How did our ancestors discover the Solar System? I wonder, how did our ancestors discover the Solar System? They did not have any telescopes to see distant objects, right? Even a planet looks like a star from a distance.

They discovered the rotations of different planets without having much technology.

1. Ancient cultures observed the sky

Night skies are naturally dark and there was no light-pollution in ancient times. So if weather permits, you can easily see a lot of stars. No need to tell about the Sun and the Moon.

Ancient people had good reasons to study the night skies. In many cultures and civilizations, stars (and also the Sun and the Moon) where perceived to have religious, legendary, premonitory or magical significance (astrology), so a lot of people were interested in them. It did not took long to someone (in reality a lot of different people independently in many parts of the world) to see some useful patterns in the stars that would be useful to navigation, localization, counting hours, counting days and relating days to seasons, etc. And of course, those patterns in the stars were also related to the Sun and the Moon.

So, surely all ancient cultures had people who dedicated many nights of their lifes to study the stars in detail right from the stone age. They would also perceive meteorites (falling stars) and eclipses. And sometimes a very rare and spetacular comet.

Then there are the planets Mercury, Venus, Mars, Jupiter and Saturn. They are quite easily to notice to be distinct from the stars because all the stars seems to be fixed in the celestial sphere, but the planets don't. They are very easily to notice to be wandering around in the sky with the passage of the days, specially for Venus, which is the brightest "star" in the sky and is also a formidable wanderer. Given all of that, the ancient people surely become very aware of those five planets.

About Mercury, initially the Greeks thought that Mercury were two bodies, one that showed up only in the morning a few hours before the sunrise and another only a few hours after the sunset. However, soon they figured out that in fact it was only one body, because either one or the other (or neither) could be seen in a given day and the computed position of the unseen body always matched the position of the seen body.

2. The Earth seems to be round

Now, out of the stone age, already into ancient times, navigators and merchants who travelled great distances perceived that the Sun rising and setting points could variate not only due to the seasonal variation, but also accordingly to the location. Also, the distance from the polar star to the horizon line also variates accordingly with the location. This fact denounces the existence of the concept nowadays known as latitude, and this was perceived by ancient astronomers in places like Greece, Egypt, Mesopotamia and China.

Astronomers and people who dependend on astronomy (like navigators) would wonder why the distance from the polar star to the horizon varied, and one possibility was that it is because the Earth would be round. Also, registering different Sun angles in different locations of the world on a given same day and at a given same hour, also gives a hint that the Earth is round. The shadow on the Moon during a lunar eclipse also gives a hint that the Earth is round. However, this by itself is not a proof that the Earth is round, so most people would bet on some other simpler thing, or simply don't care about this phenomenon.

Most cultures in ancient times presumed that the world was flat. However the idea of the world being round exists since the ancient Greece. Contrary to the popular modern misconception, in the Middle Ages, almost no educated person on the western world thought that the world was flat.

About the Earth's size, by observing different Sun positions and shadows angles in different parts of the world, Erasthotenes in ancient Greece calculated the size of Earth and the distance between the Earth and the Sun correctly for the first time as back as the third century B.C. However, due to the confusion about all the different and inconsistent unit measures existent back then and the difficulty to precisely estimate long land and sea distances, confusion and imprecision persisted until the modern times.

Ancient cultures also figured out that the shiny part of the Moon was illuminated by the Sun. Since the Full Moon is easily seen even at midnight, this implies that the Earth is not infinite. The fact that the Moon enters in a rounded shadow when exactly in the opposite side of the sky as the Sun also implies that it is the Earth's shadow on the Moon. This also implies that Earth is significantly larger than Moon.

3. Geocentrism

So, people observed the Sun, Moon, Mercury, Venus, Mars, Jupiter, Saturn and the fixed sphere of stars all revolving around the sky. They naturally thought that the Earth would be the center of the universe and that all of those bodies revolved around the Earth. This culminated with the work of the phylosopher Claudius Ptolemaeus about geocentrism.

Altough we now know that the ptolomaic geocentric model is fundamentally wrong, it could be used to compute the position of the planets, the Sun, the Moon and the celestial sphere of stars, with a somewhat acceptable precision at the time. It accounted to include the observation of planets velocity variations, retrograde motions and also for coupling Mercury and Venus to the Sun, so they would never go very far from it. Further, based on the velocity of the motion of those bodies in the sky, then the universe should be something like:

Earth at the center.
Moon orbiting the Earth.
Mercury orbiting the Earth farther than the Moon.
Venus orbiting the Earth farther than Mercury.
Sun orbiting the Earth farther than Venus.
Mars orbiting the Earth farther than Sun.
Jupiter orbiting the Earth farther than Mars.
Saturn orbiting the Earth farther than Jupiter.
The celestial sphere of stars rotating around the Earth, being the outermost sphere.

In fact, the ptolomaic model is a very complicated model, way more complicated than the copernic, keplerian and newtonian models. Particularly, this could be compared to softwares that are based on severely flawed concepts but that are still working due to a lot of complex, tangled and unexplainable hacks and kludges that are there just for the sake of making the thing work.

4. The discovery of the Americas

Marco Polo, in the last years of the 1200's, was the first European to travel to China and back and leave a detailed chronicle of his experience. So, he could bring a lot of knowledge about what existed in the central Asia, the East of Asia, the Indies, China, Mongolia and even Japan to the Europeans. Before Marco Polo, very few was known to the Europeans about what existed there. This greatly inspired European cartographers, philosophers, politicians and navigators in the years to come.

Portugal and Spain fough a centuries-long war against the invading Moors on the Iberian Peninsula. The Moors were finally expelled in 1492. The two states, were looking for something profitable after so many years of war. Since Portugal ended its part of the war first, it had a head start and went to explore the seas first. Both Portugal and Spain were trying to find a navigation route to reach the Indias and the China in order to trade highly profitable spices and silk. Those couldn't be traded by land efficiently anymore due to the fact that the lands on West Asia and North Africa were dominated by Muslim cultures unfriendly to Christian Europeans, a situation that were just made worse after the fall of Constantinople in 1453.

Portugal, were colonizing the Atlantic borders of Africa and eventually they managed to reach the Cape of Good Hope in 1488 (with Bartolomeu Dias).

A Genovese navigator called Cristoforo Colombo believed that if he sailed west from the Europe, he could eventually reach the Indies from the east side. Inspired by Marco Polo and subestimating the size of Earth, he estimated that the distance between the Canary Islands and the Japan to be 3700 km (in fact it is 12500 km). Most navigators would not venture in such voyage because they (rightly) tought that Earth was larger than that.

Colombo tried to convice the king of Portugal to finance his journey in 1485, but after submitting the proposal to experts, the king rejected it because the estimated journey distance was too low. Spain, however, after finally expelling the Moors in 1492, were convinced by him. Colombo's idea was far-fetched, but, after centuries of wars with the Muslims, if that worked, then Spain could profit quickly. So, the Spanish king approved the idea. And just a few months after expelling the Moors, Spain sent Colombo to sail west towards the Atlantic and then he reach the Hispaniola island in Central America. After coming back, the news about the discovery of lands in the other side of the Atlantic spread quickly.

Portugal and Spain then divided the world by the Treaty of Tordesillas in 1494. In 1497, Amerigo Vespucci reached the mainland America.

Portugal would not be left behind, they managed to navigate around Africa to reach the Indies in 1498 (with Vasco da Gama). And they sent Pedro Álvares Cabral, who reached the Brazil in 1500 before crossing the Atlantic back in order to go for the Indies.

After that, Portugal and Spain quickly started to explore the Americas and eventually colonize them. France, England and Netherlands also came to the Americas some time later.

5. The Earth IS round

After, the Spanish discovered and settled into the Americas (and Colombo's plan in fact didn't worked). The question that if it was possible to sail around the globe to reach the Indies from the east side remained open and the Spanish were still interested on it. They eventually discovered the Pacific Ocean after crossing the Panama Ishtums by land in 1513.

Eager to find a maritime route around the globe, the Spanish crown funded an expedition leadered by the Portuguese Fernão de Magalhães (or Magellan as his name was translated to English) to try to circle the globe. Magellan was an experienced navigator, and had reached what is present day Malaysia traveling through the Indian Ocean before. They departed from Spain in September 20th, 1519. It was a long and though journey that costed the lives of most of the crew. Magellan himself did not survived, having died in a battle in the Phillipines on 1521. At least, he lived enough to be aware that they in fact reached East Asia by traveling around the globe to the west, which also proves that the Earth is round.

The journey was eventually completed by the leadership of Juan Sebatián Elcano, one of the crewmen of Magellan. They reached Spain back through the Indian and Atlantic Oceans on September 6th, 1522 after traveling for almost three years a distance of 81449 km.

6. Heliocentrism

There were some heliocentric or hybrid geo-heliocentric theories in ancient times. Notably by the Greek philosopher Philolaus in the 5th century BC. By Martianus Capella around the years 410 to 420. And by Aristarchus of Samos around 370 BC. Those models tried to explain the motion of the stars as rotation of the Earth and the position of the planets, specially Mercury and Venus as translation around the Sun. However those early models were too imprecise and flawed to work appropriately, and the ptolomaic model still was the model with the better prediction of the positions of the heavenly bodies.

The idea that the Earth rotates was much less revolutionary than heliocentrism, but was already more-or-less accepted with reluctancy in the middle ages. This happens because if the stars rotated around Earth, they would need do so at an astonishing velocity, dragging the Sun, the Moon and the planets with it, so it would be easier if Earth itself rotated. People were uncomfortable with this idea, but they still accepted it, and this became easier to be accepted after the Earth sphericity was an established concept.

In the first years of the 1500's, while the Portuguese and Spanish were sailing around the globe, a polish and very skilled matemathical and astronomer called Nikolaus Kopernikus took some years thinking about the mechanics of the heavenly bodies. After some years making calculations and observations, he created a model of circular orbits of the planets around the Sun and perceived that his model were much more simpler than the ptolomaic geocentric model and was at least as precise. His model also features a rotating Earth and fixed stars. Further, his model implied that the Sun was much larger than the Earth, something that was already strongly suspected at the time due to calculations and measurements and also implied that Jupiter and Saturn were several times larger than Earth, so Earth would definitively be a planet just like the other five then-known planets were. This could be seen as the borning of the model today know as Solar System.

Fearing persecution and harsh criticism, he avoided to publish many of his works, sending manuscripts to only his closest acquaintances, however his works eventually leaked out and he was convinced to allow its full publication anyway. Legend says that he was presented to his finally fully published work in the very day that he died in 1543, so he could die in peace.

