What happens over time as a neutron star cools? My understanding of a white dwarf star is that, although it hasn't happened yet due to the age of the universe, eventually as it loses energy due to radiating heat and tidal forces, it will cool to become a black dwarf - essentially a black ball of carbon.

What happens when a neutron star eventually cools? What is left, and do we know what it would look like?

The cooling history of a neutron star can be divided into an extremely rapid neutrino cooling phase, followed by an indefinitely long cooling phase due to the emission of photons from its surface.

The key to understanding neutron star cooling is to realise that degenerate neutrons possess almost no thermal energy, even when extremely hot - because the Pauli Exclusion principle forbids the neutrons from losing kinetic energy and falling into an already-occupied quantum state. Any cooling processes therefore act to reduce the temperature of the neutron star very quickly. Secondly, the interior of a neutron star is almost isothermal, due to the high thermal conductivity of degenerate gases, but the temperature at the surface is smaller by about a factor of 100 or so.

When a neutron star first forms, its internal temperature exceeds 10 billion kelvin, its surface temperature would be 100 million degrees and emit hard X-rays. Nevertheless, the small size of the neutron star limits radiative losses and instead it is neutrinos escaping from the interior that can cool it by a factor of around 100 in a few thousand years.

The neutrino losses proceed by something called the modified URCA process (somewhat faster direct URCA or processes involving pions/kaons maybe important in the first seconds/minutes of more massive neutron stars, but are blocked by complete neutron degeneracy and the comparative paucity of protons in the interior): cycles of neutron beta decay and inverse beta decay on protons produce anti-neutrinos and neutrinos. Bystander particles are required to conserve momentum. The temperature dependence of this process is $\sim T^{8}$, so if the temperature falls by a factor 100, the neutrino cooling rate falls by a factor of $10^{16}$.

Meanwhile, photon cooling from the surface, whilst initially negligible compared to neutrino cooling, falls only as $T^{4}$. After around 10,000-100,000 years, when the neutron star surface has cooled to about a million degrees, photon cooling, through soft X-ray emission dominates.

If nothing else happened to an isolated neutron star, it would continue to cool, such that $T \propto t^{-1/2}$. The vast majority of the estimated 1 billion neutron stars in our Galaxy, are in this state, whereas most of the pulsars we can observe are young and in the neutrino-cooling phase.

How cold can they get? The formula above suggests their temperature halves when their age quadruples. Following this, they could cool to the temperature of the Sun in around a billion years. However, there are reheating processes that can occur.

Isolated neutron stars can accrete material from the interstellar medium. This may be very little, since most neutron stars move fast. The gravitational potential energy is partly released as heat when it impacts the neutron star surface. Then, the low heat capacity of neutrons works in the opposite direction and comparatively little accretion is required to balance the photon losses.

Secondly, neutron stars are born as fast rotators and spin down. Angular momentum must be transferred outwards from the fluid interior to the outer crust. This is not a frictionless process. Thus some of the rotational kinetic energy can keep the star warm.

Thirdly, the neutron star has a huge magnetic field that gradually dissipates. Some of that energy may end up as heat in the neutron star through simple Ohmic dissipation of currents.

The bottom line is that the thermal emission from neutron stars older than about 100,000 years has not been observed, so we simply don't know how cool they might get. Some more information and references can be found in the Physics SE answer where I discuss where old neutron stars might appear on the HR diagram.


Why don't we see the galaxy center? When we look at the sky, and see the Milky Way, we mainly see a band, with no real center.

On a more accurate picture we can guess the center of the galaxy (see this picture), but we still don't see the center as some kind of big bright "star", like on this picture.

I know there is a lot of objects (stars, planets and dust) between us and the center of the galaxy. But stars are bright and they shouldn't "hide" the galaxy center's light since they shine too.

Why don't we see the galaxy center as a "night sun"?

How much dust is there so it's blocking the galaxy center's light?

Is it because of all Oort clouds of all solar systems that we don't see the galaxy center clearly?

The main reason we don't see the bright center of our galaxy, which is composed of millions of stars, is dust. Visible light is absorbed and scattered by interstellar dust, but that doesn't mean we can't see it on other waves of the spectrum, for example, infrared light doesn't suffer as much because of the dust.

Notice on this image how bright the galaxy center looks on the infrared and the near-infrared pictures!:

Image taken from the MultiwaveLength Milky Way website of NASA's Goddard Space Flight Center.


Collimator for Dobsonian, same as Newtonian? I have been looking on line for Collimator for my Meade Dobsonian. The only Collimators I can find are made for Newtonian Lenses. It's a 8" mirror with a 1.25" lens. Will this work for my telescope?

Yes, of course it will work. A Dobsonian is merely a Newtonian telescope on a special mount.


Why do type Ia supernovas produce more iron than type II My course book on astronomy states the following.

Older stars seem have higher oxygen abundances than iron. Explanation is that back in the days when these older stars were being formed type II supernova's were common, while type Ia weren't. So later, when the type Ia was becoming more common, the - now younger - stars were formed with higher iron abundances.

Why are type Ia supernova's better for iron enrichment than type II, and were these type II supernova's in any way better for higher oxygen abundances - or just less good at producing iron (and why)?

Context

Iron has the highest nuclear binding energy of all the elements (not completely true, but sufficiently accurate in an astronomical context). So, fusion of light elements into iron or something lighter is an exothermic process - you gain energy doing it, allowing the star to function. This is what happens in the last stages of a type II supernova. The core of a massive star in its last moments of life is hot and dense enough to fuse silicon into iron. Just before the supernova explosion, there is an iron ball of about 1.4 solar masses at the centre.

The progenitor of a supernova type Ia is a binary system where a "normal" star loses mass to a compact stellar remnant (a white dwarf). Once the white dwarf Contexthas accreted enough mass to be above a limit of 1.4 solar massses, fusion starts again, completely disintegrating the compact object.

Explosion

A SN Ia completely destroys the white dwarf progenitor in a runaway fusion process.

In a SN II, the pressure on the central iron ball exceeds the degeneracy pressure exerted by the electrons in the iron atoms' electron shell. The Fermi principle in quantum mechanics states that no Fermion (such as an electron) may occupy the same quantum mechanical state as another. The pressure exerted here is so large that the electrons of the iron atoms can no longer obey it and are forced into the nucleus, where they react with the protons to form neutrons.

Iron abundance

Why do SN Ia enrich their environment with more iron than SN II? This is not so much a matter of iron production, but about how much of that iron ends up in interstellar space where it can be part of a new generation of stars. In a SN Ia, the progenitor is completely destroyed, scattering all its constituent atoms into its host galaxy. A SN II forms a compact remnant, either a neutron star or a black hole. A lot of the later, heavier fusion products end up not being carried outward in the supernova explosion but become part of the compact remnant.

Note that a lot of the heavy elements scattered from a supernova of the "exploding massive star" result from an abundance of neutrinos escaping the central explosion and reacting with the outer shell of lighter elements that is blasted away.


Observing eclipsing binaries In a matter of weeks, I will be performing my first observation using a 25-cm Newton telescope (+CCD camera) and would be grateful for advice regarding an otherwise basic research that I have decided to try out due to this occasion.

I will be observing eclipsing binary stars (type Algol), yet have never used any program or site. Just to determine which binaries will be visible at my location on a specific date at a specific time - which sites do you use for this?

Also, what should be my main goals when taking photos in terms of timing and exposition?

Later on, I will use APT for photometry and estimate the time length of the eclipse, which will be my objective.

Thank you in advance.

www.skymaponline.net might be the one you want.

Main goal when taking images in terms of timing and exposition:
1. Choose time with good sky condition (e.g., clear sky, no wind)
2. Expose long enough for good S/N ratio, but not too long that CCD is saturated.
3. If you want to measure the length of the eclipse, you can follow since it is coming in until going out. Or, you can take a couple of images at different epochs, then assume some trajectory and just doing some math.


Simple mathematical models of solar systems I’m a hobbyist programmer with a profound love for procedural content generation.  I’m working on a 2D space exploration game and while it’s not meant to simulate anything real, I’d like to give it a natural feel.  I have no real knowledge in astronomy, so I figured someone here should have an answer.

I was wondering if there are a few simple mathematical models I could use to determine some parameters for my generation algorithm.

the orbital radius of the planets
the size of the planets (is it more likely that gas giants and the likes will show up closer to the sun or further away, etc.)
the number of planets orbiting a star

Also, a few questions come to my mind too:

is it a common occurence to have a star with nothing orbiting it?  Like just a single lone star sitting there in the middle of space.
IIRC, Earth is kind of a weird thing (besides the fact that it hosts life) because our moon is way bigger than the average moons and also the only satellite of the planet.  Is there any usual correllation between the planet size/density and the size or number of moons around it?
Is an asteroid belt something common in solar systems?

Thanks a lot for your help!

I "grew up" with the old roleplaying game 2300AD's system creation rules, which were based on fairly hard science (and used the Gliese star catalog as a map!) They have stood the test of time fairly well (here is an astronomer friend's take on it).

Here is a rough sketch of how I typically make my system generators, which is based on the same logic:

Generate a number $N$ of planets in the system. At present there is no clear data constraining this, although we may imagine high metallicity stars could have more planets (but maybe not).
Generate orbits $a_i$: start with a random innermost orbit $a_1$, and then make $a_{n+1}=(1+r_n)a_n$ where $r_n \sim U(K_1,K_2)$ is a random number. This imitates the Titus-Bode law pattern and looks roughly like observed systems. I tend to use $K_1=0.1, K_2=2$ but this is guesswork. Note that this process has problems handling hot Jupiters really close - you may want to add them by hand. It also ignore resonances etc.
Generate a planetary seed size/mass. This can be done in an elaborate way by looking at exoplanet $(a,M)$ distributions and extrapolating to unobserved corners, or using some suitable skew random number distribution.
From the mass and distance to the star, estimate planetary and atmospheric composition. Basically, outside the snowline ($\approx 1.4\sqrt{L/L_\odot}$ where $L$ is star luminosity) planets will start accumulating volatiles, and having ice crusts become possible. Essentially, decide on a core density randomly.
Calculate the approximate minimum molecular weight retained based on the surface gravity. This will determine the atmospheric density. Especially, if the planet retains hydrogen it will become a gas giant: multiply up the mass a lot (the overall radius of a planet scales as $M^{1/3}$ for solid planets up to a few earth masses, and up to Jupiter size for gas giants - then degeneracy pressure will make the size level off and they become denser instead). Note that this is going to be the most fudgy step, since actually estimating a consistent atmosphere (especially exosphere) temperature, composition and pressure is really involved (and nontrivial even in reality).
Given the size, temperature zone and atmosphere classify it suitably ("ice ball", "gas giant", "venusian hothouse", "super-earth", ...).
Add random moons, rotation periods, eccentricities and whatnot, calculating their effects (like tide sizes, estimated magnetic fields, temperature variations, number of Hadley cells...) - lots of handwaving here, even when you base it on real astronomy and atmospheric physics papers.

