Filling the Roche Lobe For mass transfer in a binary system, one star must fill its Roche lobe. What determines whether a star in a binary system will fill its Roche Lobe? How will I calculate it?
I can't find any mathematical formula that shows any equation of mass transfer.

One should distinguish the "Roche limit" from the "Roche lobe," as the latter is what you need when you have two objects of similar mass, like a binary star.  The Roche lobe is an effective equipotential in a frame that cororates with the orbit of the binary (assumed circular).  The equipotential is "effective" because it includes both stellar gravities, and also the centrifugal force from the rotating frame.  The Roche lobe is not only an effective equipotential, it is the last one that is closed-- any equipotential further out connects to infinity and cannot correspond to material in equilibrium that is corotating with the orbit.  From the side, this last equipotential looks like a kind of infinity sign.  Material outside that "infinity sign" has to start moving in the rotating frame, bringing in coriolis forces and in generally leading to complicated motion.  However, if close to the Roche lobe, the gas tends to flow through the nodal point at the center of the "infinity", so tends to be transferred from the domain of gravitational attraction of one star to the other.  That's "mass transfer."

Setting up the calculation of the Roche lobe is fairly easy because you only need the two gravities and the centrifugal force at the orbital period.  But it is a difficult set of equations to solve, and even at the research frontier, is usually approximated rather than calculated exactly.  No doubt this is the source of your difficulty.

Why one star fills its Roche lobe is a consequence of stellar evolution, often as one star attempts to evolve into a red giant or asymptotic giant.  Before it can reach its giant radius, its outer layers encounter the Roche lobe, and pass along it through the central node (the "Roche point") and may ultimately end up attracted to the other star.  This is what is thought to have happened in Algol, for example, which is a binary where the star with the greater mass now started out as the star with the lesser mass (because it gained from its companion).  The rate of mass transfer is also difficult to compute, because it depends on how fast the mass-loser is evolving.  It is only its evolutionary rate that causes the "Roche overflow," so the mass transfer rate is pegged to the evolutionary rate, which is largely due to rate of fusion that adds mass to its core.


Orbital velocity of a planet - why is my calculation off by about 10%? I am not sure if I am doing something wrong, or misunderstanding Reider and Kenworthy (2016).

I'm just trying to reproduce the orbital velocities listed in Table 1. The second paragraph of Section II list a mass of the primary and semi-major axis for the planet's orbit of 0.9 solar mass and 5.0 AU. From the table the mass of the planet ranges from 20 to 100 Jupiters, which is actually quite sizable, but I'll start without using the reduced mass.

The numerical values I'm using:

$$GM_{\odot}=\text{1.327E+20} \ \mathrm{m^3  kg^{-2}}$$
$$GM=0.9GM_{\odot}$$
$$\epsilon=0.65$$
$$1 \ \mathrm{AU} = \text{1.496E+11} \ \mathrm{m}$$
$$a=5.0 \ \mathrm{AU} \ = \text{7.480E+11} \ \mathrm{m}$$

The formulae I'm using:

$$r_{\text{peri}}=a(1-\epsilon)$$

$$v^2=GM(2/r-1/a)$$

$$v_{\text{peri}}=\sqrt{GM(2/r_{\text{peri}}-1/a)}$$

I get:

$$r_{\text{peri}}=\text{2.618E+11} \ \mathrm{m}$$

$$v_{\text{peri}}=\text{2.744E+4} \ \mathrm{m/s}$$

which is  $27.44 \ \mathrm{km/s}$. But for $\epsilon=0.65$ the table below shows $29.5 \pm 0.4 \ \mathrm{km/s}$. Close but not really close enough, it's off by almost 10%.

If the mass of the planet (which is quite large) was considered, then the table would have to list a wider range of velocities, wouldn't it?

Well done you. I double checked the calculations and couldn't fault what you had done. So I contacted the lead author of the paper about it and here is the response:

"After checking the numbers in our paper, I found an error: we actually used a mass of 1.0 MSun for J1407 in our simulations, instead of the 0.9 MSun as stated. This accounts for the difference in pericentric velocities (as well as the different semi-major axes, which would be smaller in the 0.9MSun case).
We will attempt to correct this in the published version, and send a correction to arXiv."


What's the furthest distance that something could travel and eventually come back to Earth? Imagine u shoot a photon into the sky at a mirror far away in space, and you want the photon to bounce off the mirror and eventually come back to you. Considering the cosmological constant, what's the furthest that mirror could be (at the moment you shot the photon) and still have the photon get to it, bounce back and eventually return?
I know the mirror would start moving away, and the photon would have to catch up with it. I think it would have to be around 15 BLY away by the time the photon reached it (I'm not sure), but I don't know how far it would have to be at the start to be 15 BLY away by then.

There's an easy way to work out the distance to the mirror at the present time, if it moves with the Hubble flow: imagine that instead of being emitted by us, the light is emitted by the mirror image of the Earth and simply passes through the mirror to us. The farthest comoving distance that the mirror Earth could be is the same as the farthest present-day distance from which light emitted in the present day can ever reach us; assuming ΛCDM is correct, that's around 16.5 billion light years. The comoving distance to the mirror (and the metric distance to it in the present day) is exactly half that.

Working out the metric distance to the mirror at the time of reflection is trickier. If (nonstandard terminology warning) $D(a)$ is the comoving distance light can travel from scale factor $a$ to $∞$, then the scale factor at which the light hits the mirror is $D^{-1}\left(\frac12 D(1)\right)$. If all energy was dark energy and the expansion was exactly exponential then the solution would be $2$ exactly. Using central parameters from here I get $a\approx 2.09$, so the mirror distance is a bit higher, around 17 billion light years.
For what it's worth there's an exact expression for this function in a flat perfect dust+dark energy universe:
$$D(a) = \frac{c}{a \sqrt{Ω_Λ} H_0} \; {}_2F_1\left(\frac13,\frac12;\frac43;-\frac{Ω_m}{a^3Ω_Λ}\right)$$
where ${}_2F_1$ is the hypergeometric function.


In which direction does the ecliptic plane make an angle of 60 degrees with galactic plane? In which direction does the ecliptic plane make an angle of 60 degrees with galactic plane? Is it toward the galactic center? (It can have any 360 degree orientation.)

The line defining where the ecliptic plane and the Galactic plane coincide lies (approximately) between the points defined by right ascensions  and declinations  of 6h +23 and 18h -23.

This can be seen below - the plot is in celestial coordinates with the ecliptic plane marked (thicker solid line) and the Galactic plane clearly visible in the background map of dust extinction.

The second of these coordinates is close to, but not coincident, with the Galactic centre (marked Sgr A*). So the relative tilt is sort of at right angles to the Galactic centre at present.

Of course this will change over tens of millions of years as the solar system orbits the Galaxy.

Edit: Another way of specifying the direction is just to ask where the ecliptic north pole is. It is at RA=18h, Dec=66d 34m, in the constellation of Draco.

A further way of thinking about it is to consider at which points in the sky do the ecliptic and Galactic planes have their maximum angular separation. These are 90 degrees away from the two points I defined above and are almost on the celestial equator at about 0h and 12h.


Projecting an image of the sun / eclipse This might be an astronomy.stackexchange.com question, but I think it is more in line with a camera obscura.

During the last eclipse I didn't plan ahead and grabbed some lousy toy "spy" telescope we had (about 2 inches long), set it up on a tripod and pointed it at a white foam board.

I'd like to plan ahead more this time.

What can I use to project a decent image of the sun onto a foam board?  What kind of lens do I need?  Would a magnifying lens work?  How about a close-up diopter filter?  How about a 50mm lens?  (I'd like to use something I already have, or is cheap -- and I don't want to use a pin hole.)

My first thought was that a magnifying glass only focuses the sun down to a pin point (and would burn the board), but I'm recalling now that past that focal point you can project an image.

I've done this using two lenses.  In my case they were a 300 mm and a 50 mm.  I built a jig to hold the lenses in-line with each other, to allow for small adjustments in the placement, and for a tripod mount.

Put the long lens, in my case the 300 mm, in front normally.  That will form a focused image behind it.  Put the short lens backwards behind that.  This lens is used in reverse.  It takes the small image created by the long lens and projects it out what is normally its front.

The ratio of the front lens focal length to the back lens focal length is how much angular "gain" you get from this setup.  In my case, the cone of project of the sun was 6 times wider coming out of the 50 mm than going into the 300 mm.  This means you don't have to get as far away from the overall setup to get a decent size picture.

I'm not where this jig is now.  I'll try to remember to take a picture of it and post it here within the next day or so.

One thing you'll find is that the earth rotates annoyingly fast when you are looking at a magnified picture of the sun.  Use a telescope mount if you can.

Another issue is blocking out as much ambient light as you can.  The projected image of the sun is reasonably bright, but so is daylight.  Some sort of baffle over the top and the front except for a hole for the front lens helps a lot.  People can then view the projected image from the sides.


Viewing a solar eclipse through a leafy tree Viewing my Facebook feed today, my local news station posted regarding a solar eclipse taking place today:

Note the line about using a leafy tree as a filter:

  Scientists say NOT to directly look at the sun (if skies are clearing where you are), but looking at the eclipse through a leafy tree creates a natural filter!

I've heard a lot of people recommend viewing a solar eclipse by looking at the projection of the eclipse caused by light passing between leaves in a tree, in this case looking at the ground or a wall, where ever the light hits after passing through the tree.  However, the Facebook posting here seems to recommend looking up at the solar eclipse through the tree.

I suspect that they have misunderstood the projection method (or inadequately explained), and their advice is not only wrong, but potentially dangerous.

Is this method recommendable? Should it be considered dangerous?

Follow-up

Almost immediately upon having seen the posting I had commented on it noting my belief that such advice could be dangerous, including a more clear explanation of how viewing using trees should be done. The staff member who originally posted the status later took note of this and edited the status to less ambiguously explain viewing the projection. Unfortunately, this was done after the solar eclipse had concluded. Hopefully no damage was done.

It seems that the station has, in fact, given out extremely dangerous advice - simply because it was poorly phrased.

Viewing a solar eclipse is incredibly dangerous - mostly under circumstances where you're looking towards the Sun. As Wikipedia warns,

  Looking directly at the photosphere of the Sun (the bright disk of the Sun itself), even for just a few seconds, can cause permanent damage to the retina of the eye, because of the intense visible and invisible radiation that the photosphere emits. This damage can result in impairment of vision, up to and including blindness.

