Historical Lunar Tables and the Associated Error Bound Consider as an example, AstroPixel's ``Six Millennium Catalog of Phases of the Moon''---
http://astropixels.com/ephemeris/phasescat/phasescat.html
My question is two-fold:

How are these tables which give (for example) dates and times of new moons and full moons to the nearest minute, generated?

How precise is this information? From http://astropixels.com/ephemeris/phasescat/phases0001.html
we see, for example, that on Jan 13 0001, a New Moon occurred at 10:58 (UT). How much of an error bound is there likely to be on these data?

The first question (how are the tables generated?) is easy to answer. The acknowledgement on the Astropixels page says:

It is based on procedures described in Astronomical Algorithms by Jean Meeus (Willmann-Bell, Inc., Richmond, 1998).

For the second question (how accurate are the calculations?), there are 3 sources of error that I can think of:

The procedure may be for a limited time span. Meeus' book does not mention what time span the formulas are good for, but the source from which the procedure was based on (Chapront's ELP-2000/82) used the years 1900 to 2000 to determine the variables in the equations. (See NASA Astrophysics Data System - The lunar ephemeris ELP 2000.) It is hard to know what the accuracy is when going backward or forward several hundred years. Mathematically, extrapolating beyond the chosen time span could be reasonably accurate, or it could be "completely wrong". (I did not read all of the Astropixels' pages, but I know that a lot of historical research was done when calculating the solar eclipse tables. So I would anticipate some cross-referencing between the Moon phases and eclipses. After all, you cannot have a solar eclipse unless you have a New Moon!)
Time is based on the rotation of the Earth, such as GMT (Greenwich Mean Time), UT (Universal Time), or the time in a local time zone. This type of time is not consistent over long time intervals. The Earth's rotation slows down and speeds up which requires adjustments to our clocks. The motion of the planets is (more) consistent and runs at a steady rate. Astronomers call this type of time Dynamical Time (formerly known as TDT Terrestrial Dynamic Time or ET Ephemeris Time). The difference between UT and Dynamical Time is explained in the Astropixels' page Uncertainty in Delta T. The uncertainty in Delta T is over an hour for the year 2000 BCE, so the time of the phase can be off by this amount just due to the uncertainty of clock time (the Earth's "clock") versus Dynamical Time (the solar system's "clock").
Computer languages can introduce inaccuracies. For example, it may not be possible to store the number of days, hours, minutes between now and the year 2000 BCE to full precision due to the limited accuracy of storing really large number. There are approximately 1.05 billion (1.05E9) minutes in 2000 years. In order to calculate a result 2000 years ago to a precision of 1 minute requires the ability to store a number such as 1,050,000,001 and 1,050,000,002. (I am not a computer expert, but these numbers can probably be stored easily and to a higher accuracy. The potential problem is that the procedure uses values of time, time squared ($T^2$), time cubed ($T^3$), and time to the fourth power ($T^4$)! Thus, numbers can quickly become larger than what the computer can store to full precision.)

When put all together, I would expect that the times 2000 years ago (or 2000 years in the future) may only be accurate to a couple of hours.


How to estimate the inclination of a galaxy between edge on and face on? I have the images of a galaxy and the field of view and I would like to determinte the inclination of a galaxy between edge-on and face-on. Would that information be enough to do the calculation? Or what other details would be needed?

If you tilt a circular object, then along one axis the length will appear to be shortened, while along a perpendicular axis the apparent size will be unchanged.  So you can use that to determine the tilt angle (inclination).
If we define $0^\circ$ to be face-on, then when tilted at an angle $i$ away from face-on, the shorter axis will appear to be decreased in length by a factor of $\cos i$.  Thus, assuming that the galaxy is circular in cross-section, you can determine its inclination from $\cos i = \frac{\rm short\ axis}{\rm long\ axis}$.
This should work for spiral galaxies, but won't work for elliptical galaxies, which aren't necessarily circular in projection when viewed face-on.


Derivation of the formula for longitude of ascending node for a satellite I've been looking into the document IS-GPS-200H to understand how to calculate satellite location in the ECEF coordinate.

I am having problem understanding the formula to derive $\Omega$, the longitude of the ascending node (LAN) relative to Greenwich at given time $t$:

$$
\Omega = \Omega_0 + \left( \dot{\Omega - w} \right)\times t_k - w \times t_{oe}
$$

where:
$$
\Omega_0: \text{LAN relative to vernal equinox, at the beginning of the week}\\
\dot{\Omega}: \text{angular velocity for LAN, relative to vernal equinox.}\\
w: \text{angular velocity of earth, relative to vernal equinox.}\\
t_k: t - t_{oe}\\
t_{oe}: \text{ephemeris reference epoch}\\
$$
(and let us denote the beginning of the week as $t_0$ for brevity).

But if I try to work out this from scratch:

At $t = t_0$, LAN was $\Omega_0$. But since what we really need is the difference of LAN and longitude of Greenwich, we also need to know $w_0$, the initial longitude of Greenwich at $t = t_0$.
$$
\Omega(t = t_0) = \Omega_0 - w_0
$$
At the ephemeris reference epoch time $t = t_{oe}$, LAN and the earth both rotate with their respective angular momentum and hence:
$$
\Omega(t = toe) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe}
$$
As time varies from $toe$ to $t$, again LAN and the earth both rotate with their own respective angular momentum and hence
$$
\Omega(t) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe} + (\dot{\Omega} - w) \times t_k
$$
which obviously differs from the right formula by $w_0 + \dot{\Omega} \times t_{oe}$.

My question is where am I making mistakes/misunderstanding the eqution?
Explain also why we don't need to know $w_0$ or equivalent input, that would be greatly appreciated.

I think you're not too far from understanding the formula. However, there are a couple details you're missing:

From what I understand from the document you linked to, $t_0 = t_{oe}$ is the time at the beginning of the week, or reference epoch.
When computing an angle from a angular velocity, you usually need to compute it with respect to an initial angle and an initial time: when the Earth and the satellite rotate between $t_0$ and $t$, the angle they rotated of is $\dot{\Omega}(t - t_0)$ or $\dot{\Omega}_E(t - t_0)$.
The only exception to this rule seems to be the initial Greenwich longitude with respect to the vernal equinox, which seems to be computed from $t = 0$.
$w_0$, the initial Greenwich longitude wrt the vernal equinox, is $\dot{\Omega}_E t_{oe}$. From this formula, it seems like the document considers an origin of time $t = 0$ when the longitude of Greenwich wrt to the vernal equinox at this time is 0.

To summarize: the formula would have been more explicit and comprehensible written as:
$$
\Omega(t) = \Omega_0 - \dot{\Omega}_E t_{oe} + (\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})
$$
with:

$\Omega_0$ = LAN wrt the vernal equinox at $t_{oe}$
$\dot{\Omega_E}$ = angular velocity of the Earth wrt the vernal
equinox
$-\dot{\Omega}_Et_{oe}$ = Longitude of Greenwich at $t_{oe}$
$\dot{\Omega}$ = angular velocity of the satellite wrt the vernal
equinox
$(\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})$ = relative angular
displacement between the satellite and Greenwich, between time $t$ and
$t_{oe}$

That said, it seems like really bad practice to name $\Omega(t)$ the longitude with respect to Greenwich and $\dot{\Omega}$ the angular velocity with respect to the vernal equinox. This causes quite a bit of confusion, since mathematically speaking, for any linear function $f$, $f(t) = f(t_0) + \dot{f}\times(t - t_0)$


Rendering stars in 3D space - AbsMag to OpenGL scale values I am using the AMNH's Digital Universe stars.speck file to render stars in 3D space. The speck contains parsec-scale coortinates, where our Sun is at 0,0,0. It also lists the AbsMag values - i.e the luminosity from 10 parsecs away. I want to now scale my star textures in openGL, so the resulting star sizes are accurate. How could I convert the AbsMag inverse logarithmic scale to a scale value of 0-infinity, where 0.0 means the star goes away, and 1.0 is no change? Obviously can't make the Sun stay at 1, as that would make it huge.

"Accurate star size" is a problem. Obviously you do not have the resolution on your screen for accurate angular resolution of stars (they would require an absurdly fine resolution), and the dynamic range of brightness is also too low -- stars range over many orders of magnitude in brightness, far more than a screen can show. Deeply annoying. What one might want to do is to make each star have a brightness and screen size that is proportional to their actual brightness in the sky to give the same feeling as the sky would give. This is still very tough, since screens have different gamma correction. To top it off, the human eye actually has a pretty logarithmic response to light (this is why magnitudes and gamma correction make sense).

Here is a rough idea. A star with AbsMag of $M$ has an actual luminosity of $$L = L_\odot 10^{0.4(M_\odot - M)}$$ where $L_\odot$ is the luminosity of the sun and $M_\odot$ is the absolute magnitude of the sun. Things get easy if we just count luminosity in terms of $L_\odot$, making the sun one unit of luminosity.

A spot of radius $r$ on the screen with luminance $l$ radiates power as $P=\pi r^2 l$. That luminance is due to the gamma-corrected luma value $V$ the computer displays: $l = K V^\gamma$.

Assuming the radius changes with luminosity too as $r(L)$, I would try $r(L)=r_0 L^a$ where $a\approx 0.6$ (but this is guesswork) and the sun has radius $r_0$ pixels.

So trying to put this together, we get the pixel brightness as $$V = V_0 (L/ r(L)^2)^{1/\gamma} = V'_0 L^{(1-2a)/\gamma}$$ where $V_0$ and $V'_0$ are the pixel brightnesses used for the sun in this model. Basically this squashes the actual luminosity with $a$ (representing using bigger spots for brighter stars, not requiring as intense pixels) and $\gamma$ (to correct for the screen and the eye). What values to use will largely be trial and error unless you want to try to use screen photometry equipment.

So the full formula converting from absolute magnitude to pixel value would be: $$V = V_0 10^{0.4(4.83 - M)(1-2a)/\gamma}.$$


What caused instant jumps and exactly flat periods in Kepler's light curves? Last 65 days of Kepler's light curve for the infamous KIC 8462852 star is shown below from the published paper's graph. Skip the dips here, I wonder about two other features:

Day 1559, during a period which is calm but also the very brightest, the light curve instantly and permanently drops by half of a percent.
Days 1581 to 1587 the light curve is exactly flat (also days 803-809
not shown here). Is the real data for those periods lost?

