Why isn't IMBH candidate GW190521 very "chirpy"? Which types of events will produce the most chirp-like behavior? The first gravitational wave observations were noted for their "chirp-like" behavior; the objects would speed up as they got closer, radiating more strongly and therefore loosing energy faster. An example of a "chirp" is shown below.
The new PRL GW190521: A Binary Black Hole Merger with a Total Mass of 150 M⊙ seems to show the first observation of the formation of an intermediate mass black hole (IMBH). The image below doesn't seem to exhibit much chirpiness compared to lower-mass mergers. For example from  P. B. Abbott et al. (2016)

On September 14, 2015 at 09:50:45 UTC the two detectors of the Laser Interferometer Gravitational-Wave Observatory simultaneously observed a transient gravitational-wave signal. The signal sweeps upwards in frequency from 35 to 250 Hz...

Question: Why do some gravitational wave events chirp more than others? Which ones produce the best, longest, and most obvious chirp in frequency?

above: not very chirpy GW190521 from Abbott et al. (linked above)
below: "chirpy behavior" from this answer and P. B. Abbott et al. PRL 116, 061102 (2016) Observation of Gravitational Waves from a Binary Black Hole Merger.

See also this neutron star merger

All the binary mergers chirp, but the overall timescale of the event depends on the total system mass (or rather the chirp mass - see below). The more massive the system, the more rapid the evolution of the amplitude and frequency and the lower the orbital frequency when it finally merges. What you observe is also governed by the response of the detector - that is, its strain sensitivity as a function of frequency.
Thus high mass stellar black hole mergers have rapidly evolving signals and end their chirps at relatively low frequencies. Lower mass black hole mergers or neutron star mergers are much more slowly evolving and end their chirps at much higher frequencies.
The key equations (assuming circular orbits and to first order) are:
$$ \frac{df}{dt} \simeq \left(\frac{96}{5}\right)\left(\frac{G\mathcal{M}_c}{c^3}\right)^{5/3}\pi^{8/3} f^{11/3}\, ,$$
where $\mathcal{M}_c$ is the "chirp mass" and is given by $(m_1m_2)^{3/5}/(m_1+m_2)^{1/5}$, where $m_1$ and $m_2$ are the component masses. The chirp mass basically gets larger with the total mass of the system for a given mass ratio, but strictly speaking, what I said in the first paragraph about high- and low-masses means high and low chirp masses.
You can see from this formula that at a given frequency $f$ (in this case in the LIGO sensitivity band), that the rate of change of frequency is higher for higher chirp masses.
The chirp ends when the objects "merge". Exactly what this means is a bit loose, but is when the separation of the components is a small multiple of their radii in the case of neutron stars or a small multiple of their Schwarzschild radii for black holes. A simple use of Kepler's third law (not strictly applicable in General Relativistic conditions, but it gives the right scaling), and recalling that the GW frequency is twice the orbital frequency yields
$$ f_{\rm max} \sim \left(\frac{GM}{\pi^2 a^3_{\rm merge}}\right)^{1/2}\ ,$$
where $ a_{\rm merge}$ is the separation at merger and $M$ is the total system mass. If we let $a_{\rm merger} \sim 4GM/c^2$ (for black holes), then we see that $f_{\rm max} \propto M^{-1}$.
$f_{\rm max}$ is around 130 Hz for a $30+30$ solar mass black hole merger (like GW150914, which had $\mathcal{M}_c \simeq 28M_{\odot}$, so would be just $\sim 50$ Hz for a merger resulting in a 150 solar mass black hole like GW190521 (which had $\mathcal{M}_c \simeq 64M_{\odot}$).
So that I think is your answer. The sensitivity of LIGO is quite poor below 30 Hz due to seismic noise. Thus in this event, the merger and chirp is barely seen at all in the sensitive LIGO frequency window of 30-2000 Hz - just the final few orbits (in fact only two for GW190521) before the merger and ringdown phase begins at a peak frequency of $\sim 50$ Hz.
To get a more chirpy signal that is detectable by LIGO you need (a) a lower mass system with a higher $f_{\rm max}$ and (b) it needs to be close enough so that LIGO can detect the lower amplitude GWs with $f<f_{\rm max}$ that are emitted prior to the merger.


Why is the Eagle Nebula so "static"? This, of course, is a newbie question, as I am nothing more than a hobbyist. But I was quite surprised to see the recently-released "before and after" pictures of the so-called "Pillars of Creation" in the Eagle Nebula (here and here). For something that looks like such a "nebulous" cloud of gas (pardon the pun), I would have expected at least some visible sign of change in the 25 years between the two photos. But at the pixel resolution which I had available, I could not detect even the slightest difference between them. Of course, I am accustomed to terrestrial clouds which are in constant motion, so I (mistakenly?) expected something analogous at the astronomical level. Can someone provide, in layman's terms, how my understanding of this cosmological feature is deficient? In other words, how might I adjust my instinctive intuition that this object should be more dynamic? Is it just the sheer scale that I am not comprehending? (By the way: what is the distanse between the three spires, as compared to, say, our solar system?)

It appears to be static because it's huge beyond your imagination.

The distance to the nebula is 7,000 light years. Its apparent size is 7 arc minutes. Therefore its linear size is about 14 light years.

Think about that. The whole nebula is so big, it takes light 14 years to cross it. Any motion therein must necessarily be much, much slower. No wonder you're not seeing much change.

Data source: wikipedia

Calculation using Wolfram Alpha


Meteor shower from Moon's "debris" Did we ever witness a "meteor shower" which was actually dust or debris of some kind from our own Moon?

There is a class of meteorites called Lunar Meteorites and several of those have been found on earth.

Of course, it doesn't necessarily indicate whether we have actually witnessed a certain lunar meteor shower. Considering that nearly 100 metric tons of space debris falls on earth each day, one would imagine that it'll be colossally impractical to attempt mapping a meteorite to the moon, using a reference to a recorded meteor shower.

However, it might be worthwhile to take a look at this link which addresses rather interesting questions:

 How did lunar meteorites get here?

  Any rock on the lunar surface that is accelerated by the impact of a meteoroid to lunar escape velocity of $2.38 km / s$ or greater will leave the Moon's gravitational influence. Most rocks ejected from the Moon become captured by the gravitational field of either the Earth or the Sun and go into orbit around these bodies. Over a period of a few years to tens of thousands of years, those orbiting the Earth eventually fall to Earth, whereas those orbiting the Sun may also eventually strike the Earth in a few tens of millions of years after they were launched from the Moon.

 How do we know that they come from the Moon?

  Chemical compositions, isotope ratios, mineralogy, and textures of the lunar meteorites are all similar to those of samples collected on the Moon during the Apollo missions. These meteorites are rich in a mineral called anorthite and have high concentrations of aluminum and calcium. Because of some unique aspects about how the Moon formed, the lunar highlands are composed predominantly of anorthite. To the best of our knowledge, anorthite is much less common on asteroids, and on the surfaces of any other planets or planetary satellites.

  Quotes derived and paraphrased from the linked article.

An up to date catalog of all lunar meteorites found so far indicates that the heaviest lunar meteorite is a rock (pair) named Kalahari 008 & 009, weighing ~$14kg$.


Why is frequency different for gravitational waves coming from SMBHs vs small compact binaries? Why do gravitational waves emitted by supermassive black holes binaries have low-frequency range (~10$^{-9}$ - 10$^{-2}$ Hz) than those emitted by stellar mass compact binaries (~10$^{-4}$  - 10$^2$ Hz)?

Below, a NASA graphic showing the wavelength ranges of various phenomenon found here. Click for full size.

The reason for the upper limit is that gravitational force scales as the product of masses over separation squared and that the closest approach of black holes is limited by their event horizons. The lower limit is because the amplitude of the waves gets smaller and becomes undetectable at lower frequencies.

In more detail:

The frequency of gravitational waves is twice the orbital frequency.

Kepler's third law tells us that the orbital period $P$ is related to the total mass $M$ and orbital separation $a$ as
$$ P^2 \propto M^{-1} a^3 $$

The orbital frequency $\omega =2\pi/P$ so
$$\omega \propto M^{1/2} a^{-3/2} \tag{(1)}$$

The frequency of gravitational waves therefore depends on both the mass and separation of the binary system.

However, the maximum frequency occurs at closest approach, which for black holes is when their event horizons come into contact. The Schwarzschild radius is $2GfM/c^2$, where $f$ is the fraction of the system mass of each black hole. Thus the closest separation $a_{\rm min}$ is the sum of the Schwarzschild radii and is simply $2GM/c^2$, since $f_1+f_2=1$.

Using $a_{\rm min}$ for $a$ in equation (1) we see that the maximum frequency
$\omega^{\rm max} \propto M^{1/2}a_{\rm min}^{-3/2} \propto M^{-1}$
Thus the maximum frequency gets bigger as the mass gets smaller, and as supermassive black holes are a million to a billion times more massive than stellar black holes, the maximum frequencies are correspondingly smaller.

The lower limits to frequency are not really limits at all. The frequencies can be anything smaller than the maximum. However the amplitude of the gravitational waves depends on the acceleration of the masses, which in turn decreases as they become more widely separated. This effectively gives a window between where the waves become detectable and the maximum frequency.


I need to study space, planets and black holes, how do I study them on my own I need to study space, planets and black holes, how do I study them on my own, knowing that I am an Arab and I do not know English. I use translation applications. I speak with you. Can you help me with a site or YouTube channel? I don’t want books

I am sorry to say, but the best resources are still books. If it is really a no-go, read further. It is very likely, that you can find and buy astronomy books on Arabic on the Internet. First start with popular science books. Buy multiple books. To learn something, books are still somehow far better than the Internet.
Check the Arabic wikipedia, like this article.
Just think freely. Imagine a question, anything, like "How far are we from the edge of Universe?"1. My experiences is that the resources on my first language (far lesser amount of speakers than Arabic) are significantly lesser, but they exist. Enter the question on Arabic into the google search.
Any time if something is not clear, come here and ask.

1It is actually a bad question, but you don't know yet, why. Here is the answer, why. This was only an example question.