You couldn't pay me enough to so much as face toward the Sun and look at my feet and attempt to view a solar eclipse out of the corner of my eye!

There are safe ways to view an eclipse, the most common being using a pinhole camera to indirectly view the eclipse. Not only are you not looking directly at the Sun, but you can actually turn away and look at the projection.

But here's where I think the station messed up. According to CNN,

  Or let a tree do the work for you. "Overlapping leaves create a myriad of natural little pinhole cameras, each one casting an image of the crescent-sun onto the ground beneath the canopy," NASA says.

So NASA actually advocates using the leaves as a pinhole and using indirect viewing methods - i.e. looking anywhere but the Sun - as opposed to watching the eclipse through the leaves.


Do any known exoplanetary/solar bodies have "annular" eclipses similar to Earth's? This was somewhat inspired by this question but is a bit different: I am interested in if there are any known systems in which the angular size of the eclipsing object is almost the same as the angular size of the object being eclipsed from the surface of a planet, as is the case for the Moon and the Sun as viewed from Earth. This is the only way that one can get "annular" eclipses, or eclipses in which the host star's corona is clearly visible. This could be the case for some moons in the solar system (which requires that they eclipse at all), and at least Phobos in particular seems to have a similar angular size to the Sun as viewed from Mars,

but it could also be true for some known (or candidate) multiple-planet exoplanetary systems in which the periods of the two planets are not too different. Probably a perusal of the public Kepler candidate sample would be helpful here.

This will depend the position of the observer and (obviously) the relative sizes of the star and the eclipsing body.

For an intelligent observer standing on the surface of a planet the most obvious and likely candidate for an eclipsing body would be a moon of that planet. The nearer the planet is to the sun the larger that moon has to be, and the further the planet is away from the star the smaller the moon can be.

Finding planets around stars is hard enough, finding moons that will be orbiting quite closely around those planets across interstellar distances is even harder (I'm not going to say impossible, but it's pretty close) with current technology, so at the moment the answer is "None, that we know of".

Also don't forget that, according to current theories, the moon was formed when a large Mars sized body collided with the nascent Earth, the chances of another Earth/Moon type system is probably remote. This means that most natural satellites are likely to be Phobos sized bodies. Therefore to get an eclipse the planet would have to be further out from the star (so the star looks smaller) and that would tend to put it on the edge of the "Goldilocks" zone, making the likelihood of there being an intelligent observer somewhat rare.


Ideal conditions for life Is Earth the perfect place for life ? If it isn't then how could conditions be better here ? For example, the size of the Planet, our atmosphere, distanace from our Sun etc.

It is a very good place for Earth Life, but that is because we have evolved and adapted to this planet. For example, we like to have an atmospheric pressure of about 1000 mBar, and a partial O2 pressure of about 200 mBar, which is exactly what the atmosphere provides.

Earth would be a very bad place for a species to which oxygen was toxic. A species that hadn't evolved here would probably find conditions challenging in one way or another.


What is the degree of reflection of light from sun to the moon? As Moon reflects the light from the sun it must have an ANGLE OF INCIDENCE and ANGLE OF REFLECTION...
What Is the angle of reflection then?

Basically, the sunlight is reflected from the Moon at different angles. It works the same as on Earth. For instance, during the sunrise & sunset, the sunlight is reflected at a bit above 0 degrees angle, whereas at the noon in the tropical areas (in general) where Sun passes the zenith, the sunlight is reflected at 90 degrees. On Moon and other planets, it works exactly the same! This is how the angle of reflection works. Obviously, because we have a rough surface built with different rocks, we can't treat the celestial body as a mirror. However, the phenomenon is analog, as each of them has its own albedo - ability to reflect the light. Lunar albedo is 10%.
https://en.wikipedia.org/wiki/Albedo#:~:text=Albedo%20(%2F%C3%A6l%CB%88b,that%20reflects%20all%20incident%20radiation.
https://scienceworld.wolfram.com/physics/AngleofReflection.html#:~:text=The%20angle%20of%20reflection%20of,same%20point%20as%20the%20ray.

Obviously, it's worth mentioning that as the angle of light reflection is smaller, then the light beam is weaker (less intensive). It arises out of the inverse square law:
https://en.wikipedia.org/wiki/Inverse-square_law

which can be explained by the image above.
The angle of incidence is always equal to the angle of reflection, but it's perpendicular to the surface exactly, where the beam hits this surface.

http://www.mkrgeo-blog.com/the-aspect-of-light-reflection-in-the-long-distance-observations/
https://en.wikipedia.org/wiki/Angle_of_incidence_(optics)
Otherworldly, the angle of incidence is related to the plane of incidence. The plane of incidence is the "place" where an observer can see the reflected solar beam on the surface.
Along with the angle of incidence, we can mention the light glitters produced when our angle of reflection is low enough.


Did we discover how nebulae hold their shapes? As you know, nebulae such as the Eagle Nebula, the Cat's Eye Nebula, and the Pillars of Creation, have extraordinary shapes.  But have we discovered why that is so?  I've tried researching it myself however all that comes up are theories of a magnetic field, etc.

"Nebula" is quite an imprecise word; over the years, it's been applied to everything from interstellar dust clouds to supernova remnants to the birthplaces of stars to even galaxies. Wikipedia separates true nebulae into four categories: H II regions, planetary nebulae, supernova remnants, and dark nebulae. In this case, it might be better to make the distinction between nebulae that primarily surround on star/star system and nebulae that exist in interstellar space, potentially encompassing many stars.
Circumstellar nebulae
I'll classify planetary nebulae and supernova remnants as circumstellar nebulae. They're both the result of a star passing into the final phase of its life.
Planetary nebulae (which include the Cat's Eye Nebula) are the remains of the outer layers of gas given off by relatively low-mass stars, like the Sun. Balick & Frank (2002) subdivide these based on morphology, into four categories:

Round
Elliptical
Bipolar
Irregular

The names describe the nebulae's characteristics fairly well. However, they show observation bias; for instance, a bipolar nebula may appear to be found if its lobes are lined up perfectly with Earth. The exact reasons for the distinctions aren't well-know; Wikipedia cites Jordan et al. (2002) (which I think you may have already read) in saying that magnetic fields could be the cause, but there are obviously other reasons for a lack of spherical symmetry: intense rotation (possible in an AGB star), violent mass loss (unlikely in stars of this low mass), a strong stellar wind, interactions with other bodies, etc. The latter may have caused the dramatic lobes of M2-9, where interactions with a binary partner produce extreme distended streams of gas:

Image courtesy of the ESA under the Creative Commons Attribution 3.0 Unported license.
Supernova remnants fall into three basic categories:

Shell-like, which are simply spherical shells
Composite, characterized by radio emission from a pulsar (with potential shell-like structure)
Mixed (also sometimes termed "composite"), which has clear characteristics of both of the above, with X-ray and radio emissions

NASA has three pages on the different types, though it uses "composite" in place of "mixed", designating "composite" nebulae "Crab-like", after the Crab nebula. The shape of a remnant depends on which type it is; for instance, composite nebulae may be shaped by the stellar wind of the central pulsar.
Interstellar nebulae
H II regions (which include the Eagle Nebula, and of course the Pillars of Creation) and dark nebulae fall into this category. Both may be part of or interact with molecular clouds, which are the star-forming regions of the galaxy. H II regions, like molecular clouds, are shaped by the stars inside them. Radiation ionizes and even pushes away the gas surrounding young, hot stars, breaking up the nebulae. However, variations in dust density mean that different parts may "evaporate" at different rates, leading to some weird shapes - including the Pillars of Creation.
Evolution and dissipation
When I first read your question, I focused on the phrase (emphasis mine)

how nebulae hold their shapes

because nebulae are actually often extremely short-lived, on astronomical timescales. While the exact lifetimes vary, they're typically quite short when compared to stellar lifetimes:

Planetary nebulae: On the order of $\sim10^4$ years, although this may be influenced by the progenitor and the nebula's morphology (see Badenes et al. (2015)).
Supernova remnants: On the order of $\sim10^5$ or $\sim10^6$ years, although this depends on how you define the lifetime of a remnant. It can differ if you define it on X-ray emission, radio emission, or something else entirely.
H II regions: On the order of $\sim10^6$ or $\sim10^7$ years, depending on the evolution of the star(s) inside the region. Their radiation pressure will dissipate the gas cloud, so its lifetime depends on their characteristics. As Stars, Nebulae and the Interstellar Medium explains, H II regions would live longer if they weren't dissipated by the stars. Otherwise, they could survive for orders of magnitude longer.


Are there any known planets whose axis is oriented such that one pole always faces its star? Earth rotates around the sun and around its axis (A North, South axis) giving us days, nights and seasons.

Are there any known planets that rotate with an axis pointing toward its sun so that one half of the planet is always facing it (the other half is always facing away from it)? Like Uranus, but not switching which side is facing the sun every 6 months.

(Would this even be theoretically possible?)

Would this even be theoretically possible?

No. The law of conservation of angular momentum prevents this for a planet with reasonable inner structure in a gravitational field like that around the sun.

The angular momentum of the planet points in a certain direction. Changing this direction requires a torque perpendicular to the rotation axis.

If a torque existed that could force the rotation axis of the planet to always point towards the sun, the corresponding force had to have a component perpendicular to the line from the sun to the planet. The force also had to change periodically with the orbit of the planet. Obviously such a force does not exist in the planet-sun-system.


Why was this asteroid (4864 Nimoy) chosen to be named after Leonard Nimoy? The asteroid 4864 Nimoy was recently named after Leonard Nimoy. It was discovered on September 2, 1988 so it went nearly 27 years without a name. Why was this asteroid chosen to be named after him? Were they trying to think of a name for it and chose to name it for Leonard Nimoy, or were they looking for an asteroid to name after him and chose that one?

I'm not entirely sure, but I would guess that it's simply because the discoverer suggested it. After a few steps in the process of characterizing an asteroid's orbit, the discoverer gets to suggest a name (within reason) to a committee.  The full guidelines are available at the relevant IAU webpage.  (It's longer than is worth reproducing here.)

Note that there are plenty of "celebrity" asteroids, drawn from a wide range of human endeavours: Adamsmith, Balzac, Beethoven, Megryan, etc. If anything, I'm surprised that an asteroid wasn't named Nimoy sooner. There's already a Mr. Spock!


Can I make a black hole with one or two atoms? So I was watching something that said

if we compressed Earth into the size of a peanut: we would get a black hole;
if we compressed Mount Everest into a few nano-meters; we would get a black hole.