This is the simple version. It does not try to actually simulate planet formation, where important processes like migrating gas giants can really affect systems.

You extra questions:

Right now it looks like planets are very common, so a system with no planets is likely pretty unusual.

Big moons might be uncommon, or they might be ordinary. In the solar system Pluto/Charon is also a pretty even pair. I personally think double-planets may be more common than people expect. We'll see.

Asteroid belts are likely pretty common. In a way we have two, the main belt and Kuiper belt. I would expect most systems to at least have a bit of debris that has not cohered in the outer regions.


Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths' side? When Venus is at an angle with the Sun from the Earth's perspective we are able to see the reflection of the Sun on Venus. But imagine that Venus is on the farthest half of its orbit around the Sun. In that case the planet is further away but we receive his reflection on a larger plane. So could an eye see the difference whether Venus is further away in his orbit or closer to the Earth?

The apparent brightness of Venus does vary.  The distance and phase both affect the brightness, and the light doesn't reflect equally in all directions (though this effect is less significant than in the case of Mercury)

The maximum brightness of Venus occurs at a phase angle of 125 degrees. This is a crescent (0 is full, 90 is half lit and 180 degrees is when Venus transits the sun.)

The brightness curve doesn't vary extremely

In this image generated from NASA data by David Barry you can see the peak brightness is about -4.9.  There is a short central dip. The brightness drops to about -4 as Venus gets further away (even though its phase increases) and note the small rise when Venus is most distant (due to more light being reflected directly than at an angle). These occur when Venus is close to the sun in the sky, and so the brightness can be lost in the dawn or dusk sky.

Judging the difference between -4.9 and -4 is rather difficult, especially as you are not viewing it against a dark background.


Replenishing hydrogen in the core of the sun How is hydrogen supplied to the core of the sun? Shouldn't the radiative zone prevent this? Shouldn't heavier helium fill the core? How much of the hydrogen can the sun fuse?

There is very little mixing in the core of the Sun, where the stratification is fixed by radiative (rather than convective) heat transfer. The heavier helium does "fill the core", but takes about 12 billion years to do so, during which time, the concentration of helium gradually increases.

During its main sequence lifetime, most of the energy generation takes place in the central 20% of the Sun, where approximately 600 million tonnes of H is turned into He every second. (You get this by dividing the solar luminosity by $c^2$ and note that the pp H-burning chain is 0.7% efficient.) If we take this as a good average value over the main sequence lifetime (the Sun was fainter in the past, but will be brighter in the future), we calculate that $2\times 10^{29}$ kg of H is turned into He, which is 10% of the solar mass, and about 13% of its initial H.

Once the Sun evolves away from the main sequence, it burns more hydrogen in a shell around the core, and the convective envelope can mix fresh material into this shell. Given that the Sun is expected to leave behind a $\sim 0.5 M_{\odot}$ C/O white dwarf at the end of its life, then at least half of its mass will have been processed by nuclear reactions (and probably more, since some processed material is lost in the solar wind during the red giant phase).


What supernova has created the iron currently found in Earth core? Iron is generated by stars in a certain part of their life cycle. Earth contains a lot of iron inside, however it is clear that this iron could have not been generated in a star in close proximity.
If we consider the current Earth position as reference, where would have lied 4.5 billions years ago the stars that generated the iron currently found in Earth core? Can we say that in the current Earth position there was a huge star 4.5 billions years ago? How did the iron formed in these old stars reach the primordial gas cloud that turned into the solar system? For how long did it travel? Can we say that it came from a set of stars that lived (again, 4.5 billions years ago) in the area that we now call the galaxy core, for example? Can we define the set of "iron generating" areas? Can we define something like the "Earth core iron generator sphere"?

Iron is mainly made or is the product of decays from nuclear processed material inside supernovae.

As the Earth (and solar system) is around 4.5 billion years old, then the stars that manufactured the iron that is currently in the Earth's core died more than 4.5 billion years ago. Note that the solar system formed out of the gases that had been enriched by the supernova explosions of hundreds of millions of stars all mixed up together.

There are basically two categories of star that can have exploded as supernovae and disseminated this iron into the interstellar medium and from which the solar system could then have formed. The first is massive stars ($>8-10M_{\odot}$). These can produce iron and nickel in their cores as a result of silicon burning in the final stages of their evolution. There is then a brief core-collapse followed by an explosion and the resulting supernova can scatter some of this processed, iron-rich material into space.
The remnants of these ancient supernova explosions could be a neutron stars or a black holes. These are almost untraceable/unobservable now, but there should be around a billion of them in our Galaxy.

The second category is the progenitors of what are called type Ia supernovae. These are thought to arise from the thermonuclear explosion of a white dwarf star. White dwarfs are the end point of the evolution of less massive stars. Iron-producing type Ia supernovae $>4.5$ billion years ago, would have begun as stars with masses between about 1.5 and 8$M_{\odot}$. These would have burned hydrogen, then helium to produce a degenerate carbon and oxygen core. This core then simply cools and fades away as a white dwarf star in most cases. In type Ia supernovae, some event later in their lives, either mass transfer from a companion, or merger with a companion, caused the white dwarf to exceed its Chandrasekhar mass and triggers an instability that leads to the rapid total consumption of the star in a thermonuclear explosion. The products of this explosion include a large amount of Nickel, that then radioactively decays into Iron. Nothing is left of the white dwarf.

EDIT: Having established this we can begin to look at your edited question. Firstly, the gas and stars in the Galaxy are basically orbiting the Galactic centre. The orbital period at the Sun's radius is about 230 million years, so it has completed many Galactic orbits. Not only that, but it could have migrated in orbital radius as well. There are claims and counter claims in the literature and the issue is not settled. The Sun could have moved in or out by a significant fraction of its current Galactic orbital radius.

The high mass progenitors of core collapse supernovae will have been born (and died) very close to the Galactic plane. The same is not so true of type Ia supernovae, which had longer lived progenitors that could have moved significantly from the Galactic plane before exploding, and indeed would themselves have orbited the Galaxy many times. The gas expelled in a supernova explosion spreads out (over thousands of years) over tens of light years and becomes mixed into the interstellar medium. The interstellar medium is itself stirred and mixed by the energy input from these supernovae, but also due to the heating and winds of other stars, the tides of the galaxy and spiral arms. The interstellar medium appears to be quite homogeneous in terms of chemical composition, though radial gradients exist with scale lengths of order ten thousand light years.

In conclusion, what you ask is almost impossible to answer. The solar system iron almost certainly came from countless supernovae with a variety of progenitors, that would have exploded any time between almost the birth of the Galaxy 11-12 billion years ago (in fact the supernova rate was likely higher then) up until the Sun's birth. The biggest contributors would come from those stars inhabiting an annulus of many thousands of light years across, centered roughly on where the Sun was born, which is itself uncertain.


Calculate position of the Sun in ECEF I'm making a very basic orbit simulator in C# using the Helix Toolkit.

I want to position the Helix lighting object at the sun, but also be able to calculate solar eclipse by the moon and earth for a satellite (position defined in ECEF).

Upon some brief research, I'm reading that getting an accurate position calculation for the sun in ECEF is a lot of work.

Is there an algorithm that can perform this calculation for me that I can include into an UpdateSunPosition() method so that I can have a reasonably accurate position calculation of the sun in ECEF?

Perhaps I have to scratch the eclipse calculation ambitions for now, instead calculate Geographical Position of the sun to get the Declination and Greenwich Hour Angle, Call those latitude and longitude, and then just give it a satisfactory altitude.

I've noticed that people are referring to SOFA a lot when talking about this, but I'm not sure what SOFA is, or whether I can use it in C#. I'm pretty sure it doesn't operate in a .NET environment like visual studio, but please correct me if I'm wrong.

I appreciate any and all suggestions or insights anyone may be able to offer here.

SOFA is the Standards Of Fundamental Astronomy, approved by the International Astronomical Union (IAU) as being the canonical standard for calculating positions and time in astronomy. It is available in C and F77 versions as standard, and wrappers have been written in other languages, most notably Python as part of AstroPy. For your purpose, the thirdparty part of the website mentions the World Wide Astronomy (WWA) library which is C# so that would be worth a look for your application. I would recommend reading the SOFA Cookbooks, particularly the ones on Earth Attitude and Time Scales to get a better idea of how the (many) various time and co-ordinate systems fit together.


Units of integrated flux density (irradiance) along longitudes If I've taken an image of the Galactic bulge with the pixel units in MJ per steradian:

and I've looked down each column of the image and added up the values:

then do the units change? My thinking is that I'm taking something that is measured per steradian (which is a solid angle) and integrated it along a single axis, which would be in degrees. Would this make the units of my graph now MJ per radian? Or am I simply adding values in MJ per sr, so the units don't change?

You could say "MJ in bins of x degrees in l" where x is the degrees spanned by a pixel.  Or, you could divide the fluxes by the steradians in each column and keep it MJ/sr or "Mean MJ/sr within longitudinal bins". Now the width of the bins is not important as long as they are small.


What are the stars/constellations a beginner/enthusiast can easily identify? I was just wondering what are some of the stars, constellations or any other things that one can just identify by looking up at the night sky.

