Moments of inertia of planets The three principal moments of inertia of Earth and Moon are reported, for example, in T. J. Ahrens, Global Earth Physics - A handbook of Physical Constants.

I have found reports of the moment of inertia factor $\dfrac{C}{M R^2}$, where $C$ is the polar moment of inertia, $M$ the mass and $R$ the mean radius, for planets of the solar system.

Are the three principal moments of inertia for the planets of the solar system known?

The planets are very, very close to spherical (the Earth has a variaton of about 0.1% between polar and equatorial radii).

The rotational inertial for a solid sphere is

$$I=\frac{2}{5}mr^2$$

If you give us details about why you might be interested in looking at the rotational inertia of planets in greater detail, there might be more motivation for putting effort into finding an answer for you.

Just saying.


Difference in observing hot stars that have been reddened and stars that are cool and red How do astronomers distinguish hot stars, reddened by the interstellar light absorption, from the actual cold
red stars?

Reddening does not change the details of the spectrum of a star. The differences in absorption lines that are present in a hot star vs a cool star are still there after reddening. So if you have a spectrum then distinguishing the two scenarios is easy.

If you don't have a spectrum but have two colours then it is still possible to distinguish the scenarios by tracing the star back along an extinction vector in a colour 1 vs colour 2 diagram until it intercepts an intrinsic locus for (e.g.) main sequence stars. This can be definitive, but can be ambiguous depending on the types of star and particular colours.

If you just have a colour, you can't tell.


How Good Are the Upper Limits on Heavy Elements? There are between 90 and 254 stable nuclei all the way up to element number 82. In discussions and graphs about big bang nucleosynthesis nothing above lithium is even mentioned. It's a pretty safe bet that none of the heavier elements have been observed in reasonably primordial gases, since the implications would be profound enough to require mention in even popular level books about the big bang.

That said, has anyone bothered to perform searches for all of the heavier elements? As in, do we have explicit experimental upper limits on the concentrations of some or all of these heavier elements, or just a more qualitative, "No evidence seen for anything else"?

Good question! Normally, the lack of heavy elements is taken as an indicator that something was made from (close-to-primordial) gas.

So an answer could be to ask what are the lowest abundances ever measured with respect to hydrogen.

I am not completely up to date with the current record holders, but stars with iron abundances that are 5 orders of magnitude less than in the Sun have been found (Norris et al 2013). This corresponds to A(Fe)$\sim 2$ on the usual logarithmic scale where hydrogen has A(H)=12.

The constraints on other iron-peak elements are similar. Alpha elements like O, Mg are usually enhanced in very metal poor stars, so the constraints are an order of magnitude higher.


Why is metallicity important in the death of stars? I always thought that mass was the sole determinant of a star's fate. Then I saw the table here.
So why does metallicity influence a star's ability to become a black hole or neutron star? Does it have as much influence as mass?

I can't give a detailed answer; the details are buried in the depths of numerical stellar evolution models.

The thing that changes most with the metallicity of a newborn star is the radiative opacity of the gas. Higher metallicity leads to more opacity.

This has two immediate effects - it makes energy harder to get out of the stellar interior and makes it more likely that convection will take over.

Convection has the property of mixing up all the material within the convective zone. This can have knock on effects as to how long each nuclear burning phase lasts and how much material is consumed. It also mixes synthesised material from the interior outwards.

A further important effect is that mass loss from massive stars is very extensive and is due to radiatively accelerated winds. For a given luminosity, high metallicity gas is more opaque and easier to accelerate. Hence mass loss is very sensitive to metallicity and determines how massive the star is as it reaches the end of its life. This in turn has a large bearing on what the remnant will be.

There is a further feedback in that the wind metallicity is the metallicity at the surface, but this in turn can be affected by interior mixing that is in turn metallicity-dependent.

If that sounds complicated, that's because it is, and detailed numerical models are required to see how these things play out.


Is there a ninth planet? After Pluto's demotion as a planet, we have currently eight planets in our solar system. But Sun's gravitational pull can be felt well beyond Pluto, so is it possible to have a ninth planet beyond Pluto? I asked this question because we have Kuiper belt beyond Pluto, so there might be some Kuiper belt object which may qualified to be a ninth planet, isn't it possible?

According to Caltech there is possibly a planet 10 times the mass of earth in a 15-20 thousand year orbit at an odd angle. It hasn't been observed yet. There's a search for it ongoing.


Will a new planet form if Jupiter's influence on asteroid belt will diminish in a few billion years? I know that tidal forces are pushing Jupiter farther from the Sun, but I couldn't find exactly the yearly amount. In a few billion years would this effect (and subsequent decrease in gravity pull) allow the formation of a new planet from the asteroid belt or would Jupiter take the asteroids with it farther away?

Sometime after 4.5 billion years, the sun will change into a red giant star and expand to about 1 AU. Some people think it will be beyond Earth's orbit. (In any case, Earth's orbit would have changed and likely to have struck the surface of the expanded sun.) The asteroids would experience a stronger solar wind and their orbits would become more elliptical, so Jupiter's influence would be significant.


Why are radio telescopes shaped so differently than optical telescopes? Why are radio telescopes typically only a dish with a receiver above it, while optical telescopes have a primary, plus secondary and sometimes even a tertiary mirror?

In other words, why do radio telescopes have only one reflector, while optical telescopes have up to three or more?

The same wave phenomena, such as focusing, should apply in both cases. So I don't understand why the geometry would be radically different. You could just replace actual mirrors with a convenient solid, such as plastic, that reflects radio waves just as wells as mirrors reflect optical waves.

They're not different. Same principles do apply. You could have secondary, tertiary, quaternary, and so on, mirrors with instruments at any wavelength, either optical, or radio, or infrared, etc. You could also have instrumentation placed directly in prime focus (so no mirrors other than the primary) with any kind of instrument - radio or infrared or visible or whatever.

See this image of the 5 meter Hale telescope on Mt. Palomar - there is no secondary mirror in this case, the observer is sitting in a little cage at prime focus, using the primary mirror directly:

Of course, for other scenarios, the Hale telescope employs secondary and tertiary mirrors - it depends on the particulars of the telescope, the instrumentation, the experiment or research you're doing, etc.

One reason many of the large optical telescopes very often have at least a secondary mirror is that the architecture preferred in most of these cases is the Ritchey–Chrétien - chosen often for the largest professional telescopes because it eliminates coma, an aberration that is detrimental to astrometry (with coma, images of stars are not round, so it's hard to measure angular distances between them). You could use the primary mirror of such a telescope directly, sure, but being a concave hyperbolic mirror, it has strong aberrations of its own, and so requires the convex hyperbolic secondary (often a strong hyperbola, with a large eccentricity) to correct the aberrations.

The Hale telescope pictured above has a parabolic primary, so using it directly is not a problem.

Again, all of the above are not strict rules, just statistical observations.

Some radio telescopes have instrumentation at prime focus simply because it's convenient for that particular case. Other radio telescopes have secondary mirrors. Again, it all depends on what you try to achieve. E.g., the Arecibo radio telescope could be used either in prime focus, or with a secondary mirror in a Gregorian configuration - here's the image with the prime focus instrumentation and the Gregorian mirror to the left:

In the case of the Arecibo scope, the N-ary mirrors are sometimes used to correct the aberrationa of the spherical primary reflector, but that's not the only reason why they're used.

Here's a discussion comparing various architectures (classic Cassegrain versus Ritchey-Chrétien versus anastigmatic aplanat) for a large radio telescope, highlighting various design, performance and operational issues for each. TLDR: classic Cassegrain is traditional for radio telescopes, but the R-C architecture performs better and is not significantly more difficult to build; OTOH, with R-C you must always use the secondary.


In the closest stars, can you see the stellar corona? Stars appear larger than they are, could that be the stellar corona here around Sirius, making it appear n-times larger than the stellar body? NASA recently mapped the solar corona to extending 12 solar radii above the surface 1, the corona as a whole occupying 13x the suns diameter if viewed from a distance.

2

Refences:

1) NASA's STEREO Maps Much Larger Solar Atmosphere Than Previously Observed - nasa.gov

Stars other than the Sun do not have photospheres or coronae that can be directly imaged (the exception would be the red giant Betelgeuse that can just be resolved from the HST). They "appear larger than they are" as you put it because of the limitations of physical optics or turbulence in the atmosphere. There is a minimum-sized "point spread function" image that will be convolved with the intrinsic image of a star, whatever its size. Thus even a point-like source will appear to have a finite size when viewed with a telescope with a finite aperture. Images of very bright objects may also be affected by "saturation" of the detector - the counts in saturated pixels can "bleed" into surrounding pixels. All stars (bar Betelgeuse) are too small and far away to be anything but intrinsically "point-like", even when observed with space telescopes.

The image of Sirius in your question is the saturated image of a very bright, point-like object blurred by the imperfect optics (and perhaps a turbulent atmosphere) on a telescope of finite aperture. Any "corona" you see on this picture is the extended wings of the telescope point spread function. The size of Sirius cannot be directly inferred from this picture. (It is also the case that Sirius, an A-type star, does not have a corona like the Sun's, because it lacks the magnetic dynamo and strong magnetic fields that are responsible for heating the solar corona, and it lacks the very strong winds that shock heat coronae in much hotter stars).

That said, there are clever, interferometric techniques to synthesise a large telescope using extended baselines between multiple telescopes. These are capable of directly measuring the angular diameters of stars.  The assumption is that what is being measured is the angular diameter of the photosphere, since this is what dominates the light at optical and near-infrared wavelengths from most stars. In particular, the corona - a regions of tenuous, optically thin, ionised gas - emits hardly any visible/infrared light (it is of order a billion times less bright than the solar disk and due mainly to Thomson scattering of photospheric light from free electrons and dust scattering) and can safely be neglected. That is why the optical/infrared corona of the Sun can only be seen during total solar eclipses on Earth or from spacecraft with occulting disks that can totally block out the dominant light from the photosphere.

Distant stars are too far away for this occultation technique to work. The presence of any coronae around them can only be inferred from the X-ray emission due to the hot coronal gas. In some cases it has been indirectly argued that the coronal gas can extend several radii or more above the stellar surfaces (e.g. by analysing the fall and rise times of stellar X-ray flares or from the lack of X-ray eclipses in some eclipsing binary systems).


