Can dark matter be found in the shape of planets, galaxies etc.?

Planets and stars, no. Globular clusters and galaxies, yes.
Small scales
To condense into such relatively compact objects as planets, stars, and even the more diffuse star-forming clouds, particles need to be able to dissipate their energy. If they don't do this, their velocities prohibit them from forming anything.
"Normal" particles, i.e. atoms, do this by colliding. When atoms collide, they're excited, and when they de-excite, they emit radiation which leaves the system, carrying away energy. In this way, an ensemble of particles can relax into a less energetic system, eventually condensing into e.g. a star. Additionally, the collisions cause more energetic particles to donate energy to the less energetic ones, making the ensemble reach thermodynamic equilibrium, i.e. all particles have the same energy on average.
Dark matter is, by definition, unable to collide and radiate, and hence, on such small scales as stars and planets, particles that enters a potential well with a given energy will maintain that energy. They will thus accelerate toward the center, then decelerate after its closest approach to the center, and finally leave the system with the same energy as before (if it was unbound to begin with). This makes it impossible for collisionless matter to form such small objects.
Large scales
On the scale of galaxies, however, various relaxation mechanisms allows dark matter to form structure. This is the reason that in pure N-body simulations of the Universe, such as the Millennium Simulation, you will see galaxies. The sizes of these structures depend on the resolution, but are measured in millions of Solar masses.
The relaxation mechanisms include:
Phase mixing
This is sort of like galaxy arms winding up, but in phase space rather than real space.
Chaotic mixing
This happens when particles come so close that their trajectories diverge exponentially.
Violent relaxation
The two mechanisms listed above assume a constant gravitational potential $\Phi$, but as the systems relaxes, $\Phi$ changes, giving rise to an additional relaxation process. For instance, more massive particles tend to transfer more energy to their lighter neighbors and so become more tightly bound, sinking towards the center of the gravitational potential. This effect is known as mass segregation and is particularly important in the evolution of globular star clusters.
Landau damping
For a perturbation/wave with velocity $v_p$, if a particle comes with $v\gg v_p$, it will overtake the wave, first gaining energy as it falls into the potential, but later losing the same amount of energy as it climb up again. The same holds for particles with $v\ll v_p$ which are overtaken by the wave.
However, particles with $v\sim v_p$ (i.e. that are near resonance with the wave) may experience a net gain or loss in energy.
Consider a particle with $v$ slightly larger than $v_p$. Depending on its phase when interacting with the wave, it will be either accelerated and move away from resonance, or decelerated and move closer to resonance. The latter interact more effectively with the wave (i.e. be decelerated for a longer time), and on average there will thus be a net transfer of energy from particles with $v \gtrsim v_p$ to the wave. The opposite is true for particles with $v$ slightly smaller than $v_p$
You can read more about these mechanisms in Mo, Bosch, & White's Galaxy Formation and Evolution.


How do we know about the origin of ʻOumuamua?

I'll start with the second question first:

Also, do we exactly know about each and every object inside our solar system?

Each and every object? Of course not. That however is irrelevant to the question at hand. The vast majority of objects in our solar system are very, very small. Quite literally as small as dust. But because they're so small, they have essentially zero impact on the orbits of objects that are not as small as dust.

How do we know that ʻOumuamua is interstellar and not otherwise?

Because it's moving so quickly. ʻOumuamua's huge 26 kilometers per second excess means that is not possible that ʻOumuamua originated within the solar system.
I'll start with the two body problem, e.g., the Sun and some other object interacting gravitationally. The Newtonian gravitational interaction between two objects is a conic section: Either a circle, an ellipse, a parabola, or a hyperbola. The total mechanical energy (kinetic energy plus gravitational energy) for objects following a circular or elliptical trajectory is negative. Such objects are gravitationally bound to one another. Hyperbolic trajectories on the other hand have a positive mechanical energy. Parabolic trajectories form the boundary cases, where the total mechanical energy is exactly zero. Objects on parabolic or hyperbolic trajectories are not bound. They visit once and then they're gone. The excess velocity of an object on a parabolic trajectory (zero excess velocity) or a hyperbolic trajectory (positive excess velocity) is a constant of motion in the two body problem.
The N-body problem makes things a bit dicier. Long-period comets are believed to have originated from within the solar system. These comets appear because something perturbed their very long-period orbits. The perturbation make those long-period comets dive well inside Neptune's orbit. A few of those long-period comets appear to be following an unbound trajectory (non-negative excess velocity). A chance close encounter with a planet can make that happen.
This raises the question: Could ʻOumuamua be yet another long-period comet whose orbit has been perturbed to an extent that puts it on a hyperbolic trajectory? This would not be the first case of an object with an over-unity eccentricity. The answer to this question is no. A close encounter with any of the four known giant planets, or even with the conjectured Planet IX, or even multiple encounters, could not have given ʻOumuamua the excess velocity that has been observed for it.


Other than it being a Full Moon and the Moon is high in the sky, there is nothing unusual going on.


Let's take a look!

While I can't quite track down the source of this image, I've found it throughout the web, so hopefully it's roughly accurate. I'm pulling it from the Wikipedia article on the Orion Arm.

In this image you can also see the Orion Nebula sitting (relatively) cozily to the Sun.  The distance between the two is believed to be around 1350 light years.  Zeta Reticuli, on the other hand, is believed to be a mere 39 light years from Earth. Given the orientation of these two locations and the relative distances, it would be hard to place Zeta Reticuli anywhere but firmly within the Orion Arm.

Not exactly rock solid evidence, but sometimes some back of the envelope deduction will do.


Yes, a little, but not enough to matter.

Magnets can spin other magnets.   Think of magnets as a bar magnet with a north and a south, and you can think of them as linear not spherical if that helps with the imagery, because in a sense, they are linear with a distinct North and South magnetic pole.

Earth's magnetic field, for example has a distinct longitude and latitude that moves over time, but it's fixed at any given time.

For your question to work, it's easier to imagine a very powerful magnetic object flying past, and to think of the Earth's internal magnet as a compass arrow that wants to line up.   Magnets want to line up with magnetic field lines and a force would be created by a passing strong magnetic field that could orient the Earth's magnetic field to align with it.

You might think that this would turn the Earth, but it wouldn't.  Earth's magnetic field isn't "the Earth" it happens in the Earth's core, and it may happen mostly in the region where Earth's inner and outer core meet and the outer core is liquid (dense and viscous but still not solid).  The force would be applied to the magnetism inside the Earth, perhaps giving the Earth's inner core a slight tug and twirl and perhaps re-orienting the flowing charged liquid iron just outside of the core, but this force wouldn't turn the Earth as a whole, it would generate a force on the core and the core doesn't spin precisely with the Earth's surface anyway so any movement would likely not show up on the surface, for some time.

Given that a percentage of earth's magnetism is carried by the liquid outer core, then the effect on rotation would be even less, more of a re-orientation than a direct angular momentum tug.

It's also difficult to generate a magnetic field strong enough to move something as massive as the Earth's core very much anyway.   Something to keep in mind as well is that the magnetic field drops by the cube of the distance, not the square, so you'd need a very powerful magnet, very close, to have any effect at all.   Anything large enough to have that strong a natural magnetic field, the gravity would have a much bigger effect than any interaction between the magnetism.

But to answer your question, a tiny bit of angular momentum would be transferred when two magnetic fields fly past each other in space, so there would be some variation in rotation, but it would be tiny.

Magnetic fields as strong as Earth's are rare, you'd probably need a gas or ice giant planet or large rogue planet, and the interaction between passing magnetic fields is very weak compared to the gravitational interaction.  Compared to it's mass, Earth is a very very weak magnet, so using magnetism to spin it would be ineffective and slow, but but in theory, if we wanted to set up a current and generate a magnetic field around the Earth, such a project could in theory spin the Earth's inner core like a induced magnetic field spins a magnet in a generator, but that would also depend on how permanent the magnetism is inside the Earth and it might not be fixed, but instead, a product of flowing magnetic metals.   Permanent magnets are easier to rotate than charged flowing liquid . . . but I digress.

It would be an interesting calculation if somebody wanted to do the math behind this, but even without doing the math I can tell you that the effect would be teeny-tiny.


How can heavy elements exist when the universe is young compared to the lifetime of a star?

First, stars evolve at radically different rates. The higher the star's mass, the faster it evolves.  The more massive stars had plenty of time to go through multiple generations before the Sun started to coalesce.

It went even faster than it does today, because the collapse of interstellar clouds to form stars is halted mainly by radiation pressure from the new star blowing the rest of its birth cloud away and ending accretion. Since hydrogen and helium are transparent to most of the new star's radiation, the very first stars could grow larger than stars do today. (They were pretty much made of nothing but H and He (with a bit of Li), since very little of the heavy elements had yet formed.) They went through their lifetimes and turned into supernovae and neutron stars very quickly. The largest had lifetimes on the order of just a million years.

So by a very short time after the first stars lit, they were already spewing heavy elements back into space and later stars already had some heavier elements in them at formation.  The fraction of heavy elements has increased steadily since then, but the Sun formed many billions of years after heavy elements became common.


What earth organisms would survive if they arrived on Mars?

On the surface of Mars probably none, since it's too dry or too cold, or both, to stay active.

