ice deposits at Mercury’s poles.
Tracking of near-Earth asteroids to monitor impact risks.
Mapping the cloud-covered surface of Venus.
Radar imaging of the rings of Saturn, revealing new details of the ring structure.
First detection of methane lakes on Titan, a moon of Saturn.
First detection of an asteroid with a moon.

Question: How did Arecibo detect methane lakes on Titan, and image Saturn's rings? These are pretty remarkable feats from Earth for a single radiotelescope. How were they done? Can citations be found and the examples of the image of the rings and methane lake evidence be shown?

Titan "lakes":

Published Open Access in Science: Radar Evidence for Liquid Surfaces on Titan Campbell, D. B., Black, G. J., Carter, L. M., and Ostro, S. J., Science 302, 5644, pp. 431-434, 17 Oct 2003 DOI: 10.1126/science.1088969

This was a really elegant experiment! A continuous, unmodulated, circularly polarized 13 cm wave was broadcast from Arecibo towards the Saturn/Titan system, and the Doppler shift was used to isolate the returned signal from Titan.

Most of the surface is rough, so there is signal returned from areas all over Titan's disk, and since the moon rotates, albeit slowly, returned power from the "left" and "right" sides are shifted to higher and lower frequencies.

However during some observing times there was a very strong and pronounced reflection with zero Doppler shift with respect to Titan's known radial velocity, and this peak is attributed to specular reflection. Checks on received polarization confirm that while the power from the rough surface is returned in both circular polarization state, the presumed specular component is only in the expected circular polarization state.

As pointed out in @Martin Kochanski's thoughtful answer there is no determination from the radar observation that the returned specular reflection comes from methane. This is simply a presumed component of the presumed lakes, based on known information about Titan's chemistry at the time (2003).

  We observed Titan on 16 nights in November and December 2001 and on 9 nights in November and December 2002, transmitting at 13-cm wavelength with the 305-m Arecibo telescope and receiving the echo with Arecibo. Titan's rotational and orbital periods are 15.9 days, and our 2001 observations were obtained at a uniform 22.6° (∼800 km) interval in longitude. The 9 observations in 2002 did not provide uniform coverage. The latitude of the subearth track was 25.9°S in 2001 and 26.2°S in 2002, its farthest southern excursion. The round-trip light time to the Saturn system during the observations was 2 hours 15 min, and the limited tracking time of the Arecibo telescope meant that signal reception was restricted to ∼30 min per day, corresponding to 0.5° of Titan rotation (20 km of motion of the subearth point). On one night in 2001 and for most of the 2002 observations (as well as others when we were attempting ranging measurements to Titan), the 100-m Green Bank Telescope (GBT) was also used to receive the echo for the full round-trip time. These data have lower signal-to-noise ratios than those obtained with Arecibo receiving the echo, but the longer receive time corresponding to 2.1° of Titan rotation allowed more subearth locations to be studied.

Here is some of the Titan data:

  Fig. 3. The OC radar echo spectrum at 1.0-Hz resolution for the 2002 observation at the subearth longitude of 80°. The normalized cross section for the specular component of the echo and the RMS slope are 0.023 and 0.2°, respectively.

 Click for full size

  Fig. 1. Arecibo radar echo spectra from the 2001 data for five subearth longitudes on Titan. Spectra are shown for both the expected (OC) sense of received circular polarization and the cross-polarized (SC) sense. The ordinate is in standard deviations of the noise. The limb-to-limb Doppler-broadened bandwidth for Titan is 325 Hz. Four of the OC spectra show evidence of a specular component at 0 Hz.

Saturn's rings "imaged" (delay-Doppler):

From Radar imaging of Saturn’s rings Nicholson, P. D. et al., Icarus 177 (2005) 32–62, doi:10.1016/j.icarus.2005.03.023

The "image" below is not a conventional image, since the Arecibo dish has no way to spatially resolve the transverse expanse of Saturn and its rings. It is a "delay-Doppler" image, using 12.6 cm, ~500 kW radar broadcasts transmitted by Arecibo. The round-trip light time was about 135 minutes. Because Arecibo has limited steering away from the zenith (<19.7 degrees) max Saturn was only available to the dish for 166 minutes even under ideal conditions.

The vertical axis shows a delay of about +/- 800 milliseconds which demonstrates spatial resolution, but in the radial or depth direction. The horizontal axis is Doppler shift. The +/- 300 kHz shift represents the orbital velocity of particles in the rings.

While the Titan specular reflection above was done with a continuous or CW beam, the delay-doppler imaging technique requires a frequency modulation of the beam with a frequency-hopping pattern. By applying a correlation function using the known  pattern to the recorded received signals, components with different return times and different Doppler shifts can be extracted, and the results is then hzstogramed, producing the delay-Doppler image below.

