The article references a paper by Huo et al. (2004; that's the reference number "33" in your quoted text). Huo et al. merely list a redshift of "0.0103" for Arp 299 in their Table 1 (without any apparent sources); combined with the fact that the authors of the Science paper say what Hubble constant they assumed (73 km/s/Mpc), this tells us that the distance is redshift-based. That is, they used the redshift plus the Hubble constant to derive the (comoving) distance.

If you look up Arp 299 in the NASA Extragalactic Database and scroll down the resulting page, you can see a list of possible redshifts ("Calculated and Corrected Velocities") for this galaxy, ranging from 3088 km/s (which is in fact $z = 0.0103$) to 3511 km/s. Immediately below that, you can see the derived distances ("Hubble Flow Distance"), assuming $H_0 = 73$, the same value the authors say they used. The closest number to their value is the 44.4 Mpc, assuming "3K CMB" correction (which seems to match the authors' description). Why this isn't exactly the same as the authors' value of 44.8 Mpc I don't know; possible a slightly different value of the redshift was used.

At this relatively small redshift, the differences between the different distances that pela mentions are pretty minor. E.g., plugging $z = 0.0108$ (NED's "3K CMB" corrected redshift) and $H_0 = 73$ into Ned Wright's cosmology calculator gives you a comoving distance of 44.2 Mpc, a luminosity distance of 44.7 Mpc, and an angular-diamter distance of 43.8 Mpc. The differences between these are all smaller than any plausible uncertainty based on using the redshift.[*]

So, to summarize, the distance in the paper is (as pela says) the comoving distance; the method used to derive it was the redshift of the galaxy (corrected to a CMB-relative value) plus a Hubble constant of 73 km/s/kpc.

[*] E.g., if we assume a redshift uncertainty of 300 km/s, based on possible peculiar motion of Arp 299, then the distances are uncertain by +/- 4 Mpc.


A newly formed neutron star cools (via neutrino emission) to an effectively "zero-temperature" configuration in a matter of a few minutes.

What is meant by that is that it contracts to a radius where the separation between its neutrons is so small that neutron degeneracy pressure and the pressure associated with mutual repulsion between neutrons (from the strong nuclear force) becomes large enough to support the star, and since neither of these depend on temperature, the star can continue to cool off without any change in radius. In other words, the radius it has when it is at an interior temperature of $10^{10}$K (and a surface temperature of about $10^{8}$K), a few minutes after formation, and the radius it has for any cooler temperature are essentially the same. Hence neutron stars do not contract, become denser and collapse into black holes as they cool.

A neutron star cools for the first million years or so by various neutrino emission mechanisms in the interior. Thereafter, at a surface temperature of about 100,000 degrees, the dominant cooling is radiation from the surface, which then proceeds on timescales of only millions of years. The neutron star is then thought to get colder and colder (NB no old, cold neutron stars have ever been observed, but they are so small and so faint at low temperatures that they could never be found with current technology, but there should be about a billion of them out there in our Galaxy).

The future of the neutron star is therefore to be a fading cinder. The vast majority of the billion or so neutron stars thought to exist in our galaxy are probably at temperatures of around 1000-10,000 degrees, and probably maintained at that temperature by the accretion of small amounts of gas from the interstellar medium. They may also be reheated by Ohmic dissipation of their magnetic fields or viscous dissipation of their rotational kinetic energy.

Now, it is possible, that if a neutron star accreted enough material, then at some point in the future it could exceed the maximum possible mass for a neutron star and become a black hole. But most neutron stars appear to be born at masses well below this limit and it would likely take many galaxy lifetimes to accrete enough material for them to make that transition. Neutron stars born into binary systems may experience more rapid accretion. This accretion is observed (in X-rays) and could result in black hole formation on shorter timescales.


In what timeframe is W26 expected to go supernova?

There is little peer-reviewed information that gives a definitive time frame of when W26 would go supernova.  The reason for this is that we have models of stellar lifecycles, and have found candidates at each 'age'.

With that in mind, according to a very recent article: The Ionized Nebula surrounding the Red Supergiant W26
in Westerlund 1 (Wright, 2013),  (RSG = Red Super Giant)

  The presence of the nebula suggests extensive mass loss
  in the recent history of W26. Its late spectral type,
  very high luminosity and spectral variability all suggest the
  star to be highly evolved amongst the RSGs. Both the star and the nebula are comparable to the RSGs VY CMa and
  WOH G64, both of which are highly luminous late-type
  RSGs with evidence for circumstellar gas.
  W26 provides a rare opportunity to directly investigate
  an extreme mass-loss event from a highly evolved RSG.

Looking at the compared stars to W26 to see what theories, even timeframes are suggested:

According to the article Fundamental properties and atmospheric structure of the red
supergiant VY CMa based on VLTI/AMBER spectro-interferometry (Wittkowski et al. 2012), the supergiant star VY CMa is

  close to the Hayashi limit of recent evolutionary
  tracks of initial mass 25 M⊙ with rotation or 32 M⊙ without rotation, shortly before evolving blueward in the HR-diagram.

So, according to Wittkowski et al. rather than being close to supernova, could well be close to entering the next phase of stellar evolution.

According to the article Rd Supergiants in the Local Group (Levesque 2013) and Spatially resolved dusty torus toward the red supergiant WOH G64
in the Large Magellanic Cloud (Ohnaka et al. 2008), studies of WOH G64

  implies that this object may be experiencing unstable, violent mass loss.

TL:DR So, based on observations of W26 and comparable stars, there is no definitive timeframe, primarily due to these stars being close to the Hayashi forbidden region, which, according to the article Late-Type Red Supergiants: Too Cool for the Magellanic Clouds? (Levesque et al. 2007), results in them being

  unstable hydrodynamically, which we
  expect to lead to this variability and behavior.

The Hayashi track/forbidden zones in relation to stellar masses and main sequence is shown below:

Image source

to which Levesque et al. state

  Further monitoring of these stars, both
  photometrically and spectroscopically, may lead to an improved understanding of this phase
  of massive star evolution.

Suggesting that this behaviour may be a phase (albeit one of the final phases) in their evolution.  Also, at that distance, it is likely to put on a small light show, but not pose much danger to the Earth (except in the very unlikely event of a gamma ray burst).


Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument.

Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec.

Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec.

The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but actual distance varies all the time).

Let's apply the tangent of 0.04 arcsec at 55 mil km:

https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000

It's 10 km. It would only be able to see the major geographic features.

To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are placed several km apart and are coupled optically to function as a single huge mirror (well, sort of - this is more of an intuitive explanation).

It would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona.

Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work.

All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings.

Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the planetary surface, the ground provides the required rigidity. In space you'd have to... I dunno, use space magic.


Doppler Spectroscopy - Finding mass of an exoplanet

To use the Doppler shift, you need to know by how much the light is blue/red-shifted.

Suppose you have a light curve where the wavelength varies between $\lambda_1$ and $\lambda_2$. Using these wavelengths, the Doppler shift will give you an interval of radial velocities. This gives you a value for K that you can plug into your last equation to derive $M_p$.


What will the last days of Earth be like?

A lot of the answer depends on Orbital mechanics and calculating that isn't easy.   If no action is taken, Earth's orbit is still likely to change, but how much and in which direction is hard to say.

Long before the sun goes red giant it will get too hot for the Earth.  If the earth doesn't move, the sun's increased temperature will likely make the earth uncomfortable in a few hundred million years and perhaps scorching in as little as a billion years.   Warmer air and more water vapor in the air will trap heat and at a certain point, you'd get a run-away green house effect, long before the red-giant phase.    Short term, changes in CO2 in the atmosphere might slow that.   We could see ice ages before we see run-away heating, but over time, with no human action, run-away heating is likely inevitable - on a very large timescale, like a billion years or so.   Longer if the Earth drifts farther from the sun.

  what will the last few decades of Earth's life be like? Will the
  planet be heated to the point where its surface is like Mercury's
  sun-side? Will the sun's output drop low enough that it instead cools,
  like Mars? Will the sun expand and engulf the planet, or will the
  planet continue to orbit an expanded sun as a lifeless ball?

  Will the sun's expansion be gradual or rapid? How will this expansion
  affect the rocky planets nearest to it?

I'm going to answer the 2nd part first.   The sun's expansion will first be gradual, but the red giant stage will be fairly rapid.   Over the next 4 billion years or so, the sun will get some 60% more luminous, so significantly much more heat at the same orbit.   A very rough rule of thumb is 10% more luminous every billion years. (that's not for every star, just for 1 solar mass) - and that's a pretty ballpark estimate.

But just 10% more heat from the sun would raise the Earth's temperature uncomfortably, and with the added water vapor in the air and greenhouse effect, the earth might be a permanent steam room, even at the poles, at just 10% more heat from the sun.

asking about the last few decades is the wrong time frame as changes would probably happen much more slowly, though atmospheric/feedback mechanisms in the climate might make changes happen more quickly - ice age speed for example, a few thousand years or so.   Also, as Antarctica drifts away from the south pole, we might see an end to ice ages - there's lots of moving parts to long term climate predictions, but in general, you'd see gradual heating to the point where complex life might have a hard time adapting.

The suns output won't drop until after it goes red giant, it's on a steady but gradual increase, followed by a relatively fast increase at the red giant stage.

Now, if we terraform Titan, that might be a safe distance to watch the sun go red giant from and it would probably be quite a spectacular show, well, over enough time.

As to how fast it will go red-giant, fast for a star means, about 200 million years, so, from the perspective of a human lifetime, watching a star go red giant would be like watching grass grow.

A much more intelligent discussion on that than I can give, here:   https://physics.stackexchange.com/questions/25622/how-fast-will-the-sun-become-a-red-giant


Why isn't this so?

If the mass of the star is about 1/4 the mass of our sun, then the core's (where fusion occurs) temperature and pressure will never be enough to fuse anything other than hydrogen into helium. The death of this star may be something like a fire dying. The flames disappear for a while, then you may get one or 2 popping back into existence.