Cosmic events as standard candles Type 1a supernovae are used as standard candles. Are there other high energy cosmic events which can be used as standard candles. For example, collapse of neutron star into a black hole is also similar to type 1a.

In a galaxy, there are far more white dwarfs(candidates of type 1a) compared to neutron stars(I read somewhere that our galaxy contains around 200 million of them). Also, the energy released in neutron star collapse is not in visible spectrum(I don't think so that is the criteria for "standard candles"). Are these the reasons that neutron star collapse is not qualified as another type of standard candle?

Intro for the uninformed: A standard candle is an important concept in astronomy, helping to map out distances in the Universe. Since the observed flux $F$ of a light source decreases with distance $r$ by a known factor ($r^2$), if we know its intrinsic luminosity L, we can calculate the distance. For large distances, where bright sources are needed, we usually use supernovae (SNe). But the luminosity of a SN depends on the mass of its progenitor star which is not known in general. However, for a specific type of SNe — "type Ia" — it is known: This type are SNe that explode when a white dwarf accreting mass from a companion star exceed the mass threshold for explosion of $1.4\,M_\odot$.

In addition to the gravitational waves discussed by Rob Jeffries, I can mention the following candidates for standard candles:
Type II supernovae
Type Ia SNe are so similar in luminosity because they all have (almost) the same mass when they go off. But there is also evidence that type II SNe can act as standard candles. As is also to some extent the case with Ia, their lightcurves (how the luminosity changes with time) are not identical, but can be standardized using the so-called Philips relation (see e.g. Kasen & Woosley 2009).
GRB supernovae
Gamma-ray bursts as a whole are too diverse to be used as standard candles, but when an associated SN is detected, it becomes a standard(izable) candle (Li & Hjorth 2014).
Quasars
My favorite candidate are quasars. This technique doesn't rely on the Philips relation. Quasars are caused by gas accreting onto a supermassive black hole in the center of galaxies, resulting in an "active galactic nuclus" (AGN) with extreme energy outputs (easily over $10^{12}\,L_\odot$, and even up to $10^{14}$–$10^{15}\,L_\odot$; Ibata et al. 1999). It turns out that there is a correlation between the absolute luminosity of the AGN and the size of its broad-line region (BLR), i.e. the region around the quasar where fast-moving gas clouds absorb the continuum$^1$ of the quasar and emit lines$^2$, e.g. H$\alpha$ (Watson et al. 2011). The reason is that the size of the BLR is determined by the depth that the ionizing radiation from the quasar can penetrate into the BLR, which is proportional to the square root of the luminosity. The figure below (from Watson et al. 2011) shows the relation between distances ($D_L$) determined by this technique and distances obtained from type Ia SNe.

Advantages over supernovae
A huge advantage of quasars over SNe is that they don't disappear after a few weeks, meaning that if e.g. we want to refine some measurement, we can go back and observe it again at any time. Another advantage is that quasars, being so luminonous, can be detected out to much larger distances (roughly to $z\simeq4$) than SNe (which are only observed ut to $z\simeq2$).
Reverberation mapping
Since quasars are so far away, the BLR, being less than a parsec in size, cannot be resolved. But luckily, a technique call "reverberation mapping" allows us to determine the size:
The spectrum of the quasar consists of a continuum with spectral lines. Quasars vary in luminosity on rather short timescales. If we measure a quasar's luminosity regularly over some period of time, we get a so-called "lightcurve". But since the lines are created at a distance from the source, a given "bump" in the lightcurve (i.e. a temporary increase in luminosity) does't show up in the continuum and the lines at the same time. Instead, there is a delay, corresponding to the extra distance that the light had to travel from the quasar to the cloud reflecting$^3$ it.
In the figure below, blue shows the continuum, while red shows the lines. The lines are seen to lag behind the continuum, since they first had to travel from the quasar (black) to the clouds (magenta).

Since the clouds lie at a range of distances, they exhibit different time lags, effectively broadening the line:

So, in the figure above, showing observed flux as a function of time, the light in the line increased in luminosity roughly 1.5 days later than the light in the continuum. This means that the BLR is roughly 1.5 lightdays (or ~250 AU) in radius.

$^1$I.e. the continuous and relatively featureless spectrum of many different atomic transitions and physical processes.
$^2$I.e. features in the spectrum resulting from strong atomic transitions. For instance, an eletron falling from the second to the first excited states of a hydrogen atom emits a photon with the wavelength 6563 Å, called "H$\alpha$".
$^3$The light isn't really "reflected". Instead, it is the high-energy photons (UV and X-rays) of the continuum that ionize the atoms in the clouds. When the ions recombine, they emit spectral lines.


Could the 9th planet be virtual? Astronomers recently claim that there is evidence for a 9th planet. As far as I understood it, this is mainly based on coinciding orbital parameters of a few Kuiper belt objects.

Could the 9th planet be a virtual planet? I mean this in the sense that we see the gravitational influence of a multi-body system. Could the 9th planet simply be the barycenter of the Kuiper belt and its apparent orbit some form of precession?

Apart from the question, if this was a possible or likely scenario, how could we distinguish a "real" 9th planet from a virtual one, except by the obvious answer if we can observe it directly?

Forgive me, if this is a stupid or trivial question or if this was answered before. I am not a professional astronomer, yet I have a physics/engineering background. I did a quick internet search to answer this question but I simply may have used the wrong terms or sources.

The ninth planet can absolutely be a "virtual", in the way you describe it, meaning observed data indicating the gravitational influence by an object is not actually caused by such an object.

A simple visualisation of this is the case of a two-body system, where we observe two objects orbiting a common barycentre. From the observed data, one can get the impression that there is an object in the barycentre, pulling both of your observed objects against it.1

However, a barycentric way to explain the observations gets less likely as you add more bodies. Be aware that the current indications of "planet 9" is based on a mere correlation of some not-so-accurate data, from a small sample size, heavily biased by observation constraints.

  how could we distinguish a "real" 9th planet from a virtual one,
  except by the obvious answer if we can observe it directly?

That is only by its gravitational influence alone.

As a rule of thumb, incomplete and inaccurate data always imply the risk of causing virtual objects.

It should also be noted too that observations in the Kuiper belt generally can not yield how a orbit changes, due to the extremely long revolution times. observing "Interactions" are actually mostly about tracing an orbit back from its current state vectors, to find probable interaction events in the past. That is literary to ask for data artefacts and virtual objects.

1Before anyone objects about how easy it is to falsify that: Yes it is easy to determine that there can not actually be an object there, based purely on observing the two bodies. But, that requires 1. accurate data about the motion of the two objects, and 2. their mass. Our current data about the Kuiper belt fails to give good enough data on both points.


Are there auroras on Mercury? Auroras on Earth are caused by interactions between the solar wind and the magnetosphere. I've also seen a few pictures of auroras on Jupiter and Saturn.
Mercury has a global magnetic field. Does it also have auroras? If so, what do they look like?

A magnetic field is only one of the requirements for an aurora (and given the signs of emissions from excited oxygen on Venus after solar flares, it may not actually be a requirement). There needs to be an atmosphere, too. In Mercury's case, there isn't much in the way of an atmosphere. The magnetic field still channels solar wind to the polar regions, where it causes sputtering of sodium. Given this is the result of the same magnetic channeling that produces the aurorae on Earth, you could think of this as Mercury's equivalent of aurorae. The sodium forms a faint glowing tail as it escapes the planet.


How do scientists determine if a star's color is due to Doppler shift vs the star's composition vs temperature? When looking at visible light coming from, say a star, scientists can determine by the shifting of the light towards the red or the blue end of the spectrum to determine if it is moving towards or away from earth. But they can also look at the light as it falls on the visible spectrum to determine the composition of said star. How do they determine if the light landing on the red end of the spectrum, for example, means it's moving away vs the composition of the star? How do they tell the difference?

In addition, how is all this reconciled with the fact that a star's color is also based on it's temperature?

Note: Question was expanded to include temperature after comments and updated answer subsequent to my original question asking only about motion vs composition.

The colour of a star is due mostly to its temperature. Except at extreme velocities, the red-shift won't affect the light enough to change the visible colour. To detect a red or blue shift you need some marker in the spectrum that can be measured accurately.

The spectrum of light from a star will contain dark lines, called Fraunhofer lines, which are due to the absorption of light at particular wavelengths by gases in the star's atmosphere. The wavelengths at which these lines appear are known very finely, for example sodium causes a line at 589.592nm . If the star is moving, the light, along with the Fraunhofer lines is red- or blueshifted.

The presence of these line shows the composition of the star's atmosphere.
Measurement of the position of the lines allows for accurate measurement of the Doppler shift of a star.

A cool, red star which is stationary will have the Fraunhofer lines in the same position as a hot stationary blue star. The motion of the star is measured by the position of the Fraunhofer lines, not the colour of the star.


(How) Can one determine if an exoplanet is synchronously tidally locked or not? Mercury is asynchronously tidally locked so that all of its surface regularly sees the Sun. But an exoplanet which is synchronously locked to its star, could we tell that it is? It would have a hot side and a cold side. This should affect transit and occultation on some wavelengths, but does the symmetry of the movement degenerate any possibility to disentangle tidal locking from that?

Mercury seems to cool down during night time, and gas giants maybe distribute their temperature atmospherically. So their tidal locking might be hard to discern. But there are terrestrial exoplanets with much shorter orbital period which should have a cool side and a hot side permanently. That's my speculation. My basic curiosity is:

Is Mercury a freak or a standard in terms of tidally locked planets being asynchronously so? And by observational support, not just models.

No. Not with the limitations of our current technology. 99% of anything you've heard describing the state, makeup, nature of or appearance, of any exoplanet, is wishful thinking and conjecture. Our current technology is still rather limited. Today, we are able to come up with reasonable guesstimates of the sizes of exoplanets and their distances from their host stars, all with large degrees of uncertainty. That's it - size and distance - nothing else - at least for today (2017).

When we look at a star and see that its light has regular periodic dips, we measure the length of, the strength of, and the period of those dips (that is all we can see). Then we do a lot of math and guess at the size (mass) of the exoplanet and its distance from its host star. Even the size (both mass and diameter) of these planet-hosting stars are mere guesses. Only 16 of all the visible stars have ever been directly imaged (resolved to a disc).