There was a heated debate between supporters and oppositors of Copernic's heliocentric theory in the middle 1500's. One argument for the opposition was that star parallaxes could not be observed, which implied that either the heliocentric model was wrong or that the stars were very very far and many of them would be even larger than the Sun, which seemed to be a crazy idea at the time.

Tycho Brache, which did not accepted heliocentrism, in the latest years of the 1500's tried to save geocentrism with an hybrid geo-heliocentric model that featured the five heavenly planets orbiting the Sun while the Sun and the Moon orbited Earth. However, he also published a theory which better predicted the position of the Moon. Also, by this time, the observation of some supernovas showed that the celestial sphere of the stars were not exactly immutable.

In 1600, the astronomer William Gilbert provided strong argument for the rotation of Earth, by studing magnets and compasses, he could demonstrate that the Earth was magnetic, which could be explained by the presence of enourmous quantities of iron in its core.

7. With telescopes

All of what I wrote above happened without telescopes, only by using naked eye observations and measurements around the globe. Now, add even some small telescopes and things change quickly.

The earliest telescopes were invented in 1608. In 1609, the astronomer Galieu Galilei heard about that, and constructed his own telescope. In January of 1610, Galieu Galilei, using a small telescope, observed four small bodies orbiting Jupiter at different distances, figuring out that they were Jupiter's "moons", he also could predict and calculate its positions along their orbits. Some months later, he also observed that Venus had phases as seen from the Earth. He also observed Saturn's rings, but his telescope was not powerful enough to resolve them as rings, and he tought that they were two moons. These observations were incompatible with the geocentric model.

A contemporary of Galilei, Johannes Kepler, working on Copernicus' heliocentric model and making a lot of calculations, in order to explain the differing orbital velocities, created an heliocentric model where the planets orbits the Sun in elliptic orbits with one of the ellipse's focus in the Sun. His works were published in 1609 and 1619. He also suggested that tides were caused by the motion of the Moon, though Galilei was skeptical to that. His laws predicted a transit of Mercury in 1631 and of Venus in 1639, and such transit were in fact observed. However, a predicted transit of Venus in 1631 could not be seen due to imprecision in calculations and the fact that it was not visible in much of the Europe.

In 1650 the first double star were observed. Further in the 1600's, the Saturn rings were resolved by the use of better telescopes by Robert Hooke, who also observed a double star in 1664 and developed microscopes to observe cellular structures. From them on, many stars were discovered to be double. In 1655, Titan were discovered orbiting Saturn, putting more confidence on the heliocentric model. More four Saturnian moons were discovered between 1671 and 1684.

8. Gravitation

Heliocentrism was reasonably well-accepted in the middle 1600's, but people was not confortable with it. Why the planets orbits the Sun afterall? Why the Moon orbits Earth? Why Jupiter and Saturn had moons? Although Keplerian mechanics could predict their moviment, it was still unclear what was the reason that makes them move that way.

In 1687, Isaac Newton who was one of the most brilliant physic and mathematic that ever lived (although he was also an implacable persecutor of his opponents), provided the gravitational theory (based on prior work by Robert Hooke). Ideas for the gravitation theory and the inverse square law already were developed in the 1670's, but he could publish a very simple and clear theory for gravitation, very well-fundamented in physics and mathematics and it explained the motions of the celestial bodies with a great precision, including comets. It also explained why the planets, the Moon and the Sun are spherical, explained tides and it also served to explain why things falls to the ground. This made heliocentrism to be definitely widely accepted.

Also, Newton gravitational law predicted that Earth rotation would make it not exactly spherical, but a bit ellipsoidal by a factor of 1:230. Something that agreed with measures done using pendulums in 1673.

9. What are the stars and the Solar System afterall?

In the early 1700's, Edmund Halley, already knowing about newtonian laws (he was a contemporary of Newton) perceived that comets who passed near Earth would eventually return, and he found that there was a particular instance of sightings every 76 years, so he could note that those comets in reality were all the same comet, which is called after him.

The only remaining problem with the heliocentric model was the lack of observation of parallax to the stars. And nobody knew for sure what the stars were. However, if they in fact are very distant bodies, most of them would be much larger than the Sun. At the first half of the 1700's, trying to observe parallax, James Bradley perceived phenomena like the aberration of light and the Earth's nutation, and those phenomena also provides a way to calculate the speed of light. But the observation of parallax remained a challenge during the 1700's.

In 1781, Uranus were discovered orbiting the Sun beyond Saturn. Although barely visible to the naked eye in the darkest skies, it was so dim that it escaped observation from astronomers until then, and so were discovered with a telescope. The first asteroids were also discovered in the early 1800's. Investigation on pertubations on Uranus' orbit due to the predicted newtonian and keplerian movement eventually leaded to the discovery of Neptune in 1846.

In 1838, the astronomer Friedrich Wilhelm Bessel who measured the position of more than 50000 stars with the greatest precision as possible, could finally measure the parallax of the star 61 Cygni successfully, which proved that stars were in fact very distant bodies and that many of them were in fact larger than the Sun. This also demonstrates that the Sun is a star. Vega and Alpha Centauri also had their parallaxes measured successfully in 1838. Further, those measurements permitted to estimate the distance between those stars and the Solar System to be on the order of many trillions of kilometers, or several light-years.


Why does moon moves to side if I change my position directly into it? I've setup this scenario on Stellarium where the moon is aligned with East direction. I wanted to see the change of altitude of the moon if I move straight into its direction (time of observation is always the same) but instead it keeps changing its direction to the left. I expected to keep on the same azimuth and only change altitude but no, and I don't understand the geometric reason of it, for example If I walk in a straight line towards a building it won't move right or left. On the equator works as I expected.

I can approximate the scenario in question at 2020-03-15 04:23 UT.
At location 1, the Moon is at alt=34.6°, az=90.0°.
At location 2, the Moon is at alt=41.2°, az=85.1°.

We see a shift in azimuth because the Earth's surface is not cylindrical.
You can get from location 1 to 2 by moving 721 km on a constant heading of 90°.
However, the 36°S parallel curves almost 5° to the right in that distance.

Moving the same distance constantly toward the moon, along a great circle, would put you about 29 km north of location 2 with final heading 85.3°.
Continuing around the world on the same great circle, you'd cross the equator at 39.6°E, heading 54.1°.

A view of the zenith showing the equatorial grid (blue) may help to illustrate the issue.
The left half of the prime vertical (green) is due east.
The longitude difference between locations 1 and 2 is equivalent to 0.53 hour of right ascension.


How co-planar and co-rotatory are the Milky Way and its constituents? All the planets in the solar system rotate roughly in a plane that intersects the sun.
Stars in the Milky Way rotate around its centre forming a rough planar disk.
Question
Are these planes all parallel? I.e. Do planets of all stars rotate in planes parallel to the plane of the galaxy? Do they rotate in the same direction?
If so, is it
(a) because the whole galaxy evolved from a spinning mass so it naturally turned out that way
(b) because, over time, the 'wheels' that were spinning on other axes were somehow dragged into alignment
(c) Nobody knows
(d) Something else

As far as we know exoplanetary systems have orbital planes that are randomly distributed with respect to the Galactic plane. Indeed our own ecliptic plane is quite inclined with respect to the Galactic plane - as one can readily tell from the fact that the planets are not always seen against a background of the Milky Way.
Further details and some references can be found in answers to the related questions Detection of exo-planets and Do the axes of rotation of most stars in the Milky Way align reasonably closely with the axis of galactic rotation?


Does binarity play a role in spectral fitting? Most of the spectroscopic surveys like SDSS classify stars into one pure spectral type.

However, a large fraction of stars are in binaries.

Can we know whether a spectral fitting is good enough? Is there something like $\chi^2$?

In other words, is it possible we can estimate the probability that a star is in a binary based on one spectrum?

Yes you can, to some extent.

If the binary has a short period, then the chances are that the two stars have a different radial velocity. In other words, there are two sets of spectral lines in the spectrum. A cross-correlation or direct fitting technique can then pick the two stars out.

If the two stars are unresolved and have indistinguishable velocities (true for the majority of binaries), then it is more difficult. A fit to a single spectral template may be poor (and you could judge this from the chi-squared of the fit), but there needs to be a significant difference in the spectra of the two stars in order for a two-star model to adequately separate them.

There is basically a trade-off. You want the stars to be of similar brightness but to have quite different spectra. This is difficult to arrange for main sequence stars and there is normally a sweet-spot where one star is a bit cooler and a bit dimmer, but can still be seen in the combined spectra. Other combinations are easier - a hot white dwarf and an M dwarf for example have very similar luminosity but very different spectra, such that each dominates in different parts of the spectrum. Others are harder - a red giant and a main sequence star of lower mass.


What would a 220km diameter asteroid impact do to Earth? I was looking at this chart,

which shows that an order of magnitude increase in asteroid impactor diameter roughly corresponds with two orders of magnitude increase in it's rarity.  So if the K-T Impactor at 10km diameter is a 100 million year rarity impact, then a 100km diameter impact has a 10 billion year rarity (on the order of the lifetime of our planet).  On a large time scale, orbital perturbation allows for some chaotic orbits, so it might not be vanishingly likely that one of the big asteroids eventually hit earth. The asteroid 16 Psyche orbit overlaps in distance to the sun with Ceres, so it could potentially get a natural gravity assist into another orbit.
What would happen to Earth if it got hit by the asteroid 16 Psyche at, say, 40 km/s in the Pacific off the coast of Japan?
I am specifically looking for an answer to one or more of the following questions: How big would the crater be?  How much kinetic energy would be released?  What portions of that would be transferred into heat, ejecta, and deformation?  Would there be a secondary antipodal mountain formation like the Caloris impact on Mercury?  Would it cause magma oceans?  Would life on earth survive to reclaim the surface before the Sun went red giant?

Using the impact effects calculator
The energy before impact (making a some reasonable assumptions) is $3.35 × 10^{27}$ Joules = $7.99 × 10^{11}$ MegaTons TNT, most of which would be deposited in the Earth.
The impact effects calculator doesn't discuss the formation of antipodal mountains (this is not a well-understood process) but it certainly seems possible that there could be major antipodal effects. The initial crater would fill with lava, but there would not be a planet-wide magma ocean.
Would life survive? perhaps, life is surprisingly resiliant.  This would cause mass extinction, and probably wipe out much or all of complex life. But "until the sun expands to a red giant" gives us millions of years for evolution to find a way. But any answer would have to be speculative.
The average interval between impacts of this size is longer than the Earth's age.