Spiralling in Galaxy evolution Could some stars be orbiting galaxies in outward spirals instead of perfectly round circulation? Patterns of matter movement relate strongly to the galaxy centre. A special relationship between the galaxy and the central object suggest the centre may be an origin point in 'outflow'. The BH central object behaves strangely by instigating the release outwardly of massive amounts of matter. This may form stars that spiral out across the galaxy. The grand design of features in galaxies may related to general outward spiralling.

Basically, yes, it's possible.  Our moon is slowly moving away due to energy transfers from Earth related to tidal drag.  This is very slow spiral.

I do not know whether a star can experience sufficient tidal drag relative to the center of its galaxy to cause a change in orbit - and there are a few billion other objects in the galaxy producing gravitational force in other directions, making the analysis kind of messy.


What are the RAs of the boundaries between traditional zodiacal signs? I'm trying to figure out at what right ascension (RA) — not celestial longitude — the sun enters the various traditional (Western) astrological signs of the Zodiac; but I can't figure out how to calculate this.

What are the right ascensions of the boundaries between traditional zodiacal signs along the ecliptic?

TL;DR: Calculating the right ascension (i.e. converting from ecliptic to equatorial coordinates) requires some spherical trigonometry, but fortunately there are internet tools that will do this, such as here. The resulting RA coordinates are:
Aries 0h 0m 0s; Taurus 1h 51m 39s; Gemini 3h 51m 16s; Cancer 6h 0m 0s; Leo 8h 8m 44s; Virgo 10h 8m 21s; Libra 12h 0m 0s; Scorpio 13h 51m 39s; Sagittarius 15h 51m 16s; Capricorn 18h 0m 0s; Aquarius 20h 8m 44s; Pisces 22h 8m 21s.

Explanation and background

Since traditional (Western) astrology uses the same tropical year as astronomical celestial coordinates, the tropical zodiac commences with Aries at 0˚ longitude on the ecliptic, and each subsequent zodiac sign (NB not the constellation, which is different) commences exactly 30˚ further along the ecliptic. So, the zodiac division of Taurus is from 30.00˚ to 59.99˚, Gemini from 60.00˚ to 89.99˚, and so on. It's then just a matter of converting ecliptic longitude to celestial coordinates (right ascension).

Further useful information on the relationship between the ecliptic and right ascension can be found in Mike G's answer to "Effect of the obliquity of the ecliptic / tilt of the Earth on the equation of time".

Both the traditional tropical zodiac and modern astronomical coordinates are based on the northern hemisphere vernal equinox. This starting point, also known as the First Point of Aries, is one of the two points at which the celestial equator crosses the ecliptic.

When Hipparchus defined the First Point of Aries in 130 BCE, it actually aligned with the border between the constellations of Pisces and Aries. However, due to the precession of the equinoxes, the tropical zodiac's starting point has a retrograde movement along the ecliptic at the rate of one degree every 71.5 years. As a result, the First Point of Aries is currently deep in Pisces! Using the current official IAU constellation boundaries, the point of 0h RA will cross into Aquarius in the year 2597, so we have quite some time to wait for the "Age of Aquarius".

Note that not all astrological traditions are based on the tropical year. Hindu astrology, for instance, is based on the sidereal zodiac - i.e. it aligns permanently with the visible stars, regardless of the Earth's axial precession. Some modern Western astrologers also use a sidereal zodiac, in which sidereal Aries currently begins on 15 April.


What exactly was the Moon's "Evection Resonance"? Ward et. al define the evection resonance of the Moon with the Earth and Sun as:

occurring when the lunar perigee precession period equals one year.

Does this simply mean that the angle formed by the Sun-Earth-Moon when the Moon was at perigee was the same every month?
I tried looking at the wikipedia definition of Evection but just found it more confusing.
If it helps, I am curious about the lunar evection resonance because it is often cited as a mechanism to explain Moon/Earth angular momentum shedding necessary if one assumes the Giant Impact Hypothesis of Moon formation.  See Ward and Canup:

The resonance can exchange angular momentum between the Earth-Moon
system and the Earth's orbit.

The orbit of the moon is elliptical.  But the direction of the major axis of this ellipse isn't fixed. Perturbation by the sun causes the axis to rotate towards the East.  The time for the moon to go from perigee to perigee is slightly longer than the time it takes for the moon to orbit the Earth.
The direction of the major axis of the moon slowly rotates around the Earth.  Now it takes about 8.85 years to make a complete 360-degree rotation.
But early in the history of the Earth-moon system, it might have made a complete rotation in less time. There could have been a time when it took one year to make a 360-degree rotation.  This could lead to a resonance that would affect the orbit of the Earth around the sun.  In particular, it could transfer angular momentum from the Earth-Moon system to the Earth-Sun system.  This can, perhaps, explain why the Earth-Moon system has less angular momentum than would be predicted by a system that formed from a planetary collision.
The paper investigates how such a resonance could form, how it could transfer angular momentum, and how the resonance could eventually be broken.


Which celestial body's length of day is closest to that of Earth? I know the planets Mars and Eris are among bodies that have the most Earth-like mean solar days (about 24.67 and 25.9 hrs). Among moons there is Mimas of Saturn which has one of the most Earth-like days. Uranus and Neptune also have more Earth-like day lengths than many other known planets or moons. However I don't know of a body in the Earth's planetary system that would have a more Earth-like solar day than Mars (other than the Earth of course). Do you know a planet, moon or asteroid that has a daylength closer to that of Earth than Mars' is?

Methone, a tiny moon of Saturn has an orbital period of 1.01 days. This is approximately 24 hrs and 14 minutes. Ref. 1

https://www.windows2universe.org/our_solar_system/moons_table.html


Are there any stars that orbit perpendicular to the Milky Way's galactic plane? Most stars orbit in the Milky Way's galactic disc. But is it possible for one to orbit perpendicular to it?
Here on Earth since we're inside the galactic plane we can't get a good view of what the Milky Way looks like. But would the whole Milky Way be visible from a planet orbiting such a star?

The Sun and most of the other stars are in the bulging disk of the Milky Way galaxy, but about 1% of the galaxy's stellar mass is in the galactic halo.  The halo also includes 50 globular clusters and about 20 satellite galaxies according to Helmi 2008: The stellar halo of the Galaxy.  Here is a nice graphic:

Note that the Sagittarius Stream of stars is extremely close to passing the galactic poles.  If you were on a planet next to one of these stars as it passed a pole, you would probably have a glorious view of the Milky Way stellar disk.
Bodies in the galactic halo don't necessarily follow the elliptical paths predicted by Kepler, so their orbits may not be consistently perpendicular to the galactic plane.  Some stellar streams have extremely odd orbital paths like the Phlegethon stellar stream, referenced in this answer:

There is some speculation that the Halo stellar streams are the remnants of dwarf galaxies the Milky Way has absorbed.  Indeed, Liang et al. say that:

the Galactic halo has complicated assembly history and it not only
interacts but also [is] strongly mixed with other components of the Galaxy
and satellite dwarf galaxies


Planets and moons positions in cartesian coordinates? Is there freely available and reliable source, where one can find positions of planets and moons of the Solar system at given time in cartesian coordinates? (I am writing a simple gravity simulator and it would save me a lot of work with conversion.)

Horizons Ephemeris Generator

Here is a screen capture where vector option is chosen:


Why is the Color scheme of natural satellites in our solar system based of light shades of grey? What is the explanation of the color similarity that the majority of natural satellites obtain?

The colour of a natural satellite is almost entirely based off the atmosphere and surface of the satellite. The moon Io, of Jupiter is not grey and is coloured because of sulfur on its surface that create a colour. Io has a very thin atmosphere.

So that brings us to your question, why is the majority of natural satellites grey? We know that moon's colours are dependent on what they are made up of and if they have an atmosphere, so the basic answer is that the majority of moons are grey because they do not hold an atmosphere, and are made up of rock. Our moon is made out of basalts which are a grey rock. You can probably go out and research what certain moons are made of, and find out more about the rocks on them and how they affect the colour of the planet.


How much of the Universe is subatomic? How much does subatomic particles such as photons, tachyons, and electron being the biggest particles etc. occupy space? If put in a ball how big of a ball would that be?

First things first ...

  Tachyons

These are entirely hypothetical and no evidence for them exists.

  How much does subatomic particles such as photons, tachyons, and electron being the biggest particles etc. occupy space?

How long is a piece of string ?

There's no simple answer to what size an elementary particle is.  You can view them, depending on context, as having no size (point like) or being spread all over space (wave distribution).

We can't say how many free particles there are (i.e. particles not bound to other particles).

This is a lot more complicated by the current mainstream thinking that dark matter and dark energy make up most of the mass/energy content of the universe.  Whatever dark matter may be, if it turns out to be a particle or particles we don't know about yet (quite likely on existing evidence) then those particles will almost certainly exist as free particles (only gravitationally bound together) as current thinking is that dark matter does not interact much at all with other matter or itself.  And we have no idea what dark energy could be.  So with most of the mass/energy is in forms we cannot identify we can't really say anything much about what you're asking.

A very, very, very rough estimate on the amount of dark matter is about 25% of the mass-energy of the universe.

  If put in a ball how big of a ball would that be?

In a ball it would be (possibly) a huge black hole comprising anywhere from 25% to 95% of the mass-energy of the universe.  I won't even bother estimating the size of the event horizon as it's just a useless number.

However "how big a ball" could equally mean it was, as it is now, spread out over the entire universe.  So a perfectly correct answer is - universe sized.


Heavy bombardment - Why has things cooled? I was always taught and lead to believe that at the beginning of our solar system, there was a period of heavy bombardment, which is conceivable.

Why has it just stopped and we are at peace from this, and, there is no such things happening today?

According to the Nice model -- that's Nice, France (rhymes with Reese or geese), not nice (rhymes with mice) -- the solar system formed with the giant planets (the outer four) in the following order: Jupiter, Saturn, Neptune, Uranus. Outside of these planets existed a large field of asteroids (also called planetesimals), the remnants of which form our Kuiper belt.

The Nice model hypothesizes that Neptune and Uranus actually swapped places due to an orbital resonance between Jupiter and Saturn. This had the effect of moving Jupiter inward (towards the Sun) and kicking Neptune farther out. This orbital disruption also caused a thinning of the Kuiper belt, where some of the planetesimals gained kinetic energy and were knocked into the Oort cloud, and some of them slowed down and fell into the inner solar system, causing a period of heavy bombardment.