However, people do love to dream. Some scientists LOVE to embellish. Some scientists like to walk around like peacocks with their plumage spread in display, hoping folks won't see their behind behind their tail feathers.
(I got to tell you - I really enjoyed thinking up that last sentence.)


Why does shell fusion produce more energy than core fusion? Stars go from the main sequence phase to the red giant branch due to the depletion of hydrogen in the core. As a result, the star contracts and shell hydrogen fusion begins, which apparently produces much more energy. Why does this produce more energy than core hydrogen fusion? The Wikipedia article on red giants says that this is because of higher temperatures, but why does shell hydrogen fusion occur at higher temperatures than core hydrogen fusion?

After the helium flash in low-medium mass stars, does shell hydrogen fusion continue to occur? If so, then why does there a drop in luminosity from the helium flash to the horizontal branch? And if not, then why not?

Similarly to the first question, in the AGB phase, why does shell helium fusion produce so much more energy than core fusion?

Why does the early AGB phase derive most of its energy from shell helium fusion and not shell hydrogen fusion?

Ultimately this is more of an overly long comment, as I think a more satisfying and complete answer would properly explain things in a more concrete fashion—more of a "it has to do this because..." answer than my "it can do this because..." one.

The short of the answer to the first question is that helium fusion needs ~25 times the temperature that hydrogen fusion does.  The proton-proton chain initates around $4\times 10^6$ Kelvin, whereas helium fusion doesn't begin until around $10^8$ Kelvin.  So when the main stage ends and the helium core contracts and the temperature rises, the "edge" of the core can have temperatures well in excess of the minimum hydrogen fusion temperature, and so a shell around it can have temperatures well beyond it.  Fusion rates are (approximately) polynomial in temperature, with the degree depending on the reaction in question, so small increases in temperature can produce substantially more fusion.  The gravitational force is strong enough to overwhelm the pressure from the induced fusion, and so will contract the surrounding shell to temperatures in excess of the minimum needed.  This is basically what happens in the cores of main sequence massive stars (relative to, say, the Sun).  Actually, our own Sun's energy is mostly from the proton-proton chain and has a core temperature of around $1.57\times 10^7$ Kelvin, nearly four times the minimum necessary.  And still, the core needs to be nearly 10 times hotter than that to initiate Helium fusion.

For the second question, the short of the answer is that the core has undergone thermal expansion after the Helium flash, and so occupies the temperature ranges most conducive to a strong hydrogen fusion rate.  The material outside the core is now at lower temperatures and pressure, so the fusion rate is reduced substantially.  So the energy output comes principally from the core, and the helium fusion at near minimum temperatures releases less energy than the hydrogen shell at (well) beyond minimum temperatures did.  Thus the star overall produces less energy and contracts.

The remaining questions are explained in similar fashion: one has to pay attention to the sensitivity of reaction rates to temperature, and what the temperatures in those shells actually are.  The sensitivities are different for each reaction chain, and the temperatures can go well beyond the minimum necessary.


How come moon's difference in thickness between sides does not create an unbalance on its mass? I've just found out that moon's dark side is 30 miles thicker then the closer side to the Earth. So I'm wondering about moon's mass. How come it's still balanced? Supposing the far side is indeed heavier, considering More Mass More Gravity, wouldn't it have to be closer to Earth?

Sorry if I'm making stupid assumptions, I'm just a curious fela.

As is the case on Earth, the Moon's crust is less dense than the material that comprises the Moon's mantle or core. On the Earth, this means that areas with thick crust rise above sea level. (These are our continents.) On the Moon, this means that the Moon's center of figure (it's center based on surface topography) is offset by a couple of kilometers from Moon's center of mass, with the center of mass closer to the Earth than the center of figure.

This center of mass to center of figure offset does not mean much with regard to the Moon's stability. What matters much more is the distribution of mass top to bottom vs side to side vs front to back. To be stable in a tidally locked orientation, a moon's principal axes deduced from the moment of inertia tensor should be very close to the set of axes as seen from the planet.

Moreover, a very specific ordering of the moments about the principal axes as deduced from the inertia tensor needs to exist. For the moon to be in a stable tidally locked configuration, the axis of rotation needs to be more or less orthogonal to the moon's orbital plane, the moment of inertia about the rotation axis needs to be the largest, and the moment of inertia about the line from the center of the planet to the center of the moon needs to be the smallest, leaving the moment of inertia about the side-to-side as the intermediate axis.

This most definitely is the case with the Earth's Moon. The angular difference between the Moon's geometric axes as seen from the Earth and the Moon's principal axes deduced from its inertia tensor (in technical terms, the angular difference between the Moon-Mean Earth frame and the Moon's principal axis frame) is a few hundredths of a degree.

That the Moon's far side has a much thicker crust than does the near side is a third order effect. The Moon would be nearly as stable as it is now if it was the Moon's far side with its much thicker crust that faced the Earth rather than facing empty space.


Does Gold really come from Supernovas? Do we really know that the origin of gold was in a supernova explosion?

I am talking about the gold we find here on Earth.

So then how are we sure that Gold actually came from these explosions.  Did they go back some million years ago and figure out that Gold was made in supernova.

We don't need to see things to understand them.

It is now generally understood that gold is formed in neutron star collisions.  The evidence for this partly theoretical: Models of neutron star collisions suggest they would produce large amounts of heavy elements, and regular supernovae would not produce gold in the right amounts to explain the concentrations found in the solar system.

Secondly, we have observational evidence from the collision of two neutron stars, detected by gravitational waves, which allowed for scientists to analyse the light coming from this collision in detail. We can't see gold, but we can see radiation that indicates the presence of radioactive isotopes that will decay to form gold.

This gold would float in the interstellar space for a while until it forms part of the dust around a newly forming star. That dust gathers together to form planets. Most of the actual gold we find came to the Earth carried in by meteorites, as the original gold will have fallen to the Earth's core when the Earth was molten.

So we have a good model for explaining how gold can be formed in neutron star collisions and evidence for this actually happening. That is enough for this to be the generally accepted theory.


Error when calculating Alt/Az from Ra/Dec I wrote a program based on this tutorial: http://www.stjarnhimlen.se/comp/ppcomp.html to calculate altitude and azimuth of a celestial object. When I compare the calculated RA/Dec values to the ones from Stellarium, they are very accurate. But when I compare the Alt/Az values to the ones from Stellarium, there is an error of about 2 degrees! I use the following method to calculate Alt, Az:

GMST0 = Ls + 180_degrees # Ls = Sun's longitude
GMST = GMST0 + UT
LST  = GMST + local_longitude
HA = LST - RA

x = cos(HA) * cos(Decl)
y = sin(HA) * cos(Decl)
z = sin(Decl)

xhor = x * sin(lat) - z * cos(lat)
yhor = y
zhor = x * cos(lat) + z * sin(lat)

az  = atan2( yhor, xhor ) + 180_degrees
alt = asin( zhor ) = atan2( zhor, sqrt(xhor*xhor+yhor*yhor) )

Some specific example:

Test date 15.09.2018, time 15:00 UT
Planet: Mercury

Coordinates: +47.55777777° +8.89888888

What stellarium says:

RA = 11h 18m 13.26s

Dec = +6°25'08.5"

Az = +250°21'13.2"

Alt = +25°25'00.1"

What my program says:

RA = 11h 18m 14s

Dec = 6° 25' 6.59"

Az = +248° 49' 6.9"

Alt = +26° 33' 16.43"

Thanks in advance!

The algorithm on that page is very simplistic and is probably the source of a good portion of that error.  It's best to test your code using a very accurate ephemeris, even if your final goal is to have a much more simple ephemeris implementation.  DE405, or VSOP87 would be good alternatives.

A small error in GMST can have a big effect on Alt/Az computations, I would expect that the simplicity of the algorithm on the page produces a good sized error in GMST.

Another common source of error is differences in time scales.  You need to make sure you propperly account for things like leap seconds, which that page specifically says it ignores.  You need to convert UTC to Terrestrial Time by subtracting 37.0 + 32.184 (where 37 is the current number of leap seconds).

Also make sure you're not comparing different coordinate systems.  Like J2000 vs "Of Date", or coordinates adjusted for atmospheric refraction.


Are we really star-stuff from the interior of collapsing stars? Carl Sagan has said several times that we are "star-stuff".

One instance can be found in Good Reads' Carl Sagan > Quotes > Quotable Quote:

  The nitrogen in our DNA, the calcium in our teeth, the iron in our blood, the carbon in our apple pies were made in the interiors of collapsing stars. We are made of starstuff.

Question: Was most of my nitrogen really made in the interior of a star during its collapse? Was my calcium and iron made there as well, and not (for example) in an expanding shell after a supernova?

The straightforward answer is, "Yes, we are made of star stuff."

Some of it will be from the interior of collapsing stars, some will be from supernovas, some from normal everyday fusion, and some from other processes.

The answers from @HDE226868 and @RobJeffries on this question on where heavier elements come from gives good background, including this nugget:

  The split between r-process and s-process production of heavier than iron (peak) elements is about 50:50. ie They weren't mainly made in supernovae, which is a frequent, incorrect claim.

but of most importance is Rob's final point:

  The relative contributions of various sites to the r-process remains an unsettled matter. You could also read my answers on this topic in Physics Stack Exchange.

On following Rob's links I think this provides you with an excellent overall answer (and relative percentages)

  A more up-to-date visualisation of what goes on (produced by Jennifer Johnson) and which attempts to identify the sites (as a percentage) for each chemical element is shown below. It should be stressed that the details are still subject to a lot of model-dependent uncertainty.

Looking at C and N - the majority seems to be from dying low mass stars, and Ca and Fe are from exploding stars, which indicates that Carl is not far off the mark.


If "9 th planet" is ejected by Jupiter or Saturn, why does it have a very far perihelion? If the "9 th planet" is ejected by Jupiter or Saturn, I think the perihelion should be near Jupiter and Saturn, but now the predicted orbit is very far away from Netpune, why would that happen?

This is certainly a problem for any hypothesis that form planet 9 by ejecting it from the inner solar system. The issue (for those wondering) is that if you simply scatter an object out of the inner solar system, but it is retained in a bound elliptical orbit, then it should come back! The proposed planet 9 needs to have a perhelion distance beyond 200 au and at least a moderate eccentricity.

