I have a question about the wording in a paper about a planetary system and comets I was reading a paper. This is the paper:
https://arxiv.org/abs/2006.01711 (https://ui.adsabs.harvard.edu/abs/2020MNRAS.496.4979D/abstract
In the method section in the second paragraph there is a bit of wording that is confusing to me.

We will examine the dynamics of some Solar System comet analogs3, namely, 2P / Encke, 9P / Tempel 1, and 31P / Schwassmann-Wachmann 2; they are later referred to as comet A, B, and C, respectively, for convenience.

The paper is essentially about a system called 55 cancri. They do simulations about it to look at comet interaction since there is an empty space without planets where there might be comets. They say they examine 3 solar system comets.
Does that mean they just use the regular 3 planets and star in the 55 cancri system and then add 3 comets as an assumption? Or do they do 2 different simulations one for the solar system to test the numerical simulation and then the cancri system?
The confusion also arises since they mention comet analogs. I do not understand what analog means here. They also link to a table with the planets in the 55 cancri system which I do not understand how it relates to this sentence.
Can somebody explain what they mean here?

What they have done is take each of the 580 Jupiter-Family Comets (JFCs) from our solar system (from the JPL Small Body DataBase; set object kind=comets,
Comet Orbit Classes to Jupiter-family Comet) and kept all of the orbital elements the same, except for the semi-major axis which they have scaled by the ratio of the semi-major axis 55 Cnc d (5.74 au) to Jupiter's (5.2 au). So a comet with a semi-major axis of 3 au in our solar system would be $3\times(5.74/5.2)=3.312\,au$ in the 55 Cnc system - this is what they are referring to as comet analogs
All of these 580 comet analogs are being inserted into the 55 Cnc system, For each comet analog inserted, they are creating 100 clones of each analog, varying the mean anomaly. This creates almost 60,000 "comets" which are then integrated along with the 55 Cnc system planets and the outcomes of the comet clones are tracked to see how many survive, how many are ejected from the 55 Cnc and over what timeframe. To illustrate this, they have picked the 3 comet analogs of 2P/Encke, 9P/Tempel 1 and 31P/Schwassmann-Wachmann 2 as examples of JFCs which have high, medium and low eccentricity respectively (their Table 2) and find that comets which have eccentricity tend to preferentially ejected and in a shorter time due to a greater number of encounters with the planets of the 55 Cnc. This occurs much quicker in the 55 Cnc system than their corresponding counterparts in our Solar System.


Is it safe to watch solar eclipse's reflection in water? There will be a solar eclipse soon at my area... naturally I want to watch this with my son.

Is it safe for us to watch the solar eclipse's reflection in the water? There is a swimming pool near my house and I plan to watch it there.. is it safe to do so?

I remember vividly doing this with my dad when I was little, but some website says we are not supposed to watch it in a bucket of water..

So is it ok to watch it in the water?

If it's not safe to look at the sun's reflection in the water normally then it's no safer during an eclipse because even if 1% of the sun is showing this is more than enough light to seriously damage your eyes. The best way to view the sun, if you haven't ordered your eclipse glasses yet, is to poke a hole in a piece of paper and project an image of the sun onto another, like so:

There are instructions on how to do this here on the JPL website.


What are practical considerations for backyard radio-astronomy detection of black holes? Evidently, direct observation of a black hole for an amateur astronomer, such as described for what professionals do in the question "How are black holes found?" would be nigh on impossible, so the observations would be based on their effects of the surrounding matter (specifically, accretion disks and jets) and more likely detecting the radiation signals.

Given that the locations of many black holes are known, what practical considerations would a backyard radio-astronomer need to consider to attempt to do detect black holes?

Amateur radio-astronomy detection of pulsars and gamma ray bursts are discussed practically in Introduction to Radio Astronomy (Society of Radio Astronomers), this question is to see what practical considerations would be needed to extend those techniques to detecting black holes.

The resolution/error box. Radioastronomy has always been hindered by the resolution, because it is inversely proportional to the size of the telescope and making larger telescopes (even with interferometry) is not always easy. No amount of modern technology can substitute for a large effective diameter.  (When I say effective I'm including interferometry here; either way one needs a large area to build on).

Let's take a look at Reber's initial paper1 where he first mapped the sky:

In the diagram on the left, from top down, the three peaks in the contour map are Cas A, Cyg A, and finally Sgr A. The latter two have black hole origins, the former is a supernova remnant.

The resolving power of Reber's telescope here seems to have been 6 degrees, and it was 31.4 feet in diameter (and he focused on a 1.9m wavelength).

Now, by the Rayleigh criterion, the angular resolution is proportional to the wavelength divided by the diameter. As mentioned before, this is the major limiting factor for radioastronomers, and will be what holds amateur radioastronomers from making great telescopes — amateurs usually don't have acres or land in which to build a good interferometer (let alone the precision), and single-dish telescopes can't be made too large by an amateur. One may notice that I'm citing rather old observations here, on old telescopes; however given that radioastronomy technology has not changed nearly as much as size has, it should be OK to compare amateur telescopes with the smaller telescopes of the past.

Now, Cyg A was the first to be identified as a black hole, even though the radio-brightness of Sgr A was discovered at the same time. I'm focusing the rest of my analysis on Cyg A for this reason, as it stands to reason that the first confirmed BH out of the brighter radio sources would have more prominent indicators that it is a black hole.

Let's have a look at Cyg A with a better resolution:

(From this paper2, using the 5km array)

Note that the black blob in the center is the actual galaxy (probably an optical photograph superimposed on the contour map).

We can see that the lobes are less than a minute wide. (The actual galaxy is around 50 arcseconds wide)

To me, the most interesting thing that one would like to see here are the gas jets coming from the central galaxy. As mentioned in my answer here, these radio-emitting gas jets are in a steady line over thousands of light years, indicating that they come from some sort of cosmic gyroscope which has been steady for very long. However, even with the Ryle telescope the 1969 people could not get a picture of them; just a slight hint of their existence from the shape of the lobes.

Alright, so no gas jets. What else can indicate a black hole? They could try to look at the lobes themselves. They do not directly indicate the existence of a black hole, but their shape hints towards them being formed from jets (this is pretty much in retrospect).

However, with lobe dimensions of less than an arc minute, there's not much an amateur can get here, either. It's possible that a really good amateur telescope would just manage to notice that there are two lobes, but not much else as far as I can tell.

The other interesting bits would be the central galaxy itself, but it's too small. In the optical region one may have a chance of seeing Baade's "colliding galaxies" (it only looks like a pair of colliding galaxies). The gravitational effects (lensing, etc) are really only visible in the optical and beyond, for it to be visible in the radio we would need to be very lucky and have a huge radio source pass behind Cyg A — not happening anytime soon.

I'm pretty sure that a similar analysis would work for Sgr A or any other black hole candidate; the gas jets would be too small for an amateur radiofrequency resolution, and the gravitational effects of the black hole would only work well in the optical and X ray frequencies.

1. Reber, G. (1944). Cosmic Static. The Astrophysical Journal, 100, 279.

2. Mitton, S., & Ryle, M. (1969). High resolution observations of Cygnus A at 2. 7 GHz and 5 GHz. Monthly Notices of the Royal Astronomical Society, 146, 221.


Different times of sunset at the same location due to elevation Me and my friend were watching the sunset on the same location (at my country the latitude is approximately 40 degrees). I am at sea level and my friend is at an elevation. He observed the sunset with a difference of about 4 minutes than me. He is a physicist so he challenged me to find out his elevation. Any ideas on how to use these information and win the challenge?
I found some information on Wikipedia here but cannot understand a lot.
Any idea on how to solve this will be appreciated. Thanks in advance for your help.

The page you refer to is about sunrise/sunset throughout the year, but what you really care about is differences in the geometry on one day. This is sometimes expressed as different observers seeing the horizon at different distances, or sometimes as an angle, a “dip” of the horizon when seen from higher altitude.  Thinking about that geometry should get you started on working out the answer.


Determination of High-Accuracy Distances of Terrestrial Planets from the Sun I understand how in olden days the Sun-Planet distances were estimated using:-
(i) measures of planet orbital periods ($T$) from analysis of observations over several centuries;
(ii) Kepler's 3rd Law $(T^2 = k.a^3)$ applied to the planet orbital periods to determine the relative lengths of the Sun-Planet semi-major axis distances ($a$) expressed in terms of the AU (astronomical unit = distance from centre of Earth to centre of Sun);
(iii) particular Earth-Planet distances along Solar radials were measured using Parallax techniques such as that by Casinni for Mars and Transits of Venus;
(iv) From one or more Earth-Planet radial distances the length of the AU could then be determined by simple algebra.
(v) Other Sun-Planet distances could then be determined once the AU length was found.
Modern ephemerides e.g. those issued by JPL use various techniques:-

The orbits of the inner planets are known to subkilometer accuracy through fitting radio tracking measurements of spacecraft in orbit about them. Very long baseline interferometry measurements of spacecraft at Mars allow the orientation of the ephemeris to be tied to the International Celestial Reference Frame with an accuracy of 0′′.0002. This orientation is the limiting error source for the orbits of the terrestrial planets, and corresponds to orbit uncertainties of a few hundred meters.
The orbits of Jupiter and Saturn are determined to accuracies of tens of kilometers as a result of fitting spacecraft tracking data.
The orbits of Uranus, Neptune, and Pluto are determined primarily from astrometric observations, for which measurement uncertainties due to the Earth’s atmosphere, combined with star catalog uncertainties, limit position accuracies to several thousand kilometers.

from Folkner et al 2014.
I understand how spaceraft telemetry has been used to obtain highly accurate distances between Earth and the terrestrial planets.  But I am not clear about how highly-accurate distances can be determined between those planets and the centre of the Sun.
Question
Do such (Sun-planet-distance) determinations still basically rely on (the more accurate Newtonian version of) Kepler's 3rd Law relating planet orbital period and Sun-planet semi-major axis distance?
$$\frac{T^2}{a^3} = \frac{4 \pi^2}{G(M+m)}$$
What other methods or assumptions are used/involved?
EDIT - Afterthoughts
After further thought prompted by feedback from  u/atmosphericprisonescape
my question boils down to  "How are the positions (over time) of the Sun's Centre tied-in (with sub-kilometre accuracy) to the 4D space-time array of triangulated, high-accuracy, telemetry-derived, Terrestrial planet position determinations.
I guess that Kepler's 3rd Law is not invoked as such. Rather, high-accuracy (sub-kilometre) determinations of the position of the Solar centre presumably require (in addition to the high-accuracy planet positions) some specific deterministic "motivation model"; i.e. a model of motion-determining factors.
These factors would include Newtonian Inertia, Newtonian Gravitational forces and Non-Newtonian factors leading to additional orbital angular velocity (cf Non-Newtonian Perihelion Precession and General Relativity as described at  wikipedia/Apsidal_precession ).
The "motivation model" would constrain the relative positions of the Sun and planets via their involment as motion generators, reactors and enactors.