This is a standard technique and has been used to image other planets and asteroids:  See the following items and refereces within:

this answer to What causes “North-South ambiguity” when doppler radar imaging a planet surface equator on?
Why do radar maps of the surface of Venus have missing slices?
What is the physical geometry of this apparent “eclipse” of a tiny moon of Asteroid Florence?
Is delay-doppler radar imaging of NEO asteroids possible only if it spins fast enough?

  Fig. 2. Delay–Doppler images constructed from data obtained in (a) October 1999, (b) November 2000, (c) December 2001, and (d) January 2003.
  Both OC and SC polarizations were combined to maximize the signal to
  noise ratio. Note the four bright regions in each image where the delay and
  Doppler cells are parallel and where the A and B rings appear to cross over
  one another.


Does supermoon's white light splits into seven colours I took a photo of SuperMoon 2016 and it was very shiny, the light around it showed seven colours. Was it my camera's effect or was it the moonlight's true colours.

The answer by @comprehensible nicely states that whatever the effect might be, it is related to your camera. Of course the moon's light has a spectrum all the time, just like the sun, but we don't see it until we look through some dispersive material or object.

I thought I would just add a little back-up information as well.

I took your photo and rotated, cropped, and flipped it left-right on my laptop, then made 1D plots of the three color channels (R, G, B) normalized to local (R+G+B) by projecting down to the new x-axis. Note that this is very unscientifically done, but it's still interesting.

If the dispersion were diffraction, then moving away from the moon you'd expect to see the order B, G, R, B... since diffraction peaks are at larger angles for longer wavelengths. It would repeat if there were multiple orders. Of course this is messy geometry and the source (the moon) is very wide, so this is only guessing.

Surprisingly, this is the order that appears! So it's possible that there is some periodic pattern somewhere on the lens or in the camera. Since the characteristic size of one cycle is of the order of the diameter of the moon or about 0.01 radian, if there scratches or streaks on the lens the period owould be of the order of 100 wavelengths. There is an obvious periodicity in the camera's CCD (assuming it's not a film camera) but then the ray tracing would be more complicated (multiple reflections) and this simple analysis wouldn't work.

This is just speculation, and thin film interferece can also produce color effects also, and non-uniform thickness (like a soap bubble) can make color gradients as well.

Also note, our perception of color is a complex subject, and while our eye and brain together try to reconstruct a rich color behavior, the simple up and down wiggles of the intensity in the three color channels of your camera's sensor is all that is actually in the photo.

The color modulation is quite strong, so I am guessing that whatever is producing the 1 directional flair is also the thing that has the periodicity.

Why don't you try a similar experiment and point at strongly overexposed LEDs or maybe street lights (Sodium or Mercury vapor) and see if you get a clearer effect. This would be just for fun. This is just an artifact of your camera and has nothing to do with the moon.

Here's the Python script I used to plot the image after I rotated/cropped/flipped it on my laptop:

import numpy as np
import matplotlib.pyplot as plt

fname = "Moon Spectrum.png"

img = plt.imread(fname)

img1D = img.sum(axis=0).T[:3]

img1Dnorm = img1D/img1D.sum(axis=0)

plt.figure()

plt.subplot(2,1,1)
plt.imshow(img)

plt.subplot(2,1,2)
for thing, color in zip(img1Dnorm, ['red', 'green', 'blue']):
    plt.plot(thing, color=color)

plt.xlim(0, img1Dnorm.shape[1])

for lab, x, y in [('B',  58, 0.35), ('G',  80, 0.39), ('R',  97, 0.31),
                  ('B', 126, 0.36), ('G', 141, 0.46), ('R', 166, 0.40) ]:        plt.text(x, y, lab, fontsize=14)

plt.show()


What distance is considered 'close' for the 9th nearest neighbor of a galaxy? I am currently working on determining the nth nearest neighbors of a large survey of galaxies. I am not an astronomer by trade, so I am unsure of interpreting some of my results. I have discovered for the sample of galaxies their 9th nearest neighbor distances. It turns out the mode of this distribution by far is between 2mpc and 3mpc. Since I am unsure of the scale of these implications I would prefer if a well versed astrophysicist can tell me if the distance to the 9th nearest neighbor galaxy of 2-3mpc would be considered densely clustered. These galaxies are at intermediate redshift.

If galaxies were randomly distributed (a spatial Poisson process), then the probability of having $N$ galaxies inside a radius $r$ sphere is $\Pr[N]=\lambda^N e^{-\lambda} / N!$ where $\lambda = (4\pi /3)\rho r^3$. So the cumulative distribution function of the distance to the $N$th neighbour is $$\Pr[R_N<r]=1-e^{-\alpha r^3}\sum_{n=0}^{N-1} \frac{\alpha^n r^{3n}}{n!}$$ ($\alpha=4\pi\rho/3$ for brevity). Taking the derivative to get the PDF gives $$f(r) = e^{-\alpha r^3}\left [(3\alpha r^2) + 3\sum_{n=1}^{N-2} \frac{\alpha^n r^{3n-1}}{(n-1)!}\right ].$$ While one can try taking the derivative of this to get the mode (or setting the CDF to 1/2 to find the median) this becomes algebraically messy so a numerical solution is likely best. I get that the mode for the 9th neighbour is 2.44 times the mode of the first neighbour.

