Is atmospheric turbulence irrelevant for ExoPlanetary transits and radial velocity measurments? Atmospheric turbulence is known to scatter photons in a quasi-random way along their path throughout the atmosphere, resulting in lower imaging resolution than would have been anticipated by instrument-only considerations.

I have been thinking whether the same effects can play a relevant role in limiting the sensitivites for photometry in transits or for spectrometry in radial velocity measurments.

My thoughts so far:

Transits: as I'm no observer I don't know if atmospheric turbulence is actually strong enough to scatter source photons out of the line-of-sight, rendering them undetected. This would fiddle with the signal-to-noise ratio per measurment and would let it fluctuate over time.
Radial velocity: Turbulence should be able to influence a spectral measurment from the ground, if the induced turbulent broadening is significant compared to the line width that can be resolved with the instrument considered. Taking the turbulence-induced doppler shift $\Delta v/c$ as $10cm/s /c \sim 10^{-7}$ (I assumed turbulent eddy-velocities to be comparable to typical winds) as typical for Earth's atmosphere, this should be  insignificant even for a high-resolution spectrograph like HARPS that has $\lambda/\Delta \lambda \sim 10^5$.
However smaller eddies rotate faster, they could thus reach the detectability range when $\Delta v/c \sim 10^{-5}$

Here my expertise in this topic ends, and I'd hope for someone from this community to illuminate the points above. Also googling usually only points to the benefits in direct imaging.
Bonus question: Would adaptive optics always help to remedy any issues that might arise?

One can imagine turbulent eddies in atmosphere as very weak optical lenses which focus and defocus stellar radiation. This leads to image degradation (seeing) and fluctuations of flux being registered through some aperture. The latter effect is called scintillations. It is very prominent for naked-eye observations. For telescopes the averaging by large aperture reduces the magnitude of the scintillations. Nevertheless it is main limiting factor for high-precision photometry at telescopes larger than ca. 2 m. See e.g
http://adsabs.harvard.edu/abs/2012MNRAS.426..647K

As for spectrscopy, atmospheric turbulence does not affect the wavelength of radiation as long as in modern high-precision spectrographs (e.g. HARPS) actual detection takes place in vaccuum.


When we see half moon, why is it always the lower half? Even if we see a crescent moon, always the lower circumference is visible. Why we never see this?

PS: this image is vertically inverted.

Actually, you sometimes can see the Moon illuminated by the Sun from the top, and therein lies the answer to your question - during day. When looking at the Moon during night time, the side of the Earth where you're standing and looking at it is pointed away from the Sun, while this is reversed during the day. The illuminated side of the Moon is also facing the Sun, of course. So we have:

Night time on Earth: Side illuminated by the Sun is "below us" and that's where the Moon is illuminated from too.
Day time on Earth: Side illuminated by the Sun is "above us" and that's where the Moon is illuminated from too:

Blue Moon in the daytime sky (Photograph source: LPOD)

So, the direction from which the Moon is illuminated from is the same as the direction the Earth is, both illuminated by the same celestial body - our Sun. So, when you see the Sun high above the skies, and if the Moon is visible during daytime, it's going to be illuminated from the same direction from which you are, and during night time, when you'd expect the Sun "below where you're standing", it's illuminated from that direction.


If a black hole pulls every object in the universe, why there is a light around it? I've seen some black hole images and I've seen some movies about it e.g(interstellar). I wonder why it has a light around it. Even if I know that light can't escape through it. Can someone enlighten me about this? Thanks.

Well, you're partially correct here. A black hole traps everything inside its event horizon. Outside of the event horizon, matter (and energy) can still escape from its gravitational pull. This is due to the fact that the escape velocity of a black hole beyond its event horizon becomes greater than the speed of light itself.

So, the light you see around a black hole is actually just the matter and energy that is outside the event horizon. In other words, you're seeing the light and matter that is not actually inside the black hole. This "stuff" usually orbits the black hole at large fractions of the speed of light.


Is there a way to tell the difference between earth andesite from Mars I'm curios if there is a way to test if this sample might be Martian andesite.

What type of oxygen isotope (or other) test can be done to address this in some way?

As background, any information about such tests at the University of Utah would also be helpful to know.

If this is something that you have found (rather than purchased as a meteorite) the chances are very small that it is a meteorite. Even if it is a meteorite, the chances it's a Martian one are even smaller still and none have been found in the United States.

According to the Meteorites in the US page, which draws from the Meteoritical Society database, only 1821 meteorites have been found in the US over the past 200 years. Of all meteorites found in the world, less than 0.1% are from the Moon or Mars (source statement, graphs of meteorite fractions) and none of these have been found in the US, with the vast majority (99%) found in Antarctica or the African or Arabian deserts (lunar meteorites)).

There is a long "Meteorite Realities" page and a shorter, graphical "Self-Test Check-List" that it would be good to check and go through before it gets to chemical testing.

If you are determined to get testing done, the same meteorites.wustl.edu site on its page on meteorite chemical composition recommends chemical testing by Actlabs; there is more information on what they need (a 5g sample) and the type of tests to ask for on this page. Andesite is a type of basalt formed by volcanism and while it is true that most of the Martian meteorites are basalts (as discussed here), so are a lot of Earth rocks. On the chemical composition page, in plots of chemical composition such as silicon dioxide (SiO$_2$) vs total iron and magnesium oxide content  e.g. the martian meteorites (red squares) separate from most terrestrial/Earth rocks and the "meteorwrongs" (white circles) due to having higher iron+magnesium content in the form of pyroxene, olivine and ilmenite (from 'Chemistry' section of How Do We Know That It’s a Rock from the Moon?) However as noted on the basalt page:

  Unfortunately, the only way to distinguish a terrestrial (Earth) basalt from a basaltic meteorite (Moon, Mars, asteroid) is with expensive chemical and mineralogical tests. If you find a basalt, it's probably not a meteorite.