It is thought that the period of heavy bombardment ended once the orbits of the outer four planets stabilized and an equilibrium was reached. The result was the current order of the outer planets we see today (Jupiter, Saturn, Uranus, Neptune), a thinner Kuiper belt, and an outer Oort cloud. Of course, this is just a model; the Nice model itself has already undergone some revisions, even though it is relatively recent. A nice video illustrating the Nice model can be found here.

I hope this answers your question.


What is the scale of things you can see with gravitational lensing? I'm trying to understand the examples of gravity lensing (using the general relativity property of large masses to bend light like a lens).

Most of the examples I see are of some galaxy (presumably a large mass) between us and a star (or are at least vague enough to not specify.

Yet my intuitive understanding of the sky is that all the stars we see are relatively close, all entirely within the Milky Way galaxy (and many of the stars in our galaxy provide a general glow), and that other galaxies are far enough away that it is difficult to image individual stars. The only single objects large enough comparable to a galaxy would be a quasar. Is that right?

So what then is going on usually with examples of gravity lensing? I find it hard to believe the captions that say a galaxy is enabling seeing more distant stars. I would think one could only apply that concept to a star or galaxy to see something much further away and as big or much bigger. Could one really use a galaxy as a lens to see a star? I wouldn't expect a star to be behind a galaxy from us.

Also, whatever the objects themselves, what is the scale? If the lensing is of a galaxy done by a star, I'd expect the star to galaxy distance ratio to be well below 1:1000 (~width of Milky Way to distance to Andromeda). But for galaxy to galaxy or galaxy cluster to galaxy or quasar lensing what are the likely relative distances?

You are right that the stars seen on the sky are within the Milky Way. Only with a large telescope is it possible to resolve individual stars in other galaxies, and only for the nearest ones.

I don't know which sources you refer to, by I think perhaps you are confusing the different types of gravitational lensing. I cannot explain them better than the excellent review by astromax, but briefly, there are three types:

Strong lensing, where are foreground galaxy cluster (i.e. a group of $\sim$100–1000 galaxies) magnify and severely distort background galaxies,
Weak lensing, where clusters or individual galaxies distort the shapes of many background galaxies on the percent scale, which can only be seen statistically, and
Microlensing, where a single object inside the Milky Way happen to pass in front of another single object, also inside the Milky Way. These objects are usually stars or planets, and do not distort the images of the background objects, but merely increase the flux for a while. This effect has been used to find exoplanets.

Whereas type 1 and 2 are more or less static in a human lifetime, type 3 is an event that happens once only for a given set of stars (as seen from Earth).

Scales

Strong and weak lensing happen on very large scales, from a few hundred million lightyears, up to the order of the size of the observable Universe (e.g. Wong et al. 2014). While the lenses themselves are cluster of galaxies, and thus a few to $\sim10$ megaparsec across, the lenses object are typically individual galaxies. Gravitational lensing is most efficient when the lens is halfway between us and the background source.

Occurring in the Milky Way, microlensing, on the other hand, happen on the scale of a few kiloparsec, again with us-lens distance being of the same order as the lens-background object distance (see e.g. this Wikipedia list).


iteration to cover the whole sky with right ascension, declination, angle I am sure I get parts of the terminology wrong but if anyone can shed some light in the following:

I understand that for a given right ascension (RA) and declination (DEC), one has defined a ray (half-line) in the sky starting from the center of the earth towards infinity.

Now, if I also provide a angle, say d degrees around that ray, I have essentially defined a cone in the sky. My question is the following:

Which steps (RA_step and DEC_step) should I use in the following loop, as a function of d to ensure that I cover the entire sky and don't leave any patches anywhere?.

for (RA = 0 ; RA <= 360 ; RA += RA_step)
    for (DEC = -90 ; DEC <= 90 ; DEC += DEC_step)
         examine-cone(RA, DEC, d)

Imagine you are patching a plane with circles, instead of patching the sky sphere with circles (which is the same as filling the space with cones). Now, for every circle of diameter D define a square of diagonal D (D=2d). It is easier to patch the plane with squares, isn't it? So you have squares of side D/2*sqrt(2), and that is exactly your DEC_step, for each vertical sweep, and your RA_step, from sweep to sweep.

This will lead to a lot of duplicated examination (in fact you will examinate exactly the same place at the Poles once for each sweep) but you can be sure you'll not leave anything out of your squares.


Loss of atmosphere on Mars If the atmosphere on Mars was once much thicker, how was it likely lost?  Was it due to interaction with the solar wind, the small size of the planet, both, or something else, and approximately how long did it take to reach its current thickness?

The loss of the Martian atmosphere can be mostly attributed to its mass. The reason why Earth still has an atmosphere made of lighter elements is because with larger mass comes larger escape velocity, which is the speed at which an atom's kinetic energy overcomes the gravitational potential energy of its planet.

The distribution of speeds of most gasses can be described by the Maxwell–Boltzmann distribution.

This curve represents the probability of finding a particle with a given speed at a given temperature. Holding temperature constant, the above chart illustrates that lighter molecules will have higher probabilities of being found with higher speeds.
$$P(\epsilon) = \frac{2}{\sqrt{\pi}k_{B}T} \left( \frac{\epsilon}{k_{B}T}  \right)^{1/2} \exp{-\left(\frac{\epsilon}{k_{B}T}\right)} $$
where $\epsilon$ is the energy of the particle, $\frac{p^{2}}{2m}$ in the absence of a gravitational field (see this lecture for information regarding how to include the gravitational term).

Integrated over time, the lighter gasses tend to exceed the escape velocity more often than their heavier counterparts. This is why the larger planets like Jupiter and Saturn still have atmospheres dominated by hydrogen and helium.


Light pollution and apparent magnitude of objects in solar system Dust along the ecliptic scatters light, causing “light pollution” throughout the orbital plane of the solar system. If an object like, say, the sun's corona, had its brightness measured from above the ecliptic, would the measurement reflect that?

The answer is yes: dust is highly constrained to the ecliptic plane and so viewing from above the plane would result in less light that needs to be subtracted from observations of the solar corona.  The zodiacal light is quite strong in the direction of the sun because the dust strongly scatters in the forward direction. Near the Sun, its surface brightness is somewhere around 21 - 23 mag/arcsec$^2$ from our viewpoint in the ecliptic, and this would be an additive contamination to any measurement of the Sun's corona.  From the ecliptic pole, this light foreground would be much less.  Of course, a careful observer would know to also measure the zodiacal light contribution and to subtract it off, so both observers would get the same brightness for the corona roughly.  The observer at the pole would have lower noise for a comparable instrument and observing time.


Could we detect an earth-like planet around another star? Could we detect an earth-like planet (e.g. similar distance from star, and similar size) with current planet finding techniques?  If an earth-like planet were orbiting our nearest star, excluding the sun :-), would we have detected it by now?

It seems current techniques tend to find large planets orbiting close to their star, or for direct observation, find large planets further away from their star.  Likewise, we have found many very large planets very close to their star (e.g. hot Jupiters), and some rocky planets somewhat larger than earth close to stars, but not very many (if any) that are very similar to earth (e.g. distance from star and size).  It seems the current techniques available do not readily detect earth-like planets though, so it may be that there really are many earth-like planets, but we haven't seen them yet.  Here is an article on current planet detecting techniques.

http://en.wikipedia.org/wiki/Methods_of_detecting_exoplanets

The short answer is yes, the Kepler mission has used transits to detect nearly "Earthlike" planets, one of the classic examples is Kepler-22b is only about 2x the radius of Earth and has a year that is 290 (Earth) days. Kepler has detected many planets smaller than Earth and closer to the star. With more observing time Kepler could discover similarly sized or smaller planets at comparable orbits, so technologically transits can already detect  "Earthlike" planets. Practically, with Kepler's lost reaction wheel this has been difficult, but more Earth sized planet detections are likely from the K2 Kepler reboot.

Microlensing is more sensitive to planets further from stars, a next-generation space telescope like the WFIRST space telescope, which might launch in the 2020s, expects to be able to detect planets smaller than Earth at slightly larger distances from stars.

Taking a photograph or spectra, or directly imaging, a small planet far from its parent star is much more difficult, light from the Sun is around 10 billion times brighter than the sunlight the Earth reflects (a $10^{10}$ contrast), the sensitivity of current ground based instruments (like GPI) and the Hubble space telescope are closer to $10^7$.  WFIRST project will be be able to detect giant planets and may be able to detect Earthlike planets but the technology is not quite there yet.

Because of instrument stability and stellar variability, radial velocity instruments are not yet sensitive enough to pick out the incredibly small wobbles caused by a small planet in an Earthlike orbit. Future instruments, like the 100 Earths Project may get to this sensitivity in the moderately near future.


How can only 1 magnetic pole of the Sun change at a time? I read some news stories about the current Solar maximum and the flipping of the magnetic poles. They say that one of the magnetic poles has switched polarity and the other will switch in about a month. I don't really understand magnetism, but I had thought you had to have 2 opposite polarity magnetic poles. How can the Sun's magnetic poles not switch in sync with each other?

The Sun is not a point object, in fact it occupies a huge swath of space. Nor is it solid, rather the Sun is mostly fluid. Therefore, there is not one single "Sun's magnetic field" but rather the aggregate of hundreds, thousands, or perhaps trillions of magnetic fields, each of which changes in complex ways and are affected by (and affect themselves) nearby fields. Thus, the aggregate field does not always resemble the simplification to one single large "Sun's magnetic field".

Note that the page the OP links to has a nice diagram which explains the "two south poles" phenomenon. During the process of the Sun's poles swap, for a short time the magnetic positive 'pole' shifts to the Sun's equator and both geographic poles have magnetic negative poles. Then the once-magnetically-negative geographic pole becomes positive and the equator no longer acts as a magnetic pole. I've never seen this mentioned anywhere else, and I'm sure that it is an oversimplification, but it does explain the "two south poles" statement.


What makes a flying object keep moving with the Earth's rotation? When a bird is flying or when a copter is hovering or a plane is flying what force is making them rotate with the earth beneath them?

