Formation of a magnetosphere for gas giants / small stars This question got me thinking about this.

Jupiter has, and presumably super-Jupiters can have a strong magnetospheres.  A solid metallic core and rotating material around the core, creates a magnetic field and all 4 gas planets in our solar-system have magnetic fields, though Jupiter's is by far the strongest.   Source

As I understand it, a planetary magnetic field requires a solid core and a slight variation in rotation where the liquid outer core rotates in relation to the solid inner core.  The planet's speed of rotation might not be necessary, but it could be a factor.

During the hot period of formation, gas giants might not form magnetic fields but when they cool down enough to have a solid core, they probably would have them.

So, the question is, is there a rough mass for which planet-wide magnetic field formation becomes unstable, where the core of the heavy Jupiter or small or old brown dwarf takes a very very long time to solidify, say, many billions of years?

I'd think the high temperature of formation and additional heat from fusion in a Brown Dwarf wouldn't be a good situation for that kind of star-wide magnetic field to form due too inevitable convection and probably no solid core.   That kind of star would probably form multiple magnetic dipoles like our sun has, but perhaps in an older brown dwarf a single magnetic field might be possible.

White dwarfs have very strong magnetic fields, so pressure obviously isn't an issue, in fact, I think high pressure can create a stronger magnetic field.

We do not know whether Jupiter has a solid core. That is one of the primary goals of the Juno mission.

In general it is clear that a solid core is not required to generate magnetic fields. The Sun does so and magnetic fields have been measured (with average surface field strengths of around 3 kG) on very low mass stars. Incidentally, the Sun's magnetic field does have a large scale dipole component.

More recently, there have been observations of cyclotron and auroral emission suggesting that even old, cold brown dwarfs have strong, organised magnetic fields. See for example Kuznetsov et al. (2012); Hallinan et al.(2015) and references therein.

http://arxiv.org/abs/1111.7019
http://arxiv.org/abs/1507.08739

At the moment, nobody can say for sure how exactly these fields are generated, sustained and organised.

In fact you could do worse than look at a paper that popped up at the top of today's astro-ph listings by Williams et al.  The abstract begins:

"The well-studied M9 dwarf TVLM 513-46546 is a rapid rotator (P_rot ~ 2 hr) hosting a stable, dipolar magnetic field of ~3 kG surface strength."

http://arxiv.org/abs/1511.05559


how far away could we detect that Earth has life? My guess is that life bearing planets are too far apart to be detected. I think we can only find the ones within a sphere around our planet that is 100s of light years in diameter but I suspect life bearing planets may be much farther away than that.

I would like to estimate the diameter of the sphere within which we could detect life on another planet and then estimate the probability that there is life within that sphere.

For example, give our current technology what is the furthest distance would be able to detect life on Earth?
How many stars like our sun are in that sphere?
How long would it take for SETI to rule out each of those stars?

I had put off answering this question because it seems too broad without specifying what sort of detection methods are proposed. But if you answer it directly from the perspective of - if we were to take the solar system and put it at some distance from us, would we able to detect signs of life on planet Earth - then the answer is probably not.

Using current technology (and by that I mean experiments and telescopes that are available now) we would probably be unable to detect life on Earth even if observed from a distance of a few light years. Therefore there are no stars within this sphere (other than the Sun).

No planets quite like the Earth have yet been detected around another star. That is to say, none that have a similar mass, radius and orbit at 1 au (or close to it) from a solar-type star [EDIT: There is of course now a close contender in Kepler-452b, though it is 60% bigger than Earth; Jenkins et al. 2015.]. With current technology, it is just about in reach. Therefore any directed search for life on Earth has a limited number of places in which to start. If you can't detect the planet at all then there is absolutely no chance of looking at its atmospheric composition to look for biomarkers (e.g. oxygen along with a reducing gas like methane, or chlorofluorocarbons from an industrial civilisation - Lin et al. 2014). The only exoplanets for which atmospheric compositions have been (crudely and tentatively) measured are "hot Jupiters". - giant exoplanets orbiting very close to their parent stars.
A "blind" search could look for radio signatures and of course this is what SETI has been doing. If we are talking about detecting "Earth", then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio "chatter" and accidental signals generated by our civilisation.
The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): "Typical signals, as opposed to our strongest signals, fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star". Quoting from Tarter (2001): "At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...". The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches.

Hence my assertion that with current methods and telescopes there is not much chance of success. But of course technology advances and in the next 10-20 years there may be better opportunities.

The first step in a directed search would be to find planets like Earth. The first major opportunity will be with the TESS spacecraft, launching in 2017, capable of detecting earth-sized planets around the brightest 500,000 stars. However, it's 2-year mission would limit the ability to detect an Earth-analogue. The best bet for finding other Earths will come later (2024 perhaps) with the launch of Plato, a six-year mission that again, studies the brightest stars. However, there is then a big leap forward required to perform studies of the atmospheres of these planets. Direct imaging and spectroscopy would probably require space-borne nulling interferometers; indirect observations of phase-effects and transmission spectroscopy through an exoplanet atmosphere does not require great angular resolution, just massive precision and collecting area. Spectroscopy of something the size of Earth around a normal star will probably require a bigger successor to the James Webb Space Telescope (JWST - launch  2018), or even more collecting area than will be provided by the E-ELT in the next decade. For example Snellen (2013) argues it would take 80-400 transits-worth of exposure time (i.e. 80-400 years!) to detect the biomarker signal of an Earth-analogue with the E-ELT!

It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio "chatter" out to distances of 50 pc ($\sim 150$ light years) - see Loeb & Zaldarriaga (2007). This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals. A good overview of what might be possible in the near future is given by Tarter et al. (2009).


Presenting / printing Right Ascension and Declination values Are there any standards or established practices on how to write / represent Right Ascension (RA) and Declination (DEC) values in the sexagesimal format?
I have seen all of the following. E.g. for RA:

02 03 04 (assumed to mean 2 hours, 3 hourly minutes and 4 hourly seconds)
2 3 4 (same as above)
2h 3m 4s (same as above)
2d 3' 6" (assumed to mean 2 degrees, 3 degree minutes and 4 degree seconds)
2d 3m 6s (same as above)
15.33 (assumed to mean 15.33 degrees)
9.5H (assumed to mean 9 hours, 30 hourly minutes and zero hourly seconds)
8d 3.5 (assumed to mean 8 degrees, 3 degree minutes and 30 degree seconds)

Likewise for DEC (even though for DEC when the first component does not have a suffix it is assumed by default to be in degrees, not hours). Does any astronomical association publish guidelines for the printing / presentation of such values or on how software should accept user's input of such values?

There are no standards for representing right ascension and declination apart from the abbreviations RA and Dec, and the symbols $\alpha$ and $\delta$.

The use of decimal degrees, dms, °'", hours and decimals, hms etc. depend on the [author's] style and intent. For example, determining transit times from local sidereal time and right ascension is easier when RA is expressed in hours, whereas doing calculations of angular separation is easier to calculate in software when decimal degrees are used.


Do planets migrate suddenly or gradually? Current models of the configurations of the planets conclude that Neptune once migrated outwards. AFAIK this refurbishment came about by Jupiter and Saturn gradually getting into and leaving a gravitational resonance.

Did Neptune (and possibly a fifth giant planet) redirect itself in a sudden encounter event, like comet 67P Churyumov–Gerasimenko did in 1959, halving the perihelion of its orbit since then? Or did it happen gradually over maybe millions of years? Is planetary migration in general sudden or gradual? Is it a matter of a single encounter event, or an adaptation to a changing gravity landscape?

Thommes et al. (2001) ran simulations and found that, at optimal conditions (namely, a planet of ~ 10 Earth masses), migration can be complete with ~ 100,000 years. Note that this was done before in-depth research was done on the Nice model, which is very similar. However, the mechanisms are different, as are the planet masses. The difference in timescales is dramatic.

Levison et al. (2007) did explore their own model - the Nice Model. They found that it took 60 million years to 1.1 billion years for Jupiter and Saturn to break their resonance. The period of encounters and scattering lasted for 878-885 million years, followed by a period of eccentricity damping lasting for 0.3 million years to 4 million years.

What was relatively quick was the ejection of the hypothetical 5th giant planet. The change in the other giants' orbits, however, was not.

So yes, planetary migration on this scale takes a long time. A very long time.

For some really interesting results, see the graphs of semi-major axis vs. time from the various four-, five-, and six- planet models of Nesvorný  & Morbidelli (2012). There are some incredible oscillation among the orbits of Uranus and Neptune in some of the simulations, which is eventually slowly damped.

  Fig. 14.— Orbit histories of the giant planets in a simulation with five initial planets. See the caption of Fig. 1 for the description of orbital parameters shown here. The five planets were started in the (3:2,3:2,2:1,3:2) resonant chain, Mdisk = 20 MEarth and B(1). The fifth planet was ejected at $t =$ 6.1 Myr after the start of the simulation.

There's another relevant passage, if you're looking at setups with six giant planets:

  The instability typically occurred in two steps, corresponding to the ejection of the two planets. Sometimes, as in Fig. 18, the ejection of the two planets was nearly simultaneous, but most of the times there was a significant delay between ejections. This was useful because the first planet’s ejection partially disrupted the planetesimal disk and reduced its capability to damp e55, which was then excited by the second planet’s ejection. While this mode of
  instability can be important, we would need to increase the statistics (>100 simulations for each initial condition) to be able to properly resolve the small success fractions in the six-planet case.


How does one measure vector magnetic field of astrophysical object? Magnetic field strength is measured using Zeeman splitting. This is the one of the way Sun's magnetic field strength is measured. Now, how one measure vector magnetic field?
Vector magnetic field = Magnetic field and it's components.

Using spectropolarimeter one can measure vector Magnetic field.