So I would guess that these additional tests for the contents may be enough to distinguish a non-Earth basalt from an Earth one when the appearance to the eye or under a microscope is very similar (due to the similar formation mechanism via lava). However additional tests needed for trace elements may also be needed (Lunar basalts are Chromium-rich but have much lower concentrations of the alkali elements of potassium, sodium, rubidium, and cesium)


Is a spotting scope or binoculars a better choice for astronomy? I'm interested in observing the moon and planets, and maybe some nebulae and star clusters.

I have a pair of binoculars:

Nikon OceanPro 7x50 binoculars (around \$300)
Celestron 20x80 binoculars (around $100)

I'm looking at a step up, to 100mm. Specifically, I'm trying to choose between  25x100 binoculars (either a Celestron or an Orion) and a 100mm spotting scope.

The only reason I'm considering a spotting scope is that it has a magnification of 22-66x, compared to the binoculars' 25x. But I want to check if such high magnifications really work for astronomy.

Let's say I'm observing the full moon. What's the highest usable magnification, without resulting in too dim an image?

An Amazon review says that:

  I have found that while the readability improves from 22x to about 38x
  the resolution actually decreases after that and I never use it from
  40-66x, there is no point. Maybe it would have been better to have
  optimised it for 22-40x.

If the scope is limited to 40x in daytime use, I expect it to be worse at night, even observing the full moon.

I'm considering buying the scope only for the increased magnification over the binoculars, but if I end up using the scope only at 22x, it defeats the point of buying the scope in the first place. Should I stick with the binoculars?

The main point is that it all depends on the optics quality. Nikon is a good brand, but $300  for a 50mm binocular is way too much. Remember the golden rule while buying any kind of telescope or binocular: Aperture matters, not the Magnification. In a binocular, say "7x50" models, you get 7x magnification with 50mm aperture, which restricts how far you can see. But the Celestron 80mm scope with a 20x magnification is perfectly fine. It would be advisable if you can get hold of a good tripod, get it. The beauty of the heavens is unimaginable; they shall unfurl as you explore. A 80mm scope will let you see even Dark Sky Objects like Nebulas and Distant Galaxies.

In a nutshell, it would be best for you at this stage to grab the Celestron 20x80 scope, which is both better and less costly.


How do I see Deep Sky Objects in my telescope? I just bought a new telescope. I tried it out, and I even woke up at 3am when the moon set so there was light pollution. I have an 8 inch dobsonian telescope with a focal length of 1200 mm. For some reason, I only see points of light in the sky, except for the moon and Jupiter and the Orion Nebula.

My goal is to complete a messier marathon, but first I need to be able to see the Messier objects. So my question is, what is the best way to look for Deep Sky Objects with my telescope? And how much does light pollution get in the way of looking for them? For reference, I live in the northern United States.

To see Deep Sky Objects you'll definitely have to head out of whatever town you're in by at least 5km, but the further the better. Next, you'll have to have a good knowledge of where to look for them or a guide. You should be able to see a fair few once you've done that, especially if you've already seen the Orion Nebula.


What's the difference between GSR and Grf I note there are two terms about Galactic rest frame in literatures:
Galactic Standard of Rest (GSR) and Galactic Rest Frame (Grf).

Are they the same frame? If not, what's the difference?

Short Answer
They're the same.
Long Answer
Local Standard of Rest (LSR)
Let me start out by defining the dynamical Local Standard of Rest (dynamical LSR or also LSRD). This definition is taken straight from An Introduction to Modern Astrophysics, 2nd Ed. by Carrol & Ostlie, page 903 (emphasis theirs).

To investigate the motion of the Sun and other stars in the solar neighborhood, we first define the dynamical local standard of rest (dynamical LSR) to be a point that is instantaneously centered on the Sun and moving in a perfectly circular orbit along the solar circle about the Galactic center.
...
Note that once the LSR is chosen, the Sun immediately begins to drift away from it, implying that we would effectively need to redefine the reference point constantly. In reality this is not a significant problem because (fortunately) the 230-Myr orbital period of the LSR is very long compared to the time since modern telescopic observations began (and even longer compared to the lifetime of a typical research grant); consequently there has not been sufficient time for the effect to become noticeable.

Galactic Standard of Rest (GSR)
Now it is important to understand how the LSR coordinate frame is defined, because the Galactic Standard of Rest (GSR) is defined with respect to the LSR. You can read about it here, but in effect, the LSR is a non-inertial coordinate frame co-rotating with the rotation of the galaxy. The GSR is defined in the same way, but removes that co-rotation by subtracting out the rotational velocity of $220\:km/s$. In the LSR frame, another star's velocity would essentially be with respect to our Sun but in the GSR, the star's velocity would be with respect to a fixed point in the galaxy and be a summation of that star's motion with respect to our Sun and it's orbital motion around the galaxy.
Galactic Rest Frame (GRF)
The GRF is identical to the GSR, although I've heard GSR more often. In this paper, you can see they define the GRF to be

... the galactic-rest-frame (grf) proper motions ... which are those seen by an observer at the location of the Sun and at rest with respect to the Galactic center.

This is identical to the definition of the GSR above. Another paper similarly defines the GRF.

Removing the contribution of the motion of the Sun and of the Local Standard of Rest to the measured proper motion produces a Galactic rest-frame proper motion.


Will all galaxies eventually become elliptical galaxies? According to the Wikipedia article Elliptical galaxy, elliptical galaxies have a sparse interstellar medium. I know that probably sometimes two spiral galaxies collide then exponentially approach the state of being an elliptical galaxy. For a spiral galaxy that will never collide because if it hasn't collided before all the other galaxies went past its cosmic horizon, I wonder whether its interstellar medium will slowly get thinner until it's so thin that stars are no longer confined to the plane of the galaxy by friction and then as soon as a few of them start to go slightly outside the plane, they will bring with them their gravitational field causing stars to go further outside the plane and then after it gets wide enough, the interstellar medium will be even faster to get consumed by stars moving in random directions.

