Apparently two objects in Solar Eclipse image These images are taken from a mobile camera of 21 June 2020 solar eclipse.
So the thing surprising me is there are two objects.
I have shown these images to some people. Some says it's just reflection of sun in camera lenses. And some says the blue object is Venus. But we know if that is Venus then crescent should face the sun.
So I want to know what that light blue crescent object is. And if it is the eclipsed Sun then what is the other bright thing?
First image:

Second image:

The people who say it is an internal reflection are right.  This is an internal reflection inside the lens of the camera. There is no actual object there. In particular, Venus is not visible this close to the sun even during a partial solar eclipse. Moreover, Venus is currently a very very thin sliver, no more than a scrape of light at the edge of the planet, and much smaller than this image shows.
It's actually quite nice, because the sun itself is overexposed and the eclipse is not visible,  but the internal reflection is exposed quite nicely, and shows the progress of the eclipse well.


Does the given radius of a planet/moon include its atmosphere? In NASA fact sheets, such as this one for Venus, a radius of the planet is given.

Is that radius the radius of the surface of the planet, or of the envelope of atmosphere?

Put another way, if calculating the angular diameter of a planet/moon from a location, do I only have to look up this radius, or do I have to look up this radius along with the atmosphere height to determine the angular diameter?

I looked for a source without luck, so I'll just post cause I've read this answer before.   The diameter is the solid surface, not the atmosphere for rocky planets.   For gas giants it's the atmosphere up to 1 bar.

To determine Venus' angular diameter, you'd need to add something for it's atmosphere.   It's difficult to say how high it's atmosphere stops being reflective (and if anyone knows the details, please answer), but I would guess about 50 km where it's clouds reach to on it's sunny side.   You'd need to add more if you look at it through infra-red.   But I think it would be an error to add the full 250 km listed as the thickness of it's atmosphere and it's reflectivity probably trails off well before then.

A curious sidebar, Ganymede is listed as the largest moon by diameter, but because of it's atmosphere, Titan appears larger.  It's visibly larger on accurate moon charts.   Because of ice in it's upper atmosphere and low gravity, Titan's atmosphere remains reflective for a significant distance.


Why a blue spot in some hubble images of NGC 4302? In some images with coloring like in this one:

There is a noticeably bluer region in the "bottom left" of the image, and the bottom part of the galaxy on the left. Being an inexperienced galaxy viewer, I can't figure out what to make of it.

Is the spectrum of that region more complex than it seems? What is it?

That's a region of recent star formation in the disk of NGC 4302. Regions of recent star formation include short-lived, high-mass O and B stars, which are blue (because they're hot) and luminous enough to outshine the (redder) lower-mass stars that are also present. If you look carefully at the image, you can see hints of blue light from other parts of the disk (e.g., in the upper right and upper left sides of the disk). You can also see lots of recent star formation in the disk of the neighboring galaxy, NGC 4298.

If you look at the full-resolution (84 MB) version of the image available here, you can see that the blue region resolves into many blue pinpricks in the midst of a fuzzier blue haze, a clear indication that it's made up of lots of bright blue stars (or small clusters of bright, blue stars) and a larger number of somewhat fainter blue stars. You can see the same thing going on in the blue regions of NGC 4298.

There are probably other, similar regions in the disk of NGC 4302; the reason this particular region shows up so well is probably because it's located in the outer part of the disk on the side closest to us, so there's relatively little dust blocking its light.

(Although there is undoubtedly ionized gas in the vicinity of those stars, the gas is not producing enough light to show up in the image -- you would in fact see a reddish/pinkish region if it was mostly light from ionized gas.)


Assumptions necessary for the strong form of Olber's Paradox From Wikipedia, here's Lord Kelvin's statement of Olbers' Paradox:

  Were the succession of stars endless, then the background of the sky would present us a uniform luminosity, like that displayed by the Galaxy – since there could be absolutely no point, in all that background, at which would not exist a star.

This is one form of Olbers' Paradox I've heard, which is that with infinitely many stars scattered in an infinite volume, every ray must eventually intersect a star. I call this the "strong form" because it seems to me stronger than the statement that infinite stars imply an infinite amount of light hitting every point, though not necessarily from every possible direction.

The strong form of the paradox doesn't hold if each star is a point mass and there are countably many stars, since the number of rays from any given point is uncountably infinite. Or imagine instead that stars are balls with small positive radius. If there is such a star at every lattice point in $R^3$ (an assumption that is consistent with my crude understanding of the cosmological principle), I can easily find a ray that intersects no star. I start from (0, 0, 0.5) and choose the ray in the direction of (1, 0, 0.5); the z coordinate will always be 0.5 and therefore my ray will never come close to a lattice point.

Is there another, palatable assumption we can add to make the strong form of Olbers' Paradox true, perhaps an argument that the distribution of stars must be more random than what I've described?

For simplicity, let's pick a standard star, of radius $R_\odot$ (you can easily generalize this for stars of any radius by picking $R_\odot$ to be the lower bound of star radius).

Let's split the universe into cubes. It would be much easier to use concentric shells, but later on it will be practical to have everything split into components of standard sizes. Each cube is of side $\ell$, and volume $\ell^3$. Within each cube, there is a probability distribution $P(N_\odot = n)$ to have $n$ stars within that cube. We're not requiring anything of that probability distribution, except

\begin{equation}
E[N_\odot] = \rho_\odot \ell^3
\end{equation}

There is an average stellar density $\rho_\odot > 0$ that is the same everywhere.

Now trace your line of sight from $0$ to infinity. This will go through an infinite number of such cubes. We can simplify the problem by having $\ell \approx R_\odot$, so that we don't have to worry about whether the line passes through a star. The fact that this line will encounter no star is equivalent to the experiment of each of those cubes to have zero stars. As each such probability is independent, this is

\begin{equation}
\prod_{n = 1}^\infty P(N_{n, \odot} = 0)
\end{equation}

We can set this as a Bernoulli trial. Zero stars is a failure with probability $p$, more than zero star is a success with probability $q$. The probability of zero star cannot be $1$, since we assumed a non-zero density. Therefore, the probability becomes

\begin{equation}
\lim_{n \to \infty} p^n = 0
\end{equation}

While it is not impossible, the probability of that line never hitting a star is of measure $0$, and the probability of hitting one is of measure $1$.


How do scientists know that a rock found on Earth came from Mars? There are many documented cases of meteorites being reported as from Mars.

http://en.wikipedia.org/wiki/Allan_Hills_84001

While I can understand that it's possible to know if a rock originated as a meteorite. What I don't understand is how scientists can say it came from Mars with certainty.

How can they verify such a claim, and differentiate between a rock that is similar to one from Mars to one that really is from Mars.

Mars is the only planet providing rocks with a similar chemical composition and age. So the origin of the meteorite is evident.

This doesn't mean, that it's absolutely waterproof. There might have existed protoplanets similar to Mars 4 billion years ago, which since then have been swallowed by Jupiter, the Sun, or have been ejected out of the solar system. But this would be much of coincidence without leaving convincing evidence. (Here a paper about a couple of simulations.)


How significant and accessible is modern Chinese astronomy? Is China fully integrated in the international astronomical community, or does military secrecy and language problems and sanctions (ITAR) or something else stand in the way? Does China have valuable assets for astronomical observations, and a vibrant national astronomical academical community?

China has an ambitious and successful HSF and Lunar program with orbiters and a lander. They have also announced a very ambitious program for inter-planetary probes to practically the entire Solar system, to date they've flewn by the NEO Toutatis. The are also building a 500 meter radio telescope super-version of Arecibo.

Will for example US astronomers be able to use that telescope? Will they be offered samples of the upcoming Chinese lunar sample return, maybe from the lunar far side? Have lunar ground penetrating radar data from Chang'e 3 been presented publicly and led to papers in international journals? Is there any prominent Chinese astronomical journal?

China is pretty well integrated into modern astronomy -- there are regular international conferences held in China (the most recent General Assembly of the International Astronomical Union was held in Beijing in 2012), lots of Chinese astronomers at other international conferences, Chinese participation in international observatory projects (the Joint Institute for VLBI, the Thirty Meter Telescope, the Square Kilometer Array, etc.), and so forth. I'm not sure I'd say they have any really significant facilities yet (with the possible exception of LAMOST), but they have very ambitious plans, as you've noted.

(ITAR has very little to do with astronomy except when it comes to a few things like nuclear physics codes produced by places like Los Alamos.)

I don't know how much international participation is anticipated for the FAST project, though I gather the instrument development has some involvement from British and Australian institutes, so they'd likely get some time.

Yes, results from the Chang’E-3 mission have been published in scientific papers, e.g.
http://www.sciencemag.org/content/347/6227/1226.abstract

There are several Chinese astronomy journals (e.g., Acta Astronomica Sinica; Research in Astronomy and Astrophysics), but like most "national" journals they aren't that important internationally.


Difference between "planetary ring" and "circumstellar disk"? Related: Do/Can Ringed Stars Exist?

Is there any particular difference, in behavior or properties, between a planetary ring system and a circumstellar disk? Or is the only real difference a matter of what's in the middle?

This question came up as a matter in the related discussion, particularly because the (game-simulated) object that inspired the question is a brown dwarf - something between a star and a gas giant. What I observed in-game appeared to be more similar to the rings around Saturn, than to the description and image provided of a circumstellar disk.

So, are there properties of the rings/disks themselves which can differentiate one type from another? Or is the type of central body the only thing that defines them?

If the central body is the only factor, what should we call a ring around a brown dwarf, assuming one could exist in nature?

From Physics SE:

  A circumstellar disk is a set of objects that orbit about a star-like object and that collectively look disk-like in some way.

From Wikipedia:

  A ring system is a disc or ring orbiting an astronomical object that is composed of solid material

So, a circumstellar disk may have gas, dust, asteroids, etc., but it must orbit a star-like object. A ring system must have something solid--dust or moonlets--but it may orbit any astronomical object.