Can I make a black hole with one or two atoms? If yes, would it become larger and turn into normal-sized black hole?

There are two answers: yes and no.

Yes because every mass M has a Schwarzschild radius given by $\frac{2GM}{c^2}$ (where G is the gravitational constant (about $6.7\times10^{-11}$ and c is the speed of light (about 300 000 000 $\mathrm{m/s}$). If something is compressed to its Schwarzschild radius it becomes a black hole. You can do this for an atom. An atom of carbon (for example) has a mass of $2\times10^{-26}\mathrm{kg}$ so its Schwartzschild radius is
$$ \frac{2\times (6.7\times10^{-11})\times (2\times10^{-26})}{300 000 000^2}\approx 3\times10^{-53}\ \mathrm{metres}$$

So the actual answer is no as there is no feasible way of compressing an atom to this size. Of significance here is the fact that this size is so small that objects this small don't behave like small balls but as quantum mechanical objects. But a black hole is a gravitational object modeled by General Relativity, and Relativity and quantum mechanics don't work well together. In other words, we don't have a scientific model for describing how an atomic mass black hole would behave.

Stephen Hawking has shown that small black holes are unstable, so an atomic mass black hole would be very unstable, evaporating in a very short time.


Is the Big Bang inflation caused by Dark Energy The rate of expansion of the universe after the Big Bang is described as Inflation. It is stated as being initially very fast then slowed. But the expansion is still continuing.  Some say expansion is currently powered by Dark Energy. The equations unlike gravity say that the rate appears to be linearly proportional to distance, like Hooke's Law for springs. Initially the universe was very small, so if the spring constant is dependent on the size of the universe, could this explain the inflation? What experiment could be devised to test this?

The expansion caused by a cosmological constant (a particularly simple version of dark energy) is not exactly Hooke's law, it is an expansion that maintains a fixed Hubble constant. So this means $\mathrm{d}a/\mathrm{d}t = Ha$, for scale parameter $a$ and constant $H$. The solution is exponentially growing in time.  Hooke's law (without the minus sign) would have the acceleration $\mathrm{d}^2 a/\mathrm{d}t^2$ being proportional to $a$, which permits more complicated solutions, but you're right it certainly has similarities  if one picks out the exponentially growing solution.

What's more, you are right that some treatments of inflation look like that of a cosmological constant, in that they are also exponential in time, but with a much much larger H constant (like 50 orders of magnitude larger).  So if it's the same thing, then for some reason the cosmological constant must have dropped vastly in value.  It's thought to be for a rather different reason, though this is all speculation.  The common speculation is that the "false vacuum" went through a phase change during inflation, leading to the vacuum we have now, and the vacuum we have now has a very different cosmological constant.  Maybe you can think of it as a phase change in the cosmological constant itself, I don't know.  So there could be some connection, but they can't be just the same.


Calculating star radius from temperature and absolute magnitude - what am I missing? In an attempt to visualise nearby stars, I've tried to get the star radius (measured in solar radii) using the equation provided on the following site:

http://cas.sdss.org/dr4/en/proj/advanced/hr/radius1.asp

Now though the equation seems to work fine for a star like Sirius, when I plug in the values for absolute magnitude and temperature for Barnard's star (according to wikipedia, 13.21 and 3134 K respectively) I get a radius of 0.0722. The radius provided by wikipedia is more than twice that value, at 0.196.

There are many potential culprits. Could it be floating point errors in the php code? The equation looks like this by the way:

$$\text{radius = pow(5800/3134, 2)} \times \text{pow(pow(2.512, (4.83 - 13.21)), 0.5})$$

Another possibility is that the equation is too simplistic, and there are other variables I need to take into account.

Any insights would be greatly appreciated!

Michael

The page you quote is guilty of a gross simplification (or error). The luminosity ratio cannot be estimated from the difference in visual magnitudes for stars of different spectral type. You need to use the difference in bolometric magnitude, which indicates the luminosity integrated over all wavelengths. This will involve making bolometric corrections to the visual absolute magnitudes.

A star like Proxima Cen radiates much of its light outside the visible band and has a much different and larger bolometric correction than the Sun or Sirius. This accounts for your problem.

Example:

The correct calculation uses a table of bolometric corrections vs temperature. This rapidly changes for M-dwarfs in the visual band (you would be more precise using K magnitude), but is about 2.6 mag for a star at 3100 K (see Zombeck 1992). That is, the bolometric magnitude is smaller than $M_V$ by 2.6 mag. The equivalent correction for the Sun is only 0.1 mag.
$$ M_{\rm bol}= M_V - BC$$

Thus there is a 2.5 mag less difference in the bolometric magnitudes in your final equation than the difference in absolute visual magnitudes, and I get a radius for Barnard's star of $0.23\ R_{\odot}$.

The probable reason for the discrepancy between this and the radius quoted in wikipedia (actually it is the radius determined by an interferometric angular diameter measurement plus the parallax by Lane et al. 2001; a more recent measurement by Boyajian et al. 2012 gives $0.187\pm 0.001\ R_{\odot}$) is that (a) the assumed temperature is too low or (b) that the bolometric correction scale for V-band magnitudes is not very well defined for very cool stars. Both (a) and (b) are possible, especially as Barnard' star probably has a significantly subsolar metallicity.


Is there a stellar database that indicates how long ago stars in our Galaxy formed? There are several ways of determining the age of a star: its position in the HR diagram, the presence of a protoplanetary disk, it belonging to a cluster...

When did the stars in our Galaxy form? Do we know their age? Does there exists a database of stars for which we know the age, from which this average could be computed ? Is there a histogram of how long ago the stars in our Galaxy formed?

The amount of stars being formed as a function of time is shown for instance in Snaith et al 2014, but this doesn't directly translate into a histogram of the age of the stars currently in our galaxy, since the more massive ones that formed early on are gone.

There isn't really a database as you request. Finding the ages of stars is difficult. Only one star has an accurately known age - the Sun. That comes from radioisotope dating of meteorites. For other stars we must rely on models to a greater or lesser extent and we can only estimate an age if the star has a mass or is in a phase of its evolution where things are changing rapidly enough to give some handle on how old it is. For stars at the mass of the Sun or a little bit bigger, one can use evolution in the Hertzsprung-Russell diagram. Stars become more luminous as they burn through their core hydrogen and precise measurements can give an age to about $\pm 1$ Gyr. These are the data that Snaith et al. (2015) use in their paper.

So what Snaith et al. do is they attempt to constrain the star formation history of our Galaxy (or at least stars in the disk of our Galaxy) by making a Galactic Chemical Evolution model that predicts how the abundances of silicon and iron change with time and compare the results of their model with the observed silicon and iron abundances in this set of solar-type stars with reasonably well-estimated ages. Iron and silicon are used because they provide differing constraints. Iron is produced during the evolution of relatively low-mass ($1.5 - 4 M_{\odot}$) stars that live their long(ish) lives (1-10 Gyr), become white dwarfs and then some of the white dwarfs explode as type Ia supernovae. Silicon is produced by massive stars with short lives ($<0.1$ Gyr) and spread promptly into the interstellar medium by core-collape type II supernovae.

A representative lot from Snaith's paper is shown below. The star forming rate is shown in plot (a) and the match of the model to the observational data (they are trying to match the solid green points here) on chemical abundance is shown in the other plots.

They test a number of models and different assumptions, but it appears to be quite a robust result (found in other studies also) that there was a burst of star formation 10-12 billion years ago, a lull at 8-9 billion years ago and then a more constant, lower rate over the last 7 billion years. The very low rate in their model over the last 2 billion years is unlikely to be true and is probably an artefact of the age uncertainties in their data points (e.g. stars with age of $1 \pm 1$ billion years) and we certainly see plenty of star formation in the disk of the Galaxy today.

Contrary to what you say in your question this does (almost) represent the age distribution of stars in the Galaxy. That is because the vast majority of stars (90+%) are of a solar mass or lower. Such stars have main sequence life times of 11+ billion years, such that almost every one of them that was born is still a main sequence star now.

There will be a small reduction in the numbers of the oldest stars, because a small fraction of those ($<10$%) will have lived and died and also because a small fraction of the oldest stars may have escaped the Galaxy entirely (due to various kinematic processes that cause their orbital speeds to change). There is also an effect whereby if you take a volume close to the Sun, the oldest stars will be under-represented because they are distributed more broadly around the disc midplane for the same reason.

But by an large, the upper left plot in the picture is roughly what we expect the age distribution of stars to be in the Galactic disk.

There are a small population of very old stars ($12+$ bilion years), perhaps 1%, that are distributed more widely in the Galactic halo.


What does it mean 'black holes is created when center of a very massive star collapses in upon itself'? I read many times that black holes is created when center of a very massive star collapses in upon itself but what does it mean ? Does weight of center of massive star increase so much that itself can't bear weight or I can say space itself cant bear weight of such center of massive star that it collapse. I tried to read answer on google but want to know in layman language as I am not astro physics researcher or student. Thanks

This Wikipedia page on Stellar Evolution is a good starting point with lots of links to more info.

A star is a huge sphere of mostly hydrogen.

Like any large mass it has a gravitational field and that field acts on star itself.

So it would always want to collapse to as dense a sphere as possible under it's own gravity.

Now as the density increases, particularly at the center, the pressure also increases and so does the temperature.  When these are high enough the core of the star will be able to fuse hydrogen (and later other materials) into heavier elements.  Doing so releases huge amounts of energy and this is what creates all that light we see.

But the energy released also prevents the star from collapsing more.

And it can stay that way (more or less) for billions of years.  It is this energy produced by fusion at the core that balances the gravitational force (weight) of the rest of the star trying to crush the core.  So fusion that generates more energy than it requires to happen is essential to keep a star "alive".

After a long time, the hydrogen at the core runs out and the star starts to burn heavier elements at the core (helium, carbon, etc. ).  It does that until those run out and (roughly) we reach elements like Iron remaining.  At that point fusing those elements no longer generates enough energy to stop the star collapsing and it starts to collapse again, but very, very rapidly.

Exactly what happens next depends mostly on the mass of the star.  It can end up as a white dwarf, neutron star or black hole.

But the final collapse is a very rapid process and quite violent.

Once the useful fuel at the core runs out, the stars dies a spectacular death one way or another.

What is a black hole (in simple terms) ?

Gravity is a result of the way mass (or even energy) can distort space.