First of all, stars are not confined to the plane of a spiral galaxy's disk by friction with the interstellar medium. Stars are very massive objects and the interstellar medium is extremely thin (comparable to the best laboratory vacuums on Earth), so there is effectively no friction involved. (And if there were, it would cause stars to spiral into the center of the galaxy.) Collisions between gas clouds do tend to keep the gas confined to a thin disk.

In practice, an isolated spiral galaxy will gradually turn into a lenticular (a.k.a. S0) galaxy, not into an elliptical. As it uses up its gas in forming stars, star formation will then gradually cease, and the visible spiral arms will weaken and fade away; you will be left with a smooth disk of stars. (This process can happen more quickly in galaxy clusters, where the high-speed motion of a spiral through the hot, high-pressure gas of the cluster rapidly strips out the gas, leaving behind just the stars.)

Stars in spiral-galaxy disks do tend to increase their vertical motions a bit over time, due to the gravitational effects of minor mergers, spiral arms, and massive molecular clouds (not because of friction, but because the gravity of the molecular cloud shifts the orbits of nearby stars a bit). Ironically, this means that the thinning out of the interstellar medium will probably weaken the tendency of stars to increase their vertical motions, because there won't be any more molecular clouds to perturb the stellar orbits.

Given sufficient time, close encounters between pairs of stars will tend to randomize their orbits, and the lenticular galaxy will eventually evolve into something more like an elliptical galaxy. However, because the average distances between stars are so large, the frequency of close encounters is very low, and the whole process will take trillions of years.


Best way to simulate star sizes to scale in celestial sphere Using the Hipparcos catalogue, I am trying to create a celestial sphere. As all the stars are a fixed distance from the centre of this sphere, the only way to differentiate the distances and magnitude is through the sizing of each model of the star.

With the Parallax angle $p$ and visual magnitude $m_{vis}$ of each star from the catalogue, I have created the code below to calculate the radius of each star relative to the radius of the sun. I had used this website to calculate this. The problem is that some stars are too big, almost bigger than the celestial sphere itself! I am trying to keep the radius below a certain threshold.

Here $p$ is in arcseconds and $d$ is in parsecs.

$$d = 1 / p$$

$$M_{abs} = M_{vis} - \log_{10}(d^5) + 5$$

$$ T_{surf} = \frac{8540}{CI + 0.865}$$

$$R_{rel} = \left( \frac{5800}{T_{surf}} \right)^2 \sqrt{(2.512)^{4.83 - M_{abs}}}$$

$$R_{Sun} = 2.5$$

$$r = R_{rel} R_{Sun}$$

    private void positionStar()
    {
        double radius;
        cartesianPositioningCalc();
        gameObject.transform.position = cartesianPositioning;

        // Convert Plx from milliarcseconds to arcseconds (seconds of arc)
        double PlxSOA = Plx / 1000;
        // Calculate distance from equation d=1/p
        // distance d is measured in parsecs and the parallax angle p is measured in arcseconds.
        double dPC = 1 / PlxSOA;

        double absMag = Vmag - math.log10(math.pow(dPC, 5)) + 5;

        surfaceTemperature = 8540 / (CI + 0.865);

        double relativeRadius = math.pow((5800 / surfaceTemperature), 2) * math.sqrt((math.pow(2.512, (4.83 - absMag))));
        double radiusSun = 2.5f;

        radius = relativeRadius * radiusSun;

        gameObject.transform.localScale = new Vector3((float)radius, (float)radius, (float)radius);
    }

First of all, am I using the right mathematics? If I am, then how can I make sure that the radius is below a certain maximum threshold (5 units, for example.)?

If the mathematics is incorrect, please assist me on how to fix it.

Thanks!

With a slightly modified version of @Mike G's formula in their answer below:

radius = 50 * math.pow(10, (-1.44 - Vmag) / 5);

I was able to get this result:

And if I'm not mistaken, I think I can locate the Ursa Major constellation offset slightly to the left from the centre of the screenshot.

However, after uploading to nova.astrometry.net the stars and constellations are still not being picked up. Is this because the star sizing is still a bit incorrect or is this issue on the website's side?

I can use another website/app such as Stellarium to cross-check, is there a way the app can process an input image or do I try to recreate using my image on the app?

Your maths looks ok, bar the fact that $1/$parallax is a biased estimate of the distance (but that can be forgiven so long as you are using data where the parallax uncertainties are much smaller than the parallax).

Your main problem is that stars do indeed have a vast range of sizes. Thus if you really do want to show the relative sizes of the stars you have a dynamic range problem.

The conventional way to deal with this would be to use a logarithmic scale, such that each increment in size corresponds to a multiple of the one below. e.g. Make your plotted radii proportional to $\log_{10} R_{\rm rel}$.

EDIT: In response to altered emphasis of question.

It seems you no longer want to scale the stars according to their actual radius, but rather according to their brightness. The logarithmic scaling is therefore already handled by the apparent magnitude (which is on a logarithmic scale).

Therefore all you need to do is decide what your smallest and largst radius are and compare that with the brightest and faintest magnitudes you want to plot.

e.g. If your biggest star is 5 units and your smallest is 0 units, and your bright and faint limits are $m_{\rm bright}$ and $m_{\rm faint}$ respectively, then the size of an arbitrary star of magnitude $m$ is
$$ r =\frac{5(m -m_{\rm faint})}{m_{\rm bright} -  m_{\rm faint}} \ .$$

And if you want your faintest stars to be size 1 and brightest to be 5, this modifies to
$$ r =\frac{4(m -m_{\rm faint})}{m_{\rm bright} -  m_{\rm faint}} +1 \ .$$

Yet another alternative would be to make the area proportional to magnitude. Again, scaling between radii of 5 and 1:

$$ r^2 =\frac{24(m -m_{\rm faint})}{m_{\rm bright} -  m_{\rm faint}} +1 \ .$$


What was the the value of the Hubble constant at the time of the CMB's 'release' (i.e., 379,000 years after Big Bang)? What about its value roughly 9 billion years after the Big Bang, when dark energy started to 'take over' and accelerate the expansion of the universe?
Is there a timeline or chart somewhere that shows approximate, theoretical values of Hubble's not-so-constant constant throughout the lifetime of the Universe?