Spores or other dormant forms probably could survive for centuries, until radiation will gradually destroy the organic molecules necessary to get back into an active state.

But there are "Mars Special Regions", where either Earth microbes or potential Martian microbes - if some exist - cannot be ruled out to be able to spread.

Things might look quite different in the underground, especially in warm and wet zones, which may exist by geothermal heat and ground water.
There, at least lithotrophic bacteria/"chemolithoautotrophs" could survive and spread.
Those could then form a basis for a food chain of other organisms.

Plants dependent on sunlight, or animals dependent on oxygen would hardly survive.

For a list of species, see table 2 on page 895 of the above reference paper. The following species and genera are mentioned in the table:
Psychromonas ingrahamii, Planococcus halocryophilus strain Or1, Paenisporoarcina sp. and Chryseobacterium, Rhodotorula glutinis (yeast), Colwellia psychroerythraea, Nitrosomonas cryotolerans, (lichen) Pleopsidium chlorophanum.


It's liquid.
As detailed here,

To figure out whether Mercury's core was liquid or solid, a team of scientists led by Jean-Luc Margot at Cornell University measured small twists in the planet's rotation. They used a new technique that involved bouncing a radio signal sent from a ground telescope in California off the planet and then catching it again in West Virginia.
After 5 years and 21 such observations, the team realized their values were twice as large as what would be expected if Mercury's core was solid.
"The variations in Mercury's spin rate that we measured are best explained by a core that is at least partially molten," Margot said. "We have a 95 percent confidence level in this conclusion."

The NRAO has another article on it, which goes slightly more in-depth into the subject.
The official site of the Messenger mission is slightly more cautious:

However, these constraints are limited because of the low precision of current information on Mercury's gravity field from the Mariner 10 and MESSENGER flybys. Fundamental questions about Mercury's core remain to be explored, such as its composition. A core of pure iron would be completely solid today, due to the high melting point of iron. However, if other elements, such as sulfur, are also present in Mercury's core, even at a level of only a few percent, the melting point is lowered considerably, allowing Mercury's core to remain at least partially molten as the planet cooled. Constraining the composition of the core is intimately tied to understanding what fraction of the core is liquid and what fraction has solidified. Is there just a very thin layer of liquid over a mostly solid core or is the core completely molten? Addressing questions such as these can also provide insight into the current thermal state of Mercury's interior, which is very valuable information for determining the evolution of the planet.

At this point in time, though, all evidence indicates that Mercury has a molten core.
As userLTK pointed out, lower pressure inside Mercury makes it easier for the core to stay liquid at lower temperatures.


I think Aabaakawad's link gives a complete answer, but to give an astronomy for dummies answer, there's nothing about a white dwarf that causes a planet's orbit to decay at least, not directly.  Your article (I've pulled quote below the caption):

  Slowly the object will disintegrate, leaving a dusting of metals on
  the surface of the star.

That's only talking about this particular situation and there's a difference between disintegrate and decay.    This planetoid is enormously close to the white dwarf.  So close, that what we think of as normal white dwarf/planet dynamics (very cold) is no longer true.  This planetoid is slowly being vaporized.

Looking at the orbital period of 4.5 hours (about 1948 orbital periods in 365.25 days).   The orbital distance to orbital ratio relation is exponential to the power of 2/3 (this varies a bit due to eccentricity, but it's generally correct), so an orbital period 1948 times faster means about 156 times closer, and giving the white dwarf equal mass to our sun, that puts the planetoid at a bit under 1 million KM.  If this white dwarf is lighter than the sun, the planetoid would need to be even closer.  That's close to the Roche limit and would be inside it if the planetoid wasn't dense and rocky/metallic.

If we estimate the white dwarf to be about the size of the Earth, which is a common size given for white dwarfs, An Earth sized object from 1 million KM would be larger in the sky than the sun appears from Earth, and presumably quite a bit hotter than the surface of our sun too, so this isn't a tiny white dwarf in the sky from the perspective of the planetoid.   It's a blazing furnace of a sun, so hot, it vaporizes metallic gas and dust off the surface of the planet.

The article mentions this (end of page 3), that Poynting-Robertson drag see here and here, and that may be a factor in any orbital decay in this scenario.  The article is clear that there's a good deal of uncertainty on with that effect, and that only affects tiny particles, but enough tiny particles could create a drag over time. . . . (maybe).   The general scenario with this orbit is a planet scorched and as a result, is losing material.  It's likely the very high heat that's driving any orbital decay, not gravity.

Gravitational decay / orbital decay does happen, usually much more slowly.   That's probably not what's happening here.

There are some interesting orbital effects that can happen when a main sequence star goes red dwarf and later when it creates a planetary nebula, significant increases in tidal forces due to the star's greater size in the first case and increased drag in the 2nd, but at the white dwarf stage, there's no significant orbital decay effects.

Update:

  Why not Poynting Robertson drag and Orbital decay effect the planetoid
  when white dwarf was a star or even red giant? Is there any
  "interesting orbital effects" when a star undergoes red giant?.Can you
  update your answer to summarize the forces and their effect on the
  planetoid in each phase of the star.
  and also what do you mean by orbital decay? Does it have something to deal with the Roche limit.

OK, I think, having read more about it, Poynting-Robertson effect only matters when the orbiting objects are very small.   I've linked it twice above, but the simple explanation is that objects in orbit move and so any light or debris from the sun hits the moving object at an angle, not direct on.    If the object is small enough, this over time drives the dust and maybe grain of sand sized particles into the sun.   This doesn't affect larger objects, so it's not really relevant to any planets or planetoids.

As far as "interesting red dwarf" effects.   That really has to do with tides.   Using the Moon/Earth example, the Moon creates tides on the Earth, a tidal bulge towards the Moon, but because the Earth Rotates faster than the Moon orbits, this tidal bulge is always ahead of the Moon and this creates a gravitational tug on the moon that pulls it away from the Earth - very slowly.

The same thing happens with planets around stars, but even more slowly,   lets pretend it's just the Earth and the Sun - a 2 body system (in reality, with several planets it's much more complicated), but just Earth and sun, teh Earth creates a tidal bulge on the sun, the sun rotates ahead of the earth, this causes the earth to very slowly spiral away from the sun - so slowly that it might take a trillion years for the Earth to spiral away.

Now when the sun goes Red Giant, the sun is essentially the same mass but much more spread out and parts of it, much closer to the Earth and less gravitationally bound to the sun.   This creates a far larger tidal tug.   Also, as the sun expands it's orbital velocity drops, because orbital momentum is concerved, so when the Sun is Red Giant, the tidal bulge will be behind the Earth which drags it in towards the Sun.    Due to the size and proximity of the Red Giant star, this draws the remaining near-by planets towards the sun fairly quickly, at least compared to main sequence stages which, provided the sun rotates faster than the planets orbit, has a much smaller outward tidal pressure on the planets.

And when the sun goes planetary nebula, any debris in the planet's path can also cause the planets to slow down slightly - the precise process there I'm less clear on, but in general, any orbital debris creates drag and can slow down a planet's orbit.   This may be a key factor in the formation of hot jupiters, cause they can't form close to their suns but enough orbital debris can drive them in closer to their suns.  (or planet to planet gravitational interactions can too).

That's the gist of the Sun-Planet orbital relation.  When the sun is young, planets are mostly driven outwards, and young suns can have far greater solar flares and stronger solar wind.   How much that effects the planets, I'm not sure.

During the Main sequence stage, stars tend to push planets outwards (unless they rotate very slowly, in which case the tidal effect is reversed), but this tidal effect is very small and very gradual.

During the Red Giant stage, stars tend to drag planets in wards, and I assume, during the planetary nebula stage as well.   This effect is larger for closer planets.

You also asked about Orbital Decay - if you click on the link, there are examples of that.  That probably gives a better explanation than I could.   In general, Orbital decay happens very slowly unless you're talking Neutron Star or Black hole in which case the relativistic effects can cause orbital decay to happen quite fast.  There's nothing about a white dwarf star that would cause faster than normal orbital decay but a white dwarf star would lose any tidal bulge tugging that a main sequence star has, so there would be essentially no tidal outwards pressure either which could in theory speed up decay cause you've lost a small outwards pressure but you would still have any debris or space dust clouds causing a small inwards pressure.  (if that makes sense?)

That's my layman's explanation anyway.


The short answer is no. There is no huge accelerations of star systems.

Look at a pair of interacting galaxies  These are the Antenna galaxies (image copyright Star Shadows Remote Observatory and PROMPT/CTIO, from apod)

You will see the long "antennae", which are formed of stars which have been ejected from their galaxies by the interaction. But the accelerations needed are not large, because the interaction takes a long time. The acceleration nothing like $km/s^2$, The force required to accelerate a planet like that would smash it. remember 1$km/s^2$ is about 100g. But small accelerations over long periods can have big effects. The only way to cause objects to accelerate is to apply a force, and the only force that can move planets is gravity. A galaxy has a great big gravitational field, but it is very spread out, and as you note, most of the galaxy is empty space.

Planet orbits should be stable, certainly inner planets would remain bound to their stars. There could be disruption to oort clouds, but the gravitational field around a star would not change much. Similarly the star's magnetosphere would not be changed much. As above there are no large forces applied to stars or planets.