Our sun will start to fuse helium into carbon before all the hydrogen in the core is used. In the description of a heavy star, it is described as a onion with its layers with the higher fusion products occurring in the lower layers. They use this model since the iron fusion requires temperature and pressures above a certain level. Neon has lower requirements, but still greater than carbon fusion, and so on. What is not mentioned, is that there is still hydrogen fusion occurring in the center, as well as all the other fusion, helium, carbon, neon, etc.


What do we know about the characteristics of stellar objects, 13,3 billion light years away from us?

Holger,

I highly recommend you check out the NASA story on the discovery MACS0647-JD, which includes many dimensional facts (mass, distance, age, etc.) and comparisons:

http://hubblesite.org/newscenter/archive/releases/2012/36/full/

From the article:

  The estimated total mass of the stars in this baby galaxy is roughly
  equal to 100 million or a billion suns, or about 0.1 percent to 1
  percent the mass of our Milky Way's stars.

Measuring the mass of a galaxy is very tricky indeed (especially one as far as 13.3 billion light-years from earth!), but there are a couple different methods that can one can use, see here:

PSU Astro Measuring Mass of Clusters

And this Physics Stackexchange thread as well:

https://physics.stackexchange.com/questions/123/how-does-one-measure-the-mass-of-a-galaxy-and-other-such-large-quantities


Why is there no time in the distance equation using Hubble constant, red shift and speed of light?

Are you familiar with Hubble's Law? Because this is just it, in a slightly less familiar form.

Note that the distance in this formula is the proper distance, the distance between us and some source, not the distance travelled.

Other than that, the speed of light in a sense defines our conception of time and space, so it is mentioned in the formula. ;)


Planning observation: determine dates where moon is within 30 degrees of my object?

I would use JPL Horizons to calculate the position of the Moon and save the results to a text file. Then use a spreadsheet to calculate the separation between the Moon and your object. $$\cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (\alpha_1 - \alpha_2)$$ where $\theta$ is the separation, $\delta$ is the declination, and $\alpha$ is the right ascension.

Since the Moon moves about 1 degree in 2 hours, you can use that to determine what time interval to use in the output, depending on how accurately you want to find the 30 degree separation.


Does the universe expand at the same rate everywhere in the universe?

What's outside the observable Universe, we can't say anything about, but averaged over large enough scales ($\gtrsim$ a billion lightyears), it does indeed seem to be expanding uniformly.

However, the presence of mass, or more generally energy, retards the expansion. This means that on the scale of clusters of galaxies, the Universe expands more slowly, and on the scale of galaxy groups, the galaxies' mutual gravitational attraction will prevent them from receding from each other. This is also why our galaxy, Solar system, planet, and bicycles will never get torn apart (unless the cosmological constant is not a constant).

Conversely, in mass underdensities, i.e. the huge voids between clusters and filaments of gas and galaxies, expansion is increased (relative to denser regions). In fact, it has been hypothesized that that the observed accelerated expansion of the Universe is not due to dark energy, but could be an "illusion" from accidentally living in the center of a huge underdensity (e.g. Zibin et al. 2008). More recent observations seem to rule out this possibility, though


Why aren't the farthest objects close to each other?

the big bang did not happen in one place, it happened everywhere, and since then space has been expanding. so the stars that where formed shortly after the big bang where everywhere in space, but space was smaller. therefore you can see them everywhere if you look 13 billion light-years away, because they where everywhere.


Is it possible for a star to orbit a planet?

Pretty much by definition, no.

In order for "object A" to orbit "object B", "object B" needs to be substantially more massive than "object A".  In order for your red dwarf to orbit something, that "something" needs to be more massive than a red dwarf.  With the current composition of the universe, that "something" will be mostly hydrogen, and once you reach a certain level of mass (about 90 times that of Jupiter), you can't prevent nuclear fusion from starting.  The definition of "planet" is a bit fuzzy, but everyone agrees something lit by nuclear fusion is a star, not a planet.

This could theoretically change in the distant future, when there's enough non-hydrogen mass around for stellar-scale rocky bodies.


Getting atoms together into molecules requires two basic things - getting the reactants together and them providing the means to overcome the "activation energy" that is due to the mutual repulsion of the electrons that surround them.

Astrophysical chemistry is a rich and complex field of study. Most molecules are produced by a wide variety of processes. For ethanol in solar system objects, a cartoon of the process might be this:

Grains of dust (basic silicates and soot) form in the protosolar nebula; cosmic rays and high energy radiation from the protosun ionise atoms (remove outer electrons), which makes them "sticky"; these ionised atoms are concentrated onto the surfaces of dust grains, increasing their densities to the extent that reactions between them become likely; hydrogen ions are able to hydrogenate carbon and oxygen to produce CH and OH radicals; these then encounter each other in the presence of more hydrogen to form ethanol; dust grains with ethanol are then incorporated into solar system bodies like comets.

But as I said, for any individual ethanol molecule, other pathways are possible.


Determine Julian Date from Gregorian without formula

The Year 0 does not exist in BCE/CE and BC/AD notation. It jumps from 1 BCE to 1 CE directly. So if we look at the sequence of years in BCE/CE notation, we have:

3 BCE
2 BCE
1 BCE
1 CE
2 CE
3 CE

If you want to calculate the number of years between, say, January 1, 2 BCE and January 1, 2 CE, you can add the year numbers to get 4, but then you need to subtract 1 to remove year 0 from the total, giving a total of 3 years. In your particular case, this leads to $4713 - 1 + 2010 = 6722$ years.

In astronomy, we often use another year numbering system called the Astronomical year numbering, which includes a year 0 and makes things more convenient arithmetically when calculating time intervals between eras. It goes (-3, -2, -1, 0, 1, 2, etc.) To convert years BCE to the Astronomical year notation, we subtract 1 and add minus(-) in front of the year. For example, the year 4713 BCE is -4712 in Astronomical year notation. Your year interval is then calculated as $2010 - (\text{-}4712) = 2010 + 4712 = 6722$.

For the leap years calculation, since -4712 is a leap year, this simplifies matters, but don't forget to take the smallest integer greater than or equal to your result (the ceiling function). So in this case $\frac{6722}{4} - 3 = 1677.5$, and the ceiling is 1678, and that is the number of leap years between your years. Also, in your initial calculation, you had forgotten to subtract 3, because 6724 / 4 - 3 = 1678, not 1681.

Finally, the time interval between January 1 and February 28 is 58 days, not 59, because we do not count the last date as part of the interval. This gives a total of $5044 * 365 + 1678 * 366 - 10 + 58 + \frac{6 * 60 * 60 + 30 * 60}{24 * 60 * 60} = 2455256.27$.

You state that the actual answer is 2455255.77, but in fact the answer really is 2455256.27. Perhaps you typed "06:30" instead of "18:30" in the converter.


Can fish and other sea creatures survive in watery planets like Neptune or any other planets that has water?

According to Wiktorowicz & Ingersoll (2006), Neptune is too hot and too dry to form liquid oceans at the present time. This may seem counterintuitive for a planet with a far higher bulk fraction of water than the Earth which is located so far from the Sun, but the water is going to be mixed up with hydrogen, helium and various other materials in the planet. The water is going to be in a supercritical state, transitioning to even weirder forms in the extreme temperature and pressures of the deep interior.

If Neptune survives the post-main sequence evolution of the Sun, it may be possible that it forms oceans of some kind in the distant future. In any case there is very little free oxygen so life that requires it (e.g. fish) would not be able to survive. The known extrasolar Neptune-like planets are much hotter than Neptune as they are much closer to their stars, so the "too hot" problem would also apply there.

A better bet might be an ice moon such as Europa. In particular, the radiation environment of Europa might have led to oxygen being produced and dissolving in the ocean. The pressure would be extreme, though life has been found at the bottom of the Mariana Trench. The precise conditions in the Europan ocean, and the oceans on other ice moons is not well known, so there may well be additional problems. It is not clear that there would be the right availability of nutrients, for example.


What could be the utmost lowest temperature in the universe/multiverse?

What temperature means...

Temperature is the measure of the energy of particles. The higher the temperature, the more energized the particles. The more energy particles have the faster they move around. This is the particles' kinetic energy that is rising. As it rises, the particles will begin using up more space. Moving particles need more space. In a vacuum this can be measured as pressure which is the stress of thermal expansion on the closed system. In an open system the matter will expand freely. As the particles increase their speed they also move more erratically so the entropy, or measure of disorder, will also increase.

The coldest...

Now understanding all that, what would the lowest temperature, or energy state, be in the universe? The answer is a state of no energy, 0 K, or absolute zero in the Kelvin scale. It is −273.15° on the Celsius scale and −459.67° on the Fahrenheit scale. At this temperature, which is impossible to occur by only thermodynamic means, the particles are completely still and entropy drops to 0.

Temperature reference points...

The surface temperature of the sun is 5,778 K.
Water boils at 373 K.
Water freezes at 273.15 K.
The moon’s darkest craters that never receive sunlight are 33 K.
The cosmic microwave background fluctuates around 2.8 K.
The Boomerang Nebula, the coolest natural place currently
known in the universe, measures 1 K.
Absolute zero is 0 K.


Which direction does the Earth face as it travels through the Galaxy

My question is, which pole on Earth faces in the direction that the Sun moves? The South Pole or the North Pole?

The North pole is in the hemisphere of the direction of motion of the solar system through the galaxy.

The direction in which the Sun in traveling relative to surrounding stars is called the solar apex and is located at (RA)  18h 28m 0s (dec) 30°N, which is a point in the northern hemisphere.


Poisson's equation-like formula in axis-symmetric Galactic potential in terms of Oort constants

Poission equaiton:
$$\nabla^{2} \Phi=4 \pi G S_{0}$$
In cylindrical polar coordinates:
$$\frac{1}{r} \frac{d}{dr}\left(r \frac{d \Phi}{d r}\right)+\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}$$
We also know from $\ddot{\vec{r}}=-\nabla\Phi$ that $\ddot{r}-r \dot{\phi}^{2}=-\frac{\partial\Phi}{\partial r}$, so for circular orbit, we have: $r\dot{\phi}^2=\frac{\partial \Phi}{\partial r}$. This means that we can write $v_c=r^2\dot{\phi}^2=r\frac{\partial \Phi}{\partial r}$
Rewrite Poisson above:
$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}-\frac{1}{r} \frac{d}{d r}\left(v_{c}^{2}\right) = 4 \pi G \rho_{0}- \frac{2v_c}{r}\frac{dv_c}{dr}$$
Keeping in mind that  $A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$, rewrite RHS:
$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}+2(A^2-B^2)$$
which is what we wanted.