This answer to the question “Is the Hubble constant dependent on redshift?” gives the formula (a form of the Friedmann equation) for the Hubble parameter $H(z)$ as a function of redshift $z$:
$$ H(z)^2 = H_0^2 \left[ (1+z)^4 \Omega_r + (1+z)^3 \Omega_M + (1+z)^2 \Omega_k +  \Omega_\Lambda \right] $$
where the $\Omega$ terms are the fractional densities in radiation, matter, curvature, and dark energy, respectively.
Using that, plus the knowledge that the redshift of the CMB is $z \simeq 1100$, you can plug in values for the densities (I used WMAP values quoted here) and get that $H$ at the time of the CMB was about 22,000 times larger than the current value.
That answer also gives a graph of the value of the Hubble parameter as a function of time.


Why doesn't the earth's surface collapse onto itself? From what I know the sun (or any other star) nuclear fission keeps it from collapsing under its own gravity until the fuel runs out, if that's so why doesn't the earth surface like the tectonic plates collapse to the earth's core? is it because of similar processes that happen in the earth's outer core (nuclear fission)? and if so what would happen if they would once stop?

For the same (an more prosaic) reason milk skin floats over boiling milk - because it is lighter than the layer below. During the formation of the planet it went through a process called Iron Catastrophe, were a large parcel of the denser materials like iron and nickel sunk to the core, leaving the lighter materials close to the surface.

Nice graph with relative densities follow:

Fission contributes to keep our internal engine super-hot and well-oiled, allowing for our metallic core to spin and generate our magnetosphere. Eventually we'll run out of heat, the outer core will solidify and stop spinning.


Is it possible for a septuple eclipse to happen on Saturn considering there are 7 moons capable of eclipsing the Sun? The reason quadruple eclipses can never happen on Jupiter is because of the 1:2:4 orbital resonance between Io, Europa, and Ganymede. As far as I know this isn't a problem for Saturn's moons.

From Wikipedia's Solar eclipses on Saturn:

  Seven of Saturn's satellites – Janus, Mimas, Enceladus, Tethys, Rhea, Dione and Titan – are large enough and near enough to eclipse or occult the Sun, or in other words to cast an umbra on Saturn.

Of the 7 listed above, Mimas and Tethys are in a 2:1 orbital resonance and so are Enceladus and Dione, however that doesn't stop either pair (or all 4 together) from eclipsing Saturn at the same time.

So, in theory, it is possible for all 7 moons to more or less line up together. Does this mean that during eclipse season on Saturn, there can be a septuple eclipse? If so, when is the next one going to happen?

The short answer is no, septuple eclipses can’t happen.
Using a method described by Meeus in Mathematical Astronomy Morsels (p. 190), we find that while it is, in theory, possible, for the seven satellites to align, it happens only once per ~ 20 million years—and that’s among themselves only; I didn’t factor in the Sun’s position yet!
The possibility that the Sun is along the same line is $ \displaystyle \frac { \text {(angular size of the Sun as seen from Saturn)}°} {360°} = \frac {0.05} {360} $; let’s be generous and round it off to 1:10,000.
Multiplying 20 million years by 10,000 shows us that the seven satellites would line up with the Sun once in 200 billion years—longer than the life of the Universe!
Finally, the above was estimated by using circular orbits, which is not the case in practice. Plus, orbital parameters change (“chaotically”) over long timespans, so it would (at least currently) be impossible to compute the actual positions of satellites in the remote future on timescales such as the ones above.


How to be an astrophysicist? For a high-school students, what are the ways to become an astrophysicist?

What should he/she take in college?

What is the career path to become an astrophysicist?

Different education systems differ, however

At school you would take maths and physics courses, at least covering calculus.

As an undergraduate, taking (or majoring in) physics. Also probably doing some more maths and perhaps some astrophysics courses.

As a postgraduate doing Masters study in astrophysics leading to PhD research in astrophysics.  By now you could say that you are an astrophysicist.

Then working at a university or similar institute.

This makes it all sound easy, but I think I should add a warning. Getting a permanent (research) position at a university or similar institute in astronomy is highly competitive, and to get there takes a combination of hard work, high competence, high motivation, and luck. Many highly competent people end up moving out of astronomy and use their skills elsewhere, for the simple reason that society does not spend enough money on astronomy to employ all highly competent people who want to spend their life on astronomy.


Question about expansion of space time Was the expansion of space time during big bang uniform if so how come the stars and planets be formed because and matter as well as gravity would have been same everywhere

Your question is one that bothered cosmologists for a long time, how do you get clumps of matter out of a smooth initial condition, in the time allowed?  It was long understood that a perfectly smooth matter distribution could not be maintained forever, because gravity creates instabilities-- the smallest overdensity in one region would tend, over time, to pull in more matter and increase the overdensity.  But this takes time, and the CMB measurements were coming in surprisingly smooth-- so how did galaxies and stars form in less than a billion years, given that very uniform initial density?

The answer seems to be dark matter.  Dark matter increases the mass and therefore the gravity, so it reduces the time it takes for instabilities to develop and overdensities to grow.  So the simple answer to your question is that even though cosmology assumes a smooth initial density, it does not say the density is exactly smooth, and indeed the CMB tells us there are tiny density variations (you can look at the tiny variations seen by the PLANCK satellite at https://en.wikipedia.org/wiki/Planck_(spacecraft)#/media/File:Ilc_9yr_moll4096.png, for example-- the scale shows variations of 1 part in ten thousand).  Then gravitational instability increases the overdensity, but needs stronger gravity to do this in the time allowed, so that appears to be provided by dark matter.