Beyond a certain point this distortion becomes so great that everything on one side becomes trapped inside a "surface" called the event horizon.  An event horizon is what a black hole actually is - no event horizon, no black hole.

Within the black hole, space is distorted in a way that's basically impossible to describe without some mathematics.  The effect is that there is no direction you can move in and no speed you can travel at which will not lead you to the center of the black hole - the singularity.  Don't even try and visualize it, it just isn't a common sense idea.

This even works with light.

Now an event horizon is not a surface you can feel.  You can actually travel across an event horizon (but only inward, not outward !).  So this change in the way space-time is distorted is not something that is sudden or obvious to the person doing the falling.

Back to our star collapsing.

As it collapses (without energy from fusion to stop that), it becomes denser and denser.

Now if it is large enough it will collapse to a black hole, because the core will become so dense from the collapse that it will be enough to form an event horizon.

Our Sun is not large enough to do that.


What is the centre of rotation of Pluto? I know that Pluto and Charon have been found to circle round each other, hence the reason Pluto is no longer a planet I believe, which would mean that the centre of rotation is an arbitrary (!) point somewhere between the two of them. Where is it please? Can it be calculated?

I know that Pluto and Charon have been found to circle round each other, hence the reason Pluto is no longer a planet

This is not correct.

It is considered a dwarf planet because of a technical definition employed by the IAU in deciding what is a planet and is not.  This technical definition relates to an object's ability to "sweep" it's orbit clear of asteroids and other small debris.  The decision has nothing to do with Charon.

The centre of rotation is calculated using Newtonian physics.  I think what you want is the Center Of Mass and related topics.


What are the minimum element requirements for a star? What elements would be essential for a star to exist?

From what I understand our sun is made of mostly hydrogen and helium and most stars have small amounts of heavier elements like carbon, nitrogen, oxygen and iron.

But could a sun exist on a single element such as Hydrogen? Or on another gas for example oxygen or nitrogen? Or even rare gases such as Neon or Xenon?

Is it even possible for a star to exist without any hydrogen or helium?

Yes, a star can exist that's entirely hydrogen. Hydrogen is the fuel that makes stars happen unless they're very large. A star made entirely of hydrogen, so long as it was massive enough, would be very similar to the stars we see. The "metalicity" which refers to non hydrogen-helium elements has some effect on the star's rate of fusion, density and spectroscopy but it doesn't have a major effect.

When the universe was very young and almost entirely hydrogen and helium in terms of regular matter or elements, star formation was difficult because those light gasses were difficult to clump together.   The first stars were probably very large as a result, but once heavier elements began to be released in early supernovas, then star formation became easier because there were elements that could clump together and help smaller stars form.   So the other elements play a key role in star formation, but a much smaller role in the life of the star.

Very large stars fuse heavier elements, but as I understand it, fusion of heavier elements is more explosive. Hydrogen is wonderful in the way that it's very unlikely to fuse, so even large stars take millions of years to fuse their hydrogen because something like 99.9999% of the time, when two hydrogens get close enough to fuse, they just return to being two hydrogens, and only 1 in a hundred million billion billion times (1 in 10^26), credit to @PM2Ring.  It's only very rarely that the proton-proton collusion undergoes the fusion process and becomes a deuterium.

This very slow process doesn't happen with the heavier elements, so a star made entirely out of elements heavier than hydrogen and helium would begin it's life as a kind of white dwarf star, that is, the mass of a star but the size of a planet, and technically, it wouldn't be a star at all because it wouldn't be undergoing fusion and when the fusion process finally starts, it would be more like a type 1A supernova, happening very quickly and explosively.

The CNO process which creates some of the heavier elements in larger stars doesn't happen without hydrogen either, or, more accurately, free protons.

Basically, you can make stars out of pure hydrogen. Other elements would be made inside stars, but you can't make stars out of any other elements, unless you consider variations of a type-1a supernova a "star".  I personally don't.


Distance between each dish in an array telescope configuration In an array telescope configuration, like VLA and ATA, what dictates the distance between each dish?

Physically there is nothing dictating the distances between dishes.

These telescope arrays are interferometers. By combining the signal of the two dishes you can achieve a spacial resolution of a giant virtual dish that is about as large as the distance (called the baseline) between the two real telescopes.

A nice example of the power of this technique is the VLBI, which is essentially such an array only the individual dishes are actually just observatories all over the world.

For more detail on how this all works exactly take a look at the wiki pages on aperture synthesis and astronomical interferometry.


Is the conversion from proto-star to main sequence an event or a process? When a proto-star becomes a main sequence star, is that something that happens in an instant when a certain threshold is met, or is it a process that takes a few thousand/million years?

Astronomers distinguish a prototstar from a star based on whether the object is visible. A protostar is hidden by the gas cloud that surrounds it. Protostars aren't visible. At some point in their evolution (and where this occurs depends on mass and metallicity), a protostar will start clearing the surrounding cloud of gas. This process happens very quickly from an astronomical point of view. (Aside: From a human point of view, this is anything but quickly.)

In the case of very massive stars, the newly emerged star is already on the main sequence. Very massive stars are "stars" (objects with a stable mass, stable size, and fusing hydrogen into helium) well before astronomers can see them. In contrast, very small stars spend hundreds of millions of years of evolution between being a "star" (visible to astronomers) and being on the main sequence. The star is a pre-main sequence star during this long span of time.

Intermediate mass stars also spend some time as a pre-main sequence star. How much time depends on mass and metallicity. In the case of a star with a mass of ~1 solar mass and metallicity comparable to that of the Sun, the time spent as a pre-main sequence star is on the order of tens of millions of years. In contrast, the time spent as a protostar is very short, tens of thousands of years. The time taken to clear the star system of gas (the transition from protostar to PMS star) is shorter yet, hundreds to a few thousand of years.

There is no point at which one can say fusion starts. It's a probabilistic thing, with probability increasing sharply with density and temperature. What about objects at the low end of density and temperature? This is the boundary that separates red dwarfs from brown dwarfs.

There is very little difference between the smallest red dwarfs and the largest brown dwarfs. The smallest red dwarfs are quite cool. They fuse hydrogen into helium via the p-p chain, but rather infrequently. The largest brown dwarfs also fuse hydrogen into helium via the p-p chain, but here the fusion reactions are so very infrequent that the brown dwarf gets cooler and cooler. In a trillion years, we might be able to point to a marked boundary between old red dwarfs and old brown dwarfs. Right now, that boundary is rather arbitrary.


Why is the Innermost Stable Circular Orbit (ISCO) of a non-rotating black-hole 3 times the Schwarzschild radius? If anyone has a derivation of this result, it would be appreciated greatly.

There's a fairly full treatment in these lecture notes but it's not really amenable to a summary in a stackexchange answer. However they end up with the following equation for the radius of a circular orbit of a particle with angular momentum $L$ and mass (ie not a photon).

$$GMr^2 - L^2r + 3GML^2 = 0$$

For suitable values of $L$ this has two solutions for $r$ and, one inside the ISCO radius (which is $6GM$ in these units) and one outside. Further analysis shows that the inner orbit is unstable and the outer one stable. See around equations 7.55 for more details.

Photons have an unstable circular orbit at $3GM$ and the event horizon is at $2GM$.


Question about tidal forces and the Roche limit As we know, an object that is beyond the Roche limit doesn't disintegrate (obviously) because the tidal forces upon the object are weaker than the gravitational pull of the object towards its centre, so the planet remains intact.

My question is: if that's so, why does the body deform even before crossing the Roche limit?

If the gravitational pull of the particles in the object outweighs the tidal forces, wouldn't the object have to stay totally round? There couldn't have been any deformation because the tidal forces are totally cancelled out.

When we do mathematical calculations, sometimes terms cancel and we get to draw a line through them.

But in the real world cancellations and "total cancellations" don't really happen.

Forces just add. Often there's a big force and one or more little forces, and as long as the big force substantially dominates the little forces, we tend to notice only the big force. But it doesn't cancel, it just covers up the effects of the others to the casual observer.

Let's think about gravity. We are so used to moving around in 1 g of gravity that we don't notice it. You could say that our brain "cancels" out 1 g from our conscious awareness and we compensate for it automatically. We don't feel and notice the 1 g when we stand or sit or lay down, even though we make use of it to stay put.

However, get in an elevator or an amusement ride and add a little acceleration, and we notice it right away! That sinking feeling when a fast elevator starts accelerating downward for example watch (but don't listen to) the video The Equivalence Principle

The purpose of the above discussion is to discount the idea of "total cancellations". A bit of rock or soil on the surface of a body feels the gravitation of the rest of the body pulling down, the "centrifugal force pulling" it up if the body is rotating, and the gravitation of nearby bodies (e.g. Sun or Moon or planet) pulling up or down or sideways depending on orientation. It feels all of these all the time.

If the net force, the sum of all the forces, pulls up, then the body starts to disintegrate. If the net force is still down, it doesn't.

Think of it as a threshold, rather than a cancellation.

  ...if that's so, ¿why does the body deforms even before crossing the Roche limit?

As long as the body is made of material capable of deforming or flowing (which rock certainly is, especially when hot) then it will continue to move or flow until the sum of all the forces reaches zero. Gravity usually dominates, but it's the force of the compressed layer below pushing up that counters the sum of all the forces.

If the body is rotating, then the net force is weaker on the equator than at the poles, and so less pressure is needed to compensate. That means the higher pressure at the poles will cause some of the material to move to the equator.

To read more about several of the competing forces pulling on us all the time as we stand on the Earth, see:

@DavidHammen's cool table
my (sadly down-voted) calculation
all of the answers to Did the Mars rovers actually confirm the gravity of Mars?

Once you are comfortable with the idea of equilibrium, multiple forces adding and pressure, then have a look at all the interesting answers to these particular Roche limit-related questions:

Is there any orbit at which the Roche limit can be “felt”?
Roche limit - calculation with centrifugal forces
How Wide is Earth's Roche Limit?
Filling the Roche Lobe


Is phys.org/space-news reliable source? I found this page: http://phys.org/space-news/ . It seems to me pretty good page with interesting news and so on and I like that it is free. But I noticed in comments people arguing about the validity of the information and so on. Are they just "trolls" or is this page really bad source of information, if that is so, how much can I trust it?
Thanks in advance!

I've been reading phys.org for years.
It's right up there with physics.aps
and physicsworld.

All good sources that provide decent refs.
Comments are often where people who like to argue, argue. Sometimes they are useful, often not.