The problem here is of course that actual galaxies are highly clumped. That will cause a (potentially large) reduction in the median distance.

In fact, most galaxies have satellite galaxies, so unless one has a definition in the survey excluding them the answer will be that the 9th nearest neighbour is very close (in the case of the Milky Way the distance would be about 30 kpc). So any answer will depend on the criteria on inclusion in this particular catalogue.


What is "Median Stellar Mass"? Could someone explain what is "median stellar mass" and how is it calculated?
I have seen it written as:

$$\log(M)$$

or as:

$$\log\left(\frac{M_*}{M_{\text{Sun}}}\right) $$

but I do not understand what it means.

just check what 'median' means. the median stellar mass is such that half of all stars have lower and the other half hihger mass. it has nothing to do with the logarithm. I havn't seen the term (median stellar mass) in the scientific literature.

Looking at the paper you're referring to, they never use the expression "median stellar mass". Moreover, this study is not concerned about the masses of individual stars, but with the total stellar mass of a galaxy, which they donote by $M$. They then consider the typical or characteristic value for $M$ (as obtained by fitting a Schechter function to the observed distribution of $M$), and denote it by $M^\ast$. Since this is a rather large number, of the order of $\sim10^{10}$M$_\odot$, they prefer to use its logarithm to base ten: $$\log\left(\frac{M^\ast}{\mathrm{M}_\odot}\right).$$
I think the paper explains all that quite clearly and am puzzled by your difficulty.


Does a replacement exist for Dexter? (A tool that reconstructs data from a pdf of a graph) Dexter was a tool from the German Astrophysical Virtual Observatory, made by the Center for Astronomy of Heidelberg University (ZAH). It was able to reconstruct data from a graph if you provided a pdf or an image of that graph. It relied on Java, and since browsers have stopped support for Java it no longer works.
Website for reference: https://dc.zah.uni-heidelberg.de/dexter/ui/ui/custom
Does anyone know of another tool that does the same thing?
Thanks

You question reminded me of a tool I first encountered in the 80’s (I know right!) called “data thief”. Back when Mac applications had a four letter “creator code” this one had the code “DIEF” (which is the Dutch word for “thief”).
I appears to be around still - and as it doesn’t rely on a browser it may well still work (I haven’t tried in years though).
Take a look here


Were K2-141b and WASP-63b selected as the first two exoplanet atmospheres for JWST to look at during its first year of observations? If so, why? In The Observatory @Donald.McLean linked to Scientists will peer at first galaxies with James Webb telescope which says in part:

The new telescope will augment science performed by Hubble, not necessarily replace it, (Nestor) Espinoza said. He is part of the James Webb team at the organization that manages science done with such space instruments, a Baltimore-based a non-profit consortium of scientists that advises NASA.
Espinoza will serve as the principal scientist on projects in which the Webb telescope will examine two planets in distant planetary systems. One of them, K2-141b, is about 50 percent larger than Earth, while the other, WASP-63b, is a gas giant similar to Jupiter.
"Hubble can't perform the kinds of observations we need to learn about the atmospheres of these planets, or other details, since it was designed before the first planets outside our solar system were detected," Espinoza said.

Also

Many of the proposed tasks for the Webb telescope were planned and approved in the 1990s as the observatory was under initial development, said Klaus Pontoppidan, an astronomer and Webb project scientist at the institute.
"We've just added the final projects that will be included during the first year or 13 months," Pontoppidan said.

I don't know if K2-141b and WASP-63b are the first exoplanets that JWST will look at in the first round of observations, or just two examples of many. I also don't know if these exoplanets had even been discovered yet when "(m)any of the proposed tasks for the Webb telescope were planned and approved in the 1990s..."
So I'd like to ask:
Question: Were K2-141b (50 percent larger than Earth) and WASP-63b (gas giant similar to Jupiter) selected as the first two exoplanet atmospheres for JWST to look at during its first year of observations? If so, why? If not, how many other exoplanet atmospheres are planned to be studied during this period?