How would light from a blue or red star affect the way we see? How would things look if we saw them through the light of a blue or red star?
Would there be any changes?

I guess it would be influenced. Right now, our vision is already influenced by that massive blue sky above us. The human visual system corresponds to the large influx of blue light that gets scattered in the atmosphere. Having a red or a blue star would probably also have some influence, as would have a different kind of atmosphere (which is not mainly nitrogen and oxygen). The brain would probably compensate, but the rods in the eye would have a different composition.


If a hot Jupiter collides with the parent star, does it enhance or reduce the life of the star? Suppose a hot Jupiter is collided into parent star due to tidal force, does the life of the star becomes longer due to extra hydrogen, or becomes shorter due to extra mass?

The short answer is pela's comment: more mass equals shorter life.

For a bit of a longer explanation...

The only way the life of the star could be prolonged is if the new material can be transported to the core.  This requires convection.  Most stars are not fully convective, having distinct layers of either convective or radiative zones. The net effect of the collision in this case is to simply add more mass to the star as a whole while adding no new fuel to the core (or a burning shell); ergo, shorter lifespan.

But some stars do (or may) have fully convective stages, such as the main sequences of very small (aka red dwarf) stars and very large stars with enough metals for a dominant CNO cycle.  So in such a situation the material of the colliding body would be transported to (and from) the core, giving more fuel to burn.  However, the star still has the same, increased mass, regardless of where the "new mass" currently is inside the star, resulting in higher core temperature. The net effect of which is that the increase to the fusion rate exceeds the proportional mass increase.  So, shorter lifespan.

The basic rule here being that core temperatures are dictated by the gravity well of the star, which is dictated by the mass (and a few other GR contributions, but usually predominantly mass).  The more mass, the deeper the well. So adding more mass just means the core (and areas near it) are in a deeper gravity well, meaning the temperatures go up.  Stellar fusion reaction rates are approximately non-linear polynomials in the temperature; some go as far as being proportional to $T^{40}$, the 40-th power of the temperature!  It's this non-linearity which results in the fundamental relation of the short answer: more mass equals shorter life.  Increasing the mass by a given proportion increases the fusion rates by a (much) larger proportion.


Calculating object velocity at perihelion I would like to calculate the velocity of an asteroid orbiting around a star (Sun) at the perihelion of its orbit. I know the excentricity of the ellipse and its semimajor axis.

I have found that the vis-viva equation is used to calculate the velocity of an object on an elliptical orbit and that the perihelion is at distance r = a(1-e). However I (simply enough) cannot see how to mathematically combine these two pieces of information in order to get the velocity at the perihelion.

(I am not so much looking for just a formula but rather a proof/intuition regarding how to get from the vis-visa equation for velocity to a perihelion velocity equation)

The vis-viva equation is commonly written like this:

$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)$$.

For $r=a(1-e)$:

$$v = \sqrt{GM\left(\frac{2}{a(1-e)} - \frac{1}{a}\right)} = \sqrt{GM\frac{1}{a}\left(\frac{2}{1-e}-1\right)} = \sqrt{GM\frac{1}{a}\left(\frac{1+e}{1-e}\right)}$$.

The derivation of the vis-viva equation is not at all trivial and can be found here.

The product $GM$ is also called the standard gravitational parameter and for solar system bodies is often known more accurately than $G$ and $M$ separately. For the Sun $GM_☉$ is about 1.327E+20 m³ s⁻² which in different units is 1.327E+11 km³ s⁻² or about 1.0 AU³ year⁻².


Is it ever possible to see Earth's shadow on other planets? Is it ever possible to see Earth's shadow on other planets?
How would the shadow present, and what would be the magnification, equipment, or other conditions necessary to see it?
Is there any way of using such a phenomenon to help prove that the Earth is a spheroid?

First, the Moon is not a planet and therefore not applicable to this question.  Second, the inferior planets, specifically Mercury and Venus cannot have a shadow cast by the Earth because their orbits are inside those of Earth.  Notice I did not say they are between the Sun and the Earth, as they could be in superior conjunction with the Earth (opposite side of the Sun from Earth).

The superior planets (Mars out to Neptune) are candidates for the casting of shadow of the Earth by absence of light from the Sun, however we need to explain the difference between umbra (Latin:shadow) and penumbra. Let's first use the Moon as an example. The umbra (darkest part of the shadow) is the shadow cast by the Earth into space whereby an observer cannot see the Sun at all. This happens during a total Earthly eclipse of the Moon (not a lunar eclipse).  You would be in complete darkness on the Moon. If the Earth and the Moon were the same size, the umbra cast by the Earth would continue out into space past the Moon the same distance that the Moon is from the Earth. However, since the Earth's diameter is about 3.5 times that of the Moon, the umbra extends further, but not all the way to the nearest planet, Mars.  So the Earth's umbra will not even reach the nearest of planets.

However, the penumbra (Latin:partial shadow) in theory, extends out to infinity.  First, in order to have shadow, there must be a source of light.  So if you can "see" the light source, then there is also a penumbric shadow associated with that source anytime any matter comes between the observer and the source.  The term "see" in your question is the matter.  With the average unaided human eyeball, two of them in fact, there is not enough light sensitivity to detect a difference of penumbric shadow on a planet.  However, with a ultra sensitive telescope one might be able to detect changes in the intensity of light from one time moment to the next.  In fact, current research in finding planets orbiting nearby stars uses a similar method called the exoplanet transit or transit photometry method.  So, a very sensitive telescope could see the Earth's penumbric shadow when measured over a long time period, but the human eye is not sensitive enough.


Can a change in the density of interstellar and intergalactic quantum fields affect the Redshift of photons? Using our solar system's heliosphere as an analagy to what is happening in our galaxy and even the universe as a whole, can light redshift with changes in quantum fields?

The universe is constantly being bombarded by particles from stars, pulsars, quasars, black holes, etc..    Is it possible for these particles to interact with photons in a way that will cause light to redshift?

No.

Or at least such an effect has never been observed, neither in the locality of the Earth or in light detected from distant sources.

If a photon has an interaction with a quantum field (such as an electromagnetic field) this causes a scattering. Scattering would cause a blurring and dimming of distant sources. This is not observed.

Such an effect, if it existed, should also be present on more local scales. But there is no redshift related to distance within our galaxy. Nor is there any evidence of any shift, even tiny, in the photons emitted by other local sources

Thus such a proposal would require an unknown interaction of light, that causes no scattering, only affects light from outside the local group of galaxies, but is otherwise linear with distance. And one would have to explain such phenomena as the slowing of time in distant galaxies, (which an expanding universe model understands as the relativistic time dilation)

So in short: No, a change in the density of interstellar and intergalactic quantum fields does not cause the observed red-shift of photons.


Where does Love's number come from in $\delta m = \frac{1}{2}k_2M_2\lbrace\frac{R_1}{r_\star}\rbrace^3$, the formula for the mass of a tidal bulge? My question refers to page 165 of this paper. Specifically, how does one derive the equation $\delta m = \frac{1}{2}k_2M_2\lbrace\frac{R_1}{r_\star}\rbrace^3$ for the mass of a tidal bulge? A full derivation would be much appreciated, and also an explanation as to where Love's number $k_2$ comes from.

Let us begin with the basics. If an exterior perturber generates the potential W, then an originally spherical planet gets distorted. Owing to this distortion, its potential acquires an increment U which is usually called the additional tidal potential or, simply, the tidal potential.

Now let us fix a point on the surface of the planet. Let $\gamma$ be the angular separation between the planetocentric vectors aimed at this point and the perturber.

The perturbing potential W in that point can be expanded into a series over the Legendre polynomials $P_l(\cos\gamma)$. This is an almost trivial fact, see equations (9 - 11) in
https://link.springer.com/article/10.1007%2Fs10569-009-9204-7

https://arxiv.org/abs/0803.3299

In those equations, $r^*$ is the position of the perturber, while R is the surface point.

As it turns out, a degree-$l$ term in that equation contains a factor of $(R/r^*)^{l}$.

The additional tidal potential U in that point can be expanded over the same $P_l(\cos\gamma)$.

Moreover, you can write down the potential U in a point r located right above R (i.e., having the same angular separation from $r^*$ as R has).  This expansion is rendered by equation (19). In that equation, you need only the first line. Ignore the rest for now.

As we see from this formula, the degree-$l$ term of U is equal to the degree-$l$ term of W multiplied by $(R/r)^{l + 1}$ and by an additional factor $k_{l}$ named the degree-$l$ Love number.

Derivation of this formula is boring. It is a solution to a boundary-value problem for an elastic medium, and the factors $k_{l}$ emerge as some functions of the shear elastic modulus of the material of which the planet is composed.

This is how it is for static tides. In real life we, of course, have tides evolving in time, and all this machinery has to be generalised. We have to expand both W and U into Fourier harmonics and to write down all this formalism for each harmonic separately.
Then you get different lags for different harmonics involved. Also, the dynamical analogues of the Love numbers will be functions of other rheological parameters (like the viscosity).  However, the central idea will be more or less the same.  I can give you more references, if you are interested.

Jérémy Leconte in his work tried to substitute the degree-2 part of U with an equivalent potential generated by two equal masses located oppositely. You can think of this as two tidal bulges approximated with point masses. This was done for illustrative purposes solely and, in my opinion, bears no practical value.


Apparent Centering Problem When Using the Perifocal System to Draw Orbital Ellipses I am attempting to render orbital ellipses in software.  Given a set of orbital elements, I am using the Perifocal system to determine a set of points along the curve of the ellipse.  I am using the equation:
r = r cos(v) P + r sin(v) Q,
where v is the angle, P and Q are the perifocal vectors correspondent to I and J, and r is the polar equation of the conic:
r = p / 1 + e cos(v),
where p is the semi-latus rectum, e is eccentricity, and v, in both equations, is the angle.
I am determining the vectors P and Q using the following equations:
Pi =   cos (lan) cos (aop) - sin (lan) cos (inc) sin (aop)
Pj =   sin (lan) cos (aop) + cos (lan) cos (inc) sin (aop)
Pk =   sin (inc) cos (aop)

Qi = - cos (lan) sin (aop) - sin (lan) cos (inc) cos (aop)
Qj = - sin (lan) sin (aop) + cos (lan) cos (inc) cos (aop)
Qk =   sin (inc) cos (aop)

where lan is the Longitude of Ascending Node, aop is the Argument of Periapsis, and inc is the inclination.
This process yields the correct ellipse!  However, the ellipse does not seem to be centered correctly on its focus; the focus appears to be far too close to the center of the ellipse.  Here is an example:

You can see that these two highly eccentric ellipses have their shared center of mass at their centers - not at their respective foci.
Is there some step in my process that I'm missing?  I could just apply an offset to the ellipses, but that would be a hack, I want to solve the problem.
Any help is appreciated!  Thank you!
QUESTION EDITS AND CLARIFICATIONS:
This image is taken from a vantage point nearly normal to the orbital plane; there is almost no perspective here.
Second, a helpful user has pointed out an issue in the equation for Pk. The correct expression is:
Pk = sin (inc) sin (aop)
Unfortunately, this was correct in my code - so while the correction was a good one, it was not the source of my problems.

