How to find the mass of a planet not knowing the gravitational constant? So I found this problem, I know the gravity of earth $g_0$, I found the orbital speed of one satellite using this equation $\frac{GM_em_s}{(R_e+h)^2}=m_sa$
($M_e$ is the mass of earth, $m_s$ is the mass of the satellite, $R_e$ is the radius of earth, $h$ is the height of the satellite)
and by taking $h=0$ I found the gravity of earth $\frac{GM_e}{R_e^2}=g_0$ so I found $GM_e=R_e^2g_0$ and then I found the orbital speed $v=R_e\sqrt{\frac{g_0}{R_e+h}}$ and then I calculated kepler's third law. but the last questions says deduce the mass of the earth $M_e$, How can I do this by not knowing the gravitational constant?
I'm sorry for bad english, and thanks a lot

How can I do this by not knowing the gravitational constant?

Sarcasm: You could do what Henry Cavendish did, which was to measure how the horizontal deflection rather heavy balls separated horizontally by a small distance and held in place by torsion rods. The concept of the Newtonian gravitational constant did not exist in Cavendish's time, and would not exist until almost a century after Cavendish's paper on Experiments to determine the Density of the Earth was published. At the time, Cavendish envisioned his experiment as a way to "weigh the Earth".
After the fact (about 80 years after the fact!), physicists saw that Cavendish had measured something much more fundamental than the average density of the Earth. We now see that Cavendish had indirectly measured what we now call the Newtonian gravitational constant. G does not come into play in Newton's formulation of his law of gravitation.
The modern version of the Newtonian law of gravitation does use the Newtonian gravitational constant, which frees you from having to redo Cavendish's experiment.


The Sun or Our Galaxy revolves around what? Like every planet they are in orbit of something or another planet like our solar system every planet revolves around the sun, my question is: what's the name of the planet or object which the sun revolves around; if the sun is stationary or if our galaxy is in motion, around what it is revolving.

As a previous comment states; it is complicated. But in principle the Sun is just one object in a large collection of objects which we call the Milky Way Galaxy. All the objects in this Galaxy orbit the barycentre (or barycenter, if you prefer the American spelling). The barycentre is, essentially, the common centre of mass of the whole galaxy. In the same way the planets in our solar system don't actually orbit the Sun, they orbit the Solar System Barycentre.

As for the Milky Way Galaxy, that too is a member of a larger group of galaxies, and they all orbit their common barycentre.


Is there a database of all known star names/identifiers? I am looking for a list of all known identifiers used for stars. I'm not sure on the terminology. While I do want the "proper names", I also want whatever identifier is used for each star detected, even those without proper names. I am assuming there are far more of the latter.

If possible I would also like to have the coordinates and brightness (and any other interesting info) for each of these star. But just the list of identifiers would be great.

Does such data exist somewhere? Ideally I can just download a simple file (like a csv) and I can write my own software to process it without fiddling with some other software. However, it may be too large for that.

I'm not sure what to expect... 1 million entries?

The closest service to what you are describing is the SIMBAD Astronomical Database from the Université de Strasbourg/CNRS. At the time I write this post, it contains 10.8M objects and 35.5M identifiers. It does not have a single CSV you can download with this information (to the best of my knowledge, and I've asked), but there is an API and TAP service available for programmatic access.


How many stars and galaxies can be seen by the naked eye? How many of the luminous dots that we see naked are galaxies and not stars from our galaxy?

I imagine that the majority of the luminous points that we see naked eye during the night, are actually stars from our galaxy. But how many of them are other objects (other galaxies, nebula, etc.), excluding planets from our Solar System?

In the best sky conditions, the naked eye (with effort) can see objects with an apparent magnitude of 8.0. This reveals about 43,197 objects in the sky.
There are 9 galaxies visible to the naked eye that you might see when observing the sky, and there are about 13 nebulae that you might see.
Sources:

The Bortle Dark-Sky Scale - John E. Bortle
How many stars are in the sky? - NASA
Naked-eye galaxies - Wikipedia
List of planetary nebulae - Wikipedia
List of diffuse nebulae - Wikipedia


Is the universe eternal? I understand that as the universe continues to expand, eventually our supply of hydrogen will be exhausted and then it's "lights out."  Is that really going to happen or is there something "eternal" about the universe?  It matters because if the universe is eternal, and since we are a part of the universe, we too are eternal as well in some sense.

The metaphysical aspects of your question can't be answered. Our bodies aren't eternal, and astronomy can't answer questions about the soal.

Observations of the universe suggest that it will continue expanding.

As you note, eventually all the stars will run out of hydrogen, collapse to white dwarfs, which then slowly cool. What happens next is rather speculative. At this time there will be nowhere that is warm in the universe. Everywhere will cool to very close to absolute zero. Complex chemistry will no longer be possible, and there will be no life in the universe.

Some stars will fall into black holes. Others will be ejected from the galaxy. If protons decay, then very very very slowly, the matter will break down, leaving only black holes.

If black holes emit Hawking radiation, then they too will break down.

In this model, the spacetime of the universe is eternal, but nothing within it survives.

Wikipedia has a list of some of the events of the far future


What is the difference between LMC and SMC? Apart from the obvious difference as suggested by their names that the Large Magellanic Cloud is 'larger' than the Small Magellanic Cloud, what are the other differences between them?

Do give all the details you can. Thank you!

The LMC has an apparent size of about 645x550 arc mins, the SMC 320x205.

Both contain several hundred million stars each. The LMC is about 14000ly in size, and is about 10 billion solar masses; the SMC is 7000ly in size, and is about 7 billion solar masses.

The visual magnitude of the LMC is +0.28, the SMC is +2.23.

Both feature a number of interesting clusters, nebulae, and supernova remnants. Notably, LMC is home to NGC 2070, the Tarantula Nebula, and Supernova 1987A.
L
The LMC is usually considered an irregular galaxy, though it has a prominent bar, somewhat warped, and a spiral arm. The SMC is a dwarf irregular galaxy, or maybe a barred disc.

LMC is the fourth largest galaxy in our local group, and the third closest to us. SMC is the fifth? largest galaxy in the local group, and the fourth closest to us.

{In a previous version of this response I said: A number of websites proclaim that the LMC and SMC are gravitationally bound to the Milky Way. This has been known to be incorrect since 2007 when Hubble observations showed they are travelling too fast to be orbiting the Milky Way. See also.}

A number of websites proclaim that the LMC and SMC ARE gravitationally bound to the Milky Way. This was based on old information. More recent measurements have challenged this assumption, though further analysis has left open the door for the possibility that the LMC and SMC are gravitationally bound to the Milky Way. We can, however, safely say that the LMC and SMC may be gravitationally bound to the MW.


How do they know the newly-spotted gas near the center of our galaxy is molecular without knowing what gas it is? Phys.org's Mystery gas discovered near center of Milky Way links to Teodoro et al 2020 in Nature Cold gas in the Milky Way’s nuclear wind who's abstract is shown below.
Question: My reading of the Phys.org summary and the abstract tells me that the gas is known to be fairly cold and molecular, and yet it is not known what the gas is! Is it so obviously hydrogen as H2 that nobody says this explicitly? If not, how does one determine a gas is molecular without knowing the chemical nature of the gas? It seems that if one resolves vibrational/rotational features in optical, IR or radio spectra one could simultaneously identify the species, but perhaps I'm naive. (Okay, I'm almost certainly naive!)
--
The abstract from Nature:

The centre of the Milky Way hosts several high-energy processes that have strongly affected the inner regions of our Galaxy. Activity from the super-massive black hole at the Galactic Centre, which is coincident with the radio source Sagittarius A*, and stellar feedback from the inner molecular ring1 expel matter and energy from the disk in the form of a galactic wind. Multiphase gas has been observed within this outflow, including hot highly ionized (temperatures of about 106 kelvin), warm ionized (104 to 105 kelvin) and cool atomic (103 to 104 kelvin) gas. However, so far there has been no evidence of the cold dense molecular phase (10 to 100 kelvin). Here we report observations of molecular gas outflowing from the centre of our Galaxy. This cold material is associated with atomic hydrogen clouds travelling in the nuclear wind. The morphology and the kinematics of the molecular gas, resolved on a scale of about one parsec, indicate that these clouds are mixing with the warmer medium and are possibly being disrupted. The data also suggest that the mass of the molecular gas outflow is not negligible and could affect the rate of star formation in the central regions of the Galaxy. The presence of this cold, dense and high-velocity gas is puzzling, because neither Sagittarius A* at its current level of activity nor star formation in the inner Galaxy seems to be a viable source for this material.

They know there's molecular gas because they observed emission from the CO molecule in two of the previously identified (atomic-hydrogen-emitting) gas clouds. From the paper:

... we targeted two objects (hereafter, MW-C1 and MW-C2), highlighted
by red boxes in Fig. 1, in the 12CO(2 → 1) emission line at 230.538
GHz with the 12-m Atacama Pathfinder Experiment (APEX) telescope....
We mapped both clouds in 12CO(2 → 1) emission over a 15′ × 15′ field
centred on the peak of the H I emission,... These data revealed
molecular gas outflowing from the centre of our Galaxy.

CO emission is always assumed to come from molecular gas clouds which are, by mass, mostly H$_2$, though the conversion factor needed to turn a CO measurement into an H$_2$ measurement is a matter of debate (this is discussed in the paper).


Why are most lunar maria on the visible side? Is there any reason why the Earth side of the Moon has enormous plains and mountains, while the far side contains just few small ones inside largest craters?

The two hemispheres have distinctly different appearances, with the near side covered in multiple, large maria (Latin for 'seas,' since the earliest astronomers incorrectly thought that these plains were seas of lunar water). The far side has a battered, densely cratered appearance with few maria. Only 1% of the surface of the far side is covered by maria, compared to 31.2% on the near side.

One commonly accepted explanation for this difference is related to a higher concentration of heat-producing elements on the near-side hemisphere, as has been demonstrated by geochemical maps obtained from the Lunar Prospector gamma-ray spectrometer. While other factors such as surface elevation and crustal thickness could also affect where basalts erupt, these do not explain why the farside South Pole-Aitken basin (which contains the lowest elevations of the Moon and possesses a thin crust) was not as volcanically active as Oceanus Procellarum on the near side.