Other than that, you specifically asked about a planetary ring. That would mean a ring system that specifically had a planemo as the central body.

In other words, a disk around a brown dwarf could be described as a circumstellar disk (a brown dwarf is a star-like object) or a ring system (as long as the disk contained solid material). It could not be called a planetary ring, because a brown dwarf is not a planet.


Does the Sun have hard radiation? Does the Sun spread hard radiation waves around the Solar System?  If so, why are we safe here (or are we even safe)?

The Sun outputs several different kinds of things.

Electromagnetic radiation

The Sun is (partially) a black-body radiator at a temperature of near 6000 K, and therefore emits all sorts of electromagnetic energy, including UV and X rays.

UV is stopped in the upper atmosphere. X rays are absorbed by the whole atmosphere, and are pretty weak anyway.

The Sun's EM activity in terms of harsh radiation increases greatly if there's a solar flare blowing up. Even then, we're pretty safe on Earth.

High energy particles

The solar wind is basically a flux of particles (nuclei of hydrogen, etc) shooting out of the Sun at pretty high energies. These are charged particles. When approaching the Earth, the Earth's magnetic field for the most part deflects them. If you're on Earth, you're safe. Even in space, the flux is not very great; you'd be safe inside a spacecraft.

However, during a solar flare, the flux of particles increases greatly. The Earth is safe, but an astronaut en route to Mars would be in a pretty unsafe situation.


How did Bradley arrive at the +/- correct speed of light when he calculated it? After Bradley discovered stellar aberration and the corresponding constant of aberration he was able to calculate the speed of light, since he knew the speed of the Earth around the Sun.

As far as I think I understand the math behind his calculations. Yet, I believe that Bradley (and anyone else until Relativity) didn't consider the velocity of the Solar System itself. It seems (to me, that is) that Bradley assumed a Solar System at rest. If the Solar System is moving, this certainly affects the result of the calculation (especially so since at that time the speed of light was not considered constant).

I'd appreciate if someone pointed out to me where my thinking is wrong.

He didn't need to know the absolute position of the stars. The change of the appearent position  has been sufficient to get a good estimate.
For velocites small in comparison to the speed of light, the shift of the angle is still small.

For small angles the sine is proportional to the angle with 2nd order precision.
The first Taylor summands are
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - + ... $$

The 2nd order approximation $$\sin x\approx x$$ can be applied on stars perpendicular, or at least apearently perpendicular, to the plane of the Earth orbit, where the method is most accurate.

Hence the underlying reason, for which it works, is the smoothness of sine.

This argument can be adjusted to less optimal conditions.


What would happen if a rocket traveling at speed of light would collide with a planet (like earth)? Let's assume it is possible for a spacecraft to travel at the speed of light (I've read the interstellar book by Kip Thorne, apparently this is theoretically possible if you swing around two black holes)

I have watched the video where the traveling of a light particle is simulated and while planets are certainly big enough so you can adjust your route, what would happen if a rocket (or spaceship in general) would collide with an earth-like planet (i.e. a planet having a massive surface). Would we burst through the whole planet, make it only to a core, or just a few km? I know what happens when a meteor lands, but obviously they aren't traveling at the speed of light.

And one similar another question - when a spaceship is traveling at the speed of light, is there a problem with collision with smaller objects? I think you can't avoid millions of rockets in outer space, it would surely damage the spaceship, wouldn't it?

I don't know about a spaceship, but the XKCD guy wrote an interesting article about what would happen if Earth was hit by a solid asteroid, travelling at various different speeds:
what-if question: Diamond

The ship would be a lot smaller than an asteroid, so I think the damage would be a lesser version of these descriptions.  (The most extreme case described is the whole planet being vapourised and even neighbouring planets being affected by the radiation.)

As for the second part of the question - protecting the craft from minor collisions - I admit I have no idea how to work out the kinetic energy from collisions at such high speeds. But even a grain of sand would at least damage the front of the craft, and I guess some sort of thick shield would be needed to protect against erosion from interstellar dust...

Even at normal orbital speeds, Space Shuttle windows have had minor damage as this paper describes...


What does Gauss' divergence theorem say about compression of a body under self-gravitation? Below this question there is a comment that caught my interest:

  First, there's a misconception. Compressing a body does not increase the force of gravity on outer-lying masses. That's Gauss's famous divergence theorem.

I understand that one can use the divergence theorem to address (derive) the shell theorem - one consequence of quick is that a particle inside a spherically symmetric shell experiences no net gravitational force from the shell, but this seems different as it refers changes in average density of a body and gravitational forces on outer-lying masses.

So I'd like to understand better: What does Gauss' divergence theorem say about compression of a body under self-gravitation?

For reference, this answer covers the shell theorem nicely.

Gauss's divergence theorem applied to the gravitational field $\vec{g}$ is that
$$ \oint \vec{g} \cdot d\vec{A} = \int \nabla \cdot \vec{g}\ dV,$$
where the left hand side is the flux of gravitational field into/out of a closed surface and the right hand side is the integral of the divergence of that field over the volume enclose by the surface.

The fundamental definition of how mass produces gravitational field is that
$$ \nabla \cdot \vec{g} = -4\pi G \rho,$$
where $\rho$ is the mass density.

If you are at a point outside the mass then the right hand side of the divergence theorem  becomes a constant.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV = -4\pi GM.$$
Therefore the left hand side is also constant when calculated over any surface that encloses all the mass.

If you allow the mass to contract in a spherically symmetric fashion, then as the gravitational field on the left hand side is then always in the radial direction then $\vec{g}$ must stay the same at any position in space outside the mass distribution.


angles on the celestial sphere This is a somewhat simplified, and drawn / formulated in a manner more targeted to astronomy, version of a diagram and a question that I've also posted in math.stackexchange.com.

In the diagram below I depict:

the celestial sphere with the North Celestial Pole (NCP), the South Celestial Pole (SCP) and the celestial equator
a latitude circle at latitude $\delta$ (declination)
two points $A$ and $B$ on that latitude circle
the center of the celestial sphere $O$ and the center of the latitude circle $O'$

My question is:

Given angle $\phi$ between the rays $OA$ and $OB$, what is the angle $\theta$ between the rays $O'A$ and $O'B$ ? I am trying to derive $\theta$ given $\phi$ and $\delta$. Observe that if $A$ is the first point of Aries, then the angle $\theta$ is the Right Ascension of a star located at point $B$.

My assumption is that the following formula holds:

$\phi = \theta*cos(\delta)$

This answer in math.stackexchange provides the correct relationship between $\theta$ and $\phi$. And this answer supplies the proof.


Measuring the size and the geometry of Universe When we would be able to tell with surety that universe is finite or infinite? How can we exactly determine the density of the universe so, that we compare it with critical density, in order to determine its geometry? Does the infinite Universe entails infinite black holes or infinite stars?

The relative recent WMAP result shows, the Universe seems infinite.

However, the "is the Universe infinite" question has a hidden problem: it doesn't matter, how far away can we see, we can't be sure, if anything out of it is finite or not. :-) Thus, it is about a definitely unknowable thing.

The WMAP measured essentially the curvature of the Universe and it found a zero curvature (i.e. it seems to be planar). If there is a minimal curvature, below a measurement precision, and we have a cyclic, spherical geometry, then its radius must be at least 300billion light years.


What is the volume of the universe? Of course it is not exactly known how big the universe is, but I thougt that the universe is about 100 billion lightyears in diameter. But if that is true can you also make an estimation of the volume of the universe?

Based on your comments, I think your confusion comes from having seen the classic rugby ball-shaped image of the CMB.
The CMB we observe is not from the whole cosmos, but only from a thin and completely spherical shell centered on us, with a radius $R_\mathrm{CMB}\simeq45.6\,\mathrm{Gly}$ (using a Planck 2015 cosmology). Just as the shell of Earth can be projected onto a rugby-shaped figure using a Mollweide projection, so can the shell of the CMB. Here's a figure from Universe Adventure that can help visualize:

Although we (still) can't see beyond the CMB, the observable Universe is all the way out to redshift $z=\infty$, while the CMB comes from $z\simeq1100$, but the difference is not big; $R_\mathrm{obs.Uni.} \simeq 47\,\mathrm{Gly}$. Thus, the volume of the observable Universe is
$$
V = \frac{4\pi}{3}R_\mathrm{obs.Uni.}^3 = 435,\!000\,\mathrm{Gly}^3.
$$

The total Universe is probably much, much larger, and may in fact easily be infinite.


Could Black holes forge heavier elements that have yet to be discovered? Observations

The heaviest elements known in abundance in nature are forged deep within stars.
These elements are made possible by the high densities/pressures within the stars.
Black holes are known to have a much higher density/pressure than any known star.
Black holes are also known to be a phase of stellar evolution - this suggests that the original star's internal process of forging metals would persist within the resultant black hole.
Scientists have forged synthetic/ephemeral heavy metals under conditions which could hypothetically be sustained within a black hole.

Hypothesis:

Black holes forge heavier elements that have not been observed on earth. The conditions needed to sustain these elements are unique to the black hole, due to its high density/pressures. These conditions can be glimpsed, but not sustained in any experimental context

Follow up questions:

Has this been hypothesized?
Where can I find research on this topic?

The heaviest elements known in nature are forged deep within stars.

No, the heaviest elements are made on Earth in scientific laboratories, or in the extreme gravity of a neutron star's crust.

  These elements are made possible by the high densities/temperature/pressures within the stars.

Many of the larger elements can be made in supernovae and neutron star collisions, not in stars. It requires extreme conditions for these elements to form.

  Black holes are known to have a much higher density/temperature/pressure than any known star.

Black holes are actually very cold, they "absorb" any radiation that passes their event horizon.  Outside the event horizon may be some very hot material, but it is not actually so hot compared to the core of a star.

  Black holes are also known to be a phase of stellar evolution - this suggests that the original star's internal process of forging metals would persist within the resultant black hole.