I solved the issue.
Everything I posted in the OP was correct; I made a simple, stupid sign error when transcribing the expressions into code.
It should also be noted that in the OP, the expression for Pk is wrong.  The correct expression is
Pk =   sin (inc) sin (aop)
Just in case anybody is trying to pull this off themselves!
Thanks for the ideas everyone!
Correct Orbits:


Will an observer falling into a black hole be able to witness any future events in the universe outside the black hole? I know someone or something sucked into a black hole could not survive, but is the bending of light so immense that an observer falling into a black hole will be able to witness any future events in the universe? That would definitely be cool :D I have a feeling this won't work though... so could you explain in detail why? Thanks.

The confusion about finite or infinite time dilation at the Schwarzschild radius began since Karl Schwarzschild in 1915 used a space-time coordinate system where coordinate time went to +∞ when a falling object approached the Schwarzschild radius. But t → +∞ for a world line that approaches the horizon is a thing different from t → +∞ far from the black hole. An object that is about to cross the horizon is in no position to synchronize clocks with an outside world.

No, completely opposite situation to the OP’s expectations. If a falling observer will transmit a powerful beam of light out of the hole, then photons will reach increasingly far future of the exterior universe. But the future seen by the falling observer will be very limited; see http://jila.colorado.edu/~ajsh/insidebh/penrose_schw.html for graphic depiction.


Is there evidence of super-heavy elements in the x-ray and gamma-ray spectrums of neutron stars? It occurs to me, that between the surface and interior of neutron stars, gravitational pressure might produce super-heavy elements and that signatures of such elements might be detectable in the star's spectrum.

The crusts of neutron stars will contain "super-heavy", neutron-rich nuclei. This is an inevitable consequence of the high density material, the accompanying degenerate electrons (that block $\beta$-decay) and what we know about nuclear physics.

However, the only things that contribute to a neutron star's observable spectrum are materials within a few cm of the neutron star surface. It is expected that this will be made of highly ionised "iron-peak" nuclei (Fe, Co, Mn etc.) and perhaps hydrogen and helium that has been accreted from the interstellar medium.

The "super-heavy elements" won't be present until the densities are much higher than in the surface layers. In a low-density material, these exotic nuclei will simply undergo a series of $\alpha$ and $\beta$ decays or even fission. In high density materials, the electron Fermi energy can exceed the maximum possible decay energy of any beta electrons and stymie the decay process.

e.g. For a surface density of order $10^{9}$ kg/m$^{3}$ (at which point nothing very exotic is created), the electron number density will be about $3\times 10^{35}$ m$^{-3}$.
Soft X-rays emitted at $10^{6}$K, will undergo Thomson scattering with a mean free path of $1/n_e\sigma \sim 50$ nm. No (detailed) information about the interior can emerge in such circumstances.


How to calculate geocentric conjunction time and moon altitude at a given time Umm al-Qura calendar of Saudi Arabia determine the new moon if these criteria meet at 29th day:

The geocentric conjunction occurs before sunset.
The moon sets after the sun.

I've got the formula to calculate when the sun (seen by the observer) sets, considering observer's latitude, longitude, height (altitude from sea level), and the date (Julian Day). I would like to ask the formula to calculate:

When (the time) geocentric conjunction happens.
How much (the angle) the altitude of the moon (seen by the observer) at a given time is.

Thank you

I assume you have an algorithm to compute position of the Sun, since you can compute rise and set times.  You will need to compute the position of the Moon in a similar manner.  Then, convert both positions to geocentric ecliptical coordinates (ecliptic latitude and longitude) and use some numerical bisection method (root finder) to find where the difference of the longitudes of the two objects has a root (i.e., becomes zero).  The corresponding JD provides the moment of the geocentric conjunction.
Calculation of the altitude is a standard procedure - you'll need to convert your ecliptic coordinates to equatorial (right ascension and declination) and then to horizontal (azimuth and altitude).  Finally, correct for the horizontal parallax to convert your geocentric coordinates to topocentric coordinates for the location of the observer.


How to get the solar zenith from the longitude and latitude? I have real data of “Direct Normal Irradiation” and “Diffuse Horizontal Irradiation” and I should calculate “Global Horizontal Irradiation”. I have also longitude, latitude and elevation.
Could you advise me how I can get zenith from these data, then to calculate the GHI?
Many thanks,

Direct Normal Irradiance (DNI) is the the amount of solar radiation from the direction of the Sun.

Diffuse Horizontal Irradiance (DHI) is the amount of solar radiation that does not arrive on a direct path from the Sun.

Global Horizontal Irradiance (GHI) is the total amount of solar radiation. It is the sum of Direct Normal Irradiance (DNI), Diffuse Horizontal Irradiance (DHI) and ground-reflected radiation. Ground-reflected radiation is usually insignificant (compared to DNI and DHI) so for all practical purposes GHI is the sum of DNI and DHI.

Thus $GHI = DHI + DNI \times cos (Z)$ where $Z$ is the solar zenith angle.

Solar zenith angle is derived from latitude.

Solar zenith occurs at local solar noon, which is derived from longitude.


Is it safe to view a solar eclipse with eyeglasses along with eclipse glasses? In an article on Ars Technica about safely viewing a solar eclipse, they say this:

  First, what not to do: Don’t view the eclipse with your naked eye or unfiltered telescopes, binoculars, sunglasses (yeah, even if they’re really dark), camera lenses, or other optics devices. Don’t use anything that focuses light, even if you’re wearing eclipse glasses.

This may sound like a stupid question, but are eyeglasses (by which I mean, lenses to correct poor vision) safe to wear during an eclipse along with eclipse glasses? I don't know if the light focusing is strong enough to be concerned about.

Yes, it's safe :).  Your eyeglasses are just correcting your vision. The net effect is just to give the equivalent of normal vision - and in any case, you'd normally be holding the eclipse glasses in front of your eyeglasses, in the same way that front-of-scope solar filters work to reduce the incoming light to a safe level.

Even if you're longsighted (which uses converging (positive) lenses to correct your vision, the glasses are close enough to your eye that you're nowhere near the focal point - so the light doesn't get concentrated much (and probably gets unconcentrated again by your eye, anyway).

If you're shortsighted, like me, your glasses will be diverging (negative) lenses which spread the light out rather than concentrating it (and again, are close enough to your eye that's there's not enough distance for it to affect things much.


Dish antenna as parabolic mirror for OPTICAL telescope? Can I use a chrome painted dish TV antenna as a parabolic reflector for my optical telescope? Given that glass mirrors are quite expensive, I am thinking of using a Dish TV antenna and coating its reflecting surface with chrome paint to give it a mirror finish. But I couldn't find any instance of it online. I did find several examples of people using a dish antenna for making an amateur radio telescope.

I was wondering if there is a reason why we can't use it in optical telescope.

It's not possible I'm afraid.

Optical wavelengths (light) are typically of a wavelength under a micron, and an optical surface needs to be accurate to this level or better to be useful.

Radio wavelengths are typically 10-20cm or longer, and an adequate reflector can be made with surfaces accurate to a few centimetres (at a guess as I don't know the exact tolerance).

Also - I once experimented using a frequently mentioned substitute, a convex shaving mirror(!) as a reflecting telescope.  Even that was so poor it was unusable at the lowest possible magnification.

(But it is possible to make your own reflector out of reasonably stable glass - it just won't be as big as a satellite dish! Search around for "Amateur telescope making".)


My local night sky is so bright that I can only see the Moon, so would a low-cost telescope be any use? Light pollution is so bad near our home that we can't see anything other the Moon in the night sky. For instance there is a street light on all night right in our lane and there are many other housing lights in the area.
Am I right to think that with a low-end telescope I won't be able to see anything else besides the Moon?

There is hope!
You may be able to see some stars through a telescope even though you can't see them with your eye.
For extended objects like sky brightness, nebulae, and the Moon and resolved planets, magnification reduces sky brightness, while aperture increases it.
For unresolved (point-like) objects like stars and unresolved planets, as long as focus is good magnification does not reduce its brightness but aperture still increases it.
However
If you can't see stars or planets by eye, it is going to be hard to point your low-end telescope at them, and searching randomly has a low probability of success.
In this case
I recommend first getting your hands on a pair of binoculars with low magnification (see if you can borrow some somehow, don't buy them yet!), something you can hold in your hands comfortably. Then get a star chart or use a website like in-the-sky.org and going to the planetarium page to view a simulated night sky, and find a bright star or planet and note where it is relative to the horizon. You might use a map or compass to find out where your North, South, East and West are so you can compare the sky map to your local surroundings.
Once you are able to find stars or planets this way, you can ask yourself if it's worth it to get a telescope and go through the same exercise.
You can also ask yourself if a pair of binoculars might be a better investment.
Also, please read answers to I live in an area with a lot of light pollution, how can I view the stars without building an observatory? for some more advanced options.


How are we identifying chemicals in space? How do we identify chemicals in space?   For example, how did we find the cloud of methyl alcohol (aka methanol) in W3 (OH) which is 6500 light years away.  I can understand a scope being able to tell something is there, but I don't see how they determine that it is not only alcohol but the type of alcohol.
Here is a link to the W3(OH):  https://phys.org/news/2014-09-alcohol-clouds-space.html
But I am looking for a general answer - as to how we identify chemicals, any chemicals, in space.