The DE430 JPL ephemeris "memo" (https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf) has details of how it is constructed and what from. Mercury and Venus's positions are tracked by tracking the spacecraft orbiting them (to sub-km accuracy). Mars, Jupiter and Saturn are also measured from the orbiting spacecraft such as Galileo and Cassini. The outer planets plus Pluto positions are primarily done from astrometry so measuring the positions of the planets relative to the background stars, in the past through photographic plates or with CCDs in more modern times.

Positions of stars and planets are all measured in something called the International Celestial Reference System (ICRS) which is based on measuring very distant and fixed on the sky objects called quasars using a precise radio technique called VLBI.


Is there a natural process by which hydrogen is generated from heavier elements in the cosmos? we know that stars fuse hydrogen into helium starting at 3 MK; 13 MK in the Sun's core;
carbon fusion starts at above 500 million K, and silicon fusion starts at over 2700 million K for comparison;
we know fusion stops at iron, because a star has to use more energy to fuse that than it gets back;
so heavier elements are created mostly in a supernova (but also possible in small quantities by special processes like neutron capture);
finally sun-like stars end up as white dwarfs, bigger stars as neutron stars, quark stars, black holes;
and black holes ultimately convert themselves into radiation, in the far distant future when the stable black hole mass limit goes up high enough that even the most massive black holes evaporate;

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

so my question is, will it be like Stephen Baxter said, that in the future only radiation will be left in the universe?
Specifically, is there a natural process out there by which hydrogen is spewed into the cosmos, converted back from heavier elements, to regenerate the fuel for stars so that they may shine in the far distant future as well?

Of course we don't need to worry about this for the time being. This is only considering our concern with what will be 10^70 years from now.

There are a couple of relevant questions one would want to ask:

1) Do protons decay, and if so, what do they decay into? The answer appears to be no, or at least the theoretical lifetime of the proton must increase as a results of these experiments. If they do, eventually the universe could end up in a state of radiation (and dark energy, and dark matter, unless they also decay).

2) Is Hydrogen a bi-product of any natural decay process? Below is a table of all known nuclides.

As you can see, the majority of the elements (not necessarily by number or mass in the universe) do decay through some type of process. There exists a 'stable' ridge (called the island of stability, surrounded by the sea of instability) of elements which will happily exist forever.

The question is, which modes of decay produce protons (Hydrogen nuclei)? Well, there is proton decay (not the proton itself decaying), which is colored in red, though I have to admit that I don't know exactly what this refers to. Fission bi-products are gamma rays (high energy photons), neutrons, and daughter nuclei (see Decay chain). Though, I should mention that free neutrons produced from this type of radioactive decay are not long lived, decaying into a proton and an electron (this process takes on average approximately 11 minutes). By this logic, also isotopes which decay by emitting neutrons, colored purple, would also eventually produce protons. $\beta^{-}$ and $\beta^{+}$ refer to the beta decay process, where the minus sign refers to the emission of an electron and the plus sign refers to the emission of a positron (the anti-particle of the electron). $\alpha$ decay is the emission of a Helium nucleus, which is stable.

Now, given that there are ways for heavy elements to naturally produce protons, the question I would ask is what is the rate of these processes in the universe compared to fusion processes occurring at the centers of stars. I'm not sure that I could give you an answer to this question (or even point you to the appropriate material), but in principle these rates are known. I'd imagine that it'd be quite a lot of bookkeeping to get it correct.


Galactic winds/outflows: why and how are they detected via blueshifted absorption lines in spectra? I know that galaxies can eject gas due to supernovae, accreting black holes, etc. These galactic "winds/outflows" are often defined/detected observationally using blueshifted absorption lines in spectra. But why is it always blueshifted absorption -- what about redshifted absorption? Does it have something to do with the outflow velocities being measured relative to the galaxy central/systemic velocity, such that redshifted absorption implies inflow instead of outflow? But what's the geometrical picture?

Yes, a blueshift here would mean relative to the systemic velocity of the galaxy - which means that, relative to the galaxy, the material is moving at least partly in the direction of Earth.
In many cases the outflow may be a pair of jets being ejected in opposite directions out of the nucleus of the galaxy, but we are less likely to see absorption lines from the jet on the far side of the galaxy (relative to Earth) because the disk of the galaxy may be blocking our view of it, and in any case it won't produce absorption lines because it isn't between us and the rest of the galaxy.


From where does the energy for gravitational waves come from? As far as I understand, in the events detected by LIGO, about 4% of the total mass of merging binary black holes was converted to gravitational waves.

Where does this energy come from, i.e. what exactly gets converted into gravitational waves?

Is it simply the kinetic energy of the merging objects (velocities of these objects before merger are huge, up to 60% of c if I recall correctly), so does it mean emitting gravitational waves makes them orbit slower, but retain their original masses? Or do the compact objects really lose "real" mass, meaning they become lighter and in case of BHs their radius changes accordingly?

As an example, let's assume two BHs, both with 50 solar masses, orbiting each other far enough (say 1 light year) so that GWs nor kinetic energy has no significance to these initial mass measurements. During the merge, they should radiate about 5 solar masses in GWs. Would the resulting black hole have mass of 95 or 100 solar masses?

Radiating gravitational waves makes an inspralling binary orbit closer and faster. (Rob Jefferies)

The source of the energy for both increased kinetic energy, and the gravitational radiation is the same: gravitational potential energy. (PM 2Ring)

Two black holes at a distance of 1 light year have a huge amount of potential energy, about 10^48 Joules of potential energy. As they spiral, a significant amount of that energy is radiated as gravitational waves

This is real mass lost. The mass of resulting black hole is smaller than the sum of the two merging black holes, though at no point does any black hole itself become smaller.


How to get ellipticities of galaxies from SDSS using casjobs? How to get ellipticities of a list of random galaxies. I saw few programs written in SQL and they were only giving stokes parameters and isophotes values for ONE galaxy(for which we input the ObjID). I am not familiar with syntax of SQL. Is there any method to put the objid in a loop so that the program generates stoke parameters values for a list of galaxies?

Just found out this method in casjobs.Previously I came across one SQL code in Stackexchange in which you have to input one ObjID to get its stokes parameters and isoA and isoB values.In sample codes in DR7, you can edit the line which gives Feild values and coordinates to get Stokes parameters(type q_r,u_r in their place).

From stokes parametres you get the ellipiticity of the galaxy from this equation (which can be done in python.)

$e  = 1 - \frac{b}{a} = 1 - \frac{1 - \sqrt{Q^2 + U^2}}{1 + \sqrt{Q^2 + U^2}}.
$

Note that if you are downloading the Casjobs query data in CSV or excel format,You need to write python code to read the file and apply the formula
(to learn how to read csv files in python click here)


How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?
We know the Sun is orbiting the galactic center every 225 million years.
But how far are we from the galactic plane, and is the inclination of the plane of the Sun's orbit around the galactic center, as compared to the galactic plane, known?

According to this drawing

the Sun is traveling approximately in the ecliptic plane.
However, the ecliptic plane is steeply inclined to the galactic plane, so if this drawing is correct (from voyager.jpl.nasa.gov) then this would mean the plane of the Sun's orbit around the galactic center is also quite inclined.
And thus my question, what is the distance to the galactic plane, are we heading toward/away from it, and what is the inclination of the orbital plane of our Sun/solar system to the galactic plane? 3 questions I guess.

Humphreys & Larsen (1995) suggest, using star count information, a distance of $20.5 \pm 3.5$ pc above the Galactic plane; consistent with, but more precise than the Bahcall paper referred to by Schleis.

Joshi (2007) is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between 13 and 28 pc above the plane.
This paper gives an excellent review of the topic in its first couple of pages.

The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this "peculiar velocity" is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and $W=7.17 \pm 0.38$ km/s (up and out of the plane).  (Dehnen & Binney 1998)

The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from http://www.visioninconsciousness.org/Science_B08.htm to show this oscillatory motion.
As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (Matese et al. 1995), it will be roughly 30 million years before we cross the plane again.

EDIT: An update. A new analysis by Karim & Mamajek (2016) yields a distance above the plane of $17.1 \pm 5$ pc and they provide a meta-analysis of previous analyses that gives a median distance of $17.4 \pm 1.9$ pc. They also refer to a paper by  Schonrich et al. (2010) that gives the velocity away from the plane as $7.25 \pm 0.37$ km/s (though with an additional systematic uncertainty of 0.5 km/s).