No, inside the black hole everything falls, and reaches a singularity in a short amount of time.

  Scientists have forged synthetic/ephemeral heavy metals under conditions which could hypothetically be sustained within a black hole.

As above, the conditions beyond the event horizon are unlike anything we have on Earth, because there is the unavoidable singularity.

After some matter has crossed the event horizon it will certainly come to the singularity. (in the same way as you will certainly reach tomorrow) And as it gets closer the tidal effects get greater, eventually ripping the atoms apart.  The extreme gravity in a black hole will tend to pull matter apart not fusing it to larger atoms.

There may be nucleosynthesis in the accretion disc around a black hole. While the amount of high mass atoms made here is relatively small, it may be useful for the sake of detecting and distinguishing black holes from neutron stars or white dwarfs.


An hypothetical timepiece on board of a spaceship. Please Help On board of a hypothetical spaceship, I have a hypothetical clock consisting of 2 mirrors and a light beam is being reflected between the mirrors to measure the time. Now if we hypothetically exceed the speed of light; what will I see is happening to my clock on board of my ship? (a) Light can't exceed the speed of light so according to me the clock has freezed, (b) Light always passes the observer at the speed of light, so my clock is keeping perfectly time but the speed of light inside my clock has surpassed the speed of light? (c) The clock will run backwards?  I just can't wrap my mind around this problem.

On board of a hypothetical spaceship, I have a hypothetical clock consisting of 2 mirrors and a light beam is being reflected between the mirrors to measure the time.

The mirrors, clock and you are all in the same reference frame and your relative velocities are zero.

The speed the whole shebang is traveling at relative to something else is irrelevant.

So the clock on board the ship with you will calmly pass the seconds just as you'd expect an object at rest relative to you to do.

  Now if we hypothetically exceed the speed of light

We can't do that.

  what will I see is happening to my clock on board of my ship ?

Beyond the speed of light has no meaning I know of in this context.

  I just can't wrap my mind around this problem.

There's no problem : you can't travel faster than light : keep repeating that mantra and all will be well. :-)


Could we fly/drive through Jupiter? If Jupiter is made of gas, could we fly or drive through it or would its center be too dense?

http://spaceplace.nasa.gov/review/dr-marc-solar-system/gas-giants.html

We think of a gas as something very . . . well, airy. After all, air is the gas we all know and love. We breathe it and fly planes right through it with no trouble. So it makes sense to think that a gas planet must be like a big, airy cloud floating out in space.
Saturn in true color.

The bigger a planet becomes, the heavier is the material weighing down on its center. Think of how it feels to dive under water. If you are wearing a face mask, you notice that as you dive deeper, the mask presses harder and harder on your face. Also, your ears start feeling the pressure even at 2 or 3 meters (5 or 10 feet) below the surface. The pressure you feel on your body is due to the weight of the water above you. The deeper you go, the heavier the water above you and so the greater the pressure on your body. Even on Earth's surface, each square inch of your body experiences 14.7 pounds of pressure due to the weight of the atmosphere above you. If you could dive down to the center of Earth, the pressure on your body would be about 3.5 million times as great! The center of Jupiter is more than 11 times deeper than Earth's center and the pressure may be 50 million to 100 million times that on Earth's surface!

The tremendous pressure at the center of planets causes the temperatures there to be surprisingly high. At their cores, Jupiter and Saturn are much hotter than the surface of the Sun!

Strange things happen to matter under these extraordinary temperatures and pressures. Hydrogen, along with helium, is the main ingredient of Jupiter's and Saturn's atmospheres. Deep in their atmospheres, the hydrogen turns into a liquid. Deeper still, the liquid hydrogen turns into a metal!

But what's at the very center of these planets? The material becomes stranger and stranger the deeper you go. Scientists do not understand the properties of matter under the extreme environments inside Jupiter and Saturn. Many different forces and laws of nature are at work, and the conditions inside these planets are very difficult to create in a laboratory here on Earth. But you can be sure that you wouldn't be able to fly through these bizarre materials! As we now know, the gas giants are much more than just gas.


Why does the planet Saturn have numerous (62) moons compared to the rest of the planets in the Solar System? Saturn is a gas giant like Jupiter. It has everything from

  tiny moonlets less than 1 kilometer across to the enormous Titan, which is larger than the planet Mercury. Saturn has 62 moons with confirmed orbits.

  Wikipedia

Why does Saturn have more moons?

Saturn and Jupiter have many moons for quite a few reasons, one of the main ones being that they have an absolutely immense gravitational pull. During the early stages of the formation of our solar system, there would of been many planet-like objects floating around which our gas giants would have attracted. Furthermore, these planets are so far out in the solar system water would if frozen (which explains Saturn's rings of ice). Infact, we can show that the ice can form moons by looking at some of the moons of Uranus, some of them are half made of ice!

A few of the outer moons of our planets are captured asteroids. Phoebe, which is a moon of Saturn, is believed to have been a captured asteroid.

I haven't heard anything about Saturn having more moons than jupiter.


How are parachutes usable in other places than Earth? Mars landings, Moon landings. . . Wherever I see a documentary or still pictures in Wikipedia, there are parachutes in the landing of rovers. Whenever I see them, I ask myself, "How they can it decelerate the speed of rovers, when there is no / thin air?" Something needs to be there to create friction!

Parachutes have never been used on the moon, but they are viable for Mars because Mars does have an atmosphere - albiet one much lighter than that of Earth.  For that reason, parachutes cannot be the only means of slowing down on Mars --- for example, in the case of the Curiosity Rover, that's why an elaborate booster and crane combination was used (see figure).  Another approach, taken by the Pathfinder Rover, was to use large airbags to bounce on landing.

One benefit to using parachutes on Mars, is that they can be used at higher velocities --- unlike on Earth where the heavier atmosphere (and thus stronger drag force) would shred a parachute at comparable speeds.


Why does disorder (entropy) increase in the same direction of time as that in which the universe expands? Why does disorder  (entropy) increase in the same direction of time. as that in which the universe expands?

Why can't entropy decrease in the same direction in which it increases ?

Entropy is the only quantity in the physical sciences that requires a particular direction for time, sometimes called an arrow of time. As one goes "forward" in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease, Universe can seen as one big isolated system in which total energy is always constant. Hence, from one perspective, entropy measurement is a way of distinguishing the past from the future.
https://en.wikipedia.org/wiki/Entropy_(arrow_of_time) - wiki for relation between time and entropy.

https://www.youtube.com/watch?v=870y6GUKbwc - Basics of entropy.

Highest state of entropy is achieved when the system is in equilibrium, to have this equilibrium, energy entities (mass and other energy radiating sources) should be enough apart from each other that no more energy interaction(give and take) could happen and energy is distributed evenly in the universe, thus expansion.

https://www.youtube.com/watch?v=xZTb6sfHEX8- Will the Universe expand forever
https://www.youtube.com/watch?v=26_GN-uL4Jc - The Laws of Thermodynamics and                      Fate of The Universe


Why does the moon 'peel' away from the horizon in this photograph? The Himawari-8 geostationary weather satellite recently had a lunar intrusion into it's primary sensor (the Advance Himawari Imager, or AHI):

Zooming in on the moon:

I see a strange cupping (or 'peeling') of the moon away from the horizon:

My question is: what is the explanation for this 'peeling' aberration?

Note: if your answer involves the lunar terminator, please account for the following concerns:

At the time of day of the photograph, we see the Earth with ~5 degrees of longitude in night. The Earth presents approximately 17 degrees across from GEO, so the moon is $17^{\circ}/2=8.5^{\circ}$ degrees away from nadir, and so I expect the moon to have $5^{\circ}+8.5^{\circ}= 13.5^{\circ}$ of longitude in night. With barely half the moon peaking out above the Earth's horizon, the terminator is well behind the Earth.
The cupping aberration appears equal on both edges. The Sun is in the top right corner of the picture, and so I would expect a terminator related effect
to be more pronounced on the bottom of the moon than the top.

Looks like an inferior mirage to me.

This:

https://aty.sdsu.edu/explain/simulations/inf-mir/Kaplan_photos.html

Has a work through for the disk of the sun but I think the same general principles apply here as well.

The specific effect will be somewhat different since in the image from space rays are moving from space down into and through the atmosphere, and then back up again, whereas in the image below the rays terminate at the Earth's surface.

Simulation and photograph from Andrew T. Young's page Photographs of an Inferior-Mirage Sunset:

  The inferior mirage is very clearly shown in a series of sunset images taken by George Kaplan, of the U. S. Naval Observatory. Here, I describe what's seen in these images, which are actually on the Naval Observatory's website. They are such a “textbook example” of the inferior-mirage sunset that I begged for some details, and permission to show the images here.


How do astronomers know if a planet is orbiting a star, or it's simply another star in the background? For example, if astronomers photograph a "planet", which is in close proximity to a given star, then this "planet", could actually just be another "star" in the background, many light years away.

It might just be in the photograph by chance.

How can astronomers determine if this is, in fact, a planet orbiting that star or just another star many light years away?

For example, a common method to find planets orbiting stars is the transit method. ie, we can observe how the star appears to slowly decrease brightness, then increase back to normal again. This basically tells astronomers that this must be a planet.

But how do astronomers know that this isn't just the "front star" transitting a star "in the background"?

I think there are two parts to this question: How do we know that the transiting object is a planet, and how do we know that it's gravitationally bound to the parent star, rather than an interloper?

If the object is indeed orbiting a star, we should see periodic transits. If a candidate transit is observed, follow-up observations should also be able to observe it. A lack of further confirmation would be a strike against the idea that the object is indeed bound. An additional tool that's useful here is spectroscopy. If the body is orbiting, it should influence the motion of the parent star, because they'll orbit the shared center of mass. This means the star's motion with respect to an observer will change over time, causing a Doppler shift in its light. This shift can be measured throughout the body's orbit. If a periodic shift is observed - with the same period as the transits - then we have another bit of evidence for an exoplanet.