Is there another explanation, apart from the doppler effect, for the redshift observed in distant galaxies? On episode 10 of the original Cosmos TV series, Carl Sagan mentions that some scientist don't agree that the redshift observed in distant galaxies is evidence of the big bang, stating that probably the doppler effect is not responsible for this effect. What other explanations are there?

Tired light has been used as an explanation before, whereby light loses energy whilst travelling through space---a sort of drag effect.

I don't think anyone actually supports it nowadays, though.


Nodes and declinations of Moon This nice page from NASA has the possibility to create an astronomical events calendar for a certain period. I tried to do that for the year 2017 for Central Europe. What comes out puzzles me:

On July 8th the Moon is at it's south Declination. On July 12th the Moon goes through the descending node. On July 22nd it finds itself at the north Declination.

I had always though the south Declination means the southernmost point of the Moon on its orbit. Furthermore I had always thought the descending node is the point the Moon passes by going "down" in its motion. How can it reach a southernmost point and later go even further downwards?
I had been inclined to think that the north and south Declinations would be about half way between nodes (give or take due to slight changes during the motion). How can the Moon be in one of these south or north points and only 4 days later pass a node? And take again 9 more days to reach a south/north Declination again?

It would be nice if the answer could, besides clarifying these concepts (ascending and descending node, south and north Declination), point me to a reading where they as well as other Moon orbital points and parameters are clearly defined and explained.

The image above (click for full size) shows the moon's path for much of July. The celestial equator is in purple and the ecliptic is in green. Note that:

On July 8th (meaning between July 8th at 0000 UTC and July 9th at 0000 UTC), the moon does indeed reach it's southernmost declination, but is still north of the ecliptic.
On July 12th, the moon, whose declination is increasing slower than that of the ecliptic, crosses the ecliptic from north to south, it's descending node.
On July 22nd, the moon is at its northernmost declination, but still south of the ecliptic.
On July 25th, the moon, moving south slower than the ecliptic, crosses the ecliptic from south to north, it's ascending node.

As the page you reference notes, the moon's northern and southernmost declination this month are +-19.4 degrees, about 4 degrees less than the tilt of the ecliptic. Thus, the moon changes declination slower than the ecliptic, at least for this month.


How is it ensured that a parabolic shape forms while polishing a telescope mirror? I understand polishing the mirror with a successively finer abrasive can create a curvature, but how is it possible to obtain a uniform curve that forms a parabola while polishing by hand?

How is it ensured that a parabolic shape forms while polishing a telescope mirror?

While a spherical shape is fairly easy to abtain, a parabolic shape definitely isn't!

It's a combination of special polishing strokes, and feedback via control and testing.

https://stellafane.org/tm/atm/mirror-refs/strokes.html

Normal polishing strokes (center over center, 1/3 diameter amplitude) tend to make a sphere. When all goes well, the surface of the mirror tends to become spherical, or very close to it. This is not different from grinding, only more precise. When two surfaces are rubbing against each other in a ball-and-socket type of joint, any irregularities will be removed and both surfaces will "want" to become spherical, or close to it.

When making telescope mirrors, the so-called "normal stroke" is what accomplishes that in the best way.

Now here's the essential detail: no matter what other curve you're trying to obtain (parabola, hyperbola, ellipse), for the purpose of making telescope mirrors all these curves are very close to a circle, within a fraction of a micron.

So a typical process has an initial goal of making a sphere (cross-section is a circle). This is not mandatory, but it's easier to first make a sphere, and then apply "corrections" to it. You can definitely make a parabola directly, but the process is more complex.

For telescope mirrors, the parabola that's needed is basically a circle that's deeper in the center, and more shallow towards the edge. So, if you start with a spherical mirror (circular section) and use polishing strokes that tend to deepen the center in a gradual way, you can get pretty close to a parabola. This is known as a "parabolizing stroke".

Keep in mind that as you apply that stroke, the center keeps deepening. At some point the correction will become deeper than a parabola and the mirror will become hyperbolic. That's sometimes what you actually need (if you make a hyperbolic mirror for a Ritchey-Chretien telescope). Or if you stop too early, the mirror is elliptic, which is used in other designs. The parabola is the thin edge between the elliptic and hyperbolic domains.

You start with a mirror where the cross-section is a circle (spherical mirror). You begin to apply the W stroke. The circle immediately becomes a very shallow ellipse, which keeps deepening. At some point, for a very short time, that ellipse becomes a parabola. As you keep doing the W stroke, the parabola then changes quickly and now you have a hyperbola which just becomes more and more deep.

The key here is testing. You apply the parabolizing stroke for a short while, then stop and test the mirror. There are many testing techniques; the Couder mask is simple enough and very popular; or you could use an interferometer if you have (or can build) one. Is it still elliptic? Keep parabolizing. Has it become hyperbolic? Reset to a sphere and try again. Or adjust the stroke if either the edge or the center are being corrected too fast.

Apply parabolizing strokes. Test. Modify technique if necessary. Repeat. That's all there is to it.

Note: I am using jargon in the text above. "Deeper" really means "more strongly curved". "Shallower" really means "less strongly curved". The difference between circle and parabola is that the parabola is more strongly curved in the center, while the curvature at the edge is less strong. The circle has constant curvature everywhere, by definition.

This doesn't necessarily mean the center of the mirror is physically more depressed; while doing any strokes, the whole surface keeps being polished down all the time. But the relative curvatures of center vs. edge may change.

Deeper vs shallower are terms that reflect the physical process, whether the tool applies more or less pressure on a certain area of the mirror (so polishing goes faster or slower there). But the geometric attribute that the process actually changes, that we are interested in, is the radius of curvature. Where the tool applies more pressure, the radius of the curvature decreases rapidly (the surface becomes more strongly curved).

Deeper and shallower are practical terms related to process. More or less strongly curved are theoretical terms related to geometry. When you make mirrors, after a while these terms become linked in your mind.

It helps if you're familiar with the conic sections:

https://en.wikipedia.org/wiki/Conic_section

red = circle

orange = ellipse

green = parabola

blue = hyperbola


Finding the radius of a star in arcsec If i know the radius r of a star in cm, how can I convert it to arcseconds?

For example, if I have a star with r = 3.18e13 cm, and distance to the star d = 220 parsecs, what is the relation to convert the radius from cm to arcseconds.

Thank you.

Using basic circular maths: where $d$ is the distance of the star from the observer in AU, and $r$ is the star's radius in AU, and $a$ is the angle encompassed by the radius of the star in degrees:

$
r = \frac{a}{360}2\pi{d}
$

Now rearrange it to make $a$ the subject:

$
a = \frac{180r}{\pi{d}}
$

To get $a$ in arcseconds, you need to multiply the result by 3600 (because there are 3600 arcseconds in a degree):

$
a = \frac{648000r}{\pi{d}}
$

Now, convert your numbers from cm and parsecs to astronomical units:

$r$ = 3.18e13cm = 2.126 AU

$d$ = 220pc = 4.538e7 AU

Put them into the equation:

$
a = \frac{648000\times2.126}{{4.538\times10^7}\pi}
$

To get:

$
a = 0.00966327′′
$

Hope that helps!

Edit:

As Mike has pointed out, the final equation I've ended with can be further simplified if you use different units, to:

$
a = \frac{r}{d}
$

where $r$ is in AU and $d$ is in parsecs, giving an answer $a$ in arcseconds.

This is because the equation $a = \frac{648000r}{\pi{d}}$ has $\frac{648000}{\pi}$ in it - which is the definition of a parsec in AU.  Therefore, multiplying $d$ by this number to convert it from AU to parsecs, we get $a = \frac{648000r}{648000/\pi\times\pi{d}}$. Cancel out the pi on the bottom, and cancel the ${648000}$ on the top and bottom, and you're left with $a = \frac{r}{d}$ where $r$ is in au and $d$ is in pc.


Did we find any exo–dwarf-planet? We have discovered thousands of exoplanets so far. Do we categorize them into planets and dwarf planets, too?

Yes. There is no reason why one wouldn't. But we don't know any exo-dwarf planets (yet). See the answer to this question why.


Calculation of RA and DEC I'm currently stuck on a problem.
I have the RA and DEC for JD2000:

RA = 17 57 48.49980

DEC = + 04 41 36.1114

I want to calculate the RA and DEC for 01.01.2017, how can I do it?
The solution might be easy but I'm totally stuck right now.

Thanks in advance.

Simon Littlefair of Sheffield University gives formulae for adjusting for precession

$$
Δα = M + N \sinα \tanδ
$$
$$
Δδ = N \cosα
$$

where M and N are given by

$$M = 1°.2812323 T + 0°.0003879 T^2 + 0°.0000101 T^3$$
$$N = 0°.5567530 T - 0°.0001185 T^2 + 0°.0000116 T^3$$

and $T$ is defined as

$$
T = (t-2000.0)/100
$$

where t is the observation date, in years. $Δα$ is the change in RA that you would need to add, in degrees.

(formulae quoted directly from the website)

So you would have T = 0.17, M= 0.21782075093,N=0.09464464234

The actual changes in RA and Dec can be calculated from the trig formulae above.

Alternately caltech provide an online calculator


Solar wind, Earth wind and planetary winds? In February 2021, a new form of "electrified gas breeze" was announced, see blog by Tony Phillips entitled A New Form of Space Weather: Earth Wind which might a possible explanation for traces of water on the moon:

“Earth wind” comes from the axes of our planet. Every day, 24/7, fountains of gas shoot into space from the poles. The leakage is tiny compared to Earth’s total atmosphere, but it is enough to fill the magnetosphere with a riot of rapidly blowing charged particles. Ingredients include ionized hydrogen, helium, oxygen and nitrogen.
Once a month, the Moon gets hit by a blast of Earth wind. It happens around the time of the full Moon when Earth’s magnetic tail points like a shotgun toward the lunar disk.

I suspect, that all planets with atmospheres should display this kind of planetary winds (even without having a strong magnetic field), but I am sure it was never termed Venus wind or Jupiter wind since that naturally refers to processes within the atmosphere of the repsective planet. Does anybody know (publications) on the topic of gas jets errupting from solar planets with atmosphere? What order of magnitude would such a particle flow have for Venus or Jupiter?
References

H. Z. Wang et al.: Earth Wind as a Possible Exogenous Source of Lunar Surface Hydration 2021 ApJL 907 L32
Why did Venus not lose its atmosphere without a magnetic field?
Why has Venus's atmosphere not been stripped away by solar wind?