A close-ish approach, to 1 light year, would cause disruption of the oort cloud, but nothing more. Stars may have larger relative motions, but no where near enough to cause blue shifting of light into x-ray, in fact it would be hard for observers on the planet to tell which stars came from which galaxy. Andromeda is approching at 120km/s. But proper motions of stars in the milkyway can be greater than this, Barnards star has a relative velocity of 142km/s

At the end of the interaction, the star and its planet is likely to be scattered to the halo of an elliptical galaxy.


What do the months and hours mean in Tomas Filsinger's Map of The Universe

The hours show the lines of right ascension, where right ascension is one of two coordinates used to specify a point on the sky. (The other coordinate is declination, measured in degrees from the celestial equator.) Right ascension and declination are similar to longitude and latitude on the Earth.

The month and day show the right ascension of the Sun on a given date. The dotted circle passing through the zodiacal constellations (Pisces, Aries, etc.) is the actual path of the Sun (the ecliptic). Therefore, the position of the Sun on any day can be determined from the chart.


C-Type Asteroids and Carbonaceous Chondrite Meteorites

The difference between asteroids, comets and meteors is discussed in What is the difference between asteroids, comets and meteors?

Some groupings of asteroids have been correlated with meteorite types:

  C-type – Carbonaceous chondrite meteorites
  S-type – Stony meteorites
  M-type – Iron meteorites
  V-type – HED meteorites

The S- and C-type asteroid classifications originate from the taxonomy proposed by David Tholen.

Meteors are generally considered to be asteroids which enter Earth's atmosphere.
Meteorites are meteor remnants which reach the ground. So essentially, carbonaceous chondrite meteor(ite)s start out as C-type asteroids; they're made of the same stuff.


A beginner project in plotting celestial bodies in 3D coordinate space

Here you can choose your target body and center of observation. If you want cartesian coordiantes use "Vector Table".


$\pi$ is just a number. It doesn't vary with anything.

The ratio of the circumference to the diameter of a circle varies with geometry, so in the general case, this definition doesn't hold. But there are other definitions, such as
$$
\pi = \int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}},
$$
or
$$
\mathrm{The\,smallest\,positive\,number} \, x \, \mathrm{for\,which}\, \arccos(2x) = 0.
$$
The latter definition can also be expressed a power series, which is independent of geometry.


While the other answer here and the link provided by Jeremy give excellent explanations, I believe a bit more nuanced reasoning is required.

Although the theory of planet formation is currently still incomplete, it is generally accepted that planets form in a so-called proto-planetary disk as a part of stellar formation process. This is backed up by several observation of such disk and even by
directly observed planets in systems where remnants of the disk are still present (most notably Fomalhaut-b).

However, exactly how these planets form is still unclear. A popular theory is that these planets form in place and therefore keep the orientation and inclination of the disk. This beautifully explains the Solar System and specifically why all planets have roughly the same orbital inclination. But that's not surprising as the model was specifically designed to do just that! Yet the Solar System does not have to be the only possible type of planetary  system, or even the most common for that matter. Indeed the study of exoplanets has confirmed that very different planetary systems are possible.

An alternative theory is that planets are created far away from the star and migrate inwards. During this migration multiple planets can interact with one-another and be pushed out of the disk plane to higher inclinations. They could even flip over completely and become retrograde.

Now to come back to the actual question; the Solar planets probably have about the same inclination simply because they were created that way.

You are also right in saying that gravity will tend to align the planets. How this works physically is that the star can exerts a torque on the planet, which produces a tidal effect (like how the Moon creates tides on Earth). Over time this aligns the angular momentum of the planet to that of the star by pushing the planet towards the equatorial plane of the star.

As for your second question, yes, there are probably planetary systems which are misaligned with respect to the star. At least there are certainly planets that are not aligned with the equatorial plane of the star. The well named HAT-P-11b is a known example, but there are a lot more.

Note that for exoplanets the important parameter is not the inclination (which for exoplanets is the angle between the normal of the orbit and the line of sight), but rather the obliquity or tilt of the planet rotation with respect to the rotation of the star.

As for a third twist to this story, very recently there was a discovery of an entire proto-planetary disk that is misaligned with respect to the star, which opens the door to even stranger configurations.


This must be what you are looking for, "Tabulae Rudolphinae". Tycho Brahe's star catalogue and planetary data published by Kepler in 1627 .


Does the earth's atmosphere act as a spherical lens and refract light from space?

If so by how much does it "spoil" the view of stars and galaxies etc.

There are several very different issues related to your question. Let's tackle them one by one.

Atmospheric refraction

Yes, the Earth's atmosphere refracts light. One notable effect is that objects near horizon appear higher than they should be.

Therefore, the Sun (or any other object) rises earlier than it should, and sets later.

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

It affects astrometry, because it messes up the coordinates of objects near the horizon. Other than that, it does not cause great harm to astronomy.

And then there are several effects harmful to astronomy:

1. Seeing

What we call in astronomy "seeing" is just air turbulence. Because air moves around due to heat, it distorts the image continuously. This is how the Moon looks like in a telescope - notice the shifting around due to seeing (animated GIF, reload if it's not moving, or open in new tab):

If we lived in a vacuum, that image would be static, and perfectly clear. But turbulence causes the shifting. Also, it causes a blur - the image is less sharp than it could be. It's also what causes the stars to "twinkle".

It's basically the same phenomenon that you see in summer along the hot walls of a house under sunlight - the same shimmering effect of the image seen through the hot rising air along the walls.

For amateur astronomers, seeing affects high-resolution targets such as planets, the Moon, double stars. It does not affect targets that are perceived to be "low-resolution" in an amateur telescope: galaxies, nebulae, star clusters.

More reading:

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

Since seeing depends on weather, it can be predicted to some extent:

http://cleardarksky.com/csk/

2. Light pollution

Atmosphere also scatters and reflects light back to Earth. All the city lights around you are scattered by haze and dust in the air, and some of that light goes back down into your eye, creating sort of a "light fog".

As a result, objects that are not too bright are masked by light pollution and become less visible, or invisible altogether.

A bigger telescope will help you pierce through the haze of light pollution, but only to some extent. The best way to see faint objects is to put the scope in the car and drive far away from the city.

Here's a light pollution map:

http://www.jshine.net/astronomy/dark_sky/

For amateur astronomers, light pollution affects the "faint fuzzies": galaxies, nebulae, star clusters. It has no effect whatsoever on bright objects such as: planets, the Moon, many double stars.

Other ways the atmosphere can negatively impact observations:

clouds - this is an obvious one
transparency - different from clouds, it includes haze, dust, particulate matter, water vapor. It affects low contrast objects such as galaxies.


A long time ago.
The major mare forming eruptions occurred more than 3.5 billion years ago, but a paper  Ages and stratigraphy of mare basalts finds that some volcanic basalts are as young as 1.2 billion years. However, it seems that impacts have been the only active geological process for a billion years, giving rise to the observation that just as complex life on Earth got started, the moon (geologically) died.


How did early estimates of a "potato radius" set 1 eV ~ GMμ/R and get 200 to 300 km?

How do I use this to arrive at a potato radius R of about 200 to 300 km?

You can't, for multiple reasons. One reason is that the authors themselves disclaim this as a valid approach:

  The result is a radius of a few thousand kilometers for a rocky body and ~1000 km for an icy body. This was meant to be only an order of magnitude calculation. We try to improve on it below.

Another is that the authors don't show their math. On trying to replicate it, I get rather different results. The mass of a more or less spherical object is $M=\frac43\pi\rho R^3$, where $\rho$ is the mean density of the object. Plugging this into their expression yields $1\,\text{eV}\approx \frac43\pi G\rho\mu R^2$, or
$$R \approx \sqrt{\frac{1\,\text{eV}}{\frac43\pi G\rho\mu}}$$

This suggests the potato radius is inversely proportional to the square root of density and inversely proportional to the square root of particle mass. This obviously is not the case; the observed potato radius of rocky objects is about 50% more than that of ice objects, even though rocky objects have the greater density and are made of more massive molecules.

I took this as a throwaway paragraph. The key results are further on in the article.


How do we know galaxy GN-z11 is as far away as it is?

So GN z-11 is the latest "furthest away" galaxy. It has a claimed redshift measurement of $z=11.1$, meaning that we are seeing the light it emitted about 400 Myr after the big bang (dependent on an assumed set of cosmological parameters).

To get the redshift measurement, the discoverers used grism (relatively low resolution) spectroscopy in the near infrared. What they were looking for is the rest-frame Lyman alpha continuum break, which would be redshifted into this wavelength range.

The Lyman alpha continuum break is caused by the absorption of nearly all ultraviolet photons with wavelengths smaller than 121 nm. These high energy photons are capable of photoionising the neutral hydrogen present in the intergalactic medium at a range of lower redshifts. This neutral hydrogen is present in abundance at redshifts greater than 6, as it had not yet been re-ionised by quasars and starlight. The consequence is that no light is expected to reach us from rest-frame wavelengths shortwards of 121 nm, but a galaxy's light can reach us from rest-frame wavelengths that are longer than this. When one observes the spectrum, we see flux at long wavelengths which suddenly cuts off at shorter wavelengths. The wavelength of the break is $\lambda = 121 \times (1+z)$ nm, where $z$ is the redshift.