Approximate spectral type and luminosity given apparent magnitude and distance I am looking at the data acquired from the Gaia DR2 survey. I found that most of the stars had their distance and apparent magnitude catalogued, but not their spectral type and luminosity class. Is there a way to approximate this?

You can calculate the absolute magnitude of a star:
$$M=m-5\log_{10}(\frac{d}{10\,\text{pc}})$$
where $M$ is absolute magnitude, $m$ is apparent magnitude and $d$ is the distance.
Then you take a look at the HR diagram.

One can easily see, that you need two data to obtain the third one, but we have only one data (absolute magnitude). That means, that you also have to know the luminosity classification of a star:

I: supergiants
II: bright giants
III: giants
IV: subgiants
V: main sequence
VI: white dwarfs

Then you just look at the intersection of the class and absolute magnitude.

If you know radius, then this is even easier. All you need is this formula:
$$T={\Big{(}\frac{10^{0.4(4.77-M)}}{4\pi R^2 \sigma}\Big{)}}^{0.25}$$
$T$ is the temperature in Kelvins, $M$ is absolute magnitude of the star, $R$ is radius of the star, $\sigma$ is Stefan-Boltzmann constant.
With these data you look at the table of temperature and spectral classes or HR diagram.

If you are given color, then you already have an answer (spectral class is defined with color). If you are given temperature, you can look at the HR diagram.


Why dust is optically thin in Far Infrared wavelengths? What is the actual meaning of the statement 'Dust is optically thin in the Far Infrared (FIR) over most of the Galaxy'? Kindly Help

The term "optically thin" means that the optical depth is small. The optical depth is a measure of the opacity of a medium, in this case dust, experienced by light traveling through that medium, and is defined as
$$
\tau \equiv n \, r \, \sigma,
$$
where $n$ is the density of the particles in question, $r$ is the distance traveled through the medium, and $\sigma = \sigma(\lambda)$ is the cross section of the particles, which is dependent on the wavelength $\lambda$ of the photons. If $\tau\ll1$, the medium is said to be optically thin, while if $\tau\gg1$, it is said to be optically thick. The fraction of the light that is extinguished by the journey through the medium is $e^{-\tau}$, so the two cases indicate when the light is mostly transferred and mostly extinguished, respectively.

Cosmic dust is composed of particles spanning a large range of sizes. Photons interacting with the dust are either scattered in another direction, or absorbed, depending on the albedo of the dust. But in both cases, a photon has a larger probability of interacting with a dust grain which is comparable to its wavelength. Because of the size distribution, an ensemble of dust grains hence has a characteristic extinction curve. In the figure below (modified from Laursen et al. 2009), I plotted the (functional fits to the observed) extinction curves of dust in the Small (dashed line) and Large (solid line) Magellanic Clouds$^\dagger$:

The extinction is here given in terms of "cross section per hydrogen atom", but you can just think of it as an average dust grain cross section. The colors show the ultraviolet (purple) and infrared (red) regions of the spectrum.
You can see that the extinction, or the cross section, is largest for UV photons, and as you go to longer wavelengths, the cross section decreases sharply and is very small in the far infrared ($\lambda \gtrsim 30 \, \mu\mathrm{m}$; the exact definitions depend on your field of interest).

This is what your statement refers to. But in fact it's badly phrased, because the optical depth is also a function of density and distance, so for instance for a distance $r = 1 \, \mathrm{cm}$ through a dust cloud, the optical depth is $\ll1$ for all photons.

$^\dagger$Extinction curves in other galaxies look similar. In particular, the extinction along most sightlines in the Milky Way features the same bump around $\lambda \simeq 2175$ Å. The origin of this bump still isn't well understood, but may be partly due to graphites and/or PAHs.


Could someone please explain to me how I can extract velocity component values from Stellarium, for a given planet? For a project that I am currently working on, I would like to know how one could extract velocity component values ($v_x$, $v_y$ and $v_z$) for a planet from the Stellarium software. The current information panel for a planet does show its orbital velocity (which is the resultant velocity, $v$) but does not give the individual $v_x$, $v_y$ and $v_z$ values that result in that velocity.

What I essentially need, is to establish a velocity heading for the planet.

Any help (code-based/mathematical) would be much appreciated! Thanks in advance!

To follow up on the helpful comments below your question, here's a program that will do what you need, but without using Stellarium.

If you really need to use Stellarium then hopefully someone will find a way to help you.

This uses the Python package Skyfield and you can see a lot of documentation on that site and elsewhere. Right now I've specified the DE421 ephemeris, but there are many to choose from. This one only works +/- 100 or 200 years I think, others go for thousands of years in the past or future.

If you would like to use NASA JPL's Horizons web interface online, here are some instructions. It uses some of the same Development Ephemerides that Skyfield downloads for you.

Here's some output for roughly 2019-10-02 07:18:59 UTC which is the moment when you posted your question.

                    relative to solar system barycenter
               x            y           z            vx        vy          vz
Sun        -453731.5    1039601.2    451057.1     -0.015      -0.002       0.0
Earth     1.4764e+08   2.1455e+07   9.3005e+06    -4.9362     26.9295     11.6751
Jupiter  -2.2857e+07  -7.2214e+08  -3.0898e+08    12.9012      0.3382     -0.1690
Mumbai    1.4763e+08   2.1454e+07   9.3025e+06    -4.8369     26.5009     11.6749

                      relative to Earth's geocenter

Sun      -1.4809e+08  -2.0416e+07  -8.8494e+06     4.9215    -26.9312    -11.6754
Earth     0.           0.           0.             0.          0.          0.
Jupiter  -1.7050e+08  -7.4360e+08  -3.1828e+08    17.8374    -26.5912    -11.844
Mumbai   -5873.8303   -1362.2932     2072.3226  9.9338e-02    -4.2861e-01 -1.9206e-04

                           relative to Mumbai

Sun      -1.4809e+08  -2.0414e+07  -8.8515e+06     4.8222    -26.5026    -11.6752
Earth       5873.83     1362.29     -2072.32      -9.9338e-02  4.2861e-01  1.9206e-04
Jupiter  -1.7049e+08  -7.4359e+08  -3.1828e+08    17.7381    -26.1626    -11.8439
Mumbai    0.           0.           0.             0.          0.          0.

Here's a Python 3 script for that output as an example

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader, EarthSatellite

load    = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data    = load('de421.bsp')
ts      = load.timescale()

sun     = data['sun']
earth   = data['earth']
jupiter = data['jupiter barycenter']
mumbai  = earth + Topos(latitude_degrees  = 18.98,
                        longitude_degrees = 72.83,
                        elevation_m       =  20.0)
time    = ts.utc(2019, 10, 2, 7, 18, 59)

things     = (sun, earth, jupiter, mumbai)
names      = ('Sun', 'Earth', 'Jupiter', 'Mumbai')
positions  = [thing.at(time).position.km for thing in things]
velocities = [thing.at(time).velocity.km_per_s for thing in things]

for name, position, velocity in zip(names, positions, velocities):
    print(name, position, velocity)

print("relative to Earth's Geocenter")
for name, position, velocity in zip(names, positions, velocities):
    print(name, position-positions[1], velocity-velocities[1])

print("relative to Mumbai")
for name, position, velocity in zip(names, positions, velocities):
    print(name, position-positions[3], velocity-velocities[3])


How much mass per year may the black hole at the center of the Milky Way gain by capturing neutrinos? If we are immersed in a sea of neutrino, then a black hole should be constantly capturing the ones crossing its event horizon, and gaining mass.

How much neutrino mass falls inside a black hole per year? For example, the one at the center of our galaxy.

Not enough to worry about.
Most neutrinos we detect on Earth come from the sun.  The black hole at the centre of the galaxy is not so close to a star as the Earth is so you would expect the neutrino flux to be lower than on Earth,  let's assume it is similar: about $10^{11}$ neutrinos per $\text{cm}^2$ per second. But each neutrino is light, even including its kinetic energy it has only a few hundred KeV (again based off solar neutrinos for the sake of a rough estimate). Converting that to mass ($m= E/c^2$) gives the mass of neutrinos per cm² per second.  It's about $5×10^{-20}$ kg/cm²/s
The black hole is relatively small in diameter, even if you account for the fact that the effecive radius is larger than the actual event horizon radius, since things will tend to be bent towards the black hole, the cross sectional area is about 10²⁵ cm², so roughly 500000 kg of neutrinos might fall into the black hole every second.  (Although this is very much in the spirit of a Fermi estimate, this is probably an overestimate)
But the mass of the black hole is about 8,000,000,000,000,000,000,000,000,000,000,000,000kg
So 500,000kg is negligible.


Why are there no more images revealed by New Horizon? Since a few days after the flyby there are no more images published on the NASA page.

Why is NASA waiting to publish any more images for some reason?

Or is there something happened to New Horizon?

http://www.planetary.org/blogs/emily-lakdawalla/2015/06240556-what-to-expect-new-horizons-pluto.html

  The transmission of the High Priority data set will be complete on July 20, and then image transmission will pause. For nearly two months, until September 14, New Horizons will switch to near-real-time downlinking of data from other, so-called "low-speed" instruments while it transmits just housekeeping information for all of the rest of the data. No new images will arrive on the ground during this time


Radial Velocity Graph and Rossiter Mclaughlin Effect I am trying to understand the effect of the Rossiter–McLaughlin effect on radial velocity plots. Can someone help me and explain what exactly is happening in the red circle?

What's happening is that (i) the star is spinning and (ii) it is being eclipsed (by an exoplanet perhaps?)
As the exoplanet (or whatever dark object eclipses the star, the same explanation would apply to a brown dwarf or very low-mass companion) traverses across the face of the star it will eclipse parts of the stellar surface that gave their own specific line of sight velocity.
For example if we are seeing the system edge-on to the stellar equator and the planet orbits in prograde fashion with zero inclination, then it will first eclipse the side of the star that is spinning towards us. This takes a blue-shifted chunk out of the stellar spectrum, leading to a slight redshift (compared to what might have been expected). As the transit continues, the exoplanet moves across the centre of the star and then to a position blocking the redshifted side of the star, leading to a slight blueshift compared to what would be expected.
The image below (by AMitchell and shared under CC Attribution-Share Alike 4.0 International shows this situation.