How are the speed, distance and time of India's MOM and USA's Maven Mars missions calculated?

There is an extremely accurate model of the solar system and the positions of the planets and so on. So basically you have to lay Newton's equations for Mars, the probe, Earth and so on, you have an equation that models Mars's path, and you have equations for your probe, so you get some x's and y's that eventually will have to match if you want to reach Mars. From there you can then calculate the values for the initial conditions of the probe -such as velocity and direction- that will satisfy you equations.


To answer this, I interrogated the revised Hipparcos catalogue produced by van Leeuwen (2007), accessed at the CDS, Strasbourg.

There are 4559 stars in the catalogue with a Hipparcos magnitude $<6$ and which you might class as "naked eye" objects, though you'd need good eyes, a good site and plenty of patience to be dealing with stars fainter than about 5th magnitude.

Sorting these objects by total proper motion, we find there are three objects with proper motions between  4.0 and 5.3 arcseconds per year:

Star           Mag       Total PM (arcsecs/yr)

HIP19849       4.56       4.1

HIP108870      4.83       4.7

HIP104214      5.37       5.3

Let's say a human lifetime is 80 years. This means that these stars move an angular distance of between 328 and 424 arcseconds on the sky with respect to the average position of all the other stars.

The angular resolution of the human eye is about 30-60 arcsecs (for really good vision and a completely dark adapted eye). So the separation between two stars could be estimated to some fraction of this.

I think this would be very difficult though unless these stars were very close to another naked eye star, so that you could see the separation changing with time.

So, I did another search. I looked for naked eye stars that were within 10 arcminutes of another naked-eye star and which had a high relative proper motion that would make them move significantly with respect to each other.

The best candidate pair is HIP71500 (which is basically stationary) and HIP71502 (which is a high proper motion star), magnitude 5.58 and 5.85, which are separated currently by about 15 arcseconds, but after 80 years, this will have changed by 52 arcseconds. What this means is that for someone with acute naked-eye vision (is it possible to have that over your lifetime), a star that was apparently single to the naked eye will gradually become a (just) resolved visual binary.

Nice bit of trivia!


Why we define Stellar motions with respect to sun?

The Earth is a moving (actually, accelerating) platform from which we make our observations. If you want to describe the motion of a distant celestial body, then it does not make much sense to provide a geocentric velocity, because this will depend on exactly when the observations were taken (because the Earth orbits with a speed of about 30 km/s, but the velocity changes throughout the year), the position on the Earth that the measurement was taken (because the Earth's equator revolves at around 0.5 km/s, but this is different at different latitudes) and the position of the star in the sky (because the previous two effects have to be resolved in the direction of the star).

Given that, then it is far easier to quote a velocity in a frame of reference that does not depend on these things. One can use the heliocentric frame in most circumstances - that is the frame of reference in which the Sun is stationary. To do this, one has to remove the components of the Earth's orbital and rotational velocity. However, if very precise measurements are being done then it is better to use the barycentric reference frame, which is with respect to the centre of mass of the solar system (since the Sun orbits this with a speed of around 0.1 m/s).


Will universe's lakes get boiled by the CMBR?

Not really: microwaves heat mainly by using a lot of radiation, the particular wavelengths matters only for the ability to (somewhat) heat evenly.

You probably know of the Sun, a thing in the sky which shines most brightly in the optical, that is in wavelengths much shorter than the ones in a microwave. Yet it is clearly very capable of heating you and everything around you far above the ambient temperature of space.

So the wavelength of radiation is not crucial for heating (it matters a bit depending on the material), it is the amount of incoming radiation that matters. If the CMB could start to boil water because it was redshifted from 2 mm to 12 cm, it would need to be at such high intensity that water would already be boiling, because water absorbs more radiation of 2 mm wavelength than of 12 cm:


Are there any alternative explanations for the gravitational force we attribute to dark matter?

There are certainly people who study alternative (non-General Relativistic) theories of gravity. The most popular theories have so far been:

Modified Newtonian Dynamics (MOND) - which essentially postulates that Newtonian Mechanics break down on some scale, leading to the rotation curves we see in galaxies.
Tensor–vector–scalar gravity (TeVeS) - this is a relativistic generalization of MOND.
Falling into the MOND category is f(R) gravity, which alters the general relativistic expression relating the curvature of space-time to the stress-energy tensor (the 'source' of gravity - like mass and pressure). These theories relate different functions of the Ricci scalar to this stress-energy tensor.

TeVes seems to be the most promising alternative to General Relativity (it may be able to explain things like rotation curves, predicts gravitational lensing, and some other things), but there are still many problems it doesn't do well with. So far it seems like modified theories of gravity are creating more problems than they are fixing - though I must admit, the existence of things like dark matter and dark energy are non-trivial issues cosmologists must come to terms with.

When it comes to things like alternative universes and the like as the cause of our current mysteries, I'm sure these things have been proposed. The problem here is coming up with an experiment to test them. Dark matter has been named as such because we know that it doesn't emit or absorb any radiation. As far as detecting such a particle, many real experiments have been conducted and will be conducted, and is an exciting field of research.

Below shows a few of these experiments and the range of cross-section/mass parameter space it has/will probe:


Joshua, your question combines a number of complex but related astrophysics issues, so let me try to untangle them as best I can in layperson's terms, with links where you can find more detailed information.

What you're describing is known as the cosmological horizon, which defines the limit beyond which a particle's light can never reach us because of the expansion of the Universe. While gravitational waves ("GWs") are unrelated to electromagnetic waves (what we call light), it's believed GWs are propagated at the speed of light (the universal limit for how fast information can travel), so the cosmological horizon would apply to gravitational waves as well.

However, gravitational waves are different from gravity.

To put it overly simply, the general theory of relativity says that mass distorts both space and time, and gravity is what we observe of this curvature of spacetime. Gravitational waves, on the other hand, are "ripples" in that spacetime curvature caused by non-symmetric acceleration of mass.

GWs are indeed "all around us", but they are extraordinarily tiny - Albert Einstein himself doubted they could ever be detected. Our most sensitive equipment, LIGO, looked for the very largest GWs for eight years (2002-2010) and found nothing. It wasn't until a significant enhancement came on line just three years ago (Sept 2015) that the first GW was detected - from the collision of two black holes with a combined mass equivalent to 65 Suns! - and since then we've only detected 11. GWs are really hard to detect!

Since gravity is an intrinsic geometric property of spacetime, it exists even beyond the cosmological horizon: it's inseparable from the Universe as a whole. All parts of the Universe contribute to the curvature of the Universe, no matter how far away one point is from another. There's no "threshold" for gravity.


Is it possible to disprove geocentrism without telescopes?

I would argue that the Apparent retrograde motion of the planets over time, is evidence against geocentrism, and probably why Aristarchus of Samos proposed a Heliocentric model of the Solar System.


Why is $g$ tied to the oscillator strength $f$ in $\log{gf}_{\odot}$

You don't make it clear, but you may be confused about what $g$ is. It is the statistical weight of an atomic energy level.

The $gf$ value, which refers to a particular transition between two energy levels in an atom/ion, is used because there is symmetry in terms of emission/absorption processes once the statistical weight is taken into account.
$$g_1 f_{12} = - g_2 f_{21}$$

If you just quoted $f$, then you would also need to know what the appropriate statistical weight was in order to calculate transition probabilities.


Where can I download processed data of the Kepler telescope?

What you are looking for is here. http://archive.stsci.edu/kepler/search_retrieve.html
This one is good for exoplanet light-curves https://exo.mast.stsci.edu/


How are people converting intensities in Janskys to Kelvin?

This is, indeed, a result of how we measure things in radio astronomy.  It's not just interferometry, but radio astronomy in general.  The thing they're referring to is a concept called "brightness temperature".  In the low frequency limit (valid for radio astronomy), we can use the Rayleigh-Jeans approximation, which gives us the expression
$$
T_B = \frac{S_\nu \lambda^2}{2k \theta_s^2}
$$
where $S_\nu$ is the flux density, $\lambda$ is the wavelength, $k$ is Boltzmann's constant, and $\theta_s$ is the size of the source (or the resolution in this case).  As long as your source and wavelength are constant, brightness temperature and flux density can be interchanged.

The reason astronomers talk about temperatures instead of fluxes sometimes is twofold.  First, a large number of objects examined by radio astronomers are not blackbodies.  Therefore we cannot calculate their temperatures like we normally would.  Instead, we calculate what their temperature would be if they were blackbodies.  This at least gives us a value to compare them to other things.

Secondly, when discussing radio telescopes and antennae, we usually say that various parts of the antenna contribute different 'temperatures'.  We can quantify system temperatures to help us understand noise.


You need three coordinates to represent a point in 3-dimensional space. Galactic coordinates are a 2-dimensional coordinate system on the surface of the abstract celestial sphere. Typically it is easier to measure position on the celestial sphere than it is to measure astronomical distances, so usually the systems are given in those terms. For 3D you need to supplement the Galactic coordinate system with distance as a separate coordinate, this is in fact the natural choice for a 3D spherical coordinate system.

Remember that the galactic coordinate system has its origin in the same place as equatorial coordinates (right ascension, declination), so the distance measurement is the same: you don't have to translate it relative to the Galactic Centre. To measure it, you use the usual techniques used to determine the cosmic distance ladder: parallaxes, standard candles (Cepheid variables, etc.), redshifts...


How far could be an object from the Sun and still be under the influence of its gravitational field?

The Sun's gravity extends infinitely, but eventually solar objects would be unstable due to the influence of other stars. The minor planet "Sedna" has an orbit which takes it nearly 1000 AU (0.016 light years) from the sun at its furthest point (but now it is a lot closer)

It is also thought that billions of comets must orbit in the outer part of the solar system, out to 50000AU, or 0.8 light years, (or possibly further) forming the Oort Cloud. However, at such distances, they could not be directly observed. This marks the greatest distance at which orbiting solar system bodies can be found.