It has also been proposed that the differences between the two hemispheres may have been caused by a collision with a smaller companion moon that also originated from the Theia collision. In this model the impact led to an accretionary pile rather than a crater, contributing a hemispheric layer of extent and thickness that may be consistent with the dimensions of the farside highlands.

Though the far side has more visible craters, this is due to the effects of lunar lava flows, which cover and obscure craters, rather than a shielding effect from the Earth. NASA calculates that the Earth obscures only about 4 square degrees out of 41,000 square degrees of the sky as seen from the moon. "This makes the Earth negligible as a shield for the Moon. [...] It is likely that each side of the Moon has received equal numbers of impacts, but the resurfacing by lava results in fewer craters visible on the near side than the far side, even though the both sides have received the same number of impacts."

From wikipedia page and an other source.

EDIT: to complete the answer.

From space craft orbiting the moon, it has been found that the centre of mass of the moon is displaced toward Earth. An explanation of seas distribution states that lunar core, aesthenosphere and the mantle is off-centre, whereas the crust is not — so the crust is thicker on the far side (see figure). Thus the basalt would reach the surface more easily on the near side due to thinner crust (to 60 km near side, 100 km far) than on the far side.

From this source.


How to lose the moon? In this worldbuilding question, the moon is gone:

  an extra-solar gas giant passed close by Earth in which the Moon was
  knocked from orbit to become a planet in an independent orbit, as well
  as sending Mercury crashing into its parent star.

So by introducing a rogue planetoid, the orbit of both the Moon and Mercury needs to be affected, with at least the former surviving the encounter.

Is this possible? If so, what are the requirements (mass, velocity, proximity)?  Would it be possible for it to happen and affect the Earth minimally, so that it remains habital?

The Earth and moon orbit around a barycentre that is "about 3/4 the radius of Earth" away from the Earth's centre: https://en.wikipedia.org/wiki/Orbit_of_the_Moon. Removing the moon would almost certainly affect the Earth's orbit around the sun due to the change in the centre of mass, and probably a resultant change in momentum of the Earth system (such as it was). That's without even considering the gravitational tug of the planetoid on the Earth, and that of the moon as it gets snatched away. I would think it spectacularly unlikely for the interaction to leave Earth travelling along the orbit of the centre of mass of the old Earth-moon system. But, I haven't actually modelled it, so I might be wrong there.


How thick or deep does solid carbon dioxide accumulate on Mars? On Mars is there an area in which all the $CO_2$ evaporates exposing dirt near the pole? How deep can solid $CO_2$ accumulate during winter in that area during the Martian Winter?

https://space.stackexchange.com/questions/31724/solar-carbon-dioxide-collector-on-mars

1-2 m in the north and 8m in the south. https://en.m.wikipedia.org/wiki/Mars#Physical_characteristics https://en.wikipedia.org/wiki/Martian_polar_ice_caps#North_polar_cap

It is the northern cap that disappears in summer.


Is the location that published calendars use to calculate the seasons one specific location on the earth? All the calendars in my house say the fall equinox is Tuesday the 23rd. All the astronomical programs and astronomical sources say it is it is Monday the 22nd in Ohio. Since the time given is 10:29 p.m. EDT all of the US starts fall on Monday. I’m guessing the calendars use Greenwich Time. This same discussion also applies to when calendars show the full moon dates.

I know this isn't exactly an astonomy question, but I thought this site was the closest I could get.

Astronomical events such as the solstices and equinoxes occur at a specific instance in time on a specific day, often given in Universal Time.

For example, this winter's solstice occurs on 21 Dec 2014 at 23:03 Universal Time. Since most of Europe is an hour ahead, this may occur on 22 Dec 2014 at 00:03 (and subsequent hours) given in local time based on the time zone. So British and American calendars will show the solstice occurring on 21 Dec, and in countries east of Greenwich on 22 Dec.


How far apart is the dust in the Sombrero Galaxy's dust lane? In his answer to this question, TildalWave made this remark:

  I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.

Here's the photograph:

Since it almost feels like I can see individual 'pieces' of dust in the lane, it would follow that the particles must be very far apart. Just how far apart are they? Are they grouped in clumps, or relatively evenly space?

Based on 1995A&A...303..673E,

How far apart they are?

No idea. Somebody else might help to provide a density distribution.

Are they grouped in clumps or relatively even spaced?

This is what I really wanted to answer. They are in a concentric ring-like structure around the galaxy. The maximum optical depth is 0.8 with a mean value of 0.3. By the way, optical depth gives you a nice idea of how far apart they might be, not a good density distribution but you can get the physical picture in mind.


What can be learned from, or noted in this LIGO Orrery video? The Ars Technica article Physicists detected gravitational waves from four new black-hole mergers notes that

  This brings the total number of events detected by LIGO and Virgo to 11.

It also has a link to the SXS Collaboration' YouTube video LIGO Orrery

SXS stands for Simulating eXtreme Spacetimes https://www.black-holes.org/

  The SXS project is a collaborative research effort involving multiple institutions. Our goal is the simulation of black holes and other extreme spacetimes to gain a better understanding of Relativity, and the physics of exotic objects in the distant cosmos.

so quite a substantial amount of math and physics went into this work.

But as an outsider I don't really understand what I'm looking at. The video's comments just say that the surface is the event horizon:

  Computer calculations modeling the gravitational waves LIGO has observed to date and the black holes that emitted the waves. The image shows the horizons of the black holes above the corresponding gravitational wave.

Question: What does the color represent, could it be some kind of "temperature" or energy density of distorted space on the event horizon surface? Why can I see some evidence of rotation in the resulting merged object for a few rotations, then it disappears? What else (if anything) can be learned from this video?

Screen shot from the YouTube video LIGO Orrery:

The SXS consortium has a page linking  a number of their videos. From the text around the videos it seems they use colour to encode various different scalar quantities in different videos. For instance

  The colors encode the lapse function — the slowing of the rate of flow of time.

And

  Colors indicate the vorticity of the apparent horizon

and

  with the color scale giving the density of matter within the star [this is a neutron star merger, not black holes]

Unfortunately, the LIGO gallery has not preserved this information (nor  the links to published papers where more details seem to be given) so it is hard to say in this specific case. Also simply understanding what many of these quantities are needs a fair level of differential geometry.

You asked:

  Why can I see some evidence of rotation in the resulting merged object for a few rotations, then it disappears?

Since we don't know what quantity is being depicted, it's a bit hard to be very precise, but there is an established concept of "ringdown".

You may have heard the phrase "black holes have no hair" meaning that from outside everything about a (non-quantum) black hole is determined by its mass, spin and charge, with no details (the hair) remaining. This is not exactly true. What is true is that any "hair" a black hole has is quickly radiated away as gravitational waves so that it rapidly approaches the ideal state given by its mass, spin and charge. This phenomenon is called "ringdown" and is what you are in seeing in the first moments after the collision. It can be detected in the gravitational waves.


Aren't there any rock or similar firm material on/in the gas giant planets? Aren't there any rock or similar firm material on/in the gas giant planets? What happens if a rock asteroid hits one of these planets? Shouldn't the rock accumulate in the center of the planet due to its high density?

The inside of a giant planet is not like regular gas.

First of all, it is hot. You are away of how it gets hot inside the Earth (causing volcanos). It is also hot inside Jupiter, but since Jupiter is bigger, it is hotter. It would be hot enough to vaporise rock under "normal conditions".

The pressure is immense, and things stop behaving as you are used to when the pressure is very high. For example, all the atoms in a gas are pushed so close together that they are touching (like they would be in a liquid). This means that there is no boiling point and no real difference between the gas and liquid phases. At extreme pressures, hydrogen atoms are pushed so close together that their electrons can start to flow from atom to atom, forming a fluid metal.

Pressure that can push atoms so close that they can turn hydrogen to a metal mean that you can't send a probe to this region. You can't swim in a metallic hydrogen sea.

Nevertheless, heavier atoms like iron, silicon, carbon, oxygen, will tend to fall towards the centre of the planet, and so there may be a since rocks are made mostly of atoms like iron, silicon, carbon, oxygen, you could describe this as a "rocky" core.  But don't think this is a kind of solid surface that you could stand on.


A "tidally locked" double planet? First, I'd like to take the definition of a "double-planet" as two bodies orbiting each other where the center of gravity is not inside the larger body. Also, the system would have to fill other planet requirements (like emptying it's own orbital area around the star).

Now, let's say that the "binary system" is in an orbit close enough to a star to be tidally locked to the star. How would the tidal effects of the star's gravity effect the orbit of the planets around each other??

Let's take an example of a "double-planet" (of $0.075$ and $0.030\:\mathrm{M_{Earth}}$ with a separation of $1.17 \times 10^6\:\mathrm{km}$ orbiting their CM in about $450\:\mathrm{days}$. They are $0.085\:\mathrm{AU}$ from a star of mass $0.35\:\mathrm{M_{Sun}}$ and thus orbit the star in $15.3\:\mathrm{days}$. The planetary rotation/co-orbital period is much longer than the orbital period around the star, so I would think that the planets would have characteristics of a lone TL planet.

I am wondering how such a system would evolve over time. Would someone on the surface of one of the planets experience experience night and day in 15.3 day cycles? Would the difference in tidal effects of the star on the different sized planets cause the orbit to "precess" or have some other effect?

Imagine designing a long term calendar for this system!!

Your scenario isn't stable.   A simple way to explain this is to imagine that the planet's orbit each other at the same rate they orbit the star (your scenario has them orbiting even slower).

At the same rate of orbit, the synodic period essentially approaches infinity.  (see diagram of the Moon's synodic orbit).

When this happens, the inner planet is at L1 to the outer planet and the outer planet is at L2 to the inner planet.   That's ofcourse, not quite right as the different masses would have different Hill Spheres, and the mutual gravitation of two objects massive to have a barrycenter outside the more massive planet would combine into a slightly faster orbit, but it's close enough to demonstrate the instability.

The L1 and L2 points are on the border of the Hill Sphere and well outside the true region of stability.