A spectropolarimeter based on a ferroelectric liquid-crystal modulator is described. An optical system with spatial modulation of the positions of the components of Zeeman splitting is a specific feature of this instrument. In comparison to the familiar instruments, the developed spectropolarimeter utilizes the light flux more efficiently and contains only one photodetector array. An operating spectropolarimeter developed at the Sayan Solar Observatory is considered as an example. Comparative estimates of noises in different operating modes are presented.

(Ref: Kolobov, D.Y., Kobanov, N.I. & Grigoryev, V.M. Instrum Exp Tech (2008) 51: 124. https://doi.org/10.1134/S0020441208010156)


Do we know if we are overtaking or falling behind our arm? Here's our galaxy as best we know, courtesy of this Wikipedia article:

It seems that as best we know, each arm is a compressed or brighter area (rather like a wave).

Looking at that image, do we know/guess if the arm-patterns are rotating? Or do they just sit there statically, unchanging from that diagram? Which way does it rotate, cw or acw per that diagram? (If we know.)
If the pattern is rotating - how long does it take it to rotate 360 degrees? As far as we know?

And hence ..

Indeed, is the arm-pattern catching us and moving through us ... or is the arm pattern slower than us and we're moving through it?

The answer to the question depends on precisely where you are in the Milky Way. According to density wave theory, which appears to describe the formation and evolution of spiral arms quite well, the arms move relative to some static inertial frame at a global pattern speed $\Omega_{gp}$. At some defining radius $R_c$, known as the corotation radius, stars move at $\Omega_{gp}$. Inside $R_c$, stars move quicker and thus move ahead of the arms; outside $R_c$, stars move slower and thus "fall behind" the arms.

There is uncertainty surrounding the values of $R_c$ and $\Omega_{gp}$ in the Milky Way in all its spiral arms - especially since local stellar kinematics differ in each of the arms, and so too may $R_c$ and $\Omega_{gp}$ - but measurements place the Solar System slightly inside the corotation radius, meaning that we are moving ahead of the arms.

I have data from a review by Gerhard (2011). Here are some basic average parameters, for the Milky Way as a whole:

Bar corotation radius ($R_{CR}$): $3.5$-$4.5\text{ kpc}$
Spiral arm corotation radius: $8.1$-$8.4\text{ kpc}$
Bar pattern speed ($\Omega_b$): $50$-$60\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$
Spiral arm pattern speed ($\Omega_{sp}$): $17$-$28\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$

This also goes along with a local mean speed of stars in the solar neighborhood - the Sun included - of $\sim220\text{ km}\text{ s}^{-1}$.

Hence for the three questions: (1) The pattern is rotating clockwise as seen in that diagram; (2) It rotates about once every 300 million years; (3) and at our radius, we're moving a little faster than the pattern so we are overtaking it.


Could neutrinos be at least a part of Dark Matter? What we know about neutrinos - its are almost invisible the same as Dark Matter, right?
Why Dark Matter could not consist of neutrinos?

Ethan Siegel has a good discussion of this question here.  Read that for more detail, but briefly, the main issue is that neutrinos move too fast - they are a form of what cosmologists call "hot dark matter".  The high speeds mean that they wouldn't settle into dense areas (e.g. galaxies, and clusters of galaxies) as dark matter is observed to do.   Evidence from structures we see in the universe tell us that most dark matter must be cold.
There certainly can be some hot dark matter in the form of neutrinos, but if there were a lot of it, it would leave an imprint on the cosmic microwave background that is not observed, constraining this component to be about 1% of the mass in the universe.


Computing the local orbital frame vector directions in a geocentric reference frame In the context of the study of a cubesat (little satellite), I am asked to define the local orbital frame and compute the local orbital frame vector directions as a function of time :

Here the context :

Anyone could give me some tracks or links to find the answer of part 2.1 Local orbital frame ? this would be kind.

Regards

I would interpret this as the Local Vertical, Local Horizontal (LVLH) frame (see, e.g., Vallado and McClain, Fundamentals of Astrodynamics and Applications, 4th ed., p. 156). This is also known as the RIC frame (radial, in-track, cross-track) or the RTN frame (radial, transverse, normal).

The radial or local vertical vector $R$ is just the position vector from the center of the Earth to the spacecraft.
The in-track, transverse, or local horizontal vector $I$ is the vector in the orbital plane that is both prograde and perinduclar to the radial vector. It is not, in general, the velocity vector. (The in-track and velocity vectors are coincident if and only if the flight path angle is zero, i.e., you have one of a) a circular orbit, b) the spacecraft is at apoapsis, or c) the spacecraft is at periapsis). You can construct it by applying the Graham-Schmidt Process to the position and velocity vectors; $I = v - \frac{R\cdot v}{R \cdot R}R$.
The cross-track or normal vector $C = R \times I$ completes the right-handed set.


Is our universe stable or metastable? I've been reading and thinking about this subject for some time, but I can't seem to find a clear answer.

After reading Q14:

  Aspects of the Higgs boson suggest that our Universe is only
  “marginally stable,” or is perhaps in a transition phase to a more
  stable state that would result in a universe with fundamentally
  different properties.

And this article:

  With the discovery of the Higgs field, one question we can ask is
  whether the Universe as a quantum system it is in a stable, low-energy
  state, or in a metastable higher energy state?

According to the standard model of particle physics, the answer depends upon the masses of the top quark and the Higgs boson. The more massive the Higgs boson, the more likely the Universe is to be stable. The more massive the top quark, the more likely the Universe is to be metastable, or even unstable. From our current measurements of the top and Higgs masses, it seems that our Universe is metastable.

Maybe an answer already been found, but I just didn't find it, or am I not on the right level.

Is our universe stable or metastable?

Yes ... and no.
Experimentally, there's no indication that the laws of physics are changing (c.f. conservation of energy: if the laws of physics change with time, energy would not be conserved via Noether's theorem). So if you're an empiricist, you'd conclude that the universe is stable (or at least its mean lifetime must be more than ~14 billion years, the age of the universe).
However, if we believe string theory (which has no compelling experimental evidence), then we expect roughly $10^{500}$ "vacuum states", which correspond to $10^{500}$ different possible universes, each with their own laws of physics. It's possible one of these other universes have a lower "total energy" than us (this word's meaning is vacuous since if one varies the laws of physics, the word 'energy' itself might lose meaning). This means it's in principle possible that our universe will quantum tunnel to a universe at lower energy, in which case our universe is metastable.
Of course, this leads to other problems. How would one prove the multiverse exists? If a generally accepted way exists, I've not heard of it. But if you can't prove the multiverse exists, how would you prove our universe is metastable? For that matter, how do we know the electron is stable? Why can't it have a mean lifetime of, say, $10^{100}$ years (so long that we don't observe it decaying)?
At this point, we're purely in the realm of speculation. We're talking about things so far above the experimental threshold that some people will just dismiss it as pointless. Suffice to say that if the universe isn't stable and quantum tunnels to a lower energy, we won't be able to see the tunneling coming, and we're all instantly dead anyway. Whether the question is worth thinking about further depends on the individual.
tl; dr:

Experimentally, universe is stable.
Theoretically, maybe it isn't, depending on what your theory is because we don't have a widely-accepted, experimentally-backed theory.


Warm and Hot dark matter density profiles For cold dark matter, density profiles are well known and easy to find information about - eg. NFW, Burkert, Einasto, and others.

But for some reason I couldn't find explicit expressions for the density profiles for Hot and Warm dark matter.

I need to know what are $\rho_{_{HDM}}(r)$,$\rho_{_{WDM}}(r)$.

I'm interested in the differences between cold/hot/warm dark matter and why do we use Cold dark matter ($\Lambda$-CMD  model) and not Hot or Cold dark matter. I need to back my claims with formal quantitative analysis and not just a qualitative explanation.

If you can recommend about other aspects which differentiate the models, or a good review paper on the subject, I would welcome it.

There are indeed models for halos made of warm dark matter. The profiles will depend on exactly what the dark matter is and what its temperature is. I am no expert, but a starting point could be a paper by Vinas et al. (2012), who claim that the profiles are much flatter in the centre than those of CDM halos.

http://arxiv.org/abs/1202.2860

Hot dark matter is defined as such during the early universe and will be nearly uniformly distributed at the epochs of early structure formation. However at later epochs, hot dark matter with rest mass can cool down. For example, the cosmic neutrino background would have been highly uniform when it decoupled and when  most galaxies formed. But because of the non-zero neutrino rest mass and the expansion of the universe, these neutrinos are now non-relativistic and will be affected by cosmic structures and become "clumpy". The same will be true for any dark matter particle with a rest mass greater than about 0.1 eV.


Can we detect a black hole's radiation? Stephen Hawking proposed that a black hole emits radiation as it evaporates.

Black holes don't let any radiation escape their event horizon because even particles with the speed of light get trapped by their magnetic field.

Does that mean when black holes emit radiation, they just re-absorb it?

The black hole itself does not emit the radiation, but rather the space just around it. Because of quantum effects, small particles and anti-particles will spontaneously create everywhere (and later annihilate themselves in a collision). If this happens just outside the event horizon, one particle will inevitably fall into it, never to be seen again. The other particle is now 'free' and this is the radiation Hawking predicted.

  Can we detect a black hole's radiation?

Well, this radiation is way too small to be detected by itself, even if the black hole would be at our doorstep. However, the radiation makes the black hole less massive and smaller and finally it will vanish, producing a gamma ray flash which we hope is detectable, even from a very large distance.
As @JamesK notes in the comments, this evaporation will take extremely long for the 'common' black holes caused by supernovae and those found in galaxy centers, but smaller primordial black holes, formed in the very beginning of the universe, could be evaporating now and that's why we have a telescope looking for those gamma ray flashes.


What equipment do I need to measure the spectrum of a star? For an amateur astronomer, is it possible to measure the spectrum of a star?

I can easily take a photo of a star just by putting a camera on one end of a telescope and over exposing it.