Constellations are human constructs to make sense of the night sky.  When you are trying to find your way around, it helps to "chunk" stars into patterns and assign those groupings names.  When I want to point out a particular object in the sky (say Polaris, the North Star), I start by pointing out a familiar constellation (say Ursa Major, the Big Dipper).  From there, I can tell my friend to follow this or that line to get them to look where I'm looking:

With the advent of computerized telescopes and large data sets, constellations are less important for professional astronomers.  However, many stellar databases use Flamsteed or Bayer designations, which assign stars to constellations.  In order to include all stars, the sky is divided into irregular regions that encompass the familiar constellations.

So, which constellations is the Sun assigned to?  Well, from the perspective of someone on the Earth, the Sun moves through the constellations throughout the course of the year.  Or rather, Sol moves through the region of the sky where some of the constellations would be seen if its light did not drown out distant stars.  Our moon and the rest of the planets move through those same constellations.  (The Greek phrase which gives us the word "planet" means "wandering star".)

The current position of the sun against the background of distant stars changes over the course of the year.  (This is important for astrology.)  It's a little easier to make sense of with a diagram:

So perhaps a better question is:

  What constellation does the Sun belong to today?

Presumably an observer on an exoplanet would assign Sol to some constellation that is convenient from her perspective.  But from our perspective within the Solar system our sun, moon, and planets are not part of any constellation.


Sky-glow is a generic term for the diffuse light not due to the sun, moon or stars (or aurorae, I suppose). Much of it is light pollution from the ground, some is the zodiacal light from dust in space, and the coolest part may arguably be the air-glow from chemical reactions in the upper atmosphere.
So, depending on height the different components would change strength. Urban light pollution apparently declines as $\propto d^{-2.5}$ ("Walker's law") due to spreading and absorption, so it would decline with altitude. The zodiacal light would get stronger with less absorption in the atmosphere. The air glow is altitude-dependent, and would presumably intensify at first, until one is above it (like the ISS) is.


Perhaps you are looking for the backup function

  File -> Backup...

and to restore the settings

  File->Restore...


That is a lot of questions, but I can take them in order:

  It seems GW has Doppler shift too. There is a GW spectrum with emission or absorption lines like optical spectrum?

Yes, GW has Doppler shift too, as they are travelling at a finite speed, but no, there are no emission or absorption lines we can detect it from. In EM radiation those are caused by the fact that light is quantized, such that certain wavelengths can be absorbed or emitted by atoms. As far as we know, gravity is not quantized, or the effect is too small to be measured. Any irregularity in the signal is then going to be continuous, and not wavelength specific. Detecting a spectra in gravitational waves is unlikely, but would strengthen the theories of quantum gravity and the graviton.

  people take their masses derived seriously?

The method for the mass estimate is described well by @Rob Jeffries, but an even simpler point can be made: There are no other ways to measure their mass more accurately.

  What does a 100HZ GW mean?

Hertz is the frequency, 1/s. That is the number of waves per second. 100Hz is comparable to normal sound waves.

  What does the frequency stand for in the binary system?

As two objects revolve around each other, they would make two waves per orbit,resulting in a double frequency of the system's rotation.

  DO you think the team win a Nobel prize only after other facilities like LISA find GW again?

This is almost certainty ending in a Nobel prize, and they do not necessarily have to wait till after the results are confirmed by other experiments. That said, verification by peer researchers is one of the core principles of science.


The same way rotation does.  If you look at each star, it is following some kind of orbit, so it has a motion.  The only difference between a spiral and an elliptical is that the motions in a spiral are all in the same plane and the same direction, but in the elliptical, they are all over the place.  So "random velocity" just means you look at a bunch of stars-- each one is still in an orbit, so it's no surprise it stays in orbit.


Necessary steps to calculate photon's path by using null geodesic equation

The geodesic equation below is a second-order partial differential equation for the spacetime coordinates $x^\mu$, $\mu = 0 \ldots n$, $${d^{2}x^{\mu } \over ds^{2}}+\Gamma ^{\mu }{}_{\alpha \beta }{dx^{\alpha } \over ds}{dx^{\beta } \over ds}=0 \;,$$
where $\Gamma ^{\mu }{}_{\alpha \beta }$ are the Christoffel symbols, and the proper time $s$ is the parameter of the curve (i.e. how ``far along'' you are on the path). You need to solve these $n+1$ coupled equations as any such system, e.g. numerically.

If the parameter $s$ is confusing, you can reparametrise the curve as a function of coordinate time $t = x^0$, as explained e.g. on Wikipedia's page on geodesics in general relativity, $${d^{2}x^{i} \over dt^{2}}=-\Gamma ^{i}{}_{{\alpha \beta }}{dx^{\alpha } \over dt}{dx^{\beta } \over dt} + \Gamma ^{0}{}_{{\alpha \beta }}{dx^{\alpha } \over dt}{dx^{\beta } \over dt}{dx^{i} \over dt} \;,$$
which gives you $n$ partial differential equations for the spatial coordinates $i = 1 \ldots n$. (Note that there are remaining greek indices, for which $dx^0/dt = dt/dt = 1$.)


Are the stars distributed in uniform distribution, on the celestial dome, with respect to brightness?

Naked eye stars are not distributed uniformly in the sky. That is because the median naked eye star is at a distance of 440 light years, and this is far enough away that some of the details of Galactic structure start to become apparent. Most importantly, the density of stars increase towards the Galactic midplane and has a scale height of a few hundred light years and thus there is an overdensity of stars towards the Galactic plane.

I did an analysis of this in https://astronomy.stackexchange.com/a/10260/2531 , but the important plot is repeated below. It shows the normalised density of stars (per square degree) as a function of Galactic latitude. You can see the peak at low Galactic latitude and even work out that our viewing point is probably slightly above the Galactic plane, given that the peak is at about $-10^{\circ}$. The peak is still there in 3rd magnitude stars. You could repeat for a sample limited to very bright stars, but you run into number statistics trouble.

The distribution of bright stars with Galactic latitude.

EDIT: Extra mile time. Below I show the Hipparcos data, divided into 4 apparent magnitude bins. Points are plotted on a RA vs Dec Aitoff projection. Superimposed in red is a Galactic coordinate grid. There are too few stars with $V<2$ to form much of an opinion, but for fainter stars, a ring-like structure, coincident with the Galactic coordinates equator (i.e. Galactic latitude of zero, or just below, as I showed in the 1D projection above) is clearly seen.


What is the apparent magnitude of the weakest stars we can see in one telescope on the ground with a diameter of 2.4m?

You just need to add the 13.4  that you got to the 6 that you can see without a telescope and that's it.  The formula is telling you the difference in magnitude that arises from the different apertures of the eye and the telescope.


Why do stars born in a cluster finally disperse?

The boundedness or otherwise of clusters remains to be established in most cases. The vast majority of clusters become unbound and disperse at a much younger age than the Pleiades. Or they may be born unbound as you suggest.

The stars in a cluster have a distribution of velocities and there will always be a tail of high speed stars that will be able to escape the gravitational potential in an "evaporation process", even in a cluster which is globally bound (i.e. the sum of kinetic and gravitational potential energies is negative).

It turns out that this isn't usually sufficient to explain the disappearance of clusters. A more effective disruptive mechanism are the Galactic tides. Stars that stray beyond a tidal radius can be stripped from the cluster. The reduction in the cluster's mass then reduces the tidal radius allowing more stars to escape in what eventually would become a runaway process. The tidal radius for a cluster near the Sun is about $5 M^{1/3}$ light years, where $M$ is the cluster mass expressed in units of the solar mass. For the Pleiades, $M \sim 1000 M_{\odot}$, so stars that stray beyond 50 light years can be lost.

In some cases, the evaporation process causes a cluster core collapse. The escaping stars take away kinetic energy and the remaining stars become more tightly bound. This is what has happened in most of the massive and long-lived globular clusters. Less massive open clusters like the Pleiades are much more susceptible to tidal dissipation.


They come from Gliese's catalog of nearby stars (various editions and co-author), the number is the star's catalog number.


How do I see more than just points in the sky?

Dark Skies are your friend.

When Messier was searching for comets, it was before electric lighting and the nights were dark in a way that is rarely seen today.

Take a look at the Dark sky map, and try to get somewhere in the blue or black region.

Then allow your eyes to become dark adjusted, and to look with the edge or your vision, or learn to do astrophotography.

And limit your expectations. Messier was looking for comets (little fuzzy blobs) and he made a list of the fixed little fuzzy blobs that might confuse him. Most Messier objects do look like fuzzy patches. You might take an account with telescope.org to see what can be done with a telecope 16 times larger than yours.


Why don't we see "the milky way" in both directions?

I'll turn my comment into a full answer.

To put it simply, we actually do see the Milky Way all around us, even in the diametrically opposite direction from the galactic core. You can see this in the image below which is a full sky image I took from APOD.

If you look at the edges of the disk in that image, you're looking at what is actually the edge of our galaxy, opposite in direction from the core. In effect, this is precisely the image you requested because it does contain the part of the sky which contains the anti-core part of the galaxy. It is certainly not as bright, but there are still stars and dust out there. In fact, if you look very closely, you still see lots of dark splotches that mask distance stars and galaxies, indicating there is dust there.

I think the issue you may be having is that you expect many more stars to be within the outer regions of the disk than there actually are. The stellar density profile for our disk is roughly exponential, meaning there are literally exponentially more stars near the core than at the edges. If this means anything to you, the scale length for the exponential radial density profile is ~4 kpc.

To really get a good understanding of the stellar distribution, take a look at Jurić et al. (2008). They looked at (~48 million) stars from the SDSS and analyzed the stellar distribution across our galaxy (that is visible to us). You should find figures 10 through 18 of particular interest, however I'll present part of figure 16 here.

This image shows the (logarithmic) density of stars as a function radius from the galactic core. The varying shades of grey indicate varying heights above the galactic plane (numbered in parsecs). The dashed lines are various exponential decay models with differing scale heights. You can see that these star densities, even within the limited radial ranges covered by the SDSS fall off by an order of magnitude! Hopefully this helps you appreciate the significant difference between the core brightness/visibility and that of the galactic edge.