No system can be stable where the Moon (or binary planet system) orbits around each other slower than they orbit around the central star.   They need to be at least 50% to 67% closer in than that would allow.

Using the Orbital Period or "T" squared = distance ("a" semi-major axis) cubed, per Kepler's laws, the upper limit for a stable orbital period of a moon around a planet or a binary planet would be about 19%-35% the period of the planet's year.    Any longer the Moon or binary system would risk instability.

You might be able to set up a synchronous system where the planets orbited each other 4 times for every orbit around the sun, where they'd end up in the same (basically eclipse) position every perihelion with the planet experiencing greater tides (smaller and/or more fluid planet) closer to the sun at perihelion.  I suspect a 4-1 orbital period ratio is about as close as you're likely to get for any length of time for a stable binary planet-star, tidal locked system and that would be odd, but I see no reason why it wouldn't be stable.   It would only be possible around a small star where the planets were fairly close so that all 3 could have significant tidal effects on each other.   Without strong tides you don't get tidal locking.

You also can't have tidal locking with both objects when they have different orbital periods.   Tidal locking means that the rotation period is equal to the (sidereal) orbital period.


Determining needed exposure time given SNR I need help on how to estimate an exposure time for a desirable signal to noise ratio (SNR). For example, given the magnitude and a desired SNR of a star we want to achieve, how long should the exposure time be? We may assume that the noise is only due to counting statistics, and ignore dark and flat noise. Any useful function to make this estimate?

Here's a simple start, and I suggest you must include your sensor's noise -- not flat-fielding, but just the accumulated readout noise in a pixel.
You can use the Poisson SNR rule for the incoming light, so optical noise $N_o = \sqrt{S_o}$.  Just add the electronics noise, again as a simple model, $N_e = C_(readout) + k\tau$ where $\tau$ is the integration time and $k$ represents leakage current into the pixel well.   $S_o$ also equals some constant times $\tau$ , so just combine the two noise terms in an RSS manner, i.e. $ N_{tot} = \sqrt{N_o^2 + N_e^2}$ .


Identify Greek letters used in formulas Could anyone tell me what the names of the Greek letters in the screenshot are? Is the first one alpha? They are used in calculating ICRF rotations. Wikipedia uses a different font for their list of characters, so it's hard to tell.

The first one is alpha (α); the second one is delta (δ). You can study the shapes of the letters here:

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

Here's what they mean (information from a comment to your question):

  Orientation models use three Euler angles to describe body
  orientation. The first two angles are the right ascension and
  declination of the north pole of a body as a function of time. The
  third angle is the prime meridian location (represented by "W"), which
  is expressed as a rotation about the north pole, and is also a
  function of time.


What would harnessing of gravitational waves look like? We have harnessed and learned to generate and control sound waves, light waves, water waves, and electromagnetic waves in general.  What would be needed for science to be able to generate and use gravitational waves in the same way? What would be the outcome?

The mass of fundamental particles, like the proton or electron is very very small in comparison to the charge. This means that the motion of electrons in atoms can produce significant amounts of high frequency electromagnetic radiation.

Quantum effects stop electrons from falling into the protons they orbit.

To produce and control gravitational radiation in significant amounts you would need something very small, but very massive: a microscopic pair of black holes.

Such things would be rather hard to manage. In short it is beyond any conceivable current or future technology.


Chromatic aberration in lens Is there a way to prevent chromatic aberration without using an achromatic lens in refracting telescopes or in any other instruments...?

It's probably impossible to prevent chromatic aberration, since there is no such thing as a perfectly achromatic lens. but you can certainly minimize it.

As you have mentioned, for a given spectral range, the best way is to choose a lens that is said to be "achromatic" even though the correction is never perfect.

But if you want to use a lens that is not achromatic, then you can try the following to reduce the chromatic aberration:

Use a lens that has a very long focal length, for more on that see How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens? I figured out the answer there and will try to post an answer soon!
Minimize your spectral range with a filter, or at least cut out the short wavelengths (e.g. blue). The dispersion of glass is stronger at the shorter wavelengths.
If you are doing color astrophotography with a color sensor, then there may be a way to shoot the three color images separately, bringing each color into focus separately. One problem is that for color sensors used to reproduce human color perception, the red channel often has a blue bandpass to replicate the way color receptors in our eye work. See answers to Filter for RGB separation and its effect on the image and the image below. So if you are trying this, then it might be helpful to add a blue filter, at least to cut below say 425 nm.


Does the sun now 'bypass" some of the original zodiac constellations? This question got me thinking of an interview of, or a talk by Neil deGrasse Tyson that I saw somewhere in the YouTube universe, giving a question one might ask to possibly challenge literal interpretation of one's astrological sign. For some reason I can't find it now, but I believe he said that currently the sun does not actually pass through all twelve zodiac constellations any more.

I'm not looking for opinions or any particulars about Astrology, please, just the science! (if there is some) If someone is familiar with the statement, I'd like to hear if I've got it right, and if it's due to proper motion, precession, redrawing/redefinition of constellation boundaries, or something else.

The ecliptic still passes through the original 12 constellations, although Earth's precession has shifted the Sun's apparent position by roughly one constellation. That is, if your birthday is today, May 2, an astrologer would claim you were Taurus, whereas in reality the Sun is currently positioned in Aries.

Additionally, the ecliptic now passes through the constellation Ophiuchus (The Serpent Bearer). You can see the "historical" and the "true" constellations e.g. here.


Ecliptic coordinate ranges in radians In ecliptic coordinates, what are the ranges of ecliptic longitude and latitude in radians? Do they both go from $-\pi$ to $\pi$?

It is purely a matter of convention.  Longitude and latitude are normally expressed in degrees, longitude from 0 to 360⁰ and Latitude from -90 to 90⁰.  If you express these in radians, longitude would be from 0 to $2\pi$ and latitude from $-\frac\pi2$ to $\frac\pi2$.
If you are writing software, your functions should be robust enough to accept any value out of this range, for example when calculating the position of a planet, it may be convenient to have a continuous mapping from time to longitude, this would necessitate longitudes that are greater than $2\pi$.


Advice on first Telescope (refractor vs reflector), Barlow lens I am planning on buying my first telescope. I am getting an 80mm refractor and a 114mm reflector for the same price range. Which would be a better option? Further, would a Barlow lens solve the chromatic aberration issues with a f/5 refractor?

Further would a Barlow lens solve the chromatic aberration issues with
  a f/5 refractor.

I suppose in theory they could, but they would have to be designed to do so. In practice, they vary widely in quality, and either don't have an effect on chromatic aberration, or actually make it worse.

  Which would be a better option?

There is no objective way of answering this question. Based simply on price per square unit of aperture, a reflector is more economical, and therefore better. But if you're looking for a telescope that doesn't require collimation, a refractor might be best. You have to consider the portability of the specific models you're considering as well. Finally, what would you like to see with your telescope? A high-end refractor with correcting optics is, in theory, better than a similar reflector for high-quality imaging of planets. Ultimately, the best telescope is the one you're comfortable using regularly without feeling overwhelmed.


What's that ring that only now appears in black hole illustrations? If you'll notice, space time starts to deform near the black hole, but then at considerable distance from the black hole' surface, images in the background become ultra-distorted, but then as you look even closer to the black hole, the distortion starts to look normal again almost like nothing is happening except for a minor fish-eye lens effect.

If space-time distorts in a monotonic way as you approach the surface of a black hole, how does this ring of ultra-distortion exist? Why doesn't the image look even more distorted at the surface than at that ring?

This is due to photons actually going in circles around the black hole multiple times.

Far away from the BH, the image is only slightly distorted. These are photons that deviated only a little bit.

As you get closer, you're seeing photons that deviated more and more. At some point, you'll see photons that came from behind you, went around the BH, and hit your eye. Eventually, close enough to the BH, you'll see photons that did a full 360 degree turn. The BH acts like a fisheye camera.

Looking closer and closer to the disk, you get photons that did 1.5 turns, 2 turns, 3 turns and so on.

What you see is concentric ring images, each one of them containing the full picture of everything around the BH but compressed in a ring. There's essentially an infinity of concentric images, but you only see clearly the first few of them.

https://imgur.com/VZnkFxP

It should be noted that the black disk in the middle is not the event horizon. It's simply a region in space where no light comes from that could reach your eye. The actual event horizon is about 2x smaller. You could think of it as the image of the event horizon "magnified" by the spacetime distortion, but the analogy is not perfect.

A black hole is, in many ways, like a very strange "lens".


What is the hottest thing in the universe? Straight from my 7 year old to you, exactly what it says on the cover:

  What is the hottest thing in the universe?

To make it Stack Exchange-friendly, I'll add the following caveats:

it should be bounded, as in an actual compact object, or class of objects, or part of an object
it should be observable
it should be an astronomical object, ie a Quark Gluon Plasma created by collisions at the Large Hadron Collider doesn't count.

Thanks,
Bruce

Energetic neutrinos have been observed from the core of a supernova (SN 1987A). The inferred temperature at the "neutrinosphere" is about 4 MeV (equivalent to 50 billion K - ($5\times 10^{10}$ K, Valentim et al. 2017). Hence it is observable and has been observed.

The very centre of the proto-neutron star that is responsible for the neutrino emission is likely to be a factor of two or so hotter, but cannot be observed, even with neutrinos, because the "neutrinosphere" is opaque to neutrinos. By the time this "clears", the proto-neutron star is much cooler - its surface would be orders of magnitude cooler.

Arguably we could study the very core of a supernova through gravitational waves if one were to explode in our own Galaxy. Whether this counts as "observing" a hot object, I'm not sure.

In a similar vein, we have observed "kilonova" that appear to be due to the merger of two neutron stars. The temperatures generated in these events are also likely to be of order 100 billion K ($10^{11}$ K), but again these temperatures are not observed directly - the gravitational waves and gamma rays produced in these events are caused by "non-thermal" mechanisms.


Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*? Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*? Additionally, would it matter if we are exactly in the plane with the accretion disk, i.e. could we still observe the event horizon?

We don't know the answer to this (or if someone does, I'd really like to hear that answer). Our galaxy's Super-Massive Black Hole (SMBH) is an unusually quiet one, with little to no accretion disk. If one indeed exists, and it almost assuredly exists, even if it is tenuous and small, we have not yet seen it.