They aren't.  To first order, airborne objects are moving with the Earth because they were moving when on the ground.  However, what they retain is their angular speed, and since the circumference increases with altitude, an object that doesn't accelerate appropriately will appear from the ground to fall behind to the West (the Earth is rotating to the East).  This is a well-known phenomenon whose magnitude is latitude-dependent, falling to zero at the poles.


A clock travelling "faster than the speed of light" When a clock travels close to the speed of light as observed by us, it should tick slower.

Does this mean that, when using this clock to measure a second - it could travel much more than 300 000 km during a second? In fact, this clock could travel infinitely fast when viewed only from its own perspective?

Does this imply that if I traveled on a space ship, at the speed of 99 % of light, I would age 4 years, but could reach a destination perhaps 400 light years away?

My friends and family on earth would have perished, but the people I saw on the destination planet as I started my journey will have aged only 4 years as well.

If the last assumption is correct; it means that we could watch a launching space ship at alpha centauri, it could arrive here in just 3-4 days - appearing to travel much faster than the speed of light, without breaking the laws of relativity. We would just say that it was relativistic blue shift that made it seem to travel faster than the speed of light.

Let's restrict to special relativity, meaning two inertial frames moving in a Minkowski spacetime.

A clock in the first inertial frame ticks slower, when seen from the second. A clock in the second inertial frames ticks slower, when seen from the first.

Now assume, that you are fixed to one of the two inertial frames. Usually you measure velocities within your own inertial frame, meaning measuring distance and time in your inertial frame to calculate the velocity of the other frame. In this case you'll get a velocity below the speed of light.

If you measure the distance in your inertial frame, and divide this distance by the time you observe on the clock of the moving inertial frame, you'll get higher velocities, which may exceed the speed of light. But this result has only the dimension of a velocity; it is not a velocity with respect of any of the two inertial frames.

The moving observer would observe a slow-down of external events, not a speed-up.
The observer would observe a speed-up only in the sense of a relativistic Doppler blue-shift when moving towards the observed object.

The people on the destination planet age the same as on the Earth, provided they don't move relative to the Earth. It's just you, since you travel that fast, the distance gets shorter, and therefore the time for the travel also gets shorter. By the Doppler shift and space contraction you'll see them age faster.

From Alpha Centauri, it's again the space contraction for the traveller, making the distance shorter. Again we have a combined space contraction/Doppler effect looking the people on Earth aging rapidly for about 4.5 years.

For people on Earth observing the travel takes 4.5 years. For the traveller it may take just a few days. With an acceleration of just 1g you could traverse the Milky Way in about 20 years, seen from the spaceship. The same travel seen from Earth would take about 100,000 years.

More on the twin paradox on Wikipedia.


Why does the focal length of a telescope have an effect on the magnification? I previously asked that

  Is the angular resolution of a telescope irrespective of used eye-piece?

I learned that the resolution is fixed when the light enters the telescope, and that the eye-piece is used just for magnifying the obtained image at the focal plane.

Now, the magnification of a telescope is calculated as

$$
\text{magnification} = \frac{\text{focal length of telescope}}{\text{focal length of eye-piece}} \qquad (1)
$$

For example, if the focal length of my telescope is $900$ mm and I am using a $25$ mm eye-piece, the magnification I get is

$$
\text{magnification} = \frac{900\ \text{mm}}{25\ \text{mm}} = 36.
$$

However, where does the formula $(1)$ come from? Why is the magnification affected by the focal length of the telescope and eye piece?

(I am using a Newtonian telescope if it matters.)

The objective lens of a telescope forms an real image of the night sky, the size of that image is in proportion to the focal length of the objective lens. The reason for this is simple geometry: If two stars are 1 arcminute apart, and the lens is forming an image of them, then the further the image is from the lens, the further apart the images of the two stars will be.

The eyepiece then magnifies the image. When using a lens to magnify you put the object at (or near) the focal distance from the lens. The shorter the focal length of the eyepiece, the closer you can get to the object and so the larger it appears. All this is a round about way of saying the the magnification of a lens is inversely proportional to the focal length.

In the case of a telescope, the "object" being magnified is the image formed by the objective lens. Putting this together: the apparent size of the image is directly proportional to the focal length of the objective lens, and inversely proportional to the focal length of the eyepiece, and the constant of proportionality is 1 (just imagine that both the objective and eyepiece have the same focal length). So
$$\mathrm{Mag} = \frac{L_o}{L_e}.$$

This might be a useful point to mention that magnification is usually the least useful function of an astronomical telescope. Their main purpose is to gather light.


Am I seeing the constellation Orion? I am on the west coast of the U.S. and for several weeks I have seen two large, twinkling objects in the eastern sky at night with the naked eye. In between them there are 3 smaller ones that are stacked vertically and very close to one another. All 5 rise in the sky as a group as the evening progresses. Could this be Orion? Thank you.

Using the calsky.org website, this is a chart showing a view of the eastern night sky from a point near San Francisco, CA at 7pm this evening.  (The green tick mark on the horizon is due east).

Orion is located between the moon and the horizon, and will be rising quickly.


What is the physical, geometric shape of the universe? I'm not asking about theoretical ball, vs saddle, vs flat surface which is just a metaphor with 2D space.

It's hard to say as we see very little of it, and we see them in the past because light travels for so long. But what we do know is that it is inflating (not exploding as one might thing from the "big bang" naming).

How would the universe look like, if we were to freeze it in a moment, is it likely to be a ball, or rugby ball, a cone or a sort of an irregular shape?

Is it filled throughout with galaxies, dust, black-holes, or does it live on the edges of its 3D shape, and the central part is "empty"?

Does it have a giant black hole or a star in the middle around which everything revolves?

Ok, maybe you have some misconceptions.

The Universe has no center at all. It looks the same from avaery point, to wherever you look to. It is approximatedly as follows:

This image represents a very large scale box on our universe on current time (not on perceived time, based on received light). Of course, it is just a computer simulation. Each dot represents a cluster of galaxies.

So you need to imagine an infinite tridimensional space filled with filament-like structures like these. And infinite means it has no bounds, so it has no "external" shape. No ball, rugby-ball nor cone there. Also not irregular external shape, just infinite. Any of these shapes have a 2D boundary on a 3D space, but the universe has no boundary.


What does the opacity of a molecular transition mean? In particular, what does it mean when a line (e.g, emission line in stellar wind) is optically thick or thin. I know what an optically think/thick medium is, but how does this concept compare to optically thick/thin line.

I know what an optically think/thick medium is...

Okay so this isn't much more complicated. A medium or material can be optically dense or opaque at one wavelength, but fairly transparent at a different wavelength.

If you look at the dark plastic window on a remote control for a TV or other appliance, you can't see through it. It's optically dense at visible wavelengths, but it's mostly transparent between say 850 and 1050 nm wavelengths, so that the infrared LED light can still shine through it and communicate with the receiver.

So at each wavelength the medium can have a different opacity. For this window, it's roughly a step function in the visible and near-IR, high opacity in the visible and low opacity in the IR.

You can extend the same idea to one spectral line or many of them. Perhaps a medium can absorb at or near the center of the spectral line where there is a strong transition, but more weakly as you move away from the center wavelength, and then perhaps almost not at all farther away.

I the video testing Hoya UV & IR Cut Filter at first you can see the purple spot; that's actually IR light that can still be seen by the video camera. The dark plastic has high opacity in the visible and low in the IR. Then they introduce an IR filter that is the opposite, high opacity in the IR and low in the visible (you can see right through it) but it still blocks the IR.

See also the answers to What is the purpose of the IR-diode Cover on Remote Controls?


Powerful-enough telescopes to observe the Lunar Roving Vehicle from Earth? Apparently, not even the Hubble telescope (which is actually in orbit, not even on Earth) is powerful enough to observe the LVR... will we ever have the technology to observe the LVR from Earth?
(it would be my dream to be able to observe the Apollo relics with a "retail" telescope. I am no expert in optics, so have no idea if for example "computer-enhanced" zooming could work, instead of a giant mirror?)

I don't think so. In order to resolve an object one metre across (as a single pixel) at a distance of 384,400km (the diameter of the Moon's orbit), you would need a telescope mirror about 200 metres in diameter, and to achieve a resolution of 10cm (which you would probably need to form a recognisable image), you would need a telescope mirror with a 2km diameter. Currently the largest Earth based telescope has a 30-metre mirror.


Why is JD 2451545.0 = January 1, 2000, Noon instead of JD 2451558.0? I know that JD0.0 is Jan. 1st 4713 BC at Noon, a year in the Julian calendar is 365.25 days and that the number after 'JD' is the number of days so...

I want to calculate the number of days between Jan1.2000AD and Jan1.4713BC and I do

$$ \text{# of years} = 4712+2000=6712 \text{ years (skipping the year 0)} $$
$$ \text{# of days} = 6712 \cdot 365.25 =2451558 $$

So I think it should be JD2451558. An extra 13 days. What's going on here? I thought maybe the length of a year in the Julian calendar is $2451545/6712=365.2480632\text{days}$ but I can't find anything to support that.

I imagine that people would have questioned whoever proposed J2000 as JD2451545.0 and got a satisfying answer but I can't find one! Help

This Wikipedia article states (correctly) that

  "The Julian Day Number (JDN) is the integer assigned to a whole solar
  day in the Julian day count starting from noon Greenwich Mean Time,
  with Julian day number 0 assigned to the day starting at noon on
  January 1, 4713 BC, proleptic Julian calendar (November 24, 4714 BC,
  in the proleptic Gregorian calendar)"

The Julian year consists of 365.25 days while a Gregorian year consists of 365.2425 days.

Since the introduction of the Gregorian calendar, the difference between Gregorian and Julian calendar dates has increased by three days every four centuries (all date ranges are inclusive).  In addition, 10 days were added to the Julian calendar to obtain the Gregorian calendar at the time of the calendar reform. The current difference between the Gregorian and Julian calendar dates is 13 days: 14 September 2016 Gregorian date is 1 September 2016 (Julian date).

This should explain the extra 13 days in your calculations.


How does interstellar matter density vary? Answer to my question partially answers this one, about density of intergalactic matter and matter within galaxy:

  But it is mostly a hot, ionized void. How void? The density of the intergalactic medium is about 1 to 100 particles per cubic meter (you can compare it to the mean galactic density, of about a million particles per cubic meter, or that of Earth's atmosphere, of about 10^26 particles cubic meter). How hot? It can go from 10^5 to 10^7 K.