Now, the light curves of transiting exoplanets can look similar to those of eclipsing binaries - binary stars where the orbital plane of the components is aligned with the line of sight of the observer. A couple key elements can help astronomers differentiate between the two:

An eclipsing binary should display a secondary eclipse when the less luminous component passes behind the first. This is not likely to be prominent in the case of a transiting exoplanet.
The mass of the orbiting object can be determined by measuring the radial velocity of the star, as determined by looking at the Doppler shift. This is an easy way to distinguish an exoplanet from a star.

In short, an occultation of a background star by an interloper would not be periodic and would not be associated with Doppler shifts in the other star's spectrum. The mass of the orbiting body - as well as the presence or absence of a significant secondary eclipse - can help us determine its nature.

I'd like to note that we've directly imaged a number of exoplanets over the past decade. Beta Pictoris and HR 8799 are well-known examples. There are some difficulties with this, of course; planets are much less luminous than their parent stars, and so the light from the central star must be blocked during observations. Direct imaging is useful for planets in face-on orbits, where we're looking nearly perpendicular to the orbital plane. These exoplanets will, naturally, never transit their parent stars from our perspective, and so transiting planets aren't usually directly imaged planets. Additionally, direct imaging is also good for planets in orbits with large semi-major axes - which are also planets that are unlikely to yield strong transits from any position.


Energy needed to create a black hole vs energy needed to run the Large Hadron Collider? I was disappointed to hear Brian Cox's flippant reply to whether or not the LHC could create a black hole big enough to swallow the earth in this video:

https://www.youtube.com/watch?v=QFy87tFTZwY

I'm guessing that the power used by the LHC is nothing like that used when gravity crushes a star to a singularity, but ...

... how big is "nothing like"?

Your question touches on a few points.

First, yes, he was flippant, but the risk was super-low.   The simplest way to explain this is that nothing happens in CERN that doesn't happen all over the universe and in the upper atmosphere of the Earth or on the surface of all the moons, planets and stars every day.  For example, the Oh My God particle was traveling with several million times the energy of the particle collisions inside CERN.   At those speeds, accelerating the particle further doesn't add much velocity, what they're basically doing is adding energy or mass to the particle and only slightly increasing it's velocity as it's speed gradually moves closer to the speed of light.

What makes CERN useful, isn't that the high speed collisions don't happen anywhere else (they happen all the time all over the Universe).  What makes it useful is the million dollar equipment that zooms in on those collisions, recording what happens with the best cameras in the world.   We can't predict when and where a very fast moving cosmic particle will hit the Earth, but they know when and where the CERN particles are going to collide, because that's planned to happen in a specific place and time.

So, if those kinds of collisions happen all the time and no planet or star has ever been observed being eaten by a particle collision created black hole, then such events are, at the very least, extremely rare.

Add to that, Hawking radiation dictates that black holes that small should evaporate before they can eat anything.   So, when Brian Cox laughs at that question, he has good reason to laugh, because it was never a real risk and it's more of a joke question than a genuine risk.

Now, when you say "pressure" to create black holes.   The pressure inside a black hole, or, I should say, pressure inside a Neutron Star about to turn into a black hole, or the pressure inside a collapsing core of a very large star about to create a black hole - those pressures are crazy high, but if we could re-create that pressure on earth (We can't), but if we could, it wouldn't be enough to make a black hole.   Pressure alone doesn't do it, you need mass (at last about 3 solar masses) to fit within a certain size and the less mass you have, the greater the required density (and greater pressure) is needed to achieve the necessary mass to schwarzchild radius ratio.  The pressure required inside the forming black hole actually decreases as the mass of the black hole increases.  (based on simple pressure equations and schwarzchild radii).   Equal pressure in a lab on Earth wouldn't accomplish it.

What makes black holes theoretically possible is the small quantum size and possibility of extra dimensions and quantum rules that only apply to the very small, so it's a very different set of circumstances.

Collisions in CERN are measured in Electron-volts.   It was theorized that if there were extra dimensions on the quantum level, then micro-black holes might form at the energy of CERN's collisions (this would imply that black holes of this type form in Earth's upper atmosphere all the time, implying they're not dangerous).

But so far, no luck.  No micro-black-holes have been observed.  They should dissipate very rapidly but their decay streams should be recognizable so they should get noticed, but so far, none have been spotted and as a result, there's no evidence of extra dimensions.  That doesn't mean they're not there, and that micro-black-holes can't be created, but so far there's no evidence that they've been created up to about 13 trillion electron volts, the current peak of CERN's collision energy.   See here

Hope that makes sense.  Corrections welcome.


How can we ge to the exoplanet Proxima b? I would like to know please how can we get to the exoplanet Proxima b. I heard it is the potentially habitable exoplanet closet to Earth at only 4.2 light years away. So exciting!

It's habitability is uncertain.   Red Dwarf stars that small could be unkind to their planets and strip them of most of their atmospheres.   We'll get a better look at the exoplanets around Proxima Centauri when the James Webb Space Telescope is up and running.  It's scheduled to launch in 2019 and it should be able to tell us something about exoplanet's atmospheres, which would go a long way towards estimating any habitability.

Actually getting there, even with a small unmanned craft is a long ways away.   There's still significant technological difficulties with Breakthrough Starshot.  The lasers aren't set up yet for one.  For two, the aim has to be just about perfect, and three, we currently have no means of receiving a signal from a craft that far away.   We have to improve our communications tech.   Now, in theory, we know how to do it.   We could build giant satellite dishes on the Moon, for example, or orbiting in space, but that's very expensive.

Another issue with starshot is that they have no means of slowing down, so they'd pass through the solar-systems they were sent to examine in a rapid fly-by, perhaps 10% of the speed of light, with only a few hours within the solar-system.   Fly-by missions are still useful, and given the difficulty of transmitting data an orbiting mission in another solar-system might not make practical sense, because such a mission could easily gather more information than it could send back.    The recent New Horizons / Pluto mission was a fly-by and it gathered a lot of information, that took over a year to send back to Earth.

Rapid fly-bys shouldn't be dismissed though.   They can gather a lot of information that we have trouble seeing from Earth, even with newer and bigger telescopes.

More ambitious missions, and ships with their own fuel so they can break and remain in a distant solar-system are harder.   That requires much more equipment, much more mass and as a result, would accelerate more slowly.   Manned missions 4.2 light years away are so far out of reach that it makes little sense to talk about them outside of science fiction.   A manned mission to Mars is hard enough and that's over 100,000 times closer.

Starshot is many years away.   We might see Starshot launch in a decade or two - if it moves quickly.   Then it'll be another 2 decades or so for the ship to reach the Proxima system and another 4 years and change for any information to be returned.   35-45 years is probably on the ambitious estimate before we receive the first starshot data.

Most of the new information about exoplanets that we're going to get over the next few years will come from telescopes like James Webb and others that follow.


Did I see a meteor(oid) or space debris or something else? About three years ago I think, at night I've seen something enter the atmosphere in a glowing bright fireball. But I think it wasn't the fire from burning up, but because of ionization (what also happens to meteors) but I might be wrong and it may have indeed been fire. There was also a kind of a contrail behind the glowing ball (which was visible only in the vicinity of the ball). It was in November in the northern hemisphere, so it might have been a Leonid, but it didn't look like the usual meteors. It was much slower and brighter. It was cloudy that night and eventually it disappeared behind a cloud. Whatever it was, it wasn't a typical meteor nor was it an airplane (it was quite straight above me, soundless and it was something entering the atmosphere).
My guess is it was either a meteoroid that was very dense and comparably big (several feet) and a comparably slow one, or it was some retired satellite or some spacecraft debris entering the atmosphere. What do you think?
update: I estimate its speed at about 1/6 that of a typical shower meteor.

The OP's clarifying comment under the question offers an opportunity to examine further:

Do meteoroids really get that slow? I estimate its speed at about 1/6 that of a typical shower meteor.

The apparent or angular speed of an object in the upper atmosphere depends on several things, including

the actual linear speed in say km/sec
the distance from observer to the object
the cosine of the angle between the direction of motion and direction to object (i.e. the dot product of the normals of those vectors); if it's moving straight towards you the angular speed is zero.

The linear speed of meteors with respect to Earth is a function of the orbital speeds of both objects. See the (currently unanswered) question How to calculate the position of a meteor shower's radiant point based on its associated comet's orbit? and links therein. Just for example if the meteor shower is associated with a comet in a circular orbit at 1 AU with an inclination of 90 degrees (a "polar orbit") then the relative speed is
$$\sqrt{2} \sqrt{G M_{Sun}/1 \text{AU}} \sim  29.7 \sqrt{2} \text{ km/s} \approx  42 \text{ km/s.}$$
Speeds of meteors relative to Earth can theoretically be somewhat higher or much lower than this. So despite 42 being an aesthetic number let's just us 30 km/s for a typical value.
The orbital speed of a spacecraft at 100 km above the Earth before drag stars slowing it down is
$$ \sqrt{G M_{Earth}/(6378+100) \text{km}} \approx  7.8 \text{ km/s.}$$
Since both are roughy at the same distance, a burning reentry might appear to move (in angular velocity) four to six times faster than a small bit of space junk burning up.
That ratio could be reduced if the meteor was sighted near the radiant point, where it will appear to move much more slowly by a factor of the cosine of the angle mentioned above.
So in some cases a meteor might appear to move the same speed as a bit of reentering junk, but in that case the trail will be very short, it will not cross the sky. From the OP's question it sounds like the motion was much larger than a few degrees so we can rule this out.
But what if the comet associated with the meteor was moving in an orbit nearly tangent to Earth's with a similar to velocity to that of Earth at the intersection point (i.e. low inclination)?
There can't be absurdly slow slow meteors like that because of Earth's gravity. An object moving nearly the same as Earth will accelerate as it gets closer to Earth. $v_{escape} = \sqrt{2} v_{orbit}$ means even a co-orbiting object will reach almost 10 km/s by the time it enters the atmosphere. And I don't know of any meteor showers associated with comments who's orbits fit this description.
Conclusion
Most likely it was space junk. Bits and pieces reenter regularly so while it's not common to see, from time to time people see it. And soon some people may pay to see it on demand!