From going through the literature the paper by Wang et al. 2021 is citing, I am nearly certain that the term "Earth wind" must be a recent invention, perhaps by those authors themselves.
It is however correct to call the solar wind a 'wind'. This is because a wind is a pressure-driven bulk motion of a collectively coupled gas. The solar wind, at its base is driven by the enormous pressure gradients via the 1 million Kelvin hot solar corona.
While Earth's primary, hydrogen-rich atmosphere might have experienced such a phenomenon, it is incorrect in naming the current atmospheric escape as 'wind'.
But now vocabulary aside, current atmospheric escape rates at the terrestrial planets are governed by various ionic escape processes, the most important of them being polar cusp escape. So the data you are actually looking for is data on polar cusp escape from various orbiters. Fortunately, Gunell et al., (2018) have given a recent compendium on this (see their table A.1), and attempted a simple modelling of the data. There is no data available on polar escape for Mars and Venus, as this concept doesn't apply to those planets. The other escape processes are on average smaller for hydrogen, but larger for oxygen ions.
Similarly, I think there is no data available on the escape of gases from the gas giants. The magnetospheres of Jupiter and Saturn have been studied, and their particle populations characterized, but I think it is unclear how much of the magnetispheric content gets ultimately lost. There is no reliable data available on polar escape for Uranus and Neptune, as those planets were only ever visited by one singular flyby (that of Voyager 2).


Upgrade tripod on a cheap telescope I got a cheap refractor as a present - my intro to astronomy. I think it’s not great, but I’m having plenty of fun looking at the moon and planets, and who knows, maybe some other objects soon.
The thing that limits me the most, I think, is the tripod that came with it. It is super flimsy, I can hardly point it at the right thing, even focusing the telescope moves the alignment. It’s also too short, so I’m hunching down all the time.
I’m therefore wondering, might it make sense to upgrade just the mount for now? If I end up taking it up as a hobby, I can maybe upgrade the telescope, and if not, it’s less money than buying a “proper” set. Just a simple alt-az probably, but dedicated to astronomy. The refractor is super light so doesn’t need anything fancy.
Due to life situation, I’m not really in a position to join a club and explore this way, I can do my astronomy when I have 15 minutes to spare (and a clear sky!).

Unfortunately, many cheap mounts have a specific, proprietary connection to the telescope tube rather than something like rings, standard tripod threaded attachment or dovetail, so replacing them is difficult or impossible.
If you are fortunate enough that your inexpensive scope has a standard 1/4" threaded attachment, you could replace the mount with an inexpensive camera tripod with probably better results.


Is dark energy constant? Does the "amount" of dark energy in the universe change over time, or is it constant?  The dark energy I am asking about is the rate of acceleration of the universe's expansion.

As mentioned above the density of vacuum energy in empty space is absolute constant, even as the volume of a region of space grows as the universe expands. So the total energy, density times volume, goes up.

More information here.


Do the planets, asteroid belt, kuiper belt, and scattered disc lie on the same plane? If not, do any of the four groups in the title lie on the same plane as other members of the group? E.g. does the asteroid belt lie on the same plane as the planets do?

They are not exactly in the same plane, but roughly are.

For example this graph representing the centaurs, the Kuipier belt and scattered disc objects show that most objects are around the same plane, with few objects with more than 30 degrees of inclination:

This is the inclination of the asteroids:

Again, few have more than 30 degrees of inclination.

The planets are also all roughly in the same plane. Lets take the ecliptic as a reference (which is almost the same thing as Earth orbital's plane):

The Sun's equator has an inclination of 7.25 degrees.
Mercury has an inclination of 7.00 degrees.
Venus has an inclination of 3.39 degrees.
Mars has an inclination of 1.85 degrees.
Vesta has an inclination of 7.13 degrees.
Juno has an inclination of 12.98 degrees.
Ceres has an inclination of 10.59 degrees.
Pallas has an inclination of 34.84 degrees.
Interamnia has an inclination of 17.29 degrees.
Hygiea has an inclination of 3.84 degrees.
Jupiter has an inclination of 1.30 degrees.
Saturn has an inclination of 2.48 degrees.
Uranus has an inclination of 0.77 degrees.
Neptune has an inclination of 1.76 degrees.
Orcus has an inclination of 20.57 degrees.
Pluto has an inclination of 17.15 degrees.
Ixion has an inclination of 19.58 degrees.
Salacia has an inlination of 23.94 degrees.
Varuna has an inclination of 17.2 degrees.
Haumea has an inclination of 28.19 degrees.
Quaoar has an inlination of 7.99 degrees.
1992 QB1 has an inclination of 2.19 degrees.
Makemake has an inclination of 29.00 degrees.
Eris has an inclination of 44.04 degrees.
Sedna has an inclination of 11.92 degrees.

Most of those bodies have a low inclination to the ecliptic, but Pallas is an interesting exception. I don't know why Pallas has such a high inclination but I would bet on a Kozai mechanism due to Jupiter's influence.

Plutinos and Kuipier belt objects are influenced by Neptune, and part by the Kozai mechanism.

Eris is a special and curious case, it is too far from Neptune, although there is still some weak gravitational Neptune's influence there, but I don't know why its inclination is so high. This contrasts with Sedna, which have a more reasonably lower inclination.

Further, the Oort cloud is expected to be full of bodies of essentially unrestricted inclinations and possibly very excentric orbits, as a result of gravitational interactions with passing stars on the past. By the way, a passing star in a distant past might be an explanation for Eris orbit and inclination.

Note: All of the data here, as well as both of the images, I found on wikipedia.


Estimating close approach between all asteroids themselves I am currently writing a C++ program to show asteroids in 3D, and find close approaches or collisions. I got my orbital elements from JPL https://ssd.jpl.nasa.gov/sbdb_query.cgi. So far so good, with over a million asteroids drawn in 5ms with my old Quadro GPU
But the problem is the variation of the elements over time. For example, CERES:
epoch,a,e,i,om,w,ma
2459200.5,2.766089105818,.07816842657453,10.58789954719,80.27235841368,73.72488984426,205.5454154582
2459310.5,2.765760313090,.07831877879848,10.58807660401,80.26860808947,73.73699886586,229.1146825391

After only 4 months, the eccentricity and all other elements have changed as seen on the second line.
How can I compute the change in the parameters without knowing their derivatives?
Or where to get those derivatives?
I searched all JPL small bodies site and their telnet access or email, but could not find a way to download orbital elements with derivatives or a way to compute the change.
I am aware that the difference might be just a pixel on the screen but it represents hundreds of thousands of kilometres. What I want to do is study the close approach between the asteroids themselves and eventually near collision. I already did that for artificial satellites using SGP4 propagator using OpenCL running on GPU video card. I can propagate 20,000 satellites (include debris) in a few milliseconds and get results exact to 1 kilometre. Compare with Celestrak Socrate. A prediction of a 1,000-kilometre approach must be possible.
Does anybody know how?

Thanks folks for helping.
I knew about SPK files but from Horizon's telnet inteface there is a limit of 200 bodies per request.
Making over 5000 request might be possible but not very productive.
I also looked at the  DE421 file from ftp://ssd.jpl.nasa.gov/pub/eph/planets/bsp/ but it looks like the corrections for the 8+1 planets
here is the output from BRIEF, a SPICE utility: https://naif.jpl.nasa.gov/naif/utilities.html
BRIEF -- Version 4.0.0, September 8, 2010 -- Toolkit Version N0066
Summary for: de421.bsp
Bodies: MERCURY BARYCENTER (1)  SATURN BARYCENTER (6)   MERCURY (199)
    VENUS BARYCENTER (2)    URANUS BARYCENTER (7)   VENUS (299)
    EARTH BARYCENTER (3)    NEPTUNE BARYCENTER (8)  MOON (301)
    MARS BARYCENTER (4)     PLUTO BARYCENTER (9)    EARTH (399)
    JUPITER BARYCENTER (5)  SUN (10)                MARS (499)
    Start of Interval (ET)              End of Interval (ET)
    -----------------------------       -----------------------------
    1899 JUL 29 00:00:00.000            2053 OCT 09 00:00:00.000

I can always do a statistical estimate, any close approach being false but the overall count being close to reality.
Too bad to drop the project, I was getting nices pictures from it. The coloring is done against eccentricity, red=0


Why is Uranus able to support a regular satellite system? Uranus has an obliquity of 98° which means that the mutual inclination between a satellite orbiting in its equatorial plane and the orbit of Uranus around the Sun would exceed the critical angle for Kozai oscillations. This would drive the satellites to high eccentricities which would likely be extremely damaging to the stability of the system, yet Uranus does host a system of regular satellites on near-circular orbits in its equatorial plane.

This suggests something is suppressing the Kozai oscillations for the regular satellites, so what is doing this? I suspect it is linked either to the non-spherical shape of the planet or gravitational interactions between the moons, but I'm not sure which would be a more relevant factor.

The solar perturbations on most of the satellites of Uranus are on a very small scale indeed, which may explain the absence of the instabilities noted in the question.

A perturbational effect depends on the scale of the solar perturbing accelerations relative to the ordinary inverse-square attraction of the primary body. The scale factor (often designated $ m^2 $ in texts on the lunar theory such as Airy's 'Mathematical Tracts' and Godfray's 'Lunar Theory') increases with the separation of the satellite from the primary, it is also as the cube of the distance-ratio satellite-primary : primary-Sun (see detailed calculation below).

Taking as an example the most distant of the major moons of Uranus, Oberon at 583500 km:

The scale factor for the solar perturbing accelerations on Oberon in its orbit relative to Uranus is only about 1/5,000,000 of the gravitational acceleration towards Uranus felt by Oberon (calculation below).

Comparing the corresponding scale factor for Earth's moon and its solar perturbations, the factor is very much larger, close to  1/178, as is well-known.  The Moon has a fairly strongly sun-perturbed orbit relative to the Earth, but the solar perturbations on Oberon are more than four orders of magnitude smaller, really very tiny.

It is possible that the two tiny outermost satellites XVI and XVII of Uranus may be experiencing perturbations large enough to drive growth of their eccentricities, their eccentricities at 0.18 and 0.52  are very large compared with the major satellites, all with eccentricity < 0.004 (source, Astronomical Almanac 2016).

Detail of the approximate calculation:

The 'sampled' solar perturbing acceleration on a satellite is taken for present purposes to be represented by the perturbing acceleration on the satellite when it is close to quadrature with the Sun (as seen from the planet). In this configuration, the solar perturbation on the satellite is directed towards the planet, adding here to the ordinary inverse-square attraction of the satellite towards the planet.