If we skip the most dense concentrations of matter (stars, planets, generally everything solid, liquid or plasma, and border conditions like their atmosphere) how dense interstellar matter can we find? What is the concentration of matter in densest nebulae that still don't collapse into bodies like planets or stars?

And conversely, how empty does the space get at its emptiest? I could imagine only very few particles in last their short moments of travel into the center exist under a black hole's event horizon, but other than that, how empty a space can be found in the universe, and where?

The interstellar medium is a multiphase medium, and you can find (some references in this lecture and in this thesis manuscript (this one is in French, but numbers are international)):

the hot ionized medium (HIM) with density as low as 10^-3 cc (particles per cubic centimeter);
the warm ionized medium (WIM), with density of the order of 0.03 cc;
the warm neutral medium (WNM), with density of the order of 0.25 cc;
the HII regions, with densities ranging from 1 to 10^4 cc;
the cold neutral medium (CNM), with density of the order of 25 cc;
the molecular clouds, with densities above 10^3 cc, up to 10^6 cc roughly.

These different phases are due to the interplay of cooling and heating processes of its own components (Wolfire et al. 1995). The lowest density regions are hot and associated to supernova expansion bubbles. The HII regions (HII for "ionized hydrodgen") are associated with O stars (massive stars that can efficiently ionized their environment).

The highest density you can get with no evidence of forming stars are of the order of 10^4 cc (see for example this talk by Ward-Thompson on starless cores). Dense, molecular clouds that are about to collapse and form a star have a density of the order of 10^6-10^7 cc.


How is the universe bordered? This answer says that some models describes the universe as finite. How do those models describe the universe's border?

Does the border (theoretically) exist?
Is it a solid border? Do they predict what happens if matter crosses this border (maybe it simply can't)?

Your intuition is correct.  It does not really make any sense for the universe to have a border.  Would it be an unbreakable wall? What would be on the other side?  Models with a finite universe would not have a border.  The universe would be curved into a closed loop like the surface of a sphere.  In other words, if you go far enough in any one direction you come back to the starting point.


Is the white hole a myth or not? I've been doing some research on white holes, and there has not been enough information to provide the answer. Some websites are saying that they might exist, but it is not yet proven. So, is it true that they exist or not.

Well, currently, we don't have a shred of evidence that they exist.

White holes are hypothetical objects predicted by certain solutions to Einstein's theory of general relativity. To be specific, they  come from the maximally extended Schwarzschild solution. We can deal with them a little mathematically, but nobody will really believe in objects like these unless we have experimental evidence for them.

Do we? I quote NASA here

  White holes are VERY hypothetical.

and here

  There is no observational evidence for white holes.

This supports this position:

  Unlike black holes, there is no real astronomical evidence to suggest that white holes actually exist.

Also, the idea is put nicely here:

  A white hole is only a concept for higher levels of thinking. No one has every observed one and no one probably ever will.

Why? This gives an interesting answer:

  Once even the tiniest speck of dust enters the part of space-time which includes the black hole, the part which includes the white hole disappears. The universe has been around for a long time and so even if it did start with white holes, they would have all disappeared by now. (Emphasis mine)

Essentially, white holes are unstable. Even if a particle did try to enter one from a black hole via an Einstein-Rosen bridge, the bridge would collapse before the particle could cross, and the particle would hit the singularity.


Did I see the dragon capsule on Friday night? I live just south of Sydney Australia, and I went outside to watch the ISS pass overhead on Friday night. The ISS was following a smaller object easily visible to the naked eye.

Sydney time 20.43 or UTC 9.43. (Friday March 8) The smaller object was ahead of the ISS by maybe 25 degrees. Appeared to be moving at exactly the same speed and trajectory.

Was this the the SpaceX Crew Dragon capsule? Sorry if this is the wrong place to ask.

I have certainly watched the ISS and an undocked Shuttle Orbiter chase each other across the sky. Since they appeared to be of equal brightness, I didn't know which was which. Their angular separation from my POV was approximately 20°, so their actual separation would have been about 150km. It was one of the last Shuttle flights, so I was very lucky to see it.

It is possible that you saw the Dragon capsule, and it would certainly have appeared to be be fainter than the ISS. The easiest way to establish the fact is by using a program like Stellarium. This will show you if the ISS was visible at your location, given the date and time of day.


The size of the Earth Has the earth always been this size apart from a few occasional additions from time to time.

Meteorites hitting the planet now don’t become part of it and the moon when it collided just spun of into its orbit.

So how, as documentaries show, do chunks of rock stick together to form planets etc?

I get dust particles coming together to make suns eventually but in a vacuum. I don’t get how stuff with a large mass sticks together. When stuff with mass collides today it doesn’t stick together: Kuiper belt, rings of Saturn etc.

The early sun was surrounded by a disc of dust and gas.  Such a disc is not stable, it will develop parts that are are more dense and parts that are less dense.  The combination of friction (between bodies of gas and dust) and gravity (pulling together) will tend to make the denser lumps grow denser, eventually coalessing into a ball of metal, rock and gas. This "proto-Earth" was comparable in size to the Earth now.

In the process, most of the rest of the dust and gas either fell onto the ball, or was pushed out of the way. It's not true that meteorites don't become part of the planet. Meteorites do add to the mass of the planet, but this is a slow process.

After this the early Earth collided with another planet. The other planet was destroyed and the remanents formed the Earth and Moon.  By now the Earth was roughly the same size as it is now.

There was then a period when the meteorite rate increased greatly. This added a little to the mass of the Earth.  The meteorites often were at least partly icy, and this water formed the seas.  The rocky an metallic parts of these meteorites were incorporated into the crust of the Earth.

By the time the "Late Heavy Bombardment" finished 3.8 billion years ago, the Earth was almost exactly the same size as now.  There has been some gain from later meteorite impacts, and some loss due from the top of the atmosphere, but these are minor.


Dark Matter Particle Candidates Dark matter appears to dominate the matter component of the universe as compared to luminous, or baryonic, matter. Though it does not interact electromagnetically (it doesn't absorb, scatter, or emit photons), there is an ever-increasing mountain of evidence for its existence through it's gravitational interactions with stars, galaxies, and clusters, as well as it's influence on objects behind it through what's known as gravitational lensing.

My question is, what are the most promising particle candidates of dark matter, and which experiments currently exist (or will exist) to attempt to answer this question?

When you say "particle" candidates, I assume you're excluding MACHOs and RAMBOs. MACHOs are "dark" objects at the stellar scale like black holes, neutron stars, brown dwarfs, etc. RAMBOs are clusters of similar dark objects. MACHOs and RAMBOs are made of primarily baryonic matter (everyday stuff like protons and neutrons — electrons are not baryons but they can be a part of this too).

This leaves us with the WIMPs and possible other particle candidates. I don't know the details of any experiments looking for these offhand (I'll see if I can find it out and update the answer in a few days), but the major particle candidates are:

Neutrinos: It's highly unlikely that these make up any significant chunk of the DM in our universe, but they deserve a mention.
Sterile neutrinos (I'm using the term in the specific sense here, not the general sense): This type of neutrino is right handed and only interacts gravitationally (note that there can be right handed active neutrinos, too). There is some preliminary evidence suggesting that these might exist.
Axions: These are hypothetical, but if they exist (and make up the majority of DM), we kill two birds with one stone by both explaining DM and resolving the lack f symmetry breaking in the strong interaction.
Gravitinos and the Lightest Supersymmetric Partner: Both of these require supersymmetry to hold, but they are major DM candidates.


Characteristics of the first planets in the Universe? What would have the very first planets looked like, based on their most likely chemical compositions?

For example:

Were they mostly grey gas giants with atmospheres of hydrogen and helium, coalescing around the first Population II stars?
Were they blue gas giants, enriched with water vapour?
Were they brownish-red and grey, like our own Jupiter and Saturn?
Were they a heterogeneous mix of all sizes, colours, and chemical compositions that included rocky and gas giants?

What papers support research into this area?

https://iopscience.iop.org/article/10.1088/0004-637X/751/2/81/meta
https://www.mdpi.com/2076-3263/9/3/105
https://arxiv.org/pdf/1805.00023.pdf

Note: By first planets I mean the first planets that most likely existed in the Universe. In The habitable epoch of the early Universe, for example, Loeb suggests that a mere 13.5 ± 3.5 Myr after the Hot Big Bang could have witnessed the first viable supernova followed by planet formation. Another paper (can't find it) suggested that gas giants could have formed from the primordial hydrogen and helium soup before reionization, without needing any stars.

Related:

http://www.sci-news.com/astronomy/science-helium-rich-gas-giants-02905.html
https://en.wikipedia.org/wiki/Sudarsky%27s_gas_giant_classification
https://www.nasa.gov/centers/goddard/news/topstory/2003/0709hstssu.html
https://www.space.com/17441-universe-heavy-metals-planet-formation.html
https://earthsky.org/space/how-and-when-did-the-first-planets-form-in-our-universe
https://medium.com/starts-with-a-bang/ask-ethan-how-fast-could-life-have-arisen-in-the-universe-863bd17b92ca
https://www.nationalgeographic.com/news/2012/3/120326-oldest-planets-found-jupiter-big-bang-space-science/

According to Mashian & Loeb (2016) (also in ArXiv), one possibility is that planet formation may have occurred around carbon-enhanced metal poor (CEMP) stars in the early universe. The paper focuses on CEMP-no stars as the most metal-poor CEMP stars mostly fall into this category.