Are such features commonplace or another particularity for KIC 8462852?

How wide would a confidence interval typically be for a finished Kepler light curve when all adjustments for telescope movements and whatnot have been made?

There are two separate points of interest you're looking at so I'll separate this into sections.

Sudden Drop at Day 1559

As near as I can tell, this is the result of a quarterly roll of the satellite, specifically the end of quarter 17. Every 90 days, which NASA calls a quarter, the space craft does a 90 degree roll to optimize the solar panel efficiency. What this means for the data is that when this roll occurs at the end of the quarter, the photons from the star now fall on a different CCD channel which is calibrated slightly differently, causing a slight discontinuity in the data. You can read about "Quarter Stitching" here. I've extracted some relevant text and an image for posterity.

  After each roll maneuver, most Kepler targets fall on a different CCD channel and the target’s point-spread function will be distributed differently across neighboring pixels. Naturally this redistribution requires a new computation of the target mask and optimal aperture size within the Kepler pipeline, taking into account the point-spread function, new CCD characteristics, and new estimates of nearby source crowding. The operational outcome is often a different mask shape, with differing amounts of flux within the optimal aperture from both the target and contaminating sources.

Light curves extracted from all single pixels within the quarter 4, 5 and 6 Target Pixel Files for KIC 9603833 (the symbiotic star StHA 169). Gray pixels comprise the optimal apertures that yield the archived light curves. The target point-spread function is distributed across neighboring pixels differently from quarter to quarter and hence the optimal aperture varies in size and shape from quarter to quarter.

Flat line between days 1581 to 1587

I'm less certain about this one. I believe the flat line is just missing data. To keep the line continuous, they just connected the data points they had with a straight line.

Why the data is missing, I can't be sure. I know that Quarter 17 ended with the failure of Kepler's second reaction wheel so my guess is that this missing data is in some way related to these reaction wheel troubles. I do know they tried,in the past, to put the spacecraft in 10-day "wheel rest" mode in hopes that it would give the reaction wheels a break, let them get lubed up again, and be better when they restarted. Perhaps, by this stage, they had already suspected a second reaction wheel problem and were trying out a wheel rest mode. I can't find any evidence a wheel rest occurred during Quarter 17 though, so this is just my guess.


Collision of asteroid and gaseous planet - what would happen What happens when asteroids (or other small bodies) "collide" with a gas planet? In my head it would either go through it or just stay "inside" it due to the strong gravity. I tend to believe in the latter one, but if this is the case, does this mean that the gas giants are full of small bodies inside them?

Thank you.

A very well documented example of a small celestial object with a gas giant is the collision between Comet Shoemaker-Levy 9 with Jupiter.  As it was a comet that was tracked, the whole process of the collision was well documented and observed.

Though the collision occurred in July, 1994 - the first effects occurred 2 years earlier, according to the NASA Technical Report "Shoemaker-Levy 9 and the tidal disruption of comets", when the

  break-up of Periodic Comet Shoemaker-Levy 9 into multiple pieces following its grazing encounter with Jupiter

This represents one of the first stages of the collision that could occur as the celestial object approached the large planet.  In this case, this breakup resulted in Comet Shoemaker-Levy 9 to form a train of fragments to eventually collide into Jupiter - the full process at this stage is shown in the diagram below:

Image Source: Tufts University

The train of fragments was captured by the Hubble Space Telescope (shown below):

Image Source: NASA

Each fragment was designated a letter (e.g. Fragment A etc).  According to the article "The collision of Jupiter and Comet Shoemaker-Levy 9." (Zahnle and Mac Low, 1994), these fragments were in the order of about 100 m to 5 km in size, heading straight towards Jupiter.

Then to the main event - the collision of the comet fragments with Jupiter.

An example was from Fragment G, one of the larger fragments, when it collided with Jupiter (image below):

Image source: NASA

As you can see, the explosion was spectacular - about 21 fragments impacted into Jupiter over a number of days, given Jupiter's relatively fast rotation, the impacts became quite distributed on the top of the clouds of Jupiter, as shown in the ultraviolet image below (the circle shadow below is the moon Io):

Image Source Air and Space Museum

There have been several impacts into Jupiter (and presumably into the other gas/ice giants as well), another one that underwent scientific scrutiny was an asteroid 'the size of the Titanic' colliding with Jupiter in 2009, reported in the article "New evidence that asteroid, not comet, struck Jupiter in 2009", describes a comparison between the 2009 asteroid and Comet Shoemaker-Levy 9 are described as:

  The dark debris, the heated atmosphere and upwelling of ammonia were similar for this impact and Shoemaker-Levy, but the debris plume in this case didn't reach such high altitudes, didn't heat the high stratosphere, and contained signatures for hydrocarbons, silicates and silicas that weren't seen before. The presence of hydrocarbons, and the absence of carbon monoxide, provide strong evidence for a water-depleted impactor in 2009.

Overall, the article mentioned the likely process during any collisions of a celestial body with a gaseous planet, using the example from Jupiter:

  Plunging through Jupiter's atmosphere, the object created a channel of super-heated atmospheric gases and debris. An explosion deep below the clouds – probably releasing at least around 200 trillion trillion ergs of energy, or more than 5 gigatons of TNT

and the celestial object itself - vaporised, totally destroyed in the resulting explosion.


Why do zodiac constellations not look like their names? I am trying to answer the question of one of my youngest students (who is not yet in school), and although going through books, wikipedia etc. I do not have a good answer to his question yet:

Why does none of the 12 zodiac constellations look their respective name?

Joe Rao: Why Don’t Constellations Look Anything Like What They’re Named? offers the following explanation which does not convince my student:

Remember, of course, that a few thousand years ago there was not much to entertain people once the Sun went down. They would go outside under the cover of darkness, look up at the stars, and weave stories or tales about people, animals, and mythical beasts, and use the patterns of stars as illustrations.

The best answer I have so far:

The zodiac names are old, and the constellations have been looking back then more like their actual names.

I am not sure how accurate that answer is and whether there is any better explanation, that's why I am asking the simple question here.
References

History of Zodiac constellations (Wikipedia)
https://www.lpi.usra.edu/education/skytellers/constellations/

Who says they don't? Surely it just depends on how you play "join the dots"? The human brain will make out patterns where it wishes and what things look like is largely in the eye of the beholder.
The second explanation isn't true. Although the relative positions of the stars do change, they have not changed very significantly over a period of 3000 years, certainly not enough to change the basic shape of the constellations.


How to interpret this old degree notation? I'm trying to follow the calculations in this paper on page 5 (see image below). The paper is describing traditional methods of calculation for a luni-solar calendar.

The math is not hard, but I can't make sense of the notation.

I figured that the ... = 150 : 1780 (as in step 3) means that 150 is the integer result, and 1780 is the remainder.

But what is ... = 2; 19:28 (step 8)? Degree; minute:second? Okay, but then I'm lost at step 10 with 134 * sin(0:32) = 0:01. I haven't managed to get that result.

This doesn't really answer your question, but I read https://github.com/profound-labs/calculating-the-uposatha-moondays/blob/master/references/Rules%20For%20Interpolation%20in%20the%20Thai%20Calendar.pdf and noted that footnote 13 (which is referenced by equation 8, although it sort of looks like 3^13) reads:

The routine subtraction of 3 arcmins is a causing Ashadha to be month
10; but by this Wat geographical longitude correction for the sun, as
is the Tapotharam reckoning Vaisakha is month 7 (Keng subtraction of
40 arcmins for the moon (sec. C13). Tung style), causing Ashadha to be
month 9.

which suggests the calculation is being done in arcminutes, not degrees.

Unfortunately, this doesn't help, since the result is 76.5283 arcminutes, which is nowhere near 2 degrees 19 minutes 28 seconds (which is 139.467 arcminutes).

134*sin(0:32) may refer to 32 seconds of arc, and the result is 0.02 which is 1 minute or arc (0:01). Terrible use of notation here if that actually is the case.


Exactly what in Saturn has been located to within 4 km precision? Saturn has been located with fantastic precision thanks to Cassini and VLBI. The location of the 120,000 km diameter giant planet (when defining the 1 bar pressure level as its surface) is known to within 4 kilometers. This has already been helpful in the (preliminary) discovery and characterization of "planet nine".

I suppose it is the Saturn system's barycenter which has been located gravitationally. At this level of precision I suppose that the moons', especially Titan's tidal effects must be considered. But also Saturn's flattened shape and maybe even weather systems. Saturn is known to be lumpy inside, this has been discovered by the effects it has on the inner rings. Such lumps have similar effects as a moon, so I wonder if these too are being considered at the 4 km precision level, and overall exactly how the point located is defined?

I suppose it is the Saturn system's barycenter which has been located gravitationally. At this level of precision I suppose that the moons', especially Titan's tidal effects must be considered. ..., so I wonder if these too are being considered at the 4 km precision level, and overall exactly how the point located is defined?

Short answer: No.
As pointed out in the comments by @AtmosphericPrisonEscape (Oct 5 '16 at 14:42) and @userLTK (Oct 6 '16 at 9:25) - it is a barycentric measurement.
Other references:
Solid-body tides on the Moon
Is Io's orbit or rotation affected by its volcanism?
The Barycenter coordinates are an average, a mean, small perturbations are smoothed in the calculations by various factors such as: measurement accuracy, use of the International Celestial Reference Frame (ICRF), even the Cholesky whitening (Source: "Measuring the mass of solar system planets using pulsar timing" (21 Aug 2010)) used in some of the calculations.