Why would a box full of 1cm balls released into LEO be so scary to an engineer supporting the ISS?


Wavelengths and bounderies I know light and sound travel along wavelengths and in all diagrams I have seen the pictures of photons travelling up and down and forward. How does something travelling so fast stay within the confines of the wave please ? I am new to this site but intelligent enough to understand complex answers.

It would be easier if you could include an example of the kind of image you mean, but I can guess that you're thinking of those drawings where there's a sine wave drawn in space and suggesting that that's what a wave "looks like".

This plot from EMS 2- Wave Properties of Light shows "crest", "trough" and "rest position" when explaining light waves, and this might be a good example of the problem. All three imply that there is some up-and-down-ness and that's not what a light wave is.

I've seen much worse, floating sine waves with arrows at one end, or hitting people or plants, and those drawings are crazy-wrong! ;-) When we draw those sine waves, it's on a graph or a plot of something versus time or versus space on the x-axis, and the y-axis axis on the plot is pressure for sound or usually electric field for light or electromagnetic radiation.

Light or photons, or sound waves for that matter don't physically go up-and-down like that.

The plot below is somewhat better. The arrows are the amplitudes of the electric and magnetic fields. There's nothing that is actually moving up and down, it's just a mathematical plot.

 Source

  English: Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This diagram shows a plane linearly polarized wave propagating from left to right. The electric field is in a vertical plane and the magnetic field in a horizontal plane.

Found in Transverse wave


Why are telescope mirrors made of glass? Optical telescope mirrors are typically made from endlessly polished glass surfaces that have been aluminised to provide a very smooth surface of reflective aluminium.

But why the need for glass at all? Why can't mirrors be made of pure metal, or of metal deposited onto some other surface?

Is there something special about the glass grinding process that means no other surface can be made as smooth and still have aluminium deposited on the surface?

Or is it to do with the thermal expansive properties of the glass used?

you are partially correct. Glass is used for several reasons. It is a very stable material and will hold its shape well for thousands of years. Glass can also be polished to a high degree of accuracy without having defects.  Another major reason is the expansion properties are very favorable as well. Glass doesn't corrode and is easier to mold into shape than metal mirrors. Glass is relatively cheap and readily available.

Some useful references are found here ,here and here.


How will the Breakthrough Initiatives affect other astronomy observations? BACKGROUND

The Breakthrough Initiatives is in the news because of a $100 million private donation for a SETI project. It doesn't seem to be for building any new observatory (but maybe for buying computing power for handling the data for SETI purposes?). It will use two of the greatest radio telescopes in the world, Parks and Green Banks, and optical Automated Planet Finder Telescope at Lick, some part of the time during 10 years.

QUESTIONS

Does this mean a zero sum redistribution of observatory time from other astronomical purposes to SETI?
How does it work when someone can buy the usage of these telescopes? Isn't there a danger in letting popular scifi ideas decide how the great telescopes should be used? Do recruited big names like Hawkins and Drake decide how they should be used anyway??

I was fortunate enough to spend the past week at a workshop held at the Green Bank Observatory (GBO), so I can give you a partial answer based on how the Breakthrough Initiatives affects it in particular.
Telescope time
The GBO actually operates a selection of telescopes, of which the most widely-known (and largest) is the Green Bank Telescope (GBT). The GBT is the main instrument used by Breakthrough Listen at the observatory. During the second half of 2016, the GBT was oversubscribed by a factor of 2.6, meaning that only about one in every 2.6 proposals was approved outright (“filler” proposals will be accepted if short periods of time are available). This doesn’t mean that for every hour of telescope time, 2.6 hours worth of observations were proposed, although the ratio is close (2.3), but it does mean that the GBT is very much in demand - and those numbers have increased this year. The telescope is not in use on only two days annually, Thanksgiving and Christmas, although there are blocks set out each week for maintenance.
The answer to your first question, then, is yes - there is approximately indeed a zero-sum redistribution, because the telescope is extraordinarily busy. The actual data-gathering proportion may be slightly different because overhead times vary by the type of observations, and things like changing receivers or pointing at a different source do take time. However, it is true that Breakthrough Listen, like any other set of observations, takes time away from other scientists - about 20% of the GBT’s annual total observing time.
(As an aside, Breakthrough Listen operates a “backend” in the GBO computer room - a set of GPUs that analyze incoming data. There are plenty of backends for various receivers and instruments, and they do take up space - and power. We were told that the room the Breakthrough Listen computers are in wasn’t actually designed to hold that many computers, which did cause some initial cooling issues (which were quickly rectified) when these new units were installed. So it’s important to realize that the project takes up more resources than just observing time!)
Money
The financial system at Green Bank is actually a little complicated. In 2012, it was recommended that the GBT and the Long Baseline Array be removed from the NRAO, and this was implemented in 2016. In some ways, the transition is still happening - much official GBO information is still on the NRAO’s website - but the relevant point here is that in the 2017, the GBO only received about 66% of its funds from the National Science Foundation (NSF). The current agreement is that the NSF will gradually defund the GBO over the next few years.
This is why Breakthrough Listen is so financially crucial, and why it is getting so much publicity at the GBO. Without sources (and currently, without non-governmental sources) of funding, the observatory simply cannot operate. That said, the GBO is not beholden to people like Yuri Milner (the man behind the Breakthrough Initiatives) who have specific projects in mind - and indeed, that type of donor is quite the outlier. It does seek out donors and philanthropists who are willing to give money to simply ensure that the facility keeps going.1 If it can be guaranteed that these sources will continue giving without setting limits on what types of observing should be done, then things can progress as close to normal as possible.
The exact details of funding for the GBO are not (as far as I know) public, and I don’t want to speculate too much. However, I can say that the observatory is almost certainly not going to push ordinary astronomers out of the way for the sake of big projects - in the short-term. Long-term funding is uncertain, true, but that does not mean that it will be renamed the Breakthrough Listen Observatory any time soon. The number of people who have operated the GBT number in the thousands, and they don’t pay to use it. It was built using public funds and has operated largely using public funds for quite some time now. I think the GBO intends to keep operating like this, and will not turn to the alternative - subject to the interests of only a select few projects - unless matters become extraordinarily desperate. It’s possible, but not something to be immediately worried about.

1 One day last week, we noticed a helicopter landing at the airstrip. It turns out that the passenger was a VIP who was given a full tour of the telescope, guided by several observatory higher-ups. He was, as you can image, a prospective donor.


How do you calculate the effects of precession on elliptical orbits? Kepler's first law states that planets (and all celestial bodies orbiting another body) travel in elliptical orbits, which have well-known formulas that make it relatively easy to calculate the orbital elements and associated behavior. However, ongoing precession means that the orbit is constantly changing - and so the planet isn't actually traveling in the ellipse it originally set out on! You can calculate precession and its related effects (this question and answer are helpful), but is there any way to calculate how the elliptical orbit will be "deformed" by precession?

A good starting point would be <insert name of some scientist from long ago> planetary equations of motion. For example, there are Lagrange's planetary equations (sometimes called the Lagrange-Laplace planetary equations), Gauss' planetary equations, Delaunay's planetary equations, Hill's planetary equations, and several more. The common theme amongst these various planetary equations is that they yield the time derivatives of various orbital elements as a function of the partial derivatives of the perturbing force / perturbing potential with respect to some generalized position.

In general, the only words that can describe the result of this process at first is "hot mess." A hot mess did not deter those brilliant minds of old. Via various simplifying assumptions and long term time averaging, they came up with fairly simple descriptions of, for example, $\left \langle \frac{d\omega}{dt} \right\rangle$ (apsidal precession) and $\left \langle \frac{d\Omega}{dt} \right\rangle$ (planar precession). You can see some of this in the cited 1900 work by Hill below.

While these techniques are old, these planetary equations are still used today. That sometimes you do get a "hot mess" is okay now that we have computers. People are using planetary equations coupled with geometric integration techniques to yield integrators that are fast, accurate, stable, and conserve angular momentum and energy over long spans of time. (Normally, you can't have all of these. You're lucky if you get just two or three.) Another nice feature of these planetary equations is that they let you see features such as resonances that are otherwise obscured by the truly "hot mess" of the cartesian equations of motion.

Selected reference material, sorted by date:

Hill (1900), "On the Extension of Delaunay's Method in the Lunar Theory to the General Problem of Planetary Motion," Transactions of the American Mathematical Society, 1.2:205-242.

Vallado (1997 and later), "Fundamentals of Astrodynamics and Applications", various publishers.
Other than the hole it punches through your wallet, you can't go wrong with this book.

Efroimsky (2002), "Equations for the keplerian elements: hidden symmetry," Institute for Mathematics and its Applications

Efroimsky and Goldreich (2003), "Gauge symmetry of the N-body problem in the Hamilton–Jacobi approach." Journal of Mathematical Physics, 44.12:5958-5977.

Wyatt (2006-2009), Graduate lecture course on planetary systems, Institute of Astronomy, Cambridge.
The results of the Lagrange planetary equations are presented on slide 6.

Ketchum et al. (2013), "Mean Motion Resonances in Exoplanet Systems: An Investigation into Nodding Behavior." The Astrophysical Journal 762.2.


When was visible light for the first time created? "A few minutes into the expansion, when the temperature was about a billion (one thousand million) kelvin and the density was about that of air, neutrons combined with protons to form the universe's deuterium and helium nuclei in a process called Big Bang nucleosynthesis. Most protons remained uncombined as hydrogen nuclei. As the universe cooled, the rest mass energy density of matter came to gravitationally dominate that of the photon radiation.

After about 379,000 years the electrons and nuclei combined into atoms (mostly hydrogen); hence the radiation decoupled from matter and continued through space largely unimpeded. This relic radiation is known as the cosmic microwave background radiation."