Here's a silver of Andromeda spectacularly demonstrating the falloff: even though galaxy photography tends to suggest to the casual eye an evenly-dense plate:

This synthetic image referenced by Andy, from the Tycho Catalog Skymap also shows the situation clearly.


Why didn't Johannes Kepler use data about more planets?

I know that quoting Wikipedia is frowned upon here, but as there has been no other answer posted, this is what the Wikipedia article on Kepler has to say about the matter:

  He then set about calculating the entire orbit of Mars, using the geometrical rate law and assuming an egg-shaped ovoid orbit. After approximately 40 failed attempts, in early 1605 he at last hit upon the idea of an ellipse, which he had previously assumed to be too simple a solution for earlier astronomers to have overlooked. Finding that an elliptical orbit fit the Mars data, he immediately concluded that all planets move in ellipses, with the sun at one focus—Kepler's first law of planetary motion. Because he employed no calculating assistants, however, he did not extend the mathematical analysis beyond Mars. By the end of the year, he completed the manuscript for Astronomia nova, though it would not be published until 1609 due to legal disputes over the use of Tycho's observations, the property of his heirs.[Caspar][Koyré]

[1] Caspar M., Kepler, pp. 131–140;

[3] Koyré A., The Astronomical Revolution, pp. 277–279


If a Jupiter-sized sphere of a translucent liquid were to suddenly appear, what would the inner core consist of? Would the planet refract light?

A Jupiter made of fruit juice would probably allow sunlight to enter the first few metres below its surface, before being absorbed.  Just think how far light penetrates the water of the ocean. So you wouldn't be able to see the sun through it. It wouldn't act like a lens.
If you magically created it with uniform density, it would be very very far out of equilibium. The central parts would compress and release huge amounts of gravitational energy in the process.  The effect would probably be dramatic.
If you created it with a density and temperature gradient to keep it stable, you'd probably get a mantle of ice and water (including exotic high pressure ices) Depending on your exact temperature profile the water would be supercritical (it is indistinguishable from very high pressure steam, and there is no boiling point). The centre might be supported, at least in part, by electron degeneracy.
And (again depending on the exact parameter you choose) there could be a water vapour atmosphere.
As Rob says in a comment, "I doubt whether anybody has spent their time working out what happens to a Jupiter-mass of fruit juice." But we can look at a slightly smaller planet made out of ammonia and water: Neptune and Uranus are mostly water and ammonia ices, under a thick a hydrogen/helium atmosphere.


Is there a type of star that emits relatively monochromatic visible light?

In some sense, yes.
UV continuum $\rightarrow$ UV line
This is probably not the answer you're looking for, but massive stars (O and B stars) emit a very hard UV spectrum of light (a UV "continuum", i.e. a broad range of wavelengths). Since these stars don't live for long (because they burn their fuel fast), they tend to be located in the gas clouds from which they were born. The UV light ionizes the enshrouding hydrogen, but the protons and electrons quickly recombine. If the electron goes directly to the ground state, another UV photon is emitted, capable of ionizing another hydrogen atom. In most cases, however, the electron "cascades" down multiple levels, emitting photons of different energies, which may excite other atoms.
It turns out that for every 3 ionizing photons emitted by the star, 2 will eventually — after several interactions — result in the photon corresponding to the energy difference between the first excited state and the ground state; the so-called "Lyman $\alpha$" photon, with a wavelength of 1216 Ångström (121.6 nm). Although there is some broadening due to thermal motion of the atoms and, in particular, due to the resonant scattering of Lyman $\alpha$ on neutral hydrogen, the result is that most of the light from these stars is converted into a single, very narrow (of the order of a few Ångström) emission line, i.e. very close to monochromatic light.
These stars are probably rarely surrounded by planets (because the radiation pressure will tend to blow away the particles used to build planets), and even if they were, these processes happen farther away from the stars than the planets would be. But if you observe a young galaxy, whose spectrum is dominated by young stars, the Lyman $\alpha$ emission line is often the only light visible.
UV line $\rightarrow$ visible line
The Lyman $\alpha$ line is still in the UV, and thus invisible to humans. However, since light is redshifted as it travels through the expanding Universe, Lyman $\alpha$-emitting galaxies sufficiently far away will have their emission line carried into the visible range. Since the shortest wavelength we can see is 400 nm, it must be redshifted by a factor $400/121.6 = 3.3$ (i.e. $z=2.3$), corresponding to a distance of 18.6 billion lightyears (but if it's farther away than 25.5 billion lightyears, it will shift into the infrared and be invisible again). Note, though, that this is only in principle; such galaxies are far too dim to be visible to the naked eye. Too see them, you must use a telescope and a camera.


In my mid-forties, I believe I have seen Pluto for the first time with the naked eye

Pluto is something like magnitude 14. The limit for the human vision is somewhere between magnitude 6 (widely accepted) and 8-ish (highly trained observers with perfect vision in ideal conditions using special techniques - and it's a bit controversial anyway).

There's zero chance that was Pluto. It was definitely a fixed star.


Can Summer Triangle asterism be used for navigation?

Yes, it can be used, but it is not easy.

For the latitude, you usually need to know the Alt of Polaris. You can derive that if you know the Alt of two of the three, but the (spherical geometry) formulas for that are not simple.

For the longitude, you need Greenwich time and the RA of any star. Any star includes Vega, Altair and Deneb ;)


What will the mass of the new galaxy be?

The mass will be slightly less than the combine masses of the 2 galaxies since some of the stars will be hurled away.

Since the disks of the galaxies are at an angle to each other, the volume would be (roughly) the volume of the 2 galaxies as they collide. Eventually, the volume will decrease (I'm guessing to between 60 to 70%) as the stars adapt to their new environment and the central massive black holes combine.


Where do stellar classes names come from?

It seems like these letters are inherited from Draper classification system, which used letters to subdivide classes defined by Secchi. Secchi classification uses numbers I to V, while Draper did use letters in alphabet in order. Some of the letters got dropped, and classes reordered later to match physical features (temperature / spectrum) better.

I believe Secchi and Draper added categories as new types of stars were discovered (eg carbon stars and Wolff-Rayet stars seem to be added later), and thus the categorization did not remain as logical as it could've been had it been designed with full knowledge from scratch. Modern system wanted to retain (some of the) existing labels but order them based on some physical feature.

Wikipedia article seems to have pretty decent details on this too: https://en.wikipedia.org/wiki/Stellar_classification


What are the curved lines in this X-ray image of the sky?

GIFs too big to post here directly: 1, 2

NICER is all about time resolution and energy resolution and is used to collect the time and energy-resolved X-ray spectra from pulsars. It isn't an imaging telescope and so can't record "patches of the sky" like other devices can.

The curved lines are the paths that the narrow field of view (FOV) of the NICER telescope has traced out across the celestial sphere during this 22 month (intermittent) observation period.

According to this page the FOV is only about 30 square arc minutes:

Large effective area: ~1900 cm2 at 1.5 keV
Broad Bandpass: 0.2 < E < 12.0 keV
Absolute timing precision of < 300 ns
Moderate spectral Resolution: 6 < E/ΔE < 80 from 0.5 keV to 8 keV
Restricted field of view: 30 arcmin²

With limited observing time and effectively a narrow single pixel field, it can only trace out thread-like arcs across the sky.

NICER is mounted on the constantly moving International Space Station and has its own pointing mechanisms on top of that.

When you see several arcs meet at a certain point, this is an interesting X-ray source, usually a neutron star (the primary purpose of the device and the "N" in NICER), and the telescope passes over it several times in order to collect more data.

Watch the COOL YouTube video NASA'S NICER Does the Space Station Twist

Watch the COOL YouTube video ISS NICER - First Deployment & Range of Motion Checks

Watch more NICER videos here: https://heasarc.gsfc.nasa.gov/docs/nicer/nicer_gallery.html

Watch the YouTube video Unlocking Secrets of Neutron Stars with NICER

Read the presentation NICER Science Overview

You can see that it is a number of grazing-incidence curved crystals used as concentrators, with "single-pixel" X-ray detectors behind them in the photo below (from How will NavCube (actually) be important for the XCOM testing and demonstration?):

below: NICER X-ray timing telescope/concentrator array, from About NICER.


Why is there no color shift on the photo of the M87 black hole?

The picture isn't a "colour" picture - it is monochrome. i.e. It is obtained at a single microwave wavelength of 1.3 mm, and so not at any wavelength you could see (Akiyama et al. 2019).
There isn't therefore any spectral information that would reveal the expected Doppler effect. Any difference of colour in the "false-colour picture" is purely related to the intensity of the emission, not its wavelength.
If one were able to obtain coverage at multiple wavelengths then you might expect a "bluer" (i.e. shorter average wavelength) colour to be associated with the brightest regions.


Figure out the location of celestial objects relative to the Milky Way in x, y, z co-ordinates

The spherical $(r, l, b)$ can be converted to
rectangular
$(U, V, W)$ as:
$$\begin{align}
U &= r \cos b \cos l \\
V &= r \cos b \sin l \\
W &= r \sin b
\end{align}$$
Using $(r, l, b) = (770 \pm 40 ~\text{kpc}, 121.2^\circ, -21.6^\circ)$,
$$\begin{align}
U &= -371 \mp 19 ~\text{kpc} \\
V &= ~~~ 612 \pm 32 ~\text{kpc} \\
W &= -283 \mp 15 ~\text{kpc}
\end{align}$$
That's relative to the Sun.
For a position relative to the center of our galaxy,
subtract $8 \pm 0.5 ~\text{kpc}$ from $U$.


In the big bang model of the Universe, every observable thing is thought to have expanded very abruptly from a point of infinite density and zero volume.

Thats not quite true. Have a look at Wikipedia where you can read this:

"Extrapolation of the expansion of the universe backwards in time using general relativity yields an infinite density and temperature at a finite time in the past.[18] This singularity signals the breakdown of general relativity and thus, all the laws of physics. How closely this can be extrapolated toward the singularity is debated—certainly no closer than the end of the Planck epoch. This singularity is sometimes called "the Big Bang",[19] but the term can also refer to the early hot, dense phase itself,[20][notes 1] which can be considered the "birth" of our universe."

You can't necessarily extrapolate back all the way to a point singularity.