For example, when ever i look up, I immediately see Orion and can identify Betelgeuse, Rigel, Bellatrix. Apart from that i can identify some of the bright stars liike sirius , vega.
So, what other stars are there that one can easily identify and point?

In the Fall (early morning) and the Winter (early-late evening), Orion is a large an easily identifiable constellation.

In the Spring (early morning) and the Summer (early-late evening), Cygnus and Pegasus are two large and easily identifiable constellations.

Check out an application called Stellarium.  It's free and available for every current platform and has a large database.


Sunlight intensity during a day What simple mathematical function would reasonably well simulate sunlight intensity (disregarding weather) from dawn till dusk in relative numbers, 0-100 %. Are there significant differences (apart from maximum power and daylight duration) between temperate winter, summer or tropical days? The goal is more natural LED lighting for captive wild animals.

Well, there are three parameters of interest to be evaluated to get intensity of sunlight.

Sinusoidal variation of sun vector on earth's surface througout the year
The latitude variation. Which is also sinusoidal
The time variation. Which is also sinusoidal

$$ Intensity = A_0 \sin(\theta - \theta_0) \sin(\pi/2(1 - |t-12|/6))$$

Where,

$$ \theta_0 = A_1\sin(2\pi|d/365.25|)$$

Here $d$ is the day and starts from vernal equinox. $t$ is the time from 00:00 - 23:59. $A_0$ and $ A_1$ are constants. Also when Intensity is negative, obviously there it means darkness.

This is very crude approximation. It assumes circular orbit.


Delay of Moon passing a defined meridian each day On the website Lunar Synodic Curve a nice image of an Moon analemma is shown together with an explantion on the reason why for the different phases of the Moon the images have to be taken with 24h 51min apart. It is explained that this difference is due to the longer synodic month of 29days 12hrs 44mins (compared to the sidereal month of 27days 7h 43mins). Whereas the explanation that due to the additional rotation time for the moon to reach the same angular relation to the sun an additional angle of 29.1 degs has to be travelled, the last step to come to the additional 51mins each day is not clear to me. Any clues on how to calculate this from the sidereal and synodic month duration or the associated mean motions of moon and sun would be very welcome.

The Moon is moving around the Earth in the same direction it is rotating. This means that it would be moving across the sky in the opposite direction the stars are, in case the Earth stopped rotating, however, the Moon is moving in the same direction as the stars due to diurnal motion (only slower, because it also has velocity in the opposite direction). Its angular speed across the celestial sphere will be

$\omega = \frac{2*\pi}{T_{stars}} - \frac{2*\pi}{T_M} = \frac{2*\pi}{T}$

where $T_M$ and $T_{stars}$ are the sidereal period of the Moon and the sidereal day length. Substituting the values above, we get that the Moon takes about 51 minute over 24h to go around the celestial sphere, crossing the same meridian. Note that this is a rough approximation because the Moon's orbit is not circular, but it explains the value you mentioned.


Angular speed of a planet across the sky Assuming circular orbit, the evaluation of the angular speed of a planet around the Sun, just requires its orbital period. Similarly, to calculate its angular speed relative to the Earth (also seen from Sun and considering the approximate circular orbit), its synodic period is needed.

The question is: what is the angular speed of an interior planet across the sky in the conjunction with Sun and how can I deduce its formula?

There is no simple formula.

First, it's not true that the angular speed of a planet relative to the sun can be found from its orbital period.  This is because planets don't move in circles, but in eclipses and the speed of the planet changes, as described by Kepler's second Law.

When describing the motion of a planet in the sky you need to consider the elliptical motion of both the Earth and the planet, the superposition of these motions causes the planets to make looping motions in the sky, sometimes slowing down and moving backwards

The image shows how Mars would have appeared to loop and move backwards in 2003 and 2005.

The motion in the sky can be calculated, but not by a simple formula. You start with the orbital elements of the planets, solve Kepler's equation to determine the position on their elliptical orbit. Solve some trigonometry to determine the position of each planet relative to the sun, then some vector calculations and spherical geometry to find the position of the planet relative to Earth.

There is a guide online for writing a computer program to do this.


Escape velocity of Milky Way galaxy I'm wondering that what would be the escape velocity of our Galaxy?

500+ kps.  There's a pretty good answer over at https://space.stackexchange.com/questions/3948/escape-velocity-for-our-galaxy, but the question appears to fit here as well.

For a mass all located inside a circular orbit, escape velocity is a simple function of orbital velocity, but that ignores most of the mass outside that circular orbit (all of it if its distribution is spherically symmetric).  The Milky Way has a lot of mass outside the Sun's orbit, though.


Doing Correct Calculations with Binary Star Systems I've been experimenting with Binary Star System calculations but came across some issues.

I can correctly calculate the Semi-Major Axis of the Earth's orbit around the Sun by doing the equation:

Using data from Wikipedia.

a = (1.521e11 + 1.47095e11) / 2.   <- average radius of Earth's orbit (meters)
μ = G * M = 6.67408e(-11) * 1.98847e30 gravitational constant * Mass of Sun (kg)

I got ≈ 31557884 seconds which equates to ≈ a year

Now my issue is that I've been searching around for a formula that calculates the period and etc of a Binary System where two masses (such as stars) orbit around each other.

So then I try essentially the same formula, this:

I used data from the Procyon A and B (binary system)

a = Semi-Major Axis = 4.3 AU
M1 = Procyon A Mass = 1.499 Solar Masses
M2 = Procyon B Mass = 0.602 Solar Masses

these give me a T (period) of 4731227. It says 40.82 years on Wiki.

What am I doing wrong? Is it the data? is it the formula? My comprehension of the question? Thank you for the help!

The semi-major axis is 4.3 arcseconds. At a distance of 3.51 parsecs, this corresponds to 15.09 au. Using your formula I get 40.3 years, which is close enough given the 2 significant figure precision of the semi-major axis.


Theoretically what happens to a black hole that evaporates? Theoretically, does an evaporating black hole remain a black hole until it evaporates to nothing? Or does it change into something else once it evaporates to below a certain mass?

Utterly unknown.

So Hawking proposed radiation, that reduces the mass of the black hole (BH) by black body radiation.  The effective temperature is antiproportional to the mass of the BH.  In the linked paper, Hawking concludes that BHs end in "fairly small explosions by astronomical standards."

There are different solutions to the equations of general relativity that produce black holes, not only the common Schwarzschild metric.  Whether the curvature singularity or the event horizon vanish before the BH completely evaporates, depends on the internal structure of the BH.  As any mass, even very small ones, can become BHs if you compress them enough, there might be a singularity until the very end (not taking into account quantum mechanics).  On the other hand a "vanishing event horizon" is not something we should be strictly speaking about, because then it wasn't an event horizon in the first place.  You may see the looming problems with black hole remnants.

Also: according to the no-hair theorem, BHs are completely characterized by mass, electric charge and angular momentum.  If you can see a remnant, e.g. a clump of matter, could you derive its complete properties from the previous three only?

In research, a remnant seems to be generally undesired
 but possible.  This fresh and unreviewed paper jumps on the bandwagon and proposes white holes as remnants. So the debate about the end of BHs and their information theoretical implications is still a hot topic.

Disclaimer: I'm not a BH specialist.  (And I'm waiting for the time where "black hole specialist" is a craft job designation.)


What's the cheapest lens to buy for showing ISS to a flat earther? Recently I entered a discussion with someone that has a relatively popular YouTube channel which says among other things that the earth is flat. I said that earth isn't flat because you can see the ISS by yourself through the telescope. I was reminded that I'd only seen it on the internet, but not actually seen it directly myself. I was told that if I buy a telescope and film by myself they will also buy one, do the experiment, and post the footage on their YouTube channel.

So, what's the cheapest telescope that I can buy and record with a cellphone? I've seen people using a Nikon P9000 but what matters is its lens, which is already expensive, and the camera is expensive too. Maybe it's possible to buy a cheap photography lens that has both good quality and high zoom and can be used with a cellphone?

I think you'd want a focal length of 700-1000 mm, and the cheapest way to get that would be a basic Newtonian reflector of 100 mm aperture.
Examples: Meade Polaris 114mm, Celestron PowerSeeker 114EQ, Orion SkyQuest XT4.5.
There are brackets available to mount a smartphone on any telescope.

Before trying to track ISS, practice acquiring easy targets such as the Moon, the Pleiades, etc.
Then use Heavens Above to predict ISS passes visible from your location.
After you collect your evidence and your hyper-skeptical acquaintance rejects it, you can still use the scope to enjoy the planets and a few dozen star clusters, nebulae, and galaxies.

On the other hand, maybe you don't need a telescope.
ISS is visible to the naked eye, is silent, and doesn't appear to slow down as much near the horizon as aircraft do.
Seeing it disappear as it enters Earth's shadow could be persuasive.


Which part of the milky way does the image show? As a general question and more in specific regarding the image above: Which part of the milky way, planets, stars, clusters are included in the image?

How can I know which entities are included in different pictures?

I wholeheartedly recommend astrometry.net for this sort of thing. It Just Works(tm); running on your image produced this output with absolutely no hints or guidance from me:

For avoidance of doubt, I have no association with astrometry.net.


What are some of the best astronomy gatherings in the USA? I'm part of a local astronomy club but the results from that have been vastly underwhelming with a membership count of around 50-200 active people at any given time. All that being said- I guess my question is for 2019 what are the biggest and most informative gatherings in terms of astronomy? Anything you know of from lectures to imaging to stargazing would be wonderful.

More specifically:
I was considering spending the 250 dollars to go to the Northeast Astronomy Forum this April 4th and 5th and the Northeast Astro Imaging Conference. It seems to be one of the larger scale events that happens anually. Is this a good event for beginners? Heres a link to their site:

http://www.rocklandastronomy.com/neaf.html

http://www.rocklandastronomy.com/neaic.html

If youve been Id love to hear from you. If this is off-topic I apologize, but the only other place I had to ask was our clubs forum and it has virtually no traffic.