The list of JWST approved programs for Cycle 1 (12 months long, starting after the 6 month long commissioning phase following launch) is available here. If you unfold the Exoplanets and Disks science category, you will see the programs, along with the links to proposal abstracts and target lists.
Looking at the proposal titles only, the exoplanet targets for atmospheric studies  include:

HD 189733b (2 proposals)
WASP-121b
Gliese 486b
GJ1214b
TOI-421b
55 Cnc e
HD 80606b
WASP-166b
K2-141b (2 proposals)
WASP-127b
Kepler 51d

The Espinoza et al. proposal is proposal number 2159 is one of two to study K2-141b. So K2-141b and WASP-63b are just two of the exoplanets that will be studied by JWST in the first cycle, with others likely in following cycles.
In the overall summary of the JWST Cycle 1 proposal process that was made available to the community, exoplanets accounted for 928 out of the 3500 hours allocated to small programs (slide 7). The exoplanets to be studied are chosen by the individual investigation teams that submit the proposals based on what particular aspects of the atmosphere they are trying to study. These proposals are evaluated by the various exoplanet subpanels of the time allocation committees and by Space Telescope Science Institute (STScI; who operate JWST and HST) staff for scientific merit, technical feasibility, capabilities of the proposing team and optimal use of JWST, a process which is common to almost all competed observing facilities. (The science panels are supposed to only consider the scientific merit, the technical feasibility is handled by STScI staff; this process has been improved by STScI for recent HST and all JWST Cycles by employing Dual Anonymous Peer Review where the panel doesn't know the identity of the proposers and the proposers don't know the membership of the review panels)


What solar features would I see at 10x magnification? I have a pair of binoculars with 10x magnification, 50mm objective diameter, and 6.5 degree (real) angular field of view.

I'm looking into buying solar filters for it (something like these) so I can view the sun. If I do so, will I be able to see most/many sunspots and planetary transits? Or is this simply not a powerful enough instrument to be worth investing in solar filters for it?

I like the idea of being able to check out the sun sometimes, but I don't know if I'm just getting way out of my league here and would need more complicated and expensive equipment to make it at all worthwhile. Thought I'd ask some experts before buying anything.

You will definitely see some sunspots at 10x; in exceptional cases sunspots are big enough to be visible with the naked eye through the sunset haze near horizon, but many sunspots are visible in a small instrument like a 10x binoc with proper filtering.

The Sun is at low activity right now, but still a few sunspots are showing up once in a while. And the activity should increase gradually in the future.

Planetary transits - yes. However, the next transit of Venus will take place in a century or so. Mercury transits more often, and the filters will allow you to see it - a tiny dot on the face of the Sun. A bit higher magnification might be beneficial, but anyway even at 10x you should see the planet, even if it's rather small.

BTW, I recommend solar filters based on the Baader solar filter film (just google it). They are neutral-density (do not change the color of the image), and are recommended and have been tested with instruments from small binocs to dobsonian telescopes. Make sure the filters are firmly attached to the instrument, and are free of defects.


how to measure temperature of the distant star As part of small project i want to measure temperature of the distant star on basis of optical observations. so by what exact method i can do it? which instruments will be necessary? what is the precision of the experiment?

To measure the surface temperature of a star it is used is it's black body spectrum. You would have to get the light curve of the light from tat star and then by checking it's peak you could guess its temperature.

I really don't know which instruments to use to get the light curve from a distant star but you could use a tool like this one for the final conversion: http://astro.unl.edu/naap/blackbody/animations/blackbody.html


Inclination in Kepler's laws So, you can determine the location of an object in an orbit at any given time via Kepler's laws, but how do you determine the orbital inclination at any given point in time?

I am assuming that the question you want answered is how to calculate the elevation of an orbit above a reference plane given the orbital inclination with this plane. If so, please update your question to reflect this, heeding the advice given in the comments.

Kepler's first law tells us that planets move in elliptical orbits, which we can define as follows,

$$r = \frac{p}{1+\epsilon \, \cos\theta}$$

where $p$ is the semilatus rectum, $r$ is the orbital distance from the central body, $\epsilon$ is the orbital eccentricity, and $\theta$ is the angle to the orbiting bodies current position from its closest approach.

Using $r$ along with the orbital inclination with respect to a given plane, we can calculate the height above this plane at any point in the orbit using simple trigonometry.

The image above shows the relevant variables for a given orbit when the body is both above and below the reference plane. $r$ is again the orbital distance, $i$ is the orbital inclination, and $d$ is the height above the reference plane asked for in the question. $d$ is then given by

$$d = r \, \sin \, i$$


Calculating azimuth from equatorial coordinates I try to get the azimuth of an object from its equatorial coordinates using this formula:

$a = \arctan2(\sin(θ - α), \sin φ * \cos(θ - α) - \cos φ * \tan δ)$

Where
φ = geographic latitude of the observer (here: 0°)
θ = sidereal time (here: 0°)
δ = declination
α = right ascension