"In astronomy, barycentric coordinates are non-rotating coordinates with the origin at the center of mass of two or more bodies. The International Celestial Reference System is a barycentric one, based on the barycenter of the Solar System.
...
Relativistic corrections
In classical mechanics, this definition simplifies calculations and introduces no known problems. In general relativity, problems arise because, while it is possible, within reasonable approximations, to define the barycenter, the associated coordinate system does not fully reflect the inequality of clock rates at different locations. Brumberg explains how to set up barycentric coordinates in general relativity in his book "Essential Relativistic Celestial Mechanics".
The coordinate systems involve a world-time, i.e. a global time coordinate that could be set up by telemetry. Individual clocks of similar construction will not agree with this standard, because they are subject to differing gravitational potentials or move at various velocities, so the world-time must be slaved to some ideal clock that is assumed to be very far from the whole self-gravitating system. This time standard is called Barycentric Coordinate Time, or TCB.

Example: Motion of the barycenter of the Solar System relative to the Sun, 1945–1995.
In the paper "Uncertainties in the JPL Planetary Ephemeris", by Folkner, on page 47, is shown this chart of the uncertainty in right ascension, declination, and distance of the barycenters of the Earth and Saturn systems calculated for the years 1950 to 2050:

Celestial mechanics and the n-body problem take a long time to solve.
The paper associated with these measurements of Saturn is titled: "VLBA Astrometric Observations of the Cassini Spacecraft at Saturn" (1 Dec 2010), on page 2 it says:

"These observations provide positions for the center of mass of Saturn in the International Celestial Reference Frame (ICRF) with accuracies ∼0.3 milli-arcsecond (1.5 nrad), or about 2 km at the average distance of Saturn.
...
The DE 422 post-fit residuals for Saturn with respect to the VLBA data are generally 0.2 mas, but additional observations are needed to improve the positions of all of our phase reference sources to this level. Over time we expect to be able to improve the accuracy of all three coordinates in the Saturn ephemeris (latitude, longitude, and range) by a factor of at least three. This will represent a significant improvement not just in the Saturn ephemeris but also in the link between the inner and outer solar system ephemeredes and in the link to the inertial ICRF.".

I might return for an edit if there is interest in this Q&A.


carbonaceous content of galaxies? There is a phrase in a Science Daily article which appears to be wrong:

The Milky Way has a very high content of carbonaceous dust, which has been shown to be very rare in other galaxies.

Is there a theory that explains these observations?

From the introduction of the source paper, we can see that what is described is a particular bump in the extinction curve of the galaxy centered at 2175 Ångström (a unit of length used for wavelength). The strength of this bump is large compared to the Large Magellanic Cloud, and is rarer toward the Small Magellanic Cloud.

As mentioned in the comments, the main candidates for the origin of the feature are non-graphitic carbon or polycyclic aromatic hydrocarbons (PAHs). The absence in other environments is attributed to low metallicities or different radiative environments.

Basically, this is a case where we lose information in science reporting. "Very high content of carbonaceous dust" was a translation of a specific strength at a specific wavelength, and "other galaxies" was a translation of surrounding galaxies.

Source:

The 2175 Å extinction feature in the optical afterglow spectrum of GRB 180325A at z=2.25 - Zafar, et al.


Star versus Black Hole I'm making a simplified animation for college project on how a star reacts with a black hole. I had known that a black hole sucks in everything and thus should be doing the same for stars. But upon my research I found out that recently a discovery was made that a super-massive black hole ripped apart a star. Now I've always been curious about cosmology and this took my attention and also worried me for my assignment. There were two conditions with the above mentioned calamity - one, the black was a super-massive one and second, the victim star was possibly already stripped off it's outer gaseous by the same black hole. My question is that the ripping apart happens in all conditions and the sucking in thing was wrong or it happens only in mentioned circumstances?

A black hole (BH) never sucks anything in. 'Sucking' requires gas pressure, but a black hole only acts via its gravity. A non-rotating (=Schwarzschild) black hole attracts all massive objects, very much like the Sun attracts the planets.
Yet, the planets are not falling into the Sun. This is because the planets move on near-circular orbits when the centrifugal force balances the gravitational attraction. The planets orbital angular momemtum is conserved (because the Solar force field is purely radial [to good approximation]) and these orbits are stable.

The difference to a black hole is that very close to the event horizon no such stable circular orbits exist anymore. At those distances, nothing can orbit the BH without falling in.

However, an object orbiting a BH on a stable orbit (including passing trajectories) is still subjected to the BH's tidal forces. These distort the object, very much like the Moon's gravity distorts the Earth, generating tides.
In case, the trajectory passes sufficiently close to the BH and the object is rather big and fluffy, like (some) stars, the tidal forces may not merely distort the object, but rip it apart.

Such tidal disruption of stars must happen if a star passes close to a supermassive BH. This can hardly be directly observed, but the it is thought that some of the stellar matter forms an accretion disc around the BH and produces a characteristic light curve. An observed light curve in agreement with this model is often interpreted as circumstantial evidence for a stellar disruption event.


Dark Matter Detection A paper appeared at the arXiv.org (in April or May) claimed that the writers have found the gamma rays emitted by annihilation of dark matter in a satellite galaxy of the Milky Way by examining the Fermi data.  I cannot find this article anymore.  (It is possible that the writers withdrew the article.) Has someone actually detected gamma rays emitted by neutralino (or any other dark matter) annihilation?

Is it the paper David H pointed out, by Geringer-Sameth et al. (2015)? If so, then I should point out that they're not the first. Just using Wikipedia, I came across Weniger (2012) and Albert et al. (2008).

Geringer-Sameth et al.'s abstract reads (in part)

  We present a search for gamma-ray emission from the direction of the newly discovered dwarf galaxy Reticulum 2. . . . Reticulum 2 has the most significant gamma-ray signal of any known dwarf galaxy. If Reticulum 2 has a dark matter halo that is similar to those inferred for other nearby dwarfs, the signal is consistent with the s-wave relic abundance cross section for annihilation.

Albert et al. also studied a dwarf galaxy:

  The nearby dwarf spheroidal galaxy Draco, with its high mass to light ratio, is one of the most auspicious targets for indirect dark matter (DM) searches. Annihilation of hypothetical DM particles can result in high-energy γ-rays, e.g., from neutralino annihilation in the supersymmetric framework. A search for a possible DM signal originating from Draco was performed with the MAGIC telescope during 2007.

The point is, this has been done before. These folks aren't the first.

  Has someone actually detected gamma rays emitted by neutralino (or any other dark matter) annihilation?

More analysis is probably needed. I'll quote the paper itself:

  While Ret2's X-ray signal is tantalizing, it would be premature to conclude it has a dark matter origin. Among alternative explanations, perhaps the most
  mundane is the possibility that an extragalactic source lies in the same direction. Searching the BZCAT [54] and CRATES [55] catalogs reveals a CRATES quasar (J033553-543026) that is 0:46 from Ret2. Further work must be done to determine whether this particular source contributes to the emission, though we note that at spectrum radio quasars rarely have a spectral index less than 2 [56]. One of the much-discussed astrophysical explanations for the apparent Galactic Center excess is millisecond pulsars [24, 26, 57{61]. In the case of Ret2, it is the high-energy behavior which disfavors a pulsar model, as millisecond pulsars exhibit an exponential cut-offat around 2.5 to 4 GeV [26, 30, 61{64]. Alternatively, high-energy cosmic ray production could potentially arise in the vicinity of young massive stars. Upcoming photo-metric and spectroscopic analysis of Ret2 will check this possibility.

Don't jump to conclusions. They didn't. It stinks, but the answer to your question is, "Nobody's sure."


Gas Planets composition and gravitational effects It has been postulated that Gas Giants have a solid core and an obvious gaseous atmosphere.  Question:  As the gas is a state of matter and gravity draws all matter inward, is the reason the gas is not drawn towards the core and condensed into a solid because the extent of the effect that the gravity has on the gas proportional to the mass of that gas.  Additionally, wouldn't this create a series of unique gaseous layers and wouldn't the depth of those unique layers be contingent on the layers composition.  Also, could the gravitational effect on the known layers be used to determine the composition of the core.

The reason the hydrogen and helium (that make up most of the mass of planets like Jupiter) remain as gases and don't condense is essentially the same as any other gas. So I could ask "why don't the nitrogen and oxygen in the atmosphere of the Earth condense?"

The reason is that the temperature and pressure in the planet's atmospheres do not favour the formation of a solid. Hydrogen and Helium both have very low boiling points, and even under extreme pressure they tend to remain as a gas.

Further inside the planet the pressure does become sufficent to cause the hydrogen to condense into a liquid, or more properly a supercritical fluid (as at these high pressures the distiction between a gas and a liquid is blurred.  The temperatures are too high for solid hydrogen to form.

There are a couple of things I don't fully understand about your question, which may point to misconceptions. For example most of the mass of Jupiter is in the form of gas, and so most of the gravity comes from the gas, and not from the core (solid or otherwise).

There are layers in the atmosphere of Jupiter. The outer layers are actually rather similar to those in Earth's atmosphere. There is a turbulent troposphere and a stable stratosphere, with a warmer exosphere extending into space. Inside there are layers of supercritical hydrogen and helium and deep inside the pressures cause the hydrogen to behave like liquid metal.

Studying the gravity field (by careful observations of orbiters) is one of the principle ways that we can investigate the internal structure of Jupiter. This suggests that the heavier elements are not concentrated in the core, but mixed in the internal layers, suggesting that the planet has undergone some kind of extreme event in during its formation.


Could we see planets around active Stellar Mass Black Hole? This question is a followup of Is there any theoretical or empirical research about planetary systems with a black hole in their center?;

Since it is possible to detect planets around stars with the change of light they cause, I would think it should also be possible to detect them when they orbit around an active Stellar Mass BH.

From what I (can) understand from the first answer, there is no known case of planets orbiting around Stellar Mass BH. What could be preventing us to observe it?

Is there too few active Stellar Mass BH close enough for us to have a chance the system's plan is well aligned with us?
Do we only see Stellar Mass BH when they have a stellar companion feeding them, and this discards to possibility for close-by planets?
Or is it because the light we can see from them is too fast-changing for a wobble or transition to be observable with today telescopes?

Stellar Black Holes (SBHs) are really really hard to find. If there's a free-floating one in space with no companion, our chances of finding it are almost nil. However, observations have more or less ruled out the possibility that space is freely floating with huge quantities SBHs. This was one of the proposed sources of Dark Matter, namely MACHOs. We don't believe MACHOs exist at this point because lensing observations towards the galactic core suggest they don't exist in any significant quantity.