With mass-constants of the Sun, S, and of the planet, P ;  planetary semi-axis a ; and planet-satellite separation d , plus an assumption that the mass of the satellite is very small relative to the other two:

1 ** the planet's accelerative attraction on the satellite is $P/d^2$ ;

2 ** the Sun's accelerative attraction on the planet is $S/a^2$ ;

3 ** and so also (to a quite close approximation) is the amplitude of the Sun's accelerative attraction on the satellite  $S/a^2$ .

4 ** When the satellite is in quadrature, the resolved part of the Sun's vector attraction on it that does not cancel with the Sun's attraction on the planet, i.e. the net perturbation on the satellite at that point, is given very nearly by the proportions of the triangle satellite-planet-sun:  the lengths of two of the sides are approximately a and the third d. In this configuration, the resolved part of acceleration #3 in the direction of the planet, to a close approximation, is thus  --

$ (S/a^2) .  (d/a) $.

The ratio of perturbing acceleration 4 to ordinary planetary attraction 1 is thus

$ (S/P) . (d/a)^3 $.

For the Sun and Uranus, S/P ~= 22902 ,

for the Sun and Earth+Moon, S/P ~= 328901 .

Uranus is approximately 19 au from the Sun to the Earth's 1, the au is  149597871 km, the mean planetocentric distance of Oberon is 583500 km and of the Moon 385000 km.

Using these figures, the ratio

'solar perturbation on satellite : planetary attraction on satellite'

comes to ~ 1/178 for the Earth and Moon, and  ~ 1/5,000,000 for Uranus and Oberon.

The tiny outer satellites of Uranus are farther away from it than Oberon is, in a ratio of about 20.8 for the outer (XVII). Thus the perturbations on it, as a proportion of the ordinary attraction of Uranus at its distance, is $(20.8)^3$ times larger than for Oberon, making the relevant perturbation scale factor as large as about 1/550 , still smaller than for the one for our Moon, but perhaps enough for the disturbing effects to be reflected in its higher eccentricity of about 0.52.

{Update:} It turns out that the orbits of the outer Uranian moons have actually been studied: Brozovic, M.; Jacobson, R. A. (2009), "The Orbits of the Outer Uranian Satellites", The Astronomical Journal, 137 (4): 3834-42. Just one of the (high-eccentricity) outer satellites was found to be disturbed by a Kozai resonance and may be in unstable orbit. It appears that the conditions for the effect are not fulfilled for the others.


What day/night cycles, climate and seasons would experience Alpha Centauri Bb inhabitants? Alpha Centauri Bb is an exoplanet orbiting Alpha Centauri B. It is asserted that given the close distance to the star the planet should be tidally locked.

The orbiting period of the planet is about 3.2 days.

If the planet has no atmosphere (which is very possible due to proximity to the star), its dark side should experience low temperatures like permanently-shadowed areas of Mercury do.

At the same time the planet is about 11 AU from the other star, Alpha Centauri A.

This possibly means that the planet should experience quasi-day/night cycle each 3.2 days.

My questions are:

To what temperatures the dark surface of such planet could be heated? Is there possibility of liquid water?
Will the radiation of the second star be enough to provide normal day-like illumination and heating?
Will the calendar on such a planet differ sufficiently from a calendar of a planet that experiences day/night cycles from the nearest star?
Will such planet experience seasons and how they would be arranged?

This is just a partial answer (so far), but it will address at least part of your question.

  1.To what temperatures the dark surface of such planet could be heated? Is there possibility of liquid water?

Wikipedia gives a good estimate for this: 1,200 degrees Celsius. For a planet, that's pretty hot! Any water there would evaporate very quickly. In fact, it is expected that a lot of the planet's surface is molten. Now, does this apply to the back side of the planet? I would think so. While only the front side of the planet would bask in the warmth of Alpha Centauri B, the heat should dissipate throughout the entire planet.

  2.Will the radiation of the second star be enough to provide normal day-like illumination and heating?

The short answer here is no. In fact, when compared to our solar system, nothing about the radiation Alpha Centauri Bb gets is normal. It is situated extremely close to Alpha Centauri B - 0.04 AU. Given that both Alpha Centauri A and Alpha Centauri B are similar to the Sun, I think it's clear that this exoplanet will not receive a "normal" (normal here meaning similar to Earth) amount of radiation from Alpha Centauri B. Conversely, Alpha Centauri Bb is 11 AU away from Alpha Centauri A - much greater than the distance between it and Alpha Centauri B (a bit more than the distance between Saturn and the Sun). This means that the planet will receive only a tiny fraction of light from Alpha Centauri A relative to Alpha Centauri B.

  4.Will such planet experience seasons and how they would be arranged?

Seasons depend on the tilt of a planet's axis. Unfortunately, it is hard (if not impossible) to measure the tilt of the axis of Alpha Centauri Bb (I haven't been able to find any measurements). I would assume that it would experience some seasons, but I don't know how much they would extend.

As for question 3 - I'm not quite sure what you mean by "calendar." If you mean an artificial calendar, I would think that any beings on the planet (life there is extremely unlikely) would only use Alpha Centauri B to determine their years (just over three days and five hours).


Unknown moving objects Last night, I used my new Sony WX500 camera to get long exposure pictures of the sky from my backyard.  I used the "Heavens Above" app to find out when satellite flares would happen and I even got a great picture of the ISS passing overhead at roughly -2mag.  I was able to identify many flares that I had gotten pictures of when reviewing the pictures today.  However, this is where it gets odd.

Some moving "things" didn't match what was supposed to show according to the app.  Some were accidentally caught in the pictures I took, and I'm sure there aren't any flares that the app showed me.

What could they be?  Shooting stars?  Forgotten or broken satellites?

Converting comment to community wiki

Possibilities:

Aircraft lights (but only if non-blinking);
genuine satellites/old launch stages not in the database;
random ("sporadic") meteors not part of an established shower;
birds (yes, sometimes light-coloured migrating birds reflect city light from their bellies).


Does any asteroid we have ever seen have a hot core like earth.? I was wondering if any of the asteroids we have ever seen has a hot core like Earth's.

I'm thinking that Earth has a strong gravitational field and so formation would have involved a lot of gravitational energy being converted to heat, but not so much for the asteroids.

See Could a planet form without a molten state?, where Anders Sandberg calculates that a body with radius ~ 1000 km would melt from the heat of formation.

The largest asteroid, Ceres, has a radius of 473 km, so it's a bit too small to have totally melted just from gravitational accretion, but it may have picked up a little more energy from impacts that were faster than freefall. It could also generate some internal heat from radioactive decay, but it's not expected to contain a significant percentage of radioactive material.

Presumably, Ceres was plastic to some degree during formation, since it's (almost) spherical (and it's the only asteroid large enough for gravity to make it spherical). But that wouldn't have required a very high temperature, since Ceres contains a lot of water ice.

But even if it did totally melt when it formed, it's had over 4 billion years to cool down, so it's extremely unlikely to still have a hot core. However, it does emit more water vapour than expected, and it may be warm enough to contain some liquid water below the surface.


when are you able to see Halley's comet, and for how long? I've been hearing about Halley's comet for a while now, and I'm just curious as to when and where to see it, and for how long. :)

Halley's comet orbits the sun and its orbit lasts 75 years. The orbit is a long elliptical orbit. For most of those 75 years it is a cold black dot, and frozen solid in the outer solar system. But a short period during that orbit it gets close to the sun. The ice and gas begin to boil of its surface and gets blown back by the solar wind of the sun. It then starts to look like a comet.

The last time Halley's comet was close to the sun was in 1986, and so the next time will be in 2061. You won't be able to see Halley's comet until then.

(image from University of Wisconsin)

I remember 1986, and the comet was visible but hard to see. It was a very dim fuzzy dot, that was just about visible in my binoculars. This is because, unfortunately, the Earth was a long way from the comet. In 2061 Earth will be even further from the comet, however Earth will be better positioned when Halley is closest to the sun, so it should make a good show, and be clearly visible to the naked eye. Comet predictions are hard as much depends on exactly how much water vapour boils off, and this is hard to predict. Comets can have sudden changes in activity.

Moreover, Halley's comet isn't the only comet. There are others that can be as visible as Halley will be in 2061. There are also long period comets that are undiscovered until they come close to the sun. These can be very bright. Comet Hale-Bopp was a "great" comet in 1997 and was visible for many months. Another great comet could come along any time but we can't predict when.


Is there a limit to the distance of detectable gravitational wave events? This is somewhat a follow up question to this one.

This article indicates that the binaries detected have a distance between 320 and 2840 Mpc.

With the current technology, is there a limit to the distance of coalescing compact binaries that can be detected through gravitational waves? Or could we potentially see these mergers up to the edge of the cosmos (if there were any)?

Several factors influence whether a source of gravitational waves at a certain distance is observable by a certain instrument. One way to compute the limit to the distance is explained in Abadie et al 2010 and is as follows:

Distance. The amplitude of gravitational waves decreases roughly with the inverse of the luminosity distance $\propto D^{-1}$. So sources that are further away will be more difficult to observe and at some point they will be less/not visible.
Power of the source. The stronger the source the more easy it's signal can be detected. For binary systems with larger total mass $M$ and larger reduced mass $\mu$ you will observe higher amplitude waves. The amplitude of the signal $\vert \tilde{h}(f) \vert$ can be expressed as (the expression is from Abadie et al 2010 see Finn and Thorne 2010 for a derivation)

$$ \vert \tilde{h}(f)\vert = \frac{2c}{D} \left(\frac{5 G \mu}{96 c^3} \right)^{1/2} \left( \frac{GM}{\pi^2c^3} \right)^{1/3} f^{-7/6}$$
Sensitivity. The detector can be more or less sensitive. The more sensitive the detector the lower the luminosity or distance of the objects that it can observe.

The sensitivity can be expressed by the noise power density $S_n(f)$ (as function of frequency $f$) which is specific for the instrument (and you will be able to see graphs of this in many publications). A signal can be observed if it is stronger than the noise. Abadie et al 2010 use as limit a (conservative) signal-to-noise of $\rho$ =8, which means that the signal must be 8 or more times stronger than the background noise in order to be detected.