But I would like to measure the spectrum of a star. I imagine I could do some crude approximation by using different filters. But I would like the full spectrum to see what the composition of a star is.

How would I go about this?

Depending on the type of camera you use, you may be able to photograph stars without a telescope. I know that there are apps for cell phones that allow you to take long exposures, and there are some point-and-shoot cameras that have a "bulb" setting or allow multi-second exposures, and certainly an SLR-type will allow it. You'll need to stablize your camera thought.
Either way you can use either a transmission diffraction grating (glass or low-cost flexible film-type) or a wedge prism to disperse light into different colors.
If you use a free-standing camera, you can afix the dispersing element in front of the camera's lens. If you are photographing through a telescope you'll need to put it at the camera end. The disperser will usually spread light out by several degrees or several tens of degrees, so if you put it in front of the telescope lens the spectrum will be spread out way too far, unless you are using a very low-dispersion grating or prism specifically designed to be put in front of a telescope!
If the disperser partially covers the camera, then you'll get both an image of the star and of the spectrum. That's helpful to identify the position of lines, since you can measure the distance from the undispersed star to each line. If you use a grating that is not blazed or imperfectly blazed, then you'll get both +1 and -1 orders and you can measure the distance between the two appearences of a line without needing the central start, but it will be there anyway because most gratings will pass at least some of the 0th order.
Since the star light is spread out over a much larger area, the spectrum will be much dimmer than the undispersed image, so as you mention, you'll have to overexpose.
There's an example of a commercial product in the question Has anyone seen an actual spectrum of a satellite made by an amateur photographer?

I don't mean to advocate a commercial product, this is a handy example, and Tom Field is a contributing editor to Sky and Telescope magazine.

above: from here.

above: from here
See the FieldTestedSystems' YouTube video How to capture star spectra in your backyard


How does one "use" the Pan-Starrs data? I've seen that a large block of Pan-Starrs data has been released and is available on-line and how the distributors of the data expect professionals and the general public to use the data in their research. I am part of the general public and was wondering how accessible the data actually is?  For instance if i were interested in searching a small section of sky for a moving object, i would need multiple "photos" "data sets"? over time in order to see the movement.  Would special software be required? How would one even start the process?

There are two main datasets available from the PanSTARRS survey which are available from the archive at MAST/STScI. These are the Object Catalog, a (large) list of parameters such as position, brightness, shape etc (full list of catalog fields) and the Image Cutouts service which, well, cuts out sections of the images. (The survey went over the same patch of sky several times in 5 filters so there are deeper stacks available)

If you are looking for moving objects then you will need to use the "warp" single epoch images as a moving object would be averaged away in the stack (unless it was done on the object's motion). You will also need a source of positions in order to cutout the suitable postage stamps in the right places as the PanSTARRS archive doesn't support querying by moving objects or orbital elements. You would need to use something like the Minor Planet Center's Ephemeris Service or JPL's HORIZONS service to generate these positions (use 'F51' as the Observatory Code/Observers Location as this is the MPC site code for PanSTARRS1). Once you have a source of positions, you will probably want to investigate querying the Image Cutout Service via script/code; the Image Cutout Service documentation has instructions and a Python Jupyter notebook showing how to do this.

I should also caution you that one of the goals of the PanSTARRS survey, which my institution was a partner in, was to find moving objects. A lot of effort was put into the Moving Object Processing System (MOPS; paper, which I'm on, available (for free) here at IOPscience) and PanSTARRS was one of the two major sources of new Near Earth Objects (NEOs), see purpley bars in this plot at the Center for NEO Studies. I'm not saying there isn't new moving objects in there, we go back and search through the PanSTARRS data for precoveries of newly discovered NEOs that the automatic software missed on a roughly monthly basis, but bear this in mind before launching into a big project.


Measures for the degeneracy from protons? Come to think of it, the only things we hear about degeneracy is neutron and electron.
I made a generalization:

White Dwarf: Supported by electron degeneracy.
Neutron Star: Supported by neutron degeneracy and nuclear forces, and sometimes centrifugal force.
Quark Star: Most likely supported by quark degeneracy.

Now, we can see that there is no mention of proton degeneracy.

A quick search lead to the Wikipedia subsection Degenerate matter; Proton degeneracy, and the answer that yes, there is proton degeneracy, but is much less effective than other forms of degenerate matter.

My question: What is the force exerted by degenerate protons, and what equation is it modeled by?

Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same.

Since the Fermi momentum $p_F$ of a species is dependent only on number density$^1$, the Fermi momentum of the electrons and protons are the same. Pressure however$^1$, is an increasing function of $p_F/mc$ and since the mass of an electron is 1800 times smaller, the electron degeneracy pressure is always much higher than the proton degeneracy pressure.

Degenerate protons do exist. They form of order 1% of the density in the neutron fluid region (the bulk) of a neutron star. However, they contribute much less than a thousandth of the pressure and always lower than the degenerate electron pressure.

Actually, the form of the pressure equation (see below) does actually admit the possibility that the degeneracy pressures of electrons and protons become more similar as $p_F/mc \gg 1$ (aka ultra-relativistic). But by that time, the energies of the electrons and protons are easily sufficient to create neutrons and an equilibrium is set up so that neutrons always outnumber the protons by at at least 8:1 and therefore their degeneracy pressure will dominate.

(1) For those interested: The Fermi momentum is
$$p_{F} = \left(\frac{3}{8\pi}\right)^{1/3} h n^{1/3}\ ,$$
where $n$ is the number density of fermions.

Whilst ideal fermion degeneracy pressure is given by
$$P = \frac{\pi m^4 c^5}{3 h^3} \left[x (2x^2 -3)(1 + x^2)^{1/2} + 3
  \sinh^{-1}(x) \right]\ , $$
where $x = p_F/mc$.


What would be the sizes of the bodies created from Jupiter's trojans? If all trojans from the Greek camp and all the trojans of the Trojan camp were combined into two bodies respectively, what sizes and masses these bodies would have? Would they qualify as dwarf planets? Would they be greater than Ceres?

The mass of the Trojans (both camps as far as I know) is estimated at around 0.0001 Earth masses (reference) (Note that this number is debated). The density of them vary but the highest found is around 2.5 g/cm^3 (reference) Assuming that the composition is mainly rocky/icy with very little gas, the act of combining all of them should make a rocky/icy object with that mass.

The mass of Ceres is estimated to be 0.00015 Earth masses with a density of some 2.1 g/cm^3 (reference).

If we assume that the object made from the Trojans has the same density then they should roughly make an object the same size or slightly smaller than Ceres (about 1 % the mass of the Moon). Make that into two equal bodies and you have half a Ceres per group (so each smaller than Ceres) (assuming that there are the same mass of bodies in each group originally).

Given that there is much uncertainty associated with this, I would bet that the mass and size would be smaller than this.
Whether they would qualify as dwarf planets, I'm not sure. But they would have to be captured into the orbit rather than be formed there as I'm pretty sure the gravitational influence of Jupiter would prevent any coalescence and gravitational stability of the Trojans (though I could be wrong on this).


Formula to calculate the apparent magnitude of Earth from arbitrary distances I would like to calculate the apparent magnitude of Earth from various distances, say from the Voyager probe at this moment or from various stars in the sky. I know there must be a formula that I can enter in the distance from Earth (or the Sun) and get out the apparent magnitude. However, I am failing to find it as everything I can find assumes that I am on Earth and want to calculate the brightness of other things.

The absolute magnitude of the Earth is about $H = -3.99$. This means, from $1$ AU away from the Sun and the Earth, the Earth has an apparent magnitude of $-3.99$.
Voyager 1 is $152.26$ AU from the Earth, and $151.85$ AU from the Sun. Using the formula for apparent magnitude, $${\displaystyle m=H+5\log _{10}{\left({\frac {d_{BS}d_{BO}}{d_{0}^{2}}}\right)}-2.5\log _{10}{q(\alpha )}}$$
With $d_{BO} = 152.26, d_{BS} = 1, d_0=1,$ AU, and $q(\alpha) = 0.306$, we obtain that Earth's apparent magnitude from Voyager 1 would be about $\boxed{8.208}$.


How can we see stars if they are hundred years of light far from the Earth? So the right question is, is the light years counted from that second when you look up to the sky? If that so, how can we see the stars from the telescope (which I think it starts when you put your eyes on it, same as when you look up to the sky), that they say it's 5000 of light years from earth? If you say these lights are already started 5000 years of light ago, then there should be so many changes happened to these galaxies or stars in that timeline. Thus, for example, there should be no way to know, if an earth-like extrasolar planet is still earth-like.

You are right, there is no way to know, what is the current state of these objects "now" - if you think on the Newtonian mechanics.

For example, the 1054 supernova, what created the Crabs Nebula, happened actually not in 1054, but around in 6000 B.C. It is because it is $\approx$ 7000 ly away from us.

In the Special Relativity, "now" is relative. You can't say, that the 1054 Supernova happened in fact 7000 year ago: if you would see the whole story (1. the star explodes 2. the astronomers of a planet far, far away, some millenia later, detect it) from a spaceship going, for example, with half the light speed, you would get totally different numbers. There is no "objective now" in the special relativity.

For us, "now" is what we see in the telescopes. For us, the crab nebula explosion happened in 1054. We have no way to visit a star, for example, 50 light year away, now - we would need at least 50 years to visit it. (It is the theoreitcal limit - in practice, there is no technology to send anything even to the nearest star, 4.3 light years away. If there would be, it would likely require at least centuries to get there.)

However, the overwhelming majority of the stellar evolution processes is much longer. Most stars exists between 10million and 10billion years long. Of course we have no way to watch a star billions years long with the telescopes. However, we can see billions of stars, typically some decade (centuries) long, and even this is enough to understand a lot from them.

We have a strong reason to not expect a huge difference from the "current" state of most stars/extraplanets from what we can see today on them - simply because such events (change of planetary orbits, average lifetime of most stars) last million times longer.