The solution might be found in a paper by Bromley & Kenyon (2016), who address this very point. In their model they scatter a "super-Earth" from 5-15 au in the inner solar system, at an early stage of the solar system formation when the Sun is still surrounded by a gaseous disc. They run through a suite of simulation with different mass planets, different gas surface densities and different rates of evolution and dissipation of the gas disc.

The outcome is that there are sets of parameters that lead to eccentric orbits with large perihelion distances that are caused by damping by dynamical friction from the gas disc. What is required are long-lived, low-density gas discs, or more short-lived, massive gas discs that clear from the inside-out.

The same paper also summarises a number of other possibilities that could explain the large perihelion distance and eccentricity of putative planet 9. These are: (i) in situ formation from a ring of solids - though the orbit would tend to be more circular; (ii) the nearby passage of another star could perturb the orbit, provide a boost to the perihelion distance and increase the semi-major axis, perhaps not by enough, but this mechanism could be more effective if the planet was in an eccentric orbit to begin with; (iii) tides, from the Galaxy, or more likely, from the birth cluster of the Sun could have influenced the orbit of a scattered planet and produced its high eccentricity and perihelion distance.

The gist of the discussion is that whilst the authors think these other things are possible, they favour their own gas-drag model (naturally!).


Could a pair of binary black holes form within a star? After the discovery of gravitational waves by LIGO last week the team behind the Fermi gamma-ray telescope released a paper that showed a soft gamma ray burst was detected only 0.4s after the gravitational wave was detected at LIGO, and it was also located in the right area of the sky. This could be evidence of a gravitational wave source with a detected EM counterpart.

Ultimately though the Fermi group resist claiming a full discovery as no electromagnetic signature is expected form a binary black hole merger, but whats interesting is that I saw another paper today that claimed a GRB could be caused by a binary black hole merger inside a massive rapidly rotating star.

Is such a system realistic or even possible?

I doubt two black holes could form within a single star, but they might form from a contact binary. That's a pair of stars so close that they touch and share a single atmosphere. For the stars that made the black holes for the LIGO gravitational wave discovery, just consider two very massive stars that are touching or nearly touching. They would share a single gaseous envelope for an atmosphere. Each one separately formed a black hole when it died, and then their two black holes merged.


Several stars in line I can see several stars in the sky they can make a line come from the West south and going to East north direction at the time between 6.50 to 7.00 pm Indian time on date 13.1.20. What might these three stars be?

It might be orion as in this season it is pretty clear in India. Orion have three stars in a somewhat straight line.

From in-the-sky.org's planetarium at 19:00 Indian time near the center of the subcontinent, Orion's Belt is high in the sky and extends from the East-Southeast vertically towards the zenith.


Occultation of planets by other planets Two Keplerian closed orbits always intersect:

Source: Astronomy SE answer by Flater
As seen from Earth, Venus appears larger than Mercury; this makes sense because Venus is both larger and closer than Mercury (as measured by minimal distance; by average distance all planets are equally distant), but can also be readily observed by comparing solar transits for either planet (or wait for 26 July 69,163, when both transit simultaneously).
Unless prevented by orbital resonance, that means that sometimes, Mercury should be eclipsed by Venus (occultation) when, seen from the Earth, Mercury is exactly behind Venus; and Mars should pass in front of the outer planets occasionally, but I'm not sure which body appears larger from Earth (Mars or Jupiter).
How often do occultations between planets occur?

What you describe is a mutual planetary occultation (or transit), and yes, it has happened—in 1737—and will happen again—in 2133 (both dates are for Venus/Mercury events; there are other possible instances of planets passing in front of each other; for example, in 2065, Venus will pass in front of Jupiter). I would suggest giving a look at https://en.wikipedia.org/wiki/Planetary_transits_and_occultations
What is interesting with Mercury and Venus is that, considering the respective positions of these planets as well as the Earth, it is technically possible for Mercury to pass in front of Venus! The last time it happened was in 796, and it won’t happen again at least until the year 2500. See More Mathematical Astronomy Morsels by Jean Meeus or the link above for more information as well as for the frequency at which mutual planetary transits occur.
A “near-miss“ will happen on 27 April 2022 at 19:11 UT, when Neptune will be very close to Venus—about 25″ depending on where you will be on the Earth at that moment…


Looking for an equation that describes how star (x,y,z) positions are affected by expanding universe I understand that coordinates of stars are in RA,DEC. I am looking for a mathematical expression that describes how the positions of stars change due to the expanding universe.

The position of stars change very slowly for a couple of reasons, but not due to an expanding universe.

Galaxies distant from our own, are all moving away from our galaxy at a rate proportional to their distance. The equation is simple:
$$\mathrm{speed}=H_0 \times \mathrm{distance}$$

and $H_0$ is a constant of proportionality, with a value of about 70 km/s per kiloparsec. Since the galaxies are moving away, they don't change their position in the sky at all.

(The nearest galaxies are exceptions to this rule, they may be moving towards us)

Stars in our galaxy are also moving. They have their own proper motion, but the motions of the stars relative to our own are essentially random. You can look up the proper motion of stars on the SIMBAD database.

Stars also appear to wobble due to the annual rotation of the Earth about the Sun and our consequentially changing perspective. The amount of wobble is in inverse proportion to their distance (the nearest stars appear to wobble most)


Corresponding BV Value to a Hertzsprung–Russell I have an example BV value from an astronomical catalog of 1.04 (33 Psc, HD Catalog 28).

Given only the number 1.04, can that be charted on the HR, or do I need the individual components of the value and/or temperature and luminosity?

Edit:

I should explain why I'm asking. I've written an application that displays data from one of several catalogs, and they only provide the BV value. I want to add a component to the application that would show an HR diagram with that star superimposed on it. The example of 33 Psc was semi-random, as it's the third entry in one of the catalogs (Yale Bright Star Catalog).

The HR diagram is a plot of luminosity (or a proxy) versus temperature (or a proxy).

$B-V$ is a proxy related to temperature, therefore to plot your point on a 2-d plot you obviously need the other axis information - the luminosity, or its proxy the absolute magnitude. This in turn needs an apparent magnitude and a distance to the star.

It is also possible that the star has suffered some extinction and reddening by the interstellar medium, which would need to be estimated and corrected for before comparing with other stars or stellar models.

Note however that 33 Psc is a binary star, so it isn't clear what you learn by plotting its combined colour and absolute magnitude on the HR diagram.

Edit: I note that the catalogue you show in your edited question doesn't just give $B-V$, it gives the visual (i.e. $V$) magnitude. This is one component of what you need for the y-axis. The other is a distance, so that you can convert your apparent $V$ magnitude into an absolute magnitude.


What does it mean to say 'space exists?' From another question I posted, I learned that not only was matter created from the Big Bang, but more importantly, space itself.  What does it mean for space to be created?  Does it exist in the same way substances exist?  In other words, what do we mean by 'space exists?'

This is a very complex question, and perhaps more appropriate for the Physics or Philosophy stacks. I will attempt to give you an initial direction, but really I think this is too broad for this stack. I'm merely writing here because comments are too short.

In a very practical way, space is simply that which allows objects and events to have relative position, direction, and extent. There is still disagreement over whether space is a thing in itself, or merely a relationship between things, or part of a conceptual framework. See wiki

In relativity, the dominant concept is not space, but spacetime - closely related to a fundamental constant of the universe (speed of light) and to gravity. Space itself, or time itself, are less important, but the 4-dimensional metric of spacetime is far more fundamental, in a sense.

  Does it exist in the same way substances exist?

We don't know. See the wiki link for the philosophical debate.

  what do we mean by 'space exists?'

It means this universe is built in such a way as to allow relationships of position or direction. This is a bit of a circular definition, but it's the best we have.

  What does it mean for space to be created?

Actually, that happens all the time, even now. The so-called dark energy actually makes more space, continuously, even now. That simply means that, whereas a while ago there was X amount of distance between galaxies A and B, now there's X + Y amount of distance. The low-level mechanism that allows this creation is not well understood.

It could be imagined that something similar happened during the Big Bang, except it started from zero. Again, the low-level mechanisms around the origin are not well understood. To understand that we first must be capable of understanding something far more simple - the physics of the center of a black hole. To do that, we would need something like quantum gravity, which is a part of Physics that has not been invented / built yet.

If this seems overly speculative, it's because you're asking a very profound and difficult question - so difficult that it exceeds the limits of current science and even philosophy.


Satellites traveling in single file across the sky I just watched 46 (best count) satellites traveling in single file, about 3 second intervals between them. Is that possible?  I don't want to believe they were UFO's.  Time was 06:20 ET

In January SpaceX launched into space a string of 60 satellites into space. These are satellites for internet communications. Astronomers say the proliferation of the bright metallic satellites could seriously degrade the night view, interfering with both optical and radio astronom$y^1$

1.https://www.sciencealert.com/spacex-just-launched-a-third-batch-of-its-starlink-satellites

Could this be what you saw?  I don't know, I'm just saying this is maybe the case.


Is there any significant time dilation on other planets orbiting the sun? Is there significant time dilation on the other planets orbiting the sun?  Would an 'hour' experienced on earth actually be 59 or 61 minutes on other planets, depending on their distance to the sun?  What about the moons orbiting Jupiter?

It depends on what you mean by "significant."  None of the orbits of other planets exhibit relativistic effects at the level you specify.  You can estimate the overall size of relativistic corrections by squaring the ratio of v/c, where v is the speed of the orbit.  For the Earth, that number is .00000001, and it is not more than about a factor of 3 higher for any other orbit.  So it's pretty negligible, but that doesn't mean it has no consequences-- over many orbits, for example, Mercury precesses by a larger amount than can be understood without relativity.  Also, if one takes the relativistic perspective, the reason the planets follow the orbits they do is because of that tiny time dilation effect-- that small effect is enough to select the orbits we see over other possible ones.  So that's "significant," but not like a 59 minute hour.


How to calculate the maximum and minimum solar azimuth at a given location? Every Azimuth equation I have been able to locate so far depends on inputting a time, and I do not want to have to brute force the solution by simply trying all of the times. (See below for standard equations.)

My use case doesn't require great precision, within a few degrees is fine. So, ideally I'm looking for equation(s) that take only latitude (and possibly longitude?) as input and output max and/or min azimuth angles for that location.  Also, I'm only going to care about years within a century of now so I don't believe that the year to year variation will matter given my precision requirements, but if I'm wrong I'm happy to provide the current year as well.