The list of known and candidate SBHs contains less than 30 entries. The reason SBHs are hard to find is that, as stated above, they don't seem to be too common, and also that they can only really be observed indirectly through their interactions with a binary companion. Often, one observes the X-ray radiation coming from infalling matter around the black hole, taken from the binary companion, rather than observing the black hole itself.

  Is there too few active Stellar Mass BH close enough for us to have a chance the system's plan is well aligned with us?

The answer to this is probably yes.

  Do we only see Stellar Mass BH when they have a stellar companion feeding them, and this discards to possibility for close-by planets?

Again yes, we can only "see" the SBHs when they have a companion feeding them. This doesn't count the three cases where we've found SBHs through gravitational wave observations though, but all of those were far outside our galaxy. Whether a planet could exist in such a violent binary system is up for debate. There's only one known case that I know of where a planet is orbiting a similar system and that is PSR B1620-26. This is a white dwarf, pulsar pair with a planet around it. This would be completely analagous to an X-ray binary system with a black hole if only the pulsar were a bit more massive. So certainly it seems possible for a planet to be around such a system, but probably not very likely.

  Or is it because the light we can see from them is too fast-changing for a wobble or transition to be observable with today telescopes?

This certainly plays a part too. Even if you find a binary system with an SBH that is active and producing X-rays, you're still going to have a hard time teasing out an exoplanet signal, especially if that exoplanet is small (i.e., smaller than Jupiter). The transit method would be difficult due to the natural variability and hence noise in the X-ray emissions and the radial velocity method may not be sensitive enough to small planets. Such a measurement would be really hard.

As it turns out though, astronomers are thinking about these things. There was recently a paper published almost on this very topic. Check out Irma & Stefano (2017) where they discuss trying to find planets around X-ray binaries that may potentially house a SBH.


What defines a stable orbit? While looking for a way to prolong an orbit via moon power I was faced with the parameters of what would be a stable orbit by length of time an object stayed in orbit, What is the definition of a stable orbit? Is there a such an orbit since all orbits eventually degrade?

Can the moon's gravity provide enough force to keep an object in Earth's orbit?

There's no clearly defined definition of what a stable orbit is.  Orbits can last for hundreds or thousands of years, millions, billions or even trillions of years.  There's no neat and tidy definition between stable and unstable.

Hubble for example, orbits the Earth every 97 minutes, and in a few years, primarily slowed by the very slight air resistance, it is expected to fall to the Earth.   Many satellites, because their equipment doesn't last forever, are designed to gradually slow by air resistance and fall to the Earth after a few decades of operation.   Satellites can be positioned where they stay in orbit much longer simply by moving them a bit further out, but there's little benefit to that.   It's advantageous to have satellites fall to the Earth after they've served their usefulness.

Orbits around the Moon are more unstable than orbits around the Earth for 2 reasons.   One, because the Moon is gravitationally lumpy so low orbits around the Moon don't stay in place very long.  See here.

and two, because the Earth is gravitationally dominant, so the Moon doesn't have a very big hill sphere.  Celestial mechanics doesn't like satellites to have satellites.   See here and here.  Our moon could have a satellite, but it likely wouldn't be stable for very long, at least, not long in a celestial time-scale.

Trojan points, L4 and L5 are generally stable in a System where the larger body (Earth) is more than 25 times as massive as the smaller body (moon), but our Moon's orbit is too eccentric and too chaotic for lunar Trojan points to be stable for any period of time.     Earth's L4 and L5 points around the sun are more stable and there are a few small asteroids in Earth's Trojan points.  The Moon's L4 and L5 aren't stable cause the Moon wobbles too much in it's orbit.  Eccentricity and perturbations tend to destabilize L4 and L5 points.   That said, the Moon's L4 and L5 are more stable than L1, L2 and L3.

Mars moon Phobos is in a decaying orbit that may only last 10 million years or so.

Our Moon, however, even though it's slowly moving away from Earth, is sufficiently stable that it will outlast our sun.  It's stable on the scale of many billions of years.

As to your question "prolong an orbit via moon power", The moon tends to destabilize, not stabilize orbits around the Earth.

But while L1 and L2 points aren't gravitationally stable, they are still useful places for satellites because they require less adjustment.   Satalites with orbits that move around lagrange points are often used as useful locations.

Take this bizar orbit around the Moon, where the NASA satellite transfers from L2 to L1 and could, in theory, be transferred back and forth without expending too much energy.  This isn't a "stable" orbit by any means but it's an example of an unstable but useful orbit for the life of a satellite.

Source

The James Webb Telescope, when it's launched will orbit around Earth's L2 Lagrange point, which isn't stable, but it requires relatively small adjustments to stay in that order.  It's also useful because it will be nearly entirely shaded from the sun by the Earth and it works best in cold space, away from sunlight.

Source.

Those are 2 examples of useful but unstable orbits.

Jupiter's orbit around the sun, for example, will probably be stable for trillions of years.   Jupiter's moons around Jupiter?  Hard to say, with 4 Galilean moons, it's difficult to predict their orbits over long periods of time.
.


What is the oldest known prediction of constellation positions? Would someone know of a good reference on the oldest known predictions of constellation positions?

How far ahead could accurate positions be predicted?

For example, could current constellation positions be accurately predicted between A.D. 69 to A.D. 86?

No, there is no possible way that humans from any known civilization in the first century AD would have been able to extrapolate current constellation positions.

Prior to the work of Johannes Kepler and Isaac Newton, humans did not have sufficient knowledge of the basic principles of celestial movement to have any kind of basis for making such calculations.

Second, even if they had the basic principles, they would also have needed much more accurate and detailed information about the makeup of the Milky Way galaxy. The information currently available, that goes into the NASA models, is the product of a great many man-years invested in both observations and analysis of those observations.

Third, even if they had the basic principles and the necessary details with which to create a proper model, they lacked the computing power necessary to actually run the model - computation of that time being limited to pencil and paper (or perhaps an abacus).


Is metallicity low at the central region or nucleus of the Milky Way? Is metallicity low at the central region or nucleus of the Milky Way? Does metallicity decrease or increase as we move from the center out to the edges of our galaxy?

The stars in the Galactic bulge are predominantly metal-rich (by that I mean have a metallicity similar to the Sun or even a little higher).

Even though these stars are predominantly old, the bulge is thought to have formed extremely quickly and the interstellar medium from which the stars were formed would have been enriched with metals very quickly.

Here is a plot from Zoccali et al. (2009) (which I recommend reading). It shows the metallicity distribution of many stars in the Galactic bulge, measured using high-resolution spectroscopy. It shows that the highest metallicity stars are towards the middle (it is very hard to get samples right in the middle because of extinction) and the averge metallicity falls as you mover further from the centre (the samples with more negative Galactic latitude $b$.). The percentages in the plots are the estimated contamination from the disk population in each sample.

The dependence of metallicity on height and radial coordinate in the Galactic disk is still keenly debated. There is general consensus that the metallicity falls with radial distance from the Galactic centre and with height above the Galactic plane. The gradient is of order a few hundredths of a dex per kpc between a few kpc and 10-12 kpc from the Galactic centre. It may flatten beyond this. You could have a look at the discussion in Cheng et al. (2012), but there are many other attempts at parameterising the gradient and it is a technically difficult thing to do.


Sorting out Julian Day, Julian Date, Julian Day number, Julian Day Calendar, and Julian Day Table In this answer I mention day number which is 1 on the first day of each calendar year (January 1) and increments to 365 or 366 on December 31 of that year.

There was an edit proposed, which included a sentence explaining that this could be called the Julian Day. I left a message (elsewhere) to the author of the edit I rejected saying:

  Hi, I rejected your edit on this answer because Julian day is not correct. JD is monotonic and is currently 2,458,643.06

to which the author replied

  @uhoh many thanks for the comment! However, there is such a thing as Julian Day of the year. See this NASA web page: https://landweb.modaps.eosdis.nasa.gov/browse/calendar.html

Looking further I saw several similar pages: 1, 2, 3, 4, 5 Some are called Julian Day Calendar and the rest are called Julian Day Table, but each is a table of what I have defined as day numbers, 1 on January 1, and 365 or 366 on December 31.

Could someone help me sort out all of the possible, acceptable terms for the following numbers corresponding to yesterday which include some form of Julian:

Please be generous; include usages that are less common or have fallen out of use but still have some historical use.

2,458,643 (integer)
2,458,643.06  (float)
160  (day number for June 8 on non-leap years)
June 8, 2019

The Julian Date (JD) is the real number of days since 12:00 UT, January 1, 4713 BCE (proleptic Julian calendar).
The Julian Day Number (JDN) is floor(JD), the nearest integer ≤ JD.
The Modified Julian Date (MJD) is JD - 2400000.5, or the real number of days since 1858-11-17 0:00 UT.
The IAU officially approves either JD or MJD as long it's clear which quantity is given.
A number of 366 or less in a bureaucratic or informatics context is a day number or day of the year.
In his MJD explainer, the late USNO Time Service director Gernot Winkler strongly opposed calling this a "Julian" date:

The MJD (and even more so the JD) has to be well distinguished from this day of the year (DOY).
This is also often but erroneously called Julian Date, when in fact it is a Gregorian Date expressed as number of days in the year.
This is a grossly misleading practice that was introduced by some who were simply ignorant and too careless to learn the proper terminology.
It creates a confusion which should not be taken lightly.
Moreover, a continuation of the use of expressions "Julian" or "J" day in the sense of a Gregorian Date will make matters even worse.
It will inevitably lead to dangerous mistakes, increased confusion, and it will eventually destroy whatever standard practices exist.

Expressing the example date of 2019-06-08 13:26 UT in various forms:

Julian day number: 2458643
Julian date: 2458643.06
Modified Julian date: 58642.56
Day of year: 159
Ordinal date: 2019-159


What is the formulat to compute King tracking rate for a given set of topocentric coordinates? Canbury Tech has a good background on King tracking rate, which approximates the average sidereal motion of a refracted star across the sky. Most amateur telescope mounts have some sort of notion of this "average" rate, which is generally about 15.037 arc seconds per second.