It is this Lyman alpha continuum break that has been identified at an observed wavelength of 1470 nm. This leads to the redshift estimate of $z= (1470/121)-1 = 11.1$.

The details are presented in Oesch et al. (2016). The continuum break is the only thing visible in the spectrum. The authors are confident that this is what it is because their previous broadband photometry had given them an estimated redshift of $>10$ (which is why they observed this candidate in the first place).


No, the structures are not of the size of the observable Universe. In fact modern cosmolgical models rely on the Universe being homogeneous over scales above $100 \textrm{Mpc}$. This is an observational fact, which serves as a base for the so-called cosmological principle. See also a wiki article on large-scale structure, where the scale is mentioned in the section "End of Greatness".

The scale corresponds to a few hundred million light years. The largest structures, therefore, are clearly way smaller than the size of the observable Universe, which is of order of ten billion light years.

Note, that because it takes time for the light to reach us, the current size of the Universe we observe is larger than its apparent size due to cosmological expansion.


Actually, it is rotating relative to your perspective. (The other answer relating to liberation occurs over a longer period of time).

Here is a simulation from Sky Safari for later tonight (moonrise) and tomorrow (moonset) for my location. Note how the angle of the lit hemisphere rotates between the pictures:

Moonrise:

Moonset:

It looks like this for me because I am at 43 south. If I was at the equator, the moon would appear to have rotated 180 as it travelled from horizon to horizon.

It is easier to think about on the equator. Say the moon rises, and you consider the bit at the top to be "up". It continues on up, right over your head, and down to the horizon... But on the way down, the "up" side of the moon is now looking down. It was "up" until the zenith, but then it is "down" all the way to the horizon. It doesn't rotate at all, but the result is a 180 degree flip in appearance from horizon to horizon.

Or consider the perspective at a pole, while the moon is up. The moon travels right around the sky, always facing the same direction, with no apparent rotation (and while it is up, no moonrise or moonset, it is up all day).

Latitudes in between see an in-between effect; some apparent rotation of the moon between moonrise and moonset.


How (un)stable are the Lagrangian points 1, 2 and 3?

All five Lagrange points are unstable

L1, 2, and 3 are "saddle points" in the effective potential formed from the combination of gravity and the centrifugal force of a rotating frame of reference. An object which is in front or behind in the orbit would tend to approach the Lagrange point, and then move away, either towards or away from the sun, following the lines in the picture.

However note that in the picture, the lines around the Lagrange points are not close together, this means that the motion towards, or away from the point is very slow (in the rotating frame) An object near the Lagrangian point would not be "ejected" at high speed. Any cosmic dust that was near the Lagrangian point would drift slowly away.

However although the L4 and L5 points are unstable, there are quite stable orbits that (in the rotating frame) orbit around the L4 and L5 point, allowing for Trojan satellites.


I have not the qualification to answer the question in its whole but the question is interesting (I worked on Be Stars which are episodically surrounded by an decretion disk and which rotates at nearly critical velocities. The phenomenon in Be stars is different from accreting stars. The only consequences of subcritical velocity is a flattened envelope and modification of its inner structure and of the oscillations modes found in these stars
(if you have time and curiosity, a good example of flattened star with Keplerian rotating decretion disk is Achernar, a Be star observed using interferometry --> Have a look at Meilland et al. 2007: www.aanda.org/articles/aa/pdf/2007/10/aa4848-06.pdf)

Anyway...

I found this paper about critically-rotating accretors. May be you will find answers to your questions here or in its references (use nasa ads site for your query: http://adsabs.harvard.edu/).
http: //arxiv.org/pdf/1306.1348v2.pdf
It seems that in the introduction, there is some answers to your questions about reaching the critical velocity.

The accreted mass can increase the rotation rate until the star reaches the critical velocity.

It is said : "For a typical 6 + 3.6 M⊙ system, with initial period
Pinit = 2.5 days, in the absence of spin-down mechanisms, only
3 per cent (0.12 M⊙) of the total amount of matter transferred
by RLOF (more than 5 M⊙) is enough to spin the gainer up
to the critical rotation."

But we still don't know if the gainer can really reach the critical velocity. Some papers are dealing with break-up mechanims which does not allow the gainer to reach critical velocity: spin down by tides, magnetic breaking, limitation of the accretion angular momentum through interaction with the accretion disk, stopping of the accretion mechanism...

I am sure that you will find many papers on nasa ads that will give you answers to your questions.


Which star with a Bayer designation is the farthest from the Sun?

See important caveat to this answer below

At a distance of 2992 light years, Tau Canis Major is the furthest of the 1513 stars that have a Bayer designation, with several others almost equally distant. I computed the full list of Bayer-Sun distances: https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-bayer-distance.csv

Important Caveat

I used the HYG catalog at https://github.com/astronexus/HYG-Database but it may be inaccurate. In particular, Wikipedia gives Tau Canis Major's distance as 5,000 light years, although I couldn't find this number in the source provided (https://doi.org/10.1088/0067-0049/199/1/8), although I only skimmed.

The far from canonical https://www.universeguide.com/star/taucanismajoris notes that 2992.32 is a revised figure, down from 3197.68, but, of course, the figure may have been revised again.

Methodology

After downloading the HYG database file, this one liner creates bc-bayer-distance.csv:

zcat hygdata_v3.csv.gz | perl -F, -anle 'if ($F[9] < 100000 && length($F[27])>0) {$F[9] *= 3.26156; print "$F[9], $F[27] $F[29]"}' | sort -nr | tee bc-bayer-distance.csv

A brief explanation:

The 9th field (starting our count with 0 as Perl does) is the distance, but there's a note on the HYG git page: "A value >= 10000000 indicates missing or dubious (e.g., negative) parallax data in Hipparcos". It turns out the actual number is 100,000 (I've submitted a correction), so we ignore distances greater than 100,000.
The distance is in parsecs, so I multiply by 3.26156 to convert to light years.
The 27th field is the stars Bayer letter designation without the constellation. If it's empty, we ignore the star.
The 29th field is the stars constellation name, which, combined with 27th field, yields the Bayer designation.


Why is the opening in the Anglo-Australian Telescope's dome so small?

This is a two part windscreen designed to minimize the effects of windshake on the telescope and to avoid the deterioration in image quality that wind would cause. The AAT is in a tall 6 story dome on a pretty exposed part of Siding Spring Mountain and so is likely more affected by wind gusts. Initially there were issues with the mount being too flexible and having an lower resonant frequency than expected, which was in the range that can be excited by wind buffeting (a mention in the bio of Harry Minnett, a CSIRO engineer who worked on the AAT). This may have also lead to a more extensive windscreens but windscreens, particularly upper ones, are fairly common and exist at several other observatories. As an example, the pointing limits of the Isaac Newton Telescope page discusses the sky area that is available with and without lowering the upper windshield.


Does anybody know how to get Sun-Jupiter Lagrangian points L4 and L5 RA and Dec?

The L4 and L5 points are unstable orbits. The Trojans move in perturbed (ie slightly non-keplerian) orbits that move them around the L4 and L5 points. At any time you will find Trojans in a pair of lobes between about 30 and 90 degrees ahead of Jupiter.

You can find individual asteroids using NASA Horizons ephemeris
For example 624 Hektor, the largest Trojan (and placed in the Greek Camp around L4) will be found at RA 20 47 54.08, Dec -28 04 20.1 on Feb 10 2018. And 884 Piramus (in the Trojan Camp, L5) will be at RA 10 43 51.39 Dec +01 04 11.3.

(Jupiter is between them at RA 15 18 55.32 Dec -17 07 47.1)

Neither are easy to find, with a Magnitude of 15 or 16, due to their greater distance than main belt asteroids, and very dark colour. The image below shows the distribution of Trojans around Jupiter.


Why are so many sun-like stars found in multiple systems? How do planets form in such a situation?

You are correct - when we look at stars in the sky, about half of them are actually binary systems, with a range of separations. The frequency of binarity is strongly mass dependent; the 50% figure is appropriate for stars like the Sun, but it appears to be closer to 100% for high-mass O-stars and as small as 20% for the lowest mass M-dwarf stars.
There is also a change in the separation (and corresponding orbital period) distribution. The distributions of separations looks "log-normal" (like a Gaussian when you plot log separation on the x-axis). For solar type stars the peak is at around 50-100 au, but binaries become more compact (maybe 10-30 au) for low-mass stars. There is a tail of binaries with much smaller and larger separations.

These observational data are explained (or are in the process of being explained) as a combination of nature and nurture. The basic process of star formation results from the collapse and fragmentation of molecular clouds. If a collapsing fragment has significant angular momentum then the most energetically favourable way it can continue to collapse whilst conserving angular momentum is to sub-fragment into two pieces orbiting each other. This is thought to be the most basic binary formation process. Other processes include the formation of low mass companions through instabilities in cicumstellar disks or interactions between young stars in dense protostellar environments. Exactly how this leads to the mass dependence of the binary frequency and separation distribution is a topic of contemporary research; there is no definitive answer I can give.

After that I think there are two basic ways, or rather two basic locations, in which planets could form in a binary stem. Either around one or other of the stars or as circumbinary objects. The likelihood of each will depend how and when the binary formed, the binary component separation and mass ratio.