Standard Solar Equations Based on NOAA's Solar Calculator

Note: The sunrise and sunset results are theoretically accurate to within a minute
 for locations between +/- 72° latitude, and within 10 minutes outside of those latitudes.
 However, due to variations in atmospheric composition, temperature, pressure and conditions,
 observed values may vary from calculations.
...
Please note that calculations in the spreadsheets are only valid for dates between 1901 and 2099,
 due to an approximation used in the Julian Day calculation."

$\Xi = $Latitude (+ to N)

$\Phi = $Longitude (+ to E)

$\omega = $Time Zone (+ to E)

$d = $Date

$\tau = $Time (hrs past local midnight)

$\upsilon = $Julian Day = $d + 2415018.5 + \tau - \frac{\omega}{24}$

$\sigma = $Julian Century = $\frac{\upsilon - 2451545}{36525}$

$\rho = $Geom Mean Long Sun (deg) = $\left(280.46646 + \sigma
\left(36000.76983 + 0.0003032\sigma\right)\right) \bmod 360$

$\xi = $Geom Mean Anom Sun (deg) = $357.52911 + \sigma\left(35999.05029 - 0.0001537\sigma\right)$

$\mu = $Eccent Earth Orbit = $0.016708634 - \sigma\left(0.000042037 + 0.0000001267\sigma\right)$

$\lambda = $Sun Eq of Ctr = $\sin\left(\xi^{rad}\right)\left(1.914602 - \sigma\left(0.004817+0.000014\sigma\right)\right) + \sin\left(2\xi^{rad}\right)\left(0.019993 - 0.000101\sigma\right) + 0.000289\sin\left(3\xi^{rad}\right)$

$\kappa = $Sun True Long (deg) = $\rho + \lambda$

$\iota = $Sun True Anom (deg) = $\xi + \lambda$

$\theta = $Sun Rad Vector (AUs) = $\frac{1.000001018\left(1 - \mu^2\right)}{1 + \mu\cos\left(\iota^{rad}\right)}$

$\eta = $Sun App Long (deg) = $\kappa - 0.00569 - 0.00478\sin\left(\left(125.04-1934.136\sigma\right)^{rad}\right)$

$\zeta = $Mean Obliq Ecliptic (deg) = $23 + \frac{26 + \frac{21.448 - \sigma\left(46.815+\sigma\left(0.00059 - \sigma0.001813\right)\right)}{60}}{60}$

$\epsilon = $Obliq Corr (deg) = $\zeta + 0.00256\cos\left(\left(125.04-1934.136\sigma\right)^{rad}\right)$

$\delta = $Sun Declin (deg) = $\left(\arcsin\left(\sin\left(\epsilon^{rad}\right)\sin\left(\eta^{rad}\right)\right)\right)^o$

$y = $var y = $\tan\left(\left(\frac{\epsilon}{2}\right)^{rad}\right)^2$

$\Gamma = $Eq of Time (minutes) = $4\left(y\sin\left(2\rho^{rad}\right)-2\mu\sin\left(\xi^{rad}\right)+4\mu y\sin\left(\xi^{rad}\right)\cos\left(2\rho^{rad}\right)-0.5y^2\sin\left(4\rho^{rad}\right)-1.25\mu^2\sin\left(2\xi^{rad}\right)\right)^o$

$\gamma = $True Solar Time (min) = $\left(1440\tau + \Gamma + 4\Phi - 60\omega\right) \bmod 1440$

$\beta = $Hour Angle (deg) = $if\left(\frac{\gamma}{4}<0\right) \{\\
\frac{\gamma}{4}+180\\
\} else \{\\
\frac{\gamma}{4}-180\\
\}$

$\Omega = $Solar Zenith Angle (deg) = $\left(\arccos\left(\sin\left(\Xi^{rad}\right)\sin\left(\delta^{rad}\right)+\cos\left(\Xi^{rad}\right)\cos\left(\delta^{rad}\right)\cos\left(\beta^{rad}\right)\right)\right)^o$

$\alpha = $Solar Azimuth Angle (deg cw from N) =
$if\left(\beta>0\right) \{\\
\left(\arccos\left(\frac{\sin\left(\Xi^{rad}\right)\cos\left(\Omega^{rad}\right)-\sin\left(\delta^{rad}\right)}{\cos\left(\Xi^{rad}\right)\sin\left(\Omega^{rad}\right)}\right)^o+180\right) \bmod 360\\
\} else \{\\
\left(540-\arccos\left(\frac{\sin\left(\Xi^{rad}\right)\cos\left(\Omega^{rad}\right)-\sin\left(\delta^{rad}\right)}{\cos\left(\Xi^{rad}\right)\sin\left(\Omega^{rad}\right)}\right)^o\right) \bmod 360\\
\}$

Thanks to Ralf Kleberhoff for Pointing me in the right direction.

Step 1: Find Summer Solstice Date

For the accuracy needs of this question simply finding the hemisphere and picking the most common date will work. (June 21 in the Northern Hemisphere and December 22 for the Southern Hemisphere). If greater accuracy is required. There are many resources available.

Step 2: Find Sunrise and Sunset Times on Solstice Date

This is straightforward. However, for completeness. (Note: 90.833 is the solar elevation angle for sunrise and sunset.. However, 90 is sufficient for this accuracy and simplifies the equations.)

$ha = \arccos\left(\frac{\cos(90^o)}{\cos(\Xi^{rad})\cos(\delta^{rad})} - \tan(\Xi^{rad})\tan(\delta^{rad})\right)^o =  \arccos\left(-\tan(\Xi^{rad})\tan(\delta^{rad})\right)$

$solarnoon = \frac{720 - 4\Xi - \Gamma}{1440}$

$sunrisetime = solarnoon - \frac{4ha}{1440}$

$sunsettime = solarnoon + \frac{4ha}{1440}$

Step 3: Solve Azimuth Equation for Sunrise and Sunset Times

Simply plug the $sunrisetime$ and $sunsettime$ into the equations provided in the question.


Is sundial time entirely dependent on solar azimuth? I've visited several "how sundials work" sites and can't seem to get a
clear answer to this: is "sundial time" just a linear function of
solar azimuth? More specifically:

When the sun is due south (northern hemisphere), it is sundial
noon. All sites I've visited agree on this.
When the sun is due west (azimuth 270 degrees), I say the sundial
time is 6pm, a quarter turn/day from noon. However, I can't find a
site that actually says this, and some sites seem to disagree with
this. Same for it being 6am sundial time when the sun is due east.

I know there are different types of sundials, but had always assumed
they would give the same sundial time. Is that not true?

As a note, the sundial I describe above can be fairly inaccurate at
times, which makes me question whether I'm correct.

EDIT (to clarify question): Forgetting entirely about clock time for a
moment, suppose I build a sundial. When the sun is due west, my
sundial reads 6pm. When the sun is due east, my sundial reads
6am. When the sun is due south, my sundial reads noon. My question:
have I built my sundial correctly? The type of answer I'm hoping to
get:

No. When you build a sundial, it should read (something else) when
the sun is due west.
There are many different types of sundials. Depending on type,
your sundial may or may not read 6pm when the sun is due west.
It depends on your latitude: sundials are
latitude-specific. There's no such thing as a global sundial.
Yes, your understanding of a sundial is correct. Sundials can be
off by as much as (some number/formula) from mean solar time,
depending on your latitude. They can be as much as (some
number/formula) from clock time, depending on your longitude and
time zone.

The sundial translates the position of the sun to the time of day, so it depends on the path the sun takes across the sky, this is called the ecliptic.

Because the earth's rotation is tilted with respect to its orbit around the sun, the ecliptic shifts across the sky during the year, which is also the cause of seasonal change on earth. Yet the highest altitude of the ecliptic will always be due south, which means noon is well defined every day of the year.

Now to answer the question; if the day length (from sun-rise to sun-down) is exactly 12 hours,  the sun will rise at an azimuth of 90 degrees and set at 270 degrees. However, this happens only twice a year, namely at the vernal equinox (first day of spring) and the autumnal equinox (first day of fall).

At the longest day, the summer solstice, both sun-rise and sun-down azimuths will be at their highest shift northward. On the other hand, during the shortest day, the winter solstice, the azimuth-shift will be maximally southward. This last bit applies to the northern hemisphere, for the southern hemisphere the seasons and therefore day lengths are reversed.

So in short, you have to account for what place on earth your sundial is at and what time of year it is.


What causes celestial bodies to move like they do from Earth's surface? As you probably know, the planets have pretty predictable orbits around the sun but what exactly causes the complex movement of celestial bodies in the sky from Earth's surface? Mars is an excellent example of this.

Mars example

The apparent motion of planets is complex but predictable.

The apparent motion is due to the combination of three different motions:

The rotation of the Earth,
the motion of the Earth around the Sun,
and the motion of the planet around the Sun.

The Planets move in ellipses, slightly perturbed by other planets.

These motions are predictable, and their combination is straightforward trigonometry. The resultant motion is complex, the planets appear to loop relative to the stars, but entirely predictable. We know exactly where the planets will be in the sky for any time for many thousands of years in the future or the past.


Is the gravity on Kepler 39b higher than on the Sun? According to the formula for gravitational acceleration $a = GM/r^2$ where $G=6.674 \times 10^{-11}$, Jupiter with mass 1.89813 × 10^27 kg and radius 69,911,000 m gives a gravity around 25m/s^2. The Sun with mass 1.989 × 10^30 kg and radius 695,510,000m will have a gravity around 270m/s^2.

But using the data for Kepler 39B, mass 18 times that of Jupiter 3.416634 × 10^28, and radius about 5/4 of Jupiter's 87,220,000m, I get almost 300m/s^2, higher than the Sun's.

Is this right?

I did and redid it, and still concerned that I am doing something wrong.

You are probably overlooking why the radius is what it is, or equivalently why the planet is denser than the Sun.

The Sun is undergoing fusion in its core, which releases energy, which causes it to expand (though it achieves an equilibrium state, as expansion decreases rate of fusion which leads to contraction, which leads to accelerated fusion which leads to expansion...).  Jupiter-like bodies, however, don't produce heat from fusion (except maybe some deuterium fusion very early in their lives); they slowly cool and contract over time.  So such a body can compress into a denser state, allowing it achieve higher surface gravity (give or take difficulties in defining "surface" for a gaseous body).