The science that studies the problem you described, is called "spectroscopy". Every element absorbs some of the wavelengths. Same is with molecules. The spectrum is like the footprint of the molecule.
When you observe the light from emitted from some object, you see the continuous spectrum, emited from the light sources (stars, etc.). But it has some black lines present. These are the wavelengths that are absorbed by the molecules and elements. Then you compare it with different spectrums on the Earth (where we can identify the substance) and when you find the same spectrum, you know with a extremely high probability that this is the substance you are searching for.
Example for methanol: Astronomers observed the object with spectroscopy and found the spectrum with some black lines present. They compared it with various other spectra on the Earth and found out, that methanol has same spectral lines. Both measurments can be showed with a graph:
https://webbook.nist.gov/cgi/cbook.cgi?Spec=C67561&Index=1&Type=IR
Spectral lines are normally showed with a spectrum (with colors) and some black lines, but this graph shows the dark lines more physically (not all is absorbed) on a graph as deep valleys.
Be aware! There may be redshift present in distant objects. That means, that the lines would shift to the red part of the spectrum. (Similar to Doppler effect.)


What is so special about the Ultima Thule passed by New Horizon spacecraft? I read that the New Horizon spacecraft passed by the Ultima Thule. I also got to know that it is a flat, snowman-shaped rock.

What is so special about it?

Ultima Thule is the first small Kuiper Belt Object (KBO) we have visited. For the first time, we can see a KBO as more than a tiny point of light in a telescope.
it seems to be flat, rather than roughly round. This is really unusual (we've never seen that shape before in planets, moons, asteroids etc.).


How entropy increases over time According to 2nd law of thermodynamic,  in an isolated system entropy increases over tume. In an open system entropy might seems decreases but combined result of all related events results in overall entropy increase.

Now lets say, a suffled deck of playing cards. If I sort the cards it seems entropy decreases, what's then the related  events that increases overall entropy. I am asking just to clear some of my confusions.

While more of a physics than astronomy question. Entropy as a whole increase in the universe due to statistical probability. By rearranging a deck of cards looking at the back of them, no matter what order, it is the same macrostate with different arrangements of the cards (microstates). A better example would be if you decreased entropy by picking up cards from the floor and put them together. In that situation, while in your small area the cards lost entropy, you expended energy to do that task. The energy you expend is enough to make the entropy of you and the cards remain constant and arguably increase. However, if you then take into account the entire universe, the rest of it is increasing in entropy at a large rate anyway. This means no matter what you do, entropy will increase overall due to other effects.


Difficulty in using the telescope as a projector I have been trying, but completely failing, to copy the arrangement for viewing the sun mentioned on this website:  https://astronomyconnect.com/forums/articles/2-three-ways-to-safely-observe-the-sun.21/

The telescope shown does not appear to have a finderscope or any other obvious means of finding/tracking the sun.  This set up is actually far harder to do than it would appear in the photo.  However, I think it is worth the effort as it doesn't risk blinding the observer.
Finding the sun is not our issue because our telescope moves.  We punch in what we want to look at and the telescope moves there.   We can use it manually, but have found that the sun is surprisingly difficult to find without looking through the finderscope.   We are not willing to do that for obvious reasons.
I have been trying to figure out a way to have the board portion grab hold of our telescope instead of set on a table.  So when the telescope moves, the board moves with it.  Orion does sell a sun-filter, which I have purchased, but I am not willing to set a precedent with my homeschoolers where they look at the sun through the telescope - even with a sun filter.
Does anyone know of another website with instructions to use the telescope as a projector that uses a telescope similar to mine?

Yes this is a bit more of a challenge! The first scope in the photo is long and thin and before the big square cardboard shadow-maker is put on it can be easily pointed at the Sun by moving the telescope until its shadow becomes a small circle, the minimum outline of the scope. If it's off a little bit that long tube's shadow will immediately change shape.
But with a short squat form factor your telescope won't be so easy. Start with the smallest magnification largest focal length eyepiece and at least try to use the shadow method of looking at the shadow that your telescope makes behind it (on something flat and held perpendicular to the Sun's direction, and move the scope around trying to make the most circular profile shadow. For this method, instead of a finder scope you can use an empty long thin tube to make a shadow.
Wait, you have a finder scope!
Eyepiece projection will work just as nicely for your finder scope as your main scope, so BEFORE you use it for the Sun, check that the finder is aligned to the main scope EITHER AT NIGHT or using TERESTRIAL OBJECTS on the ground. DON'T POINT IT AT THE SUN if there's any chance that you will look through it.
Those have to be distant objects so parallax doesn't lead to misalignment.
Once your finder scope is aligned to your main telescope, use eyepiece projection to get a nice centered image from it. At that point your main scope should be really close.


Astronomical ABCD...! Today, I thought to make a list of Astronomical objects, A to Z.

It goes as follows :

A - Asteroid
B - Black Hole
C - Comet
D - Dwarf Planet
E - Elliptical Galaxy
F - Falling Star
G - Globular Cluster
H - Hypergiants (Stars)
I - Interstellar Medium
J - Jets (Astrophysical)
K -  $\color{red} ?$
L - Lenticular Galaxy
M - Meteor
N - Nebula
O - Open Cluster
P - Pulsar
Q - Quasar
R - Rogue Planets
S - Satellites
T -   $\color{red} ?$
U - Universe
V - Variable Star
W - White Dwarf
X - X-ray Star
Y - Yellow dwarf star
Z -    $\color{red} ?$

But I couldn't think of something well-known from the letters K, T and Z.

I tried to avoid using name of something specific object for ex. K - Kuiper Belt.

Can you please help me in completing this list?

Thanks!

Seems a little contrived, essentially you just have to dig deep enough to find a specific enough category that matches the letter. But here you go :

K - KBO (Kuiper Belt Object) , if you still don't like that you could use

K - Kreutz Sungrazer (A type of comet, https://en.wikipedia.org/wiki/Kreutz_sungrazer)

T - T-Tauri Star (A type of variable star, https://en.wikipedia.org/wiki/T_Tauri_star)

Z - Z Camelopardalis star (https://en.wikipedia.org/wiki/Z_Camelopardalis_star redirects to Dwarf Nova, scroll down)


IAU Dwarf planet definition Point 2.b - is not a satellite I am a bit confused with point 2.d of the IAU's resolution on dwarf planets.

(2) A "dwarf planet" is a celestial body that
(a) is in orbit around the Sun
(b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape
(c) has not cleared the neighbourhood around its orbit
(d) is not a satellite.

http://www.iau.org/static/resolutions/Resolution_GA26-5-6.pdf
I notice that a lot of websites exclude this point all together? Some of
them even rather combine it with point 2.a ? Like over here
Feels a bit redundant, since orbits the sun implies not orbiting a planet?
Or am I missing something here?

Or am I missing something here?

Yes. You are missing the word "Sun". The IAU did not want to tackle the problem of how to categorize exoplanets. There are only eight planets in the entire universe, plus a good number of (five and counting) dwarf planets. An expoplanet orbiting Alpha Centauri is not a planet or a dwarf planet, per the IAU definition. Some hypothesize that the early solar system contained five giant planets, but one of them was expelled. That rogue giant planet, if it exists, is not a planet per the IAU definition.

You are treating the word "orbit" as if it's a mutually exclusive term. It's not. The Earth's Moon, for example, orbits the Earth and it orbits the Sun. But since the Moon is a satellite of the Earth, it's not a planet or a dwarf planet. Side note: the concept of what distinguishes a "satellite" from a non-satellite is not defined.


Which star / galaxy is moving away from us the fastest? I know that we have measured the rate a lot of stars and galaxies move away from us using the doppler shift, and I know that the further a star / galaxy is the faster they accelerate away from us due to space inflation. I'm wondering, did anyone categorise which star / galaxy is moving towards us / away from us the fastest? How fast are they going in relation to our solar system?

When a galaxy recedes from us, the light we see from it is redshifted. For galaxies at cosmological distances, this redshift is fundamentally different from a Doppler shift; whereas the latter is due to a velocity difference between the emitter and the receiver, a cosmological redshift is due to photons traveling through an expanding space$^\dagger$.
Hence, as @uhoh comments, your question is equivalent to asking "Which galaxy has the highest measured redshift?". Redshift is arguably the single most important concept in astronomy, and indeed we do catalogue the redshift of all galaxies, if possible. For our adopted cosmological model, the cosmological redshift can be translated to both a recession velocity, a distance, and an age of the Universe when the light was emitted.
The answer is GN-z11 (Oesch et al. 2016) which has a redshift of $z=11.09$. This corresponds to a distance of $d = 32.2\,\mathrm{Glyr}$ (i.e. billion light-years), and hence, by Hubble's law, to a recession velocity of
$$
v = H_0\,d = 670\,000\,\mathrm{km}\,\mathrm{s}^{-1},
$$
or more than twice the speed of light. Furthermore, the light we see today was emitted when the Universe was only 410 Myr (i.e. million years), or 3% of its current age.
You may think that "twice the speed of light" violates the theory of relativity, but this velocity is not a velocity through space. Both our galaxy (the Milky Way) and GN-z11 move through space at modest velocities of a few 100 km/s. The recession is merely due to space expanding, and space is allowed to expand at whatever rate is wishes.

$^\dagger$"Fundamentally different" might be a too strong statement, since the cosmological redshift can be interpreted as infinitely many infinitesimally small Doppler shifts. However, a hypothetical scenario that emphasizes the difference between the two types of redshifts is the following:
If an emitter and an observer are stationary wrt. each other when the emitter emits a photon, then start moving away from each other while the photon is traveling, then stop again before the observer receives the photon, the observer would measure zero redshift.
On the other hand, if space is static when the emitter emits the photon, then while the photon is traveling suddenly expands by a factor of, say, four, and then is static again when the observer receives the photon, the observer would measure a redshift of $z+1 = 4$.


Can two neighboring galaxies move apart at steady speed? While I was trying to understand the three models that obey Friedmann's two assumptions of a non-static universe, I came across a line that says and I quote "It (referring to Big Bang) starts at time zero and eventually the galaxies are moving apart at steady speed." The question is - is it possible for two massive space bodies to move apart at a 'constant' speed? What do you say on account of the gravitational forces acting between the galaxies? And does it mean the universe will only expand and never come to rest? P.S. I am assuming 'steady' to mean 'constant'.