However, King actually produced a formula that would allow the rate to be precisely calculated for a given set of sky coordinates. Modern microprocessor controlled mounts ought to be able to make use of this technique, but many don't. As the author of telescope drive software, I'd like to incorporate this feature into my work, but I don't seem to be able to locate the forumula. BBAstroDesigns has a calculator for this on their web site, but the don't show the formula anywhere.

Can anyone enlighten me as to how to compute this value?

Too long for a comment. Possibly helpful resources:

http://www.bbastrodesigns.com/lib/
http://www.bbastrodesigns.com/lib/coordLib.js (file with actual formulas)
http://canburytech.net/DriftAlign/Equations.html
http://canburytech.net/GoQat/index.html (free software that implements this function)
"A manual of celestial photography: Principles and practice for those interested in photographing the heavens Hardcover – 1931" and it's 1988 reprint (ISBN 0933346468 for the latter, both available on Amazon), the original source.


Why is our solar system "tipped" about 63° with respect to the plane of our galaxy? Our own solar system is "tipped" by about 63 degrees with respect to the plane of the galaxy. Has it ever been researched or is there any scientific theory which could explain the reason why our solar system and solar systems in general are not "in-line" to the galaxy's plane? What exactly determines the orientation of solar systems?

What you could think at first, regarding the orientation of any planetary system, is that it should be roughly in the plane of the galaxy, simply by angular momentum conservation.

But, when you take a look at observations, you see that protoplanetary disks orientation is not what you would expect, with no preferential orientation (protoplanetary disks are embryo of planetary systems, that makes them interesting). In the following figure, the orientation corresponds to the inclination between the line of sight and the rotation axis of the disk.

Why there is this distribution of orientation?

The angular momentum scenario is nice but far to simple: star formation occurs in gas clouds in the interstellar medium, and these clouds are known to be turbulent (Larson, 1981). Turbulence simply disrupts the gas, and is dominent over the global angular momentum of the cloud. Actually, you can even test that with numerical simulations of star formation: put an initial angular momentum consistent with observations, and some turbulence (subsonic or slightly supersonic, also in accordance to observations), and you will get a misalignement of the rotation axis, due to turbulence.


Do narrow lines in the spectra of O- and B- type stars always indicate magnetic fields? I was reading a paper on the differential emission measures of a set of hot O- and B- type stars. As the authors discuss in Section 3 (page 959), two stars, $\tau$ Sco and $\theta^1$ Ori C, have narrow emission lines compared to other stars of the same spectral type. There is a good amount of evidence that both stars are young and have magnetic fields; the stellar winds travel along the field lines and collide at the magnetic equator, producing shocks and, indirectly, X-ray emission. This is in contrast to the mechanism behind shock formation in other O- and B- type stars, where shocks come from the line instability transition.

Now, the plasma at the magnetic equator is roughly stationary because of the collision; this means - if I'm interpreting things correctly - that there is little to no broadening of the lines, and so they are comparatively narrow. Other spectral characteristics and observations support this model.

I'm wondering two things:

Has the presence of such comparatively narrow lines in the spectra of O- and B- type stars been observed in cases where there is no magnetic field present? Very few O-type stars have significant magnetic fields, and theory predicts that none should, so these stars are the few exceptions to the rule.
If so, are there other mechanisms that could be responsible for this in cases without magnetic fields? I'm grasping at straws for ideas, like low collisional broadening.

I talked with my advisor, and we briefly discussed colliding wind shocks (CWS), which form at the interface between the stellar winds in a binary system. It's thought that these shocks contribute only a small amount to the total X-ray production of the system (see Gagné et al. (2011)), though, so any narrow lines produced in that plasma - if any exist - wouldn't have much influence over the measured spectra. So that seems to be ruled out.

Do narrow lines in the spectra of O- and B- type stars always indicate magnetic fields?

The narrow lines don't necessarily require a magnetic field, but any star that large will have one.

I'm wondering two things:

Has the presence of such comparatively narrow lines in the spectra of O- and B- type stars been observed in cases where there is no magnetic field present?

O-type main-sequence stars have high metallicity, they will have a magnetic field. Tau Scorpii and Theta$^1$ Orionis C info at their links. The Phys.Org article: "Strong magnetic fields discovered in majority of stars" says:
"An international group of astronomers led by the University of Sydney has discovered strong magnetic fields are common in stars, not rare as previously thought, which will dramatically impact our understanding of how stars evolve.".
This Wikipedia article: "Solar Dynamo" could use some citations, but here's what it says:
"The solar dynamo is the physical process that generates the Sun's magnetic field. A dynamo, essentially a naturally occurring electric generator in the Sun's interior, produces electric currents and a magnetic field, following the laws of Ampère, Faraday and Ohm, as well as the laws of hydrodynamics, which together form the laws of magnetohydrodynamics. The detailed mechanism of the solar dynamo is not known and is the subject of current research.
Mechanism
A dynamo converts kinetic energy into electric-magnetic energy. An electrically conducting fluid with shear or more complicated motion, such as turbulence, can temporarily amplify a magnetic field through Lenz's law: fluid motion relative to a magnetic field induces electric currents in the fluid that distort the initial field. If the fluid motion is sufficiently complicated, it can sustain its own magnetic field, with advective fluid amplification essentially balancing diffusive or ohmic decay. Such systems are called self-sustaining dynamos. The Sun is a self-sustaining dynamo that converts convective motion and differential rotation within the Sun to electric-magnetic energy.
...".

If so, are there other mechanisms that could be responsible for this in cases without magnetic fields? I'm grasping at straws for ideas, like low collisional broadening.

Our Q&A: "Different width of spectral lines for different groups of stars" has Rob Jefferies answer.
This seems well explained in section 3.4 of "Magnetically Confined Wind Shocks in X-rays - a Review" (22 Sept 2015), by Asif ud-Doula and Yael Nazé, on page 10:
"3.4. Structure of confined winds, as revealed by high-resolution spectra
High-resolution spectra can yield a wealth of information. With current instrumentation, line widths and shifts can be evaluated with precisions down to a few tens km s$^{−1}$ in the most favorable cases (a few hundreds km s$^{−1}$ more typically). Furthermore, the comparison of lines from H-like and He-like ions and of components of $f ir$ triplets of He-like ions constrain the temperature and location of the emitting plasma. However, such measurements are currently only possible for the brightest X-ray sources, so that few magnetic massive stars have been investigated
in this respect (τ Sco - Mewe et al. 2003; Cohen et al. 2003, θ$^1\!$ Ori C - Schulz et al. 2000; Gagné et al. 2005a, b, HD 191612 - Nazé et al. 2007, HD 148937 - Nazé et al. 2008, 2012, β Cep - Favata et al. 2009, IQ Aur - Robrade and Schmitt 2011).
Within noise limitations, the X-ray lines of magnetic massive stars were found to be symmetric, and globally unshifted. This agrees well with MHD models. In the case of θ$^1\!$ Ori C, global fitting however suggests small variations in velocity (Gagné et al., 2005a): from −75 km s$^{−1}$ when the star is seen pole-on to about 100 km s$^{−1}$ when seen edge-on. This change needs to be confirmed as the errors are large but also because one cannot exclude a stochastic variation when only a single observation per phase is available. If further observations provide evidence that velocity varies with phase, then refinement of models will be needed, as no such changes are currently predicted (Gagné et al., 2005a).
Reported widths of X-ray lines largely depend on object and ion considered. The narrowest widths, so far, were found for β Cep, whose lines are dominated by instrumental resolution, yielding only upper limit on intrinsic widths (<600 km s$^{−1}$, Favata et al., 2009). Larger widths, full width at half-maximum (FW HM) ∼ 600 − 800 km s$^{−1}$, were reported for ions with high ionization potential (Mg, Si, S) in τ Sco, θ$^1\!$ Ori C, and HD 148937, three stars with more rapid winds than β Cep. Such widths are much smaller than observed for “normal” O-type stars (FW HM ∼ v∞), where lines arise in embedded wind shocks distributed all over the wind, hence cover a larger velocity range. They indicate formation in slowly moving plasma, in agreement with the confined winds scenario. However, most MHD models predict even narrower lines (Gagné et al., 2005a).
Furthermore, lines from ions with lower ionization potential, notably oxygen, appear broader (FW HM ∼ 1800−2000 km s$^{−1}$ Gagné et al., 2005a; Nazé et al., 2007, 2008). These lines are associated with cooler plasma, which could have a different origin than hotter plasma. For example,
the dominant hot plasma in θ$^1\!$ Ori C is thought to arise in confined winds while the cooler one could arise in embedded wind shocks as in normal O-stars (Gagné et al., 2005a). This dual origin could be supported by the different temperatures derived from the different ions (Schulz et al., 2000). In Of?p stars, however, the spectra are dominated by the cooler component, i.e. confined winds emit soft X-rays in these objects (see above), but it cannot be excluded that current errors, which are large, are somewhat blurring the picture.
In He-like triplets, the forbidden line is suppressed when the density is high or the UV radiation is intense. In the case of massive stars, the latter effect is the most important one and, thanks to dilution with distance, enables us to locate the emitting region. In τ Sco, θ$^1\!$ Ori C, HD 148937, and β Cep, the start of the emitting region is found to be close to the photosphere, at radii r ∼ 1.5 − 3 R$_∗$ for the first three stars and r ∼ 4 − 6 R$_∗$ for the latter case. These values are only slightly lower than the corresponding Alfvén radii of these stars, they thus appear qualitatively compatible with MHD simulations. In IQ Aur, the forbidden line was found to be normal, suggesting a formation radius larger than 7R$_∗$ (Robrade and Schmitt, 2011) - despite a large value, this also appears compatible with the supposed location of confined winds in this star.".
Other reading:
"The surprising magnetic topology of $\tau$ Sco: fossil remnant or dynamo output?" (7 Jun 2006), by JF Donati, ID Howarth, MM Jardine, P Petit, C Catala, JD Landstreet, JC Bouret, E Alecian, JR Barnes, T Forveille, F Paletou, and N Manset
"The magnetic field and confined wind of the O star θ$^{1\!}$ Orionis C" (26 Jan 2006), by G.A. Wade, A.W. Fullerton, J.-F. Donati, J.D. Landstreet, P. Petit, S. Strasser
"X-ray hardness ratios for stars of different spectral types" (2006), by Meurs, E. J. A., Casey, P., & Norci, L.
Wikipedia - Zeeman Effect
"Magnetic Fields of the A-Type Stars" (30 Apr 1958), by H. W. Babcock