NEAF is pretty big (and probably the biggest).  There's also Southern California Astronomy Expo (SCAE).

For observing, there is the Texas Star Party in west Texas.  There is the Winter Star Party in the Florida Keys (that site for that event was severely damaged by a hurricane a couple of years ago, had to relocate up-state, but are now back in the keys.  This past year the had to limit attendees because the camp wasn't fully restored.

A little closer to home for you (in Pennsylvania)... is the Cherry Springs Star Party (there is also the Black Forest Star Party at the same park ... different months).


Moving object seen through telescope, what is it? I was using my telescope to look at a nebula, intell I seen a star like object that was moving at a moderate speed in a straight line and eventually dimmed out. My first thought that it was just a satellite but I could only see it through my telescope which makes me think that it was to small to be a satellite. I'm not sure.

Comments formed into community wiki

Satellites can and have been seen through telescopes. Possibly it is just a very small or faint satellite which requires a telescope to see and is too difficult to detect by the naked eye. The description of a "moderate speed in a straight line and eventually dimming out" sounds like every satellite. A piece of space junk is also possible. Lots of second stage boosters and other rocket parts floating around up there that are small enough that you would only notice them through a telescope

You might be able to identify it by going to the Heavens Above site and providing your location, the date & time.


What are the odds that the Sun hits another star? The Sun moves around the Milky Way disk in the same direction as most of the other stars in our galaxy (prograde).  But there are a number of older stars in the galactic halo that move in retrograde orbits, and some of them pass between the Sun and the center of the galaxy.  What are the chances that the Sun will collide with another star, say in the next 5 billion years?

TL;DR: Virtually zero.
The distances between stars are HUGE and stars are tiny compared to the astronomical scales of distance between neighboring stars. The sun, is about 0.0000001 or one ten-millionth of a light year. The probability of a star (to be generous, say a $10 R_\odot$ star) colliding with the Sun is tiny.
Every star has a different velocity vector and different positions in space-time. The probability that their positions will intersect in a relatively short timescale is almost zero. Imagine throwing a tiny dart, and hitting a bullseye the size of an atom from several miles away. That still is larger than the probability of the Sun and some other star colliding in the near future (5 billion years is a tiny moment of time when it comes to events like these).
According to Wikipedia, it takes about 30 trillion years for a star to undergo a close encounter with another star. Even if the encounter is less than 1 AU, the stars would probably just pass by and not do anything special. For a complete collision and merge, you'd need the distances to be much, much smaller, about $0.1 R_\odot$.

Years from now
Event

3×1013 (30 trillion)
Estimated time for stars (including the Sun) to undergo a close encounter with another star in local stellar neighborhoods. Whenever two stars (or stellar remnants) pass close to each other, their planets' orbits can be disrupted, potentially ejecting them from the system entirely. On average, the closer a planet's orbit to its parent star the longer it takes to be ejected in this manner, because it is gravitationally more tightly bound to the star.

Source: https://en.wikipedia.org/wiki/Timeline_of_the_far_future
Also, according to Wikipedia:

While stellar collisions may occur very frequently in certain parts of the galaxy, the likelihood of a collision involving the Sun is very small. A probability calculation predicts the rate of stellar collisions involving the Sun is 1 in $10^{28}$ years.

Source: https://en.wikipedia.org/wiki/Stellar_collision#Stellar_collisions_and_the_Solar_System
Back to crude math, assuming an uncertainty of 2 AU, the probability given these parameters is equal to:
$$\dfrac{5 \cdot 10^9 \text{ years}}{3 \cdot 10^{13} \text{ years} \cdot 10^{24} \text{ stars in the universe}} \cdot \dfrac{0.1 R_\odot}{450 R_\odot} = \dfrac{1}{27,000,000,000,000,000,000,000,000,000,000} = \dfrac{1}{2.7 \cdot 10^{31}}$$
Warning: This is only a very crude estimate, don't take this as a "real, super-mathy" answer. But I still hope this helps.
Edit: That equation above is still probably too big. Asked a question about this: If two stars collide, what is the probability that they merge to form a single star?


Is the CMB the photons that were created at the birth of the atom? Is the CMB that we see the same, unchanged energy from this moment? Does the radiation of the CMB exist as Photons and are we only witnessing the photons that have not been absorbed or destroyed? Will there be a time we can no longer witness the CMB because it has all been absorbed? How does red-shift factor in to this?

The CMB consists of photons that were emitted very shortly before and during the recombination of electrons with hydrogen and helium nuclei. [Side note - why is it called the epoch of recombination when the atoms were never "combined" to begin with?]

Those photons were then able to travel, unabsorbed across the universe until they encountered our microwave telescopes 13.7 billion years later.

Although the photons are unabsorbed, the expansion of the universe causes a redshift as they travel. Their current wavelengths are approximately 1100 times larger than when they were emitted and the universe is almost completely transparent to photons of this wavelength - that is there is very little chance of a photon being absorbed even if it travels across the diameter of the observable universe.

There are scattering processes that can deviate the photons or even increase their energy (and decrease their wavelength) via inverse Compton scattering (the Sunyaev-Zel'dovich effect), but this does not affect the CMB spectrum a lot and is difficult to observe.

Our current ideas about the future of the universe are that it will continue to expand at an accelerating rate. The CMB photons will become even more redshifted - to radio wavelengths and beyond - and the universe will become even more transparent to them.

Related: Future redshifting and effect on the 'pitch' of CMB radiation


Dataset for machine learning MK stellar classification I would like to create program for automatic Morgan-Keenan stellar classification using machine learning. For that, I need dataset of stars with known absolute magnitude, temperature and luminosity type (0, Ia, Ib, ..., VII). I found some datasets, (e. g. kaggle.com), however they contain only few hundreds stars and not all star types are represented.
Is there any large enough dataset (at least 1000 stars) in which all types of stars are represented, from hypergiants to dwarfs, and with all the information mentioned?

That I know there is the XHIP catalog via VizieR and you can enter a range in UMag (or B- and V-band) for example of -20 .. 20. Check the box SpType and Tc to get also the spectral type and temperature (double check the literature for the quantities you need)
On the column on the left titled Preferences, set the quantity max to unlimited (it's the number of data) and the box just below indicates the format to download the data, if you want *.csv select CDS Portal and click on Submit.
You will be redirected on the CDS portal and simply click on Save and then on MyData, it will show you a list of the dataset you saved and you can select the format of your file (csv, fits, etc.) and then download.
With the simple range of the V absolute magnitude between -20 and 20 you get more then 100k stars
Edited: I noticed that downloading the data through the CDS portal does not give you the quantities that you check, but some fixed one. Maybe I can suggest to select ascii text/plain and then download the page.


Why do we use the Planck length? We are not able to quantize spacetime and the theory of  loop quantum gravity has been proved wrong but we still use the Planck length in astrophysics. Why?

First off, you're much more likely to see parsecs, light years, or astronomical units than Planck lengths in astrophysics.

However, for those who like to work in a system of natural units, the Planck length is the obvious choice for the unit of length. This choice has nothing to do with quantizing spacetime. It instead results from a desire to make conversion factors such as the speed of light $c$, the reduced Planck constant $\hbar$, the Boltzmann constant $k_b$, and the gravitational constant $G$ all have a numerical value of one or some other nice number such as $
\frac{1}{4\pi}$.

One way to look at this: Compare the conventional units used in the US and a few other backward countries. These countries have to use $F=kma$ rather than the $F=ma$ used in the metric system. The $k$ is a conversion factor that results from having those conventional units being an inherently inconsistent system. From a modern perspective, force is the product of mass and acceleration; no conversion should be needed.

That said, that the mass-energy equivalence equation expressed in metric units is $E^2=\bigl(m c^2\bigr)^2 + \bigl(pc\bigr)^2$ is a sign that the metric system, too, is inconsistent. With natural units, this expression simplifies to $E^2=m^2+p^2$. Energy, mass, and momentum all have the same units; no conversion is needed.


Position of specific stars in the sky as a function of time I'm a beginner in astronomy and I'm trying to figure out how to model the positions of stars in the night sky mathematically (ie a function which spits out azimuth and altitude as a function of time). I was wondering if anyone knew of any such function, I understand it would vary with your position on earth.

Furthermore, does anyone know anything about the mathematics of star movements, I'm trying to learn all I can.

Thanks!

It's difficult to answer this question with any confidence, since it is so wide ranging, and the original poster has not (yet) provided any clarification.

However, it seems likely that the book "Astronomical Algorithms" by Jean Meeus, published by Willmann-Bell, will provide the answer(s).  Failing that, the publisher's website contains several other reference works on this topic. I have no affiliation with the publisher or author.

Finally, as the author of the question is a beginner, I wonder whether in fact a simple introduction to astronomy, of which there are hundreds, might be a better place to start.  Jumping in by trying to model the positions of stars looks hard, particularly as there are plenty of good applications already available, some of which are free.


I'd like to become an astronomer. What experience do I need? I think I'd like to become an astronomer. What degrees and training/experience do I need to become one? I've heard I should take science and mathematics, but I don't know which specific courses I should take and which degrees I would need to become an astronomer.

Well, if you want to become a professional astronomer, there is no obvious answer, since education systems can be quite different depending on your country. But astronomy & astrophysics is physics in the end, so a good training and strong knownledge in physics are definitely required (that involves math for sure). And in countries/universities where you can find an astronomy-oriented degree (at master level most of the time, before that it does not really make sense), you won't be surprised that most of the lectures are actually just physics or math.

Then, to get a PhD in the field is the way to go.

As for the positions, you can find some statistics in this paper released by ESO (the European Southern Observatory) for jobs in astronomy in the 2009/2010 hiring season. Bottom line is there were 65 opened positions worldwilde, that cannot be consider as "a lot".


In sky surveys, what does inverse variation really mean? How do we use inverse variation to remove noise? I am trying to work on data from a .fits file and it has an IVAR header. How can I use inverse variation information along with the actual signal data to get better results or reduce noise?

It is inverse variance. A larger inverse variance means the value in that pixel is less uncertain (in a relative sense).