Usage of the standard solar system (astronomical) symbols Note: this is a follow-up question based on answers I've received in TeX SE on Packages for standard solar system (astronomical) symbols on the off-chance that some here may have actual hands-on experience with both TeX and these astronomical symbols.

Recently, I have been looking for specific LaTeX packages for generating the following symbols of the solar system.

Photo source: https://solarsystem.nasa.gov/resources/680/solar-system-symbols/
I have tried using the packages mathabx and wasysym. But I have found that they both have some issues and I need to use both of them to generate the above symbols. I have discussed it on TeX SE here and got this solution:
\documentclass[a4paper,12pt]{article}

\usepackage{amsmath,amssymb}

% wasysym redefinitions
\newcommand*{\wasyfamily}{\fontencoding{U}\fontfamily{wasy}\selectfont}
\newcommand*{\astrosun}{{\odot}}
\newcommand*{\mercury}{{\text{\wasyfamily\char39}}}
\newcommand*{\venus}{{\text{\wasyfamily\char25}}}
\newcommand*{\earth}{{\oplus}}
\newcommand*{\mars}{{\text{\wasyfamily\char26}}}
\newcommand*{\jupiter}{{\text{\wasyfamily\char88}}}
\newcommand*{\saturn}{{\text{\wasyfamily\char89}}}
\newcommand*{\uranus}{{\text{\wasyfamily\char90}}}
\newcommand*{\neptune}{{\text{\wasyfamily\char91}}}
\newcommand*{\pluto}{{\text{\wasyfamily\char92}}}

% mathabx definitions
\newcommand*{\mathabxbfamily}{\fontencoding{U}\fontfamily{mathb}\selectfont}
\DeclareFontFamily{U}{mathb}{\hyphenchar\font45}
\DeclareFontShape{U}{mathb}{m}{n}{
      <5> <6> <7> <8> <9> <10> gen * mathb
      <10.95> mathb10 <12> <14.4> <17.28> <20.74> <24.88> mathb12
      }{}
\newcommand*{\Sun}{{\text{\mathabxbfamily\char"40}}}
\newcommand*{\Mercury}{{\text{\mathabxbfamily\char"41}}}
\newcommand*{\Venus}{{\text{\mathabxbfamily\char"42}}}
\newcommand*{\Earth}{{\text{\mathabxbfamily\char"43}}}
\newcommand*{\Mars}{{\text{\mathabxbfamily\char"44}}}
\newcommand*{\Jupiter}{{\text{\mathabxbfamily\char"45}}}
\newcommand*{\Saturn}{{\text{\mathabxbfamily\char"46}}}
\newcommand*{\Uranus}{{\text{\mathabxbfamily\char"47}}}
\newcommand*{\Neptune}{{\text{\mathabxbfamily\char"48}}}
\newcommand*{\Pluto}{{\text{\mathabxbfamily\char"49}}}

\newcommand*{\astrosymbolsA}{\Sun\Mercury\Venus\Earth\Mars\Jupiter\Saturn\Uranus\Neptune\Pluto}

\newcommand*{\astrosymbolsB}{\astrosun\mercury\venus\earth\mars\jupiter\saturn\uranus\neptune\pluto}

\newcommand*{\astrosymbolsC}{\Sun\Mercury\Venus\Earth\Mars\jupiter\saturn\Uranus\Neptune\pluto}

\begin{document}
\begin{align*}
   \text{mathabx: } \astrosymbolsA \\
   \text{wasysym: } \astrosymbolsB
\end{align*}
Taking \jupiter, \saturn, \pluto  \, from `wasysym' and others from `mathabx':
\begin{align*}
        \Sun\Mercury\Venus\Earth\Mars\jupiter\saturn\Uranus\Neptune\pluto
\end{align*}

\end{document}

Still, they don't exactly match the symbols of the first photo.
I guess the MWE of LaTeX codes is related to TeX SE. But I have not found any better way to cross-advertise questions of interest in multiple topics. That's why I share the codes here too.
Here in Astronomy SE, I want to know how these symbols are used in the astronomy communities/academia.

How is the frequency of using them in the technical/research publication?
How do publishers handle them when submitting articles?
Do they use any other dedicated/updated LaTeX packages available for generating these symbols?

As Anders Sandberg wrote, the Sun, Earth, and Jupiter symbols are used for putting things in context, for example M☉ for solar masses or M⊕ for Earth masses. However, the astronomical community has stopped using the other symbols early in the twentieth century. Astrologers still use them, but not astronomers.


Catalogs for type K stars and their planetary systems? I am doing data collection on all type K stars, and their planetary system so I may develop an algorithm for a planetary formation simulation with the Unity game engine.
Where could such a catalog of stars be found?

There are many star catalogues, and from your question I cannot easily tell which would be the best one for you. You probably need one which contains positions of the star and their spectral type, maybe further properties as well.
Since you mentioned planetary systems, I think you might actually want to look at NASA Exoplanet archive (and filter it for K-type central stars).


What is the gravitational lensing focal distance of a white dwarf star? I tried looking this up, but I couldn't find any formula on gravitational lensing distance.   I know that our Sun's is about 550 AU, though further distances work too, as it's not a single focus due to the gravitaional field diminishing over distance from the focusing body.

Is there a reasonably simple formula for calculating distance for a gravitational lens.   I'm specifically curious for white dwarf stars as there's one just 8 light years away and they see like a good object with a good lensing but not super tight focusing like a neutron star or black hole.

For example, if a telescope was built using Sirius B as a focus, how far would the telescope have to be and how powerful might it be (Maybe how powerful should be a separate question but I'll leave it here for now?

Would Sirius B's binary orbit be a hindrance or a benefit, allowing a greater area of focus?

Pure curiosity.   I don't expect we'll get there anytime soon.

The gravitational focus you are talking about is actually a minimum value, defined by parallel rays of light from a very distant star just skimming past the Sun as they are bent according to General Relativity.