Your plot shows this happening the other way around: a blueshift followed by a redshift. This can be explained if the exoplanet orbits in retrograde fashion, so it obscures redshifted, followed by blue-shifted stellar light.


Observed lunar distance data? Where can I find free/open data for the observed (not calculated/theoretical) distance between the Earth & Moon?

I want to see the observed distance of the Moon, hour by hour.

Found laser ranging data here: http://www.geoazur.fr/astrogeo/?href=observations/donnees/lune/brutes

Here you can search for data for an arbitrary time period.

The data is what they call "MINI" format, which is hard to read, its basically a long string of numbers.

Here is a sample line:

5120160113152419452625024340653926601301910034002705017 087323+04325 5320a0702

Luckily there is a specification for this format here: http://www.geoazur.fr/astrogeo/observations/donnees/lune/mini-format.html

The spec says the time-of-flight for the laser is charcters 24-37 from each line, measured in .1 picoseconds. So for the above line, the laser's round-trip flight time is 24340653926601 (.1 ps).

The data does not contain the distance, so to calculate the distance from the flight time, I do the following:

Divide 24340653926601/2 to get 1-way flight time in .1 ps.

Multiply the result 1.2170327e+13*.1 to get ps.

Multiply the result 1.2170327e+12*1.0e-12 to get seconds.

Multiply the result 1.2170327*299792458(the speed of light) to get the distance in meters: 364857224.599


Darkness in Outer Space I am trying to imagine how dark outer space is. Looking at images of cosmonauts in space proved not very useful, as the exposure makes space look completely dark.

Thinking about it, I would imagine an object would be quite well illuminated, being bombarded by provided it's not in the shadow of a planet or some other celestial body. Is that correct? How about light scattering? Is cosmic dust abundant enough for the purpose?

EDIT: @zephyr made me realize that the quantity of light is a function of position, so let's say the question is with regards to the Milky Way. Still, I would appreciate some comments on what's the situation outside the Milky Way too.

Just to put in a simple answer:

if you mean (A) for astronauts just in orbit (at the space station and so on) or on the surface of the moon.  Yes, it is incredibly bright.  Just like on Earth in the Sahara at noon!

A point of confusion is that THE >> SKY << IS BLACK, but it is incredibly bright.  On Earth in daytime, the sky happens to be blue, and it is incredibly bright.  On Moon in daytime, the sky happens to be >> black <<, and it is incredibly bright.  (Same deal in orbit as on the Moon.)

(If you are, say, explaining to a young person "is there daytime/nighttime on the moon" that's the salient point to explain: on the moon, confusingly, the sky is black during both daytime and nighttime.  But everything's just as bright on the moon in daytime as on Earth in daytime, there's the same amount of light, i.e. you can trivially read a book or see the ground, just the same as on Earth.)

If your question is about (B) in the solar system - say "on Uranus" or the like.  There is remarkably less light there than on Earth, the sun is much smaller.  Many young people don't realize this; everything's basically dark all the time from about Saturn outwards!

Outstanding page explaining this:

http://www.astronoo.com/en/children/sun-apparent-size.html

Finally if your question is about either (C) inside our galaxy but not near a star, or, (D) midway between the galaxies:

That's too difficult.


How much magnification is needed to see the planets of solar system? I have a 3inch Newtonian reflector telescope with 300 mm focal length. I can use highest magnification of 75x using a 4mm eyepiece. But in 75x I can't see the details of Jupiter what was expected. Instead I see a little blurry image. Now I would like to know how much magnification is necessary to see a good details of Jupiter and other planets. And one more question: Is there any way to improve the vision of my 3 inch telescope?

You're probably asking the wrong question - which I am going to answer anyway, and after that I am going to answer the question you should have asked instead.

As a general rule, there isn't much point in pushing the magnification above 2x the diameter of the instrument, measured in mm. 3 inch, that's 75mm, that's 150x max. Beyond that limit, even under ideal skies the image is large but blurry.

After that, seeing (or air turbulence) pushes that limit further down. Your aperture is small enough that it almost never suffers from seeing, but larger instruments are often affected. It varies greatly with time, place and season. There are times when a 12" dobsonian, that in theory could do 600x, is clamped down by seeing to 150 ... 180x. There are times when you could take a 20" dobsonian all the way up to 1000x - but that's very, VERY rare, it's the stuff of legends.

Assuming average seeing conditions and instruments of usual size (refractors of 3...4" aperture, reflectors 6" or larger), here are some rules of thumb:

Jupiter is seen best under mid-high magnification. It's rare that more than 200x is beneficial. This is because it's a very low contrast object, and additional magnification comes at the cost of less contrast, which makes things worse.

Saturn works best at high-ish magnification, bit more than Jupiter but maybe not much more. Around 200 ... 250x usually works. It depends on what you do - if you're trying to see the ring divisions, push it a bit higher.

Mars can use the highest magnification that you could generate, given the instrument and the conditions. It's a very small object, contrast is not bad, so crank it all the way up. Most instruments are limited by seeing when observing Mars.

Moon is the same as Mars.

As you can see, magnification is never an issue for you. More magnification will not make it better. In fact, more magnification always means the image is more blurry, not more crisp - it's always a compromise between size and blurriness that decides the optimal magnification.

Don't worry, everyone begins thinking that more is always better. Soon enough, experience shows them what's really going on.

That being said, I believe it's not magnification that's giving you trouble, but the general condition of the optical stack that you're using. These are things that are extremely important, and yet are ignored by many, many amateurs - and the results are not optimal. Here are a few things that you should investigate:

Collimation

Is your scope collimated? In other words, are all optical elements aligned on the same axis? The likely answer is no. It makes a huge difference in the scope's performance, especially for planets. Here's a collimated scope, compared to the same scope out of collimation:

Further information on Thierry Legault's site, which is extremely informative.

A series of articles and documents regarding collimation:

http://www.cloudynights.com/documents/primer.pdf

Gary Seronik: A Beginner’s Guide to Collimation

Gary Seronik: Collimation Tools: What You Need and What You Don’t

Gary Seronik: No-Tools Telescope Collimation

Note: Some telescopes (e.g. pretty much all refractors) do not require collimation; they are collimated from factory and hold collimation pretty well. But most reflectors (SCTs, all newtonians including dobsonians, etc) do require this periodic maintenance.

Thermal equilibrium

At 3" aperture, this is probably not a big issue, but there's no reason why you should add another problem to the existing ones. Your scope should be at the same temperature as the air around it, otherwise its performance decreases. Take it outside 1 hour before you start observing, and that should be enough for you.

Larger telescopes (around 10" ... 12" and larger) should use active ventilation for better cooling (a fan on the back of the mirror). More details here:

Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 1

Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 2

In your case, simple passive cooling for 1 hour should be enough, but it's worth reading those articles.

Focal ratio

A 3" scope, at 300mm focal length, that's an f/4 instrument. That's a pretty steep f/ ratio. Most eyepieces will not do well with such a blunt cone of light, and will start to exhibit aberrations that blur the image. Only very expensive eyepieces work well at such low focal ratios - things like TeleVue Ethos, or Explore Scientific 82 degree eyepieces.

Try and keep the planet in the center - most aberrations are lower there. Even very simple eyepieces do better in the middle of the image.

Look at the stars. Are they tiny and round in the center, and large and fuzzy at the edge? Those are aberrations from various sources (eyepiece, primary mirror, etc).

Coma

Of course, at f/4 even the best eyepieces out there cannot do anything about coma - an aberration coming out of any parabolic mirror, which becomes pretty obvious around f/5, very obvious at f/4, and a major problem at f/3. Again, coma is zero in the center of the image, and increases towards the edge.

A coma corrector is used in some cases, such as the TeleVue Paracorr, but I strongly recommend that you DO NOT use one - I suspect your instrument is aberrating in ways that overwhelm coma anyway. Jupiter would not be too blurry even at full f/4 coma at the edge. This paragraph is for informational purposes only.

Coma should become a concern with large telescopes, using high quality optics, with a focal ratio of around f/5 and less. E.g., you have a 20" dob with an f/4 mirror, then you should worry about coma - provided that collimation and so on are taken care of.

Optics quality

An f/4 parabola is not super easy to make at any size. I've made my own optics, and the lower the f/ ratio, the more difficult the process is. Many small, cheap telescopes are made in a hurry, and the difficult focal ratio poses additional problems - as a result, many manufacturers do a poor job. There are even cases where the primary mirror is left spherical, with disastrous results.

This is something you can do nothing about. If the primary mirror is bad, then that's just the way things are. An optician might try to correct it, but it's a difficult process, and quite expensive. I only added this here so you are informed.

This is what I would do in your case:

I would take the scope out 1 hour before observing, every time.

I would try and learn how to collimate the scope. I would try to figure out a few simple collimation techniques, and a few simple tests. I would spend a few days / weeks practicing that. I would keep reading about collimation.

When collimation is at least partially under control, I would learn how to properly focus the scope. Seems simple, but it can be tricky. Use a bright star, and try and make it as small as possible. Use the Moon when it's visible, and try and make it crisp and clear. Do not try this with a miscollimated scope, since it's pointless.