CEMP-no stars are thought to form from material polluted by supernova ejecta from Population III stars. The main grain species that forms in these ejecta is amorphous carbon. From section 2 of the paper:

  Tailoring the
  SN explosion models to reproduce the observed elemental
  abundances of CEMP-no stars, they find that: (i) for all the
  progenitor models investigated, amorphous carbon (AC) is
  the only grain species that forms in significant amounts;
  this is a consequence of extensive fallback, which results in
  a distinct, carbon-dominated ejecta composition with negligible amounts of other metals, such as Mg, Si, and Al, that
  can enable the condensation of alternative grain types; (ii)
  the mass of carbon locked into AC grains increases when
  the ejecta composition is characterized by an initial mass
  of C greater than the O mass; this is particularly true in
  zero metallicity supernova progenitors, which undergo less
  mixing than their solar metallicity counterparts (Joggerst,
  Woosley, & Heger 2009)

In this case, the earliest planets may have been carbon planets. Carbon planets with bare graphite/amorphous carbon surfaces (the carbon planet equivalent of the Moon or Mercury) would be very dark. Atmospheres would modify this appearance. As the paper notes, hot carbon planet atmospheres are expected to have spectra dominated by carbon monoxide. Cooler carbon planets would likely support various kinds of hydrocarbon chemistry that would modify their appearance, possibly producing reddish tholins as a result of photochemistry depending on what else is around.

Mashian & Loeb predict that the maximum distance at which planets form would depend on the carbon abundance:

  We find that carbon planets may be orbiting iron-deficient stars with carbon abundances [C/H] ∼ -0.6, such as HE 2356-0410, as far out
  as ∼20 AU from their host star in the case where fcond = 1.
  Planets forming around stars with less carbon enhancement,
  i.e. HE 0107-5240 with [C/H] ∼ -1.6, are expected to have
  more compact orbits, with semi-major axes r < 2 AU. If the
  carbon condensation efficiency is only 10%, the expected orbits grow even more compact, with maximum semi-major
  axes of ∼5 and 0.5 AU, respectively.

So far this is just hypothetical: we don't know of any planets around CEMP stars, nor of any carbon planets but this may suggest that it is not completely pointless to search for them.


Unidentified color changing object in sky Yesterday, I was just star gazing and I noticed some star-like object moving with a high speed somewhere near the Orion's Belt and exhibiting all the seven colors i.e VIBGYOR and after a few moments, it disappeared. Any plausible explanations for this observation?

Probably a satellite flare

It sounds like a satellite flare.

Coming, showing up clearly for a few moments, moving, then going away, are characteristic of a satellite flare, like Iridium flashes.

Check Heavens Above or any other similar site to see if any predicted flares match with your location and time.

The changing colours can be explained with "Twinkling".


How does time work beyond the cosmic event horizon? The faster you move through space, the slower you move through time - the calculations show that as you approach lightspeed, time comes to a standstill.

How does time work beyond the cosmic event horizon? Space is stretched in every direction, meaning the further away something is, the faster it moves away from us.

But what about when that speed is more than $3 \cdot 10^8 \frac{m}{s}$ due to the distance away from us it is? What happens then?

First, let's clear up a few misconceptions:

The Hubble sphere

The speed of light as an upper limit is valid in special relativity (SR). In general relativity (GR), which must be used to describe the expansion of the Universe, although locally (i.e. where SR is a good approximation) you cannot exceed the speed of light, there is no limit to the relative velocity of two objects.

By Hubble's law ($v=H_0 d$), all galaxies that are more distant from us than the "edge" of the so-called Hubble sphere $d_\mathrm{H} = c/H_0 \simeq 14.4\,\mathrm{Glyr}$ (billion light-years) recede from us faster than $c$. This is no hindrance for us to observe them; we regularly observe galaxies much farther away (see e.g. this and this answer).

The event horizon

The cosmic event horizon (EH) is the largest distance a galaxy can be at, and still emit a light signal that may reach us in a finite time. It is currently roughly 16 Glyr away, i.e. a bit farther than the Hubble horizon. We also regularly observe galaxies farther away than the EH, but we see them in the past (as we see everything else), and the light they emit today will never reach us (not because space expands, but because the expansion rate accelerates).

The particle horizon

The particle horizon (PH) is the largest distance from which light has had the time to reach us since the Big Bang. Due to the expansion of the Universe, this is currently 46.3 Glyr away. Galaxies at this distance recede from us at 3.2 times the speed of light. We don't see any galaxies here because we look back in time to just after the Big Bang where they hadn't yet formed (and because the Universe at that time was foggy), but the most distant galaxy yet observed, GN-z11, is 32.2 Glyr away and recedes at $2.2c$.

Time dilation

However, you're right that the galaxies' recession from us makes their time go slower, as seen from us. Not by the SR factor $(1 - v^2/c^2)^{-1/2}$, but by the GR factor $(1+z)$, where $z$ is their observed redshift.

Galaxies receding from us at $v=c$ have a redshift of $z\simeq1.5$, so when we observe physical processes in such a galaxy — e.g. the decline time for the brightness of a supernova — it's slower by a factor $\simeq2.5$. GN-z11 is at redshift $z=11$, so things there happen slower by a factor 12.

The farther you look, the slower things happen, and if we could look arbitrarily far, you could in principle see the event right after the Big Bang in slow motion. You cannot see "zero o'clock", which would have an infinite redshift. And you can never, not even in theory, see anything that's farther away than it's light has had the time to travel, so not even in theory does it make sense to try to assign a time dilation factor to these events.

As far as we know, there are galaxies outside the PH. As time goes, the distance to the PH increases, both because space expands, and because light fro more distant region will reach us. But as soon as that light is able to reach us, the object that emitted it is inside the PH, and thus has a finite time dilation factor.

But remember that the time dilation is only from our point of view. If you could magically teleport yourself there right now, time would pass normally for you there. So nothing particular happens there.


What is wrong with this measurement of the synodic period of Mercury? I'm measuring the synodic period of Mercury using Stellarium. When measuring the synodic period one needs to choose a reference point to start the measurement, and I choose the point when Mercury is on top of the Sun. This is the start date:

And this is the end date:

Inputting these two dates into a calculator gives a synodic period of 104 days. However, as one can see from Stellarium's data, Mercury's synodic period is actually 115.88 days. This measurement is off by more than 10 days.

Why is there such a large difference? It's not large in the absolute sense, but it seems well above the error margins. For example if instead I shift the start date to 9 April 2021, Mercury is discernibly not between us and the Sun.

Mercury's orbit is highly eccentric: 0.21 according to Wikipedia. Therefore, the actual time between repeating occurrences will vary depending on the year. If you were to perform your calculations for many periods, the average should approach the value given by Stellarium.

The theoretical synodic period, using the sidereal period of Earth and Mercury, is 115.9 days.


How significant is a planet's density to the formation of life? I'm doing some writing set on a fictional earthlike planet that for reasons needs to be substantially larger than Earth but can't have substantially higher gravity. I've been able to compromise with a planet that's 6x the circumference, having a surface gravity of 1.5G, yielding a density of about 4700kg/m^3.

This is quite a bit (~15%) less than Earth and Venus, but (20%) more than Mars. It would proportionally have less heavy metal, meaning less iron for a protective magnetic field. I would like this planet to have some basis in reality - could such a planet be earth-like, and is there a definitive range of planet densities where life as we know it could exist?

You could probably get away with replacing molten iron, 7.87 g/cc, in the core with aluminum, 2.70 g/cc, and still generate a substantial magnetic field: Molten Metal Magnet

  It’s easy to create a magnetic field by using a battery to force an electric current through a loop of wire. But Earth’s core, a rotating mix of iron and nickel with internal flows driven by the passage of heat, has no battery and no wires. Instead, it creates magnetism by means of self-sustaining feedback. Liquid metal moving through a magnetic field generates a current, similar to that induced in the moving coil of an electric generator. That current in turn generates the magnetic field. This “self-generation” mechanism can dramatically amplify the small, random fields that always exist in magnetic materials.

Sodium was used in the linked article, however any molten metal or conductor, should do, and sodium, at 0.97 g/cc is just not dense enough to sink to a planet's core.
You'd have to do some hand waving over "an iron poor/aluminum rich Bok globule", but I think you could get your planet, with a sufficient magnetic field to fend off solar winds.


Formation of "hot Jupiters" before star ignition I'm watching a BBC special on the formation of solar systems, and they're covering the topic of "hot Jupiters" that exist near their host stars. The prevailing theory, according to this 2015 special, is that the planets form farther out, then migrate in. This addresses the issue of their gas being driven off by the host star.

Isn't also possible that these planets formed shortly before the ignition of the host star? The mass that will form the star is already mostly collected, and could hold bodies in orbit, but the heat has yet to begin, and wouldn't drive off gas.

Hot Jupiters can and do form before the hot star "ignites", as you put it.

The conventional core-accretion theory, by which gas-giant planets form, requires them to accrete from a disk of gas and dust for about 10 million years. Coincidentally (or perhaps not a coincidence?), the empirically determined lifetimes of circumstellar gas and dust disks around young stars is also about 10 million years.

Meanwhile, the pre-main-sequence lifetimes of newly-born stars (which I define here as the length of time it takes to contract from a large ball of gas down to a much more compact configuration, with a central temperature high enough to initiate hydrogen fusion), range from about 10 million years for stars a little more massive than the Sun to 100 million years or more for stars that are less massive.

I think, though, that the information that you are missing is that there really is no period of time where the "heat has yet to begin" whilst the protostar is surrounded by a gas/dust disk. A protostar is actually more luminous when it is younger and becomes fainter as it contracts towards its main sequence configuration. The energy that powers this luminosity is derived from gravitational potential energy. The inner portions of the gas/dust disk are always too hot to form hot Jupiters in situ and so all current models focus on how hot Jupiters can be "moved" from where they were formed to where they are found now.


How to plot curvilinear coordinates in the frame of a Cartesian telescope field of view I am wondering what projection is used to produce plots like this:

(Image taken from here, author: NASA/Ames/JPL-Caltech, Image credit: Software Bisque)

More generally, I am trying to figure out the telescope pointings (in az/el or an earth-centered-earth-fixed coordinate system preferably, RA/dec if the former are not possible or easy) for a tiled array of square images positioned next to each other, similar to how the Kepler CCDs are laid out.  I'm thinking that if I could know what projection is used to produce that plot, then I could step in x/y to wherever I want the images to be and then de-project to convert to RA/dec.

I have tried a gnomonic projection but I found that, with increasing distance from the survey center-point, more and more overlap occur between adjacent frames.  Things looked really good close to the center-point though.  I realized that a gnomonic projection isn't actually providing what I want, since I can move across the projection area a large amount (infinity in the limiting case) and still end up with az/el pointings that are on the horizon, despite moving over an infinite number of frames.