The key word here is "eventually."
The strength of the gravitation felt between two objects is inversely proportional to the square of the distance between the two bodies. As a result of this, raising an object from a nonzero distance from object A to an infinite distance from object A requires a finite amount of energy.
In a Keplerian/Newtonian trajectory, the velocity of an object solely under the influence of gravity is governed by an exchange between gravitational potential energy, and kinetic energy.  As a result of that, no matter how long their mutual gravity acts, there's a limit to how much kinetic energy gravitation can take from the object.
If the equations are balanced such that the object would come to rest at an infinite distance, the speed that the object has is called "Escape Velocity."
If the object is moving faster than escape velocity, it would wind up with a finite relative velocity at an infinite distance, and this value is called Hyperbolic Excess Velocity.
For instance, if the Sun and Voyager I were the only objects in the universe, and there were no other forces between them other than mutual gravity, the relative velocity between the two of them would never fall below 16.6 km/s.
A similar case holds for any two objects, regardless of mass; If they're moving relative to each other higher than their mutual escape velocity, considering only the  two bodies, and only the gravitation between them they wind up asymptotically approaching a finite relative velocity.


Is there a place in the known universe that there would be no visible light from other stars/galaxies? Is there a place in the universe where, with the naked eye, you wouldn't see any stars, galaxies or light emitting phenomena? From what I understand at some point into the far future in Earth's location, we wouldn't be able to see anything. Is there a place like that could be like that now?

p.s. Not sure what the appropriate tags to put on this question. Feel free to add/change tags as needed.

If you don't want to look at the stars and galaxies, you can do two things: You can travel so far that they're too far away to see, or you can block your vision.

Go far away

On large scales, i.e. above a billion lightyears or so, the Universe is observed to be roughly homogeneous. On smaller scales, however, matter is distributed in the so-called cosmic web, where galaxies lie in filaments and sheet, meeting each other in knots (where the largest clusters are found). In between these overdensities, there are underdensities known as voids. These voids are virtually free of any luminous matter, and since they're up to several million lightyears across, you could place yourself in the middle of such a void and only see darkness.

As an example, consider a void of 100 Mpc (~300 million lightyears) in radius, corresponding to a distance modulus1 of $\mu = 35$. If a Milky Way-like galaxy (absolute magnitude2 $M=-20.5$) were at the border of the void, its apparent magnitude3 would be $m = M+\mu = 14.5$, invisible to the naked eye.

A small telescope, however, would enable you to see galaxies from here.
The limiting magnitude4 of the naked eye is roughly 6–7 (though some claim to be able to see $m=8$ objects), so the Milky Way-like galaxy would be roughly 8 magnitudes too faint to see. The gain in limiting magnitude when using a telescope is roughly $g = 5\log(D_\mathrm{tel}/D_\mathrm{pupil})$, so with your 6 mm pupil, a telescope with a diameter of
$$
D_\mathrm{tel} = D_\mathrm{pupil} 10^{g/5} \\
= 6\,\mathrm{mm}\times10^{8/5} \\
\simeq 200\mathrm{-}250\,\mathrm{mm},
$$
would enable you to see Milky Way-like galaxies from the center of the void.

Note though that no matter how far you are from galaxies, there will always be some radiation around, if nothing else then at least the CMB. Of course, being microwaves, this is not visible to the naked eye.

Block your vision

Apart from going down in your basement and turn off the light, could there be "astronomical" places where you would be unable to see anything? As David Hammen comments, at the surface of a cloud-enshrouded planet or moon, you could face total darkness (at least on the night-side of the planet). But perhaps you could also, as Wayfaring Stranger comments, go inside a dense, interstellar cloud.

Bok globules are small ($R\sim10^4\,\mathrm{AU}$), dense ($n\sim10^{4\mathrm{-}6}\,\mathrm{cm}^{-3}$) nebulae of gas and dust. A sightline from the center of such a cloud would result in the apparent magnitude being extincted (i.e. increased) by up to several tens in visual light. The largest extinction through the center of a Bok globule that I've been able to find — but I'm not expert on this — is "FeSt 1-457" with a visual extinction of $A_V = 41$ (Kandori et al. 2005), so from the center and out is roughly $A_V\simeq20$. That means that the fraction of light from outside sources that make it into the center is
$$
f = 10^{-A_V/2.5} \sim 10^{-8},
$$
which is not a lot. However, Bok globules are found in the vicinity of young stars which tend to be bright. An O star has an absolute magnitude of $M\simeq-4$. Such a star located right outside the cloud (say, at a distance of $d=2\times10^4\,\mathrm{AU}=0.1\,\mathrm{pc}$) would have an apparent magnitude of $m=M+\mu+A_V\simeq6$, and thus be just visible to the naked eye, albeit very faint. And if it were just a little farther away, you wouldn't be able to see it.

I assume that you're human, but if you are in fact a cichlid, you should be able to see infrared light (Meuthen et al. 2012). In the IR, the extinction is much smaller than in the visible, so hiding in a Bok globule won't help you.

1The distance modulus $\mu$ is a logarithmic way of expressing distances: $\mu \equiv 5\log(d/10\mathrm{pc})$.

2The absolute magnitude $M$ is a (logarithmic) measure of the luminosity $L$ of an object: $M=-2.5\log(L)+\mathrm{constant}$. Note the minus sign; the brighter an object is, the smaller $M$ is.

3The apparent magnitude $m$ is a logarithmic measure of how bright an object looks to an observer, and so depends on the distance to the object: $m = M + \mu$. Stars visible to the naked have roughly $m=0$–$6$. Again, the smaller the number, the brighter the object is. An extremely distant galaxy may have $m=30$, while the Sun has $m=-27$.

4The limiting magnitude is the largest magnitude (i.e. faintest object) visible.


Could Planet Nine be a tiny black hole or an exotic compact object such that it escaped detection? What would its bounds be? Planet Nine (a variant of the old X-Planet and Tyche-Thelistos hypothesis) proposes a super-earth planet at the edges of the solar system with a mass of around ten Earths. However, given the interest in exotic objects and gravitational variants, maybe it's interesting to study "exotic" scenarios:

Tiny black hole. Using the famous online calculator http://xaonon.dyndns.org/hawking/, for ten earth masses, we get a 89 mm black hole. Temperature about 2 mK! How can we put bounds on this scenario?
An exotic compact object. Of course, we have no evidence so far of general relativity modifications of gravity, but if not a black hole but another "thing" (a gravastar, a quark star, a preon star,...), I find hard to estimate how could we be sure to give up experimentally these examples.

Of course, the real thing is to think something less exotic (I am not sure how to differentiate the P9 hypothesis from a swarm of thousands of ETNO dwarf planet-like or planetesimal-like, but the real issue is how to test these alternative models and if these could affect what we know about the origin and formation of the solar system and its actual current structure, any hint to this in the answer to above exotic cases would also welcome).

I think the answer is included in my answer to Could a black hole pass quiescently through the Oort cloud? The mini black hole would be detected via its accretion from the interplanetary medium, probably at X-ray wavelengths.

A ten Earth mass black hole is a ten thousandth (or so) less massive than the Sun, but if it were in a distant orbit, it could be moving slower with respect to the accreting material than the 10 km/s I assumed in the answer cited above.

The net outcome is that this might still be an X-ray source well above the detection limits of all-sky surveys. What's more, it would move with time, so if followed up would show appreciable proper motion.

The variables here are the distance to the black hole and the speed at which is travels with respect to any medium it is travelling through.


Is gravitational wave frequency always equal to double the orbital frequency? If the binary does not evolve into merger stage (i.e.: it is still a steady binary), does the gravitational wave frequency have to be 2*orbital frequency? Could the frequency be, for instance, 2/3*orbital frequency and steady, rather than a chirp? And if not, could the frequency 2/3*orbital frequency be the result of interactions with other waves of this binary? The frequency range is about several milihertz, low frequency.

The gravitational waves (GWs) from a perfectly stable, circular binary system are completely dominated by GWs at twice the orbital frequency.

If the binary orbit is eccentric then the GWs have a frequency spectrum; the waves are a combination of discrete frequencies at integer multiples of the orbital frequency (e.g. Wen 2003).
At low eccentricity, most of the power is still at twice the orbital frequency, but for eccentricities greater than about 0.3, the peak frequency moves to higher and higher multiples.

Despite mmeents objections, I believe what I've written is (approximately) true for the circumstances posed in the question - i.e. a non-evolving binary, where the GW energy losses are small, the binary components are thus widely separated and can be treated as point-like, and the rate of periastron precession is small compared with the orbital frequency (note that if the eccentricity were really high, $e \sim 1$, then the latter may not be true). Closer binaries may have significant higher order multipole emission and modifications to their frequency spectrum caused by the decay of the orbit, precession of their periastron, their spin etc.


Where can I find a set of data of the initial conditions of our solar system? As the title suggests, I'm in need of a set of data of our solar system. Similar to this http://bima.astro.umd.edu/nemo/archive/#iau25, but more specifically, the Dubinsky Milky Way-Andromeda data.

I'm creating a n-body simulation for school and can't seem the find any initial particle conditions that I could use to simulate our solar system in the software I'm developing. I need initial positions, velocity and mass.

Any idea where I could find this?

Horizons Ephemeris generator can give you a planet's position and velocity vectors at a specified time.

This is one set of possible options:

Clicking Generate Ephemeris on that page will give you position and velocity vectors:

Above the position and velocity vectors are the Julian date as well as the more conventional date.

Wikipedia can give the masses of the sun and planets.


Do neutron stars shine? Ignore pulsars and the like, just a neutron star in empty space – does it emit light and neutrinos?

Neutron stars do indeed shine and some like this one have been observed in visible light. Like white dwarf stars, if they were scaled up to be the size of the Sun, they would outshine the Sun and most other stars in the galaxy. It is only their small size that make them hard to observe.

As for neutrinos, it seems they do too. This link might help explain neutrino activity in / from neutron stars further.


Why are tides biggest during the equinoxes? I've read that during the equinoxes, tides are at their maximum. Shouldn't this only apply to latitudes near the equator? I mean, when the solar declination is near to 0º and we have a full or new moon, the combined effect of the Sun's and Moon's gravity should create the highest tides, but only near the equator (see in wikipedia the article Tides, which says that equinoctial tides are maximum)

Is this true for other latitudes?

Only a syzygial tide during/near an equinox is the strongest. This means, there must be either new or full moon. In general, syzygial tides are strong because three bodies (Earth, Moon, Sun) align near one line, and tidal effects of Moon and Sun on Earth become (nearly) collinear and sum to the maximal possible magnitude. The Moon’s orbit is inclined to the ecliptic by about 5° only (cos 5° ≈ 0.996), so two tides are aligned almost perfectly on a Moon’s syzygy during whichever season. The difference is that, during/near an equinox, this line also lies in the equatorial plane and rotational motion of Earth’s surface/hydrosphere/crust can direct the stuff to move along this tidal line to a maximal range possible. At least, on the equator.