I made two JavaScript functions to implement this calculation:

  obliq = deg2rad(23.44);  // obliquity of ecliptic
  lat2  = deg2rad(0);  // observer's latitude
  lmst  = deg2rad(0);  // siderial time

  function equatorial(lat, lon) {
    // returns equatorial from ecliptic coordinates
    dec = Math.asin( Math.cos(obliq) * Math.sin(lat) + Math.sin(obliq) *
          Math.cos(lat) * Math.sin(lon));
    ra  = Math.atan2(Math.cos(obliq) * Math.sin(lon) - Math.sin(obliq) * Math.tan(lat),
          Math.cos(lon));
    ra += 2 * Math.PI * (ra < 0);
    return [dec, ra];
  }

  function horizontal(lat, lon) {
    // returns horizontal from ecliptic coordinates
    coords = equatorial(lat, lon);
    dec = coords[0];  // δ
    ra  = coords[1];  // α
    alt = Math.asin(Math.sin(lat2) * Math.sin(dec) + Math.cos(lat2) *
          Math.cos(dec) * Math.cos(lmst - ra));
    azm = Math.atan2(Math.sin(lmst - ra), Math.sin(lat2) * Math.cos(lmst - ra) -
          Math.cos(lat2) * Math.tan(dec));
    azm += 2 * Math.PI * (azm < 0);
    return [alt, azm];
  }

I cannot see any error, but I get strange results for azimuth (a) as can be seen in this table (the other values seem correct):

   λ       δ         α         h         a
    0    0.0000    0.0000   90.0000    0.0000
   15    5.9094   13.8115   75.0000  246.5600
   30   11.4723   27.9104   60.0000  246.5600
   45   16.3366   42.5357   45.0000  246.5600
   60   20.1510   57.8186   30.0000  246.5600
   75   22.5962   73.7196   15.0000  246.5600
   90   23.4400   90.0000    0.0000  246.5600
  105   22.5962  106.2804  -15.0000  246.5600
  120   20.1510  122.1814  -30.0000  246.5600
  135   16.3366  137.4643  -45.0000  246.5600
  150   11.4723  152.0896  -60.0000  246.5600
  165    5.9094  166.1885  -75.0000  246.5600
  180    0.0000  180.0000  -90.0000  248.3079
  195   -5.9094  193.8115  -75.0000   66.5600
  210  -11.4723  207.9104  -60.0000   66.5600
  225  -16.3366  222.5357  -45.0000   66.5600
  240  -20.1510  237.8186  -30.0000   66.5600
  255  -22.5962  253.7196  -15.0000   66.5600
  270  -23.4400  270.0000   -0.0000   66.5600
  285  -22.5962  286.2804   15.0000   66.5600
  300  -20.1510  302.1814   30.0000   66.5600
  315  -16.3366  317.4643   45.0000   66.5600
  330  -11.4723  332.0896   60.0000   66.5600
  345   -5.9094  346.1885   75.0000   66.5600
  360   -0.0000  360.0000   90.0000   68.3079

Does anyone see the error? Thank you.

lat2 = 0 and lmst = 0 place the vernal equinox at the zenith, so most points on the ecliptic should have azimuth either obliq degrees north of east or obliq degrees south of west. Try other values for lat2 or lmst and see if the resulting azimuths make sense.


Star like light moving in the sky, what could it be? I was camping this weekend away from the city, so my friends and I decided to spend one of the nights watching the sky, since it's impossible to see anything around where we live. After a little while, of staring up, I saw what looked like a star moving. It moved fast, but not a long distance, and various directions and back. It was about the same size of the other stars near by. It was not as bright as other stars though. I stopped looking up for a little bit, to make sure my eyes were seeing things correctly. But then I looked up again I continued to see the light moving. I thought I was going crazy, so I pointed it out to my friend. It took him a little while, but after a few minutes he also saw it. His girlfriend also saw it a few minutes later. My friend brought up something about satellite reflections and other theories, but we have no idea what we saw.

So what could it be?

There was a object, apparently flying above you, that you couldn't identify. By definition this is an unidentified flying object. However this does not imply that it was an extra-terrestrial spacecraft.

UFO reports can be explained by a combination of:

Not recognising a known natural object, such as Venus, or unusual clouds.
Planes, drones, Chinese lanterns or satellites.
Optical illusion, dreams or the effects of drugs.
Deliberate falsification.

There is no real way to decide what it was you saw. There are a couple of points in your report that are odd. You describe it as the same size as stars, but less bright. Stars have no visible size,  and have many magnitudes of brightness. Your report suggests that you are someone who doesn't regularly observe the sky, and so is likely to be more susceptible to optical illusions and tricks that eyes can play on us.


What are the equations for an orbit with time as parameter? I am preparing a talk on the Equation of Time, and I want to produce an animation of the movement of the earth around the sun, in an exaggerated ellipse, to illustrate the speeding up and slowing down in different parts of the orbit. Ideally I would like to have x and y as functions of time, e.g x = f(t) and y = g(t). Accuracy is not the prime consideration, a reasonable approximation will do.

I will use a Processing sketch to produce the animation.

Can anyone please tell me what these equations are?

I have done a lot more research into this and found an answer. It comes down to solving Kepler's Equation, and my main references are:
Kepler's equation and the Equation of Centre which gives an iterative method for solving the equation, and Determination of Position in Orbits which has the formulae for calculating orbital coordinates.