  However, the problem with this assertion is that for something to expand it first needs a locality in which to do such, which brings us to my question.

There's no problem with space expanding. Don't think of space as nothing, Einstein didn't, and nor do modern physicists such as Nobel prizewinner Robert B Laughlin.

  How can scientists believe that "everything" was condensed and then suddenly expanded, since such is automatically an oxymoron?

Because there's good scientific evidence that the universe is expanding. And when you "wind back the clock" you find that all the galaxies were in the same place 13.8 billion years ago.

  Expansion is a physical property of the universe and yet if "everything" had no where to expand, how could it have expanded? There was "nothing," was there not?

No. Space expanded, and space isn't nothing. As to what caused the big bang to occur, I don't know. Nor do I know what the universe was like 13.8 billion years ago. But I know of no infinities in nature, and I would challenge the alleged point singularity at the centre of a black hole and refer instead to the frozen star interpretation. In similar vein I would challenge the alleged point-singularity 13.8 billion years ago.

  Even if we drop the big bang model and talk about "everything's" being in a singularity point, the question still stands. Where was the singularity point?

I think you shouldn't drop the big bang model, but instead you should drop the singularity.

  Asked differently, into what is everything expanding?

Nothing. I know this is hard to imagine, but try to. Space is expanding, and there is no space beyond it. So there is no beyond it. As to whether this means you end up coming back round full circle or the universe is some "hall of mirrors" or something else, I don't know.

  (which is a rhetorical question I ask to implicate space-time as necessarily existing irrespective of any theoretical expansion and likely so vastly that one might as well assume that space-time is infinitely large).

No, one might not. Because the universe doesn't expand for nothing. It expands because space has a kind of innate "pressure", as per the energy-pressure diagonal in the stress-energy-momentum tensor. And if the universe was infinite, the pressure would be counterbalanced at all locations. An infinite universe could not expand.

  When thinking about the big-bang model, I can't help but conclude that a mere analogy was originally used

For an analogy, get a stress ball and squeeze it down in your fist. Then let go.

  The issue though is still extension. As far as I understood it, the big bang model states that space itself was also condensed along with everything else in the observable universe to a point of singularity.

See what I said above. Big bang cosmology doesn't actually say everything was condensed to point singularity.

  All the matter had to expand somewhere

There wasn't any matter initially. Remember the wave nature of matter, and pair production. We can quite literally make matter out of electromagnetic waves. These wave propagate through space, and are effectively wave sin space. So we make matter out of waves in space. So start with space, and do somehting to it such that it rings like a bell and is full of waves and starts expanding. As to what, I don't know.

  How does the theory allege where everything is yet at the same time assert that prior to the singularity point, there was literally nothing?

It doesn't. It goes back as far as a very dense universe. The stuff you hear about a point singularity and creation ex-nihilo is popscience, not big bang cosmology.


In general relativity, the curvature of spacetime is an interpretation of  the model for gravity: Starting with the assumptions of gravity being equivalent to acceleration, and the speed of light being constant, a mathematical framework that describes gravity is developed. The mathematical conclusion of this model is that the distances between points is dependent on the distribution of mass and energy. A simple way of describing this is to say that spacetime is curved.

In GR, the apparent force of gravity is explained by the curvature of spacetime. It is reasonable to ask if there is a more fundamental theory that can explain the curvature. String theory provides one possibility. In string theory it is possible to deduce the existence of a particle that would be a graviton, and the from that the field equations that describe gravity. In other words, string theory can predict exactly the same mathematical formulae that arise from curved spacetime in GR. However, on very small scales, string theory makes a series of corrections to the field theory, suggesting that the curvature of spacetime is actually an approximation, which is valid on the large scale, to a string theory of gravity.


Throughout the year, are all parts of the night sky visible from any given point on Earth?

At any one time, only half the sky can be visible, and half is below the horizon.

At the poles the sky appears to rotate about the zenith, and so no new parts of the sky ever become visible.  At the North pole, you can never see any stars in the Southern hemisphere.

On the equator different parts of the sky become visible and over the year all the stars can be seen (but stars very close to the pole will be very hard to see, as they are too close to the horizon.)

For someone who is at (for example) a latitude of 40 degrees North will not see any stars that are more than degrees south of the equator; 90-40=50; and so won't see the southern cross, the Magellanic clouds or other features of the Southern sky. Similarly an observer in Tasmania won't see Polaris, Cassiopeia, or the "big dipper" as these are features of the Northern sky.


What is the "lunar irradiance" received by the Earth from the full Moon?

You can work in magnitudes, but the magnitude scale is logarithmic.  Instead you can use luminosity.  The moon has a luminosity that is 400,000 times less than the sun, so the maximum theoretical "lunar irradiance" is about 0.0034 watts per square metre.
In practice, the luminosity at the surface is less than this, with Wikipedia quoting "how bright is moonlight" typical values on the Earth's surface of 0.1 lux, equivalent to 0.001 watts of visible light per square metre.


Will the "fabric of space time" rip from expansion?

There is no "fabric" to rip.

Spacetime can stretch as much as you want (and in ways you cannot visualize - hence we use mathematics).  There is no way to do anything more than distort spacetime.  It can (at least in theory) be distorted is very odd ways (odd in terms of human experience) but there is no distortion that would "rip" or "tear" it.  It is always there.

Regarding the expansion of the universe, think about it like a having two points some fixed distance apart.  As the universe expands they become further and further apart, but it's not them moving, it's the scale factor of spacetime getting larger.  The universe gets larger because the scale factor increases.  In the past it was smaller and things were closer together.

There is a theoretical idea called "The Big Rip", but basically this just says that eventually the scale factor (which in this model keeps increasing) gets so large that everything is isolated from everything else, down to the level of elementary particles.  But spacetime itself does not rip - it's still intact, it's just that the scale factor has become huge.

There are models of the expansion of the universe that say the opposite - that eventually the expansion stops and the scale factor stops increasing and starts decreasing and we end up with everything in a Big Crunch.  Again no spacetime was hurt in the making of this model. :-)


Can a bullet be fired on the moon and sent it into orbit?

Almost. (EDIT: actually yes, see below)

The speed of an object in orbit depends on the radius of the orbit, and the mass of the body being orbited.

The lowest orbit possible is where the bullet is just grazing the surface, so r = the radius of the Moon. M is the mass of the Moon, and G is the gravitational constant. Let's do the math:

wolfram alpha calculation

The result is approx 1.7 km/s.

A .50 BMG Saboted Light Armor Penetrator (SLAP) reaches 1.2 km/s muzzle velocity. So even a very powerful round such as this one moves a bit too slowly.

It is conceivable that a tricked-out .50 round, with an extra load of powder, and possibly shot from a specially prepared rifle, could reach the 1.7 km/s speed needed; that would be a 40% increase in velocity over the SLAP round.

EDIT: According to this article saboted .224 caliber bullets loaded into .300 RUM (.300 Remington Ultra Magnum) cases can reach a speed of 1.7 km/s (the average they measured over the whole test was a little less than that, but the highest values exceeded 1.7).

In that case, if fired at the horizon, it would do a circular orbit, would go round the Moon and hit the shooter in the back 1 hour and 47 minutes later, provided it doesn't hit a mountain or something en route, and provided the orbit is along the equator.

If you're not shooting along the equator, it will still do an orbit, but its motion relative to the surface of the Moon will be a weird spiral and it will not return to the starting point (the orbit itself will still be a closed loop, a circle, but the Moon spins underneath it, so seen from the Moon it appears like a spiral).

The circular orbit (standing on a hill, shooting at the horizon) requires the smallest muzzle speed to complete the loop. Any other orbit would require a greater muzzle velocity from the rifle. The more elongated the orbit, the higher the velocity.

Plus, if you shoot higher than the horizon, the bullet would hit the ground before it reaches you. You have to stand on an elevated point (a hill), and make sure the rifle does not point at the ground either forward or backward. This is because the return trajectory of the bullet after completing one loop is aligned with the rifle (the rifle's barrel is tangent to the trajectory), and if the butt of your rifle points at the ground, on the return trajectory the bullet would have to come out of the ground, which is impossible.

A laser mounted on the rifle must not point at the ground no matter whether the laser is pointing forward or backward, otherwise the orbit gets clipped by the ground.

TLDR: Stand on a tall hill or a mountain, shoot while the rifle is exactly horizontal. Use a top-tier extremely high velocity round, custom-loaded with extra powder, with a high caliber sniper rifle. Anything less would not work.

  I know you need oxygen for the ignition

You don't. Gun powder contains its own oxygen. You can fire guns in a vacuum just fine. Primers also go off just fine in a vacuum.

It should be noted that the important number here, 1.7 km/s, is the orbital velocity near ground level on the Moon. It must not be confused with the Moon's escape velocity (the speed at which an object breaks away from orbit and keeps going away into space forever, never to return), which is higher.


Possibility of certain conditions on a planet

The basic condition is you'd need is a more massive, hotter star than our sun.  Not necessarily huge, but 1.5 - 2.5 times the mass of the sun should be enough because that corresponds to considerable extra luminosity and UV rays, so you could move the planet far enough away than Earth is from the sun and avoid any risk of tidal locking or slowing, so the planet could retain fast rotation period.

And you'd need a very thin atmosphere so the planet doesn't retain much heat under shadow.

If the star is bright enough in the sky that you'd get a burn in seconds under direct light, then shade probably wouldn't be cool unless it was permanent shade, as you noted, perhaps a cave.

If there's no atmosphere, you'd probably get colder temperatures than what you mention in shade, so some atmosphere, but a thin one.   Maybe somewhere between 1/10th of an ATM give or take and very low greenhouse gas, so basically very little water, CO2 or CH4.    1/10th of an ATM is just a very ballpark guess.

The only part I don't see happening is 3rd degree burns in seconds.   3rd degree burns by photon radiation from a star would probably require longer than that, unless the planet's surface was hot enough to melt rock.   Bad sunburn in minutes, sure.   3rd degree burns in seconds with your other criteria sounds unlikely to me.


What is a "differential chemical abundance"?

If the line emission is optically thin, then the measured line flux is proportional to the abundance $F_{\nu} \sim n$ of a species along the line of sight (if it's only one species).