The planets would likely form in the same way that they do around single stars. If the binary star components are originally formed close together, there could be a circumbinary disk from which a stable planetary system formed. Both circumbinary disks and circumbinary planets have been observed.

Disks around individual components are truncated by tidal forces at a radius roughly corresponding to a third of the binary separation. Inside this radius it may be possible to form planets. Thus widely separated binaries could have planets orbiting the individual stars.


Could black holes be creators of dark matter?

Ultra-high energetic cosmic rays are thought to be caused by black holes as one option. Those energies should be sufficient for the formation of (hypothetical) heavy supersymmetric particles, which should decay to stable (hypothetical) neutralinos, candidates for (hypothetical) WIMPS.

Assuming this theoretical framework, black holes (or similar dense objects) may cause the generation of dark matter. But the amount of dark matter formed that way won't explain the amount of dark matter needed for the Lambda-CDM model. Also dark matter forming this way should be hot (fast moving). Lambda-CDM needs cold dark matter (CDM).


Will the 2016 Perseid meteor shower be visible in Sydney, Australia?

The International Meteor Organization calendar for 2016, page 12, gives the Perseid radiant declination as +58$^\circ$ and says "regrettably, it cannot be properly viewed from most of the southern hemisphere."
At latitude -34$^\circ$ the radiant never rises above the horizon, so there may be fewer Perseid meteors than sporadic non-Perseids.
This does not change year to year as the AG article seems to suggest.
What does vary is interference from the Moon; this year it's full on the 18th, so the article is correct to recommend the morning or two before the predicted peak on August 12-13.
Also check your local weather forecast.
The predawn hours, e.g. 2 hours before sunrise, are generally best for meteors because the morning side is forward as the Earth orbits the Sun.


What kind of triangle is formed by three unequal masses in a circular restricted three body orbit?

These questions are answered by the same references listed in my previous answer.

In the Newtonian limit, an equilateral 3-body solution exists for any combination of masses. (However, it is stable only if the inequality in the previous answer is satisfied.). This equilateral configuration orbits the center of mass, which is generally not in the center of the equilateral triangle. Each of the bodies follows a circular orbit with radius (see e.g. 1212.0754

$$ r_1 = a\sqrt{\nu_2^2+\nu_2\nu_3+\nu_3^2}$$
$$ r_2 = a\sqrt{\nu_1^2+\nu_1\nu_3+\nu_3^2}$$
$$ r_3 = a\sqrt{\nu_1^2+\nu_1\nu_2+\nu_2^2}$$

and frequency

$$\omega = \sqrt{M/a^3}, $$

where $a$ is the length of the sides of the equilateral triangle, $M$ the total mass, and $\nu_i = m_i/M$.

The situation changes when one accounts for relativisitic effects. When one accounts for the leading (post-Newtonian) corrections, then  a circular restricted three body solution still exists for general masses (with a smaller region of stability than in the Newtonian case.). However, the triangular configuration is no-longer circular (unless all three mass are equal or two masses are 0). Keeping the distances to the center of mass $r_i$ as in the Newtonian case, the sides of the triangle are now given by (again see  1212.0754)

$$ r_{ij} = a(1+\frac{M}{a}\epsilon_{ij}+ O(\tfrac{M^2}{a^2}) )$$

with

$$\epsilon_{12} = \frac{1}{24}[(\nu_1-\nu_3)(5-3\nu_2)+(\nu_2-\nu_3)(5-3\nu_1)]$$
$$\epsilon_{23} = \frac{1}{24}[(\nu_2-\nu_1)(5-3\nu_3)+(\nu_3-\nu_1)(5-3\nu_2)]$$
$$\epsilon_{31} = \frac{1}{24}[(\nu_3-\nu_2)(5-3\nu_1)+(\nu_1-\nu_2)(5-3\nu_3)].$$

The angles of the triangle can be calculated from these lengths if one is so inclined.


Why didn't we see those campfires on the Sun until now?

tl;dr they're a bit too small for SOHO and Stereo and only visible in EUV
The "campfires" are described as a few hundred km across,

The smallest of those campfires are about the size of a European country, according to Berghmans.

which means that from Earth (or SOHO or Stereo) they will subtend from one to a few seconds of arc (a few micro-radians). They are observed in extreme UV which is blocked by the atmosphere, so only space-based instruments could see them.
SOHO's imaging instrument the EIT has about a 5 arcsecond resolution.
Stereo seems to be comparable to SOHO (2000 pixels across the sun is about 500km/pixel at the centre of the disk).
Solar orbiter has a resolution of about 1 arc-second and is at half the distance, so has effectively 10 times the resolution of the other probes.
I don't know about baloon and sub-orbital instruments, but I could speculate that achieving sub-arcsecond resolution in EUV while not having an aperture so large that the incoming light fries the sensors might have been hard and expensive and required recent developments in materials and sensors coming from the semiconductor industry.


The text is trying to suggest that you do ra._degrees. How could the message be phrased better to make clear that the suggestion is to change the attribute name you are asking for from .degrees to ._degrees? Would a leading . in front of the name help?


Historical lunar occultation of Venus seen from Sicily in the winter?

If the Moon is 9 days past full, then it can't be occulting (or even at its closest approach to) Venus.  Since Venus is an interior planet, it never gets far from the Sun in the sky as seen from Earth; its maximum elongation (maximum angle between Sun and Venus) is 47°.  So to occult Venus, the Moon would also have to be that close to the Sun in the sky.

However, when the Moon is 9 days past full, it is still more than 70° from the Sun in the sky (verified with calculation of lunar position and phase from JPL Horizons; the "SOT" angle given there is "Sun-Observer-Target" and thus the Sun-Moon angle as seen from Earth). The Moon doesn't get within 47° of the Sun until about 11 days past full.

On a more literary note, the passage "nine changes of the watery star" (from The Winter's Tale) seems to be typically interpreted in commentaries to mean nine months have passed, i.e. each "change" is a full cycle of lunar phases.

EDIT:  If you want to search for historical occultations in general, it looks like the Windows software Occult 4 should be able to do it.  The PDF guide has some tips for setting it up.


Due to the atmospheric refraction the Sun rises earlier and sets later as you correctly wrote. Refraction however is perpendicular to the horizon. So if the apparent sun rise is earlier, than the true (actual) position of the Sun will be more to the north east (smaller azimuth) than when the true sun rise occurs. As refraction is perpendicular the apparent sun rise will also be more to the north east and hence the azimuth will have a lower value (88-89° instead of 90°).

NB. This is of course only true from the northern hemisphere.

NB2. The Sun is not to scale in the figure below.


Star like light moving in the sky, what could it be?

There was a object, apparently flying above you, that you couldn't identify. By definition this is an unidentified flying object. However this does not imply that it was an extra-terrestrial spacecraft.

UFO reports can be explained by a combination of:

Not recognising a known natural object, such as Venus, or unusual clouds.
Planes, drones, Chinese lanterns or satellites.
Optical illusion, dreams or the effects of drugs.
Deliberate falsification.

There is no real way to decide what it was you saw. There are a couple of points in your report that are odd. You describe it as the same size as stars, but less bright. Stars have no visible size,  and have many magnitudes of brightness. Your report suggests that you are someone who doesn't regularly observe the sky, and so is likely to be more susceptible to optical illusions and tricks that eyes can play on us.


Will Earth at some point stop rotating anti-clockwise, stay still, and then rotate clockwise?

No. This would require some force imposed externally. Momentum is a wonderful thing.


What part of the milky way do we see from earth?

Fraser Cain:

  We’re seeing the galaxy edge on, from the inside, and so we see the
  galactic disk as a band that forms a complete circle around the sky.

  Which parts you can see depend on your location on Earth and the time
  of year, but you can always see some part of the disk.

  The galactic core of the Milky Way is located in the constellation
  Sagittarius, which is located to the South of me in Canada, and only
  really visible during the Summer. In really faint skies, the Milky Way
  is clearly thicker and brighter in that region.

If you want to know more, watch this video from Fraser Cain which explains it in details:

https://www.youtube.com/watch?v=pdFWbEwsOmA

You can find your answer in the first minute


Why does the focal length of a telescope have an effect on the magnification?

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.


That equation is the Roche limit. It pertains to objects that are held together by self-gravitation only. No chemical bonds allowed! For example, an artificial satellite in low Earth orbit is almost inevitably going to be well within the Roche limit for that satellite. For example, the Roche limit for the ISS is almost halfway to the Moon. The ISS isn't being torn apart by tidal forces.

The kinds of bodies that this limit does apply to include rubble pile asteroids and comets. For example, comet 67P/Churyumov-Gerasimenko, currently being investigated by the Rosetta spacecraft, has some rubble pile characteristics to it. An even better example is comet Shoemaker–Levy 9, which was torn apart into more than 20 pieces before it collided into Jupiter.


The Sun will be at your zenith a few days after 21st March, after it crosses the equator heading northwards* going towards summer, and a few days before 21st September, before it crosses the equator heading southwards* after summer has ended.

*  from an Earth-centric viewpoint. We know, of course, that the Earth is rotating about its axis and revolving about the Sun, and that Earth's axial tilt is what causes the seasons and the apparent motion of the Sun across the sky both through the day and through the seasons.


What is the mass and momentum of a relativistic jet?

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

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


How to send information from the far side of the Moon to the Earth?