The general formula for such lensing is that light is bent through an angle (in radians) of
$$\alpha = \frac{4 GM}{c^2 r},$$
where $M$ is the mass of the lens (assumed to be a point or spherically symmetric mass)  and $r$ is the closest approach of a light ray to the lensing mass.

To work out where a ring of rays will be focused is just a bit of trigonometry.
$$ d_f \simeq \frac{r}{\alpha} = \frac{c^2 r^2}{4GM}$$

This focal distance is a minimum because it would be larger for a ring of rays that passed the lens with a larger value of $r$.

For the Sun as a lens you use $M=2\times 10^{30}$ kg and $r=6.9\times 10^{8}$ m, and calculate $d_f = 540$ au.

White dwarf stars have a similar mass (actually most are about 60% the mass of the Sun, but Sirius B is almost exactly a solar mass), but have radii about the size of the Earth - i.e. a hundred times less than the Sun.

This means that the value of $d_f$ is going to be about 10,000 times less than 540 au. You can use the formula above to calculate it for any combination of mass and radius.

To use the telescope, you place detectors at your chosen focus and observe the bright "Einstein ring" of a distant source that is exactly behind the lens. The magnification factor (the increase in the amount of light collected from the source) then is $4\alpha/\theta$, where $\theta$ is the angular size of the source without the lens.

For a white dwarf, the magnification at the minimum focus would be 100 times larger, because $\alpha$ is 100 times larger.

Note that the size of the image is modified by the ratio of the focal length to the source distance.
$$ x_i = x_o \frac{d_f}{d_o}$$
Thus the image of a distant object will be 10,000 times smaller than if using the Sun, which is much more convenient!

e.g. Observe an Earth-like planet at 10 ly at a focus of 630 au (= 0.01 ly) from the Sun. The image diameter will be 12.5 km. That's a lot of CCD detectors! Using a white dwarf at a focal length that is 10,000 times smaller gives an image just 1.25 m across.

All this assumes that the telescope is perfectly pointed with the source right behind the lens. Any relative motion has to be corrected or the image will move through the focal plane very quickly (like a planet viewed with high magnification through a normal telescope).


Has the iron core of Mars really solidified? In the Nova episode "Origins: Earth is Born", Neil DeGrasse Tyson states, "But Mars is just a fraction the size of the Earth, so it cooled more rapidly. And as it cooled, its molten iron core hardened. As a result, Mars stopped generating its magnetic shield." (February 28, 2004)

see: http://www.pbs.org/wgbh/nova/space/origins-series-overview.html#origins-earth-born

This comment from Tyson is consistent with what I have long believed. But I wonder.

The melting point of iron (Fe) is around 1538 °C.

see: https://en.wikipedia.org/wiki/Melting_points_of_the_elements_(data_page)

The the geothermal gradient for Mars is estimated to be around 6 °C per km.

see: What is the temperature 55 km beneath the surface of Mars?

If the estimate of 6 °C per km is the ballpark, then I would expect that iron is molten as a mere 256 km of depth. At the core of Mars (over 3000 km in depth) I would expect the temperatures to be far higher than 1538 °C.

It seems to me that it is impossible for iron to be in solid form at Mars' core.

You're mostly missing an important point to do with the geothermal gradient.

The geothermal gradient used in that question, and applied to a $55\:\mathrm{km}$ depth is not consistent all the way through Mar's interior. They cite the Earth's geothermal gradient (see image below) and from that page you can see how the temperature profile changes with depth. There is a much stronger gradient towards the surface. Thus you can't naively apply the surface geothermal gradient through to the core. Rather, to get the core temperature, you'd need to integrate the gradient function.

Earth's Temperature Profile


Has anyone back-calculated previous close encounters between the Apophis asteroid and Earth? The Apophis asteroid was only discovered in 2004, and shares an orbit similar to Earth's orbit, leading to many close encounters on the scale of decades.
Most of the stories about this asteroid are about predicting future encounters between Earth and Apophis.  It occurs to me that you can use the same models that predict future encounters to show past encounters as well.
Has anyone done this, and shown any past events when the asteroid has come extremely close to the Earth?

The JPL Small Body Database lists Apophis close approaches dating back 100 years before discovery.
Three fairly close ones were:

1907-04-13, 0.029 au
1949-04-14, 0.028 au
1990-04-14, 0.033 au

While it's possible to run a dynamical integrator arbitrarily far backward or forward in time, any given pos(t), vel(t) state is only a point in a cloud of possibilities.
Observations (MPC database, NASA news) constrain that cloud; close encounters accelerate its dispersion.
Specific predictions before ~1900 or after ~2050 should not be regarded with much confidence.


How feasible would it be to get to a planet orbiting directly opposite Earth? Imagine that in addition to Earth, there was a second planet, say of one Earth mass, orbiting the sun - but on the opposite side of the Sun, and at the same orbital radius.

My question is - given today's technology, would it be feasible to try to launch something to reach that planet? A rover, a capsule, or something of the sort is what I have in mind.

There are a couple reasons I initially don't see a solution. First off, the escape velocity of the Earth is about 11.2 km/s. The Earth's orbital velocity around the sun is nearly 30 km/s. This means that if I were to launch a rocket in the opposite direction of Earth's orbit to reach this "second Earth", not only would I have to counter this 30 km/s orbital velocity, but I would have to provide enough thrust for double this velocity in order to maintain a stable orbit around the Sun at the radius that Earth orbits at - but in the opposite direction. I'm not quite sure if rockets can get up to velocities that high - I know that the Juno probe is moving at around 73 km/s, but I believe that was after a series of maneuvers involving a gravitational slingshot.

Perhaps there is another solution? One that popped into my head was to launch it at an angle to the plane of Earth's orbit, but I don't quite know how that would work out, given that unless some serious trajectory corrections were made, the rocket would not arrive anywhere close to the planet - as the planet would have moved in the time the journey took.

Maybe it involves taking a path that is closer to the Sun, and more elliptical? At this point I'm just throwing out everything that comes into my head.