Jupiter with a mobile phone and Celestron Astro FI 102mm Maksutov My name is John, I am new here and now that I am retired I decided to get myself my first ever telescope, why I never got one before I have no idea...probably too busy.
So I thought I would try and get some pics last night from a cloudy  Ross on Wye, Herefordshire. England. UK. Using my mobile as mentioned in the manual.
I appreciate they are not like the amazing ones here but it is my first try. I take it the large black disc is probably a reflection of the mirror? If not can you please tell me what it is. Why does this happen?
I would like to get a different eyepiece other than the 10mm and 25mm I would like a zoom can anyone suggest a reasonable one?
Anyway here are my attempts. Thanks

Congratulations on your purchase.  The first pictures dont' show anything much. Just a out-of-focus blur.
The last one shows Jupiter and three of its moons.  I've overlaid the image onto a simulated image from stellarium (at about 10pm BST, the moons move pretty quick so you need an exact time!):

You can clearly see which moon is which, and why Europa is invisible (it's in line with Jupiter)  Look back tonight and the moons will have moved.  Ganymede will have crossed to the other side, Europa will be visible. Take a series of observations and you can rediscover Kepler's third law.
Jupiter itself is overexposed, which is why you can't see any detail. There is quite a trick to getting the exposure right. Astrophotography is a rabbit hole that goes very deep.
You don't want a "zoom". You've got a couple of eyepieces.  You don't see any extra detail with a more powerful eyepiece, you just magnify the blur.
But well-done.  You've imaged a solar system body with a telescope.  Galileo would have been proud.


Can an event horizon form with the observer already inside? I am on a planet orbting a star around the center of the milky way (assume no supermassive blackhole there). By some mechanism, mass starting from the outer part of the galaxy gets pushed inward until the schwarzchild radius is reached for that mass (.25 ly according to Wikipedia). I'm under the impression that an event horizon would form from the perspective of someone looking at our galaxy from the outside. But how would it look like from my perspective at the center of the galaxy? Would I view everything as normal? I know things would start collapsing with so much concentrated mass, but would I know I was inside an event horizon?

The event horizon is the boundary in spacetime between the black hole interior and exterior. If you exist before the black hole forms and you end up at the singularity, then you necessarily passed through it at some point. You can't get inside without crossing the boundary. From the perspective of some foliation of the spacetime, the event horizon encloses no area when it first appears, then it expands outward at the speed of light or greater, eventually passing by your location.
Nothing physically happens at the event horizon; you have to know global properties of the spacetime to work out where it is. There would be no sign that you'd crossed it.
The experience from the inside would be like a big crunch on a smaller scale (so a small crunch, I guess). The singularity is in the future, not the spatial center.


Question about mars polar caps I want to see Mars polar caps by myself. Is it feasible to make a picture of mars like this at home or to achive this very expensive telescope is required?

P.S. Maybe someone can give me a link, where it is shown how polar caps are changing during the seasons? Many pictures or animation.

Regarding the original picture from the question

No, it is not feasible to get a picture like this. According to this webpage the photograph you posted was taken with the Hubble Space Telescope's Wide-Field Planetary Camera 2.

The telescope itself has a 2.4 m (7.9 ft) mirror and the focal length of 57.6 m (189 ft) and is placed outside the Earth's atmosphere. You'd need a large instrument with adaptive optics to get near this quality (assuming you can't afford to place a telescope in orbit).

That said, it is possible for you to capture the polar ice caps on Mars using an amateur telescope. The quality of these will largely depend on the equipment you have and the location (atmospheric turbulence is very unforgiving if you use large magnifications) as well as the imaging technique.

Here's a gallery of planet pictures taken by amateurs. You can find some photos of Mars in which the polar caps are visible.

Regarding the Mars pictures you found on the Sky & Telescope website

To answer your questions about this image, I see no reason to think it wasn't taken with an 8" telescope. However, there must have been more to it than just snapping a picture.

According to the author, the photographs were taken using a Celestron 8" SCT telescope with a Meade 3x Telextender and Expanse 120 Camera.

We're talking about a focal length of 2000 mm with 3x extender, which gives us 6000 mm. Not bad.

The Expanse 120 is a CCD camera. Now, I'm not sure that's the case with these specific pictures but one way to reduce the impact of atmospheric turbulence is to record a video of a planet and stack the individual frames. Many of them will look very bad due to turbulence but occasionally, you'll get a better quality image. Selecting the best frames from a video and combining them into a single image can help you get a lot more detail.

Have a look at this guide explaining how consumer-grade computer webcams can be used. It contains a neat comparison of a single frame and a stacked image.

Please note that I'm merely speculating as to which techiniques could be used to obtain such images. I can see that the author posted their email on the S&T site and that there's a comments section as well. Feel free to ask them yourself.


What's the influence of a tilted orbital plane, when observing an exoplanet transit? I made an illustration to explain what i mean : If we assume we have two similar planet/star systems (similars in size/mass/period..) but tilted differently relative to us.
How can we predict that differences in dimming measurements (time & intensity) are caused by the tilting of orbital plane? (and not by the diameter of the exoplanet)

The depth of the transit tells you the size of the planet (relative to the star) in both cases.

The difference in total eclipse width tells you the impact parameter (how far from the centre of the disk, the transit crosses). This in turn depends on the orbital inclination and the ratio of the stellar radius to the radius of the orbit.

A second constraint on inclination and the ratio of stellar radius to orbital radius comes from the ratio of total transit duration to the time when the planet is fully eclipsing the star. A more inclined transit has a shorter period of total eclipse but longer ingress and egress.

To get a full solution then requires some estimate of the stellar mass, which leads via Kepler's third law to the orbital radius.

There are also some important additional complications due to limb darkening and eccentricity that mean, in practice, this problem is not solved analytically, but by fitting a detailed model to the light curve. For instance, the transit depth becomes smaller for transits that don't cross the centre of the stellar disk, because of limb darkening.

The best picture I can find is copyrighted material, but you can find it here from Carole Haswell's book on Transiting Exoplanets. The inclinations range from 90 degrees to 80 degrees.


Will the SKA's total bandwidth still exceed the Earth's internet's bandwidth? In the early 2010's there were many popular articles that said that the total bandwidth of the the Square Kilometer Array's optical fiber network for raw data from each dish would be larger than the Earth's internet. Here is just one example:

Wikipedia's Square Kilometre Array
:

  It would require very high performance central computing engines and long-haul links with a capacity greater than the global Internet traffic as of 2013.3

As you can see that link is broken.

The SKA project has several Precursors, pathfinders and design studies but the SKA proper has not been built yet, and perhaps is not even finalized.

Still, with available data, by the time it might be built (2025?) is it likely to still exceed the Earth's internet's bandwidth?

Will the SKA's total bandwidth still exceed the Earth's Internet's bandwidth?

No.

First some conversions:

1024 Gigabytes = 1 Terabyte
1024 Terabytes = 1 Petabyte
1024 Petabytes = 1 Exabyte
1024 Exabytes = 1 Zettabyte

From the SKA Telescope website: "The Operational Model" (.PDF), on the webpage: "Info Sheets", this explanation and graphic is offered:

"The data journey

Basic data products will be produced and stored in Cape Town for SKA1-mid and Perth for SKA1-low. From there, they will be delivered to a global alliance of SKA Regional Centres for further processing and archiving and access by the user community.

New data models will need to be developed as current software packages for radio astronomy data reduction don’t have the capabilities to handle the SKA’s large bandwidths and Field of View dataset.".

"Major dates

SKA1 OPERATIONAL CONCEPT DOCUMENT - Operational Principles

late 2027 - Commencement of Key Science Projects (KSPs)

late 2026 - Commencement of PI-led programmes

2024–2027 - Key Science Project (KSP) planning & proposals

2022 - Start of Observatory & Science commissioning

2020 - Start of construction activities

2017–2018 - Prototypes deployed at the telescope sites"

See also: "SKA Community Briefing – 18 January 2017", page 12:

Nothing will flow to /dev/null.
Source: Cisco - Global - IP Traffic: https://www.cisco.com/c/m/en_us/solutions/service-provider/vni-forecast-highlights.html#

Globally, IP traffic will grow 3-fold from 2017 to 2022, a compound annual growth rate of 26%.

Globally, IP traffic will reach 396.0 EB per month by 2022, up from 122.4 EB per month in 2017.

Global IP networks will carry 13.0 EB per day by 2022, up from 4.0 EB per day in 2017.

...
Internet Traffic

Globally, Internet traffic will grow 3.7-fold from 2017 to 2022, a compound annual growth rate of 30%.

Globally, busy hour Internet traffic will grow 4.8-fold from 2017 to 2022, a compound annual growth rate of 37%.

Globally, Internet traffic will reach 350.8 EB per month by 2022, up from 96.0 EB per month in 2017.

Global Internet traffic will be 11.5 EB per day by 2022, up from 3.2 EB per day in 2017.

Global Internet traffic by 2022 will be equivalent to 1 trillion DVDs per year, 88 billion DVDs per month, or 120 million DVDs per hour.

By 2022, the gigabyte equivalent of all movies ever made will cross the Internet every 1 minutes.

Global Internet traffic by 2022 will be equivalent to 193x the volume of the entire Global Internet in 2005.

...

No, it's not even close.


What is the maximum number of planets in the habitable zone? At present, it seems like we are alone in our solar system. How many earth-like planets would fit into the habitable zone of the solar system?

The orbits should be stable for quite a time, i.e. for life to evolve, let's say 4.6 billion years.
The habitable zone may be estimated from 0.725 to 3.0 astronomical units, according to Wiki.