Jupiter's magnetosphere interacts with the one from Saturn? Jupiter's magnetosphere extends "seven million kilometers in the Sun's direction and almost to the orbit of Saturn in the opposite direction" and Saturn's magnetosphere extends towards the Sun on average 22 Rs (Rs=60,330 km is the equatorial radius of Saturn).
My question is: do these magnetospheres actually touch each other when the planets are on the same line?

The simple answer is yes, because magnetic fields extend to infinity.

Your question should ask if a significant interaction is present. My question would be, what is a significant interaction?


Inability of Hubble to clearly resolve nearby celestial objects The Hubble space telescope performs incredibly detailed and stunning imaging of celestial bodies that are many, many light years away, and yet when it attempts to image objects within our own Solar System, such as of Pluto and Charon, the results appear out of focus and smeared. Why is this so?

I can only guess that the objects within the Solar System have a higher relative intrinsic motion than objects light years away, and thus Hubble can't track them so precisely? Or maybe Hubble's optics are "fixed" at an "infinity point" beyond "nearby" objects? Or...?

This has to do with the angular resolution of the Hubble telescope and the ratio between the distance of an object in space and its size in space. The galaxies that the Hubble telescope can see are bigger in size than they are far in light years away compared to pluto from earth.

Take the galaxy NGC 5584 for example:
It spans 50,000 light-years and it's 72 million light-years away which gives us a ratio of 0.00069

Then for pluto:
Pluto spans 2400 km in size and 4675 million km away which gives the ratio of 0.00000051

Take for example a clear view at a mountain with trees on it. You would see the mountain fairly clearly: you can see it has trees on it, perhaps snow, and other aspects. But if you look at any given tree on that mountain you can't make anything other than it's observable color and that it's most likely a tree. The relationship of distance in size is what gives you a good view of the mountain but not a tree on the mountain; the trees size to distance ratio is much smaller than the mountain's size per distance ratio.


Density of hydrogen between galaxies From watching a video on youtube, it states that Dark Matter is 6 times more abundant than known matter. I was wondering if it was possible to have hydrogen clouds between the universes galaxies. The density would need to be less than inside universes galaxies (since star formation would be possible).

Is it possible that hydrogen clouds be a major component of Dark Matter?

Hydrogen clouds don't even make up a small component of dark matter, because hydrogen is not dark. The image below depicts the emission spectrum of hydrogen in the visible regime (a.k.a., the Balmer series).

The emission spectrum of dark matter on the other hand would be completely black.

Dark matter is gets its name from the fact that it doesn't absorb or emit light, i.e., electromagnetic radiation. From an elementary particle standpoint, this means dark matter particles can't interact with photons via the electromagnetic force at all, so dark matter particles must be electrically neutral, which electrons and protons certainly are not. No electrons are protons means no hydrogen atoms.


I live in an area with a lot of light pollution, how can I view the stars without building an observatory? I live in an urban area with a lot of light pollution and very little garden space. On a clear night I can see most of the sky and the brighter stars are visible.

I can't build an observatory - I don't have the space (let alone the money required) - so what can I do to minimise the effects of light pollution so I can see more of the fainter stars and planets?

telescope.com has a quick paragraph on astronomy. Below I summarise the important points with a few of my own suggestions thrown in.

Dark & dirty places

Set up on grass or dirt, pavements and buildings radiate the heat again at night and the air flow created by this can distort your image. If possible you might be able to make use of a public park.

Failing the above you might be better off going to the outskirts of the city and finding a nice dark hill and avoid as much of the light pollution as possible, but obviously this depends on your situation.

Telescope

This doesn't require a special telescope, you should be able to manage just fine with what your budget allows for.

Though a computerised mount can aid you plenty in locating objects, as you have fewer stars to use as way-points in light polluted areas.

Here is an example I found earlier - yes they are quite pricey.

Note: This picture and the mount in the link are NOT the same mount.

Try to set up out of direct view of a source of light, i.e street-light or porch light. This is because you will suffer from a lack of contrast, making it harder to view the objects. So dark spots are good!

Light Filters

A good, fairly cheap method*, for negating light pollution that I have found is to use a light filter, these usually just screw on over the telescope and filter out light of that wavelength.

However, a very important point to note is that light filters will also filter out light from stars, galaxies and star clusters, making them appear duller.

Light filters actually decrease 'sky glow' which is, in part, due to the upper atmosphere.

What these filters are best for is looking at nebulas

Nebulas emit light of a different wavelength that are not dulled by the filter and they pass straight through.

If you are not hoping to look for nebulas then I do not suggest a light filter, as this will also dull the stars.

Examples of which can be found here

*cheap relative to buying a go-to mount that is


Where did the hydrogen come from in a type II supernova? Type II supernovae have hydrogen. Where did the hydrogen come from if that is the first element used up in the star's life cycle? Also if our solar system was seeded from an ancient supernova where did the sun's hydrogen come from?

Good question - the answer is that stars are not in general well mixed - or rather, the nuclear-burning core is not well-mixed with the rest of the star.
That means that a star will finish hydrogen burning even when about 80% of the hydrogen in the star is still available, but is situated outside the core and cannot be mixed into the core (or at least not very efficiently).
By the time a type II supernova explodes, a large fraction of its envelope (the region outside the core) is still hydrogen.
The Sun's hydrogen comes from the big-bang. Hydrogen is not manufactured in stars, it is a raw material that was created in the first minutes of a hot big bang 13.7 billion years ago. The same is mostly true for helium, but the heavier elements in the Sun are the products of nucleosynthesis of the billion or so stars that lived and died in our Galaxy before it was born. The material that they eject through stellar winds and supernovae of all types is thoroughly mixed into the interstellar medium.


How do we have photos of galaxies so far away? A possible answer for this is that, light emitted from the galaxies travelled a billion miles all the way to earth, where the hubble space telescope picked up this light through its sensors, and was able to construct an image of the galaxy

but if this is true, and galaxies are billions of miles away, shouldn't the light particles emitted from the galaxies be scattered all over the place? after all they have been travelling from millions of years, and have probably collided with asteroids and other foreign objects. What were the chances that about 95% of the photons actually reached earth, giving us a very detailed image.

Consider the andromeda galaxy which has a distance of 1.492 × 10^19 mi from earth. If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy, evident from the photo below?

Shouldn't like half of the galaxy be missing since photons could have hit other objects, and "never have reached earth"?

There are two reasons that often — but not always — light from galaxies millions and even billions of lightyears away make it through the Universe and down to us:

Particle number and particle size

First, the intergalactic medium (IGM) is extremely dilute. The number density of particles out there is of the order $n\sim10^{-7}\,\mathrm{cm}^{-3}$, or roughly 26 orders of magnitude lower that the air at sea level! That means that if you consider a tube from Andromeda to the Milky Way with cross-sectional area of $1\,\mathrm{cm}^{2}$, it will contain roughly one microgram of matter (thanks to Rob Jeffries for catching a factor $10^6$ error).
Second, even if a photon comes close to an atom, it will only be absorbed if its energy matches closely some transition in the atom. Since most of the atoms are ionized (and thus should be called plasma instead, but in astronomy the distinction if often not made), there are no electrons to absorb the photon. The photons are more likely to interact with the free electrons via Thomson scattering, but the Thomson cross section is immensely small $(\sim10^{-24}\,\mathrm{cm}^{2})$, so even if you consider the CMB photons — which have traveled through the Universe almost since the Big Bang — only around 5% of them have interacted with electrons on their way.

In other words: The amount of transmitted light depends on two factors: 1) The amount of matter along the line of sight, and 2) that matter's ability to absorb the light. In the IGM, both are tremendously small. When the light enters the interstellar medium (ISM) inside our galaxy, it may encounter denser clouds with atoms that are able to absorb the light. But usually (although not always) "dense" is still very dilute compared to Earth's atmosphere.

Mathematical expression

In general, if a beam of light traverses a region of particles, each with a cross section $\sigma$ (measured e.g. in cm$^2$), passing $N$ particles per area of the beam (measured e.g. in cm$^{-2}$), then the opacity of the medium is given by the optical depth $\tau$, defined by
$$
\tau \equiv N \, \sigma.
$$
The transmitted fraction $f$ of photons is then
$$
f = e^{-\tau}.
$$
In general $\sigma$ depends on the wavelength, and thus part of the spectrum may pass unhindered, while another part may be completely absorbed.

The figure below (from here) shows the spectrum of a quasar lying at a distance of 22 billion lightyears, i.e. $10\,000$ times farther away than Andromeda. You see that there are several thin absorption lines (caused by intervening hydrogen clouds whose densities are a factor of 10-100 higher than the IGM), but still most of the light makes it down to us.

Because the light we see from this quasar was emitted so long ago, the Universe was considerably smaller at that time, and thus the density was larger. Nonetheless, only a small fraction is absorbed. The farther away the light is emitted, the longer ago it was, which means smaller Universe, and higher density, and thus the more light is absorbed. If you consider this quasar (from here) which lies 27 billion lightyears away, you see that much more light is absorbed in part of the spectrum. Still, however, much light make it through to us.

The reason that it is only the short wavelengths that are absorbed is quite interesting — but that's another story.


What's the difference between a Newtonian telescope and a regular telescope So, recently I was shopping for telescopes and I decided on the following telescope:

https://www.amazon.com/Celestron-21049-127EQ-PowerSeeker-Telescope/dp/B0007UQNKY/

I had exactly $180 to spend, and that telescope fit the bill. However, I also noticed that it had a variety of options in sizes, mainly the following:

60MM EQ Refractor
70MM EQ Refractor
80MM EQ Refractor
114MM EQ Reflector
127EQ Newtonian Reflector

What is the difference between all these sizes/types, how does a "Newtonian Reflector" act differently than a "regular" (for lack of a better term) reflector and what are the benefits/drawbacks?