Most CCD or photographic images are equivalent to a gnomonic projection.
I suppose this is true of the Kepler CCDs; while the Schmidt camera focus surface is curved, the CCDs have individual field flatteners.
If the projection is not exactly gnomonic, the difference is not very great in a small field.

As you found, a single gnomonic projection doesn't work so well for areas much wider than 90°.
TheSky user guide
says it uses orthographic projection for fields smaller than 60°.
The slightly curved RA lines and slightly oblique intersections near the corners of the chart in question are consistent with that.
Other projection choices are also valid, e.g. stereographic projection is common in maps showing a hemisphere at once.

In any case you'd need to transform the individual image tiles to the same projection as the composite overview.
This will change the tiles' shapes or sizes, but the relation between map coordinates and sky coordinates will be consistent.


What defines the plane of an accretion disk around a black hole? The question Do we know at which angle the Event Horizon Telescope will look at the accretion disk of Sagittarius A*? about just how close to edge-on we would be viewing any accretion disk around the black hole in the center of our galaxy, and @Zephyr's excellent answer have got be thinking about the geometry and angular momentum.

While the angular momentum vector of a black hole will likely reflect that of all the matter that fell into it , would the plane of any accretion disk around a black hole reflect the black hole's current equatorial plane, or a plane defined by rotation of whatever material is currently falling in to it?

Another way to formulate this question might be to propose an incredibly unlikely Gedankenexperiment. If I inserted a black hole into the center of a rotating disk of material so that the black hole's axis was in the plane of the disk of material rather than parallel to its rotational axis, would a well-defined accretion disk form all all, and if so, what would determine the orientation of that plane?

Let me give an answer based on my intuitive understanding of the thing...

 Initial conditions

The initial accretion disk movement is defined by the objects falling into it, the black hole's mass and its spin of course.

If you take the situation where a single star falls on a quiet black-hole, the accretion will appear along the orbit of the infalling object.

Credit: NASA, S. Gezari (The Johns Hopkins University, Baltimore, Md.), A. Rest (Space Telescope Science Institute, Baltimore, Md.), and R. Chornock (Harvard-Smithsonian Center for Astrophysics, Cambridge, Ma.) and live on youtube...

 Equilibrium in an active galactic nucleus

Now on the long run, if additional matter is constantly coming in, at first thought I would think

the accretion disk momentum will be exchanged with the one of the black hole through the Penrose process,
Infalling matter momentum will reflect the overall host galaxy momentum - especially if the galaxy is a spiral


Why doesn't the Sun explode? I know stars explode because of the fuel causing a fusion compound pushing it apart and the fuel runs out and it 'bounces' for lack of a better term.

Given the fact that more massive stars are supposed to explode more quickly: why doesn't the Sun explode considering its massive size?

There are two things to discuss here: (a) why the Sun does not explode; and (b) why the Sun will not explode.

(a) An explosion occurs when the timescale for the energy release by some process is much shorter than the timescale on which a system can adjust to damp the energy release process. In the present day Sun, nuclear fusion is a very slow process: on average it takes many billion years for a proton to fuse with another. This timescale is quite temperature dependent, so you might have thought the centre of the Sun might heat up quickly, leading to a runaway "explosion". However, an increase in temperature leads to an increase in pressure that would expand the Sun, reducing the core density and temperature and decreasing the rate of nuclear fusion again. The timescale for the Sun to react in this way is just millions of years, so this acts like a thermostat that keeps the reactions under control.

(b) Stars more massive than the Sun burn through their hydrogen and other heavier fuels, and end up with an inert iron core from which no further energy can be extracted. The core subsequently collapses and the gravitational potential energy released by the core collapse powers a supernova explosion. The reason that the core does not quietly collapse into a black hole (at least in the supernovae we see!) is that the core "bounces" when neutrons in the core are squeezed very close together. This provides a quantum mechanical effect called degeneracy pressure that resists a complete collapse.

Ironically, it is degeneracy pressure in the solar core that prevents our Sun reaching the supernova stage. After burning through its hydrogen and helium the core of the Sun would consist of carbon and oxygen; but it will never become hot enough to commence fusion of these elements, because electron degeneracy pressure will support it from contracting and getting any hotter. Only more massive stars will ever attain the core temperatures required to ensure progress towards a terminal iron core.


Do the orbits of satellite galaxies show the influence of dark matter? I understand how galaxy rotation curves show the influence of dark matter, or something else that would produce similar behaviour, such as Modified Newtonian Dynamics (MOND). Do the orbits of satellite galaxies show the same influence?

In MOND terminology, the internal gravity of the satellite galaxy is not relevant to this question. The question would be if the gravitational effect of the host galaxy on the satellite was Newtonian or MOND?

The short answer is "Yes, the orbits of satellite galaxies definitely show the influence of dark matter (or something like MOND, if you prefer)." For galaxies outside the Local Group, you can only measure their radial velocities from Doppler shifts (i.e., velocities along our line of sight to the galaxy), but you can still look at the distribution of these velocities for satellites around some central galaxy and work out the approximate amount of mass required for the satellites to remain bound -- and thus you need dark matter. (The first hints of dark matter, going back in the 1930s, were from a larger-scale version of this: Zwicky and others measured the radial velocities of galaxies in galaxy clusters and found they were too high to be explained by the visible mass of the galaxies in the clusters.[*])

For satellites of our own galaxy, it has become possible, in the last twenty-five years or so, to measure proper motions (motions across the sky, perpendicular to the line of sight) as well as their radial velocities and so we can measure their actual 3D space velocities. This paper by Fritz et al. (2020) uses proper motions from Gaia observations for 45 satellite galaxies to estimate the mass of the Milky Way's dark-matter halo. They find this to be about $1.5 \times 10^{12} M_{\odot}$, consistent with other measurements.[**] (Note that unlike the case for measurements of gas clouds orbiting in a disk -- the original data indicating dark matter in spiral galaxies -- you can't assume the orbits of satellites are circular, so the calculations involved are more complicated, but they still give roughly the same answers.)

[*] Later observations with X-ray satellites demonstrated that galaxy clusters contain significant amounts of normal baryonic matter in between the galaxies, in the form of very hot, dilute, X-ray-emitting gas. However, even when you add this to the stellar mass in the galaxies, you still don't have enough mass in the cluster to explain the radial velocities.

[**] This value can be compared with a baryonic mass for the Milky Way of about $5 \times 10^{10} M_{\odot}$ in the form of stars, plus maybe another $1 \times 10^{10} M_{\odot}$ in the form of gas.


Calculate planet's surface temperature by distance from star If the planet and star in question are very similar to the Earth and the Sun, how can I calculate the planet's surface temperature by knowing the distance from the star?

This question was originally asked on the Worldbuilding stack exchange, and someone suggested I ask it here.

Neat and tidy calculations for this aren't possible.

The math behind effective temperature of a planet can be found here and that's a straight forward calculation where you can input distance, solar energy, planet's albedo and get an average temperature.   Earth's effective temperature calculation comes to -21 °C, which obviously isn't accurate, but that's as far as the calculation can take you.

Feedback mechanisms:

With essentially no change to distance from the sun, and only relatively small changes in tilt and eccentricity, Earth's temperature can fluctuate from a current 15 °C global average to about 5 °C as it goes in and out of ice ages.

Feedback mechanisms like albedo (ice cover is the key driver, but also deserts, surface area of the oceans), CO2 capture, which increases with colder oceans, can amplify small changes, leading to significant changes in surface temperature.   An orbital change, which should account for maybe 1 degree C, or an increase in CO2, which, by itself, captures less than 1 degree C, through feedback mechanisms, can lead to much larger temperature changes.

More on feedback mechamisms here.

Land placement and Ocean circulation

5 or 10 million years ago, Earth's average temperature was about 18-20 °C and that change was driven, primarily, by the closing of the Isthmus of Panama which affected ocean circulation.  Just the closing of the Isthmus made ice ages possible.  That's a pretty small change in the grand scheme of things, whether your planet has a near equatorial passage where oceans can flow or whether it doesn't, but that one thing can mean more than 10 degrees.

As Antarctica drifted over the south pole (the last 30-35 million years or so) glaciers formed and the earth cooled, because as ice forms, more sunlight is reflected into space and oceans levels drop.  Oceans having the lowest albedo and ice the highest.  Also, as Antarctica driffs away from the south pole over the next 20-30 million years, Earth is expected to warm, unless another change occurs, like the continued absorption into land and the oceans of CO2.

Earth can only enter an ice age when there's land near the poles.   The glacial period of 440-460 million years ago most of the land was over the south pole.  The sun was about 4% less luminous and the earth had much more CO2 (though CO2 levels did drop during that time, they didn't drop close to where they are now) and the orbit of the earth, 440 million years ago has some unknowns, but a large land mass over the south pole is thought to have played a key role in making that glaciation possible.  The tipping point of ice formation at the poles is particularly sensitive and a significant variable.

So, even with very similar to earth like planets.  Just where the land is and how well the oceans circulate and whether the planet can form glaciers and sea ice, or not, can fluctuate perhaps 20 degrees C in global average temperature.  How much CO2, how much active volcanism, how much reflective material in the upper atmosphere - all variables.

Even with identical earths, the feedback mechanisms make it difficult to predict what Earth's temperature would be if you pushed Earth 1 million or 1/2 million miles further from the sun.  The calculations for surface temperature wouldn't follow a simple mathematical formula.   Now, you'd still get colder as you moved the planet away, but one push might give you 1-2 degree and the next push might give you 5-10 and the one after that 0-1.  There would be no way to do a clean formula.


Are micrometeorites cosmic dust? Are micrometeorites just dust grains at high relative speed, or is there some difference between them and cosmic or interstellar or solar nebula dust? Dust as in the clouds which obscure parts of the Milky Way in our night sky.

It might seem like the way to figure this out would be to test the composition of micrometeorites and see what objects this matches up with. This, however, is not a viable technique, as Genge et al. (2008) explains:

  In interplanetary space dust particles are inherently transported sunwards by P-R light drag (Dohnanyi 1967) and once captured by the Earth their deceleration occurs at high altitudes allowing fragile micrometeoroids to survive atmospheric entry (Love and Brownlee 1991). The parent bodies of MMs, therefore, are likely to differ from those of meteorites.