Around 2013, there was a gas cloud designated G2 that was expected to be accreted which would have given us a lot of insight into the black hole and its structure. Unfortunately there was little to no activity observed. A Nature article about this gas cloud goes into more detail and talks about the accretion disk, as quoted below.

  The cautiously predicted flare-up would be a valuable probe of the environment immediately surrounding the black hole, about which little is known. What we do know is that the average rate of mass accretion driving the emission from Sgr A* is exceptionally small, thus explaining why our Galactic black hole is extremely dim. That we can detect it at all is due largely to its proximity — it is about 100 times closer to Earth than the nearest supermassive black hole in another galactic nucleus, that of the Andromeda galaxy. We also know that the accreting gas is heated so much that the individual gas particles almost all approach the speed of light, and that the observed emission emanates from a magnetohydrodynamic maelstrom. But the basic parameters of this small-accretion-rate regime are still being sought: does the accreting gas form a well-defined disk or an outflowing jet, as it often does in the more prominent cases observed in many other galaxies? Or is it better described as a more isotropic inflow comprising stochastically distributed density fluctuations?


Why search for water? Most of the news about extraterrestrial search focuses on possibility for existence of water and temperatures like earth. Why do scientists feel this would lead them to extraterrestrial life?

After all, we are searching for something that doesn't exist on earth. Why do we assume it would have behavior like us? Maybe the matrix inside their cells is of acetone or something else. Water may not play the role in their lives as it does for us. Maybe they can sustain high temperatures, maybe they can live in extreme cold. Why should we look for earth-like planets, isn't there a baseless (of course it's not baseless, there must be some logic which I want to know) assumption that the aliens are human-like (or any species on earth)?

A lot of different "alternative biologies" have been considered, but if you analyze their required chemistry, this always falls apart at some point, some mechanisms necessary for life are simply impossible. Replace carbon with silicon, increase ambient temperature, and you're getting a lot of interesting reactions that could be conductive to life. But instead of CO2, which is a gas, you have SIO2 which is a crystal. You can't breathe crystal. It's awfully insoluble, so no simple hemoglobin counterpart. And so on. Won't work.

Water and carbon based biology is one most plausible. There are odd variants, like bacteria thriving on sulfite, manganese and copper oxide, but they still require water to live, and still use carbon as the fundamental building block.

While others might exist, chance for them to form is much lower than for one similar to ours.  And since we haven't designed even one fully plausible alternate biology that doesn't involve water we don't even know what sort of detectable signatures it would produce. So we search for something we know might yield results; hint at possibility of life somewhat similar to ours - instead of a wild goose chase after the unknown.

Actually, there exist one "alternate biology" which is fully plausible even if its early origins would be Earth-like: Silicon based, artificial intelligence, robotic physiology, electrical power, proliferation through fully automated manufacturing processes. It's unlikely to produce any detectable chemical signature (at least distinguishable from half of the natural celestial bodies out there), but we are fairly sure its electromagnetic spectrum - radio waves signature - would be quite specific. Well, The SETI project is on it.


If oxygen was abundant in Neptune, would there be combustion? Since Neptune/Uranus have high percentages of methane, wouldn't it be highly likely that there will be combustion (triggered by the lightning storms or any other factor) if oxygen was abundant?

Will the hydrogen combust as well?

"At high altitudes, Neptune's atmosphere is 80% hydrogen and 19% helium" (Wikipedia). No significant abundancy of free oxygen to react with.

A source of oxygen could easily made burn on Neptune, like a source of hydrogen on Earth.
Or take a sample of Neptun's atmosphere. It would easily burn in Earth's atmosphere.

Hydrogen oxygen combustion is sufficiently exothermic (483.6 kJ/mol of O2) to sustain burning even at the low temperatures on Neptune, provided the atmospheric pressure is high enough: With a gas constant of about 8.3 J / mol K, the gas can easily be heated up to the point of combustion by the energy released by the combustion.

Assuming oxygen would be provided, just sufficiently to burn up hydrogen and trace hydrocarbons:
After the explosion and condensation Neptune would get a huge ocean of water, later freezing to water ice, with an atmosphere of helium, and some less abundant gases like CO2, nitrogen, and argon, traces of water vapor. Traces of hydrogen and methane could survive the presence of oxygen. Hence some oxygen would remain in the atmosphere. Most of the CO2 would be resolved in the water ice.

The average density of Neptune would increase, it would shrink a bit, the core would be compressed and heated by adiabatic compression.

Before combustion Neptune's gravity is strong enough to hold an atmosphere of hydrogen, the easiest gas to escape. Hence the denser planet after combustion wouldn't improve its capability to keep an atmosphere significantly: The escape velocity of Neptune is 23.5 km/s, its surface temperature is below 100K. With Boltzmann constant $k=1.38\cdot 10^{-23}J/K$ and $E=3kT/2$ we get $$E=3\cdot 1.38\cdot 10^{-23}J/K\mbox{ or }4.14\cdot 10^{-21}J\mbox{ at }100K$$ for an average $H_2$ molecule with a mass of about $3.35\cdot 10^{-27}kg$. Hence for Neptun's escape velocity the kinetic energy of a hydrogen molecule is $$E=0.5\ mv^2=0.5\cdot 3.35\cdot 10^{-27}kg \cdot (23.5\cdot 10^3 m/s)^2 = 9.25\cdot 10^{-19} J.$$ Hence the energy needed to escape is more than 200-times the kinetic energy of the hydrogen molecules; no way to escape, 2000K or more would be needed to let escape some hydrogen over time (to get below a factor of about ten between the two energies). Photolyic dissociated atomic hydrogen would still need more than 1000K to escape.

The probability to catch asteroids or comets would be reduced due to the reduced radius, if the total mass is adjusted to the mass before combustion, since asteroids would miss instead of hit the planet.


speed of light measurement While searching for different methods of speed of light measurement, i came across one of the method of fizeau discussed below which i cannot fully understand.

In short,in Fizeau’s apparatus, a
beam of light was shone between the teeth of
a rapidly rotating toothed wheel, so the
“lantern” was constantly being covered
and uncovered. Fizeau had a
mirror, reflecting the beam back, where
it passed a second time between the
teeth of the wheel.

I do understand the idea i.e. to measure the time during the course of light "from the wheel to mirror and then back to the mirror".But where are the mirror,lantern and wheel  located? What kind of wheel is that? Does light pass through the holes (or teeth) in the wheel or light gets refracted through the wheel? Finally if a group of light particles goes through one of the hole (or teeth),then how do we know that the same group of light particles came back through which hole (or teeth) after its journey,since there are soo many light particles leaving and entering different holes (or teeth)?

Here is a diagram of his setup:

See the Wikipedia page for a discription.


What is the equation of state for a relativistic fluid/gas? Say we have a relativistic fluid/gas, as we have in some astrophyical systems.

Now let us write:

$e$ - energy density in the fluid's rest frame.
$P$ - pressure in the fluid's rest frame.
$n$ - number density in the fluid's rest frame.
$m$ - mass of the particles.

I know that for the non-relativistic case we have:

$$e=nmc^2+\frac{1}{\hat{\gamma}-1}P$$

where $\hat{\gamma}$ is the adiabatic index. $\hat{\gamma}=1+\frac{2}{f}$ for a gas with $f$ degrees of freedom.

For the ultra-relativstic case we have:

$$e=3P$$

My question is what is $e(P,n)$ for a relativstic case (which is the general case of the 2 limits shown above)? I would also like to know how to derive it.

Is the following way the correct way to do it ? :

The number density of particles is:
$$n=\int_{0}^{\infty} n_p(p) dp $$

The pressure is:
$$P=\int_{0}^{\infty} \frac{1}{3} p v(p) n_p(p) dp $$

The energy density is:
$$e=\int_{0}^{\infty} \epsilon(p) n_p(p) dp $$

where:

$$n_p(p)= (2s+1)\frac{1}{ e^{({\epsilon(p)-\mu})/{k_B T}}+(-1)^{2s+1} } \frac{4\pi p^2}{h^3}$$

Here $s$ is the spin of the particles, for electrons $s=\frac{1}{2}$.

$$ \epsilon(p)=(m^2c^4+p^2c^2)^{\frac{1}{2}} $$

$$ v(p)= \frac{d\epsilon}{dp}=\frac{p}{m}\left(1+\left(\frac{p}{mc}\right)^2\right)^{-\frac{1}{2}} $$

From calculating the three integrals above we can finally obtain $e(P,n)$.

Can anyone confirm this is the proper way to do it, or am I missing something here?
It seems as if those integrals cannot be solved analytically - is this
true?
Perhaps in this case there is no explicit formula for $e(P,n)$?

Your equations seem correct. Note, you can also get $v$ without having to differentiate, from $E=\gamma mc^2$ and $p=\gamma mv$.

Here are some notes on relativistic fluids as related to stellar interiors, including derivation of these equations and their solutions in the non- and ultra-relavistic limits http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf

Correct, the integrals cannot be solved analytically (only in the two limits) so there is no general analytic formula


How old are the Martian blueberries? The martian blueberries (only blue in false color) found by the Opportunity rover in Meridium Planis are formed by slow evaporation in mineral-rich liquid water. They obviously formed early in the Martian history, but do we know more precisely when these blueberries formed?

Let's call them "hematite spherules".
In principle, things on the Marsian surface can last a long time, billions of years https://sci.esa.int/web/mars-express/-/55481-the-ages-of-mars and https://science.sciencemag.org/content/343/6169/1247166.full state that a mudstone sample was deposited before 1.6Gy.
Unfortunately, the answer so far is we do not know. As longs the formation processes of the spherules aren't known, everything is open. If they formed in mid-air during a meteorite impact they may just be a few years old (https://www.sciencedirect.com/science/article/abs/pii/S0032063314000245) (*). But if they formed in a similar process as those on earth they may be older than 3 billion years (https://advances.sciencemag.org/content/4/12/eaau0872.full).


Calculate Local Sidereal Time I am trying to follow this guide and move to a programming language
but I have doubts.