The inverse variance can be used as a weighting factor - it directly gives you level of significance for the measurement. It can also be set to zero for measurements that should not be used (for whatever reason).

An example would be finding the mean value of a set of pixels. A standard mean would apply equal weight to all pixels. A weighted mean would be calculated as
$$ \bar{x} = \frac{\sum_i x_i w_i}{\sum_i w_i},$$
where $w_i$ here would be the inverse variance of each pixel.


Calculate Distance To Stars I was just watching a lecture from Carl Sagan. He talked about figuring out the distance to the stars; it got me interested in learning more about the subject.

As far as I know, the Inverse square law and parallax can be used. Can anyone expand on these? Specifically with regards to what I could do to measure the distance from Earth to Proxima  Centauri.

The currently accepted answer is not relevant for finding the distance to a star like Proxima Centauri.

Here's how parallax works. You measure the position of a star in a field of stars that are (presumably) much further way. You do this twice, separated by 6 months. You then calculate the angle that the star has moved against its background stars. This angle forms part of a large triangle, with a base that is equal to the diameter of the Earth's orbit around the Sun. Trigonometry then tells you what the distance is as a multiple of the distance from the Earth to the Sun. [In practice you perform many measurements with any separation in time and combine them all.]

The "parallax angle" is actually half this angular displacement, and a star is said to be 1 parsec away if the parallax angle is 1 second of arc. So 1pc is 1 AU/$\tan (\theta) = 3.08\times10^{16}$ m. The larger the parallax, the closer the star.

The Gaia satellite is currently mapping the entire sky and will estimate tiny parallaxes
with precisions of $10^{-5}$ to $10^{-4}$ arcseconds (depending on target brightness) for about a billion stars.

Parallax - as illustrated at http://www.bbc.co.uk/schools/gcsebitesize/science/21c/earth_universe/earth_stars_galaxiesrev4.shtml

Now in reality, it is a bit more difficult than this because stars also have a "proper motion" across the sky due to their motion in our Galaxy relative to the Sun. This means you have to do more than two measurements to separate out this component of motion on the sky. In the case of Proxima Centauri the motion against the background stars due to proper motion is larger than the parallax. But the two components can clearly be seen and separated (see below). It is (half) the amplitude of the curved motion in the picture below that corresponds to the parallax. The proper motion is just the constant linear trend with respect to the background stars.

HST images of the path of Proxima Centauri against background stars. The green curve shows the measured and predicted path of the star against the background field over the next few years.

Parallax measurements work best for nearby stars, because the parallax angle is larger. For more distant stars or those without a parallax measurement, there are a battery of techniques. For isolated stars, the most common is to attempt to establish what type of star it is, either from its colour(s) or preferably from a spectrum that can reveal its temperature and gravity. From this one can estimate what the absolute luminosity of the object is and then from its observed brightness one can calculate the distance. This is known as a photometric parallax or spectroscopic parallax.


Question about the relevant physical parameters to build a backyard observatory I live in a big city: São Paulo, Brazil. I would like to build a simple telescope dome to do some amateur sky obseravtions. But the whole project lies on the viability to see something interresting more than just some stars. Due to high luminosity of cities this is a good question to ask.
So, which physical parameters are relevant to decide if a backyard observatory is a good holiday DIY project?

Many astronomers are using electronic eyepieces to overcome light pollution. You may want to see what these devices can do in your area'

I use several of the Mallincam cameras in my observatory.

https://www.mallincam.net/

https://groups.io/g/MallinCam/topics

Some photos of my observatory and assistants

https://www.flickr.com/photos/142615862@N04/sets/72157672437929982

More home observatory sites

http://obs.nineplanets.org/obs/obslist.html


Is there a limit to the size of a black hole? The mass of a star is limited to about a couple of hundred solar masses. The fusion rate scales strongly with density (which is why the most massive stars have extremely short lifetimes) so if a star was massive enough (above the Eddington limit) then the radiation pressure would blow it apart.

Now a black hole also has radiation, but by definition it is not capable of escaping due to gravity. But is there in some way a limit of the size. Imagine that all the matter in the universe formed a black hole. Should that be possible or is there a law which forbids creating it?

General Relativity does not limit the size of a black hole, and we have good (indirect) observations of black holes with a mass of more than a billion suns.

Black holes are not object that are held together by gravity, and which could be ripped apart if forces opposing that gravity were increased. They are (in GR) singularities: a complete collapse of the matter to a point. There is no "object" inside a black hole that could be emitting radiation.

Now if all the matter in the universe were to be a black hole, then that singularity would be in all our futures. We wouldn't "see" the singularity because it is in the future, never in the past. In fact a universe in which is one black hole could look rather like the one in which we are living.


What's a good step up from 7x50 binoculars? I own a Nikon OceanPro 7x50 pair of binoculars, but I find that when I look at the moon, I would like more magnification.

My requirements are:

costs less than $400
weighs less than 10kg
the length is less than 1-2 feet
much higher maginification than 7x (I already have a pair of 7x binoculars, as I said)
at least an 100mm objective I (I already have a device with a 50mm objective)

What kind of device should I go for? A spotting scope? Binoculars? A telescope of some kind?

Ideally, I'd go for binoculars, so that I can see with both eyes, but I'm told binoculars don't come with objectives larger than 100mm, for my budget.

I could buy a spotting scope, like this, but Wikipedia says that spotting scopes are not designed for astronomy.

I live in an apartment, so I can't use a big, heavy telescope, like a 4-feet 20kg thing. I don't have room in my apartment for such a big telescope, and I can't carry it over the threshold into the balcony.

So, what kind of device should I consider, and what are the tradeoffs? What kind of astronomical objects will I be able to see?

For higher magnifications, you are going to get into the territory where a tripod becomes a must. I mention this because you didn't say anything about a tripod, so the implication may not have crossed your mind. Unfortunately, the cost of a tripod will eat into your budget.

I am inclined to think that your best option would be a telescope. (Unless you also want to view terrestrial objects, then a cheap spotting scope might be what you want.) Maybe your best bet, to get a good one, would be to ask around on forums like cloudynights or iceinspace that have some support from astronomers local to you, let them know you are looking to buy, and maybe you can get one second-hand off someone who is trading up. You might get a better telescope for your money than trying to buy new.

If you want a smaller length telescope, you may be put off something like a newtonian reflector or another simple 2-mirror reflector. Still, try one out (maybe you can attend a local star-party and see some up close, even if there isn't a telescope store handy). A 114mm newtonian on a tripod would meet almost all your requirements including budget, even purchased new. It will be a bit less than 2' long; slightly bigger newtonians like 130mm start getting longer than 2'.

Otherwise, what might fit the bill is a small cassegrain-style reflector - these usually also have a corrector lens, making them a catadioptric type telescope - such as a Maksutov-Cassegrain. The extra manufacturing cost of making the lenses pushes up the price of these well past the simple reflecting telescopes, however.

Refracting telescopes with an objective of 100+mm are probably in your "too big" basket, and you're also not likely to find one with that budget.

You use different eyepieces to provide different magnifications using the same telescope. If you buy a telescope as a kit, it will come with at least one eyepiece, and it's likely the magnification provided will be more than 10x (divide the focal length of the telescope by the focal length of the eyepiece to find out).


Dobsonian telescopes: do collapsible scopes have any drawbacks? When choosing between a collapsible and a non collapsible 12” dobsonian scope what drawbacks if any do collapsible dobsonian scopes have?
If I transport any scope it will need collimating. Obviously a collapsible scope will store better.

I agree with Aaron F about storage space, but transport space is different; a non-collapsible tube wouldn’t fit in my car, while my 10″ collapsible-tube telescope fits nicely in the trunk and could fit on the back seat, whether “lying” on it or “standing” as it wouldn’t hit the car’s ceiling.
The only drawback I have heard of, aside from collimation that you would have to do anyhow (nevermind solid tubes holding it better; after enough road, it’s gone no matter what kind of tube you have), is that different thermal expansion of the truss tubes and their holders sometimes makes it difficult to expand or collapse the tube when it’s cold. Also, be careful when storing or transporting your collapsible tube, as the truss bars can potentially end up in positions that slightly bend them, making the expansion or collapse more difficult even if it’s not cold.


Converting FITS (NuSTAR) coordinates to WCS I'm trying to convert some pixel coordinates I have into WCS coordinates, ideally into a WCS region for use in some further analysis.

So far I've been able to load and parse a NuSTAR FITS file, do some analysis to make my selections and get the pixel coordinates for the image. For example, after my analysis I would end up having selected a rectangle of x values from pixels 480 to 518, and y values from 478 to 516.

After that I tried to use the astropy WCS module to convert them, but it doesn't seem to find the required data in the header to actually do the conversion and just says that pixel 480 gives coordinate 480. That or I'm just doing something wrong.

So I looked through the FITS header myself, found that some keys (TCRPX, TCRVL, TCDLT) give the reference pixel, reference pixel degree coordinate and pixel axis scale values.

But now I'm really stuck since what I need to get is a region file like those produced by ds9, in the format of:

# Region file format: DS9 version 4.1
global color=green dashlist=8 3 width=1 font="helvetica 10 normal roman" select=1 highlite=1 dash=0 fixed=0 edit=1 move=1 delete=1 include=1 source=1
fk5
circle(19:58:21.593,+35:11:57.46,10")

To then run this through the nuproducts tool.

TL;DR: Does anybody know how to convert the NuSTAR FITS physical pixel coordinates to FK5 coordinates?

EDIT: Wrote some code to extract the coordinates. It converts the pixel to degree fine, but converting the FK5 $\alpha$ angle to sexagesimal doesn't work correctly for... some reason.

EDIT2: Everything works now, here's a notebook where I run through it all in case anybody runs into a similar problem. I go through extracting the right-ascension and declination in degrees from the pixel coordinates given by the FITS file, then converting these degrees to sexagesimal (h:m:s) angles. From my tests it seems to work quite well, results are a tad off  due to what I assume are some floating point arithmetic errors.

You can upload your FITS files to astrometry.net (or use an API) and get the coordinates. You can optionally get new FITS files back with the coordinates included in the metadata. From there you will be able to proceed with your processing.