As for variations in terrestrial latitudes, namely mid-latitudes vs equator, comparison between syzygial tides during solstice and equinox far from the equator is a mechanical, not astronomical problem. Tidal bulge should be stronger during solstice, but projection of velocity to the tidal line will be greater on equinox. A detailed analysis of particular body of liquid (and its shore) is required to compare.

In contrast, a quadratural (where Earth–Moon and Earth–Sun lines are perpendicular) tide during an equinox is as weak as any quadratural tide. The solar tide basically cancels a part of the lunar tide. A quadratural tide during an equinox might even be the weakest, since lunar tide outmatches the solar one, but on an equinox and Moon’s quadrature simultaneously the Moon definitely lies away of the equatorial plane.


Is there available space images taken by the south pole Gregorian telescope? I searched the internet for images taken by the South Pole Telescope but without successes in finding it.

Edit:Some images

The South Pole Telescope (SPT) was funded, in part, by the NSF. This means their data must be made public as that is almost always a condition of funding by the NSF.

If you go to the SPT's homepage, they link to their Data Management Policy. Within this policy they state:

  The collaboration – in particular the P.I. and data managers at the University of Chicago – bear the responsibility for making data public. Public data products are hosted on University of Chicago machines and on the NASA Legacy Archive for Microwave Background Data Analysis (LAMBDA) server.

I went over to NASA's LAMBDA server and found that SPT data was hosted right here. Look to the left-hand side menu to see the links to raw data for the SPT-SZ and SPTpol instruments, individually.

The University of Chicago hosts their data right here (props to called2voyage). It seems that they provide publications and include any data used in that publication rather than just hosting all the RAW data like the LAMBDA server does. Use whichever is most convenient for you.


What is a 'hybrid' eclipse? As mentioned on the NASA eclipse website the eclipse on the 3rd of November 2013 will be a "Hybrid" eclipse.

I have looked at the diagram on that page but can't see what is different about the eclipse that would explain the word hybrid. This terminology also confuses me:

  It is one of the rare hybrid or annular/total eclipses in which some sections of the path are annular while other parts are total.

A hybrid eclipse is one in which it is a total eclipse in some places and an annular eclipse is seen in other places.

As you may know, a total eclipse is one in which the moon completely covers the disc of the Sun; this allows the solar corona to be seen.  Usually the central disc of the Sun is so bright that the corona cannot be seen.  An annular eclipse is one in which the Moon is not quite "large" enough to completely block out the Sun - instead, a ring of the Sun's central disc can be seen around the Moon. This ring is too bright for the corona to be seen.  Photos of both eclipse types appear below.

Annular Eclipse (credit: Fred Espenak at www.mreclipse.com):

Total Eclipse (credit: Fred Espenak at www.mreclipse.com):

The type of eclipse is determined by the geometry of the Sun, Earth, and Moon.  Since the Sun and Moon are not constantly at the same distance from the Earth, their apparent sizes will differ. When the Moon is closer to the Earth it appears larger, and when the Sun is farther away it will appear smaller.  For this particular eclipse, the geometry is such that the western part of the eclipse's center line will see an annular eclipse, but the rest will have a total eclipse.


What is difference between horizon and comoving distances? I wonder if there a difference between horizon and comoving distances in cosmology ?

I mean both correspond to a null geodesic $(ds^2=0)$ right ?

$$d_{horizon} (t) =a(t) \int\limits_{0}^{r} \frac{{\rm d}r'}{\sqrt{1-kr'^2}}= a(t) \int\limits_{0}^{t}\frac{ c \, {\rm d}t'}{a(t')}$$

with $a(t)$ the scale factor.

Is the difference just that for a comoving distance the integration bounderies changes ? like $r_e$,  $r_0$  and $t_0$, $t_e$

Thanks in advance for help

The comoving distance is a distance. You can calculate the comoving distance from your head to your toes (though it's more useful in cosmological contexts). It is defined such that today it coincides with the actual, physical distance you would measure if you froze the Universe and laid out meter sticks. In 11 billion years from now, when the Universe has doubled in size, physical (cosmological) distances have increased by a factor of two, but the comoving distances are, by definition, the same.

A "horizon" is a term used for a (non-physical) boundary, e.g. the "particle horizon", which is the boundary between the observable and the un-observable Universe. Or the "(cosmological) event horizon", which marks the boundary between regions from which we may at some point receive a signal emitted today, and regions from which we cannot.

Just like on Earth it makes sense to talk about "the distance to the horizon", you can say, e.g., "The comoving distance to the particle horizon is 46 billion lightyears", or "The comoving distance to the event horizon is 17 billion lightyears".

As you say, to calculate the comoving distance you use the Robertson-Walker metric (assuming here a flat Universe for simplicity)
$$
ds^2 = -c^2 dt^2 + a(t)^2 \big[ dr^2 + r^2 d\Omega^2 \big],
$$
and set $ds=d\Omega=0$ to measure along a radial null geodesic. Then $dr = c\,dt/a(t)$, or
$$
r = c \int_{t_\mathrm{em}}^{t_\mathrm{obs}} \frac{dt}{a(t)},
$$
where $t_\mathrm{em}$ and $t_\mathrm{obs}$ is the time when the light is emitted and observed, respectively.

To calculate the comoving distance to the particle horizon, you set $t_\mathrm{em} = 0$ and $t_\mathrm{obs} =$ today, because the particle horizon corresponds to light emitted at Big Bang and observed today.

To calculate the comoving distance to the event horizon, you set $t_\mathrm{em} =$ today and $t_\mathrm{obs} = \infty$, because the event horizon corresponds to light emitted today and observed in the (almost) infinitely distant future.


Is it possible to affect Earth's orbit so that, say, aphelion happens sooner? As part of its natural cycle, the sun is currently drifting away from the Earth, it's predicted to reach its farthest point by July 4th in 2020, at which point the distance between the Sun and Earth will start to close.

Do humans have enough power to control earth's orbit so that the sun starts its trip back sooner? I just can't wait for the heat to come back.

Humans don't have this capacity.  The Earth has a kinetic energy of about $10^{33}$ joules, relative to the sun.  Even if we put the total energy that humans use per year into the Earth's orbit (about 10^{24} joules) we still only have one-billionth of the kinetic energy of the Earth.

Moreover, the annual changes in temperature have very little to do with the distance from the sun, they are caused by the tilt of the Earth to the sun. For people in the Northern Hemisphere, the greatest distance to the sun occurs in the middle of summer when the temperature is usually at its hottest.

While we can't change the orbit of the Earth very much, we have been much more successful (!) at changing the composition of the atmosphere. The build-up of CO2 and other greenhouse gasses is warming the Earth significantly, and we can expect significant harmful consequences from this warming. We might like "warmth" but we don't like the rising sea levels, droughts, floods, and other effects of global climate change.


Is the Hubble Ultra-Deep Field image representative of the entire sky? The Hubble ultra-deep field image is a photograph of an area of sky equivalent to a 1mm x 1mm piece of paper held a metre away from the eye (one thirteen millionth of the entire sky). It contains an estimated 10,000 galaxies.

If galaxies were uniformly distributed, and the HUDF were representative of the entire sky, that would mean there were approximately 130 billion galaxies, which falls between estimates I have read of between 100 and 200 billion as the total number of galaxies.

However, my understanding is that galaxies occur in clusters, which would seem to imply that some areas of the sky should be more densely populated than others.

So is the HUDF representative of the entire sky, or are some areas much more and much less densely populated?

Short answer: both.

HUDF is representative on the type of galaxies we can find at that distance (and remember that distance also means time ago).

On the other hand, we do not really know how the galaxies are grouped beyond certain level, but we think they are grouped on so-called filaments, leaving enormous empty voids.

In this sense, HUDF's density does not represent the entire sky.

http://cosmicweb.uchicago.edu/filaments.html


Stellar age determination - code I'm trying to determine the age of some stars. I have many parameters that characterize these stars: $T_{eff}$ ,log ${g}$ ,$[Fe/H]$, $V$... I've tried to use the isochrones package but so far no success. I was wondering if any of you might have any recommendations!

I ended up using PARAM - http://stev.oapd.inaf.it/cgi-bin/param

Cheers,


How to find inverse steradian from $\text{arcmin}^{-2}$ for density of galaxy I am using a code on EUCLID future mission.
The original author of this code has set a value for the density of galaxy equal to :

ng = 354543085.80106884

I think this is expressed in inverse steradian. I think that EUCLID mission has a $30\text{ arcmin}^{-2}$ value for density of galaxies.

Is the conversion correct betweeen $30\text{ arcmin}^{-2}$ and $354543085.8010688 \text{ sr}^{-1}$ ?

and how to do this conversion ?

Indeed, I would like to calculate ng with a density of $48 \text{ arcmin}^{-2}$.

Regards

There are 60 arc minutes in a degree and $180/ \pi$ degrees in a radian.

So 1 radian is 57.2957795 degrees or 3437.746771 arc minutes.

As long as you are talking about units, you can square any of these angular units to obtain solid angle units.

So 1 steradian is 3282.80635 square degrees, or 11818102.860 square arc minutes.

Let's try your theory:

1/354543085.801 steradians is 9.259259E-06 square degrees, or 0.0333333 square arc minutes.

The answer is that it is 1/30th of a square arc minute, so YES it is 30 inverse square arc minutes!

Then 48 per square arcminute is:

$$48 \times 60^2 \times \frac{180^2}{\pi^2} \approx 567268937.282 \ \ \text{sr}^{-1}.$$


Why wouldn't massive torrents freeze on Mars? I am currently reading the Cambridge Guide to the Solar System. In chapter 8, section 8.7, they explain that massive torrents of water created outflow channels on the surface of Mars in the past. The book goes on to state that these torrents would have been unstoppable and

Such discharges would not freeze even under
present conditions on Mars.

Yet the atmosphere of Mars is under the triple point of water, so why is this the case?

... so why is this the case?