What means the strings A, B and C in this manual?
http://aastro.org/wp-content/uploads/2010/11/Calculation-of-Local-Sidereal-Time.pdf

A highly authoritative source is Explanatory Supplement to the Astronomical Almanac 3rd ed., Urban & Seidelmann, 2013.

Chapter 1 gives the introductory explanations, without all the refinements if you want millisecond accuracy. Page 12 explains that

local sidreal time = Greenwich sidereal time + east longitude

(This book assumes the reader is able to convert between angle notation and time notation whenever necessary.)

If accuracy of 0.1 second is sufficient, the formulas given by the US Naval Observatory should do the job. Note that for subsecond accuracy you will have to determine UT1, which can be up to 0.9 second different from the commonly available UTC. You will have to look up the difference each time you do the calculation because the difference cannot be predicted more than about 6 months into the future.


Could the dinosaurs have seen the asteroid that killed them? Wikipedia says the Chicxulub impactor is thought to have been a 10-15 km diameter object. Would it have been visible to a (human*) naked eye before impact? And if so, would it have appeared like a star that grew brighter and brighter each night?

* I know, there were no humans at the time.

The answer is yes; for a few nights prior to the impact (assuming they had eyes with a similar sensitivity to our own and could look up!). It could also be much longer than this if the object was a cometary body.
Details:
Impacting solar system objects would have relative closing speeds from around 11 to 72 km/s.
We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which probably precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid Vesta. This has a diameter of around $a=520$ km, gets as close as $d=1.14$ au from the Earth and has a maximum brightness of about $m=5.2$ apparent magnitude (and is hence just visible to the naked eye) and an observed flux $f = f_0 10^{-0.4m}$, where $f_0$ is a zeropoint for the magnitude scale.
Thus the flux $f_n$ received by a near-Earth asteroid of diameter $a_n$, at a distance $d_n$ from Earth (in au) and with the same reflectivity would be
$$ f_n = f\left(\frac{a_n}{a}\right)^2 \left(\frac{1+d}{1+d_n}\right)^2
\left(\frac{d}{d_n}\right)^2$$
The magnitude of the dinosaur killer would then be
$$m_n = m -2.5\log (f/f_n)$$
To be an at all conspicuous naked eye object, $f_n \geq f$. If we assume the dinosaur-killer had $a_n=10$ km, then
$$ d_n^2(1+d_n)^2 \leq 0.0022$$
An approximate solution is obtained by assuming $d_n \ll 1$ and thus we find
$ d_n \leq 0.047$ au or 7 million km.
Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus hitting the Earth about 3 days after becoming a naked eye object. Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun or had a smaller albedo than Vesta.
Another possibility is that the object is of a cometary origin with an icy composition. If that were so then it could be much brighter as a result of sublimation, outgassing and having a bright cometary nucleus and tail. he answer would still be yes, but the visibility period could be weeks (comets are rather unpredictably bright).
It thus seems to me that there is a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a few nights, but probably not much longer than that unless it was a comet.


Can you find your own location in space by measuring angles between stars? Imagine you have 3 star points in 3D space A,B and C.
From your unknown position P in space you can measure the angles APC, APB, BPC and you know coordinates of A,B and C and therefore the distances between AB,BC,CA. Is there a way to calculate your own position P? Imagine you are floating in space and you're not on a sphere like earth.

Perhaps this is a frequent problem in astronomy and am wondering if there are any techniques to solve it. If you think there is, please point me in the right direction!

Greetings,
Matlab M.

The related observation in astronomy is stellar parallax:
the observer knows the Earth's position relative to the Sun, measures tiny annual oscillations in the direction to another star, and estimates the distance to the star.
A star at a distance of 1 parsec (pc) appears to shift by 1 arcsecond (4.85 μrad) per astronomical unit (au) of lateral displacement.
Since parallax is inversely related to distance, the relative error of a parallax-based distance estimate increases with distance.
For any given star, the position uncertainty ellipsoid is shaped like a needle pointing toward the Sun.

Aboard an interstellar spacecraft with limited observational facilities, it may not be practical to measure angles between nearby stars directly.
Instead you might individually measure their positions relative to several distant reference stars near them in the sky.
For best results, you would need to account for the stars' proper motion over the duration of your journey.

After doing that data reduction, you can compute angles APB, BPC, APC fairly precisely.
However, the Sun distances SA, SB, SC have significant uncertainties which propagate to other quantities computed from them, e.g. distances AB, BC, AC.
You can estimate distances PA, PB, PC by the law of sines, but these are uncertain by similar amounts.
Alternatively you can estimate PA, PB, PC by comparing apparent magnitudes to absolute magnitudes, which are subject to the same uncertainties.
Then you can find the intersection of spheres around A, B, C to estimate position P, but still no more precisely than distances SA, SB, SC.
P is not so much a point as a probability distribution cloud which expands as you move away from the Sun.

In the real world, our most distant spacecraft Voyager 1 is only 145 au (0.0007 pc) away.
Within about 400 times that distance, or 1/5 the distance to α Cen, the Sun is the brightest star in the sky by far and would remain your primary point of reference.


Why is there a matter/anti-matter asymmetry in the universe? After the big bang occurred what biased the formation of particles over anti-particles? Why are particles more common than anti-particles?

Currently, there's no single explanation for this.

The simplest explanation is that there isn't any imbalance — there are large antimatter bodies in the universe; just separated from us. However, these would have to be pretty far away (otherwise the boundary annihilation would be detectable). Theoretically, this requires antimatter to have clumped early on, which is rather unlikely/impossible.

A far more likely explanation is that charge-parity symmetry is violated. In other words, flipping the charges and parities in a reaction may not lead to the same dynamics — the outcome (when the charge/parity is flipped back) or distributions of outcomes may not be the same. Usually, we expect that if we flip the charges on both sides of a reaction, the reaction proceeds in the same manner (with same probabilities, etc). There is evidence that suggests that this may not be true. If this is indeed the case, then the matter-antimatter imbalance is not that hard to explain.


What is the angle between the equator of the sun, and the plane of the Earth's orbit I want to know the angle between the plane that passes through the centre of sun and is perpendicular to its spin axis, and the plane that contains the orbit of earth around sun?

Is this a fixed value or it is variable over time (and if so, how does it vary) !

The plane that contains the orbit of the Earth is known as the "ecliptic". The rotation of the sun is tilted by 7.25 degrees to the ecliptic, and this value does not vary over time.

The rotation of the Earth is also tilted, by 23.45 degrees to the ecliptic, it is this angle that causes seasons.

An old paper Position of the Sun's axis describes how the Earth crosses the solar equator on June 4th and Dec. 6th each year.

Nasa has a factsheet, that has this and other information about our nearest star.


What is faster than a supernova explosion? Is a supernova explosion really fast or is it just fast, metaphorically speaking? Can you cite the fastest known phenomenon in astronomy (neutron stars, pulsars) besides supernovae in terms of explosions?

EDIT: yes, i mean the shockwave from supernova

I am assuming you are talking about how long the event lasts. See question on this site "How long does a supernova explosion last?". The answer is that if you are talking about how long it brightens then dims (its light curve), then it typically takes a couple of days to get to full brightness then weeks to months to fade. Core collapse can be a fraction of a second (keep in mind there are different kinds of supernovae).

The "fastest" pulsar found so far spins 716 times a second. If you really want fast, how about the standard for atomic clocks? The used transition is between two ground states of the element caesium that are very close together and lasts about $1\times 10^{-10}$ seconds. There are even better atomic clocks now.


Compute Planet's Apparent Visual Magnitude In a hypothetical solar system there exist:

a sun of radius $r_s$ and absolute visual magnitude $V$
the Earth, with radius $r_e$ and distance from the sun of $1\,AU$
another planet, wit radius $r_p$, albedo $a$, distance from Earth $d_e$ (in units of $AU$), and distance from the sun $d_s$ ($AU$)

Assume that the second planet is magically lit up entirely by the sun (so it does not have phases when viewed from Earth).

Now how can I compute the apparent visual magnitude of the second planet with respect to Earth?

If there are other parameters required, please tell me.

Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.
Now, the Sun has a luminosity of $L_{sun}$ such that the Solar flux as seen by the planet is
$$
 f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }.
$$
The cross section of the planet is about $A = \pi r_p^2$, so for a given albedo $a_p$ the reflective luminosity of the planet will be
\begin{aligned}
 L_{planet} &= a_p f_{planet} A \\
            &= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }.
\end{aligned}
Given that we know the absolute magnitude of the Sun, the absolute magnitude of the planet follows as
\begin{aligned}
 V_{planet} &= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\
            &= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun}
\end{aligned}
The apparent magnitude of the planet as seen from Earth can then be calculated as
$$
 m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5
$$
where the distance between Earth and the planet, $d_{e-p}$, should be in parsecs and of course depends on the orbital phase of the two respective planets.
So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.


How was Cassini able to measure and use the Earth-Mars distance to get the Earth-Sun distance? How was Cassini able to measure and use he distance between Earth and Mars to get the Earth-Sun distance? How were the ratios of Earth-Mars and Earth-Sun calculated then? Given Kepler's laws all we need to now find is orbital period? How did they manage to do this without distance?

Giovanni Cassini used parallax with a friend, from two different sides of the Earth.

It was generally accepted that the stars didn't move relative to each other or relative to the motion of the Earth.   The planets moved, the stars didn't.

Cassini was before Newton but after Kepler, so he had access to Kepler's 3rd law which is the law of periods, which tells you the ratio of Mars' distance to the Sun and Earth's distance to the Sun.     (Estimates like that could also be done with angles before Kepler), But, if Cassini could measure a single distance, for example, Earth to Mars when Mars passes at it's closest point, then the other distances can easily be worked out.

Cassini sent a friend and fellow astronomer, Jean Richer, to French Guiana and they both took the position of Mars at the same time from different sides of the world, nearly 12,000 km distance.

Mars' distance at closest pass is about 54 million km in distance, so the variation in angle from one side of the Earth to the other is a ratio of 1 to about 4,500, which is a bit under 1 minute in angular variation.

I don't know how Cassini and Jean Richter were able to measure to such a high degree of accuracy, and whether there was some lucky guessing in there, or if they had equipment that good (if anyone knows, please feel free to make a more detailed answer).  But that's the gist of it.   Mars is in a different place relative to the background stars when viewed from different sides of the Earth and by measuring how much Mars moves from those two points on Earth, an estimate of distance can be made.