Basically, I'm looking for some sort of general explanation here, with a rough idea that I could build on. And if it isn't possible with today's technology, what do we lack?

Venus has roughly the same gravitational potential as earth and the Venera probes successfully and softly landed there. Similarly, the Apollo space craft landed safely on earth on return. In both cases aerodynamic friction with the atmosphere can be used to get rid of excess velocity in order to enter orbit and eventually land. Thus landing will work cheaper, if there is atmosphere. But at the expense of much more fuel also without (see moon landing again... you'd need stronger descent rockets than Apollo for an exclusively rocket - driven landing)

Getting there is not a big deal either. Look at the Stereo mission of two satellites which trail earth's orbit more and more to get a better understanding of the sun, gaining more and more separation from earth while (nearly) staying at 1 AU: You get to another position in earth's orbit by just lifting your orbit slightly and waiting for earth to go and the other to catch up (or go slightly towards the sun and be faster yourself).


What's the difference between semi-major axis and apoapsis? The semi-major axis is half of the diameter of the major axis of an ellipse by definition. So how does that make it any different from apoapsis, the furthest point an object is from a parent body in its orbit?

Both are radii along the longest axis, surely.

If the semi-major axis of an ellipse is a and the eccentricity is e,
then the distance from the center C to a focus F1 or F2 is c = ea.
Apoapsis is a + c and periapsis is a - c.

Ellipse with e ≈ 0.78:

For the Earth's orbit around the Sun, a = 149.6 million km and e = 0.0167, so

c = ea = 2.5 million km
aphelion = a + c = 152.1 million km
perihelion = a - c = 147.1 million km


What do these statements about interstellar magnetic fields really mean? Quanta Magazine's The Hidden Magnetic Universe Begins to Come Into View is a fascinating review of a rapidly evolving field in astronomy.
It contains some statements that I feel are important to understand, but I don't.

However, once a “seed” magnetic field arises from charged particles in motion, it can become bigger and stronger by aligning weaker fields with it. Magnetism “is a little bit like a living organism,” said Torsten Enßlin, a theoretical astrophysicist at the Max Planck Institute for Astrophysics in Garching, Germany, “because magnetic fields tap into every free energy source they can hold onto and grow. They can spread and affect other areas with their presence, where they grow as well.”
Ruth Durrer, a theoretical cosmologist at the University of Geneva, explained that magnetism is the only force apart from gravity that can shape the large-scale structure of the cosmos, because only magnetism and gravity can “reach out to you” across vast distances. Electricity, by contrast, is local and short-lived, since the positive and negative charge in any region will neutralize overall. But you can’t cancel out magnetic fields; they tend to add up and survive.

Questions:

What does the phrase "...magnetic fields tap into every free energy source they can hold onto and grow..." refer to in the astronomical context of interstellar space?

What does the phrase "...only magnetism and gravity can “reach out to you” across vast distances. Electricity, by contrast, is local and short-lived, since the positive and negative charge in any region will neutralize overall. But you can’t cancel out magnetic fields; they tend to add up and survive." refer to in the context of interstellar space? Is it the lack of free magnetic monopoles, or the dearth of polarizable ferromagnetic material?

What does the phrase "...magnetic fields tap into every free energy source they can hold onto and grow..." refer to in the astronomical context of interstellar space?

Interstellar space is filled with charged particles/dust which have their own magnetic fields (e.g., their own dipole moment or any field induced by the charge moving). When you apply a larger, more global magnetic field to these small charges, very often an alignment occurs (e.g., the particles dipole will align with the larger field, or the motion will suddenly change to move along the field line). This causes these smaller fields to add to the larger field, making it even larger. This is a foundational assumption for academics which study light polarity in terms of dust grain alignment via Radiative Torque (RAT) - namely that the background global magnetic field of interstellar space aligns charged grains with that field, thus further contributing to the overall field.

What does the phrase "...only magnetism and gravity can “reach out to you” across vast distances. Electricity, by contrast, is local and short-lived, since the positive and negative charge in any region will neutralize overall. But you can’t cancel out magnetic fields; they tend to add up and survive." refer to in the context of interstellar space? Is it the lack of free magnetic monopoles, or the dearth of polarizable ferromagnetic material?

If you take a large, galactic magnetic field and it passes through an interstellar dust cloud that is full of charged particles, those charged particles will align with the field and overall add to it, making the galactic field stronger in that area. The only thing that could weaken the magnetic field is something which could cancel it.
Electric fields can easily be cancelled because an electric field being radiated out by a proton can easily be cancelled by the "opposite" field being radiated out by a bunch of nearby electrons (hence the concept of the Debye length). Magnetic fields are harder to screen and, as you say, because the things that could screen them just don't exist in space.
I would also mention that if monopoles even existed, it would require that you could have two separate flavors of monopoles such that they could cancel each other out much like the positive and negative charges of electrons. A logical pick for flavors would be a "north" monopole and a "south" monopole, but I could imagine one constructing a more complicated, yet valid system. Since we've never seen a monopole, any monopole-based models and purely hypothetical at this point.


What is the application of pseudorandomness in astrophysics? What are the advantages of using pseudorandom numbers in Monte Carlo simulation instead of random numbers? What are other applications of pseudorandomness in astrophysics? Many thanks.