The problem here is that $F_{\nu}$ needs to be measured relative to a baseline. If that fails, e.e. when the stellar signal is too noisy, one doesn't know what this should be. But a measurment at two different wavelengths $\nu$ and $\nu'$ that lie reasonably close together, can at least yield a difference in abundances $F_{\nu}-F_{\nu'} \approx n-n'$ relative to this unknown base level.
This allows then to construct an atmospheric model, up to the unknown base level, which one can try to fix with other physical arguments.


I was wrong, hot systems are where members have velocities which have magnitudes in all directions in the same order (causing them to have roundish shape), while cold dynamical systems are where random motions do not have the same average magnitude in all directions (ie disc galaxies).


Why doesn't the IAU consider Quaoar to be a dwarf planet?

It probably is a dwarf planet. (It almost certainly is a dwarf planet.)

The naming procedures at the IAU are that "Objects that have an absolute magnitude (H) less than +1 [...] are overseen by two naming committees, one for minor planets and one for planets.  [...] All other bodies are named by the minor-planet naming committee alone." source—wikipedia

Quaoar has an absolute magnitude of +2.4, so it has its name approved by the minor planet committee, and that committee doesn't rule on whether it is a dwarf planet or not.

It's just procedural.

So it's not that this is a particularly controversial object: It's big and round enough to fit the criteria for a dwarf planet. It just doesn't have the stamp of approval from the planet naming committee, and they don't get out bed for anything with an absolute magnitude of more than +1.

As observed in a comment, officially naming objects or putting an official box-label on a particular object may not get a particularly high priority in the scientific context. What an object is referred to is what gets established in debate and an "official" discussion and label is only be granted in particularly controversial cases of great general interest


It is unfortunate that the usual poor journalism labels the growth of the black hole as "inexplicable" and then further down in the article refers to some possible explanations.

The basic problem is a growth timescale one. Radiation pressure introduces a negative feedback, such that there is a "theoretical" maximum for spherical accretion called the Eddington limit, which occurs when the quasar is radiating at its Eddington luminosity. The shortest growth timescale is achieved if the efficiency with which mass is converted to luminosity is low; but if it's too low we wouldn't see the quasar at all. This is the crux of the problem. You can look at this Physics SE answer for some more of the details.

The thing is there are ways and means by which this limit can be exceeded - non spherical accretion for one - so there are lots of ideas about how this can be achieved. Another possibility is that you start off with a seed black hole that is pretty big to begin with, perhaps as a result of a merger. Or the quasar could have been less efficient in the past and is more efficient now, which is why we can see it.

Is there enough matter? Well, yes, galaxies have masses that can be much bigger than the mass of this black hole. They are rare, but of course so are > 1 billion solar-mass black holes, and these tend to be the only ones that we can see at distances of >10 billion light years.

One way of assessing the feasibility would be just to ask what a freefall timescale would be. If you have say $10^{11}\ M_{\odot}$ in a sphere of radius 10 kpc (I am just using typical sorts of numbers for a big galaxy), then the average density is $5\times 10^{-22} kg/m^{3}$ and has a freefall time $\sim (G\rho)^{-1/2}$ of 200 million years. Of course there are other problems, like shedding angular momentum, but it looks like this timescale is short enough for gravity to do its thing (in the absence of radiation pressure).

Of course the short answer to your question is that yes, there must be enough time, because this is just the latest in a population of such objects. We know that quasars with supermassive black holes have formed within a billion years after the big bang.


Is Jupiter really shepherding the main belt, and Neptune the Kuiper belt?

Orbits aren't knife-point balances of speed, distance and direction. For two objects separated by distances much larger than their diameters, almost any combination of the three where the speed is below the local escape velocity results in a stable elliptical orbit whose specific parameters are determined by the combination.
Ceres (and all the other objects in the solar system) are affected in their orbits by the gravitational influence of Jupiter (and everything else, for that matter). However, because the orbital periods of Jupiter and Ceres aren't low-denomination multiples of each other, the influence of Jupiter, while present, remains small, and the orbit remains pretty much the same, year to year.
Where the periods are a low-denomination multiple of each other, however,  the gravitational interactions reinforce repeatedly over billions of years, and can serve to both alter orbits such that the resonance is no longer maintained (see the Kirkwood gaps in the asteroid Belt), or in specific situations, lock an object into maintaining that resonance.
With Pluto and the Plutinos, the fact that their periods are clustered around 3/2 Neptune's orbital period is a direct result of Neptune's gravitational influence over time, rather than in spite of it; The regularity in the amount and the direction of Neptune's perturbations to their solar orbits keeps them near the 3:2 resonance, similar to the way Jupiter keeps the Jovian Trojans near the 1:1 resonance and the Hildas near the 2:3 resonance.
Note that all the orbital parameters of all the bodies in the solar system are changing over time (For example, the perihelion of Jupiter precesses approximately 6.55 arcseconds per year). The Keplerian Orbital Elements of the planets that you may find online are based on measurements taken at a specific time, and define orbits that assume that the planet and the Sun are the only bodies in the universe.


Olber's Paradox (Heinrich Wilhelm Olbers, 1758–1840): If the Universe were infinite, or there were infinitely many stars, then no matter which direction you look, you'd see a star. Thus, the night sky would be as bright as the Sun. Therefore, the Universe is not infinite, and does not contain an infinite amount of stars.

Also, the light from distant stars does 'get weaker' because of the inverse square law; this is the same reason lights look dimmer when they're further away.

I wonder if gravitational redshift has anything to do with it, too. Anyone?


Movement of the stars relative to my position

The Milky Way consists of numerous faint stars near our galactic equator.
Many of these are visible in binoculars.
As time lapse videos show, they move across the sky along with the bright stars.
The light and dark patches in those digital images are easier to see visually near the galactic center in Sagittarius than in the opposite direction near Orion.

Due to its oblique angle, the galactic equator appears to go through a special cycle as everything revolves around the celestial pole.
From mid-northern terrestrial latitudes, it runs overhead NW-SE when Orion is highest, low in the north when Virgo and the Big Dipper are highest, and overhead NE-SW when Cygnus is highest.
We see no more than half of this cycle on any given night, which may explain the illusion you perceived.


Double Planet Orbiting Wide Binary Star?

Generally speaking, there's 2 types of planet systems in a binary - see pretty picture.   Source.  The writer is a sci-fi writer so the entire article might be of interest to you.

Due to binary star tidal forces, there are some setups that are unlikely but in your example with a 100 AU distance between the stars, an orbit around the larger star is reasonable.

A double planet system is less likely.   It's unlikely to form on it's own during planet formation as that requires too much planetary angular momentum during formation.

It's possible, but also unlikely to form by giant impact, as that's more likely to leave 1 planet and 1 moon.  I've read (but can't find an article right now) that there's a giant impact size ratio and it's in the planet-moon range, not planet planet.   Much less than 1 to 1.  Pluto-Charon is 9-1 and Earth-Moon 81-1.   A giant impact is also unlikely to create an Iron rich core for both objects.   It's not a good way to create planet-planet.

That leaves a 3rd possibility, also unlikely, but perhaps the most likely of the bunch is planet capture.    Planets can form in Trojan points in the same orbit (Theia).  The difficulty with orbital capture is that the velocity needs to be just right and capture's are likely to be significantly elongated orbits, which, maybe, over time, perhaps with a 3rd asteroid, could even out to slightly more circular.   This is very improbable, but it might be the most likely way to form a double planet.

My 2 cents, as a layman.


Just to provide a source, the Ionization potential for a hydrogen-like atom is given by:

$$\chi_n = \frac{(13.6\:\mathrm{eV})Z^2}{n^2}$$

This is the energy necessary to ionize an electron in a hydrogen-like atom that "begins" at the $n^\mathrm{th}$ energy level. This equation is positive because you need to input energy to the electron to break it out of its potential well. Note that you may see a nearly identical equation for the energy of the electrons which will be negative. That implies the electrons themselves have negative energy because they're in the electric potential well and you need a positive amount of energy to ionize them.

Now, when you look at the Saha equation, you see a negative sign in the exponential term: $\exp(-\chi_i/kT)$. It was already established that $\chi_i$ is positive and so also are $k$ and $T$ simply by their nature. The negative sign comes from the derivation and meaning of this exponential term. Describing the full derivation would involve an entire lecture, but to point you along the right track, the exponential term is known as the Boltzmann Factor and ultimately comes from the fact that particle energies are very often distributed according to the Boltzmann distribution. That negative sign is simply an artifact of the Boltzmann distribution concerning how particles like to probabilistically arrange their energies. If that sign was a positive, the universe would just blow up so be careful with your signs!


Let me try to add some numbers to Steve's answer.

The Sun's luminosity is about $L_{\odot}=4\times10^{26}\text{ J/s}$. Now, if we assume that the majority of that energy comes from nuclear fusion, we have
$$L_{\odot}=\frac{dE}{dt}=\frac{d(mc^2)}{dt}=c^2\frac{dm}{dt}=c^2\dot{M}_{\text{nuc}}$$
Therefore, we can write the mass-loss due to nuclear fusion as $\dot{M}_{\text{nuc}}=L_{\odot}/c^2\sim4\times10^{9}\text{ kg/s}$.

Now, the Sun also loses mass via the solar wind. I believe the mass loss rate is a few times about $\dot{M}_{\text{wind}}\sim3\times10^{-14}M_{\odot}/\text{yr}\sim2\times10^9\text{ kg/s}$. Coronal mass ejections, solar flares and similar phenomena are often accounted for in this figure; at any rate, they don't add a significant amount.

This figure should increase as the Sun moves into the red giant (and especially asymptotic giant branch) phase of its life; it winds will increase to become much more significant that they are today. However, fusion will also increase, and the Sun's luminosity will increase by a factor of 1000 or 10000; therefore, the mass loss due to fusion will also increase. However, I think mass loss due to winds will increase more significantly, possibly by a factor of $\sim10^6$ - but I'll check that.

At any rate, an older star will have a lower mass than a younger star. For a Sun-like star, this would be hard to measure over the span of a human lifetime - remember that it only loses about $10^{-13}$ solar masses each year. Long-term records could make it possible, though, through various observations of the Sun and its effects. Centuries or millennia of data-gathering do tend to yield results.