Cassini's estimate was, it turns out, surprisingly accurate, within about 7% of the actual distance.  His astronomical unit was 140 million km or 87 million miles.   His method works, though the elliptical orbit of Mars makes it a bit harder.   The real problem, as was often the case in early astronomy, was how accurate the measurement was.

Short article here

Longer article here

Also discussed here, under 1671-1673

The 2nd article says the angle is 20 arc seconds.   When I do the math, I get slightly about 46 arc seconds.   So, I invite correction if I made a math error.     20 arc seconds is even harder to measure, so any details on Cassini's margin of error would be welcome too.     I suspect there was concern that his method lacked accuracy because the Venus transit 100 years late seems to get a lot more attention as the first accurate measurement of the orbital distance of the planets.


Why are solar telescopes built on lakes? The telescope big bear is located in the Southern California.
According to the wikipedia page, water can cool the observatory.
Is it better to build it on a cold high mountain with stable and clear atmosphere?
What is the difference when building a solar and an optical telescope?  Optical telescopes are usually not build on lakes.

To amplify andy256's comment, the problem that solar telescopes face is that heating of the surrounding ground during the day gives rise to turbulence in the air near the ground, making the observing conditions worse (think of the heat shimmer just above the surface of hot pavement or a hot road -- that's turbulence bad enough for your naked eyes to notice). Since water has a high thermal inertia, it doesn't heat up as much as land does, so you get less turbulence. (Another approach is to put the telescope on top of a tower, above the worst of the turbulence.)

Note that Big Bear is at an altitude of 2,000 meters, so it is at a moderately high altitude.


Cycles that are longer than the year, but less than 1000 years Wondering what primary astronomical cycles there are that are greater than a year (which is a cycle around the sun), but less than 1000 years-ish (on that scale). There is the cycle of the moon phases too, and the cycle of the earth making a full rotation (a day). But I wonder if there are any cycles at a bigger-than-a-year scale, but not too big. I guess there's also the cycle of a stars life (but that is kind of a big scale, on the order of billions of years), but I am more wondering in terms of rotation what sort of cycles there are. I guess there is another one, Axial Precession, but that is a bit large such as on the order of tens of thousands of years. Maybe there are some cycles of the planets in which they are visible from earth, or of their moons, or of comets, or of nearby stars or that kind of thing.

If there were ones around each of these points that would be nice: 2, 5, 10, 20, 50, 100, 500, or 1000 years sort of thing.

Basically I am looking for "interesting" or "popular" cycles rather than just picking any of them (since every planet and moon has an orbital period). I am looking for ones that (a) we can observe most easily either directly without a telescope, or with a simple telescope/binoculars, and/or (b) have significance of some kind historically perhaps. If there is one that is historically significant or historically popular (say something Newton or Kepler talked about for example) but is difficult to see, that would be okay too.

For example, I vaguely remember hearing people mention "hey these x and y planets just aligned, it only happens every x years", so did a quick search and found this saying Jupiter and Mars align every 2-3 years. That is pretty interesting, since it is a small timeframe, bigger than a year, and you can see it pretty easily. I would like to find a few others though, spread over the scale (as mentioned above) from 1-1000 years. Here is another interesting one, a "Super Blood Blue Moon" as NASA is calling it, which hasn't happened since 1866 (so let's say ~150 years). Not sure if that will cycle every 150 years like that, but something that does cycle similar to this is also exactly the sort of "cycle" I am looking for.

The Moon does not follow the ecliptic precisely like the Sun does; rather, it follows a path inclined of about 5.14°. The two points where the ecliptic and the Moon's path coincide are called 'nodes', and they travel around the ecliptic in a period of 18.66 years. This is a combination of the orbital period of the Moon, also called the sidereal month, about 27.32 days, and the slightly shorter draconic month, the average period between the moment where the Moon traverses a node, which is 27.21 days.

Once every 18.66 years, the ascending node coincides with the vernal equinox, and the Moon will reach declinations of about +28° and -28° (much more than the Sun, which reaches ±23°); this is called a 'major lunar standstill'. 9.33 years later, the opposite is true, and the Moon will only reach ±18° during a 'minor lunar standstill'.

According to Wikipedia these moments have had historical significance, all the way back to the Bronze Age:

  This time appears to have had special significance for the Bronze Age societies, who built the megalithic monuments in Britain and Ireland. It also has significance for some neopagan religions. Evidence also exists that alignments to the moonrise or moonset on the days of lunar standstills can be found in ancient sites of other ancient cultures, such as at Chimney Rock in Colorado and Hopewell Sites in Ohio.


Where is the center point for the Supergalactic coordinate system? I'm trying to build a 3D visualization of the Supergalactic coordinate system and couldn't find any reference to where the center point should be (the Sun, galactic center, the Earth, etc).

Also I already tried to look into some articles and books, but they only refer to the zero point related to the galactic coordinates.

Where should it be?

Could I use this model as the reference for the Supergalactic Coordinate system, starting to count the SGL (L) and SGB (B) from the yellow line?

It is still unclear to me if (l=137.37 b=0) indicates just the direction of the center of the Supergalactic Coordinates System.

Note: The Sun is being used as the center point.

Plano Galactico translates into Galactic Plane.

Centro Galactico translates into Galactic Center.

According to Wikipedia, the origin of the system is:

The zero point (SGB=0°, SGL=0°) lies at (l=137.37°, b=0°). In J2000 equatorial coordinates, this is approximately (2.82 h, +59.5°).

In the same article there is a recent referred paper on this topic, which explains [The zero point] is one of the two regions where the SGP is crossed by the Galactic plane.

Perhaps you already know this website, which shows many good pictures. I was trying to find a better figure, but I guess you have to push your imagination according to the definition.

EDIT: According to this, SG coordinates are similar to the galactic ones, so that the North Supergalactic Pole is defined in Galactic coordinates, which means that they have the same origin (i.e., the Sun).


Is it possible to overcome the problem of blind spot(s) of current gravitational wave detectors? If gravitational wave pass through specific points (which are know as blind spots) current GW detectors aren't able to detect the passed waves so over the time will we be able to completely overcome this problem or will it be there always upto certain extent?

The blind spots are caused by the way the detectors work. They are sensitive to a gravitational wave (GW) changing the relative path length along interferometer arms at right angles to each other.

Gravitational waves come in two polarisations (plus and cross). These polarisations cause alternate perpendicular expansions and contractions in space, but are rotated by 45 degrees with respect to each other (plot from Kalmus 2009).

A GW source would normally be a mixture of both. The sensitivity to each polarisation depends on the orientation of the interferometer arms with respect to the source direction. For example, a source that is directly "overhead" will only be detected in the polarisation state that lines up with the arms and not in the other, because it would make the arm lengths change by the same amount. e.g. Imagine your detector is lined up with the x and y axis indicators in the picture above, then only the plus ($+$) polarised waves would be detected.

However, if the waves come from a source in the plane of the interferometer, then neither polarisation causes a relative difference in the arm lengths if the source lies along the bisector of the two arms or along an equivalent line at right angles to this. Thus there are 4 blind spots on the sky. Note that if a source is perfectly polarised (e.g. an edge-on merging binary system), then there will be addditional blind spots.

There are several solutions: make your interferometer 3 dimensional (i.e. add a third arm at right angles to the other two); make a triangular interferometer (this is what is planned for space interferometers); build another interferometer with a different plane/orientation to the first.

As an example, the plots on the left in the picture below show the sensitivity of LIGO (Washington and Louisiana) to unpolarized GWs at the top. These two interferometers are nearly lined up and nearly in the same plane, so you see the 4 blind spots clearly. The middle plot on the left shows how things improve when you add VIRGO (in Italy) - there are essentially no blind spots in the combined instrument, and then in a planned 6-detector network at the bottom. (Plot from Andersson et al. 2013).


Please check my Mars photo Hi built my own newtonian telescope for the first time. Up until that point I had never used a telescope.
Can you please look at the photos attached and tell me if this is what Mars should look like as I'm disappointed somewhat.
Specs are as follows:

Picture taken with smartphone through the eyepiece and no filters were used or edited.
TMB Planetary Series 2.5 mm 3.2 mm eyepieces
10" (25 cm) Newtonian with f = 1470 mm

uncropped original

(Much of this echoes what antlersoft says in their answer)

For a phone photo through the eyepiece that looks about right to me!
The size... the brightness... both are as I expect.
What you could try is to use the manual mode of your phone's camera and set the ISO down to minimum (100) and the shutter speed down to something like 1/60s.
Take a few shots, pick the best one, and make a zoomed+cropped version.
You should be able to see a point of white on the south pole and a hint of dark patches.
If the seeing's bad then you'll get awful results no matter what, so wait until the view is nice and crisp.
Now, when you look through the eyepiece, how does it look to your eyes?
Is it too bright? If so then keep watching the planet and your eyes will adjust and the contrast will improve.
Is it too blurry? If so then back off the magnification a bit until it's not blurry any more. A smaller crisp image is much better than a larger blurry image - your eyes will be able to pick out small details in a tiny image with time and practice.

Here's a lucky snap I got a few weeks ago:

And here's a zoomed and cropped version:

And here's one from one minute beforehand with automatic camera settings:

These were taken through the eyepiece of a 10" F/5 Dobsonian - very similar to yours. (I can't remember which eyepiece I had in at the time, I'm afraid! I think it was a 4.5mm but it could have been an 8.8mm with 3x Barlow lens, or even a zoom lens plus the 3x Barlow...all I do remember is that the seeing was exceptionally good that night and I was making the most of it!)
The good one was ISO 100 and an exposure time of 1/60s. The bad one was ISO 100 and an exposure time of 1/17s.
Hopefully these photos demonstrate that everything has to come together to be able to get a halfway decent photo: the seeing, the collimation, a steady hand, the camera settings... :-)


Can dark matter decrease the Jeans length? I am wondering if there are any models of interstellar cloud collapse out there which take dark matter into account.