Bonus Points: Did I choose the "highest-quality" option for the price?

Side-by-side comparison of all sizes:

First off, I would not recommend any of these telescopes.  Of the ones you listed, the 114EQ is the least bad choice.  But, due to the poor quality of its mount, it's the better of the bunch.

Ok, before I get into the nuts and bolts, let me first offer a couple pieces of advice:

1.) Join an astronomy club or society.  In the US, membership usually runs $50 or less per year.  Most clubs offer guidance and advice to new members.  They also often conduct frequent star parties.  These are events where those who have them bring out telescopes and everyone spends the evening observing together.  At such events, you can see a lot of different types of telescopes of all different designs, prices, manufacturers, etc... and talk to owners about them.  Clubs also have other programs that can help those new to the hobby learn and better enjoy their observing time.

2.) The best telescopes I can recommend in the under $200 price range (which is where these seem to be) are the AWB OneSky 130 (also known as the SkyWatcher Heritage 130P outside of the US) or the Meade Lightbridge Mini 130.  Both are 130 mm Newtonian telescopes with fairly good optics for their price range and stable mounts.  Meade makes a 114 mm version, and so does Orion.  I would not recommend any telescopes smaller than that, as there's little they can show you that a set of binoculars in the price range wouldn't do just as well or better.

Now, your initial question was about the differences between them.  Please bear with me, I tend to be verbose and try to be thorough (in other words, grab a snack and settle in for the read).

First of all, let me clear up a common misconception many people who are new to amateur astronomy tend to have.  The primary purpose of an astronomical telescope is not to magnify, but to gather light.  The problem with astronomical objects is less one of apparent size (this is a secondary concern, and not unimportant, but not as critical) and more one of brightness.  For example, the Andromeda Galaxy, also known as object 31 in the Messier catalog (M31), has an an angular size of about 3° by 1.5°.  The full moon is about 1/2° wide - so M31 is a lot bigger in apparent size.  But it's very faint in comparison.  If you know where to look and have fairly dark skies, you can actually see it with your naked eye at times, but not very clearly.

All the light you see comes into your eye through your pupil.  The pupil of an average adult eye, when fully dark adapted, has a diameter of about 6 or so mm. That's about 28 or so square mm of light gathering area.   A modest 80mm telescope (80mm being the aperture) has an area of over 5,000 square mm - over 150 times as much light gathering area.  All that light is then funneled down and concentrated on your eye.  This makes dim objects much brighter.  Magnification helps, but it doesn't matter how big an object appears if you can't see it at all.

The light gathering ability of the telescope is dictated by its aperture - the size of the main optical element collecting light (either an objective lens or primary mirror).   As such, it's not really adjustable (well, you can adjust it downward by blocking the aperture, but you can't increase it).   Magnification, however, IS adjustable.  The magnification power of a telescope is calculated by dividing the focal length of the telescope by that of the eyepiece.  For example, if you have a telescope with a focal length of 1,000 mm and an eyepiece of 25 mm, you get 40x magnification. If you swap out the eyepiece for a 10 mm eyepiece, you get 100x.  Theoretically, you can magnify as much as you like, though you can only magnify so much before the image is a blurry mess (this is also a matter of aperture - larger apertures allow for higher magnification before blurriness sets in).

Thus, the most important consideration when choosing an astronomical telescope is usually the aperture of the telescope.

When most people who are inexperienced with telescopes hear the word telescope, they think of a refractor.  A refractor, or refracting telescope, uses a lens to collect light.  This lens is known as the objective lens (though sometimes referred to, erroneously, as the primary lens).  When light is passed through a substance, such as glass, it is slowed slightly by the substance.  An objective lens is shaped with a convex (outward) curvature.  When light passes through the thicker, center part of the lens it is slowed more than that in the thinner, outer portion of the lens,  This causes the light to refract, or bend. The symmetry of the lens causes the light to all bend toward a point along an axis through the center of the lens.  The point where the light rays meet is known as the focal point.  All the light it collects is concentrated here.  The eyepiece sits here to gather the light and reconstruct the image for viewing.

Refractors were the first telescopes made, invented in the early 1600's in the Netherlands.  People often mistakenly consider Galileo the inventor of the telescope.  In fact, the man most frequently credited with the invention is Hans Lippershey (though two other men sometimes are given credit).  Galileo was only the first to use it to examine the night sky.

Even in his day, however, there was a problem with the design.  When you refract light, you cause the colors to separate.  This is exactly how a prism works.  In fact, in a real sense, a lens IS a prism.  This is because different wavelengths of light, what we percieve as colors, are slowed at different rates.  Shorter wavelengths, which are bluer light, are slowed more than longer, redder, wavelengths.  What this means in a telescope is that the actual focal points of different colors of light are separated with bluer light coming to a focus a bit closer in than redder light.

We call this phenomenon chromatic aberration, or CA, and it can be seen in telescopes as ablueish blur or fringe around objects you're viewing.  It degrades the quality of the image.

There are two main methods for reducing its effects.  The first is to increase the focal length of the telescope. A lens with a longer focal length has less pronounced curve: the difference between the thinnest and thickest parts of the lens are less dramatic than in a lens with a shorter focal length. This means there's less dramatic bending, which leads to the focal points of the colors being closer together.  If you look at early telescopes, like those used by Galileo, they tend to be very long indeed with high focal ratios.  The focal ratio is the ratio of the focal length to the aperture.  My 8" (203 mm) Schmidt-Cassegrain Telescope (SCT) has a focal length of 2,030 mm, making its focal ratio f/10.  My 8" imaging Newtonian has a focal length of only 800 mm, making its focal ratio f/3.9.  I believe some of Galileo's early scopes had ratios like f/20 - the tube was 20 times longer than it was wide.  Mostly this was to reduce chromatic aberration (though it was also to increase magnification)

The other method of reducing CA is by using multiple lenses with different shapes and made of different glass materials.  The most common option is to use two complementary lenses that work together to shift the focal points of bluer and redder light to bring them closer together.  This is called a doublet or achromat.  Using three lenses makes a telescope a triplet and and is often called an apochromatic refractor.  Sometimes even more lenses are used.  But this comes at a price. Each lens modifies the light, and while correcting, can also introduce distortion.  It also requires 2 surfaces (front and back) per lens to be precisely configured with a specific curvature.  This is costly and time consuming.  Lastly, lenses are heavy, and with a refractor, they are mounted at the front, which puts the weight at the front, and makes larger and longer such scopes much more difficult to properly maneuver. The largest refractor ever made was only about 49 inches, and the largest used for research was only 40 inches.

Galileo and his contemporaries understood the problem and that it could be solved by using a mirror instead of a lens.  Unfortunately, in those days, mirrors were much more difficult to make.  They were made from solid metal, which is much more difficult to precisely form and polish.  Most often,they were made of a metal known as speculum metal, which tarnishes very easy.  It wasn't until Sir Isaac Newton came along about 60 years later that a reflecting telescope was first created.

Newton used a mirror with a parabolic curvature to gather light.  It sat at the end of an open-ended tube. Light gathered by this mirror was bent and directed toward a secondary mirror, which, mounted at a 45 degree angle, would send the light out the side of the tube to the waiting eyepiece.

Because a mirror of this kind is all metal, there's no glass to pass through, so the glass does not separate the colors of light.  This eliminated chromatic aberration (well, there was still some in the eyepiece, but not much).  But, again, mirror technology was a problem.  Still, a few people adopted the design, and even built very large reflecting telescopes.  William Herschel's telescope was about 48 inches in diameter.  It was much easier to maneuver such a telescope, as the weight was all at the rear of the instrument (not to say it was easy, but putting a 4 foot glass lens at the end of the 40 foot tube would have been exceedingly difficult to work with - very, very top-heavy).  Herschel's mirrors, however, tarnished easily.  He actually had multiple mirrors, so one was being polished while another was in use, that he could just swap out as needed.

In the mid 19th century, the process of depositing a thin layer of metal, originally silver, but now usually aluminum, onto glass was developed.  Glass, while heavy, is much easier to configure into a precise shape.  Once done, the reflective coating could be deposited onto the glass mirror blank.  Because the coating is on the front-side of the mirror, light doesn't pass through the glass, just bounces off the coating.  This made Newtonian reflecting telescopes much easier to produce.  Only the one surface, the primary mirror, has to be precisely configured, and this made them more cost-effective.  And the weight and balance issues made making larger and larger telescopes easier.

Newton's design wasn't the only.  Laurent Cassegrain and James Gregory both designed, but probably never constructed, reflecting telescopes relying on mirrors with a hole in the center.  The light is collected by the primary, sent to the secondary, and sent back toward the primary to pass through the hole.  In Cassegrain's design, the secondary is flat.  In Gregory's, it was curved.  Both allowed for a shortening of the overall physical length of the instrument.

In more recent times, the use of mirrors and lenses together lead to the development of the Catadioptric telescope.  The most common such designs are the Schmidt-Cassegrain Telescope (SCT) and Maksutov-Cassegrain Telescope (Mak).

Just as refractors suffer from Chromatic Aberration, Newtonian telescopes suffer from an aberration called Coma.  Coma appears as distortions of the stars around the edge of the field of view.  They appear to lengthen and have comet-like tails (hense the name, coma).  As with Refractors, the effect is reduced by increasing focal length.  In modern telescopes, corrective lenses are often used.

Catadioptric telescopes resolve this partly by using a different kind of mirror shape, a spherical mirror.  The curvature of the parabolic mirror used in a Newtonian design varies along its profile.  The curvature of a spherical mirror is a constant curve.  This makes them easier to manufacture and reduces or eliminates coma, but it introduces another problem: spherical aberration.  Catadioptrics overcome this by using the combination of lenses and mirrors to correct the distortions.  However, the numerous surfaces then require precise configuring, which increases the cost to manufacture.