Eyeballing the calculation, you can see that the sun is 10 times the radius of Jupiter and 1000 times the mass, so it's surface gravity should be about $1000/10^2=10$ times greater than Jupiter's, which is what you find.  But Kepler 39b is 18 times more massive than Jupiter and only marginally larger. That's a big uptick in density. So that it's surface gravity would be about $18/(1.25)^2=11.52$ times stronger than Jupiter's (also what you find).  And $11.52>10$, so Kepler 39b has higher surface gravity.


Type I delta-Cepheid star, calculation of apparent maginute and luminosity I am curious to see if I have calculated this correctly:

  A Type I delta-Cepheid star with a one month (30day) period is put at the distance of the star Vega.
  a)  Determine the apparent magnitude this star would have.
  b)  Determine how many times more luminous the delta-Cepheid star is compared to Vega.

My workings:

a) Absolute Magnitudes given for Type I delta-Cepheid star:
$$M=-2.8log\left(P\right)-1.43$$

using the information given $P=30$

$$M=-2.8log(30)-1.43=-5.58$$

Absolute magnitude is given by
$$m-M=5log\left(d\right)-5$$

using information for calculate value of $M$ and that vega = $8Pc$

$$m=5log\left(8\right)-5+5.58=5.10$$

b)
Apparent magnitude is given by
$$m_1-m_2=-2.51log(\frac{F_2}{F_1})$$

as the two start are at the same distance then I will be left with just luminosity.

so adapting the above equation and subbing in the required values

$$0-5.10=-2.51log\left(\frac{L_c}{Lv}\right)$$

Rearrange

$$10^{\frac{5.10}{2.51}}=\frac{L_c}{L_v}$$

$$107.61=\frac{L_c}{L_v}$$

so therefore

$$L_c=108L_v$$

where subscript c is delta-Cepheid  and subscript v is vega

Is this calculation correct?

Pay attention to your signs: Remember that the value of $M$ is negative so on the second equation when you move $M$ to the right hand side, is $ 5 \log()−5- M$ giving you an  apparent magnitude of $\sim -6$ which is acceptable as cepheids are the brightest stars in the galaxies


How do you go from transit times to Earth-Sun distance? In the early 18th century, Halley described a method to determine the value of the astronomical unit by observing a transit of Venus from multiple locations on Earth. How do you go from the transit time at two different locations to the distance between the Sun and the Earth? I found many descriptions in rather vague terms of this method, but what is the math behind it?

We know, precisely, the ratio between the Earth-Venus and Venus-Sun distances. Kepler's laws give us this.
We know, in terms of apparent size in the sky, the precise size of the Sun in degrees - just by looking at it.
Placing Venus between the Sun and the Earth, we can look at it from one point in the Earth (call it $N$) and see Venus against the Sun at a point I'll call $n$. Repeating from another place (call it $S$) at a different latitude, we get another point, $s$. If we know how far apart these two points are in the sky, we can use the relative distance of Venus to tell us the angle, at Venus, of the lines $N$-Venus-$n$ and $S$-Venus-$s$. Since we know the distance in kilometres between $N$ and $S$, that means we can solve the triangle and get the distance in kilometres between Venus and $N$ or $S$. (I am tacitly omitting a few steps here, such as the fact that the "distance" between $N$ and $S$ is a north-south distance and not a great-circle distance along the surface of the Earth). Having the distance between Venus and $N$ or $S$, we can multiply that up (since we know the ratio accurately) to the distance between the Sun and $N$ or $S$.
So it remains to find the distance in the sky between $n$ and $s$. Direct measurement cannot be done accurately enough. Accordingly we use transit time as a proxy. A line going through the middle of the Sun is longer than a line slicing across the top or bottom of it. So transit time tells us how far north or south of the Sun's diameter the point $s$ or $n$ is travelling.
Three transit times are actually needed, not two. To see this, suppose that the times for $s$ and $n$ were equal. All we would then know, strictly speaking, is that they were symmetrically disposed about the diameter - not how far north or south of the diameter they were. But since we know how fast Venus travels across the face of the Sun we can calculate the duration of a hypothetical transit along the diameter, and that figure can then be combined with just two actual observations.
To summarise:

Transit durations give latitudes on the Sun's disc.
Latitudes on the Sun's disc give an angular distance between two transit points.
– The ratio of orbit sizes transforms this into the angle of a long thin triangle based at Venus and joining the two observation points on the Earth.
– Trigonometry gives the Venus-Earth distance in terrestrial units.
– The ratio of orbit sizes, again, gives the Sun-Earth distance in terrestrial units.


What kind of a visual trajectory would an asteroid have if it were to hit Earth? Let's say that Peter has a set of superb telescopes and he's watching the sky every night, for 3-4 months. He may not know it just yet, but there is an asteroid heading towards Earth, on a crash course. He is too lazy to plot an orbit, because he doesn't believe he could die like that.

Over the months the Earth travels quite a lot around the Sun, resulting in various amounts of parallax, depending on the distance. The location of the telescopes are also a factor, and so is the (rotation of the) axis of Earth.

To simplify this and to get rid of Earth radius parallax, Peter is a researcher on the South Pole with those mighty telescopes.
Over the course of months, what kind of a trajectory would a doomsday asteroid make? As it gets bigger/brighter (which is also relative, because the asteroid itself can rotate)/closer, how would its position change, compared to the stars in the background?
Would the change slow down or accelerate? Would it be a small piece of a spiral (I don't mean complete spirals, just the fact that it wouldn't be a parabola or a circle)? What do you think Peter would see if he animated its movement during the course of months on a virtual sky sphere?

The National Geographic article Asteroid Called ‘Spooky’ Will Buzz Earth on Halloween has an animation of the trajectory of asteroid 2015 TB145 (nicknamed Spooky because it looks like a skull).

16 minutes apart visual timelapse:
https://upload.wikimedia.org/wikipedia/commons/e/ec/2015_TB145_discovery.gif

There is no simple answer to this question other than: you have to do the math.

Most often you would see the asteroid brightening more rapidly and moving across the sky faster as the days pass, but not necessarily.  It could arrive directly along our orbital path and just brighten but show no motion in the sky.  It could start out lit up fully (full phase) and then arrive with only partial lighting from the sun, so it does not necessarily brighten.

You generally would not know the phase, unless you calculate it, because you will not be able to resolve it until it gets really close.


What is in the brightest area of the night sky? When I look up in the night sky, I assume the brightest area, commonly known as "The Milky Way", is the Galactic Center of our home galaxy.

But then I stumbled upon this illustration, which seems to suggest what I'm seeing is (and by far) not the Galactic Center. In fact, from this picture it looks like the brightest area might be the closest spiral arm (on the opposite direction of the galactic center).

I know this illustration is not meant to be precise, so it's probably not the closest spiral arm either.

So, what am I really looking at when I stare at the brightest spot of the night sky? What percentage of it would be stars, planets, and nebulae? And what's the farthest object I can see in that direction with my unaided eyes?

All quoted text in this answer is from image captions in the Wikipedia article on the Milky Way.

  360-degree panorama view of the Milky Way (an assembled mosaic of
  photographs) by ESO

  This magnificent 360-degree panoramic image, covering the entire
  southern and northern celestial sphere, reveals the cosmic landscape
  that surrounds our tiny blue planet. This gorgeous starscape serves as
  the first of three extremely high-resolution images featured in the
  GigaGalaxy Zoom project, launched by ESO within the framework of the
  International Year of Astronomy 2009 (IYA2009). The plane of our Milky
  Way Galaxy, which we see edge-on from our perspective on Earth, cuts a
  luminous swath across the image. The projection used in GigaGalaxy
  Zoom place the viewer in front of our Galaxy with the Galactic Plane
  running horizontally through the image — almost as if we were looking
  at the Milky Way from the outside. From this vantage point, the
  general components of our spiral galaxy come clearly into view,
  including its disc, marbled with both dark and glowing nebulae, which
  harbours bright, young stars, as well as the Galaxy’s central bulge
  and its satellite galaxies. As filming extended over several months,
  objects from the Solar System came and went through the star fields,
  with bright planets such as Venus and Jupiter. For copyright reasons,
  we cannot provide here the full 800-million-pixel original image,
  which can be requested from Serge Brunier. The high resolution image
  provided here contains 18 million pixels.

  Here is a schematic map of our POV in the Milky Way galaxy.

  Observed (normal lines) and extrapolated (dotted lines) structure of
  the spiral arms. The gray lines radiating from the Sun's position
  (upper center) list the three-letter abbreviations of the
  corresponding constellations.

  A "God's view" map of Milky Way as seen from far Galactic North (in
  Coma Berenices). The star-like lines center in a yellow dot
  representing the position of Sun. The spokes of that "star" are marked
  with constellation abbreviations, "Cas" for "Cassiopeia", etc. The
  spiral arms are colored differently in order to highlight what
  structure belongs to which arm. H II regions are marked as dots
  colored in the same color as their spiral arm. They come in three
  sizes, measured by the excitation parameter U: small - U > 200 pc cm-2
  medium - 200 > U > 110 pc cm-2 large - 110 > U > 70 pc cm-2

It turns out we are in an arm -- the Orion-Cygnus Arm. The much brighter part of the Milky Way from our POV is in the direction of the galactic center, but the actual nucleus around the supermassive black hole is obscured by dust. If it was visible, it would be quite bright. What we are seeing that is bright is mostly the pseudobulge, or galactic bar formation, in the middle of the galaxy. We are looking at the galactic bar almost end-on, so it resembles a sphere from our POV.


What is the mass and momentum of a relativistic jet? Relativistic jets from accreting black holes are powerful radiation sources, but carries some ionized matter along. What are the estimates or observations of the mass or momentum flow in such jets?

I am not expert in that field, but I found a number and a reference:

A paper by E. Memola et al. on Theoretical thermal X-ray spectra of relativistic MHD jets assumes a mass flow of $\dot{M}_{\rm jet}=10^{-8}M_\odot {\rm yr}^{−1}$ for a central object of $5 M_\odot$
The paper The theory and simulation of relativistic jet formation: towards a unified model for micro- and macroquasars by David L. Maier (see also arXiv/astro-ph/0312048) looks like it may contain more insights, but I did not go through it in detail.