And, instead of uploading everything, you can also install a copy of the astrometry.net package locally if you're running Linux.


Data on the composition and orbits of asteroids I am interested in the composition of asteroids from the perspective of determining potential targets for mining.

I expect there already exists some data that can be used to identify reasonable targets for mining.

The specific mining method I am looking to evaluate is to send a "collector" out and "grab" a target, bringing it back into LEO for mineral extraction.  Because of this I would like to identify targets that are < 20m diameter. The ideal material to mine/extract would be a precious metal of some kind. This would give the most "bang for the buck". (The buck in this case being the effort rather than Delta V).
Considering the ideal material, I suspect a list of "M" type asteroids and their properties would be the data I am after.

I saw a question that asked about using IR Spectroscopy to determine the surface composition of asteroids.  That gave some good insight - but no indication of where such data currently exists. A table of this data would be a reasonable starting point.

I am also wondering if we can determine anything about the relative density of an asteroid by the gravity it exerts on nearby objects.  If that data is available anywhere that would also be an interesting place for me to look.

I am also interested in trying to determine the orbits of any target asteroids, so where would I look for data on the orbits of those that are selected?

The JPL Small-Body Database Search Engine is a good place to start.
If diameter < 20 m, then probably absolute magnitude H > 25.
Fortunately we know of over 4000 asteroids in that size range, mostly after close approaches to Earth.
Unfortunately the orbit and magnitude are all we know about most of those, due to limited observational data.
JPL SBDB lists only about 40 asteroids with known Tholen spectral type M, all with perihelion q > 1.5 AU and diameter > 30 km.
The picture is a little brighter for type S, with over 300 known, a few with diameter < 2 km and minimum orbit intersection distance < 50 LD.


Why is our Sun hotter than UY Scuti? As Wikipedia says the surface temperature of UY Scuti is 3,365 K and the sun is 5,778 K

Why is the Sun hotter than a supergiant star like UY Scuti?

UY Scuti is a red supergiant star.

When stars start to run out of hydrogen fuel, their cores start to collapse, causing the core of the star to heat up, and heavier elements start to be used as fuel. This means that the core of the star is hotter and producing more energy.

The effect on the outer layers of the star is to cause them to expand, and as things expand, they cool. The outer visible layer of the star cools (to about 3500K)

So the energy output of the star increases, but the temperature of the outer layer decreases. As you might expect this leaves the star unstable, and UY Scuti undergoes semi-regular variations in brightness.


Timeline of falling into a blackhole For some "average sized" black hole, how long would it take for a distant (spacesuit-wearing!) astronaut to:

be able to see the gravitational lensing with the naked eye
clearly feel the gravity gradient
reach a lethal gravity gradient
cross the event horizon

Ultimately I'm interested in the subjective experience of a human falling into a black hole. Would it be drawn out and unpleasant? Or is it that by the time the gradient would be painful, there'd be barely any time at all before annihilation?

Interestingly enough, Neil deGrasse Tyson, an american astrophysicist, wrote a book titled Death by black hole.

In this video he goes on to explain the details of what you asked, the subjective experience of the astronaut being sucked into the black hole. It is definitely painful!


How would humans with appropriate equipment navigate the surface of Saturn's moon Titan on foot? Taking into account the physical characteristics of Titan (e.g. surface gravity, atmospheric pressure), what would be the most efficient method of motion for an astronaut to navigate the surface of Titan on foot?

Would it be a leaping or hopping motion? Or would it make more sense to walk somehow? I'm unsure how the friction of the ground would accommodate for that.

For the purpose of this thought experiment, lets assume that humanity has the necessary equipment to put an astronaut on Titan and to have protective clothing to enable reasonable safety from environmental hazards on the surface.

Thanks in advance!

I wonder if this would be better for worldbuilding or space travel, but I can touch on the basics.

You'd need oxygen and protection from the cold, but a space suit on Titan would be less restrictive, not more restrictive than the ones used on the Moon.   The pressure isn't too much, so that actually helps and cold temperature is much easier to protect from than a vacuum.    They'd probably need some kind of full body wrap to avoid exposure and frostbite, and goggles but that's not terrible hard.   Electric heating or just insulation would both be options.

But it's safe to assume that even if they carry an electric heating device, their suits would still be less bulky and less restrictive than the ones the astronauts wore on the Moon.

Hop or Walk?

Very likely hop.   I'm going to borrow from this Quora answer, because I agree with it and it makes logical sense.   Walking is kind of a controlled fall.  Walking in 0.14 g, means your body doesn't fall forward in pace with every step, so walking would be weird and hoping would be easier.

Even on Mars, astronauts might hop due to the lower gravity.  If I was to make a wild guess, for comfortable walking I think it would begin somewhere around 50% or 60% Earth gravity.   Less then that, the slower rate of falling with each step and the lighter body weight, I think hopping is the way to go.

The human brain and inner ear begins to stop working somewhere around lunar gravity.   There are videos of astronauts on the moon falling forward and looking clumsy, but they were very fit individuals.  In low enough gravity the human brain loses a sense of which way is down.  (I read that, I'll see if I can find a link).

I would think, if it was for an extended time there, astronauts on Titan would wear lead boots, to help with their sense of which way is down and also for helping maintain musculature, but that's just speculation on my part.  There are other ways to maintain musculature.

Titan's surface might/probably gets soggy as well and humans aren't buoyant in methane lakes the way they are in water, (so maybe the lead boots aren't a good idea).   Falling into a lake or falling through ice into a like might happen on titan, but they'd be wearing full body wraps and have oxygen, so falling into a methane lake wouldn't necessarily mean death, they could probably get fished out and rescued.

Long story short, hopping is likely the way to go, and perhaps, either some kind of snow-shoe feature or perhaps an ice/jagged edges feature if parts of Titan's surface are slippery.   I'm far from an expert in such things, this is just my best guess.


Is the Alpha Centauri star system moving closer to us? The closest star system after our sun is Alpha Centauri, a three star system with Proxima Centauri, a red dwarf, slightly closer to us than the other 2, around 4.3 light years. I tried Wikipedia and other articles that describe Alpha Centauri and there is plenty of information, but no where could a find an answer to said question.

My presumption is that if the distance is measurable, so should the variance in this distance.

Yes, it's moving closer.

Per Wikipedia.

Source.

In about 30,000 years Alpha Centauri will be 3 (and change) light years away, then it will start moving further away.

If anyone wants to provide a more detailed explanation on stellar movement and charts, please feel free.


What explains this puzzling radio signature from a fireball? September 21, 2017 a bright fireball over Holland was reported by hundreds of people. Several all-sky cameras recorded it as well as a special radio beacon set up to capture radio signatures of meteors - the BRAMS beacon at Dourbes, Belgium.

I'm puzzled by what I see in the radio spectrogram from one of the BRAMS stations at Ophain, Belgium :

The orange blob in the center is the radio signature of this fireball.
The horizontal line is the beacon signal (0 Hz), time runs horizontally, frequency runs vertically, positive above and negative below the beacon line.
What is displayed are Doppler shifts.

The track for this fireball has been computed since and relative to the radio receiver at Ophain and radio beacon at Dourbes it is receding from both.
However, the Doppler blob is above the beacon line which is only possible when the fireball recedes from Ophain but approaches the beacon, approaches both, or approaches Ophain but recedes from the beacon - NOT when it recedes from both!

The only way I can make sense out of this is to assume that the ionized
gas trail from the fireball reflecting the radiowaves was NOT receding from the receiver and transmitter beacon - only the fireball was?!

Or are there any other explanations?

I checked with the people who run the BRAMS project and they confirmed my reasoning, and added another possible explanation.

Even if the fireball moves away from the receiver AND transmitter, the ionization trail it leaves behind can expand and be blown by mesospheric winds towards the receiver and/or transmitter. It's this trail that mostly reflects the radio waves.

And theoretically it's still possible but unlikely that the observed Doppler signal was coming from another meteor trail unrelated to this particular fireball, but coinciding with the exact time frame the fireball appeared in.


How to tell a pulsar is rotation-powered or accretion-powered? We need to calculate P/Pdot?

A pulsar may switch between theses two states.

For a rotation powered pulsar, it can pulse and, the pulsed Lx can be much larger than the luminosity given by its rotation?

If Lx is very small, smaller than the luminosity given by its rotation, how to tell it is rotation-powered or accretion-powered? There are not necessarily X-ray pulsations

You can see I am asking a very tricky scenario.

The spin behaviour of the two types of pulsar would usually be quite different. The $\dot{P}$ for a rotationally powered pulsar is always positive, and higher order time derivatives of $P$ are quite small. This is because the rotational kinetic energy is powering the pulsar emission and the neutron star continually spins down as it loses rotational energy.

Accretion powered pulsars can have very variable spin down or spin up characteristics, because they are powered by mass transfer and accretion in binary systems, and are influenced by a variety of factors affecting the accretion flow and how it couples to the pulsar magnetic field. The rates of spin down can be much higher than can be plausibly be accounted for by magnetic spin down as in a rotationally powered pulsar. Spin up cannot be accounted for in a rotationally powered model and neither can sign reversals or dramatic variability in $\dot{P}$.

Additionally, an accretion-powered pulsar would necessarily need to be in a short period binary system - therefore there would be a very obvious periodic modulation in the pulse period caused by the doppler shift as it travels in ts orbit.


How to find maximum and minimum right acsension and declination based on the telescope's location? If I am to apply at a telescope for viewing objects, how to determine (easily) what is the minimum and maximum possible RA and DEC visible from that telescope during that time of the year. For eg, take the telescope at paranal
(24.6258°S, 70.4310°W) and if I observe anytime from October to March.

In the Southern hemisphere, the maximum declination you can see is
90-L where L is your latitude. The minimum is -90, since you can see
the south celestial pole.

As the earth rotates and revolves, you can see any right ascension
within those declinations.

The only exception is that you ordinarily can't see stars that are too
close to the Sun. However, the Sun moves enough between October and
March that all right ascensions will be visible at night sometime
between October and March.