After a few months, when I gain confidence that the scope is in better shape, very well collimated, very well focused, I might try to borrow a better eyepiece from a friend. I said borrow, not buy. Something like a 3 ... 4mm eyepiece, good quality, that would give me a comparison for the existing eyepieces. This ONLY makes sense with a scope that is in perfect collimation, perfect temperature, perfect focus. If an improvement is seen, then get a better eyepiece - but do not spend hundreds of dollars for an expensive eyepiece that will then be used in a tiny cheap scope. Second-hand eyepieces often work exactly as well as new ones.

If you know someone in your area who makes mirrors, see if they agree to put your primary mirror on the Foucault tester, and assess its condition. But beware: the results might be very disappointing. Or not. You kind of never know with these little scopes.

EDIT: After the scope is collimated and so on, you could try to increase magnification by using a 2x barlow with your eyepieces, but do not expect miracles - the image will be bigger, but probably rather "mushy". More magnification is not always better, there's always a trade-off.

Good luck, and clear skies to you!


First observation that the Sun and Jupiter (and friends) move around a common barycenter? Answers to the question How did Kepler determine the orbital period of Mars? describing careful observations centuries ago got me thinking.

What was the first analysis of observations that directly demonstrated that the Sun and the outer giant planets were moving around a common barycenter rather than all planets rotating around a fixed Sun?

I am not sure if First observation that the movement of a planet or asteroid in its orbit was affected by another planet? is a completely separate question, or if this confirmation happened all at once, but I've currently asked it separately.

This idea would be a direct reduction from Newton's laws of motion.  In his Principia he states "Hence the common centre of gravity of the earth, the sun, and all the planets, is to be esteemed the centre of the world".  So that would be the first evidence of the Sun moving outside of a geocentric universe.  Of course he wouldn't have known about Uranus and Neptune at the time.


Celestron Astromaster 114eq Star pointer finder scope help required Just got a Celestron Astromaster 114eq as a christmas present and I have tried setting up the Star Pointer in the daylight. I did not use the red dot method as in the instruction manual, instead, I tried to do this by viewing a object through the main scope (placing the object in the center of the main scope) and then I made some slight adjustments to the Star Pointer finder scope aligning both circle and dots on the finder scope using the two adjustment screws.

This seems to work fine and finds the targets quite well,but when it turns dark and I turn on the the red dot in the finder scope and aim at a star I have yet to manage to focus in on it. Should the method I used to set up the scope finder or do I need to set up the scope finder in the dark?

Any help would be appreciated.

For an astronomical object, a 1000 mm focal length objective forms an image 1000 mm away. However, the image of a terrestrial object focuses farther out: 1010 mm for a target 100 m away, 1020 mm for a target 50 m away. When switching from a terrestrial target to an astronomical one, the eyepiece must move closer in.

Aligning the finder on a terrestrial target is bound to produce some parallax error. After focusing the telescope on a bright planet or the Moon, aim the telescope at Polaris and fine-tune the finder.

Finally, even with good finder alignment, sometimes you will have to sweep to locate your target. It gets easier with practice.


Question on Equatorial and Galactic coordinate systems I'm currently working with proper motions and have needed to convert my galactic longitude and latitude into right ascension and declination (ie from galactic to equatorial coordinates). This is fine since NED have a tool to do this, however once I convert into Cartesian coordinates I do not get the same results as I had for the galactic-cartesian conversion.

I understand that the conversion from galactic to cartesian is

x = Dcos(l)cos(b)

y = Dcos(b)sin(l)

z = Dsin(b)

where D is the distance to the galaxy in question in my case.

Also for the equatorial to cartesian transformation

x = Dcos(RA)cos(DEC)

y = Dcos(DEC)sin(RA)

z = Dsin(DEC)

I have done this for both my galactic coordinates and converted equatorial coordinates, but I do not get the same x,y and z. Should this be happening?

Yes, because the Milky Way galaxy and the Earth's equator are in different planes, currently about 63 degrees apart. To avoid confusion, rename the galactic Cartesian coordinates to (u,v,w). The transformation from (u,v,w) to (x,y,z) can be expressed as multiplication by a rotation matrix.


Since Proxima Centauri orbits the other two Alpha Centauri stars, will it eventually be farther away from us than they are? And, if so, will we have to change its name to Distal Centauri ;-) ?
Granted, its orbital period around the other two is about 550,000 Earth years, so we have some time....

In about 26,000 years, Proxima Centauri will no longer be the closest star. Instead, Alpha Centauri AB will be the closest, but only for about ~7000 years, soon replaced by Ross 248 for 10000 years, and Gliese 445 for 6000 years. Then, Alpha Centauri will continue to be the nearest star for another ~28000 years, then briefly becoming Ross 128. Refer to the diagram below (sourced from Wikipedia).

Correct me if I'm wrong, and feel free to edit my answer if it needs improvement.


I have some questions about star formation Q2 What does the phrase 'containing several distinct knots of collopsing material' mean?

Q3 In this context, does the word 'such clouds' mean protostar? (I'm not an English-native speaker.)

Q4 In the context, does the word 'a spinning disc' mean an Accretion disk ?

Q1 And the last... Could you explain 'spiral density wave' to me if you can?

I'll answer your questions in order 1-4:

The term "spiral density wave" refers to the arms of a spiral galaxy, which have a somewhat higher density than the medium in between (though not much higher). A gas-forming cloud traveling through such an arm experiences tidal forces, which perturbs the cloud and may cause local regions of the cloud to exceed its Jeans mass and collapse.
The term "knots" means clumps of higher density inside the cloud that has begun to collapse.
"Such clouds" refers to the cloud out of which a proto-star is forming. That is, it doesn't refer to the proto-star itself, but to the cloud around it. However, since the proto-star will accrete mass for a while, part of the cloud will eventually become part of the star.
The gas forms a disk around the proto-star, and loses angular momentum through heating+radiation, thus decreasing in distance to the proto-star and eventually accreting onto its surface. So yes, you can call it an accretion disk, but note that accretion disks also exist in other contexts, e.g. the accretion disk around a quasar. Also, once the central tempereture of the proto-star ges hot enough to ignite nuclear fusion and the proto-star becomes a "real" star, radiation pressure will tend to disrupt the disk, blowing the rest of it away (how much depends on the radiation field, which in turn depends on the mass of the star).


Are there any physical phenomena that destroy information? Are there physical phenomena, such as wave function collapse or black holes, which irrevocably destroy information about their prior states?

This is one of the current questions of cosmology/physics.  So far all evidence points to the negative inside a black hole so far as quantum numbers go, but it's not clear what you're considering as "prior state."

A simple counter-example is a camera. Once the photons are absorbed by the film or solid-state sensor, all phase information is lost.  Is that what you were looking for?


What does "Active Weather Patterns" mean? I just read that Cassini discovered titan had "active Weather patterns".

What does that mean?

It means that Titan has weather (driven by methane rather than water) and that it's weather changes with the seasons. Cassini has been observing Titan for almost half of a Titan year, which is 29.457 times longer than is our year. Titan's weather patterns (where it's cloudy, where it rains) has visibly changed over this time, just as weather patterns here change from summer to winter / winter to summer.


Is the earth bombarded equally in all directions by neutrinos? Perhaps it is hard to known for sure because neutrinos are very difficult to detect, although they do go through the earth in very large numbers. But are they passing through the earth equally from all directions or is it really depending on particular events in the universe. It's probable that the sun generates a lot of them, so perhaps in the daytime we receive more of them than at night. But besides the sun's radiation, are the neutrinos coming from everywhere? Or is it possible that the majority of them comes from one direction, and perhaps even has an influence on the earth's movement?

There are only two types of neutrino source that are "bright" enough to be reliably detected. The sun and nearby supernovae.

The source of solar neutrinos is nuclear fusion, which is also the source of most of the star's energy. Neutrinos also spread out in all directions, so their intensity follows an inverse square law. So the amount of neutrinos is proportional to the brightness of the star. With current detectors no star is bright enough to be observed except for the sun. Other stars do produce neutrinos, and stellar neutrinos come from everywhere (probably more from the milky way) but there are not enough of them to be detected.

Supernovae in the milky way and neighbouring galaxies produce ridiculous amounts of neutrinos, and a spike in neutrinos was observed from SN1987A the closest recent supernova.

Since the sun is the brightest source of neutrinos you might think that the Earth would block out neutrinos during night. However, neutrinos pass right through the Earth almost without noticing. The Earth is transparent to neutrinos. So we detect as many neutrinos at night as during the day.

One thing is certain, there is absolutely no effect on the Earth's rotation or anything else from neutrinos, they just pass right through.


Solar / lunar positions in ECEF using SOFA C libraries I'd like to calculate the positions of the sun and moon in the ECEF coordinate system using the SOFA (http://www.iausofa.org/current_C.html) libraries given a Julian day input.

So far, I've got this process for the sun:  (1) convert Jd (UTC) to TAI, to TT, then to TDB, then (2) calculate earth position/velocity in heliocentric coordinates (iauEpv00), then (3) negate the terms to get ECI.

(a) I'm not sure if the process above is correct, especially since calculating TDB time requires a "dtr" parameter, which apparently you have to have TDB times to estimate using iauDtdb().

(b) I'm not sure how to convert the resulting vector from ECI to ECF, which is probably just a time-based rotation of the longitude, but how much time is elapsing?

(c) Not sure where to even start with the moon.

A few responses to my own question:

(a) TDB time isn't critical because it only varies from TT by a few milliseconds.

(b) The resulting vector was not ECI but the vector from the earth to the sun in the celestial frame.  I needed to build the celestial to terrestrial rotation matrix using iauC2t00a() (assuming zero polar motion since it's generally less than 20 meters), and then rotate the resulting pv (position, velocity) vector-pair into the terrestrial frame using iauRxpv().

(c) (Still no clue on this one.)