This signal-to-noise ratio is determined by an integral of the ratio of
the frequency-domain waveform amplitude $\vert \tilde{h}(f) \vert$ and the noise power density $S_n(f)$.

$$\rho = \sqrt{4 \int_0^{f_{ISCO}} \frac{\vert \tilde{h}(f) \vert^2}{S_n(f)} \text{d}f }$$
where $f_{ISCO}$ is the frequency of the innermost stable circular orbit of the binary system

In that article (Abadie et al 2010), the limit of the distance for the detection of wave events from binary blackholes with mass $10 M_{\odot}$ was estimated at 2187 Mpc which is pretty close to the distance of 2840 Mpc estimated for GW170729 (which is heavier).

Note that the limits for binary neutron stars are more often reported and are easier to find. For instance in Moore et al 2015 you can read in more detail about the increase of the limit for LIGO from 80 to 100 Mpc in the recent years. The first image shows plots of $S_n(f)$ as function of $f$ and of $D$ as function of time (during the experiments improvements have been made and the distance was changing).

Rate of occurances. When certain events have a higher probability to occur then it may also be more likely to observe them at a far away certain distance. The computation for these rates include the aspects of distance. Also as stated by Abadie et al

  The real detection range of the network is a function of the data quality and the detection pipeline, and can only be obtained empirically.

In Abbott et al 2016 a computation is performed to determine the probability to observe a particular event at a certain distance. The distance, for $40-40 M_{\odot}$, ranges up to a roughly $z=0.6$ (or using $d \approx z c / H_0 \approx 0.6 \times 3 \times 10^5 / 74.2 \approx 2.5 Gpc$), which is plotted in the last figure of that reference.

Conclusion: the observation of GW170729 at about 3Gpc is about the limit of the current instruments

References

Predictions for the rates of compact binary coalescences observable by ground-based gravitational-wave detectors J Abadie et al 2010 Class. Quantum Grav. 27, 173001
Gravitational waves from a compact star in a circular, inspiral orbit, in the equatorial plane of a massive, spinning black hole, as observed by LISA Lee Samuel Finn and Kip S. Thorne 2000 Phys. Rev. D 62, 124021
Gravitational-wave sensitivity curves C J Moore et al 2015 Class. Quantum Grav. 32 015014
Supplement: The Rate of Binary Black Hole Mergers Inferred from Advanced LIGO Observations Surrounding GW150914 B. P. Abbott et al 2016 ApJS 227 14


Was Guth inflation propelled by what we now call "Dark Energy"? The more powerful our telescopes become, the more we are able to see earlier and earlier remnants of Guth Inflation. Assuming that it makes sense to associate a location, BB,  in space-time where the Big Bang occurred, then test particles (luminous or otherwise) have been decelerating away from BB ever since. When viewed through our telescopes, i.e.: when viewed in time reversed order, a test particle decelerating away from BB would appear to be accelerating away from us. We usually attribute this acceleration away from us as propelled by “Dark Energy”, and any study to understand Dark Energy is equivalent to a study of what propelled Guth inflation in the first place.
I am sure that I am not the first to think of such a straight forward interpretation of Dark Energy, but I do not see it in any of the popular literature on Cosmology. This implies that my thinking is probably incorrect. Can somebody point out the flaw?

One flaw is saying that the big bang occurred at a location. It didn't: the whole universe was once part of the big bang. So the stuff about deceleration being seen as acceleration isn't true.

Secondly, shortly after the big bang, the universe behaves "ballistically" (although following the rules of General Relativity, not Newtonian gravity). It was initially decelerating and the influence of dark energy was negligible.

But in the popular $\Lambda$CDM model (the prevailing model at the moment), the dark energy density is constant, so as the universe gets bigger it becomes more influential and the universe began to accelerate a few billion years ago.

None of this has much to do with inflation, which, if it happened, happened in the first tiny fraction of a second, when the universe was dominated by some energy density, similar in effect to dark energy (in that makes space expand), but much, much stronger. Exactly what this energy density was is still speculative, and inflation remains a "bolt-on" to the big bang model to explain why space is flat and why different parts of the universe appear very similar.


What is the bulk Lorentz factor? When we are talking about relativistic motions, say from a relativistic jet, what is meant by the term "bulk Lorentz factor", and the bulk speed, $\beta$?

I think it is referring to the speed and Lorentz factor $(\beta = v/c$ and $\gamma = [1-\beta^2]^{-1/2})$ of the gas as a whole. Within the gas, there could be particles moving with a variety of velocities.

So if you pick up a ball of gas at 10,000 K (ouch) and throw it at 100 m/s then the bulk speed is 100 m/s, but obviously the particles in the gas have their own individual velocities.


What part of the EM spectrum was used in the black hole image? The EHT has published the first image of a black hole.  It is the event horizon of the singularity in M87 against its accretion disk.  I've been through the reports in the popular press but none mention the light wavelengths this was taken at.  Is it an image in visible light, infra-red or longer?

According to the EHT website The observations were done using radio telescopes observing at a wavelength of 1.3mm. Visible light or infrared from the accretion disk would be nearly impossible to detect at such large distances. The radio measurements from all the synchronized telescopes were then combined and converted into an image.


Which space telescope is the most distant? From the earth, and from the sun. Only interested in active, operational ones.

Currently New Horizions is temporarily hibernating; it's last activity was two months ago. So I'm going to post a supplementary answer here because it is "operational" in the sense that it still works and will be used again, even though it is not "active" at the moment.

The most recent and farthest-from-earth telescopic observations that I know of are from the New Horizons spacecraft and it's Long-Range Reconnaissance Imager (LORRI), a 208 mm (8.2 inch) diameter Ritchey–Chretien telescope equipped with silicon-carbide optics and a cooled silicon CCD imager with a resolution of about 1 arcsecond:

The Long-Range Reconnaissance Imager (LORRI) is a long-focal-length imager designed for high resolution and responsivity at visible wavelengths. The instrument is equipped with a 1024×1024 pixel by 12-bits-per-pixel monochromatic CCD imager giving a resolution of 5 μrad (~1 arcsec)] The CCD is chilled far below freezing by a passive radiator on the antisolar face of the spacecraft. This temperature differential requires insulation, and isolation from the rest of the structure. The 208.3 mm (8.20 in) aperture Ritchey–Chretien mirrors and metering structure are made of silicon carbide, to boost stiffness, reduce weight, and prevent warping at low temperatures. The optical elements sit in a composite light shield, and mount with titanium and fiberglass for thermal isolation. Overall mass is 8.6 kg (19 lb), with the optical tube assembly (OTA) weighing about 5.6 kg (12 lb), for one of the largest silicon-carbide telescopes flown at the time (now surpassed by Herschel).

In December 2017 LORRI captured the images below of two Kuiper belt objects. From the February 9, 2018 NASA News item New Horizons Captures Record-Breaking Images in the Kuiper Belt:

NASA’s New Horizons spacecraft recently turned its telescopic camera toward a field of stars, snapped an image – and made history.
The routine calibration frame of the “Wishing Well” galactic open star cluster, made by the Long Range Reconnaissance Imager (LORRI) on Dec. 5, was taken when New Horizons was 3.79 billion miles (6.12 billion kilometers, or 40.9 astronomical units) from Earth – making it, for a time, the farthest image ever made from Earth.
New Horizons was even farther from home than NASA’s Voyager 1 when it captured the famous “Pale Blue Dot” image of Earth. That picture was part of a composite of 60 images looking back at the solar system, on Feb. 14, 1990, when Voyager was 3.75 billion miles (6.06 billion kilometers, or about 40.5 astronomical units [AU]) from Earth. Voyager 1’s cameras were turned off shortly after that portrait, leaving its distance record unchallenged for more than 27 years.
LORRI broke its own record just two hours later with images of Kuiper Belt objects 2012 HZ84 and 2012 HE85 – further demonstrating how nothing stands still when you’re covering more than 700,000 miles (1.1 million kilometers) of space each day.

Here is a 25-Feb-2018 screen shot of a map of the Solar System with New Horizon's current position plotted. You can read more about New Horizons at:

Where is New Horizons

NASA's New Horizons main page

Instrumentation on New Horizons is extensively described/explained in Spaceflight 101's Instrument Overview.
The LORRI telescope is described for example in the ArXiv preprint Long-Range Reconnaissance Imager on New Horizons (click "PDF"). It has a spectral range of about 350 to 850 nm. The limiting magnitude for high precision positional measurements is about +11.5 for a short exposure, but with some binning and a roughly 10 second exposure that increases to about +17:

For optical navigation, LORRI is required to be able to image a star of visual magnitude V=11.5 at SNR>7 in a single 100 ms exposure, with full width at half maximum (FWHM) >1 pixel. It is not desirable for too great a fraction of the energy from a point source to be imaged onto a single pixel, because stellar images become too undersampled. LORRI has a 4×4 pixel binning mode, for which its limiting magnitude requirement is V>17 in a single exposure of 9.9 s. This 4×4 pixel-binning mode will be used to search for the target KBO and to perform optical navigation on approach.

below: Image and caption from here:

With its Long Range Reconnaissance Imager (LORRI), New Horizons has observed several Kuiper Belt objects (KBOs) and dwarf planets at unique phase angles, as well as Centaurs at extremely high phase angles to search for forward-scattering rings or dust. These December 2017 false-color images of KBOs 2012 HZ84 (left) and 2012 HE85 are, for now, the farthest from Earth ever captured by a spacecraft. They're also the closest-ever images of Kuiper Belt objects.


Is star a necessary phase in black hole lifecycle? It now seems general fact that when a massive star core collapses at the end of its lifecycle it forms a black hole, can a black hole be formed skipping the star phase altogether?

There are speculated objects known as "primordial black holes" which could have been present at the origin of the universe, or formed in its early stages via presently unknown processes, but there is no observational evidence that such objects exist. Also, some models for forming supermassive black holes involve the coalescence of such a huge amount of matter that you would not call it a star, unless some kind of supermassive superstar.

Alternatively, the mass could form stars first, and then coalesce into the supermassive black hole, but those stars themselves would not necessarily need to collapse first. Indeed, since much of the mass that goes into stars goes into lower mass stars that don't make black holes, it seems likely that a significant fraction, or even a majority, of the mass in supermassive black holes was not in a black hole prior to falling into the supermassive black hole.

Finally, since we don't know much about dark matter, one presumes it might be possible for dark matter to form black holes via some presently unknown mechanism, but there is no evidence dark matter would do that, and the idea that dark matter might itself be already in the form of black holes has not been supported by observational searches.

So I'd say our current expectation is that supermassive black holes are largely made of matter that did not collapse in a massive star, and stellar-mass black holes could be mostly or entirely matter that did.