If dark matter has local perturbations or density fluctuations, then that may be a large enough effect to cause collapse of an interstellar cloud. This might increase the mass and density of a cloud enough to overcome the kinetic energy of hydrogen atoms leading to collapse.

Are there dark matter density variations at the scale of interstellar clouds?

This is an area of active research. The current Cold Dark Matter (CDM) paradigm predicts bottom-up formation of halos, whereby smaller halos coalesce in to the larger halos we observe indirectly today through X-ray measurements of clusters. This suggests that small dark matter halos wouldn't survive to the present day.

There has been research in to the possibility of mini haloes surviving, but interactions with dense baryonic structures, such as stars, as well as streaming by an irregular galactic potential tend to wipe out any small overdensities in the distribution.

However, if such mini haloes did exist in the early universe they could have contributed to the collapse of primordial gas to form Population III stars. Simulations have been carried out to investigate such a model, with Population III stellar masses comparable to those expected $(\sim100 M\odot)$.


VSOP for exosatellites? (Io, Titan, Ariel, etc) VSOP87 (and the subsequent VSOP2000 and VSOP2013) provide reasonably
accurate planetary positions while still maintaining a small file size
compared to the Chebyshev polynomial files.

VSOP includes our moon, but has anyone created something similar for
other planets' moons (such as Jupiter's Io, Saturn's Titan, and
Uranus' Ariel)?

I'm thinking about trying to create something like this and wanted to
check that it hadn't already been done.

Yes, there are several I know of for the Galilean moons, less so for other satellites.  One very good place to start is from the same organization that created the VSOP series, now called IMCCE.  Here is a link to their satellite ephemeris.  According to the Explanatory Supplement to the Astronomical Almanac, most of the other moons of Jupiter behave to chaotically to create a similar theory and numerical integration is required to produce accurate results.


Why do some planets have rings? Some planets, specifically Jupiter, Saturn, Uranus, and Neptune in our solar system, have planetary rings.

Why do some planets have rings? How are they made and from what? Most importantly, will I be able to observe the rings on any planet with an amateur telescope?

Rings are made up of tiny (and not so tiny) pieces of rock and ice that are in some way the bits "left over" from the formation of the planet. The theory involves the Roche limit - and is that particles that are already within this limit can't accrete into a larger body because of the tidal forces involved.

Another theory is that they are formed when a moon comes closer to a planet than the Roche limit, the tidal forces cause it to break up and form a ring. Though the presence of "shepherd" moons in the rings of Saturn does hint that this may not be major source of material.

Both explanations, to me, imply that you'd only get major ring systems around larger (gas giant) planets, though it doesn't preclude rings around smaller (rocky) planets. This seems to be borne out by our solar system where the gas giants have rings whereas the rocky planets don't.

Source

As to being able to see them, you should be able to see the rings of Saturn with an amateur telescope which has a 50 - 100 powers of magnification. With binoculars you'll probably see a misshapen blob.

Source


How to understand this multi-day exposure photograph of Jupiter and the Moon This answer includes a stunning photo of the Moon and Jupiter "Trails of the Shifting Moon" by photographer Jin Lu.

The caption reads:

  From the photographer: "I spent 4 days capturing the trail of the moon from the balcony of my house. This is a composite of 90 pictures, showing the shift of the path of the waxing Moon from crescent phase to near first quarter. Random floating clouds decorated the trails. The interval between each day's sequence of pictures ranges from 1.5 to 2 min. Besides the Moon, the most obvious celestial object in the composite is Jupiter. The trail of Jupiter is from pictures of Jan 9, the day the moon appeared closest to Jupiter in the evening sky. Although Jupiter looks closer to the Moon on Jan 8, that is not how it actually appeared then." - Jin Lu

I'm confused by two things:

The Moon is shown for January 7, 8, 9, 10 and the caption says "The trail of Jupiter is from pictures of Jan 9". I'd expect Jupiter to have contributions from all four days, not just one day. How can that be?
"Although Jupiter looks closer to the Moon on Jan 8, that is not how it actually appeared." If I measure the distance in pixels from the top-left Jupiter to the top-left Jan 8 and 9 Moons, the Jan 8 Moon is closer. How could that not be true in reality then?

Looking toward the western horizon from 33°N latitude, hour angle increases in the 11-to-5 o'clock direction, and declination increases in the 8-to-2 o'clock direction.
The Moon was at -2.8° declination on Jan 8 and +2.2° on Jan 9, while Jupiter was near -2.0° on both nights.
Where they overlap in hour angle, the Jan 9 Jupiter images appear closer to the Jan 8 Moon images.

For the Moon to appear at similar hour angles on consecutive nights, the photographer must have started taking images ~40 minutes later each night, increasing Jupiter's hour angle ~10° per night.
On Jan 7 and 8, Jupiter was probably still above the ~20° high frame at the ends of those short series of exposures.
When the Jan 10 series began, Jupiter was likely already behind the building.

The scenario is reproducible in Stellarium if the location is in the southwestern US and times are evenings in January 2011.


What do 5-sigma, r, and lowercase-delta mean in the physical descriptions of major telescopes? The typical 5σ point-source depth in a single visit in r will be ∼24.5 (AB). The project is in the construction phase and will begin regular survey operations by 2022. The survey area will be contained within 30,000 deg2 with δ<+34.5∘, and will be imaged multiple times in six bands, ugrizy, covering the wavelength range 320--1050 nm. About 90% of the observing time will be devoted to a deep-wide-fast survey mode which will uniformly observe a 18,000 deg2 region about 800 times (summed over all six bands) during the anticipated 10 years of operations, and yield a coadded map to r∼27.5.

I believe 'five-sigma' refers to standard deviations in a Gaussian distribution (statistics), but am unclear what it has to do with a 'point-source depth'.
And I am even more confused by 'r' and 'lowercase-delta'....

The quote appears to be from the Ivezić et al. 2019 paper about LSST.
In this context, σ quantifies the noise level in the images, and r is a photometric passband of ~620±70 nm wavelength.
In single r-band images, they expect stars of AB magnitude 24.5 to stand out 5σ above the background level.
δ refers to declination; they plan to cover the sky from the south celestial pole to 34.5° north of the equator.


How low can the sun be seen in the sky? I'm interested in knowing how low the sun could ever get in the sky and still be seen.

More technically: what's the minimum possible elevation of the sun at which it can still be seen above the visible horizon from a terrestrial vantage point?  And where on Earth would you go to make such an observation?

Some clarification:

elevation is angular distance measured from the astronomical horizon (the great circle that is 90° from the local vertical).

For example, I imagine from a mountain top near the ocean, you might be able to view the sun at an elevation angle that is negative, below the astronomical horizon but above the visible horizon, and possibly by a decent amount.  So then I thought Mt. Everest might be the spot, but locally I'm not sure how low the visible horizon is, because the surrounding area is at a high elevation too.  So maybe I'm just looking for the highest angle possible between mountain peek and sea level, in other words, where is the lowest apparent visible horizon, but the sun may not necessarily ever rise or set in that direction.

I also looked around for a map of local visible horizon angles (measured relative to the astronomical horizon), but couldn't find one.  I suppose such a thing could be constructed from a topographical map of the Earth, right?  Maybe this would be the best way to "solve" this if nobody already knows.

I'm hoping the answer can ultimately be turned into a factoid of the following format (looking to fill in the unknowns in this):

  "From the top of mount something-or-other, looking west, the visible horizon is x° below the astronomical horizon.  From this vantage point, just before sunset, the edge of the sun could be observed to be f(x) sun diameters below eye-level."

The lowest exposed point on land is the dead sea shores, at -413metres below sea level. so theres a good chance you could find a spot that lines up correctly to give you a view of a piece of the horizon well under sea level. I dont believe that elevating yourself is the key factor here though. At Around 35000 ft you can start noticing the curvature of the Earth, the horizon being 200 some odd miles away (Pythagorean theory -theorem?- can be used for this, several sources on google explain the steps to find your equation, and far better than I can. ) but the distance to the horizon grows rapidly. So you would certainly  want to find the sweet spot of height /distance, too high and you wont know when the land has ascended back to sea level, as the terrain will become more vast and harder to judge detail or distance. Too close to the shore and your only seeing a few miles but the angle of sight wouldnt be the best possible for finding the lowest horizon. You would want to work out the math and find a spot that puts the dead sea shores as your visible horizon, bc if your standing there, watching the sun rise, the sun is rising over a horizon that isnt the lowest accessible location...


Will the Magdalena Ridge Optical Interferometer be able to image extended objects like the surface of the Moon? Inspired by several questions:

When will a moon landing site be visible via telescope?
Could the E.H.T. produce an image of the human artifacts on the moon?
Picture of equipment left on the Moon?
Were the Apollo lunar activities observed from Earth?

If one wanted to resolve 1 meter or smaller detail on the surface of the Moon from the surface of the Earth (about 2.6E-09 or 0.5 mas) at say 1 micron wavelength one would need a baseline of order 400 meters.

The longest current optical baselines are only 40 to 80 meters and the longest one currently under construction is the Magdalena Ridge Optical Interferometer which

  will have ten 1.4 m (55 in) telescopes located on three 340 m (1,120 ft) arms. Each arm will have nine stations where the telescopes can be positioned, and one telescope can be positioned at the center.

This is sufficient to have of the order of 1 meter resolution at the Moon's distance, but being optimized for star-like sources it's not clear if it will be able to image extended objects like the surface of the Moon.

Question: Will the Magdalena Ridge Optical Interferometer be able to image extended objects like the surface of the Moon, or is it designed only to separate a few star-like objects, e.g. binary stars or star + planetary systems?

note: If information on this specific observatory isn't available, it would be certainly informative to extrapolate from existing imaging work from long baseline optical interferometers. The Moon presents a big challenge since its surface brightness extends over quite a large solid angle, so pinholes at the focus of each telescope in the array would generate a lot of diffracted/scattered light, whereas imaging star-like objects against a dark field would be less susceptible.

According to this site The 10 telescopes will be optically linked together in order to make images of astronomical objects with unprecedented detail. The interferometer will have a resolution 100 times greater than the Hubble Space Telescope and will be able to make accurate images of complex astronomical objects many times faster than other existing interferometric arrays.