You need a satellite that has a view of the landing site and the Earth. There are several possibilities for doing this. According to the NASA Space Science Data Coordinated Archive, the communications are relayed through the satellite Queqiao which is in a halo orbit around the Earth-Moon L2 point.
You could also have a satellite in lunar orbit, transmit data from the lander while it's over the far side, transmit data to Earth while it's over the near side.
Or relay via a satellite in interplanetary space, etc.


Why are there so many seemingly blue-shifted galaxies in deep space

You cannot gauge the redshift of a galaxy by looking at a false colour image. The images taken through different filters are stacked and colourised to suit. You can say that the blue galaxies are indeed bluer than the red galaxies, but there is no absolute scale with which to judge redshift by eye.

Secondly, there is no detail in the NASA web page, but the ACS and WFC3 cameras have near infrared capabilities. So I would think that this image is a visual false-colour image of information that extends well redward of what the eye can perceive. So even the things that look blue might have a spectrum that peaks at redder wavelengths, whilst anything that looks red might actually be infrared!

However, beyond this, in order to judge what the redshifted appearance of a galaxy would be, you would need to know what the galaxy looked like with zero redshift. That is, light that is emitted in the ultraviolet could be redshifted into the visible part of the spectrum. It is entirely possible that visual colour of a redshifted galaxy would not change very much at all if the galaxy emitted lots of UV light in its frame of reference or equally, if it emitted lots of very red light that was then redshifted out of the telescope's sensitivity range.

Finally, there could be some genuine astrophysics going on. Many distant galaxies are bluer than nearby galaxies because they are undergoing intense star formation. Massive star forming regions emit copious UV light that is redshifted into the optical.

Finally, finally! Many of the galaxies in the picture will be quite close and will not be very redshifted.


How are the ESPRESSO Echelle spectrograph's calibration "lines" produced?

You're right to think that this is a calibration technique. In use here are laser combs that use self-interference of short pulses that are reflected multiple times inside their own cavity to generate the spacing that you see.

The spacing then is mainly limited by the inverse of the pulse length, and thus by current laser technology. Sharpness of individual lines is usually increased by putting a Fabry-Pérot interferometer in sequence after the laser. For additional precision and estimation of laser stability, one uses often some secondary lamp with theoretically or laboratorially well-known lines (Th-Ar for example).

The technology demonstration for ESPRESSO was done with its precursor, HARPS.
Also A site with more details, the lowest poster looks pretty good in explaining the basics, I find.


Are there recent estimates of Kuiper belt mass?

Surprisingly, I also have not been able to find any generally recent (5 years or less ago) information about this topic.

I did find a research paper from 2007 (Here). You may want to check out page 11 in Section 4.3. It states that the maximum mass of the Kuiper Belt is 0.1 Earth masses. They also talk about how it was estimated and such.

I haven't been able to find anything else besides Wikipedia articles (which you probably aren't looking for).

Very strange how there hasn't been a lot of research on this. An intriguing topic, I must say.

That's all I could find... Hope that helps...?


Why do Black Holes in the middle of galaxies not suck up the whole galaxy?

You shouldn't think of black holes as "sucking things in". Black holes interact with matter through gravity, just the same as any other object. Think of our Solar System. All the planets orbit around the sun because it has a lot of mass. Since the planets have some lateral motion (they're not moving directly towards or away from the sun), they circle around it. This is known as conservation of angular momentum.

When talking about gravity, all that matters is the mass of the objects involved. It doesn't really matter what kind of object it is*. If you were to replace the sun with a black hole that had the same mass as our sun, the planets would continue on the exact same orbits as before.

Now, the black holes at the centers of most spiral galaxies do accumulate mass. Some of these black holes have accretion disks around them. These are swirling disks of gas and dust that is slowly falling into the black hole. These gas and dust particles lose their angular momentum through interactions with gas and dust nearby and by radiating energy as heat. Some of these black holes have very large accretion disks, and can generate huge amounts of electromagnetic radiation. These are known as active galactic nuclei.

So, long story short, black holes don't "suck". They just interact with things gravitationally. Stars, gas, and other matter in the galaxy has angular momentum, so it stays in orbit around the center of the galaxy. It doesn't just fall straight in. This is the same reason the Earth orbits around the Sun.

*Disclaimer: When you talk about things like tidal forces, you do need to take into account the size of the objects. But for orbital mechanics, we don't need to worry about it because the distances between the objects are generally much larger than the objects themselves.


Do the "schwarzschild metric" or the "isotropic metric" fit the Event Horizon telescope data on Messier 87 better?

One of the things that can be confusing for beginners to GR is that there is no physical significance to the choice of coordinates we make. For example when studying static black holes we can use Schwarzschild coordinates, isotropic coordinates, Gullstrand-Painlevé coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates and lots of others that I can't think of offhand.

But regardless of what coordinates we choose it's the same metric, just written in different ways. You ask about the Schwarzschild metric and the isotropic metric as if they were different things, but it's the same metric just written in different coordinates. It would be better to call them the static black hole metric written in Schwarzschild coordinates and the static black hole metric written in isotropic coordinates but, well, life is too short and even experienced physicists will just say Schwarzschild metric and isotropic metric to save time.

The point is that the coordinates are not directly observable. We choose some coordinate system and then do a calculation to compute something that is observable e.g. the trajectories of photons. And the results will come out the same. After all the photons don't care what coordinate system a physicist chooses to compute their paths.

The reason we have all these different coordinates is that it's often the case that some calculations are easier in some coordinates than others. For example the Schwarzschild coordinates are the closest to our intuitive grasp of the geometry around the black hole. If we were considering simple questions like the time dilation for an observer hovering near the black hole I'd use Schwarzschild coordinates to do this because they make the calculation easy. But Schwarzschild coordinates don't work well at the event horizon. If I was explaining why light can't escape from a black hole I'd use Gullstrand-Painlevé coordinates.

To return to your question, I'm not familiar with the method the JPL team use for computing trajectories in the Solar System so I can't comment on why the calculations would be easier in the isotropic coordinates than in the Schwarzschild coordinates. Presumably the computational method they use is easier in isotropic coordinates, but the observables that they calculate will come out the same regardless of which coordinates are used.

In any case neither coordinates are useful for the EHT calculations because Messier 87* is rotating not static and the spacetime geometry around it is described by the Kerr metric not the Schwarzschild metric. The Kerr metric can also be written in lots of different coordinate systems - I don't know which coordinate system the EHT team used.


Why is the "green" comet C/2014 Q2 (Lovejoy) red in this picture?

Mainzer et al. 2014 characterize the performance of the reactivated NEOWISE.
Having run out of cryogenic coolant for the original WISE's 12 and 22 μm bands, it only detects in the 3.4 and 4.6 μm bands.
The comet looks red in the false-color infrared image because, relative to the 4.6 μm "W2" band, it emits less in the 3.4 μm "W1" band than stars do.
Figure 3 of the paper shows a similar image in which an asteroid also looks red.

Section 4.1 discusses comets, attributing the 4.6 μm band detection to a combination of reflected sunlight and thermal radiation from dust, 4.26 μm emission from CO2, and a little 4.67 μm emission from CO.
They estimate how much is due to gas vs. dust by comparison with ground-based observations near the same times, as in Figure 17.


How can I check if we can observe a certain star (given its equatorial coordinates) from a certain location on Earth (given its DMS coordinates)?

You can check whether the star is anti-circumpolar i.e. if it never rises above the horizon (if its upper culmination altitude is less than zero). If it is, you will never see it from that location. If it is not, at some point during the year you will be able to see it in conditions with minimal light pollution if its upper culmination altitude is around zero (and, of course, the conditions do not have to be as good if the upper culmination altitude is greater, so you may want to increase the upper culmination bound by a few degrees to make sure you can actually see the star).
A way to justify being able to observe a star whose upper culmination altitude is big enough is to state that the Sun goes along the ecliptic across a fixed celestial sphere during the year, and at some point must be on the other side of the celestial sphere compared to the star. That day, when the Sun sets, the star will be visible.
Sirius, in particular, can be seen from Brussels tonight at around 2 AM:


Do neutrinos have as much information as photons do?

This is a very broad question and though a comprehensive answer lies outside the scope of this simple Q&A format, I give you a couple of examples where "neutrino telescopes" would revolutionary.

There is a predicted cosmic neutrino background, analogous to the cosmic microwave background. Neutrinos decoupled at 1 second after the big bang, filling the universe with neutrinos that should have now cooled to a temperature of about 1.9 K. Confirming this would be a(nother) spectacular validation of the big bang model. However, there is a wrinkle; now that we know neutrinos have mass, it turns out that these neutrinos are non-relativistic at the present epoch. That means they are capable of being diverted and concentrated by gravitational structures - potentially making them fantastic probes of such structure and perhaps more sensitive than photons. So neutrino telescopes would make a fantastic contribution to our understanding of cosmology.
Neutrinos arise from energetic processes in the cores of stars. These processes are otherwise invisible, we certainly can't see them directly with light and can only probe them indirectly using techniques such as asteroseismology or looking at the mixing of chemical elements from the core to the surface. Neutrinos potentially tell us much more; for instance giving a direct estimate of the number of nuclear reactions taking place per second. Neutrino emission is also the predominant way that cooling takes place in supernovae, and in hot neutron stars and white dwarfs. A neutrino telescope could therefore be fundamental  for a deep (literally) understanding of stellar evolution and particularly the late stages of stellar evolution.