Each of these designs has its pros and cons. There isn't one "all-purpose" option that's best for everything.

As far as amateur telescopes go, however, the Newtonian design tends to be the most popular.  This is largely due to the average cost per inch of aperture.  For example, an 8" Newtonian can be found in the neighborhood fo $300 USD for the telescope tube alone.  An 8" SCT tube alone is likely to cost about three or four times that.  Very few 8" refractors are available, and those that are would likely cost several thousand dollars.

Now, as to the telescope you chose.  The Celestron PowerSeeker 127EQ is actually not a true Newtonian.  It is what we call a Bird-Jones (or Jones-Bird, depending on who you talk to) Newtonian.  A true Newtonian has a parabolic primary mirror and a flat secondary mirror.  The only other optical surfaces are in any eyepieces used.  The Bird-Jones Newtonian uses a spherical primary mirror.  To counter the spherical aberration inherent in the design, a small correcting lens, which is not much more than a Barlow lens, is included in the optical path (usually mounted inside the focuser mechanism).  This allows for a more compact design (the PowerSeeker 127 is only 508 mm (20 inches) long, but has a focal length of 1,000 mm (almost 40 inches).  But it also decreases the quality of the view.  The optics on this telescope are pretty poor.  Worse still, it is extremely difficult to collimate.

All telescopes require the optical elements to be properly aligned to produce the best image.  This is collmation.  In the case of refractors, it rarely needs adjustment.  Most SCT's require it only occasionally, and Maksutovs rarely.  But Newtonians require it frequently.  Newtonians will have a set of adjustment screws or knobs on the back, behind the primary mirror, and on the front as part of the support that holds the secondary mirror (usually referred to as the spider).  Most telescopes use a 3-screw system for adjusting the tilt of the mirrors.  It can be a chore at first, but with a little practice, it becomes fast and easy to do.  It also requires a tool to insert into the focuser.  Traditionally, this is a collimation cap or Cheshire, which work like an eyepiece, but do not have lenses.  They help you visualize the alignment of the optics so that you can adjust it properly.  You can also get a laser collimator which uses a low-power laser to guide you.  But none of these work with a Bird-Jones scope.  The correcting lens makes them very, very difficult to collimate.

Furthermore, the mount on the PowerSeeker line, particularly the 127EQ, is not very stable.  It wobbles and shakes and makes for difficult observing.

Sadly, the reviews on sites like Amazon make it out to be a great instrument.  But these reviews themselves are flawed.  Amazon's servers use review rankings to determine the display order of search results.  If an item has higher ratings in its reviews, it appears toward the top of the list of results when you search.  Most buyers don't look past the first page, and rarely much further than 2 pages.  So if a seller wants to sell his or her product, it needs to be on the first page, preferably right at the top.  This leads a lot of companies and individual sellers to push customers to add reviews.  Some companies also falsify reviews (though it would be difficult to prove it) in order to game the system.  As such, the reviews on Amazon (and many other sites) are nearly useless to the consumer.  And the PowerSeeker telescopes are a prime example.

On multiple other sites I spend time on, this telescope is outright hated.  It is sold by Celestron (a chinese-owned company) to unsuspecting buyers.  In the long run, I think it causes more harm to amateur astronomy than good - because users often end up discoraged by its poor peformance and cheap construction.

If you've already purchased it, I'd strongly recommend returning it and getting something different (e.g. the two scopes I listed at the beginning).  If you haven't yet, then please do not.  The two telescopes I listed are significantly better in nearly every way for a similar price.  Even the slightly smaller 114 mm versions are better.  But the PowerSeeker line is nearly entirely composed of low-quality equipment that is not worth purchasing.

But even before purchasing, joining a club is the best thing you can do.  Some clubs even have loaner telescopes that members can borrow to learn before they buy.

I wish you luck with whatever you end up purchasing.  Clear skies!


What is the difference between the Yarkovsky effect and YORP effect? As far as I understand, both the Yarkovsky effect and the Yarkovsky–O'Keefe–Radzievskii–Paddack (YORP) effect exert forces upon bodies as a result of thermal photon radiation. How do both effects differ from one another?

The Yarkovsky effect is the thrust on a small object in space that has been heated by sunlight, created from radiant energy. The YORP effect is when the thrust produces rotation, likely due to uneven heating or differences in surface projections (IE: A mountain acts as a better solar collector than a flatter portion of a spherical object on the opposite side).

See: "The Yarkovsky and YORP Effects" (Feb 4 2015), by
D. Vokrouhlicky, W. F. Bottke, S. R. Chesley, D. J. Scheeres, and T. S. Statler:

  "The Yarkovsky effect describes a small but significant force that affects the orbital motion of meteoroids and asteroids smaller than 30 − 40 kilometers in diameter. It is caused by sunlight; when these bodies heat
  up in the Sun, they eventually re-radiate the energy away in the thermal waveband, which in turn creates a tiny thrust. This recoil acceleration is much weaker than solar and planetary gravitational forces, but it can
  produce measurable orbital changes over decades and substantial orbital effects over millions to billions of years.

  The same physical phenomenon also creates a thermal torque that, complemented by a torque produced by scattered sunlight, can modify the rotation rates and obliquities of small bodies as well. This rotational variant has been coined the Yarkovsky-O’Keefe-Radzievskii-Paddack (YORP) effect. During the past decade or so, the Yarkovsky and YORP effects have been used to explore and potentially resolve a number of unsolved mysteries in planetary science dealing with small bodies.".


Direction of major axis of moon's orbit Does the major axis of the moon's orbit point in the same direction (with small deviations) throughout the year? or does it migrate along a path so that after a million years it might be 90° away from where it started?

As pm-2ring already commented, the direction of the axis has a period of about 8.85 years.

It's worth noting, though, that in reality the moon's orbit does not correspond very closely to an ellipse, because it is rather heavily perturbed by the Sun.

Relative to this reality, the concept that the moon's orbit around the earth has a 'major axis' results from treating the real orbit approximately as if it were really an ellipse.  The approximations can be made in several different ways, and so in principle there can be a number of different approximating ellipses.

In effect the question presupposes the (idealised) rotating mean ellipse which results (in concept) when nearly all of the perturbation effects are treated as if they were averaged out -- except for the mean rate of axial rotation and the current position of the mean axis.

This idea of the mean ellipse is the one that underlies the 'mean elements' given in tables that express the fundamentals of the moon's motion, such as the 1954 Improved Lunar Ephemeris, which was slightly adapted from Brown's theory and 1919 Tables of the Motion of the Moon.

A more recent estimation of the position and rotation of the idealised mean ellipse, derived from lunar laser ranging via the JPL ephemeris DE200 (official basis of the Astronomical Almanac from 1984-2002), is found in J L Simon et al. (1994) "Numerical expressions for precession formulae and mean elements for the Moon and the planets", Astronomy & Astrophysics 282, 663-683.

According to this source, the mean longitude of the moon's perigee in arc-seconds (counted from the mean equinox of date) (see Simon 1994, p.670 at b.3)

$    =  300071".6752   +14648449".0869 t   -37".1582 t^2 $
$           -0".044970 t^3   +1".8948*10^-4 t^4 $

at any time defined by t, where t is the time-interval counted from the J2000 epoch of 2000 Jan 1 at 12h TT (= JD 2451545.0), in units of Julian centuries (36525 days) of Terrestrial Time).

The negative quadratic term in the expression above shows that the rate of rotation of the mean axis is very slightly slowing down; in 20 years it now rotates by about 1 minute of arc less than it used to do in 20 years around AD 1600 in the age of Tycho and Kepler.

Besides the mean approximation to the moon's orbital ellipse just described,
there is also the (variable) osculating ellipse. The largest periodic variations in the osculating longitude of perigee, relative to the mean perigee, are given in the same 1994 paper by Simon et al. at p.671, Table 4, 5th column.

Last but perhaps not least is the librating-and-rotating ellipse approximation which results when the rotating ellipse is taken to include the effect of the largest of the solar perturbations, now known as the evection.  This is described in older textbooks such as Godfray's Elementary Treatise on the Lunar Theory, see esp. pp.69-71 in the 1871 edition here.


Fractional contribution of secondary star What do we mean by "the luminosity fraction" and "the fractional contribution of the secondary star"  of binary stars? can anyone suggest any method to calculate them theoretically?

It just means that in a star that is an unresolved binary system (i.e. the light you receive comes from both stars), the secondary contributes a fraction of that light.
This fraction could be expressed as a fraction of the total luminosity or a fraction of the flux at a particular wavelength or in a particular waveband.
How do you work it out theoretically? Well you would have to know exactly what kind of stars the primary and secondary were. e.g. If you knew they were both main sequence stars and you knew their masses, you could use a main sequence mass-luminosity relation to turn a mass ratio into a luminosity ratio. If you knew their temperatures, you could model their spectra to estimate the primary Vs secondary contribution at any wavelength.


How to calculate the pixel size and beam size of an image? I have an image with some sources in it. I need to get the pixel size and beam size of that image. How can I calculate it?

If you know the angular size of your image (e.g., 10", 1' etc), then pixel size  (in "/pixel units) is just image size / number of pixels.
For beam size, depends a lot on the wavelength-range of your observations (for instance sub-mm/radio observations are conceptually very different from UV/optical), but to have an idea of the beam/psf size, you can fit a 2D gaussian/moffat profile to the emission of a point-like object (i.e., a star) in the field of your image.

Note that most of the time these info are contained in the header.


Help stabilizing telescope First time astronomer here...  (pardon any less than accurate terminology)
My son received a secondhand telescope, a Celestron Powerseeker 114AZ.
Last night I calibrated the Starpointer, and was able to see Jupiter and it’s moons as well as Saturn and it’s rings.
However, after I position the laser in the Starpointer guide, the tube slightly sinks into a different rest position. To accommodate this, I ended up having to predict this movement by setting the laser guide a little higher than the target and let it sink into resting position. I also notice that the optical tube has a little bit of vertical wiggle on the azimuth bracket, which makes the altitude adjuster less accurate. The top hinge of the tripod legs seem a little wiggly too.
Is this normal movement or should I find a way to tweak, adjust, and/or tighten the parts of or the entire assembly?