How deep is the "A" Fraunhofer line in the solar spectrum? Is it from the Sun's or Earth's atmosphere? Question: Where can I see - or even download - a fairly high resolution (~0.1 nm) solar spectrum at zero air mass - in other words from space without absorption and other features from the Earth's atmosphere? Of particular interest is the near infrared around 750-780nm. Is the "A" Fraunhofer line due to absorption by oxygen in the Sun's atmosphere, or in Earth's?

The background discussion below shows some examples, but these are not directly from scientific sources and I have a hunch they are from observations on Earth's surface below the atmosphere.

Background:

Studies of photosynthesis on Earth using spectrometers in satellites measure solar flux induced fluorescence from foliage during the day, taking advantage of certain bands in the solar spectrum that are relatively "dark". GOME-2

For example, the Fraunhofer line labeled "A" below is produced by absorption by oxygen, but it is not clear to me if it is the oxygen in the Suns atmosphere or the Earth's atmosphere that's producing this line.

The images below are spectra of sunlight with various instruments of widely different resolutions, and because the absorption is narrow, the apparent depth is a function of resolution rather than reality.

above: a low resolution spectrum from here

above: An spectrogram of the Sun from this NOAO website (N.A.Sharp, NOAO/NSO/Kitt Peak FTS/AURA/NSF) with the "A" line near 760nm appearing near the top, a little right of center. The original data is from the "Fourier Transform Spectrometer at the McMath-Pierce Solar Facility at the National Solar Observatory on Kitt Peak, near Tucson, Arizona" but seems to be formatted and colorized to look like an Echelle spectrogram.

A video describing the use of the dark band for Earth science can be seen at https://youtu.be/1XilneV3cJI.

Where can I see - or even download - a fairly high resolution (~0.1 nm) solar spectrum at zero air mass - in other words from space without absorption and other features from the Earth's atmosphere?

You can download a normalized solar spectrum with 0.1 nm resolution from here. However this is not at zero airmass.

A solar spectrum with zero airmass can be found on this website with the direct link to their data in an excel sheet here. This spectrum appears to be at a 1 nm resolution.

  Of particular interest is the near infrared around 750-780nm. Is the "A" Fraunhofer line due to absorption by oxygen in the Sun's atmosphere, or in Earth's?

If you look up the Fraunhofer A line (really it's a "band"), you'll see it is caused by molecular oxygen, namely $\mathrm{O_2}$. This should tell you right away that it is caused by our atmosphere as molecules do not easily exist in the atmosphere of the Sun. In our Sun's atmosphere, most species are ionized forms of single atoms such as $\mathrm{O\:IV}$.

In Astronomy, spectral lines which are caused by the Earth's atmosphere and not by the source object being observed, are referred to as Telluric lines (from the latin word tellus meaning Earth). I found this source which describes, in detail, the various telluric lines that you may encounter.


Why are solar systems heliocentric So this may be an incorrect assumption, but from my knowledge solar systems are heliocentric and there are always suns. (I realize to be a solar system there must be a sun semantically.) But it seems to me that planets and stars are just bodies of mass, so are there cases of solar systems where the star is just one of the bodies of the system and orbits around a super massive planet? And similarly are there 'solar systems' without stars, where there are a group of planets orbiting a massive planet at the center of the system?

Generally, astronomical bodies rotate around their common center of mass, not the one or the other body.

If one of the bodies is much heavier, the center of mass is much nearer to its center, sometimes even inside the larger body. For example, Earth and Sun both rotate around a point that is much nearer to the Sun's center than to the Earth's center, as the sun is much heavier. Looking from the outside, it seems as Earth rotates around the Sun, while the Sun just wobbles a bit in sync. But the reason for that is exclusively the masses of the bodies, not their being-a-star or not-being-a-star.

Your theoretical example of a super-heavy planet and a lighter star would result in the star seemingly rotating around the planet; however, here you need to look at why a star is a star:
Stars are created by gravity that is compressing the mass to a level that starts fusion, and thereby a star. In your example, either the 'light star' wouldn't make it into a star (and be a light body/planet), or the 'super-heavy planet' becoming a much brighter star (or both of it). That reverses the roles effectively.

Looking at it this way, the bodies in a system rotate always around the center of the mass, which is near the largest mass, which will be the first star of the system (if any).
So your first constellation is not possible.

The second is well possible - if neither of the bodies has enough mass to become a star, they will be still rotating around each other the same way, with the heaviest body being nearest the center. Such a system would be dark and basically invisible from some distance. We wouldn't be able to find them.


Do all orbiting bodies eventually collide? If two celestial bodies are in orbit, will they always eventually collide if not acted upon by outside forces?

Yes.

Two bodies in orbit around each other will inevitably collide. The reason for this is that the system will give off energy in the form of gravitational waves. This effect is commonly cited in binary neutron star systems, where the two stars are isolated and close together. One of the most famous of these systems is the Hulse-Taylor binary.

The time it will take for the objects to collide can be calculated:
$$t=\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{(m_1m_2)(m_1+m_2)}$$
where $r$ is the initial radius, $m_1$ and $m_2$ are the masses of the bodies, and $c$ and $G$ are the familiar constants, the speed of light in a vacuum and Newton's universal gravitational constant.

However, tidal acceleration could offset some of the effects.


Studies of isotropy from different ends of the visible universe? Have there been studies to compare structures that are 26 billion light years distant (i.e. from one side of the observable universe to the other) to search for faint differences?

The most distant objects which can be seen with telescopes are around 13.39 ly away, with a redshift $z=11$. Telescopes see a roughly uniform Universe, although some very large structures were already identified.

The Cosmical Microwave Background has a $z \approx 1000$.

It was measured for its possible minor anisotropies by the COBE, and the Wilkinson Microwave Anisotropy Probe (WMAP) and Planck satellites.

It measured the temperature of the 2.7K Cosmical Microwave Background with $\approx 10^{-5} K$ precision. The technology was not easy.

There were two results:

First, there is a global dipole anisotropy showing that we are moving with $\approx 368 \frac{km}{s}$ to the Hercules constellation. It is the nearest thing what could be seen, as if we had found an absolute reference frame.

Second, substracting this from the results, another, lesser anisotropy were found:

The distribution of the anisotropy, combined with the result of other experiments, helped to get

the most precise measurement of the age of the Universe (13.7 Gy).
the most precise measurement of the curvature of the Universe (0, below measurement precision).
that the expansion of the Universe is accelerating.

Also the polarity was measured, which has shown that the re-inozation happened earlier as it was previously believed.

This is the best result what we have until now.


What azimuth description systems are in use? For a time, I thought Azimuth is always direction in degrees, 0 for local, geographic north, 90 for east etc.

Then, trying to observe moonrise I downloaded an app that gave azimuth in values between (if I recall correctly) -180 and 180, and I think "0" corresponded to south. I tried to read up on that and found very little fragmented information. I found other resources that frustrated me with numbers greater than 360, or provided values commonly associated with "West" for azimuth of moonrise and generally it took a bit of searching to find something with plain "degrees from north"

Could someone either post or provide a link to a comprehensive guide to various azimuth angle systems, when and where they are in use, and how to recognize which one is used?

There are many indeed.

It may be specified from South westwards(0° to 360°), or from North eastwards and westwards(180° to -180°).

The most popular one, as used by Roy and Clarke in "Astronomy : Principles and Practices" is to measure the azimuth from North eastwards 0° to 360°. For a southern hemisphere observer, this system changes to from South eastwards 0° to 360°.

With so much confusion, how do professional astronomers go about it? They simply specify the system they are using in a given document. Doesn't cause much trouble since the differences in all of the systems is by an angle of 180° or 360°. It just causes sign changes in all formula.


Is it possible to move a planet out of its orbit? At least a lighter planet? As observed by many researches our sun will be exhausted in the distant future. So does anyone think that driving our planet (or any planet) closer to the sun for more light and heat is a solution for life to survive on earth? Or would moving it out of the sun's gravitational field and taking it to some other solar system with a healthy star be better, possibly with some kind of thrusting mechanism that we may invent in future?

I am aware of the sun's gravity and the mass of the earth.

Yes. Even with todays technology we can get started.
In this article by Korycansky et al 2001 it is suggested to use rockets to move an asteroid (like a larger version of the NASA ARM Asteroid Redirect Mission) to a cycling orbit between Earth and Jupiter. The asteroid would pass near infront of the Earth in order to give the Earth a slight gravity assist which moves it outwards. The asteroid then passes near behind Jupiter to itself get a gravity assist. This transfers energy from Jupiter's to Earth's orbit. I've also heard of a proposal to use Venus instead.


What is the gain of a telescope? What is the gain of a telescope? How would I go about calculating the gain from a flatfield image?

Gain in electronics can have different meanings:

Gain is a measure of the ability of a [...] [device] to increase the power or amplitude of a signal from the input to the output port. [...] The term gain alone is ambiguous, and can refer to the ratio of output to input voltage (voltage gain), current (current gain) or electric power (power gain).

For optics, I found the light-gathering power the most appropriate measure, which relates the apparture of the telescope with the one of the human eye:

The light-gathering power of an optical telescope, also referred to as light grasp or aperture gain, is the ability of a telescope to collect a lot more light than the human eye. Its light-gathering power is probably its most important feature. [...]
The gathering power $P$ compared against a human eye is the squared result of the division of the aperture $D$ over the observer's pupil diameter $D_{p}$ [...] An example gathering power of [a telescope with] [...] $254\,{\rm mm}$ [aparture] compared to an adult pupil diameter being $7 \, {\rm mm}$ is given by:
$$P = \left(\frac {D}{D_{p}}\right)^2 = \left(\frac {254}{7}\right)^2 \approx 1316.7$$

Edit: After the question changed slightly, I guess I should add a bit on "how to calculate gain from an image": Gain requires always two measurements, e.g. the image from your telescope and a base of comparison, e.g. the human eye, that's why I found the aperature-based-gain definition appealing.
If you have a datafile with intensities from your telescope - basically an image in grayscales depicting the spatial light intensity distribution - you would have to compile an image of the same area of the sky which serves you as a base of comparison.