In terms of what can be measured; you can measure the flux of neutrinos and the neutrino energy spectrum. There is also of course the timing information that you might get from transient events like supernovae. There is also the question of flavour oscillations and the mix of neutrino types that are observed. I don't understand enough about this to make any predictions about how astronomically useful this property is - but I bet it will be.


What is the meaning tiny bundle of energy smaller than an atom

What the video is clumsily alluding to is that the universe was very much smaller than it is today. However, if the universe is infinite, then it was still infinite at all times in the past. All one can say is that the density was much higher.

Sometimes commentators are more precise. One can define a size scale for the currently observable universe and ask how that volume of space evolved with time. This of course assumes an accurate knowledge of cosmology and the relevant cosmological parameters.

A ball park number (see here) is that the universe has got bigger by a factor of $10^{30}$ since the end of the inflationary epoch (at $t \sim 10^{-32}$ s; which is not fully understood yet). Thus the current observable universe, of size $10^{27}$ m, was contained within a diameter of 1 mm at the end of the inflationary epoch. Prior to the inflationary epoch it must have been compacted into a much smaller volume than this - and I think this is the origin of the statement in your question.

The reason that your source talks about energy is because that is in what form the density of the universe was at that epoch. It was only later that this energy started to "freeze out" into various forms of matter. It is now commonly supposed that the Higgs mechanism or some form of it, is what gives elementary particles the property of a rest mass that might serve as a definition of "matter". The Higgs mechanism kicks in at the epoch of electroweak symmetry breaking (at temperatures of about $10^{15}$ K), giving paticles mass. Prior to this the density of the universe was in the form of massless entities that you might refer to as "energy". This symmetry breaking occurs after the end of the inflationary epoch at $t \sim 10^{-12}$s, and therefore when the currently observable universe was larger than 1 mm. So prior to this, and certainly when the currently observable universe was "smaller than an atom", the contents of the universe could be described in term of energy (without rest mass).


No, the sun and all other stars do not have the same composition and density within them.

The composition varies with depth. Most stars overall composition reflects the interstellar medium from which they form. Convection, settling and nuclear processes then lead to layers of different elements at differing depths.

Fusion occurs at the core of our sun (where pressure and temperature are greatest) therefore there is a greater abundance of Helium at this depth (through conversion of hydrogen). Other stars that have entered the red giant phase will also have many more heavier elements. Very late stage stars are thought to fuse these elements progressively, leading to layers of decreasing atomic number elements stacked on top of each other.

Convection can also lead to mixing of elements between layers. The diagram below shows theoretically how these convective layers differ over a range of stellar masses

The density increases as you move closer to the center of the star. Note how the density is less than that of water for up to half the sun's radius!


Which solar system objects have the least stable solar orbits?

The least stable orbits are likely the temporarily captured orbiters of Earth and other planets. These are bodies that have been captured by the Earth/Moon gravity well and move from solar orbits to terrestrial orbits. They normally have complex orbits that take them well beyond the orbit of the Moon and most don't last long.
Most such objects are small, typically less than 1 m in diameter. There are probably about 1000 such bodies with sizes of 10cm and up orbiting the Earth. Normally, they are undetected but occasionally a larger object will be discovered.  Most recently, 2020 CD3 was captured in the winter of 2015-16 and escaped Earth's gravity in May 2020.
The orbits of such temporary orbiters are not stable, they normally escape the Earth/Moon gravitational well after a few years, and return to heliocentric orbits.


Why does time pass faster in moon than the earth?

Okay, this question is actually quite simple and requires some background on the relations between gravity and time.
The property that can be used to answer this question is :

Time is affected by gravity in the following way : Imagine you are on an exoplanet. The gravity there is twice that gravity here on Earth. Now, assuming that we take Earth Time as standard, the Exoplanet's time should be slower than Earth-Time.

Basically, the property states that the greater the gravity of the place, the greater will be the time dilation effect.
Here, for your question, we can apply the same principle. Since Moon-Gravity is 1/6th of Earth's, time should pass faster there.
@MarkAdler has given an excellent answer here, on the Stack Exchange for Space Exploration. To quote his answer :

Time passes about 0.66 parts per billion faster on the Moon than on Earth, due to not being in as strong a gravity field. The dilation, including all the effects of being further from the Earth's gravity field, being in the Moon's gravity field, and the velocity of the Moon, could easily be measured in sum by comparing good atomic clocks on the Earth and the Moon.

He has also posted some maths on the same, so I recommend checking out the link above :
Hope this answer helps,


As an example, look at Earth's only confirmed Trojan

By Phoenix7777 - Own work Data source: HORIZONS System, JPL, NASA, CC BY-SA 4.0, source
Now, to understand what is happening here. The yellow dot is the sun. The blue dot is the Earth.  Although the Earth is orbiting the sun, the "camera" is turning so that it appears that the Earth is roughly stationary (it wobbles slightly due to the eccentricty of the earth's orbit)
The pink dot is the asteroid 2010 TK7.  It has an orbit that is eccentric, and so sometimes it is much closer to the sun than the Earth, and at other times it is further, but its orbit takes nearly exactly one year.  However the exact shape of the orbit changes, sometimes it moves around the sun in slightly less than a year, so it starts to catch up with the Earth, but as it nears the Earth, the Earth's gravity tends to pull it forward, and out, causing it to slow down and move away from the earth.  The whole cycle takes hundreds of years.
Note, the actual shape of the orbits are elliptical about the sun, the odd shapes is a result of the camera turning at one revolution per year.
This kind of orbit is said to librate about the L4 point. It is called a "tadpole orbit" Trojans don't have to remain exactly at the L4 point, they can stably orbit in a tadpole orbit around the L4 point.


What does the opacity of a molecular transition mean?

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?


Can't pulsars and stars be used for gravitational wave measurement?

How LIGO, LISA, etc. Detect Gravitational Waves

The point of instruments like LIGO and LISA is to measure time-varying changes in the distance within different arms of the instrument. In the case of an arm oriented in the direction of an incoming gravitational wave (GW), the length of the arm will increase and  decrease, while an arm oriented perpendicular will remain unchanged.

The method for measuring this is to split a laser beam, send each half down the different arms, and then recombine them and look at the interference pattern. If the arms have identical lengths, then the beams will destructively interfere. If, however, one of the arms becomes longer (or shorter), the recombined beams will be out of phase, and you will no longer have perfect destructive interference.

The key to being able to measure this sort of thing accurately is two things:

Arm length, so the beam traveling down the altered-length path has time
to accumulate a large enough phase shift for the interference pattern to
change enough to be measured. Longer arms lead to larger phase shifts,
which is why LISA aims to have arms that are 2.5 million km in length.
The wavelength of the light: the shorter the wavelength, the smaller the
actual shift in arm length needs to be to make the beam shift appreciably
in phase. If the recombined beams differ in phase by 0.001%, then the change
in the interference pattern will be very hard to detect and measure. But if they differ
in phase by 50%, then the difference will be obvious. Shorter wavelengths
make for larger phase shifts.

If you want think of the interferometric measurement as a kind of timing measurement (that is, we are, in a sense, measuring the difference in time it takes light to travel down the two arms), then optical light is vastly better than radio for "timing" measurements.

Pulsar Timing

As you pointed out, there are projects that use pulsars to try to detect gravitational waves (e.g., pulsar timing arrays).

These work by measuring time delays in the arrival of radio pulses from millisecond pulsars. However, it's important to note the limitations: in order to measure the arrival time accurately, you have to observe a pulsar over a period of several minutes (or more) and add up all the pulse arrivals to get a single, accurate value. (You are not measuring changes between the arrival of one single pulse and the next.) You then repeat this measurement several weeks or months later to get another arrival time. Since you're taking measurements weeks or months apart, you can only detect variations on that same time scale, which is why pulsar timing arrays are hoping to detect GWs with periods of months to years (e.g., from binary supermassive black holes). The kinds of GWs that LIGO detects -- with periods of fractions of a second -- are completely beyond what pulsar timing arrays can do.

We cannot use optical light from stars, because we know of no star that produces the exquisitely regular periodicity of a millisecond pulsar.

As a final comment, your scenario of photons that "arrive regularly at 15 photons every microsecond onto a CCD" has the problem that Poisson noise (a.k.a. "shot noise") means that you wouldn't actually get 15 photons every microsecond; you'd get around 15 every microsecond (a typical sequence might look like this: 16, 15, 18, 15, 14, 14, 15, 23, 16, 12, ...). With that much variation, it's really hard to detect subtle variations in intrinsic signal strength.


Do Noctilucent clouds cause problems with telescopes?

Noctilucent clouds are not a problem for space telescopes because their orbits are always more than 85 km. The Hubble Telescope orbits at about 570 km.  Noctilucent clouds are a problem for ground-based telescopes (although only at high latitude sites,> 50$^{\circ}$), especially if you are trying to get accurate photometric brightnesses.  However, typically you can see them reflecting city lights or moonlight, or you can tell from the large variance in your photometry counts that they are there and you should try again another night.


Can some stars not emit any energy in the visible spectrum?