A "Celestron Powerseeker 114AZ" Google turns up old reviews at Amazon.com and .ca. Most are somewhat critical of the mount.
In all honesty that is to be expected at this level. Many amateur astronomers are unhappy with the amount of wiggle in mounts worth many thousands of dollars. A little tweaking might help just don't use much more than finger pressure on the tools.
The telescope is supposed to be OK optically. Putting the optical tube on a dobsonian mount might be a good father and son astronomy project.
For what it is worth that 114 telescope is probably ten times better than my first telescope. I was as proud of that thing as any kid could be. As adults we tend to forget those sorts of memories.


Is there a reason why the IAU differenciates objects in the SS orbiting the sun into a particular classification but it doesnt for satellites? The IAU has different classifications for objects in the solar system orbiting the sun, they can be according to their shape and if they clear their neighborhood of other material around its orbits , planets, dwarf planets or Small Solar System bodies. But for objects in the solar system orbiting a planet or a dwarf planet they are all satellites, no matter if they are rounded or if they have an irregular shape like an asteroid. Is there any known reason why this is so? Why is it needed to differenciate its shape when the object it's orbiting the sun, but it isnt needed when the object it's orbiting a planet? (or dwarf planet)?

History, and recent history at that:

At one time (1980) things seemed clear enough: There were planets (Mercury-Pluto), Asteroids (mostly between Mars and Jupiter) and Satellites.  No one care much about whether a body was spherical or not, since it was pretty clear that this was all there was.

.
.
.

Then the Kuiper belt happened, and in particular it became clear that Eris was big enough to be rounded... Now why should not Eris be a planet? Could we have tens of rounded icy objects orbiting beyond Neptune?  How many would we expect grade-schoolers to memorise?

The decision was taken to classify solar system bodies into:

Planets (round and not shareing their orbit with anything comparible in size)
Dwarf planets (round, but have not cleared their orbit)
Minor planets (not round)
Satellites (not orbiting the sun)

This is a matter of convenience for humans. It does not reflect any fundamental reality.

For convenience, there is no need to distinguish between round moons and non-round ones, so no distinction is needed (for humans)


Why didn't Johannes Kepler use data about more planets? In Astronomia Nova (1609) Johannes Kepler used observations of Mars in order to refute circular orbits. Throughout Astronomia Nova he hardly even mentions other planets than Mars, Earth and the Sun (and the Moon in a different context). Were there any astronomical reasons for him not to continue, or even beginning, his investigation by using data about other planets?

Tycho Brahe had collected corresponding data about the other planets too, right? 20 years of data is enough for Saturn to make 2/3 of its period. Jupiter and Saturn have almost twice as frequent oppositions as does Mars. And Mercury has twice the eccentricity and its much shorter period should help getting data about its apsides when it is best seen.

Would it had helped him to use several planets during the process as he describes it in Astronomia Nova?
Was it simply the calculation capacity which limited him from dragging several planets into it?
Or could he somehow conclude that it is enough to solve for Mars, to know that his physical conclusions of interests are valid for all 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


Could there be life beneath the surface of Mars or moons? I realize radiation and extreme temperatures would probably destroy life on the surface of most planets and moons, but could life exist beneath the surface (e.g. like earth worms on earth)?

That's one of the big questions. ESA scientists, at least, think it's worth looking for underground life. See ExoMars mission.

More likely than earth worms are microbes, since some microbes on Earth live under similar conditions as presumed for underground Mars.


Apsidal Precession: What Am I Getting Wrong? In Precession of Mercury’s Perihelion from Ranging to the MESSENGER Spacecraft (https://ui.adsabs.harvard.edu/abs/2017AJ....153..121P/abstract), one finds the precession of Mercury’s perihelion due to the oblateness (quadrupole) of the Sun to be:
$ \displaystyle \dot \varpi_{J_2} = \frac {3}{2} \frac {nJ_2}{(1-e^2)^2} \left( \frac {R_\odot}{a} \right)^2 \left(1 - \frac {3}{2} sin^2 i \right) $
(equation 3, p. 2). A few lines down, we read that it amounts to about 0.03″ per century—later refined to 0.0286″ per century.
However, when I do the calculation, with $ a $ = 57.90905 Gm, $ e $ = 0.20563, $ i $ = 3.38°, $ R_\odot $ = 696,342 km (all four from Wikipedia), $ n $ = 4.09°/d (calculated from formula 33.6 in Meeus 1998), and $ J_2 $ = 2.25 × 10⁻⁷ (paper’s abstract), I get 0.01855″ per century.
What am I doing wrong?

The value obtained is in degrees per day (which I didn’t know); this is simply multiplied by 3600 to convert to seconds per day, then by 36525 to convert to seconds per century.


Formation of Cosmic Microwave Background It is said that the cosmic microwave radiation (CMB) was formed when the universe was 379,000 years old. How is this calculated?

We observe the temperature of the CMB as a ~2.7 K blackbody, but that's the redshifted version we observe. The CMB is also know as the "surface of last scattering" at the point of  recombination when nuclei and electrons combined to form neutral atoms the universe went from opaque to transparent.

This happens at a temperature of ~3000 K. From this we can estimate the redshift (z~1100) of the CMB which corresponds to an age given our cosmology.


Instrument aperture sizes on Hubble Telescope When looking at the diagram of the focal plane of the Hubble Space Telescope I noticed that the aperture sizes of most of the scientific instruments are very small. They occupy a very tiny fraction of the focal plane.

I understand that for a spectrograph (like the COS) you need a very small aperture to look at point sources. But why the Advanced Cmaera for Surveys or Wide Field Camera 3 also have so small apertures? Most of the focal plane is not used and it must be very tedious for the guidance system to move the target object exactly into the small aperture of the selected instrument. Even though the WFC3 is called "wide field", its field of view is less than 3 arc minutes.

What were the reasons behind selecting these aperture sizes? Is HST designed to observe details of small targets? Are the sizes optimized for majority of observations?

On the other hand the HST is taking images of large objects like nebulae and usually it is necessary to stitch a number of individual frames to create the whole image of a bigger object. For example the Carina nebula (stitched from 48 frames) or the Crab nebula (24 frames).

EDIT: You can't have a very wide field, and simultaneously a very high resolving power, in an instrument with a large focal ratio (or any instrument, really), unless you use sensors with an unrealistic pixel count. Details below:

The linear size of the focal plane is large - one document says it's the size of a "dinner plate". But the instrument is an f/24, which is a large focal ratio. At that ratio, a big linear size in the focal plane doesn't mean much in terms of angular size.

From what I've seen, the Wide Field Camera has a total field of 4.7 arcmin, or 280 arcsec according to this document:

http://asd.gsfc.nasa.gov/archive/hubble/a_pdf/news/SM2-MediaGuide.pdf

Anyway, yes, that's a small field, if you compare it with ground-based amateur astrographs, which can have a field as wide as several degrees of arc.

But the Hubble telescope has a very high resolving power - 0.05 arcsec. Let's take your information and assume the WFC has a total field of 3 arcmin.

3 arcmin / 0.05 arcsec = 3600 pixels in theory, as a linear size of the WFC

If you wanted the WFC to have a bigger total field, you'd have to either sacrifice resolving power, or use a sensor with an enormous pixel count, which is not feasible.

The WFC uses several sensors, each one of them 800 x 800 pixels.

TLDR: Wide field, or great resolving power - pick one. They chose resolving power, and they compensate the narrow field by doing multiple passes.


Why is $g$ tied to the oscillator strength $f$ in $\log{gf}_{\odot}$ The $\log{gf}$ value of an element in a star, where $f$ is the oscillator strength, or probability of transition in an atom, is a scientifically pursued quantity with relation to the model atmosphere of the star. Why is the oscillator strength tied to $g$? And what is the motivation for including this quantity in a stellar model?

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.


Core collapse supernova vs Neutron Star I am reading about massive stars and their final consequences. what I understood is massive stars after throwing out their outer envelop with masses > 1.4 solar Mass turn into neutron stars. But I also know that Massive stars can collapse to form a type II supernova. I need to understand how we determine which stars go through which final fate. And what initial mass they have to start their life with for both fates.

I ma be completely wrong with my understandings. Your clarifications will help a lot.

To clearify: In your question you distinguish between a star forming a neutron star when its core exceeds the Chandrasekhar limit and a star undergoing a Type II supernova.
These are, however, the same events.

To summarize: Generally an isolated star can die in one of basically four ways. With increasing probability with increasing mass these outcomes are:

1. Formation of a white dwarf: This is the outcome for stars with a initial mass below ~10 solar masses and includes our sun. This includes about 97% of all stars, as small stars are the most abundant.
At the end of its lifetime, when it has burned though all the hydrogen accessible to its core, it will start to burn gradually heavier elements in its contracting core. Up to which element this can go depends on the initial mass of the star. During this process the outer layers of the star begin to expands and it enters the red giant phase. At the end of which they shed their outer layers which then form a planetary nebular. All that is left after that is the cooling core of heavier elements. The white dwarf. This white dwarf is held up against gravity by degenerate electron pressure. This means the electrons are so close together that, to contract further, they have to be pushed in an higher energy state to not violate the Pauli exclusion principle.
An example would be the beautiful Ring Nebular where you can easily see the white dwarf in the middle.

2. Supernova and the formation of a Neutron star:
The core of heavier stars ($9-100 M_{\odot}$) are, too, mostly held up by degenerate electron. During the lifetime of the star more and more heavy elements collect in a growing core. With increasing core mass/density, however, the electrons reach higher and higher energies. At some point these energies become high enough for electrons to be "pushed" into protons, becoming neutrons and releasing lots neutrinos in the process. This reduces the degenerate pressure until it can no longer support the core against the gravitational pressure of the outer layers.
This happens when the core exceeds the mentioned Chandrasekhar limit. This is what initializes the core-collapse of heavier stars.