What would happen if a rogue planet hit one of the planets in our Solar System? As there are millions of rogue planets in the Milky Way, what would happen if one hit a planet in the Solar System.

For example, if the rogue planet hit Pluto, would Pluto fall in to the Sun due to to its gravity? What kind of catastrophic damage could it do in the Solar System?

It would depend entirely on the size, composition and trajectory of the two objects coming into the collision. They could annihilate each other leaving a mess of debris. Or one or both could survive the collisions in some form. And each of the surviving objects could be flung out of the solar system, flung into a spiral into the sun, flung into an elliptical orbit, or thrown into other objects. The space between planets is big so odds of direct and immediate collision of the "planets like billiard balls" variety you see on some sci-fi aren't likely. But anything the size of a planet would be seen well before any impact.


Is the Oort Cloud actually opaque, or is the "cloud" used euphemistically because minor planets inside of it are too dim to be seen? I did read several of the Oort Cloud questions/answers which I found very helpful and instructive, but this question still persists for me:

Is the Oort Cloud actually opaque, or is the "cloud" used euphemistically because minor planets inside of it are too dim to be seen?

Well, given that we can see the stars and that the Oort "cloud" is closer than the nearest star, then the answer to your main question is obviously that the "cloud" is not opaque.

I think it is called a cloud because it consists of many individual, small "particles" that don't interact with each other. Also, the term cloud avoids giving the impression that the Oort objects are arranged in any disk-like configuration - they should be much more spherically symmetric. The word cloud also avoids giving the impression that the objects fall between two relatively tight orbital radius limits. i.e. The Oort cloud is not like the Kuiper belt.


Brown dwarf magnetic activity I have not been able to find an answer to my question through google, and therefore ask it on this medium.

Do brown dwarfs have spots and faculae similar to other heavier stars?

So far, opinion is divided, as evidence has not been found to prove either way. Back in 2015, a study published in Nature, and linked via the Verge, here suggested that no, brown dwarfs do not exhibit star-like behaviours, but are in fact much closer to planets (due to them being below critical mass for hydrogen fusion in their cores)

All planets that have magnetic fields within our Solar System have aurorae, so Hallinan and other researchers believed that if brown dwarfs were more like planets, they too would exhibit aurorae.
...each time LSRJ 1835+3259 rotated — every 2.84 hours — it pulsed a radio emission and an aurora appeared at a location on the brown dwarf's surface.
..."In the case of the brown dwarf, the aurora is so much more powerful than anything we've seen in our Solar System," says Hallinan. "Jupiter's aurorae are 1,000 times more powerful than the Earth's aurorae. Well this brown dwarf has aurorae that are at least 10,000 times more powerful than Jupiter's aurora." The aurorae's colors are also a bit more diverse, emitting red along with blue and green.
...it strengthens the argument that brown dwarfs are more closely related to planets, kind of like "beefed-up Jupiters." "Typically, people tend to think of these things as failed stars," says Nichols. "My preferred explanation is they’re more like very successful planets, and this study demonstrates more planet behavior than stellar behavior."

However this 2017 article in the Astrophysical Journal which looked at the same data from LSRJ1835+3259 suggests otherwise, identifying the magentic data not as aurorae, but as effects more like that of our own sun. A nice summary of the article on Howstuffworks.com:

As the object rotated, a powerful magnetic region came into view, more powerful than the magnetic fields associated with sunspots that we observe on the sun. Sunspots are magnetically active regions that can trigger coronal mass ejections, solar flares and produce powerful streams of solar wind — all of which can generate powerful geomagnetic storms on Earth.
In an interview with New Scientist, the researchers point out that LSR J1835+3259 is very young (approximately 20 million years old) and the powerful magnetic field could be interacting with the object's protoplanetary disk (if it has one). But if this active magnetic region is long-lasting and representative of its global magnetic field, LSR J1835+3259 is way more "star-like" than we give brown dwarfs credit for.
So rather than calling brown dwarfs "failed stars," perhaps we should call them overachieving planets or magneto-dwarfs.


How is this the last supermoon of 2019? Media coverage of the "super worm equinox moon" (*eyeroll*) has stated that this is the last supermoon of 2019.  Since supermoons normally happen every three or four months, how can there be big gaps?

It feels like phenomena like supermoons, which are basically due to the relative phase of orbits, shouldn't have sharp transitions like this.

A supermoon does not occur every three or four months. There may be 2 or 3 consecutive supermoons (that is, separated by 1 month) that occur at about the same time each year.

If you ignore the precession of the Moon's orbit, then there is one time of the year when the Full Moon and perigee occur (point 1 in my diagram below). Two weeks later, the New Moon occurs near apogee (point 2). At any other time of the year, the Full Moon occurs at a different point in its orbit around the Earth, so it is farther from the Earth than at perigee. Six months later, the Full Moon is occurring at apogee (point 4), and no one cares about that!
 (not to scale!)

Full Moon at perigee (closest to the Earth)
New Moon at apogee (farthest from the Earth)
New Moon at perigee
Full Moon at apogee

The Wikipedia article on the Supermoon has a nice graphic showing the Full Moon and distance from the Earth. Depending on how close to the Earth the Moon needs to be to be "super", you can see that there is a "season" when the supermoon occurs. (I have copied the image here, and added a dashed line at 360,000 km to show which moons might be "super" and which ones are not.)

Now, if only people would care about the Super First Quarter Moon. Then we would be celebrating on May 12, 2019! (Not really. That is farther than the 360,000 km criteria.)


Do Stars in Galaxies also move in elliptical orbits? There is a gravitational force between every pair of objects, because of which planets revolve around the Sun in elliptical orbits.

My question is this: As planets move in elliptical orbits with the Sun at a focus so do the stars in galaxies also orbit, and if then what are the foci of their orbits. Since the gravity of black holes is very high are black holes at any of the focus of a star's orbit?

Planets orbit the sun in approximately elliptical orbits. The orbits of the planets are perturbed by each other, particularly by Jupiter. So the shape of orbits are very nearly, but not exactly elliptical.

A simple elliptical orbit is found when there is a single dominating mass, for example in the solar system the Sun is much more massive than the rest of the solar system put together.

Stars in galaxies do move. However, they don't move in simple elliptic orbits for the simple reason that the mass of a galaxy is not concetrated at a single point. There is a supermassive black hole at the centre of the galaxy, but although it is massive, its mass is dwarfed by the mass of the rest of the galaxy (it has a mass of about 4 million suns, the galaxy has a mass of 600000 million suns.) The stars move in orbits around the centre of mass of the galaxy, to which the black hole contributes only a very small amount. The shape of the orbit is a "wavy circle". It is not elliptical, and so doesn't have foci.


Which structure is represented in this video (if any)? I know this is not supposed to be precise since it is just a movie but I would suppose it is based on real things. In the Contact movie intro
https://youtu.be/EWwhQB3TKXA?t=1m I think the Milk Way is shown at around 2:12, I am wondering what is the structure shown around 2:32. Does that a representation of anything real? Virgo maybe?

I think it is supposed to be a representation of the Centaurus A/M83 Galaxy group.

https://en.m.wikipedia.org/wiki/Centaurus_A/M83_Group

These are a bunch of galaxies between 12 million and 15 million light years away.

The big central dust lane in Cen A is very recognisable.

I guess the sequence then continues through the Virgo cluster and supercluster which are roughly in the same direction.


What is happening with these solar particles detected near the Sun that is so newsworthy? Phys.org's Team identifies low-energy solar particles from beyond Earth near the Sun links to the open access ApJ article Properties of Suprathermal-through-energetic He Ions Associated with Stream Interaction Regions Observed over the Parker Solar Probe's First Two Orbits.

It quotes "Dr. Mihir Desai, a mission co-investigator on the Integrated Science Investigation of the Sun (IS☉IS) instrument suite, a multi-institutional project led by Principal Investigator Prof. Dave McComas of Princeton University.. IS☉IS consists of two instruments, Energetic Particle Instrument-High (EPI-Hi) and Energetic Particle Instrument-Low (EPI-Lo)" as saying:

  With EPI-Lo, we were able to measure extremely low-energy particles unexpectedly close to the solar environment. We considered many explanations for their presence, but ultimately determined they are the smoking gun pointing to interactions between slow- and fast-moving regions of the solar wind that accelerate high-energy particles from beyond the orbit of Earth. Some of those travel back toward the Sun, slowing against the tide of the outpouring solar wind but still retaining surprisingly high energies.

I'm having trouble parsing this very long single sentence. Why are these "extremely low-energy particles" also "still retaining surprisingly high energies"?

The Phys.org article talks about Stream Interaction Regions (SIRs) and the formation of turbulent corotating interaction regions (CIRs) that produce shock waves and accelerated particles, not unlike those produced by CMEs.

Question: I think it is saying that particles from the Sun go beyond Earth and then are turned around by turbulence and move back into close proximity of the Sun, at which time they have "extremely low-energy" for some reason. Is that at least a rough understanding of what's happening?

If I understand your question correctly, it is about EPI-Lo which a part of the Integrated Science Investigation of the Sun (ISʘIS) . The picture is also taken from there:

EPI-Lo stands for Energetic Particle Instrument-Low energy which indeed sounds confusing at first sight. The essential definition can be found in the instrument paper:

EPI-Lo measures ions and ion composition from ∼20 keV/nucleon–15 MeV total energy and electrons from ∼25–1000 keV. EPI-Hi measures ions from ∼1–200 MeV/nucleon and electrons from ∼0.5–6 MeV.