Redshift and Velocity relation Let us say that we have an object so its total velocity is defined as

\begin{equation} v_{tot} = v_{pec} + V_{rec} \end{equation}

Where

\begin{equation} V_{rec} = H_0r \end{equation}

and \begin{equation} V(z) = \frac{cz}{1+z}[1+\frac{1}{2}(1-q_0)z - \frac{1}{6}(1-q_0-3q_0^2+j_0)z^2] ~~(4)\end{equation}

for small z.

So my first question is what is the $z$ value here? Is it the observed redshift or the cosmological redshift?

Also, the relationship between observed and cosmological redshift is given.

\begin{equation} 1+z_{obs} = (1 + z_{cos})(1 + z_{earth})((1 + z_{sun})(1 + z_{source})(1 + z_{gravity})    \end{equation}

If we are using the cosmological redshift then by using above equation we can write,

\begin{equation} z_{cos} = \frac{1 + z_{obs}} {(1 + z_{earth})((1 + z_{sun})(1 + z_{source})(1 + z_{gravity})}-1    \end{equation}

So is this what we put in (4)?

Edit: For the source you can look here https://arxiv.org/abs/1907.12639 Eqn(16) and (18)

In Equation (16) of the paper you link to, $z$ is the observed redshift. In the first paragraph of section 2.2

  The heart of the method is to use a measured redshift, z, to infer a velocity, v(z)


How accurately could late 50s - early 60s humans have mapped the solar system? What if early SETI investigations had focused on our own solar system, rather than distant stars?
Given the technology that existed around the 1950s, how much surveillance would humanity be able to do? This is in the era between Sputnik and Apollo 1, so various governments and scientific organizations would have had different equipment and personnel to devote to the task. So this change of focus would have impacted the way various governments, scientific organizations and civilian groups approached the search for intelligent life in space, and altered the focus of technological development.
What were the near-space observation capabilities of radio telescopes and related devices at that time? For example if Cornell University's Project Ozma or Ohio State University's Big Ear were used to look more closely at our own solar system, rather than distant stars, what would they have been able to detect or confirm?
If answers can address what kinds of information mid-20th-century technology could have detected, what level of detail, how often various regions could be scanned etc - that would be wonderful. I would also love recommendations about other scientific developments from that era which might be relevant to this topic.
(Note: I'm doing this research for an alternate-history science fiction writing project. I've edited the question to only ask about real, historical science, since the goal is a narrative that accurately represents the past as much as possible.)

I recall hearing Professor Lovelock about 10 years ago, describing a discussion with his students in the 1960s - what observation would indicate life on another planet? They concluded that spectroscopic analysis would show that the chemistry of the planet was inconsistent with the planet's age. A slower than expected rate of entropy would indicate the presence of life, because the chemistry of Earth life uses photosynthesis to build giant molecules which store energy by absorbing solar radiation
Spectroscopic analysis was used to examine of the atmosphere of Mars in the 19th century, many decades earlier than the context of your question. Before the space race began, scientists had already seen enough to rule out the possibility of life on other planets in the solar system
Lovelock's discussions about extra-terrestrial life led to his Gaia hypothesis, a set of models which demonstrate that Earth's current atmosphere and ocean chemistry emerged from billions of years of life on Earth


Can you see city lights on the Moon from Earth? This is the opposite of another question. That question is about whether you could see cities on Earth if you were standing on the Moon.

Let's there are cities on the Moon and you're standing on the Earth on a clear night. Could you see the city lights?

If you're looking at Earth cities from the Moon, your line of sight is not affected by atmospheric turbulence because the Moon is airless. But if you are on Earth, then you have to look through the turbulent atmosphere to see faint lights. So you won't have to put up with the turbulence that makes stars look blurry or twinkley from the Earth.

But if you're on the Earth, any city lights on the Moon might look blurry or twinkley from our atmospheric turbulence. If it helps, imagine yourself at the top of a mountain so there is less atmosphere to look through and hence less turbulence.

In this picture of the Moon (and Venus in the background), imagine cities on dark part of the Moon's surface facing Earth. Those city lights should be easier to see than cities within the lit crescent.

It is possible but unlikely. Here is a really good 'Science 2.0' article about the possibility (http://www.science20.com/robert_inventor/could_you_see_moon_city_lights_or_a_greenhouse_from_earth_just_for_fun-157480). Essentially, you likely wouldn't see the light on Moon settlements because there would need to have many thousands of bright lights and windows- both of these would be impractical on a moon base (windows let in radiation and are expensive, and bright lights waste valuable power). Furthermore, in order to reach a magnitude of brightness to constitute as visible, there would have to be a very large amount settlements concentrated together.


What are some formulas that are associated with the Schmidt corrector? I want to fully understand how an SCT works. Thus, I figured out I can use some formulas that describe how an incident ray that is parallel with the axis would act after refracting off a Schmidt plate. I presume that the formulas would depend on the vertical and horizontal distances from the center of the plate. My goal now is to make a ray tracing model of an SCT with certain specifications. Please make your answer as simple as possible.I will be thankful if a nontechnical source is provided at the end of the answer.

The specific purpose of a Schmidt Corrector plate is to have an equal but opposite spherical aberration to the primary mirror they compensate for. So any formula you try to come up with will depend on the primary.

In fact the information on SCT's shows how it is made using physical forms, not using formulae.

From revolvy.com:

  A thin glass disk with a perfectly polished accurate flat form was placed on a heavy metal pan. The upper edge of the pan was ground at a precise angle or bevel based on the coefficient of elasticity of the particular type of glass plate that was being used. The glass plate was sealed to the ground edge of the pan, then a vacuum pump was used to exhaust the air until a particular negative pressure had been achieved. This caused the glass plate to warp slightly. The exposed side was then ground and polished to a perfect flat. When the vacuum was released, the plate sprang back until its bottom surface was again plane, while the upper surface had the correct figure.

If, however, you do wish to work out the SCT parameters from formulae, this page on telescope-optics.net has the fairly complex info you need to use this formula:

$$z=\frac{(\rho^4-\Lambda\rho^2)d^4}{4(n^\prime-n)R^3}=\frac{(\rho^4-\Lambda\rho^2)D}{512(n^\prime-n)F^3}$$

It's not an easy page to work through - but for someone who hasn't needed complicated maths for some years I could just about handle it.


snapshot of solar system I am working on a little project where I want to simulate the solar system to teach myself working with python and OOP.

I want to calculate the orbits "real", from tangential velocities and gravitational accelerations. For this, in order to not only get circular orbits, but the real, eccentric orbits, I need kind of a snapshot of the solar system with positions and tangential velocities at those positions of the planets (and their moons, but I highly doubt that exists).

EDIT: I think something like the velocities at the Apoapsis should be enough to derive a reasonably right orbit from it, right?

I don't even know if something like that exists, or where to get it from. Does anybody happen to know something of this?

You can get a position and velocity snapshot easily using the Python package Skyfield. It downloads a JPL Development Ephemeris and then interpolates the positions for whatever time you choose for your snapshot.

You can also use JPL Horizons to produce one or thousands of positions of solar system bodies.

This answer describes how to set up Horizons to save positions to your computer, and includes a script that can read the format of the saved tables. It also happens to be a simulator that does what you are talking about - statrs with a given position and propagates the orbits.


What is tilt anisoplanatism for Laser-guide-star usage in adaptive optic systems? How does tip-tilt mirrors correct such issue? What is the physical origin of the error?

Isoplanatism  commonly refers to a region of angles over which a ground-based telescope observes effectively the same atmospheric turbulence (e.g. an "isoplanatic patch"), such that a laser guide star provides effective correction of atmospheric seeing.

an-isoplanatism refers to a lack of isoplanatism, or a way in which  the science target and the reference laser guide star differ.

A single laser guide star does not provide knowledge of tip or tilt errors since the up-going beam experiences tilts in addition the down-going signal; these two sets of tilts are entangled and in the case where the transmitter and receiver are identical the laser guide star appears stationary. See a detailed overview and history of this problem the introduction of Ragazzoni
 1996, http://adsabs.harvard.edu/full/1996ApJ...465L..73R. This is commonly solved by observing a "natural guide star" position, since its light only passes through the atmosphere once, and using that signal to maintain the position of the target star by changing the angle a small ("tip-tilt") mirror.


Activity of M dwarf stars Why are M dwarf stars the most active (in terms of starspots and faculae) of all stars?

The magnetic activity is driven by a dynamo that relies on a combination of rotation and convection.

M dwarfs have large convection zones or are fully convective beyond spectral type M4. They also have much longer spin-down timescales than G- and K-dwarfs; billions of years rather than tens of millions of years.

This means that a large fraction of cool M dwarfs are rotating fast and have deep convection zones.

This is the basic reason, but it is still a very busy research area.


Is it possible that the Universe is expanding due to additional Big Bang events? Is it theoretically possible that the Universe is expanding due to additional Big Bang events that have occurred at the same location that the original Big Bang occurred at?

Perhaps there has been an ongoing series of smaller Big Bang events over the past 13.7 billion years, which although were not as massive in scale as the original Big Bang, have created a large amount of new space-time and new matter that has resulted in the Universe expanding.

You may need to revisit the theory behind the Big Bang. The Big Bang is more or less a description of what the early universe looks like if you view expansion in reverse. The single "Big Bang" at the beginning of our universe is the only "Big Bang" event predicted by expansion. Also, the Big Bang did not happen in a single place, it happened everywhere in the universe. Finally, the Big Bang did not create space-time, and the expansion of the universe is not the creation of new matter. The Big Bang describes a time when the metric of space-time was extremely dense, and expansion is the metric of space-time increasing as below:

The distance between points in space increases, but the same mass-energy is there as before.


What kind of telescope should I get my husband? I'm very much an amateur when it comes to telescopes. My husband wants one for his birthday. I'm willing to spend up to $900 on one. Can any of you gurus give me advise as to which one I should buy?

Thank you so much in advance!

Video Astronomy or as some refer to it as Live or Near Live Viewing is a growing field of interest that lies between traditional viewing with an eyepiece and Astro Photography.