The question now is, when you count back starting with the cosmic background radiation which is now in far red (2,7K) at what time was there for the first time visible light, or was is right after the creation of atoms about 379,000 after the big bang?

The best model we have is that of electro-weak unification.

The photon is an quantum of the electromagnetic field, and as long as there has been an electromagnetic field, excitation of the field have existed.

The "weak" nuclear interaction is another field, (it has the effect of changing up quarks to down quarks, and so is responsible for beta decay of radioactive material). Quanta of the weak field are rather different from photons. They are charged and have mass, but at very high energies (a million billion degrees or so) the electromagnetic and weak fields are unified into a single "electroweak" interaction.

But after the universe had cooled (it took sometime between $10^{-36}$ and $10^{-32}$ seconds) the weak and electromagnetic fields split, and the first true photons formed.

These photons didn't get very far before colliding with other particles that were being formed. As you note, it took nearly 400000 years until the universe cooled enough to become transparent.

Now you mention visible light. The photons formed after the split would have had very high energies, we would call most of them gamma rays if they existed today. But as photon of all energies could be formed, a small fraction would have been in the visible range.

At recombination, when the universe became transparent, the temperature was about 3000K, at which temperature most photons are in the infra red range, but enough are visible to have given the universe a very bright orange-red colour, similar to the colour of a lightbulb.


Change of orbit with change of mass? In the science fiction book 2010, monoliths increase the mass of Jupiter until it becomes a star.

https://en.wikipedia.org/wiki/2010:_Odyssey_Two

  The Leonov crew flees Jupiter as a mysterious dark spot appears on
  Jupiter and begins to grow. HAL's telescope observations reveal that
  the "Great Black Spot" is, in fact, a vast population of monoliths,
  increasing at an exponential rate, which appear to be eating the
  planet. By acting as self-replicating machines, these monoliths
  increase Jupiter's density until the planet achieves nuclear fusion,
  becoming a small star.

Obviously fiction.  But my question is about orbits and the mass of orbiting bodies.  For planets, whose mass is negligible compared to their star, I understand one can ignore mass for purposes of calculating orbits.

But if Jupiter were by scifi magic made as massive as a star, would its orbit remain the same regardless?  Or if it were made to stay on its same orbital path, would it have to move more quickly, completing its orbit of Sol in a shorter time?

What you said :

  monoliths increase the mass of Jupiter

What you quoted :

    these monoliths increase Jupiter's density

Increasing density is not the same as increasing mass.

By squashing stuff together you can increase it's density without affecting it's mass at all.

  But if Jupiter were by scifi magic made as massive as a star, would its orbit remain the same regardless?

No, nor would any body in the solar system, including the Sun, remain undisturbed.  It would be a massively complex change.


If dark matter bends light, how do we know the stuff in the sky is where we think it is? We measure movement, position, and many other things of an object in space because of its light and what we can measure with it. But as far as I know there's supposed to be a HUGE amount of dark matter in space whose mass and size we don't know, and because it has mass it has gravity and it can bend light.

I know physics can take into account the gravity of stars and huge stuff in space, but how can they be certain of any of their measurements (specially the position) if they don't know how that light has deviated from a straight line?

The local dark matter density is actually quite tiny, on the order of $\rho\sim10^{-19}\text{ g/cm}^3$ (see e.g. Bovy & Tremaine (2012)). This means that there is roughly $0.001$-$0.01M_{\odot}$ of dark matter per cubic parsec - a staggeringly small amount. 1000 cubic parsecs would contain about one solar mass of dark matter - and that's a cube 10 parsecs in length on each side! Now, the distribution of dark matter in galaxies is not homogeneous - it follows, roughly, a Navarro-Frenk-White profile, decreasing in density from the center of the galaxy - but on the scale of parsecs (and certainly in the Solar System), we can consider it to have roughly uniform density.

On small scales, then, we have approximate homogeneity and low density. This means that any gravitational lensing effects from dark matter should be extremely low or self-cancelling, arising only from inhomogeneities containing large clumps of dark matter. However, such clumps are unlikely to form solely through dark matter's interaction with itself (if we discount the MACHO hypothesis, which, as far as I know, is not currently favored).

On intergalactic scales, however, dark matter can have some effects. Weak lensing is a commonly-observed phenomenon in galaxy clusters, which may have extremely high fractions of dark matter. There are several techniques currently used to model the mass distribution of the lensing galaxy (see the KSB+ method) and to reconstruct the image and position of the original galaxy via deconvolution (see Chantry & Magain; a visual example is given here). I'm not too familiar with either technique, though, so I can't give you a good overview.

Even large-scale lensing has large mass requirements. zephyr pointed out that the foreground object that created the Einstein cross contained $\sim10^{10}M_{\odot}$ of dark matter (van de Ven et al. (2010)). That is enormous!


Is there any planet or satellite in the solar system in the process of getting tidally locked? According to

Tidal locking

  The effect arises between two bodies when their gravitational
  interaction slows a body's rotation until it becomes tidally locked.
  Over many millions of years, the interaction forces changes to their
  orbits and rotation rates as a result of energy exchange and heat
  dissipation. When one of the bodies reaches a state where there is no
  longer any net change in its rotation rate over the course of a
  complete orbit, it is said to be tidally locked.

Is there any planet or satellite in the solar system in the process of getting tidally locked?

The Moon is in the process of tide-locking the Earth...
Tidal interactions between the Moon and the crust and oceans of Earth have been slowing down the Earth's rotation since the moon's formation. In the past 4-odd billion years, the Earth's rotational period has gone from about 6 hours to about 24. These tidal interactions are also responsible for the Moon receding from Earth, and given enough time, is projected to result in the orbital and rotational period of the mutually-tide-locked Earth-Moon system stabilizing at about 47 days long.
But the Sun might run out the clock...
How long is this tide-locking projected to take? My quick googling on the answer comes up with a number of different estimates, from 7.5 to 50 billion years.
According to Wikipedia: Tidal Locking:

Earth is not expected to become tidally locked to the Moon before the
Sun becomes a red giant and engulfs Earth and the Moon.


Is there a map of the galaxies? I see that there are many maps of a single galaxy, the Milky Way, showing the positions of the stars within our galaxy.

I could not find a map of Galaxies, however. Is there such a map? In lieu of that I suppose I could make my own map if I knew celestial coordinates of all the nearby galaxies and the distances to those galaxies. Is there such a data table readily available?

A map of all galaxies gets kind of unwieldy, like a map of all stars in the milky way or a map of every house in the country, or every grain of sand on a beach . . . you get the idea.

Start here - Local Group

Source

Then Virgo Supercluster

Source

Then local superclusters

Source

And an article, even if it's a summary it's very much worth reading, with a more recent map with a "ginormous" supercluster that includes the milky way.

and here's another, more info here.

The largest of these maps is some 520 million light years across, so this is just a tiny part of the entire known universe, which (depending on how you measure) is either 27.5 billion light years across or 84 billion light years across.

It's worth pointing out, I haven't actually answered your question, and that was deliberate, cause I think the nearest 520 million light years is enough and it's all we have a really good picture of anyway (as far as I know).  The farther out you go the more holes and inaccuracies there will be in the map.

There's also some closer stuff that's blocked by the Milky way so we can't see it, like the great attractor.  No worries about superclusters eating our galaxy one day because dark energy expansion keeps these super-clusters from merging most of the time.  Andromeda will merge with the Milky way in about 4 billion years though but we're not expected to ever merge with the great attractor, even as it pulls our galaxy towards it, the distance between us is growing.


Why did this meteor appear cyan? A meteor recently flashed over Queenland.

The meteor has been described by various people as appearing cyan or aqua, so the blueness of the image can't be due to a camera issue. The colour doesn't appear to be anything found along the black body spectrum either.

What (chemical/physical process) would cause this meteor to appear cyan?

 click for full size, Source

  ‘I saw it come right down in the sky’: the meteor over south-east Queensland on Saturday night. Photograph: Craig Turton/AAP

The colour of a meteor depends mostly on its chemical composition emission spectrum. To simplify magnesium meteors are noted to have an emission spectrum which results in blue to green colour. Changing factors like adding additional metals and the speed of the meteor could results in a cyan meteor compared to a blue one.

Source:
NASA on meteors


Are there examples of the effect of shock waves on chemical evolution in of interstellar medium or circumstellar medium? Are there any examples of calculations or observations of suspected effects of shock waves on chemical evolution in interstellar medium or circumstellar medium?

Shock waves in the interstellar medium (ISM) can be caused by a spiral density wave, supernova shock, or turbulent flow collision. These waves sweep, compress and heat the ISM, thus modifying which chemical reactions can occur. These physical conditions can harbor the chemical reactions to create complex molecules such as NH$_3$, CH$_3$OH, H$_2$O and more.

Based on a chemical/dynamical model, Bergin et al. (2004) investigates the formation of molecular clouds behind shock waves. In the $10^6$ years after the shock wave has passed, the temperature of the ISM is rather high (10 000 K) and molecules are photodissociated. As the ISM cools down, H$_2$ can start to form, followed by CO (which needs to be shielded from photodissociation by the H$_2$). Within 20 million years after the passage of a shock wave, atomic gas has been turned into molecular material.

The chemistry that occurs due to a supernova shock could even explain where some molecules found in comets come from. During a shock, water vapor is formed through reactions with hydrogen and oxygen. Bergin et al. (1998) shows that H$_2$O and CO$_2$ formed in the shock can deposit on grains.


Helioseismology of the Sun, what is actually measured? Wikipedia's article Helioseismology outlines several ways to analyze, data but is fairly thin on the nature of the measurements themselves.

Are there hundreds or thousands of high resolution spectroscopic measurements of the Doppler shift of some emission or absorption line distributed across the disk, or can this be done using low resolution spectroscopy to generate some variation in temperature or brightness?