It may in fact NOT be the case!
Interestingly Phys.org's new item Early Mars was covered in ice sheets, not flowing rivers: study just showed up. It begins:

A large number of the valley networks scarring Mars's surface were carved by water melting beneath glacial ice, not by free-flowing rivers as previously thought, according to new UBC research published today in Nature Geoscience. The findings effectively throw cold water on the dominant "warm and wet ancient Mars" hypothesis, which postulates that rivers, rainfall and oceans once existed on the red planet.

and says later:

Grau Galofre's theory also helps explain how the valleys would have formed 3.8 billion years ago on a planet that is further away from the sun than Earth, during a time when the sun was less intense. "Climate modelling predicts that Mars' ancient climate was much cooler during the time of valley network formation," says Grau Galofre, currently a SESE Exploration Post-doctoral Fellow at Arizona State University. "We tried to put everything together and bring up a hypothesis that hadn't really been considered: that channels and valleys networks can form under ice sheets, as part of the drainage system that forms naturally under an ice sheet when there's water accumulated at the base."

and links to the new paper in Nature Geoscience Valley formation on early Mars by subglacial and fluvial erosion


How thick can planetary rings be? This arose from a comment on Worldbuilding.

We have data from four planets in the Solar System with rings, which doesn't make for a very good sample size. Observations of exoplanets could change that, but I expect that for now, any answer will simply be theoretical.

How thick can planetary rings be? I assume it would depend on the amount of material, but I don't know.

There is an explanation for why rings flatten out here. The general mechanism is that particles collide, and gets a very uniform momentum. Thus, any set-up giving unusually thick rings is in essence "cheating".

Here are some ways:

Moons can cause spiral waves in the rings, giving them more of a structure in the z direction. The ones known in Saturn's rings has a modes amplitude of just 10-100 m, but larger Moons can easily increase that.

Another way is simply having massive rings. Then they can not get more flattened, as there are no more empty space to remove.

A tilted ring relative to the Planets orbit around the star is going to experience tidal forces, as long as the radius of the rings is some notable fraction of the planet's orbital radius. From the context that sparked the question, that is not a suitable mechanism though, along with the possibility of having a so low density that particle collisions are rare.

However, more promising:

The halo ring of Jupiter is estimated to be around 12500 km thick (about the same as the diameter of the Earth), and are very fine dust kept from condensing into a disc by both the magnetic fields of Jupiter, and by iterations with the Galilean Moons.

We have four planets with rings in the solar system, so the sample size is quite small. Applying some small-sample-size statistical methodology, in this case an unusual application of the German Tank Problem, we can give a rough but realistic maximum thickness of a ring:

$$N \approx m+\frac{m}{k}-1$$

Where $m$ is the highest observed value, and $k$ the sample size.

Modified slightly to get a non-integer version that makes some sense, we get:

$$\text{max}_{\text{thickness}} \approx 12500 \; \mathrm{km}+\frac{12500 \; \mathrm{km}}{4} \approx 16000 \; \mathrm{km}$$

By no means a very certain limit, but at least about what can obtain from what we know.


Are there any good DSO API's for python out there? I was wondering if there was good DSO database API out there for python so I could fetch things like a distance away from Earth or radius etc. I tried looking however couldn't find the exact thing I was looking for.

Given you haven't specified which kind of DSOs you want or fully given the list of properties, it's difficult to recommend a specific source.

I would suggest searching for catalogues on VizieR, which allows you to search with keywords and/or various predefined categories. Bear in mind that you may need to use several different catalogues for the various types of DSOs (galaxies, globular clusters, supernova remnants, etc.) and not all of them will have all of the data you need. In particular, distances are not present in many catalogues. In the case of galaxies, you may need to convert from redshift or recession velocity depending on the source you're using. The VizieR results indicate sky coverage of the catalogue in the oval-shaped icons, which will give you some idea of which are all-sky catalogues and which are surveys of a small region of the sky.

Once you've found a catalogue on VizieR that contains the information, you can use astroquery to retrieve the data (details of the VizieR query interface), which imports the data in a form suitable for processing using astropy.


What variables are needed to calculate simple horseshoe orbit times? EDIT This was NOT a duplicate of Horseshoe orbit cycle times.

But that other question has been deleted, regardless. My original question asked for the answer to the equation(s), and provides many of the variables.  But since no one was apparently willing/able to answer it, this new question asks only for the form of the equation(s) itself and a list of variables involved, so that I can either provide more information on a future question, or try and solve the equations myself (though I doubt my math ability to do that, hence the original question, if I know what equations and variables are involved I can at least make an attempt myself).

What variables need to be known in order to calculate a horseshoe orbit cycle time?

In other words: What is the equation, and what do the variables in that equation represent (speed? mass? semi major axes? etc.), to calculate the time it takes for an object to go from Point A on that image, through Points B, C, D, and E and back to A, along the light blue line in the image above (image also available here)?

For example, Earth and Cruithne complete their cycle in 770 years, while Janus and Epimetheus complete theirs in 8 years.  I'm confident that semi-major axes are key factors in the equation, but I don't know what other variables are included, or how the variables relate to each other in the calculation.

Also, I understand that these orbits are not stable, in the long term, and that my example of Cruithne is especially complex, as it will actually change orbit types periodically, and will likely be ejected from the solar system entirely, or impact the Sun or Jupiter, at some point.  But those complexities are for another time.  For this question, I just want to know the equation in its simplest form.

That depends on the accuracy you want to work with. To zeroth order, as outlined in Murray & Dermott, "Solar System dynamics", Chapt 3., you can do the following:

The zero-velocity contours that are plotted in your image will not be coinciding with particle orbits to infinite precision, but they're a good zeroth order approximation for objects with low eccentricity w.r.t the star ($v_{r}/v_{\theta} \ll 1$)
A particle on those orbits is on a regular Keplerian orbit of radius $r_{\rm H}$, out of reach of the gravitational influence of the planet at $r_{\rm P}$. Thus, to get the largest "chunk" of traveltime of one part of the horseshoe orbit, either inner or outer part, you can work out with relative velocities and the assumption of keplerian velocities.
Caution is required as to which orbital time you're interested in: If $a$ is Earth's distance to the sun, and the horseshoe is $d$ away from a perfectly circular orbit, thus alternating between distances $a\pm d$, then the orbital time in the rest-frame of the sun will be $v_{-}=\sqrt{\frac{GM}{a-d}}$, for a large number of orbits until close encounter, and after that $v_{+}=\sqrt{\frac{GM}{a+d}}$.
Therefore, the simplified case that the satellite is on its own Keplerian orbit is true for most of the time. Armed with this knowledge, we can approximate the Planetocentric recurrence time as $t_{\rm rec}=\frac{2\pi a}{v_{K} - v_{-}}$ as a simple catch-up time between objects on different orbits.
The relative velocity $v_{\rm rel} = v_{K} - v_{-} = \sqrt{\frac{GM}{a}}-\sqrt{\frac{GM}{a-d}}$ can be expanded in the limit of $d/a\ll1$ into $v_{\rm rel} \approx \frac{1}{2} \sqrt\frac{GM}{a} \frac{d}{a}$ and thus I derive from this $t_{rec}\approx \frac{4\pi}{\sqrt{GM}} \frac{a^{5/2}}{d}$. As expected, the cycle time diverges for $d=0$, as in the co-orbiting case with Earth this time must be infinite. An upper limit on $d$ cannot be derived from this, one would need to turn for the full solution for that.

Just out of curiosity, I've plugged in a few values into this formula and written something quick in python:

import numpy as np

#Basic physics quantities
G      = 6.678e-8 #cgs units
pi     = 3.141592
navo   = 6e23 # particles per mole
sigma  = 5.67e-5   #erg cm-2 s-1 K-4
kb     = 1.38e-16  #erg/K
km     = 1e5 #kilometers in cm
mearth = 5.98e27  #g
msun   = 2.0e33   #g
au     = 1.49e13  #cm
yr   = 365*24*3600
rearth   = 6370e5
rjupiter = 74000*km

#
# Returns the approximate horseshoe-cycle time in the reduced 3body problem
# Masses of bodies: m0>>m1>>m2
# Semimajor axis distance is from m0 to m1, radial distance is a(m0->m1)-a(m0->m2)
#
def hs_cycle(mcentral, semimajor_axis, radial_distance):
    return 4*pi/np.sqrt(G*mcentral)*semimajor_axis**(5./2.)/radial_distance/yr

#
# https://en.wikipedia.org/wiki/(419624)_2010_SO16 around the Sun
#
# Quoted cycle time ~350 years, with d=0.004 AU
#
print("Predicted 2010_SO16 cycletime [years] = ", hs_cycle(msun, 1*au,0.004*au), " predicted = 350 yrs")

#
# Janus/Epimetheus around Saturn
#
# a = 151410 km, d = 25 km, as stated in https://en.wikipedia.org/wiki/Epimetheus_(moon)
# Quoted cycle time = 8 years (from comments)
#
print("Predicted Janus/Epimetheus cycletime [years] = ", hs_cycle(95*mearth, 151410*km,50*km), " predicted = 4 yrs")

#
# 3753 Cruithne
#
# a = 1 AU and semimajor axis difference from https://en.wikipedia.org/wiki/3753_Cruithne
# Quoted cycle time = 770 years
#
print("Predicted 3753 Cruithne cycletime [years] = ",hs_cycle(msun, 1*au, (1.0-0.99774)*au), " predicted = 770 yrs")

and the results I get are

Predicted 2010_SO16 cycletime [years] =  495.7747141830971  predicted = 350 yrs
Predicted Janus/Epimetheus cycletime [years] =  11.542076781209305  predicted = 8 yrs
Predicted 3753 Cruithne cycletime [years] =  877.4773702355546  predicted = 770 yrs

So the formula may be off by up to a factor of ~2. This is for sure simply because reality is more complex than a simple circular-orbit approximation, but also due to the quality of the values used.
Wikipedia isn't well known for researching particular values well. I took those that I found there. For SO16 that was particularly confusing to select, so I took those two that were mentioned in the same line of text, hoping they would come from the same source.

Anyone finding more consistent values, is free to comment.


Sloan digital sky survey galaxies I am trying to get some data from the Sloan digital sky survey (SDSS). This task turned out to be challenging for me as someone with only little knowledge about astronomy and maybe someone here can help me. What I need is data for all galaxies of the SDSS with their optical spectroscopy and position.

I suppose I need:

redshifts
RA (right ascension)
dec (declination)
classifications of all SDSS spectra

from the Optical Spectra Catalog Data

I found a script to convert ra/dec/redshift to x/y/z coordinates.
Will I need "the associated photometric position based matches" and "stellar parameter (SSPP) results" to get RA and dec? I have no idea what these are.

Use the Optical Spectra Query Form. You'll want to set PRIMTARGET to "GALAXY". Here's the list of possible parameters it can return http://skyserver.sdss3.org/dr10/en/help/docs/QS_UserGuide.aspx#Spectroscopy. Note, there is a query limit of 500,000 rows, so you'll have to break up the query since you'll have more spectra than that.