Even heavier stars with hotter cores are increasing held up not only by degenerate pressure but regular thermal pressure. For those stars the collapse is additionally promoted when photodisintegration of heavy elements saps the core of more and more thermal energy and leads to its further contraction.
(See the first pages here for a more detailed description of how the SN is triggered)

Independent from how the collapse started, it continues until at some point the degenerate neutron pressure and most importantly the repulsion of the strong nuclear force kicks in.
At this point the core-collapse suddenly halts and the proto-neutron star becomes virtually incompressible. This all happens so fast, that the outer shells basically free-fall onto the proto-NS and "bounce off". Together with the energy provided by the intense neutrino flux, this drives away the outer shells and leads to the explosion of the star. This is a core-collapse supernova. All that is left is a expanding supernova remnant and a central neutron star, a pulsar. The composition of the remaining outer layers at the time of the explosion determines the type of the SN.
Most famous example: The Crab nebular.

3. Core-collapse SN with black hole formation:
At higher masses it can happen that not all of the outer layers can escape the gravitational well of the proto-NS. Some part of it can fall back onto it. During this process the NS can gain mass again and eventually exceed the Oppenheimer-Volkoff-Limit. When this happens even the pressure from the degenerate neutron gas is not sufficient anymore. At this point there is nothing that can stop the gravitational collapse to a black hole. The accretion of the additional material again releases huge amounts of energy and can lead to the formation of jets. This is what is suspected to be a Hypernova and can be accompanied by a gamma ray burst (GRB).
See below for and animation of a jet braking though the expanding supernova shell and producing a GRB:

4. Core-collapse with direct BH formation:
An alternative, less violent way for forming a BH is a direct collapse. This mostly happens for very heavy stars.
In this scenario the proto-neutron star is unable to initialize the ejection of the outer layers and just collapses to a BH. No huge explosion, no supernova remnant.
This is sometimes called a failed supernova. And we have actually seen such a star just disappearing:

5. Pair-instability supernova and complete disruption:
For stars with masses between about 140 and 260 $M_{\odot}$ (see respective gap in the figure below) temperatures in the core can become hot enough that thermal radiation reaches gamma ray energies. Specifically photons can reach energies high enough to produce e$^-$/e$^+$ pairs. This process transfers thermal energy into the mass of the formed particles. This leads to a contraction of the core. At that stage the core has not jet fused many heavy elements and consists mostly of Helium, which now undergoes rapid fusion as the core contracts. The released energy complete disrupts the stars leaving no compact core and only an expanding shell. Such heavy stars mostly formed in the early universe as the first generation of stars.

Which one of the scenarios 2 - 4 will happen to a star with a zero-age main-sequence (ZAMS) mass of more than 15 solar masses is hard to tell and an active field of research. It all depends on ZAMS mass, binarity, metallicity, angular momentum, and magnetic fields, among others properties of the progenitor star. Current computational capabilities are just not enough to produce reliable 3D magneto-hydrodynamic simulations with good temporal and spacial resolution and scale.

But from here there is a rough estimate of what we think is the most probable outcome for different masses and metallicities:

And from here there is a set of actual 1D simulations of the supernova 1987A showing all kinds of different outcomes for different masses and initial models of the star:
(Although 1D simulations tend to have a hard time to initialize explosions)

How complex these simulations are can also be seen in nice 3D videos like this one.

So you see, to determine which stars go through which final fate we have to torture huge supercomputers and even then we still have a hard time coming up with definitive answers.


Is there a name for the maximum in local solar elevation angle? It came up in conversation today that we were nearly at 21st June, and the days would shortly be getting, um, shorter.

We were trying to think of the the term that means "When the sun is at the highest point it can attain in the sky during the year", as opposed to during the day, (which is the solar noon). We have terms like Perigee and Apogee, it seems (to me) that there should be a name for this as well.

Is there?

I think this is an indirect answer, but the name for this event is solstice. In the strictest definition, the solstice is

  either of the two times in the year, the summer solstice and the winter solstice, when the sun reaches its highest or lowest point in the sky at noon, marked by the longest and shortest days.

By definition, the solstice is the exact point in time when the Earth's axial tilt is pointed as close to (or as far from) the Sun as possible and by extension, the Sun is at its highest (or lowest) point in the sky during the year. If you're in the northern hemisphere, this point is the summer solstice, whereas it is the winter solstice for the southern hemisphere.


Change of time in vernal equinox by altitude from the Sea level vernal equinox for a very high tower ($h_1>3000m$) in a city ($\lambda , \phi$) in altitude $h_1$ from the sea level is 8:00 Local time. Does this time change if we go to another altitude in that same position ($\lambda , \phi$), for example altitude $h_2$?

In other way, does the time of vernal equinox change with altitude?

In other way, does the time of vernal equinox change with altitude?

No.

The equinoxes are defined as the points in time at which the Earth's equatorial plane intersects the center of the Sun. From an Earth-centric perspective, the Sun appears to crosses the equatorial plane from south to north at the March equinox, and from north to south at the September equinox.

Altitude, latitude, and longitude, and length of day do not come into play. In fact, days are slightly longer than are nights almost everywhere on the Earth on the dates of the equinoxes. This is a result of the Sun's angular diameter and atmospheric refraction, both of which mean we can see the Sun when the center of the Sun is a bit (0.83 degrees) below the horizon.


How did scientists come to the conclusion that it is space that is expanding? How did scientists come to the conclusion that it is space that is expanding and not galaxies moving away from each other as in a giant explosion?

The simplest explanation relies on the statistical unlikelihood of us being at a special place and time in the cosmos.  Let's say, for a moment, that we see galaxies moving away from us because of some cosmic explosion that set them in motion.  In that case, we would expect to see the galaxies in one direction moving faster than the galaxies in another.  Looking towards the center of the explosion, we would see galaxies on the other side moving much faster than if we looked in the other direction.

This, however, is not what we see.  What we observe is all galaxies moving away from us, with a speed proportional to their distance from us.  That means that the further away they are, the faster they are moving.  It doesn't matter what direction we look in; a galaxy 500 Mpc away in one direction will have a recessional velocity approximately the same as one 500 Mpc away in the other direction.

This forces us into one of two conclusions:  either we are at the exact center of the universe, which is statistically impossible, or space itself is expanding, so everything is moving away from everything else.  The animation below shows how this would work.  From any given point in space, it would look like everything is getting further away from you.

Explaining the recession of galaxies due to an explosion also raises many questions:  What caused the explosion?  Why would it start at one particular point?  What were things like before the explosion?  All of these are questions which do not have accepted answers.  Instead, the expansion of space ties directly into the idea of the Big Bang.  The expansion of space itself is what drove the Big Bang.  It is not that the universe is expanding into something.  Instead, space itself is expanding.  However, the topic of the Big Bang is a different subject, so you can look up information on that yourself.


How are heavier elements such as carbon and silicon distributed within the Sun? In a previous question I asked about the source of carbon and silicate dust that Solar Probe Plus will encounter in its close flyby of the Sun. It seems likely that most sources would include infall from outside the solar system or fragments resulting from collisions inside the solar system.

But the Sun is believed to be about 0.4% carbon and 0.1% silicon by mass (see  here and here). I believe these represent the relative abundances during the formation of the Sun, but do not account for nucleosynthesis that early in the Sun's life.

Is the distribution of carbon and silicon in the Sun is believed to be uniform, or does it have a radially dependent distribution? Is there be any kind of gravitational settling, i.e. heavier atoms tending to move toward the center as happens in planets?

Since we can only directly measure elemental abundances directly near the surface using spectroscopic techniques, would the radial distribution come from some model or calculation?

A page from the Institute for Advanced Study links to data from various modifications to the Standard Solar Model. The newest given there is from Bahcall1 et al. (2005), which I'll use as an example. The authors' calculations depend on data from the Opacity project. Data from two models are available through the IAS link: the BS05(AGS, OP) model and the BS05(OP) model; the difference lies in the main thing you're interested in: heavy element abundances. BS05(AGS, OP) has, for instance, a central2 $^{12}\text{C}$ mass fraction of $7.79\times10^{-6}$, while BS05(OP) has a central $^{12}\text{C}$ mass fraction of $1.05\times10^{-5}$. All of the central data from the major thermodynamic variables (e.g. pressure and temperature), as well as the mass fractions, are well within an order of magnitude in each model.

That said, the general heavy element trend is the same in both models. $^{12}\text{C}$ increase by two to three orders of magnitude from the center. $^3\text{He}$ (rather than the more common $^4\text{He}$, which has a mass fraction given by $Y$) increases by one to two orders of magnitude. $^{14}\text{N}$ actually decreases by two orders of magnitude, while $^{16}\text{O}$ remains relatively constant. The model gives no data for silicon.

Other new models agree (though not all), to less than an order of magnitude or so, with the data from Bahcall et al. Graphical representations can be found in e.g. this paper:

I'm unsure of the reason behind the $^3\text{He}$ spike; that may warrant a follow-up questions. My one guess is that this could be due to its role as an intermediate nucleus in the second step of the p-p chain. However, the same trends for each element are followed, although the changes in mass fraction are not constant at all radii $r$.

As for your question about how the central figures were derived, the answer is that helioseismological measurements and neutrino fluxes are some of the best indicators of composition. This was used as the basis for the BS05(AGS, OP) and BS05(OP) models, as well as, in fact, most variants of the Standard Solar Model.

1 This is Bahcall's page, so, naturally, these are some of his papers. However, the results appear to be consistent with the calculations of others.

2 Technically, this is from about 0.0016 solar radii, but this distance step is comparatively small.