If there are neutron stars, would most stars be considered "proton stars"?

Protium is a proton + an electron.

Under enormously high pressure, it's energetically favorable for electrons to merge with protons and become neutrons - see here.

  are stars mostly protons

By mass, yes, at least before they get too old.

The mass of the universe is more complicated, but anything solid that we think of as matter is made of atoms, which are by mass, mostly protons and neutrons (you can break it up further than that if you like, but that's best for another question).

Most hydrogen has no neutrons so any hydrogen rich object (the sun, most young stars, gas giant planets) are by mass, mostly protons.  That's no longer true when a star gets close to the end of it's life and has burned much of it's hydrogen.

Jupiter, by mass is roughly about 80% protons.  The sun, because it's been turning hydrogen into helium for about 4.5 billion years, is roughly about 67% protons by mass.   The Earth, mostly other elements, Oxygen, Silicon, Iron, etc, is a about 50% proton by mass.

  Would stars like our Sun be considered "proton stars"?

I suppose you could use that term, but I don't see any benefit to it.    It's not too different than calling the sun a "hydrogen" star.  All stars start out as hydrogen stars.


At what cosmological redshift $z$, does the recession speed equal the speed of light? How is it calculated?

From Friedmann Equation, distance as a function of redshift is:

$$d(z)=\frac{c}{H_0}\int_0^z \frac{dx}{\sqrt{\Omega_{R_0}(1+x)^4+\Omega_{M_0}(1+x)^3+\Omega_{K_0}(1+x)^2+\Omega_{\Lambda_0}}}$$

The Hubble-Lemaître Law:

$$v=H_0 \cdot d$$

We want $\boxed{v=c}$ The distance that fulfils this condition is known as Hubble Distance, (or Hubble Radius, or Hubble Length):

$$d_H=\frac{c}{H_0}$$

Combining both, we obtain the condition:

$$\int_0^z \frac{dx}{\sqrt{\Omega_{R_0}(1+x)^4+\Omega_{M_0}(1+x)^3+\Omega_{K_0}(1+x)^2+\Omega_{\Lambda_0}}}=1$$

For $\Omega_{R_0}\approx 0 \quad \Omega_{K_0}\approx 0 \quad \Omega_{M_0}\approx 0.31 \quad \Omega_{\Lambda_0}\approx 0.69$

The condition is:

$$\int_0^z \frac{dx}{\sqrt{0.31(1+x)^3+0.69}}=1$$

Searching by trial and error, we find that the value of redshift that fulfils the condition is: $$z=1.474 \approx 1.5$$

I hope I am not breaking any rules by repeating here the solution I wrote on Physics StackExchange
I did it because the creator of the thread asked me to do it, there.

Best regards.


Does a black hole necessarily contain a singularity?

Yes, an event horizon will form once escape velocity exceeds the speed of light. It does not necessarily require a singularity inside it.

However, there is no known physics that would allow any non-point-like object to exist inside an event horizon. It is hard to see how any traditional forces would be able to support an object in such case, and for quantum mechanics we simply do not have the theory that'd be valid in such high gravitational field.


The limit to the neutron star mass is at least $2M_{\odot}$, since at least two have measured masses as large as this (e.g. Demorest et al. 2010).

The answer to your question is the same as the answer to what is the maximum mass of a neutron/quark star, since if you compress matter of any sort, this is ultimately what it will become.

The answer to this question is also unknown and depends on the uncertain equation of state (the relationship between pressure, density and composition) at ultra-high densities, but must be somewhere between the $2M_{\odot}$ I mentioned above and an upper limit of around $3.5M_{\odot}$, which is imposed by General Relativity and an equation of state where the speed of sound equals the speed of light (the hardest possible equation of state).

Whether these highly compressed objects are "solid" is also a matter of debate and research. The conventional picture of an old-ish white dwarf is that it is indeed a crystalline solid. The bulk of the interior of neutron stars of moderate mass is almost certainly a fluid of neutrons along with some protons and electrons. The neutrons may solidify at very high densities and this is thought to be one of the "harder" options for the equation of state.

To answer a comment by @zephyr - cold in this context means that the Fermi energies of degenerate fermion species are very much higher than $kT$. White dwarfs and neutron stars (older than a few seconds) are "cold", despite having interior temperatures, in the case of the latter, of a billion degrees. Making them even colder does not change the size/density of the object but does enable the interior of a white dwarf to crystallise.


Do elliptical galaxies look similar to globular clusters?

Your question is not very well defined since you don't say what observational properties you wish to compare. From your emphasis on distance, I assume that you want to compare the visual appearance of elliptical galaxies with globular clusters of the same angular extent on the sky.

In which case ellipticals and globulars are easily distinguishable with the aid of a telescope (I assume you do not mean naked eye objects, since the nature of globular clusters is pretty difficult to make out with the naked eye, but if you do mean that then of course most smudges on the sky are indistinguishable with the naked eye). The reason for this is that true elliptical galaxies contain factors of $\sim 10^4$ more stars than even the most massive globular clusters and are about 100 times the physical diameter. Thus, if they had the same angular extent, an elliptical galaxy be $\sim 100$ times further away, would look much smoother and less grainy than a globular cluster, because it contains $10^4$ times more stars per unit area, but would have around the same surface brightness.

Of course there are such things as dwarf spheroidal galaxies, but here the problem is the other way around. They have about the same number of stars as globular clusters, but have diameters a factor of 100 larger. Thus if they had the same angular extent they would look similar to a globular cluster in terms of "graininess", but since they would be 100 times further away, they would be $10^4$ times fainter!


It is true that a surprisingly large number of stars are smaller (and thus less massive) than the Sun. However, the stars that are bigger than the Sun are often much bigger.

Look at this chart:

Image courtesy of Wikipedia user Jcpag2012 under the Creative Commons Attribution-Share Alike 3.0 Unported license.

Notice how small the Sun is compared to some of the other stars. It's tiny! It is indeed a small star - in technical terms a main sequence dwarf.

However, despite its size, it is clear that there are many more stars less massive than the Sun that there are stars more massive than the Sun. Why? There are two reasons:

Lower-mass stars live longer.
More low-mass stars can form in a given region than high-mass stars.

Encyclopedia of Astronomy and Astrophysics

The distribution of masses can be quantified in an initial mass function, typically given in the form
$$\xi(m)=km^{-\alpha}$$

When you integrate this over a range of masses, you can find how many stars are within that range. Not surprisingly, this number gets lower and lower as you slide the endpoints to more massive stars. You can see this decrease from the fact that $\xi'(m)<0$, so long as $k>0$ and $\alpha>0$ - which is assumed by the model, according to empirical data.


According to Lodders (2003, https://arxiv.org/pdf/1010.2746 ) the relative abundance of helium to hydrogen is $A({\rm He})=10.925$, on a logarithmic scale where the hydrogen number abundance is 12. So this would mean a helium to hydrogen ratio, by number, of $10^{10.925-12}=0.08414$. i.e. 8.4% (your source uses 10.93, not 10.99, hence a very slightly different percentage).
Why have the two systems? The appearance of the spectrum - absorption lines, emission lines etc. depends on their number density. So it is relative numbers of the elements that count. On the other hand when you are doing structural calculations on stars, it is usually the mass fractions that matter.
The relationship between the two?
The relative numbers of helium and hydrogen are easily translated into relative mass fractions of helium ($Y$) and hydrogen ($X$):
$$\frac{Y}{X} = 0.08414 \times 4.0026/[(1-0.08414)\times 1.0078] =0.3649$$,
where $4.0026/1.0078$ is the ratio of atomic masses of helium to hydrogen.
Now there is also a small fraction by mass of heavier elements $Z\simeq 0.014$, where $1 = X + Y +Z$.
If we substitute $X= Y/0.3649$ and $Z=0.014$, then
$$ Y \left(1 + \frac{1}{0.3649}\right) = 1-Z$$
$$ Y = 0.264$$
To sum up:
$$Y = \frac{1-Z}{\left(1 + \frac{1.0078(1 - 10^{A({\rm He})-12})}{4.0026\times 10^{A({\rm He})-12}}\right)} = \frac{1-Z}{\left(0.7482 + 0.2518\times10^{12-A({\rm He})}\right)}$$


Are the distances between two celestial bodies the distance of their centers?

Distance is an ambiguous term that can mean both centre-centre and surface-surface depending on the context it is used.

In cases where the size of the objects is trivial compared to the distance, 'distance' is almost exclusively used for the "distance between centres" of the two objects. This is the case for Sun-Earth, Earth-Moon and almost any other pair of significant solar system object. The alternative meaning must be specified if intended, like: "the distance from the surface of the Sun to Earth"
It may however also mean the "distance between surfaces", especially when the two objects are very close to each other. This is often the case with binary asteroids.

For  an orbital spacecraft, the word 'distance' is mostly not used when referring to the central object. The terms 'altitude' or 'orbital radius' are preferred. Altitude is the distance to the surface (or rather, to a theoretical sealevel. It may also refer to the actual surface though, so be careful). Orbital radius is the distance to the centre of the Earth.

As a side note "average distance" in an orbit is referring to the semi-major axis as that is the number used for calculating the orbital period and different velocities.


Can we track matter through time by looking at different depths in space?

Would it be possible to look deep into a certain part of space and
  time to find some galaxy that contributed to the matter that makes up
  the Milky Way today?

No, that's not possible. If we could do that, it'd mean that the matter traveled from there to here faster than its light got here, and matter can't travel faster through space than light does.

All we can do is look at similar galaxies to the Milky Way at earlier times. And because of the expansion of space those galaxies are now even further away from us than they were when they emitted the light that we're seeing now.

Galaxies develop (mostly) in isolation from one another, apart from the occasional merger or collision between neighbouring galaxies. Intergalactic distances are fairly huge, so it takes vast amounts of time for matter to travel from one galaxy to another, and matter is mostly bound by gravity to the galaxy it's in. Galactic escape speeds are pretty high, although the occasional star does get flung out of the galaxy by cataclysmic events like supernova explosions. But even then, such rogue bodies mostly end up in intergalactic space. The odds of them ending up in a different galaxy are pretty slim.