Could one use perturbation theory to answer this?

This question not only makes you think about the temperature, pressure but also the impact of gravity. Not moving away from topic, I must say that the region you are saying about is quite large in the view of metrics but quite short again if we really sense gravity.

The distance of a planet from the sun is maintained by gravity, but we must also not forget that distance between two planets is also controlled by the same. You must appreciate gravity because most of the features on earth are because of the same thing.

Now, if you want to calculate it more precisely you should know each and every postulates, theorems and properties shown by the space in nourishing our Earth for habitability.

Perturbation may give you an idea of adjustments, that how our earth is adjusting day by day for its decreasing distance from sun or increasing size of the sun. Talking according to theory of perturbation there may be more than 2-3 planets in the habitable zone. But still you cannot annoy the word "more" in that.

It shall depend mainly on the following factors : -

Size of Planets
Distance between each of them
Distance from sun accordingly
Their mass (priorly)
Speed of revolution around sun

I know this information is not much for your question but this is enough according to recent knowledge about space. We even don't know the drastic changes happening in the space....the dimensions of space or how vast it is....whether it is measurable or not.....and if yes then are the measurements precise. Each answer tackles you more to have questions....But don't stop, just keep asking!


What is the RGB curve for blackbodies? I created a program to convert the temperature (in Kelvin) of a blackbody to RGB color. However, it is slightly inaccurate, and the deviations increase for values greater than 10000K and less than 1500K. Is there an accurate mathematical function that generates RGB values for given temperature, ranging from ~500K to 20000+ K?

Ok, here's my take on calculating the color of a blackbody, or any spectrum in fact:
Disclaimer: I'm not a color theorist, and there may be more accurate methods. But the result, shown in the bottom, looks about right.

Spectrum
First note that since color is a function of the relative intensity in various wavelength bands, it doesn't matter whether we express our spectrum in flux, flux density, intensity, luminance, or something else that only differs by constant values.
Let's consider a spectrum $S(\lambda)$ as a function of wavelength $\lambda$. If you're interested in a Planck spectrum, first calculate this from the temperature $T$:
$$
S(\lambda) = B(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_\mathrm{B}T} - 1},
$$
where $h$, $c$, and $k_\mathrm{B}$ are Planck's constant, the speed of light, and Boltzmann's constant, respectively.
CIE 1931 color space
The CIE 1931 color space was the first attempt to quantify the relation between the distribution of wavelengths and the color perceived by humans. It makes use of the so-called color matching functions $\bar{x}$, $\bar{y}$, and $\bar{z}$, which are functions of $\lambda$, seen here:

Credit: Wikipedia
You can find tabulated values of these functions on the internet.
The color (which is 3D because humans have three different cones) of any spectrum can be mapped to a 2D position $\{x,y\}$ in the CIE 1931 color space chromaticity diagram seen here:

Credit: Wikipedia
Tristimulus values
The pair $\{x,y\}$ are normalized versions of the tristimulus values $\{X,Y,Z\}$ defined by
$$
X = \int S \, \bar{x} \, d\lambda\\
Y = \int S \, \bar{y} \, d\lambda\\
Z = \int S \, \bar{z} \, d\lambda.
$$
(Actually they're defined using the spectral radiance), but this is equal to the intensity to within a constant factor.)
The lower-case $\{x,y\}$ are then given by:
$$
\begin{array}{rcl}
x & = & \frac{X}{X+Y+Z}\\
y & = & \frac{Y}{X+Y+Z}\\
z & = & \frac{Z}{X+Y+Z} = 1-x-y,
\end{array}
$$
and the last of these relations is the reason only two values are needed.
XYZ to RGB
To go from $\{x,y\}$ or $\{X,Y,Z\}$ to RGB is a linear transformation which can be conducted by matrix multiplication. There are many variations here; my understanding is that is has to do with whatever device you're using to display you colors, but there might be more to it.
The sRGB color space assumes that the $\{X,Y,Z\}$ set has been scaled to $\{0.9505, 1, 1.0890\}$ for the CIE illuminant D65 ("white"); that is, $Y\equiv 1$.
With the above values, this new set can be calculated as
$$
\begin{array}{rcl}
Y & \rightarrow & 1\\
X & \rightarrow & x/y\\
Z & \rightarrow & z/y,
\end{array}
$$
and the RGB values are then, using the conversion matrix from here which cites this,
$$
\left[
\begin{array}{l}
R\\
G\\
B
\end{array}
\right]
=
\left[
\begin{array}{rrr}
 1.656492 & -0.354851 & -0.255038 \\
-0.707196 &  1.655397 &  0.036152 \\
 0.051713 & -0.121364 &  1.011530
\end{array}
\right]
\left[
\begin{array}{l}
X\\
Y\\
Z
\end{array}
\right]
$$
Gamma correction
I'm not 100% sure, but I think this set of RGB values needs to be gamma-corrected, which is a non-linear correction that account for the fact the the device displaying the colors shows different relative luminances than the human eye perceives. At least my result, shown in the bottom, looks about right, whereas it doesn't if I don't perform this correction.
The gamma correction is given by the function
$$
\gamma(u) = \left\{\begin{array}{ll}
12.92 u                 & \mathrm{if\,} u \leq 0.0031308 \\
1.055 u^{1/2.4} - 0.055 & \mathrm{otherwise}
\end{array}\right.
$$
and the perceived RGB values should then be
$$
\begin{array}{rcl}
R & \rightarrow & R\times\gamma(R)\\
G & \rightarrow & G\times\gamma(G)\\
B & \rightarrow & B\times\gamma(B)
\end{array}
$$
These values are mostly in the range $[0,1]$, but if they fall outside, they're usually clipped. If you want your values to be in the range $[0,255]$ you just multiply them by 255.
Result: the color of the Universe through time
I implemented the above algorithm in Python, as a part of a larger code called timeline that calculates various properties of the Universe as a function of time after the Big Bang. You can find it here on GitHub.
With that code I calculated the color of the Universe as a function of time after the Big Bang, and hence as a function of temperature. Until star formation kicks in around 200 million years after the Big Bang, the color is given by the temperature of the cosmic microwave background.
When stars arrive, the average cosmic spectrum becomes more complicated and can't be calculated (yet) from first principles, but must be observed. I haven't found such observations at early times, but Baldry et al (2002) presents the spectrum at various times after an age of 11 billion years, at which time the Universe is a dull "cosmic latte". So in this figure, I grayed out the color between $t=200\,\mathrm{Myr}$ and $t=11\,\mathrm{Gyr}$:

But please note the disclaimer in the top of this answer :)


Is there any man-made satellite orbiting our moon? I am just wondering whether any man-made satellites are currently orbiting our Moon. If there is one, would I be able to see it using my telescope?

The Lunar Reconnaissance Orbiter is still up there. At a compact 2000kg, it's likely too small to see via earth based telescope. Mission Page

I think the Chinese orbiter is back on earth now, but there may be others.


How does the Large Binocular Telescope resolve so well in both orthogonal directions simultaneously? The Forbes article Ask Ethan: How Does Very-Long-Baseline Interferometry Allow Us To Image A Black Hole? includes an example of optical interferometry from the Large Binocular Telescope, shown below.

  The occultation of Jupiter's moon, Io, with its erupting volcanoes Loki and Pele, as occulted by Europa, which is invisible in this infrared image. The Large Binocular Telescope was able to do this owing to the technique of interferometry. Credit: LBTO

related: When will the next series of mutual eclipses of Jupiter's moons begin?

The LBTO consists of only two telescopes 8.4 m (330 inch) wide mirrors, with centres 14.4 m apart, so I would expect to see higher resolution in one direction than the other.

However in this GIF the resolution seems to be the same in both directions. While I see high spatial frequency ringing in the horizontal direction, I don't see any difference in resolution.

Since the event is so fast (probably minutes or seconds) there is no chance that the natural rotation of the baseline during the course of a night could have helped.

So I'd like to ask How does the Large Binocular Telescope resolve so well in both orthogonal directions simultaneously?

Open image in new window for full size:

Source

  The LBTI (green structure in the center of the frame) between the two 8.4m mirrors of the LBT.  High resolution image Credit: LBTO - Enrico Sacchetti

The simple answer is that it isn't resolving in both orthogonal directions equally well. The horizontal dimension is the binocular dimension, and from looking at your animation, it looks to have about ~3 times the resolution. The horizontal banding, I'm pretty sure is not ringing, and is in fact, representative of additional information.

This article does a nice job of explaining what is being seen, and it has a picture, both of a simulated depiction of Io from one of the two telescope mirrors, and the combined image.

The thing you should notice from the above image is that the 8.4 m telescope picture is at the same resolution as the vertical axis of your animation.

What is going on with the interferometer image?

By increasing the separation between telescopes, we increase the angular resolution. Just about everyone with a casual interest in astronomy will have learned that fact in this past week.

But the other thing you do, is introduce interference patterns. The two telescopes essentially act like a double slit experiment.
In the vertical (non binocular) dimension, the diffraction can be treated like a single-slit experiment. The angular intensity formula for a single slit is as follows:
$$ I(\theta) = I_0 sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$
where $b$ is the diameter of each mirror, and $\lambda$ is the wavelength of the light.

The horizontal (binocular) dimension acts as a double slit. The double slit formula is as follows:
$$ I(\theta) = I_0 cos^2(\frac{\pi d sin(\theta)}{\lambda})sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$
where $d$ is the distance between the centres of the two telescopes.

When you combine these two together, using the telescopes dimensions that you quotes (b=8.4, d=14.4), you come up with a pattern remarkably close to what you actually see from the telescope.

On the left, screencapture from above animation, on the right, the predicted double slit intensity.

Fringe removal

It seems that the animation you saw is based off unprocessed images from the LBT. Obviously, they have methods for removing the fringe bands. As to how, I have no idea. I found a paper that discusses interferometry in depth and they mention:

  What we therefore have is a series of fringes, whose amplitude is given by the Fourier transform of the source intensity distribution.
  In practice, steps are usually taken to get rid of the fringes using a
  phase rotation whose rate is known (as both B and s are known). This
  is done in optical interferometers by use of accurate delay lines to
  compensate for the path difference, and in radio interferometers by
  the insertion of electronic delays.