I'm interested in a basic appreciation of the nature and granularity of the data (e.g. 0.1 angstrom in wavelength, 1 second of time, 1 minute of arc, etc.) and what kind of instruments collect said data.

The answer doesn't necessarily have to be long, just something to get me started.

The two main methods to detect the solar p-mode oscillations, which have a period of about 5 minutes (so frequency of ~3 milliHertz) are line of sight Doppler shift and change in irradiance.

The Doppler method works by measuring the intensity of light in (normally) two narrow regions either side of a prominent spectral line in the Solar spectrum and using this change to measure the velocity shift caused by the oscillations (once the solar rotation is removed). The ground-based GONG instrument uses the Ni I at 676.8nm, BiSON network and the GOLF instrument on the SOHO spacecraft both use the sodium D lines. After isolating the light around a suitable spectral feature, then you can either use a Michelson interferometer to get a velocity image of the shifts across the surface (GONG and MDI on SOHO) or a disk-averaged velocity, treating the Sun as a star (BiSON and GOLF on SOHO). The difficulties are in keeping everything stable so that the measured shifts only come from the Sun and not the instrument. For example GONG note that their optical elements are stabilized in an oven to the order of 0.00001 K.

The irradiance method simply involves measuring the disk-integrated intensity of the Sun and watching for changes caused by the oscillations. However fluctuations in the Earth's atmosphere make this very difficult to do from the ground and the most successful has been the VIRGO instrument on the SOHO spacecraft.

The imaging instruments have a resolution of about 2" (MDI) to 5" (GONG) so return velocity data at approx. 800,000 to 100,000 locations across the Sun. This allow the study of oscillations with very high degree (up to 1000) which have very small scale on the surface. The disk-integrated "Sun-as-a-star" instruments by comparison are sensitive to the low order oscillations.


Why can't we travel faster than light? And is the theory of relativity absolutely flawless? Why can't we travel faster than light? And is the theory of relativity absolutely flawless?

No scientist really considers a theory flawless, but as theories put together by mankind go, it is one of the best ones put out there.  One of the things that makes for a great theory is not only being able to describe what's already happened, but to make predictions about new stuff, and general relativity did that with predictions of things like gravitational waves that took 80 years for technology to get good enough to see.

General relativity also passes an aesthetics test, where some people feel that the most fundamental laws are beautifully simple and elegant, and though as far as we know, there is no scientific reason that says the Universe is simple and elegant, most of the best theories turn out to be that way, including general relativity.

So if you acknowledge that relativity handles all the many many tests that have been thrown at it, the answer to your first question has to do with the Lorentz factor, which goes as $ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $, so you see that as your velocity $v$ gets closer and closer to the speed of light $c$, the bottom of the fraction gets closer to zero and the Lorentz factor blows up.  Since the kinetic energy of something moving really fast involves the Lorentz factor, that means as you approach the speed of light, our energy blows up, and to actually travel at the speed of light, you'd need an infinite amount of energy.


How was the Messenger spacecraft used to measure Mercury's orbital precession to such accuracy? Could this have been done using radar? This comment under an answer to Path of Mercury and general relativity mentions that the Messenger spacecraft was used to measure the precession of Mercury's orbit to such accuracy that the tiny component due to the Sun's $J_2$ could be resolved from other components including relativistic effects and the gravitational pull of other solar-system bodies. The value of this component is given there as "about 0.03 arcseconds/century".

That rate corresponds to a motion of the apses of about 500 meters per year, which I roughly estimate translates to an error in Mercury's position that accumulates at about 100 meters per year.

That seems to be within the a range that could be doable with an extended series of radar measurements. For comparison, in the 1960's Venus' diameter was measured to +/- 3000 meters way back in the 1960's and a precision measurement has such importance to general relativity validation.

Questions:

How was the Messenger spacecraft used to measure Mercury's orbital precession to such accuracy?
Why couldn't this have been done earlier using a series of radar observations, or could it have?

The measurements discussed weren't done by Messenger, they were done by round-trip delay-doppler radar ranging using the coherent transponder of Messenger's Radio_science_subsystem.

Park et al. (2017) discuss separating the effects of the solar quadrupole moment from the partially degenerate effects of the post-Newtonian $\beta$ and $\gamma$ parameters (where $\beta=\gamma=1$ in General Relativity). The "general" General Relativistic correction to the perhelion prececssion itself is given by:
$$\Delta \dot{\omega} \propto (2 - \beta + 2\gamma) $$
and the effect caused by the solar quadrupole moment is only 0.07% the size of this. Therefore just observing the precession does not yield an accurate value of $J_2$ without assuming the value of $\beta$ and $\gamma$, and even then would need precisons in position measurements of hundreds of metres over decades.

Park et al. instead show that separating $J_2$ from $\beta$ and $\gamma$
 requires a precise measurement of a periodic modulation (i.e. it isn't an accumulating error) in the argument of perhelion caused by the General Relativistic perturbations, which translates to modulations of order of 5-10 km in position. These also depend linearly on $\beta$ and $\gamma$ only, and if the amplitudes of these modulations can be measured to better than $\sim 0.07$% ($\simeq 3.5-7$ m) then $J_2$ could be determined independently of $\beta$ and $\gamma$.

Given that the position of the spacecraft can be tracked to a lttle better than 1 metre, this is possible. $J_2$ is indeed measured to about 4%, and half of this uncertainty is due to remaining degeneracy with the $\beta$ parameter and uncertainty in the adopted $\gamma$ parameter.

This suggests that precisions of the order of a few metres were required for this experiment and I don't believe this is possible by radar ranging to the planet itself.

Indeed on p.181 of the 1993 version of "Theory and Experiment in Gravitational Physics by Will, it is said that

  Unfortunately, measurements of the orbit of Mercury alone are incapable at present of separating the effects of relativistic gravity and of solar quadrupole moment in the determination of [the change in the perihelion precession rate]

and then on p.183

  The accuracy required for such measurements would necessitate tracking of a spacecraft in orbit around Mercury


Extragalactic Magnetic Fields - General Information That I'm Missing I have started studying Extragalactic Magnetic Fields, and mainly focusing on Biermann Battery Theory. I have gathered much information on other Dynamo Theories, as well as about our current Observational and Theoretical status on Magnetic Field Genesis.

What I'm looking for though, is a general "picture" on Extragalactic Magnetic Fields, as a whole. Faraday Rotation, AGN Jets, are basically the topics I have tracked, but I'm pretty sure there are a lot of others.

So I'm asking, if someone has studied in-depth extragalactic structures and magnetic field, possibly could guide me on some of the main and hot topics around them. This way I would have more knowledge on the topic which would be great for me!

I mean just Topic-Titles, would be great!

On this question, the Widrow (2002) review is still pretty up-to-date. You can also have a look on the introduction of Beck (2013), a recent PhD thesis on the subject with a straightfoward introduction to the subject.
The main questions remain the same for decades now:

What is the origin of the magnetic seed fields?
How to amplify magnetic field up to observed values in the local Universe?
How to get large scale magnetic field, in particular if the magnetic seed fields are produced locally?

Origin of the magnetic seed fields
Where do we start? That is no easy question, with a diversity of answers. One of the most standard ones is the Biermann battery you mentioned (one significant problem with Biermann batteries is the timescale needed to amplify such small fields; see below) in the early Universe. Another one is that the magnetic field can be generated at small scales by first generations of AGN and supernovæ and then distributed at larger scales (dynamo can help but distribution at larger scale can still be an issue; see below). Lastly, you can have some exotic processes to generate seed fields in the first instants after the Big Bang, typically linked to inflation processes and such.
Magnetic field amplification
If there is an initial magnetic field — which is not so easy to demonstrate — there are few mechanisms that could amplify magnetic fields efficently enough to explain the observed magnetic field. The most standard one is the $\alpha-\omega$ dynamo, combining a twist of the magnetic field lines in a differentially rotating disk with turbulent motions in the ISM that allows a regeneration of the magnetic fields (supernovæ explosions, stellar winds and other instabilities could do the work). Another possible mechanism is the magneto-rotational instability (MRI), that can easily generate a significant, large-scale field from almost any given seed field. There are other candidates such as Parker instability, or other flavors of dynamos.
A central question then is of timescale: if these mechanisms can generate magnetic fields, the needed timescale to amplify a seed field to the current value may be far larger than the age of the Universe.
Distribution
Magnetic field is everywhere, not only in the ISM and at larger scale in galaxies, but also in galaxy  clusters, in the IGM and so on. If the origin of the magnetic seed fields is of cosmic origin (exotic processes in the early Universe), you can explain the observed distribution; if not, things get tricky again. Turbulence, galaxy collisions and mergers may be of help, but the question is still largely open.


What is the object in this photo? What object is in this photo?

View coordinates: $47.25103 \ \ 38.81697$

Time: $2013$-$11$-$23 ~21$:$00 +4 ~UTC$

Sony $A580, \ 50mm f/1.4 15"$

Your guess was correct. It is the Andromeda Galaxy, M31.

Here is a map of the part of the sky near zenith at the place and time you provided: Sky map for Taganrog, Russia on 11/23/2013 5:00:00 PM UTC. Even the rotation is small. The sky map is rotated approximately 30° counter clockwise relative to the photo. You were approximately facing south when taking it.

On your picture there is a part of the Andromeda constellation. I will use only a small bottom part of your picture (below the imaginary horizontal line going through M31) to identify the constellation.

The bright star to the left of M31, very close to the edge is μ Andromedae (marked 37 μ And on the sky map). It is the star between Mirach (out of photo) and M31.
On the bottom of the right part of the photograph you see three stars on a imaginary vertical line: ψ, κ and ι Andromedae.
Between ψ and κ, to the right there is λ Andromedae (marked 16 λ And and connected with μ on the sky map).

The bottom of the three stars in the upper-left corner of your photo is φ Andromedae (connected by the vertical line with 37 μ And on the sky map).

The 1 ο And on the sky map is not on your photo.