Why do we see galaxies with their actual spiral shapes? Since the actual position of stars is relative, and we see the light they emitted long time ago, and one galaxy has thousands of millions of stars, and those stars can be separated by hundred of thousands of light years, why don't we see galaxies as abstract forms or lines in the sky  as the light of every galaxy star arrives to us at different times instead of well formed spiral shapes? It's like if those millions stars were forming the galaxy as a single light spot, instead of millions of spots separated by light years.

This has been in my head for decades, but I didn't find the question (not even the answer, but the question) anywhere, so either there must be an really easy answer to it or a really complex one or it's a glitch in the Matrix.

Basically, because the relative speeds of stars within galaxies are much less than the speed of light.

Imagine a structure with dimension 100,000 light years (about the size of a galaxy). Now suppose the components of that structure move with speeds relative to each other of around 100 km/s (or 0.03% of the speed of light). In the time it takes light to cross the structure (100,000 years of course), the components within it will have moved by just 30 light years (0.03% of the size of the structure).

There would therefore be no significant "blurring" of our snapshot of a galaxy caused by the differing light travel times from its structural components.

The other thing you may have not understood, is that the lives of stars are generally much longer than 100,000 years. So the stars that emit the light on the far side of a galaxy are still in  existence 100,000 years later, when their light passes stars that are present at the near side of a galaxy.


If gravity precedes the formation of a solar system, where did the mass come from that caused the gravity? In my class we are studying objects in our solar system and this question seemed to just pop up. And since I cannot answer this, I've been really frustrated for quite a while now and would like some help on understanding this.

The start of it all is usually something called a Giant Molecular Cloud, a particular kind of nebula which is denser than usual and cooler than usual.  A GMC will typically be 10x to 1000x the mass of the Sun.

That mass is enough to cause the outer layers to fall in on the inner parts and the GMC starts to collapse and shrink. (Diffuse though it is, the mass exerts the same force on the outer edges as it would if the entire mass was concentrated at the center.)

The cloud collapses and frequently fragments into a bunch of smaller collapsing blobs centered on especially dense sections of the cloud. At this point it looks a lot like the Oriion Nebula. The young stars light up and blow away the remaining bits of the GMC, and in the end, you have a cluster of young stars with planetary disks looking something like the Pleiades.

There never was a central mass -- none is needed.  The GMC collapses under its own self-attraction.


How big would the supernova when a neutron star collides with another neutron star? If two neutrons stars that are orbiting were to collide, how big would the supernova explosion be?
Would a new black hole or a new supernova remnant be formed?

When two neutron stars collide, the resulting supernova is called a "kilonova," producing much more energy than a regular supernova. Additionally, heavier elements are formed, such as gold, platinum, etc. Other effects of this collision include a gamma-ray burst and/or gravitational waves. The object formed after the event can either be a heavier neutron star, or a black hole (if the mass of the merger exceeds the maximum mass of a neutron star, about 3 solar masses).
As of right now, we have not detected a neutron star merger that is visible to the naked eye; almost all of them occurred in different galaxies, billions of light-years away. I hope this helps.


Replacing Meade ETX with computerized mount I've been using a Meade ETX 90 for a few years now. Since it doesn't have the new computerized mount (with goto), I am looking into buying one to swap out. Where might I find this and is there a more technical term than just "computerized mount" that is used?

Note: This is an old scope that did not originally come with the ports that the autostar controller now goes into (on the newer models).

Update: I ended up getting a newer Meade ETX and plan to replace my current mount with the computerized one. Also, if you check out cloudynights.com you might be able to find one. I have seen one on there before.

Meade make the 497 auto star hand controller, which plugs in where your existing hand controller goes. This turns your mount into a fully computerised one. Alternatively you should be able to use a laptop to control the mount via the aux ports. It’s all in the manual.

If your telescope did not originally come with the auto star hand controller, here is a quote from weasner.com:

  Q. How do I upgrade a non-AutoStar ETX to an AutoStar capable ETX?

  Short answer: you can't. Meade does not offer an upgrade for the
  original model ETX (also known as the ETX-90RA or ETX Astro M). But
  you can mount the OTA on a different mount (see the Telescope Tech
  Tips page) or (if still available) get a JMI computer system (see the
  "NGC-microMAX/NGC-MAX" on the Accessory Reviews - Miscellaneous page).
  Or you could sell it and by a new AT model.


When are the planets occulted by the sun? There are numerous lists available of past and future planetary transits and occultations, but I can't find any that list transits behind the sun.

Do the planets in our solar system ever pass directly behind the sun, from Earth's perspective? Is there a special name for this kind of occultation, and is there a list of these events?

The search for sungrazing comets requires looking very close to the Sun for faint objects. This requires a chronograph and for faint objects is better done above the atmosphere. The SOHO spacecraft near the Sun-Earth L1 point does this nicely.
Here is a GIF I made for this answer. In addition to some comets you can see Venus on its approach to occultation by the Sun in 2016. Seeing the Pleides so close to the Sun is exciting as well!

These LASCO C3 images from SOHO were downloaded sohodata.nascom.nasa.gov/cgi-bin/data_query. The square frame is about 15.9 degrees wide.

You can see many trajectories for 2016 in http://sungrazer.nrl.navy.mil/index.php?p=transits/transits_2016 (shown below) and replacing 2016 with other years works as well.
The problem with this page is that while it may flag likely occultations, it does not unambiguously predict them, so you may need to use another source for accurate predictions.

Here's a quick calculation for 2016 using the Python package Skyfield. It seems both Mercury and Venus were able to hide behind the Sun.

Jackpot!  Mercury, Jupiter, Saturn, Mercury

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos, EarthSatellite

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180
Rsun              = 696392. # https://en.wikipedia.org/wiki/Sun

load    = Loader('~/Documents/fishing/SkyData')
ts      = load.timescale()
data    = load('de421.bsp')

names   = ('Mercury', 'Venus', 'Mars', 'Jupiter', 'Saturn')
objects = [data[name + ' barycenter'] for name in names]
earth   = data['Earth']
sun     = data['Sun']

ts      = load.timescale()
days    = np.arange(1, 366, 0.1)
times   = ts.utc(2016, 1, days)

observations = [earth.at(times).observe(obj) for obj in objects]
sunobs       =  earth.at(times).observe(sun)
sundist      = sunobs.distance().km
separations  = [obs.separation_from(sunobs) for obs in observations]

sunhalfangledegs     = degs*np.arctan2(Rsun, sundist)
sunhalfangledegsmin  = sunhalfangledegs.min()

if True:
    plt.figure()
    for i, (name, sep) in enumerate(zip(names, separations)):
        plt.subplot(5, 1, i+1)
        angle = degs*sep.radians
        minangle = angle.min()
        plt.plot(days, angle)
        plt.ylim(0, None)
        ymin, ymax = plt.ylim()
        ymax = 5*max(ymin, minangle, sunhalfangledegsmin)
        plt.ylim(0, ymax)
        plt.plot(days, sunhalfangledegs, '-r', linewidth=0.5)
        plt.title('observed Sun-Earth-' + name + ' angle (degs)')
    plt.show()


How many planets does Omicron Persei have? Omicron Persei is a binary star. I did not find information on how many planets it has and if we can actually determine it currently. However, I do know that most stars have planets orbiting around them and so I assume there are such planets.

For those who are interested, Futurama featured the planet Omicron Persei 8 - the 8th planet in that system. I was curious as to if there is an 8th planet.

Currently http://exoplanet.eu/ (which lists more exoplanets than the official NASA archive) lists no known planets of Omicron Persei. So currently we know of none, it is outside of the Kepler field so has no Kepler candidates either.

Though just because none are known that does not mean it has none, it may be that their mass and or orbital plane is such that it is very difficult to identify planets or no one has looked with a suitable instrument.


Doesn't gravity attract objects in space until they collide? If the formula to calculate the force of gravity between two objects is:

$$F = GM_1M_2/r^2,$$

why do planets stay in orbit? Or is there another formula at work?

When an object is in orbit, there are two factors at play, not just one. The first, as you mention, is the force of gravity pulling the objects together. However, each object also has a momentum component which is generally (in the case of circular orbits) perpendicular to the direction of the gravity.

If we look at the common situation of a small-mass object orbiting a large (massive) object, then we can ignore the perpendicular velocity (momentum) component of the larger object and reach a simplification: The smaller object is continually pulled towards the primary but perpetually 'misses' due to its own perpendicular momentum.


What Types of Radiation Emanate in the Future and are Perceived in the Present? Years ago I came upon a scientific text which mentioned different types of radiation and how they are perceived. One such form of radiation was described (from our perspective and understanding of space-time) to originate from the future and travel backwards through time.

I cannot for the life of me remember what this type of radiation was called and I would love to know it.

Nothing that is known to exist

Relativity does not allow for a massive particle to travel at the speed of light, but it doesn't prevent a particle from travelling faster than light. Such a particle has been called a Tachyon. No such particle has ever been observed. There are good reasons for believing that they don't exist.

Such a particle would be extremely strange. The mass of such a particle would not be positive, it would be imaginary! However you could not use such a particle to send a message to the past. The particles can't be localised, which means that you can't detect them as being "at" a particular place at a particular time.

In conclusion, such particles have never been detected, probably don't exist, and couldn't be used to send messages.

read more at http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html


How far would EGSY8p7 be away now? Apparently EGSY8p7 is the object with the longest light travel distance, 13.2 gly or a redshift of z = 8.68 (Wikipedia).
So the light took 13.2 billion years to travel to us from that object, but we know that the universe has expanded since (I believe with an expansion rate that varied over time). So given our current theories of how the universe expanded, is it possible to calculate the theoretical distance of that object to us right now?

I assume it must be close to what is believed the radius of the observable universe (46.5 Gly) but how would one go about to calculate this?

30.4 billion lightyears.

The current distance — i.e. the distance that one would measure if we froze the Universe and started laying out measuring rods — is called the proper distance, or physical distance, in astronomy. By definition, it corresponds today to another often-used term in astronomy, namely the comoving distance. While the former increases with time, the latter is defined in a coordinate system that expands with the Universe, and so is constant at all times.

That is, the comoving distance to EGSY8p7 was the same when the Universe was half the size of what it is today, but the proper distance was half of this.

For galaxies so distant as EGSY8p7, we don't measure distances directly. Instead we measure the redshift of some spectral absorption or emission lines, and then use a model of the expansion of the Universe to convert this redshift to a distance.