But I have no idea what that means. Perhaps some boffin from Physics.SE would be nice enough to answer.


Why do some artificial satellites decrease and increase in brightness multiple times? I just observed a satellite almost directly above me travel for a few minutes while constantly dimming and then getting brighter. It did this for about 3 minutes until it faded out. I also observed a satellite with similar behavior on July 4th. I was thinking the satellite might be rotating, but I do not know of any satellites that rotate. Then I thought it could be a piece of space junk, but I don't think they get big enough to be visible from the ground. Does anyone know why the brightness of some satellites change while others remain constant? I am in the Northwest of the US. (Northern hemisphere)

I do not know for sure, but you are viewing the satellites by the reflection of sunlight. As the satellite moves across the sky, the Sun-satellite-you angle changes, therefore there is no reason to expect that the brightness should stay the same.

As for why it should fade and brighten again - if you imagine the satellite to consist of a number of surfaces, these surfaces may essentially cast a beam of light onto the Earth, which move across the Earth's surface. The satellite will brighten as one of these beams crosses you. I suppose this could happen several times for the same satellite.

The extreme examples of this are the Iridium satellite that "flare up" to magnitude -8, before fading in a matter of seconds.


How do we differentiate between a star or planet and noise introduced due to equipment? I understand that differentiating between celestial bodies and noise due to equipment is a fundamental problem in astronomy. I also understand that we solve this problem by looking at something called flux density and continuum and line spectra but how do I understand these terms ??
Please consider I'm well versed in university level physics so I understand how spectra will help but don't understand which quantity's flux density do we look for in astronomy. Please help me understand these terms and hence solve the noise vs star/planet problem.

You can differentiate between planet (Did you mean asteroid? Planets are quite bright.) and noise quite easily. Just take three pictures of the sky and they will be probably quite noisy. If some spot changes its position on all three pictures, then this is some near object (astronomically near). You can additionally verify it if its trajectory is line or not. If it is not, then it is probably just noise, but that isn't always the case.
Differentiating between a star and noise is harder. You have to take more pictures and hope that the noise will cancel out.
About flux density: this is the measure for light over some surface and it is proportional to the number of photons that hit the sensor.


Naive star filter visible at night in certain country I have a HYG catalogue and I'd like to filter out stars visible at night and in Europe.

Doesn't have to be exact. I was assuming that it'll be all stars with positive declination, but I wasn't sure about RA values.

Thanks, Michal

In the answer https://physics.stackexchange.com/a/26787 you can find an example how to convert celestial coordinates to horizon coordinates an than check, if the altitude is above the horizon


Does the orbits of stars in a binary system have a common focus, which is the center of mass of the stars? I know that when a planet orbits a star (and assume the star is stationary because its mass is >> the mass of the planet), the orbit is an ellipse, and the star sits at one of the foci of the ellipse. What I am wondering is if instead of a planet surrounding a star but a binary system where the two stars's mass are of the same order of magnitude, are their orbits ellipses and share a common focus at the center of mass?

Intuitively I think it is true because this is a result generalized from the planet surrounding the star case but strictly speaking I have a hard time to mathematically prove it.

Elliptic orbits with the focus at the centre of mass (the barycentre) are the solution to the "two body problem".

The wikipedia page on the central force problem shows how the two body problem can be reduced to solving two central force problems. Therefore the elliptical solution to the central force problem also solves the two body problem, and in moreover Kepler's laws apply to such bodies.


What observational data convinced Stephen Hawking that Cygnus X-1 was indeed a black hole and caused him to break into Kip Thorne's office? It's generally easier to be certain an object is a star than a black hole. HDE 226868 is one of our site moderators and a blue supergiant star. The latter has a 5.6 day companion that supplied a magazine subscription to Kip Thorne:

  According to the updated tenth-anniversary edition of A Brief History of Time, Hawking has conceded the bet due to subsequent observational data in favor of black holes. In his own book, Black Holes and Time Warps, Thorne reports that Hawking conceded the bet by breaking into Thorne's office while he was in Russia, finding the framed bet, and signing it. (While Hawking referred to the bet as taking place in 1975, the written bet itself (in Thorne's handwriting, with his and Hawking's signatures) bears additional witness signatures under a legend stating "Witnessed this tenth day of December 1974". This date was confirmed by Kip Thorne on the January 10, 2018 episode of Nova on PBS.)

Question: What observational data convinced Stephen Hawking that Cygnus X-1 was indeed a black hole and caused him to break into Kip Thorne's office?

From The Unknown Universe: What We Don't Know About Time and Space in Ten Chapters  by Stuart Clark:

  By 1998, Hawking was ready to concede the bet. He thought that although the specific evidence for Cygnus X-1 had not changed, there was now so much evidence for other black holes dotted around the galaxy an the wider Universe that there was no more room for reasonable doubt. Besides, [...], he had bigger fish to fry and a new bet to concentrate on. So Hawking paid for Thorne's prize.

Although Kip Thorne himself apparently still had some doubts until 2011  [see here]:

  "For forty years, Cygnus X-1 has been the iconic example of a black hole. However, despite Hawking's concession, I have never been completely convinced that it really does contain a black hole - until now," said Thorne. "The data and modeling described in these three papers at last provide a completely definitive description of this binary system."

The three paper he is referring to are:

Reid et al., “The Trigonometric Parallax of Cygnus X-1”
Orosz et al., “The Mass of the Black Hole in Cygnus X-1”
Gou et al., “The Extreme Spin of the Black Hole in Cygnus X-1”


What are some good telescopes for a programmer? I’ve never purchased a telescope and I was considering buying one. However, I wasn’t sure what to look for. I love to tinker and so a telescope that I can program with libraries such as SPICE would be great. Any ideas?

One project you could work on is an Arduino-based Go-To telescope system. You could use, in theory, any scope that can accommodate the addition for this purpose. Also, it's my understanding that any Go-To system that can be connected to a computer can receive custom scripts you could write. At least, I know it's true for the NexStar system.

In my opinion, it's best to start with a simple Dobsonian telescope from a reputable dealer, without any programmable components. It will force you to learn the sky well before you start observing. And it will serve as an entry-level introduction to the world of telescopes.


How did astronomers first come to understand that the Sun's core was a gas? I was reading a very old book (around 1850) on astronomy and the author explains Sun spots as a phenomena where you can see the black planetary like hard core of the Sun through two different atmospheres.  It got me interested that somewhere between 1850 and now it became understood that the Sun's core is a gas.  How was this determined conclusively and approximately when in history?

Well simply said: All data fits the premise that it should be a gas.

But what data?

First of all, in the late 1960's the 5-minute solar oscillations on the surface of the sun have beend discovered. Those have been recognized as surface features of internal pressure waves with penetration depth $d$, wavenumber $k$ and their relation roughly $d \sim  \frac{1}{\sqrt{k}}$.
So from the measurment and simple existence of certain wave-ripples on top of the photospheric plasma-ocean we can learn about the interior of the sun. To relate the strength of certain wavemodes to the interior structure of this plasma-ball we use usually a reverse-engineering approach called modelling. We take what we know about gases, energy transport etc. and calculate from this how a solar-oscillations-spectrum on a modelled sun would look like. Then compare, scratch your head, change the modell and repeat until it fits. More mathematical details, if you are interested can be found in the excellent lecture notes on Helioseismology from Christensen-Daalsgard.
So all this gave us a pretty good look into the sun's interior and it was quite clear, that the observed solar parameters like, size, energy output, surface temperature must lead to certain Pressure, Temperature and Density values for the core corresponding to a fusing plasma.
But those theoretical predictions would imply the core-plasma to create neutrinos from nuclear reactions, which famously (wiki: Solar_neutrino_problem) were off by a factor of about 3. With the discovery of neutrino oscillations (wiki: Neutrino_oscillation) this problem was also solved in that the sun didn't produce too few neutrinos, but the produced neutrinos would simply change their types. After some distance from the sun those changes would redistribute the numbers from 3:0:0 for electron-myon-tau neutrinos to 1:1:1.
This problem persisted until approx. the year 2000.


What escape velocity would quark stars have? Quark stars are hypothetical compact stars that are denser than neutron stars and maybe the last stage of upholding matter before stars that collapse into a singularity. Neutron stars have escape velocities up to half the speed of light, and the escape velocity of quark stars must be even closer to c. Could it be that some or all black holes are quark stars whose escape velocity is just above 1 c, rather than singularities?

The escape velocity (in terms of Schwarzschild coordinates) from the surface of a quark star will be given by
$$ v = c\left(\frac{r_s}{r_q}\right)^{1/2}, $$
where $r_s$ is the Schwarzschild radius, $2GM/c^2$, and $r_q$ is the "radius" of the quark star. Strictly speaking, $2\pi r_q$ is the circumference of the quark star.
Since $r_s < r_q$, this does not exceed the speed of light.
However, this applies to objects launched radially outwards. The escape velocity is higher at other angles. Not even light can escape if launched tangentially at $r = 1.5 r_s$ and quark stars could be smaller than this.
If the quark star is spinning (and it would be) the escape velocity becomes even more complex.


Do most astronomers think that Andromeda will collide with Milky Way? Most astronomers say it is likely to happen. Is this true? Are there any scientists that reject this?

Relatively high-resolution measurements were made with the Hubble Space Telescope over a 8 year period. From these measurements, astronomers determined that the collision was a certainty.  It is predicted to occur in about 4 billion years. I suppose there may be some astronomers who disagree with the measurements or calculations but I can find no mention of that. The results were reported in an article in "Nature" (http://www.nature.com/news/andromeda-on-collision-course-with-the-milky-way-1.10765#/b1) and published in "Astrophysical Journal"  Both are well-respected publications. The "Astrophysical Journal" reference is below.

Sangmo Tony Sohn; Jay Anderson; Roeland van der Marel (Jul 1, 2012). "The M31 velocity vector. I. Hubble Space Telescope proper-motion measurements". The Astrophysical Journal. doi:10.1088/0004-637X/753/1/7.