Light that is not light

That's meaningless.  All light is electromagnetic radiation.  A finite part of the infinitely large range of the electromagnetic spectrum is visible light.  So you should talk about EM radiation and to discuss the visible spectrum just say visible spectrum.

Stars emit lots of energy at frequencies that are outside the visible range.

  Dark Matter

Dark matter does not mean that it's just not radiating in the visible part of the spectrum.  It means it's not radiating in any part of the EM spectrum much.  If dark matter exists (and that's not certain) it won't interact with most matter much at all.

If dark matter were in star-like objects, it sounds very unlikely that there would not be any EM radiation as a result of the interactions that would result in a star-like object.  We'd detect it if it were there.  So I don't think that's likely.


Which approximations are required for the synodic period formula?

The formula :-

\begin{equation}\tag{A}
\frac{1}{T_{syn}} = \frac{1}{T_{1}} - \frac{1}{T_{2}}
\end{equation}

for the synodic period between two planets orbiting the Sun is derived in a similar way to the derivation of the Moon's synodic period  :-

\begin{equation}\tag{B}
\frac{1}{T_{syn}} = \frac{1}{T_{sid}} - \frac{1}{T_{e}}
\end{equation}

where $T_{sid}$ is the sidereal perod of the Moon and $T_{e}$ is the sidereal period of the Earth.

Formula (B) is derived by making the following simplifying assumptions about the motions of the Sun, Earth and Moon :-

Earth has a circular orbit around the Sun of constant speed and a fixed period $T_{e}$
Moon has a circular orbit around the Earth of constant speed and a fixed period $T_{sid}$. (Thus the moon's path wrt the fixed Sun frame is an epicycle, ie a small circle whose center moves along the circumference of a larger circle, in this case the Earth's orbit).
The above two orbits are confined to a common plane (the Ecliptic Plane)
From the direction we are viewing (ie `North') both Earth and Moon orbit in the anti-clockwise direction

and (A) is derived with similar assumptions about the planetary orbits.

(See this answer for the derivation of formula (B) for the Moon - for the planets the derivation of (A) would be virtually the same).

In reality the orbits are not perfectly circular and they are not in the exact same plane. The circular orbit approximation means the speed of the planet is constant, but in the case of an elliptical orbit the speed varies.


Why do scientists assume they can measure the shape of the universe if it is also widely believed to be infinite?

Yes, the universe is believed to be infinite in size. That's what you get if the curvature is zero or negative, assuming a simple topology. The curvature has to be positive for a finite universe, once again, assuming a simple topology, and no weird stuff like edges.

Now it's possible that the universe has a very small positive curvature, so that it's finite, but it's so large that it looks flat to us.

However, it's reasonable to assume that the observable universe is representative of the whole thing, and not just coincidentally a region of anomalous curvature. Of course, that's impossible to verify, but if the curvature were significantly greater outside our observable patch we'd expect the curvature to be a bit higher near the edges than it is in the middle of the patch, and we don't see that in the data.

But if the global curvature equals the maximum positive curvature consistent with the 0.4% margin of error of the WMAP, BOOMERanG, and Planck data, then the radius of curvature of the whole universe is (currently) around 150 times larger than the radius of the observable universe. And of course, in the future it will continue to expand.

That figure comes from How Big is the Entire Universe? by astrophysicist Ethan Siegel. That article has a great explanation of curvature, with lots of helpful diagrams.


Recently, the most distant object in the solar system was discovered. How was this accomplished?

The discovery announcement by the Minor Planet Center indicates that 2018 AG 37 ("Farfarout") was between 25.0 and 25.3 magnitude. The positions are based on observations taken with the 8 m Subaru telescope and 6.5 m Magellan-Baade telescope.
As a comparison, this ancient article by Sky & Telescope (What Is The Faintest Object Imaged By Ground-Based Telescopes?) indicates the Subaru telescope limit is around 27.7. ("Ancient" in terms of CCD cameras and computer processing. The article is from 2006.) Note that this same article indicates that a 16 inch (400 mm) amateur telescope should be able to capture images to 24th magnitude -- just 2x to 3x brighter than Farfarout!
Also, the Subaru Exposure Calculator indicates an exposure of 1 minute will capture a 25.3 magnitude object using default input (except for the Moon, which I set to New Moon).

Thus, it seems almost "easy" to capture an image of it.
Finding an unknown, faint object is another issue. I do not have any information on this aspect of the question, but I feel safe that computer searches are quite adept at finding such things. Once the first image (or pair of images that show motion) is found, other images are often found or taken to facility the calculation of the orbit. The time span of the images is only two (2) years which is short for a body with a period of 1000 years (estimate). The Wikipedia page for 2018 AG37 currently lists large uncertainties for the average distance (86 +/-68 AU) and period (798 +/-941 years). Those uncertainties will be reduced as a longer arc of the orbit is measured.


Are quasars simply AGNs that are viewed from a particular angle?

There is total confusion in the answers you refer to.
Quasars were first defined because of their star-like appearance - quasi-stellar objects or QSOs.
The fact that they are Galaxy-sized objects with point-like appearances means they are very luminous, dominated by central nuclear emission and far away (as confirmd by their high redshifts).
It was later recognised that quasars are just the high-luminosity end of the AGN zoo.
Quasars can also be viewed at a variety of angles and have distinct properties as a result. So there are radio-quiet and radio-loud quasars, broad absorption line (BAL) quasars and optically violently variable (OVV) quasars. The distinction between these types is thought to be dominated by viewing angle, mass of the central black hole and the accretion rate. The nomenclature can also vary depending on what wavelength region you are talking about (radio, optical, X-ray etc.).
These objects have lower luminosity counterparts amongst AGNs - FR I and FR II AGNs, Seyfert I and IIs and blazars (or BL Lac objects) - again, whose differences are largely explained by viewing angle.
I don't think these days it is helpful to think of quasars and AGNs as different objects. They are actually similar and it is just a question of the scale of the accretion process going on at their centres that is different and covers a continuum rather than a bimodal distribution.


Globular clusters vary in size quite a bit. They contain from a few hundred thousand to over a million stars. Probably best way to determine size is to measure its extent across the sky by telescope and then calculate that distance in km. One could also model the size by assuming the number of stars, the average size of the stars, and the spacing based on this size.  One could build a globular cluster of 300,000 stars or so and determine the approximate size.

You ask if 120-150 ly is the average diameter of such globular clusters. I'm confused by exactly what you mean but you have sampled a small range of sizes considering globular clusters can be well over 300 ly in diameter.

From Wikipedia, and based on comment from @AtmosphericPrisonEscape, you have "Almost all globular clusters have a half-light radius of less than 10 pc, although there are well-established globular clusters with very large radii (i.e. NGC 2419 (Rh = 18 pc) and Palomar 14 (Rh = 25 pc)).[64]"


Resource Recommendations to systematically develop a foundation via Self study in Astrophysics/Astronomy?

High School Senior

Your first mission is to concentrate on getting a good set of results from high school.  That really means not spending your time chasing astronomy and astrophysics yet.
Yes, this is what your parents and teachers would say, but you know, if you ever want to have the time, resources and maybe the chance to study and/or work in these fields you absolutely are going to need the very best high school results you can manage.  And even if you don't weant to study a B.Sc. or higher then you still need the resources and that means a decent job and career prospects.  Money is important, or as the saying goes "No bucks, no Buck Rogers”.
So focus on that first.

a systematic order

For self study this is not needed.  I'd suggest not being systematic outside of a specific goal, like having to pass exams set by other people.
You need to differentiate between Astronomy and Astrophyics.
Astronomy is a kind of "catch-all" term for every possible thing you could study or observe about anything that goes on  outside of our atsmosphere and planet.  You can include geology, biology, chemistry, physics and just about anything else that can be termed a science, as well as engineering (you have to build stuff to do stuff - rocket science and telescopes of many types, etc.).  Astrophysics is then a subset of Astronomy focusing on the application of physics to astronomy.
Now you want graduate level or graduate level "foundation" as a target.  That means you want a B.Sc. level course.  As you want Astrophysics you basically need a B.Sc. level course in physics.  That's a lot of study.  You'll cover topics like:

electromagnetism
basic quantum theory
basic special and general relativity
basic thermodynamics
mechanics
optics

Keep in mind a systematic B.Sc. course on basic physics for astrophysics typically takes four full years.  You have to learn to think about physics problems and get a feel for how things are done.  That's arguably more important than the actual details of any particular theory.
Don't be distracted by glamor subjects like quantum theory and relativity.  Mechanics, Electromagnetism and Thermodynamics are very, very important.
So let's focus on astrophysics and understand that you have to start with physics and what resources will help.
Libraries
Great things, not loved or used enough.  One set I personally really useful was the Schaum Outline series as a launching point and if you can get anything from your library, borrow them.  You don't need to swallow books whole - read what you need and borrow again when you need more.  There are lots of books out there and for physics and we all have favorites (and ones we'd avoid ever reading again).  Boring old textbooks like e.g. Halliday and Resnick are as good a place to start as any.  My first exposure to quantum theory proper was by a book borrowed from my local library.
Videos
There are multiple courses on physics online.  A couple to consider :

Yale physics courses online
Leonard Susskind physics lectures

There's also excellent YouTube channels like :

PBS Spacetime
FermiLab
Sabine Hossenfelder

By no means the only ones.