From the FLRW metric, the resulting distance today of an object at redshift $z$ is given by
$$
d_C = \frac{c}{H_0} \int_0^z dz' \frac{1}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_k(1+z)^2 + \Omega_\Lambda}},
$$
where $c$, $H_0$, and $\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}$ are the speed of light, Hubble constant, and density parameters of the various components of the Universe, respectively.

For $z=8.68$ and with the latest cosmological parameters from Planck Collaboration et al. (2016), this integral evaluates to $d_C = 30.4\,\mathrm{Glyr}$.

For $z\rightarrow\infty$, you get $d_C = 46.3\,\mathrm{Glyr}$, which is "the edge of the observable Universe", also known as the particle horizon.

Note though that EGSY8p7 no longer holds the redshift record. The galaxy GN-z11 (Oesch et al. 2016) has a redshift of $z=11.09$, corresponding to a distance of $d_C = 32.2\,\mathrm{Glyr}$.


What bright body is next to the Moon in these photos of the ISS transiting the lunar disk? The question What are the chances that the ISS transits the moon as seen from your house (or headquarters)? links to the SpaceX Instagram-posted photos of the ISS transiting the Moon's disk as seen from SpaceX headquarters.

There's a bright object next to the Moon, but I'm not sure exactly when this happened. Instagram says "last weekend" which would appear to be the last weekend of July in 2018.

What bright object is this, next to the Moon? Planet? Star?

https://www.instagram.com/p/Bl8xjJFFrHM/

Given the appearance of the moon, that must have been photographed on the weekend of the 21-22 (the moon was nearly full on the 28-29)

The object is Jupiter, and from the position I'd say the photograph was taken at about 0400 UTC on the 21st July 2018.

I think Antares can be dimmly seen towards the left of the photo, and perhaps alpha and beta Libra area visible above and below the moon, but I can't see any other stars.


What is the intensity distribution of visible light over the solar disk? What is the distribution of the visible spectrum over the solar disk?

Images of the Sun using the visible spectrum suggest that the light emmitted from the edge is almost equal to the light emmited from the center of the solar disk.

Is this correct and is there a scientific source that proves that? I was not able to find something citable.

What is the intensity distribution of visible light over the solar disk?
  What is the distribution of the visible spectrum over the solar disk

Hopefully I've answered all the different questions.

The core of the sun is extremely hot (27 million degrees Fahrenheit) compared to the surface (10,000 degrees Fahrenheit) and the atmosphere (36,000 to 2 million degrees Fahrenheit). Source: JPL NASA - "The Sun's Layers and Temperatures". To our eye the sun generally appears to have a single color and brightness due to dynamic range and color mixing.

The AM0 spectrum and intensity looks like this:

Each element emits a different multitude of wavelengths. The distribution across the surface, and out into the atmosphere, varies.

For example here is a set of images taken at different wavelengths:

That answers your question in this comment:

  "Ah thank you, that is exactly what I was looking for. But how do images like these get produced? Are they adjusting the brightness to be equal througout the image before releasing it? – Jhonny007".

Images like those are produced by placing filters over the imaging sensor.

Limb darkening is caused by the different temperatures of the different layers. The center of the disk appears bumpy due to granules. Using a greater dynamic range it looks like this:

This image is a composite of 25 separate images spanning the period of April 16, 2012, to April 15, 2013. It uses the SDO AIA wavelength of 171 angstroms and reveals the zones on the sun where active regions are most common during this part of the solar cycle.

A closeup of a sunspot (a colder region, thus darker) looks like this:

The website Schurs Solar Imaging, and webpage: "Image Gallery of Solar Phenomenon - White Light Features" has a great set of photos, and one clearly showing limb darkening.

Adjusting the contrast of an image from the University of Virginia's webpage: "ASTR 1230 (Majewski) Lecture Notes" creates an image that shows limb darkening.


How are binary star systems created? I don't know how common it is for a system to have two stars (or perhaps even more) but how do they arise? Is that due to the stellar accretion disc, or the composition of the stellar nebula?
Or are most of them created by a collision of stars?

There are two main theories for the formation of binary stars - one accepted, and one mainly deprecated.

The fission hypothesis. The fission hypothesis states that the binary system forms after the collapse of the original gas cloud into a protostar. Angular momentum is conserved, so as the extremely large cloud slowly contracts, it spins faster. After enough time, the protostar could form a dumbbell shape, one part of which eventually breaks off. There are now two blobs, and thus two protostars, which will evolve into a binary star system.
The fragmentation hypothesis. The fragmentation hypothesis states that the gas cloud fragments earlier on, due to some instability or cooling/heating effects. Each of the two fragments then evolves separately, forming a binary star system.

The fission hypothesis is no longer favored. It cannot explain the existence of systems with certain mass ratios or separations. The fragmentation hypothesis, however, can, and is accepted. It is also possible for a binary system to form through the capture of one star by another, but that requires a third body, and probably cannot explain the high fraction of binary systems in the galaxy.

Basically, then, the main theory to date is that the protostellar nebula breaks up, and then both fragments collapse independently.


moonless night and lunar phase How to define moonless night? There is no moon at all during some night?
How to calculate and know whether the moon would appear during one night?

If the lunar phase is full moon, is it possible that the moon will not appear ?

A moonless night is, as you suspect, a night in which the Moon does not appear visible in the sky. This happens once per month, when the Moon is near the Sun. Due to the proximity of the Moon and the Sun in the sky, at that time the Moon is the smallest sliver possible, and therefore not a full moon.

This is because it is actually the Sun that illuminates the Moon, and when the Sun and the Moon are in the same direction in the sky we are seeing the non-illuminated side of the Moon. Note the direction of the sunlight in this image:

Obviously, the direction of the sunlight is the direction of "up" during the day. If you look at the horizon slightly after sunset or slightly before sunrise, you might actually catch a glimpse of the sliver of Moon before it set or rises slightly after or before the Sun.


Measurement precision of celestial events The local weatherman declared that fall starts at exactly 10:21am (local time). Ignoring the fact that seasonal boundaries have multiple definitions (meteorological, astronomical, etc.), what he perhaps means to have said is that the fall equinox occurs at that moment.

Presumably, that is the moment at which Earth's axis of rotation is exactly perpendicular to the Sun-Earth line.

My question is: to what degree of precision can this be measured? The weatherman's statement was to the minute. But is it actually measured to a finer precision? If so, plus or minus how many seconds?

What about the the solstices? Can they be measured to the same degree of accuracy? Are there factors which uniquely affect the accuracy of either measurement?

This doesn't answer your question, but has two screenshots, so that's something :)

If you run http://ssd.jpl.nasa.gov/?horizons with these settings:

part of the output looks like this:

This shows the Sun crosses the equinox point between the 8th and 9th second of 14:21 (which is 10:21 your local time). HORIZONS will even let you compute at half second intervals but not beyond that.

If you use the http://naif.jpl.nasa.gov/naif/tutorials.html library, you can compute the time down to the nanosecond or better, but this doesn't really help because we don't know the Sun's position, the Earth's position, or even the direction the North Pole is pointing to that level of precision.

So, not an answer, but some hopefully helpful comments.


How does a cross dispersed spectrum look like in a reality (from echelle spectrographs)? I am an analytical chemist with some interest in amateur spectroscopy. Since astronomers use echelle spectrographs to study the high resolution spectrum of the stars, someone suggested to post the query here. An echelle grating produces a spectrum with several overlapping orders. One can then use a cross-disperser, which could be a prism or a grating, which can separate the overlapping orders and then we obtain a very high resolution spectrum. I want to know how a real echelle spectrogram looks like -as if we were looking at the 2D spectrum with our naked eyes?
I have a small echelle grating (79 grooves/mm, 75$^o$ blaze) and tried to visually generate a 2D spectrum with the help of a plastic transmission grating. The light source was an ordinary fluorescent lamp Pictures are posted below. The experiment is too elementary.  A horizontal slit made by two blades as an entrance on the shorter rectangle of a shoebox. The light from the slit falls on the reflective echelle. On the larger side of the shoebox, I made a window to take a picture.
Picture 1: A vertical slit made in a cardboard box, the long side of the echelle is perpendicular to slit An expected overlapping order spectrum seen.

Picture 2: A horizontal slit made in the cardboard box, now the long side of the echelle is parallel to the slit

Now, if we hold a plastic transmission grating (Edmund) in our hand, with its rulings perpendicular to the echelle, and view the echelle through that transmission grating, I can photograph the following 2D spectrum.
Picture 3: Hold a transmission grating with its rulings crossed with the echelle and view picture no. 2.

Unfortunately, this looks like a replica of a spectrum rather than separation of orders.
**The main question: How should one hold the transmission grating in order to visually see the separated orders. Picture no. 3, was supposedly, separation of orders but it does not look like it.
If this were a true 2D spectrum we should see some increasing distance between Hg green lines. All separated orders look the same resolution. Is there any other simple way to see a true cross dispersed spectrum with eyes using a simple cardboard type box? The key problem seems to be the issue how should we hold the transmission grating in order to see a true 2D spectrum unlike Picture 3.**
EDIT
After Prof.ELNJ's suggestion in the answer, I wanted to confirm if this is what he meant. The grating has an arrow with a pencil (from ThorLabs) which shows the blaze direction because by viewing in this direction one can see a "rainbow" of colors. So the arrow is towards the viewer, and I hold a transmission grating "crossed" with the echelle ruling.

Rough view of what appears to the eye

You are doing it incorrectly if you are trying to cross-disperse your Fig.2. You should be cross-dispersing your Fig.1.
A "cross-dispersed" spectrum requires requires the dispersive elements to be at right angles.
Your Fig.1 shows the dispersed spectrum from the echelle with overlapping orders.
To separate the orders you view Fig.1 with the cross-dispersing element, but with the cross-disperser dispersion direction placed along the y-axis of your pictures. A schematic would be something like this, where the grating rulings are indicated with lines. (I believe this picture is originally from Optical Astronomical Spectroscopy, by C. R. Kitchin).

The slit orientation would be parallel to the rulings on the echelle, but you need to make the slit short enough that the orders can be separated by the cross-disperser.
That should produce something like that shown below, if you illuminate it with a continuum source. Exactly what you see will depend on the dispersive power of the cross-disperser (should be low and often, prisms are used), how large your CCD is and how long your slit is.