Eyepieces are substituted with cameras and the viewing is done on a screen. In general the cameras are more sensitive than the eye and often reveal more detail and color.

That being said it is a field that some love and others hate. The individual needs to decide for themselves. There are extremists in both groups. An open mind helps but have a few grains of salt on hand just in case. :)

In general, the electronic viewers prefer telescopes with a faster f ratio than the eyepiece viewers. Because of the increased camera sensitivity, many claim smaller aperture telescopes can work as well as those with double their aperture. Better to find out what type of viewing is more interesting before buying any equipment.

My suggestion is to see what the electronic viewers are doing with their equipment.

Many electronic viewers broadcast their viewing sessions on the NightSkiesNetwork. Anyone can join in on these broadcasts at

https://www.nightskiesnetwork.ca/

Most broadcasts originate in North America but they could be from anywhere in the world. Check for clear weather in the Eastern time zone at nightfall in the East. If the weather looks clear take a look at the NightSkiesNetwork site then look again occasionally as nightfall passes across the continent. It is all weather dependent. In general you will find a broadcast sooner rather than later.

Most viewers of the NightSkiesNetwork started out with traditional eyepiece viewing and like astronomers everywhere are more than willing to give people advice whether it be for viewing with an eyepiece or a camera.

I use several Mallincam cameras, but there are other cameras available.

Google "Video Astronomy" or "Electronic Assisted Astronomy" for more information.

For what it is worth I viewed with eyepieces for fifty years before going electronic. In the ten years since going electronic I have had no desire to use my traditional eyepieces. Even when my eyesight was at its best and with telescopes up to 14" aperture I have never seen as much as I do now with an electronic eyepiece on my 5" and SCT's 10". That may make me sound like an extremist, if so be prepared to deal with a few more like me. :)


Assume that a star has a uniform mass density. Then how does the temperature scale as a function of the distance to the center? Assume that a star has a uniform mass density. Then how does the temperature
scale as a function of the distance to the center?
Note: I am not looking for an answer that involves derivations or integrations.
Here is what I did:
The star is in equilibrium so the pressure at distance $r$ must equal the gravity force at distance $r$. Pressure is given by $P=nKT$, and so $P\times Area=F_g$.
Then $nKT \times 4\pi r^2=F_G$. The mass of the part of the star contained within radius $r$ from the center is $\frac{4}{3}\pi r^3\times m_sn$ where $m_s$ is the mass of each particle and $n$ is the number of particles per volume.
Then we can rewrite $nKT \times 4\pi r^2=m(\frac{G\frac{4}{3}\pi r^3\times m_snF}{r^2})$ and simplify to $KTr=\frac{m}{3}Gm_s$. Isolating $T$ we get that $T=\frac{mGm_s}{3Kr}$.
This seems to imply that the temperature decreases the further we are from the core, however I am not sure what $m$ should stand for in $F_g=mg$, should $m=m_s$, i.e be the mass of a single particle? Or does $m$ itself depend on the radius in which case the answer I got is invalid.
Can someone verify that the answer I got is right?

I am not sure what you mean by looking for an answer without derivations? Of course you have to solve the maths to get a mathematical answer.

The handwaving answer is that the star has to be hotter in the middle, because at constant density the pressure depends on temperature and you need a negative pressure gradient to support the weight of the gas above.

The answer is:

The equation of hydrostatic equilibrium is
$$\frac{dP}{dr} = -\rho g,$$
where $\rho$ is the mass density and $g$ is the gravity at radius $r$, which by Newton's shell theorem is due to the total mass interior to $r$.

This equation is easily derived by considering the force equilibrium on a slab of gas in the star (your error above was in not recognising that it is the difference in pressure across the slab that provides the upward force).

If the density (and composition and ionisation state) are constant, then this equation can be written as
$$ \frac{\rho k_B }{m} \frac{dT}{dr} = -\rho \frac{4\pi G r^3 \rho}{3r^2},$$
where $m$ is the mass per particle and a perfect gas has been assumed.

Integrating with respect to radius and assuming $T(r=0)=T_0$, we obtain
$$T = T_0 - \frac{2\pi G\rho m}{3 k_B}r^2$$

Finally, if the star has radius $R$, then this will be radius at $T=0$, since nothing could be supported beyond that. In which case
$$T = \frac{2\pi G\rho m}{3 k_B}(R^2-r^2)$$

Or in terms of mass $M$
$$T = \frac{GM m}{2 k_B R^3}(R^2-r^2)$$

Constant density is of course a poor approximation. The above treatment gives a central temperature of 7 million Kelvin for the Sun, assuming $m\simeq 10^{-27}$ kg.


Is Naboo's core possible in real life? In Star Wars, the planet Naboo has a plasma core, instead of a molten one.

Can a planet really have a plasma core?

The mass will pull into the shape of the sphere if the body is massive enough. The body will then basically be in hydrostatic equilibrium. A planet like Naboo couldn't exist b/c the denser materials would move to the centre, and the lighter elements would then form an atmosphere around the solid core.

Plasmas require really high temps and a terestrial plantet couldn't compress gases to the temps required to completely ionize the gas. That and what I said about the way the planet would form. There's just no way that a shell of rocky material could be supported by gas pressure in that way. The whole thing would be really unstable.


How can satellites be safe in the atmosphere when there are so many asteroids coming in? How  can artificial satellites be safe in the atmosphere when there are hundreds of tons of dust and asteroids coming into earth's atmosphere everyday?

What is an orbit of an artificial satellite?
At what approximate distance (or which orbit) from artificial satellites do these asteroids, dust, rocks annihilate?

Space debris is a big problem for space industry. Natural bodies like dust or asteroids are not the main challenge, although probes are frequently hit by fine dust. The dust causes tiny craters on the surface of the probes or can change rotation or orbit slightly. This can mostly be corrected, e.g. by wheels fueled from solar panels.

The most concerning objects are remnants of human-made objects circling around the earth. With each collition of two such objects clouds of fragments form, leading to ever more collitions. Those fragments are tracked by radar, and probes try to avoid collitions by adjusting their orbits. But sometimes they are hit and destroyed.

An orbit is the path along which a satellite travels around e.g. Earth.

A satellite must remain outside the lower parts of the atmosphere, otherwise it will be slowed down and fall to the ground. The lower limit is somewhere near 200 km above ground. A satellite travels with about 8 kilometers per second around Earth, when in a lower Earth orbit.

Many satellites are in the geo-stationary orbit, about 36,000 km above ground. There Earth's atmosphere is neglectably thin, only some minor magnetic field can be found there; that part of the atmosphere is called magnetosphere.

The further away a satellite orbits the Earth the slower it travels to compensate Earth's gravity, because Earth's gravity grows weaker with distance from the Earth.

At present the probability of collisions, which destroy satellites, is very low for a single satellite. Actually there are only a few documented cases: "only one major incident has occurred: the 2009 satellite collision between Iridium 33 and Cosmos 2251", of the above wikipedia article. And "China was widely condemned after their 2007 anti-satellite missile test, both for the military implications as well as the huge amount of debris it created" of the same Wikipedia article.

The central problem is the growing number of fragments to be expected for the future, as a consequence of more satellites in the orbit, and the expected chain reaction of fragmentation.

Details of surface degradation of satellites due to impacts of dust particles can be read e.g. here. Most of that dust ist human-made. Surface degradation has consequences, e.g. for solar panels.

See also a related question at StackExchange.

Regarding the second part of the question:
The dust particles collide with the satellites and evaporate from the impact heat. So the annihilation distance is more or less zero.

When entering the earth's atmosphere "they usually disintegrate at altitudes of 50 to 95 km " (Meteorid). That's at least 100 km below the lowest satellite orbits.


Why do Earth and moon move apart but binary black holes move closer? According to the accepted answer on Is the moon moving further away from Earth and closer to the Sun? Why?, the moon is receding from Earth because tidal forces and friction cause energy to be lost.

However, according to LIGO's website,

  As the two masses rotate around each other, their orbital distances
  decrease

as they lose energy radiating inspiral gravitational waves.

Why do the bodies move apart in the Earth-moon case but closer together in the black hole case?

If both opposing phenomena are present but a different one is stronger in the two cases, what determines the fate of a system?

Here is how the tides move the moon away from the Earth:

The moon orbits the earth, and there is a difference in gravitational force between the the side of the Earth nearest the moon, and the side far from the moon.

This difference in force tends to pull the Earth into a oval shape with its long axis pointing towards the moon.

But the Earth is also spinning, and this spinning moves the axis of the oval forward, so the oval doesn't point towards the moon, but a little ahead of it. So there is a bump on the Earth, and it is permanently a little in front of the moon. This bump has mass and it pulls the moon towards it, So the moon is being pulled forwards. The moon pulls the Earth back. So the Earth's spin is slowed, but the moon gains energy and moves a little further from the Earth.

The moon has also been slowed down to the extent that the same side always faces Earth, and it will remain locked like this.

Gravitational waves, on the other hand, imply an emmision of energy from the binary system, and as energy is lost, the black holes spiral in.

There are no tides on black holes because there is nothing there to be pulled into an oval shape. The event horizon is not a solid surface. The mass of a black hole is concentrated entirely at the singularity, there is no structure to be deformed into a bump.


Why is there so little nitrogen in the Martian and Venusian atmospheres? Why don't our neighbors have much nitrogen?  You would think that, without 'nitrogen-fixing' organisms and such, there might be more.....

Nitrogen, with a molecular mass of 28 atomic mass units, is too light to have remained in Mars's atmosphere. Carbon dioxide, with a molecular mass of 44 amu, could (and does) exist on Mars, but it is rather sparse.
Venus's atmosphere appears to contain a small amount of nitrogen when viewed on a percentage basis. "Only" 3.5% of all of the gases in Venus's atmosphere is nitrogen. But that's not a particularly good metric. Venus has so very, very much carbon dioxide in its atmosphere that the CO2 dwarfs every other constituent. When viewed in terms of mass, Venus has a good deal of nitrogen in its atmosphere, four times more nitrogen than the Earth has in its atmosphere.