Here is the Processing sketch. If you are not familiar with Processing, it is a language developed for graphic designers, but much more versatile than just that, based on Java and hence C++. It expects two functions, one is setup(), called once at the start up of the sketch, and a second draw() that is repeated over and over until the sketch is stopped. In case you wonder: the whole is encased in a main() before it is compiled. For more details see the Processing home page.

int wsize = 400;
float rp = 9.0; // radius of planet symbol
float rs = 15.0; // --"-- sun
// Constants of orbital ellipse
float a = 150, b; // major and minor axes
float e = 0.5; // eccentricity
float halfa = a/2.0;
float c; // focal distance from centre
float E, E0; // eccentric anomaly, temporary values in iteration
float v; // true anomaly = polar coord angle
float M; /* mean anomaly which is effectively the ..
            .. time variable round the orbit*/
float r, x, y; // polar coord radius, rect coords
float k, cosE, cosv; // useful intermediate variables
float twopi = 2*PI;
float PIby2 = PI/2.0;
float Minc = twopi/250; // time proxy increment

void setup()
{ size(wsize,wsize);
  //e = sqrt(((a*a)-(b*b))/(a*a)); kept for reference
  b = a*sqrt(1-e*e);
  c = e*a;
  k = 1-e*e;
  M = 0; // Mean anomaly increases linearly with time
}

void draw()
{ background(255);
  translate(wsize/2,wsize/2); // move origin to centre of display window
  ellipse(0,0,2*a,2*b); // show orbit
  spot(c,0,rs,#FFF303); // draw sun
  translate(c,0); // move origin to focus
  // Calculate Eccentic Anomaly by iteration
  E = M; E0 = 1000; // make sure while starts
  while(abs(E-E0) > 0.000001)
  { E0 = M + e*sin(E);
    E = E0;
  }
  // calculate polar coordinates of planet
  cosE= cos(E);
  cosv = (cosE-e)/(1-e*cosE);
  v= acos(cosv);
  r = a*(1-e*e)/(1+e*cosv);
  // calculate x,y coordinates of planet
  x = r * cosv;
  y = -r*sin(v);// negative makes it go anti-clockwise
  if (M > PI){y = -y;} // show second half of orbit
  // println(v+" : "+x+" : "+y);
  // line(0,0,x,y);
  spot(x,y, rp, #03ADFF); // draw planet
  M += Minc;
  if(M > twopi){M = 0.0;}
}

void spot(float x, float y, float w, color tint)
{ fill(tint);
  ellipse(x,y,w,w);
  fill(255); // restore white background
}


How would Earth's magnetic field change if the outer core was solid? A widely accepted theory for how the Earth's magnetic field is generated is the dynamo theory.  Dynamo theory describes how molten magma convection currents containing metal are locally spun the same direction by Coriolis forces, producing a magnetic field.

If Earth's outer core was solid, this effect would go way.  However, the metals in the interior would still be moving (though all together) due to the Earth's rotation.  Would Earth still produce a magnetic field? If so, how would it compare to our current field?  Would it still help shield the atmosphere from charged particles?

We have an excellent reference of what would happen to the Earth's magnetic field if it had a solid core: we can look at Venus' magnetosphere. Its internal magnetic field is dominated by the remanent crustal magnetization and its external magnetic field by the induced currents and their magnetic field in the ionosphere. The magnetic field mostly misses any dynamo part due to Venus' very slow rotation (even though it also has a similar internal structure like Earth).
Thus as a result, the Earth's magnetosphere would shrink considerably had it as little rotation and/or no liquid outer core. With only a solid core remaining, it would also only be able to produce some small remanent magnetization which is usually MUCH less than any self-feedback dynamo like Earth, Jupiter, Sun,... are able to produce.
Erosion of the upper atmosphere would be much stronger and we could enjoy polar lights about continuously and even at low lattitudes as we'd loose considerable magnetic field strength, especially of the dipole moment.


What is the radius of observable universe- 46 billion LY or 200 yottameters? I have found two answers for the radius of the observable universe.  Wikipedia (and other places, including this site) say approximately 46 billion LY.  I am researching this to study for the National Science Bowl, and their question database says 200 yottameters, which is about 23 billion LY (half the size that wikipedia says). There are other places which cite 200 yottameters as the radius.  Which is the more accurate answer?

This radius of the observable universe is measured by the comoving distance which is the same in every direction. Using CosmoCalc (a common tool for astronomers measuring cosmological parameters) you get 46 Gly for the comoving radial distance.


What should I read next after Neil deGrasse Tysons "Astrophysics for people in a hurry"? I really liked the clever way the author explained stuff in this book but I already understood most of it. Neil himself states in the book that if you are not in a hurry and want to know more about the topic the are plenty of bigger books you can look into. So my question is: what is sort of a next step after this book on this topic (astronomy, astrophysics)?
Thanks.