The paper also addresses the scientific background of the measurements of SEPs, which are defined to have the following energy range:

They consist of protons, electrons and HZE ions with energy ranging from a few tens of keV to many GeV.

In short: The EPI-Lo measures the lower part of the SEP events.
Concerning the newsworthyness, I can only guess. Why SEPs are worth studying is nicely stated at Wikipedia:

They are of particular interest and importance because they can endanger life in outer space (especially particles above 40 MeV).
Solar energetic particles can originate either from a solar-flare site or by shock waves associated with coronal mass ejections (CMEs). However, only about 1% of CMEs produce strong SEP events.

I hope that my remarks answer (at least partially) your questions.


Could Pluto and Charon have extra-Solar origin? Pluto and Charon seem to have surprisingly young surfaces, considering that resurfacing events were expected to be rare for them.

Could the explanation be that they actually are young? Formed later than the Solar system at some other star and then ejected and captured? Or that they are ancient but have extra-Solar origin and have little cratering because they have spent billions of years as a vagabond planet system in empty interstellar space? Sedna is believed to be captured.

The surface of Pluto is not just young, but very young, and also differentiated. The 'heart', Sputnik Planum, may be currently active, with flowing glaciers, and nitrogen snow. This area may have be resurfaced during Pluto's perihelion. Your theory cannot account for the very young age of the surface.

Next capture of a rogue planet is not easy. It would need Pluto to interact with the outer planets, to slow it down to enter solar orbit (otherwise a rogue planet would just pass through the solar system and come out the other side) A captured rogue would be unlikely to have the relatively low eccentricty of pluto.

Finally, while we don't know how common rogue planets are, they are probably not common, and given the massive distances between stars, the chances of any rogue just happening to have passed anywhere near the solar system in the last billion years in the right direction to be captured to Pluto's orbit is very low.

Conclusion. Pluto formed around the sun. A capture theory does not explain the observed young surface of Pluto, and modelling suggests a capture would be very unusual.


Does the focus of terrestial telecopes need adjustment for the distance of astronomical objects? My rudimentary experience of photography and telescopes suggests that with commonly affordable optics any object beyond the range of hundreds to thousands of feet will be in adequate focus at the proper "infinity" position. For example the advice here for overcoming difficulties in seeing the focus of a star on a DLSR screen to instead focus on a brighter and distant city light: https://photo.stackexchange.com/a/23973.

However an interlocutor recently offered this advice (collated from a series of comments):

Apparently you haven't actually used a camera or a telescope. The focus adjustment is there for a reason. Any astronomer will tell you focus changes are needed between stars, stars farther away,  planets, and the Moon. ... If focused on the Moon, you need to refocus slightly when looking at stars. When looking at Jupiter you need to refocus if the target is the Moon. Close stars vs far away stars, the same thing, a need to refocus. ...  To the tune of: Jupiter and it's moons, which are close to Jupiter , need focus adjustment between them. Either Jupiter or it's moons will be in focus. Ask any astronomer. That's a difference and it's well past the supposed "infinity focus" point.

I wouldn't be wholly surprised if retargeting a telescope could require a focus tweak to correct for slight changes of the mirror distances (etc), or from thermal shrinkage across a cooling evening, but such effects would probably be invariant across objects, whereas as described in that quote targetting back on the original object would presumably require the focus to be restored to the original setting.

Is there a distance-dependent component for focus in astronomical observation, and if so what sort of equipment does it affect in practice?

This is incorrect in any operational sense, and no astronomer will tell you that:

  Any astronomer will tell you focus changes are needed between stars, stars farther away, planets, and the Moon. ... If focused on the Moon, you need to refocus slightly when looking at stars. When looking at Jupiter you need to refocus if the target is the Moon. Close stars vs far away stars, the same thing, a need to refocus. ... To the tune of: Jupiter and it's moons, which are close to Jupiter , need focus adjustment between them. Either Jupiter or it's moons will be in focus. Ask any astronomer. That's a difference and it's well past the supposed "infinity focus" point.

It's true that Jupiter and the Moon are at different distances and thus geometrical optics says that the focus needs to be adjusted, but this is only true in mathematically in an idealize instrument. In any real instrument, it is not possible to discern any difference between the various focus points.  They are all at infinity.

So one size really does fit all.

With exceptional quality optics, it's possible that a distant streetlight might require a very, very slightly different focus than the Moon or Jupiter. Nearby streetlights, sure.


What does the 'amplification factor' of gravitational microlensing mean? At first I assumed the amplification factor meant an increase in brightness. But since the equation: $$A(u)=\frac{u^2+2}{u\sqrt{u^2+4}}$$ is independent of mass, this make no sense to me. Wouldn't everything be infinity magnified (if its mass independent empty space should work, right)? Even if its simply not valid for a zero mass, everything would be amplified by an arbitrarily high amount any time even a neutrino passes arbitrarily close to the right $u$?
But if its not brightness, I don't know what signal is being amplified.

Gravitational lensing is just geometry, as is optical lensing. The amplification factor describes the increase of the area of the image (at constant surface brightness).

BTW, the amplification factor does depend on the mass $M$ of the lens, since $u=\theta/\theta_E$ and $\theta_E$ depends on $M$. Here $\theta$ and $\theta_E$ are the relative angular distance between source and lens and the radius (angle) of the Einstein ring for the lens, respectively.


Speed gaseous planets Why do gaseous planets spin faster than telluric planets? I've seen thats because of primitive nebula but I didn't understand why? Could someone help me? I would appreciate a lot!

There isn't such a great difference between Earth and Mars (at about 25 hours) and the gas/ice giants (between 10 and 20 hours). Earth had a day of about 6 hours in the early Solar system, but has been slowed by lunar tides. Small planets will not accrete gas in their early formation, and so will not pick up angular momentum from accretion discs, which can explain why Jupiter and Saturn have rapid rotation rates.

Mercury and Venus are slowed by Solar tides. In the case of Mercury, it is in a resonance between its orbit and the solar tide. In the case of Venus, there are significant atmospheric tides, that slow it down (It actually spins backwards, and the full explanation is incomplete.)

So the answer is three-fold: Distant gas giants pick up lots of momentum in their formation, they are not much affected by Solar tides, and the differences between them and the rocky planets are not actually so great anyway.


Is it only possible to look at solar systems with stars at least as old as ours to be able to find intelligent life? Assuming the speed it takes to development intelligent life is the same in all solar systems, would we need to look at exoplanets with stars as old or older than ours to find intelligent life? A young star would be assumed to have been surrounded by planets that are relatively new and haven’t had time to develop intelligent life. Visiting those planets with young stars relative to the sun would most likely only give us planets with prehistoric life.

Assuming the speed it takes to development intelligent life is the same in all solar systems,

We don't know if this is true.

We don't know if:

Life has started on more than one planet in the Universe.
How long, on average, it takes for life to start.
Whether life normally evolves past the "bacterial slime" stage to form complex multicellular life, and how long this takes.
Whether multicellular life normally develops intelligence, and how long this takes.
How long intelligent life survives, on average.

So we don't know what time it takes to develop intelligent life. However, planets are the same age as their stars, so it is reasonable to suppose that older stars are more likely to harbour intelligent life.

Moreover, visiting any extrasolar planets is too hard with current technology. Stars are just too far away.


Neutron star, aging proccess Is there any information about the future of the neutron star?
What I am interested what happens to it as the time passes. Say a billion years. Would it still be a neutron star? Would it collapse to black hole? Would it simply dissipate?

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

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

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

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

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


why there is a circle of light surrounding Sun and Moon? One day when I was playing basketball I saw a circle around the bright Sun. It is a perfect circle with a radius of about 20-30 km.

Why does this circle of light form? I have seen the same also around the moon. But this does not happen every day. Why ?

Most probable you have been experiencing a weather phenomena: Ice crystals in the sky reflect or bend the light to form a halo.

So, this depends on the temperatures and moisture content of the atmosphere and cannot be seen every day.


Stars orbiting supermassive black holes I read an article on a group of stars orbiting the supermassive black hole Sagittarius A. Did these stars get caught by the black hole's gravity like planets capture moons, or were they born from the remnants of the star that went supernova and turned into black hole?

They are captured.

The formation of supermassive black holes is rather uncertain. They may have formed from core collapse of very ancient stars  (and then grown) or they may have some other mechanism of formation in the early universe. However we do know that the first formed early on in the history of the universe. They are as old as the galaxy, or perhaps even older.

On the other hand the stars that orbit the black hole are of regular age.
They are not particularly old. Therefore the black hole formed long long before the stars that orbit it. And so those stars must have been captured by some mechanism.

Star formation doens't normally happen in supernova remnants. A supernova can provide the shock that starts stellar formation in molecular clouds.


white-dwarf merge in binaries Two white dwarfs with a total mass $>1.4\:M_{\odot}$ merge. It may trigger a type Ia SN explosion.

How about the three cases below?

Two low mass WDs merge, total mass $<1.4\:M_{\odot}$.
A low mass WD + a low mass neutron star, total mass $<2\:M_{\odot}$ but $>1.4\:M_{\odot}$.
A low mass WD + a neutron star, total mass $>2\:M_{\odot}$.

If any of the above can result in a SN, have we observed this kind of SN?

I don't think any of those would give a supernova, because the first two don't cross an important mass boundary, and the last one might make a black hole from a neutron star, but that's not going to be much of a supernova because it's already very compact and there's not going to be much to have a core bounce off of, nor is the white dwarf massive enough for a type Ia.  They should all produce some significant emission, depending on how fast the mass is introduced.  Perhaps somewhere between what we would call a supernova, and something more along the lines of a classical nova (for the first one) and quasiperiodic X-ray flaring (for the next two).   They could get pretty bright, and I don't know if the nomenclature has individual names for each one of those, but they don't quite sound like supernovae.