It is, however, possible to analyze micrometeorites and group them in various categories, and determine the parent bodies. The two main types of parent bodies are thought to be comets and asteroids. Distinguishing between the micrometeorites of these two groups, however, can be difficult. Many large micrometeorites are from main-belt asteroids (see Genge & Grady (2002)). However, Nesvorny et al. (2010) suggests that many micrometeorites (and large amounts of cosmic dust in general) are from comets.


How to calculate the thickness of the ice crust on a frozen ocean planet/moon? How do I calculate the thickness of the upper ice layer on cold ocean worlds like Europa, Enceladus, Ganymede, ...? I'm asking this for a programm I'm currently writing.

Given/Known is: mass, radius, average density of metal (assume Fe) core, rock (assume MgSiO3) layer, ice (assume H2O) layer; internal heat from tidal heating, radioactive decay and residual formation heat; external heating from solar radiation (assume equal constant heating distributed over the entire surface or ignore it if it is irrelevant/minuscule); pressure of the atmosphere above (assume pure N2 with variable pressure or no atmosphere)

Wanted is: thickness frozen (crust) and molten layer

Bonus stuff: If easy adaptation of the formula for the calculation of the rocky layers crust thickness is possible I would appreciate it. Should it be possible, an adaptation to find the rocky crust thickness and the ice crust thickness on wolds where both are needed would be helpful.

Thanks for any answer and time spent of solving this.

Based on assumptions, if you are ok with approx solution then you can solve heat transfer equation in steady state.

Heat transfer equation

Ignoring the time - long time has passed to attain equilibrium.

Heat flux assuming outward and hence internal ocean would be sum of all internal heating mechanism.

Easiest would be if somehow you can come up with surface temperature value. Then you can solve conduction equation through ice for temperature and  weight for pressure and just see at what temperature and pressure ice will go into liquid. Simply asking what is ice layer thickness to maintain a internal heat flux and given surface temperature.

You need one boundary condition. Other way would be little harder, starting at centre with outward heatflux. You need to know both phase and composition since density and cp both changes with these conditions.


Is Mars expected to go through the tail of Comet Siding? This question will likely be answered "in retrospect" in about 8 days.  So it's a great opportunity to "lay it out there" and then let reality prove you right/wrong.

On the 19th of October, Comet Siding is expected to pass Mars very closely (1/3 the distance of the moon from Earth). My assumption is that this means its nucleus.  But comet tails can be very long (100s of millions of miles).  For the sake of this questions let's keep it to "obviously observable" tail and coma.

Additionally "going through" may need a definition.  So let's say that if a line drawn from the outer (obviously observable) coma closest to Mars and the end of the (obviously observable) tail were to intersect with-in 20 miles of the surface of Mars, then we would say that Mars has "gone through the tail" of the comet.

So is Mars expected (or even more fun, do YOU expect Mars) to go through the tail of Comet Siding on the 19th of October?

If so, what will be observable from the surface of Mars? Large number of shooting stars?

NASA is preparing to hide its satellites (repositioning them to the "far side" of Mars)  from comet dust traveling at 35 miles/sec.    Is there any idea of how long the encounter will last?

Short answer: Sort of

Long answer:

One week before the encounter, Wikipedia says no. The comet's nucleus will pass by Mars at a distance somewhere of about 139,000 km from the center of Mars - way too far for any predicted collision. The main tail of the comet, too, will most likely miss Mars by about "10 Mars diameters" - roughly 64,000 km. However, small quantities of dust particles could (and most likely will) stray away from the main tail and hit Mars, Deimos and Phobos. This graphic gives a good picture of the encounter:

However, Mars may pass through the comet's coma and be hit by any dust particles there.

It will certainly be observable from the Martian surface: The comet will have an apparent magnitude of -6 at its peak. It will be quite a show, but nothing extraordinarily special.

The damage to orbiting spacecraft around Mars will likely be minimal. 90 minutes after the comet passes Mars, the worst of the dust particles will hit the spacecraft. The tiny barrage will only last about 20 minutes, though, and there is a low probability of damage. In fact, the various orbiters will take pictures of the comet and try to analyze it to learn more about its properties.

I hope this helps.

Other sources:

http://mars.nasa.gov/comets/sidingspring/

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

Update:

As of about an hour after the encounter, Mars still exists. Read this article for some not-so-juicy details.


Exoplanets, Protoplanets, and planetesimals So, most of the mass of the solar system (that isn't inside the sun) is inside Jupiter. And most of whats left after THAT is inside Saturn. So, all the rocky planets, moons, and asteroids in our solar system are a tiny minority of the mass.

Now, we're looking at other solar systems for Exoplanets, and mostly finding gas giants. But not every system has one that we can detect- even some of the nearby systems. Which implies that there are plenty of systems without a Gas Giant.

Unless I'm misunderstanding things, the amount of raw material in those systems should be fairly similar to ours. So what, realistically, would those systems have in terms of planets? Without the massive gas giants to suck up all the mass, they'd presumably have lots of smaller objects, right? Massive asteroid belts, large numbers of small planetoids? Are there any models of what those systems are likely to look like?

Whether or not gas giants form is largely to do with whether rocky/icy cores are able to accrete large quantities of hydrogen and helium, which dominate the mass of the protostellar disc.

Terrestrial planets form from the tiny fraction of the disc that is able to form solids.

There is no indication that systems without gas giants have more "solids" and most of the raw material in the disc is either accreted onto the star or expelled from the system in either case.


Angular distance/apparent separation between AE Aurigae and Orion's belt The title perhaps gives the question away, but would anyone know/be able to roughly approximate the angular distance/apparent separation between AE Aurigae and Orion's belt.

I have the value of $42^\circ$, from visual estimation but would like a way to calculate this.

Given the cords of AE aurigae:

RA 05h 16m 18.1s DEC +34° 18' 49"

And Alnitak in Orion's belt:

RA 05h 40m 45.52666s DEC  −01° 56′ 34.2649″

I get an angular separation of 34.83°

By this formula:

$\cos(A) = \sin(\mathrm{DEC}_1)\cdot \sin(\mathrm{DEC}_2) + \cos(\mathrm{DEC}_1)\cdot \cos(\mathrm{DEC}_2)\cdot \cos(\mathrm{RA}_1 - \mathrm{RA}_2)$


Why does the sky look like it has three suns in this video? Check out @natalie9280’s Tweet:

  long story short, friday morning at around 6:40am i looked outside and i saw three suns.
  i checked to see if there was an eclipse and i could[n't] find anything.
  twitter help pls

  [VIDEO: three progressively brighter "suns," left to right in a closely spaced row]

The ghost images in the video resemble the Sun and are separated by only 2 or 3 degrees (4-6 Sun widths), so we can rule out atmospheric phenomena.
Sun dogs or parhelia have shapes different from the Sun and appear 22 degrees away from it.

The effect is more likely due to looking through a double-paned window at an oblique angle.
The bright image on the right is the Sun viewed directly.
The image in the middle is a reflection by the inner glass surfaces facing each other, offset due to the gap width and the viewing angle.
The faint image on the left is a further reflection of the middle image.


Is 486958 Arrokoth (2014 MU69 aka Ultima Thule) the only solar-system object determined to be binary by occultation? The asteroid 486958 Arrokoth was discovered to be binary by astronomers lugging a bunch of telescopes around (I think) camping with them. The timing offered additional astrometry for the orbit determination before New Horizon's flyby, and it also served to put an upper limit on any dust or debris cloud around the object that might suggest a farther flyby distance be considered.

Their hard work certainly paid off and an outline of the object was reconstructed, showing that it was likely a binary; two circular objects. This posed a bit of a mystery because there wasn't any periodic variation in brightness detected that would suggest rotation, but that's a different story.

Is this the only solar system object that was discovered to be binary (or higher) via occultation?

above: reconstructed occultation timing from several nearby sites from this answer below: series of images from New Horizons during flyby from this question

1.3 MB GIF: https://i.stack.imgur.com/Ci213.gif also viewable in YouTube

Actually, the very first binary asteroid was discovered by stellar occultation in 1980.  The evidence was rejected by the IAU, and wikipedia credits the first binary asteroid discovery to the Galileo spacecraft finding the moon of 243 Ida in 1993.
The first moon of the well known asteroid 216 Kleopatra was observed through stellar occultation in 1980.  From this article:

As occultation czar David Dunham would later report in Sky &
Telescope, nine different observing stations visually timed blinkouts
of the 9th-magnitude star SAO 128066. Collectively these timings
traced chords across an irregular, elongated silhouette measuring
roughly 60 by 80 miles (95 by 130 km). "But there was an unexpected
secondary occultation too!" Dunham wrote. "Gerald Rattley and Bill
Cooke, observing from sites 2,000 feet apart near Loma Prieta,
California, independently and nearly simultaneously saw the star
disappear for about one second."

Dunham concluded that they had discovered a satellite of 216 Kleopatra. At that time, however, it was unknown if asteroids had satellites, and their evidence was not accepted.  There was some stigma associated with the occultation methods.   In 2008, the Keck II telescope finally verified the moons of 216 Kleopatra:

In 2011, one of Kleopatra's moon's size and orbit was shown to be consistent with the original 1980 occultation observations, referenced in this paper, and reconstructed here with the original chords:

Even though there were two independent chords that occulted the star, the IAU's former Minor Planet Director Center called the evidence "shoddy", even decades later.  In 1980, he had refused to accept that asteroidal satellites existed.
Interestingly, even after vindication 30 years later, stellar occultation evidence for binary asteroid systems are still met with skepticism.  For example, the wikipedia page for the asteroid 911 Agamemnon says a 2012 occultation is merely 'suggestive' of the Asteroid having a satellite.  To me, the stellar occultation of the light intensity measurements from this paper, is far more than just suggestive, but instead clear evidence of a satellite of 911 Agamemnon:

The large steady dip is the main body of the asteroid passing over the V magnitude 8.0 star HIP 41337 = SAO 60804 = BD= +37° 1857 = HD 70920, spectral type K0.  The small dip is the satellite passing over the star.
Note:
It's strange to me that the discovery of binary asteroid systems is met with skepticism since the method is essentially the same method often used to detect extra-solar planets (the transit method explained nicely in this answer).