I liked Black holes & time warps by Kip Thorne. Here's the link. While being easy to understand, he does go into some details regarding black holes and the warping of time, which I think are interesting to the casual reader. If you have a science degree, you might not learn much new informations, but it's a good read.


How accurate are NASA's eclipse predictions? NASA's predictions for the August 2017 solar eclipse show that about half of Kansas City will see a total eclipse and the other half won't (they'll just see a "near" total eclipse, perhaps even Bailey's Beads?).

I thought this would be a great way of testing how accurate NASA's eclipse path prediction is: have people in Kansas City report, with location, whether they saw a total eclipse or not (of course, many people in Kansas City will probably head into the total eclipse zone just to see it, but hopefully few will remain outside it).

It then occurred to me that someone must've already done this for previous eclipses (I know it's done for lunar occultations of other stars).

However, googling only tells me that NASA's 2017 path predictions are more accurate than ever before, but not actually HOW accurate they are.

Where can I find data about accuracy of eclipse predictions?

The path of totality, which eclipse maps show to be 70 miles wide, is actually narrower than that, by up to 1 mile, according to some experts.
An article by the Kansas City Star discusses this:

Those maps, provided by NASA and others, show a crisply defined, 70-mile-wide path of totality where the moon will block 100 percent of the sun. But they are not as precise as they appear, at least on their edges.
The southern edge of the path as shown on the maps could be off by as little as the length of a football field or as much as a half-mile, eclipse mapping experts say. Likewise for the northern edge, meaning the path of totality might be just 69 miles wide.
“This is an issue. This is really an issue, but it’s not advertised. … Yeah, all the maps are wrong,” Mike Kentrianakis, who is the solar eclipse project manager for the American Astronomical Society and who routinely consults with NASA, told The Star.
Xavier Jubier, a French engineer whose calculations have been used to create the interactive Google maps of the eclipse, confirmed to The Star by email that the actual path of the totality is slightly narrower than the 70 miles shown on current maps.
Ernest Wright, who created maps and other multimedia presentations on the eclipse for NASA, said he thought the map might be narrower by about 100 meters, slightly longer than a football field. He also said it’s possible that Kentrianakis is correct in his estimation that the path is narrower by a half-mile or more.

The article explains the reason the path of totality is narrower than the eclipse maps indicate is because they use a 41-year old value for the radius of the Sun, which is now known to be too small a value:

Wright explained that eclipse maps are made based on what is known about the relative sizes and positions of the moon and the sun. “We have really good information about the orbit of the moon, the positions of the sun, the positions of the Earth. All of that is really well nailed down,” Wright said. “In order to get more accuracy, we need to take into account the mountains and valleys on the moon, and the elevations on the Earth. And we’re starting to do that, as well.”
The size of the moon, in fact, has been measured to within a meter, and its position in the heavens has been measured to within a centimeter. “But the last sort of uncertainty might surprise you,” Wright said. “It’s the size of the sun.”
Jubier said that the current maps are accurate using the the 696,000-kilometer radius and other standards agreed upon in 1976 at a meeting of the International Astronomical Union. “This is perfectly accurate but we know it does use a solar diameter that is not large enough. Why don’t we change the value(?)” Jubier wrote. “Well simply because the IAU (International Astronomical Union) has not yet approved a new value. This is part of the research we’re doing and for which we’re looking for funding.”
He continued, “So technically speaking if the Sun is larger than the adopted IAU value, and we know it is, the eclipse path is necessarily narrower and our tools can simulate this, yet the standard maps for the public will still retain the currently adopted solar radius until a new value has been accepted."


Orbital period for a 1 light year object? How long would the orbital period be for an object the mass of the moon orbiting the sun at 1 light year in a circular orbit?

To answer this, let's visit Kepler's third law which tells you the orbital period of a body, if you know how far away it is.

$$\frac{P^2}{a^3} = \mathrm{constant}$$

As it turns out, if you use the correct units, for the orbital period, $P$, and the semi-major axis, $a$, namely years and astronomical units (AU) respectively, then that constant is equal to 1! You have our moon orbiting the sun at one light year, so this needs to be converted to AU. Google tells me that $1\:\mathrm{light\:year}$ is $63241.1\:\mathrm{AU}$. For this specific case, you get that

$$P = a^{3/2} = 63241.1^{3/2} = 15,903,734\:\mathrm{years}$$

Notice that this formula didn't depend on the mass of the moon! (Technically the mass of the moon is in there, but it's pretty negligible and so doesn't matter much.)

You may be scratching your head over the units $-$ how can a timescale squared equal a distance cubed?. Technically there's some trickery going on because that constant isn't unitless and I've implicitly assumed those units are hanging around, without explicitly saying so. Rest assured, the units do make sense for this equation.