Is radiation from neutron stars delayed by time dilation? I understand a neutron star to be the densest stuff that can exist without becoming a black hole, the densest thing which directly gives off radiation we can detect. At the event horizon of a black hole, time stands still, in our frame of reference.

Does there exist neutron stars which are very close to getting an event horizon?

Does this "neutron star horizon" cause time dilation, a zone from within of which movement is very slow in our frame of reference?

Could it for example be that the first light from a neutron star itself reaches us billions of years after the light of the supernova of the "classical" star which created the neutron star and its event-horizon like relativistic effects?

A typical neutron star of $1.5M_{\odot}$ is thought to have a radius of around 8-10 km. This is only a factor of 2 larger than the Schwarzschild radius for a similar mass black hole.

We know that more massive neutron stars do exist. The current record holder is around $2M_{\odot}$. Most equations of state (the adopted relationship between pressure and density) for dense nuclear matter suggest that more massive neutron stars are smaller and therefore must be even closer in radius to the Schwarzschild radius.

So the premise of you question is basically correct. It is certainly true that when you deal with neutron star spectra you do have to apply significant general relativistic corrections to measured temperatures and the same corrections would need to be applied to any temporal variations.

Thus a time-variable signal from a neutron star surface will appear slower to an observer on Earth.

For the last part, I suspect that the scenario you propose is extremely unlikely. Rhoades & Ruffini (1974) first established that there must be a maximum mass for a neutron star under GR conditions, even if we allow the equation of state to harden to the point where the speed of sound is the speed of light. This maximum mass is around $3.2M_{\odot}$.  This sets an upper limit to the possible value of $GM/Rc^{2} \leq 0.405$ (see p.261 of Shapiro & Teukolsky, Black holes, white dwarfs and neutron stars). This in turn sets an upper limit the possible gravitational redshift (and time dilation factor) of 2.29.

Beyond this point the neutron star is unstable and will collapse to become a black hole. In reality the limit is probably a bit tighter than that because most proposed equations of state result in neutron stars becoming unstable at finite densities and at masses quite a bit lower than $3.2M_{\odot}$.

So I think the most time dilation you are ever going to see from a neutron star surface is a factor of $\sim 2$.


Would it be possible to equip an asteroid to collect space junk in Earth's orbit? We all have heard about the problem with space junk in low earth orbit. NASA has a plan to put an asteroid in orbit around the moon. Would it be possible to equip propulsion systems onto an asteroid in order to maneuver it into an elliptical orbit around the earth, with the goal of collecting this junk?

This is plausible, and might even be a good idea if used right.

First off, NASA has been working on plans for an asteroid redirect mission, called ARM. While it remains to be seen whether this will be approved, and they plan to put it into lunar orbit, this is a hypothetical scenario anyways so I'll ignore that.

Now, putting an asteroid into earth orbit is a little bit of a difficult subject. Sure, you could do it, but messing up would net you a whole lot of problems. I presume it would have to go into high earth orbit if it goes into orbit at all.

This rules out using the asteroid directly like a satellite, but that would have been impractical even if it was sanctioned by everyone with enough influence to have an opinion on earth. From there the satellite would be best used as a base of operations, from which you could send drones to collect junk and keep refueling stations. I'd also include a telescope or two, but that's personal preference.

If you want a list of targets though, here would be a good place to look.


Which star with a Bayer designation is the farthest from the Sun? The key to finding the answer seems to involve processing data from astronomical databases, which I'm not sure how to do in this case. I'd prefer to limit the answer to the original list of approximately 1500 stars, but answers that rely on a list that contains stars with more recently acquired designations could be interesting to consider as well.

See important caveat to this answer below

At a distance of 2992 light years, Tau Canis Major is the furthest of the 1513 stars that have a Bayer designation, with several others almost equally distant. I computed the full list of Bayer-Sun distances: https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-bayer-distance.csv

Important Caveat

I used the HYG catalog at https://github.com/astronexus/HYG-Database but it may be inaccurate. In particular, Wikipedia gives Tau Canis Major's distance as 5,000 light years, although I couldn't find this number in the source provided (https://doi.org/10.1088/0067-0049/199/1/8), although I only skimmed.

The far from canonical https://www.universeguide.com/star/taucanismajoris notes that 2992.32 is a revised figure, down from 3197.68, but, of course, the figure may have been revised again.

Methodology

After downloading the HYG database file, this one liner creates bc-bayer-distance.csv:

zcat hygdata_v3.csv.gz | perl -F, -anle 'if ($F[9] < 100000 && length($F[27])>0) {$F[9] *= 3.26156; print "$F[9], $F[27] $F[29]"}' | sort -nr | tee bc-bayer-distance.csv

A brief explanation:

The 9th field (starting our count with 0 as Perl does) is the distance, but there's a note on the HYG git page: "A value >= 10000000 indicates missing or dubious (e.g., negative) parallax data in Hipparcos". It turns out the actual number is 100,000 (I've submitted a correction), so we ignore distances greater than 100,000.
The distance is in parsecs, so I multiply by 3.26156 to convert to light years.
The 27th field is the stars Bayer letter designation without the constellation. If it's empty, we ignore the star.
The 29th field is the stars constellation name, which, combined with 27th field, yields the Bayer designation.


Which spiral arm of the Milky Way is Kepler-62 in? I am trying to find out which spiral arm of the Milky Way the star Kepler-62 and its planetary system are in. All I can find by Google searching is that it is part of the Lyra constellation, but I don't know if that entire constellation is within 1 spiral arm or not, nor which arm it is in.

Is Kepler-62 in the Carina–Sagittarius Arm, or in the Orion Arm, or...?

The only map picture I can find that is somewhat related is this (I assume that the yellow light area is the area in which Kepler-62 was sighted by the Kepler spacecraft): http://upload.wikimedia.org/wikipedia/commons/b/be/LombergA1024.jpg

Thank you!

The Kepler space telescope had a field of view along the so called Orion arm, or spur, of the galaxy. The same structure which we ourselves are inside. It basically looks along a line where the star density is the greatest within the distance that it could detect transiting planets. Less than 3% of the galactic diameter, I think. Note that the illustration your question linked to seems to overestimate that distance. One link and another link. Kepler-62 and all another Kepler stars are certainly in that FOV. It would not be able to see transiting planets beyond the local arm. However, if the Kepler K2 mission gets going, it will use several different field of views.


Possible occultation of HIP 99314 and Ganymede We all are of course looking forward to see the great conjunction of Jupiter and Saturn. So I looked up the timeline of close approach in Stellarium and I saw that there might be an occultation with Ganymede and HIP 99314 (JD = 2459204.532313, date = 2020-12-21, 00:46:31 UTC). The star has a magnitude of around 7.5 and Ganymede has a magnitude of around 5.47 (combined magnitude is 5.31 -> change in magnitude is 0.16), so it might be interesting to see the perceivable dim (maybe not with the naked eye, but maybe on camera). Unfortunately, I don't think that Stellarium is built with that precision, so I am asking your opinion about this situation.
There are 2 options:

Stellarium is built with that precision and this event is even more rare than previously thought.
Stellarium is wrong: there are other data that don't support its results.

Help is very appreciated, but let's hope for the best.
Thank you!

For what it's worth, I just got a shot of what looked like a 5th Galilean moon near Ganymede tonight (from South Florida). Here's a quick, unprocessed image. I thought I was going crazy seeing an extra moon (in the correct plane) that kept showing up. But looks like it's HIP 99314 (2nd dot to the upper left of Jupiter).


Can the difference between a star and a galaxy which are point sources be detected? Obviously a star would be a point source. A galaxy should be an irregular blob if close, but if it is far away then it would seem that a galaxy too would be just a point source.

Given that the star and galaxy were both only detectable as point sources can astronomers tell them apart with redshift? By some other method?

A followup question...

What percentage of galaxies if our universe can we only detect as point sources?

To distinguish galaxies from stars, you can use the spectrum. Roughly, stars have a black-body like spectrum with features depending on the absorption and emission on the line of sight and in the chromosphere of the star.

Galaxies on the other hand of a spectrum that is the composite of tons of stars. The spectrum will for example be much wider (ranging from smaller to larger wavelengths) because of the diversity in the spectra of stars.

Take a look at http://www.atnf.csiro.au/outreach/education/senior/astrophysics/spectra_astro_types.html if you want a quick overview of the differences.

I don't have a precise number about the number of galaxies we see as point source, but the answer varies greatly from one instrument to another one. If you try to observe a galaxy using radiotelescopes in interferometry, you can resolve much better scales than an Earth-based small visible telescope, etc…


Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind? Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind?

Yes, the ions are said to be "swept out" by the solar wind, such that they reach the speed of the solar wind.  That speed varies a fair amount, however, because there is both a "fast" (about 600 km/s) and "slow" (about 300 km/s) solar wind.  It is really the magnetic fields entrained in the wind that sweep out the charged ions, since each ion has such a low mass it is easily affected by those fields.  Note the dust particles from a comet, which is the tail that we see, are much more massive and are not affected by the solar wind, so they are moving out much more slowly and often make a noticeably curved shape.


Would the heliosphere disappears if our sun becomes a white dwarf? Just curious I like to know if one fine day our Sun turns into a white dwarf at the later stage of its life, would the magnetic bubble, heliosphere gets thinner and eventually disappear? also at this moment is it changing in size?

Yes the heliosphere will disappear. As you know, the heliosphere exists due to the solar wind, which consists of charged particles that follow the magnetic field lines starting from the Sun and extending far into the universe (of course at some point they return to the Sun, as the loops need to be closed, but it's so far away that we usually assume it to be at infinity). Now when the Sun goes Red Giant and then loses its outer shells (planetary nebula), it will become a white dwarf. Then there won't be any wind originating from its surface and therefore there won't be any "magnetic bubble" left.
The heliosphere stays roughly the same in size as long as the Sun remains in its Main Sequence stage, as a star remains relatively stable during this time of its life.