And: The Interferometer will take delivery of the second telescope enclosure in February 2020 and the second telescope in August 2020. They expect to fully incorporate the second telescope by the end of 2020, which will allow the instrument to produce “fringes,” using the proprietary fringe-tracker called ICoNN.

There is also this headline:

  How America's Spooks Seek to Spy on Distant Satellites The
  intelligence community has plans for a telescope network that can see
  not just a blob in orbit but details such as a satellite’s solar
  panels.

from here.

That's one reason why the U.S. Air Force, which wants to monitor its own orbital assets and presumably those of others, is funding MROI. "They want to know: Did the boom break or did some part of the photovoltaic panels collapse?" says Michelle Creech-Eakman, an astronomer at the New Mexico Institute of Mining and Technology in Socorro and project scientist on MROI. But if the facility succeeds, its biggest impact could be on the field of astronomy, by drawing new attention to the promise of optical interferometry, a powerful but challenging strategy for extracting exquisitely sharp images from relatively small, cheap telescopes.

Radio astronomers have had it easier. The long radio wavelengths mean data from separated dishes can be recorded, digitized, time-stamped by an atomic clock, and combined later for analysis. But optical interferometry is far trickier: The short wavelengths of visible light, running at terahertz frequencies, cannot yet be digitized by any electrical system. So the light must be merged in real time, with nanometer precision.

From here.

Also Reference 3.

http://www.mro.nmt.edu/about-mro/interferometer-mroi/


What causes hot temperature extremes? Recently I looked into the hottest(and coldest, but in another question) natural places. I find this to be a "cloud of searing gas that is surrounding a swarm of galaxies clustered together five billion light-years away in the constellation of Virgo". This place has a temperature of 300 million degrees Celsius, which is super hot even compared to the temperature at our Sun's core. A quick search why gives:

  A group of galaxies collided violently with another swarm of galaxies at a speed of 2,500 miles per second.

    https://www.skymania.com/wp/hottest-spot-in-universe-found/

However, what mechanism exactly causes this temperature? Friction? Or something else? Would stars there be fusing stuff from the outside?

Within that galaxy cluster, things become chaotic as these galaxies are flung around but kept in orbit by their strong gravity. And as Steven Linton mentioned,

  the galaxy's organized motion is turned primarily into disorganized
  motion within the material that is stuck together, which is what we
  know as heat. So the atoms of gas in these clouds collided and bounced
  off in all directions, where they hit other atoms, until the two
  clouds had formed one, with huge random internal motions (i.e. very
  hot).


Why is there so much methane in space? In school I learned that methane is organic matter and that it is a possible product of crude oil refinement. However, recently I read that Uranus has huge methane storms and the Saturn moon Titan even has lakes consisting only of methane or ethane.

Now, I'm wondering, how does all that methane get out there?

Is it the result of organic processes or is it created in the cores of suns like other elements? Do we even know?

Methane is also a pretty complex material. Do we know sources of more complex molecules other than methane in such quantities? Except life on earth?

Often for molecules to form in interstellar space, dust is used as a catalyst. The reason is that in typical interstellar environments, densities are so immensely low that even for just two atoms to meet, the probability is so small that formation time scales are very long. For 3+ atoms, the chance decreases rapidly. Instead, an atom can stick to a dust grain and wait for ages until other atoms stick. The atoms slowly "crawl" around on the surface of the dust grain, eventually meet and make bonds. If the formation of a bond releases energy (is exothermic), the molecule can be ejected from the grain surface. This process is call adsorption.

The dust also helps shielding the molecules from stellar radiation which would otherwise easily destroy them. This is why molecular clouds are also very dusty. But actually UV irradiation helps with the formation of very complex molecules by ionizing less complex molecules that subsequently can make bonds with other atoms and molecules.

If the environments are extremely dense, as when a star dies and ejects its gas either as a supernova or a planetary nebula, molecules can also form. This is probably also how the dust itself is formed, although it might also be formed later on (this is currently debated; the problem is that supernovae are so powerful that their shock waves tend to destroy dust shortly after it forms, and planetary nebulae are created from stars that live so long that they can't explain the abundance of dust in the very early Universe where they wouldn't have had the time to live their lives).

As Stan Liou and LocalFluff says, methane is actually an "easy" molecule to form, both because it's rather simple, and because its constituents, hydrogen and carbon, are the most and the fourth most abundant elements in the interstellar medium, respectively.

In fact, far more complex molecules are regularly found in interstellar space, as can be seen on this list.


What is the naming convention for temporary moons? Is there a naming convention for temporary moons? Or do they follow the naming conventions of other small bodies?
In the case of 2006 RH$_{120}$, 2015 HP$_{116}$, 2020 CD$_3$ and 2020 SO, the first number is the year. I assume the following letter+number combination somehow indicates the time of year of discovery and/or the number of similar bodies discovered prior?
How were YX205B9, WT1190F, S509356 and XC83E0D named?

These are just the standard minor planet designations.
Provisional designations
As described on the MPEC page on New- and Old-Style Minor Planet Designations, the new-style provisional designations for minor planets (like 2006 RH120) are in the form year, letter indicating the half-month, letter indicating the sequence within the half month, and number indicating the cycle of the second letter (no number = 1st cycle, 1 = 2nd cycle, etc.)
The letters for half-months omits the letter I and doesn't use Z as there are only 24 half-months in the year. The second half-month in a month starts on the 16th for all months.
The letter indicating the order is the alphabetic sequence, again skipping I but this time including Z.
Old-style provisional designations were used in the 19th century and are not relevant here.
Temporary designations
These are assigned by observatories according to their own rules, rather than by the MPEC. I think three of the ones you list (YX205B9, WT1190F and XC83E0D) were assigned by the Catalina Sky Survey, but I haven't seen any documentation of what their internal rules for assigning identifiers is. These objects were confirmed to be artificial before they had sufficient orbit determinations to be assigned minor planet designations, so never received provisional designations.


Deriving phase of a pulsating star from list of Julian date values of observing brightness I want to convert the list of Julian date values of observing brightness of a Cepheid variable into its phase.
The equation I found in the Internet is the following:

$$\mathrm{Phase}= \frac{\mathrm{HJD}-T_0}{P} - \mathrm{INT}\left[\frac{\mathrm{HJD}-T_0}{P}\right]$$
What is the $\mathrm{INT}$ in this equation? I only know that $T_0$ is the one reference point among the list of Julian date values and $P$ is the period.

Also, what kind of criteria exist in determining the reference point?

INT refers to the floor function. So $\mathrm{INT}(x)$ is the largest integer not exceeding $x$.

In this case, the function is used to get the fractional part of $\frac{\mathrm{HJD}-T_0}{P}$, i.e. the part after the decimal point. So if for example $\frac{\mathrm{HJD}-T_0}{P}=3.4$, then $\frac{\mathrm{HJD}-T_0}{P}-\mathrm{INT}\left(\frac{\mathrm{HJD}-T_0}{P}\right)=0.4$.


How old would I be if I travelled 1000 light years in one year If I were to travel 1000 light-years, and witness only 1 year elapse in my frame of reference, how much time would have passed in the frame of reference of the Earth?

The answer is sort of trivial. If you travel 1000 ly so fast that in your own reference frame it takes one year, then you will have aged by one year in your own reference frame. To do so, you will need a speed of almost the speed of light, so in the reference frame of Earth, you will have spent just a tad more that 1000 yr to travel 1000 ly.

In general, the time dilation is given by the Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$ so to be exact, your speed must be
$$
1000^2 = \frac{1}{1 - v^2/c^2} ⇒ \\
     v = (1-10^{-6})^{1/2}\,c = 0.9999995c
$$
so your journey will take
$$
t = \frac{d}{v} = 1000.0005\,\mathrm{yr},
$$
i.e. 1000 years, 4 hours, and 23 minutes in Earth's reference frame.

"Realistic" acceleration

As David Hammen comments below, this assumes that your spaceship accelerates instantaneously to $v$. There are infinitely many ways to achieve that speed.
The proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $d$ when traveling at a constant acceleration $a$ is
$$
\tau = \frac{c}{a} \cosh^{-1} \left( \frac{ad}{c^2} +1 \right).
$$
Solving for $a$ yields the acceleration needed. The most pleasant way would arguably be accelarating at $a\simeq 19.2\,\mathrm{G}$ for half a year, and the decelerate at the same amount for the rest of the journey. A more pleasant way in the beginning, but less pleasant in the end, would be to accelerate at $a\simeq9.6\,\mathrm{G}$ for one year, and then crash into planet WASP-142b, which lies at a distance of roughly 1000 ly.

Your journey, as measured by people on Earth, would then take
$$
t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right),
$$
which works out to roughly 18 and 36 days more, respectively, than in the instantaneous case. The reason that it doesn't differ that much is that you actually reach relativistic speeds pretty fast at this acceleration.

The largest G-forces can be with endured in the "forward direction", i.e. corresponding to lying on your back and accelerating upwards (accelerating along the direction of your spine tends to break it, and accelerating "backward" makes your eyes pop out). According to Wikipedia, "acceleration pioneer" John Stapp withstood 25 G for 1.1 seconds, so you might want to send him instead of yourself.


Why do stars seem not to move relative to each other? Whilst I understand why stars move across our skies,(Earth rotation) I struggle to understand why they do not appear to move relative to each other.

To explain:  Take the best know constellation the Plough.  It is a familiar site to most people.  But it has been there apparently un-changed for years.  If we (I mean earth, moon, stars and everything else ) are hurtling through space, then surely changes should be apparent unless we are all travelling in the same direction at exactly the same speed.

They do move - just far too slowly for you to detect by eye even over several human lifetimes.

  Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space

Douglas Adams, "The Hitch-hikers Guide to the Galaxy"

Even the closest stars are a very, very, very long way away so their apparent movement relative to each other is going to be very small.

You can see the same effect when looking out of a moving vehicle through the side windows. You see the objects closest to you rushing past, but objects on the horizon appear to move much more slowly. Now, scale that up by many orders of magnitude to interstellar distances and you'll see why the stars don't appear to move relative to each other.

There's software that simulates the night sky and you can run time backwards and forwards - if you wind it far enough you'll see the constellations change.