What Crayford focuser to buy for a B&L Criterion 8000? I'd like to get a dual-speed Crayford focuser for my newly acquired a B&L criterion 8000 (8") as the mirror shift is pretty bad.
I'm really not sure which one to buy because I'm quite confused about the dimensions and potential adapters I need, and don't know how standard this config is.
Let's call [A] the 4 screws-piece at the output of the optical tube, and [B] the threaded adapter that I'm holding in my hand in the 3rd picture. The ruler is in inches.
My questions:

a) where would a Crayford Focuser like this one go? (edited after feedback) Should it replace part [A] or part [B]?
b) what is part [B] officially called? and what diameter is it referred to by? (I measured 1.9in but can't seem to find anything like that on the Internet)
c) how standard is this mount? Should I worry about the dimensions for what might replace Part [B]?
d) can anyone comment with potential focusers that you think would fit this build? I'm looking for dual speed and < $150

Thanks!

Note that I am aware these aren't the best scopes ever made but I do wish to keep it for a while.

Item B is called a visual back.  The new focuser would just replace this, and screw onto item A.  The focuser you link to is designed for a refractor, but you need one for a SCT, like this
https://www.firstlightoptics.com/skywatcher-focusers/dual-speed-2-crayford-focuser-for-sct-telescopes.html
There are lots of different makes and prices.
Regarding the sizing, I'm not sure about an old scope like the Criterion: it's probably the same as the other popular current models, but you might do better with cloudynights.com to get that confirmed.


Big Bang / Big Crunch cycle? I've been reading some of Stephen Hawking's work on the theory of the expansion (and ultimate contraction) of the universe and was curious if anyone has every hypothesized that the universe has possibly gone through this cycle before (or many many times before).

I know it would be impossible to determine because the idea of the universe contracting back to the size it was at the point of the Big Bang (or at the end of a Big Crunch) suggests that all of the laws of physics would break down but has anyone hypothesized and written work on this. I'd be curious to read more about the idea if so.

I figured I might as well expand my earlier comment into an answer, as I've cobbled together some more information.

You're talking about the cyclic model, which states that the universe goes through cycles of Big Bangs and Big Crunches. We live in the middle of one of those cycles. Albert Einstein was one of the first to investigate it, but his efforts came to nothing.

The theory has been reincarnated, most recently in 2002 by Paul Steinhardt and Neil Turok. They propose that the cosmological constant has changed over time, which could make the accelerating expansion of the universe fit into the theory. Another good article on their idea can be found here. Roger Penrose has also come up with a theory - Conformal Cyclic Cosmology - along similar lines as Einstein's. You can use this as a starting point for more reading.

A related topic is the Big Bounce, which attempts to address the Big Bangs and Big Crunches.


What did LIGO Actually See? (Gravitational waves discovery) I am trying to find an original video/image of what LIGO actually saw, but all I can find is artist renditions of gravitational waves.

The actual image isn't much. I was able to find it from Science, and this is all it is:

It's just a ripple, seen at slightly different times from two different observatories. The shift fits perfectly by shifting it by the speed of light difference in their locations. Thus is the proof of gravity waves.

It should be noted that the reason there are two instruments is to provide a cross check against other vibration sources. Each observatory works by detecting vibrations on a 4 km scale, down to a very small order of magnitude (1/10,000 the width of a proton). When the two are compared, then one can assume the signal must have come from a non-local source, which only Gravity Waves fit that definition.


How do a planet's moons and a planet's rings interact? I am making a video game which features a planet and its moon. And I thought I would add rings too.

But that got me wondering... how do a planet's moon and a planet's ring influence each other?

Does a moon clear out all the debris in its orbit? In other words: does it make a gap in the ring? If so, how large is the gap in relation to the moon's radius?

If not, does the moon travel along as all the other matter in the ring? Would it not make the debris fall to the moon, thereby growing the moon?

Lastly, are both the rings and moons typically in the same plane?

First, there are different types of rings. Using saturn as an example: there are icy particles, dust bands and more. These interact differently if they were to hit a moon, for example enceladus.

Every little particle in a ring is a „mini-moon“ and obeys the same laws (see kepler) as the big ones.
They technically all have their own orbit.

The main rings of saturn have a high enough particle density to exert gravitational influence on each other.

There are hypothesis that suggest that rings are made of shattered moons (like through collision) or another one that says that the rings are made of particles that „failed“ to make a moon in the first place.

So now, what happens when rings collide with moons is dependent on the type of ring.
Enceladus is a small icy moon in the so called E-Ring of saturn. This ring is small and made of water/ice, just like the moon. Because the ring is so very small and such a water/icy „cloud“ would  not really stay together, it is believed that it is „refilled“ by enceladus constantly.
The surface of enceladus was observed to be smooth, not really having craters one might expect, but this could also be due to geological activity. Enceladus also „geysirs“ out water all the time, so this might be how the ring is kept.

Edit: I recommend checking out this link: https://courses.lumenlearning.com/suny-astronomy/chapter/planetary-rings/