The primary advantage of pseudorandom numbers is speed (as Cristiano mentioned) and convenience: they're what are available in almost any software package or library, whereas true random number generators are not. As long as you use a pseudorandom number generator (PRNG) with good statistical qualities and are not worrying about true cryptographic randomness, this is perfectly acceptable.
In addition to Monte Carlo simulations, pseudorandom number are used for many different applications, including all sorts of physical simulations (cosmological, stellar evolution, planet formation, etc.), statistical analyses (bootstrap resampling), initial conditions for fits, updating parameters in genetic algorithms, sample selection, etc.
A personal example: I've worked on two different studies where we were making classifications of different aspects of galaxies, based on their individual isophote plots or radial surface-brightness profiles. Rather than worrying about the possibility of subtle biases based on possible knowledge of individual galaxies ("Oh, it's NGC 1097, I'd expect that galaxy to be..." or "Well, the list is alphabetically ordered, so the middle galaxies will have NGC prefixes, and those will mostly be really luminous galaxies, so..."), we set up a blind analysis system, where the galaxies were presented to us in a random order, without names (with the actual identifications automatically recorded where we couldn't see them until the whole thing was over). The randomness of the order was, of course, really pseudorandom order.
This is how things like Galaxy Zoo work, for example: the individual galaxies that citizen scientists classify are (pseudo)randomly selected. (And for one of the projects I mentioned above [Kruk et al. 2019], we set up the classification process using the "Zooniverse" framework that Galaxy Zoo developed).
A separate advantage is that if you save the number used as the seed for the pseudorandom number generator, you can use that to regenerate the exact set of pseudorandom numbers used in the analysis/simulation/whatever, which can be very useful for debugging purposes, and also for allowing others to replicate your results. (E.g., you run some kind of relatively short/simple simulation, and derive some results and figures from that. How can someone else trust that your results and figures accurately represent the simulation? Provide them with the code and the PRNG seed, and then can generate the same simulated/random-sampling results and check. Plus, they can vary the seed and make sure you didn't cheat -- or fool yourself -- by reporting atypical results. Here's a Github site for a recent paper of mine that includes both the code for a survey simulation and the seed number I used for the figure in the paper.)


In what timeframe is W26 expected to go supernova? All the news-articles say that it is expected to "eventually" explode. Which doesn't really tell me much. When is "eventually"? Also, considering this is the largest star in the known universe and 16k light-years away, will this be visible? Will it be dangerous?

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

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

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

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

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

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

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

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

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

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

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

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

Image source

to which Levesque et al. state

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

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


How loud was the asteroid that killed the dinosaurs? Asteroid impacts on the Earth can be very loud when they break up in the atmosphere, like Chelyabinsk, captured in this video:

The Chelyabinsk asteroid break-up was estimated to be about 180dB at 3 miles, and was detected as far as 9000 miles away.
If asteroids don't break up, presumably they are extremely loud on impact as well.  How loud was the asteroid that killed the dinosaurs either on atmospheric entry or impact with the ocean?

Some estimates which I found worth sharing:

For fun, I searched for Chixculub TNT equivalent and e.g. ScienceDaily claims

The energy released by the impact that blew out the Chicxulub crater was equivalent to about 100 million megatons, many orders of magnitude greater than the nuclear explosion at Hiroshima, a 15-kiloton blast.

This seems to match up what WolframAlpha uses to rank TNT equivalents.

There is a Noise Prediction Calculator which estimates the safe distance from a blast:

This calculator can be used to predict the distance from an open detonation at which 140 decibels (dB) could be expected to be achieved. The 140dB level is widely used as a "safety cutoff" for exposure to impulsive noises while using hearing protection.

Plugging in $10^{17} {\rm kg}$ (of TNT) in that formula (which is probably a rather crude estimate) gives ${460\,320 \rm km} \approx 1.4 \cdot (2 \pi a_\oplus)$ (with $a_\oplus$ being the Earth's radius), meaning more than equatorial circumference of our planet. In other words, no where on Earth was it safe according to UN safety standards.

So let us approach it the other way round: What is the loudest possible sound?

Strictly speaking, the loudest possible sound in air, is 194 dB. The “loudness” of the sound is dictated by how large the amplitude of the waves is compared to ambient air pressure. A sound of 194 dB has a pressure deviation of 101.325 kPa, which is ambient pressure at sea level, at 0 degrees Celsius (32 Fahrenheit). Essentially, at 194 dB, the waves are creating a complete vacuum between themselves.
You can go louder than 194 dB, but that’s not technically a “sound” anymore. The extra energy starts distorting the entire wave, and you end up with something that’s more a shockwave and less a soundwave. At that level, sounds don’t pass through air — they push the air along, producing pressurized burst (shockwaves).

On the shockwave of Chicxulub, I found the page at the Lunar and Planetary Institute insightful:

The Chicxulub Impact event produced a shock wave and air blast that radiated across the seas, over coastlines, and deep into the continental interior.  Winds far in excess of 1000 kilometers per hour were possible near the impact site, although they decreased with distance from the impact site.  The pressure pulse and winds would have scoured soils and shredded vegetation and any animals living in nearby ecosystems.  An initial estimate of the area damaged by an air blast was a radius 1500 kilometers.  There are several factors that can affect this estimate, so the uncertainty might be better reflected in a range of radii from ~900 to ~1800 km.  The travel times are quite short, so this effect would have occurred in advance of any falling debris ejected from the Chicxulub crater.

Equipped with all this, we can now try to answer "How loud was the Chixculub explosion?" even if you did not specify where you measure loudness: At the radius of $\approx 1\,500 {\rm km}$ from the epicenter, the shockwaves "fades" out to become the loudest possible sound of 194dB.


Converting the log (Stellar Luminosity/Solar Luminosity) and the log of the plus and minus variation There are some stars in the NASA Exoplanet Archive that I am interested in but I am confused by the physical meaning of the terms.

For example, there's a star BD+15 2940 with a log10 of Solar Luminosity of 2.01 +/-0.75 where the variation is also a log.

So is the linear equivalents 18.2 to 575.44 with most likely value 102.33?

I went to the citing paper Nowak et al. 2013and they quote it as log L*/LSun 2.01 +/- 0.75 as per the NASA archive.

You're completely right, except that the authors, as far as I can see, don't specify whether the central value (i.e. $\log L = 2.01$, or $L = 102$) is the most likely value (i.e. the mode), or the average, or the median. My guess is it's the average, but especially for logarithmically distributed values, my opinion is that the median would makes more sense to quote, since in that case you'll get the same result whether you take the median of $L$ or of $\log L$.


What are the units of distance in this subtended angle calculation? I'd like to find the distance at which a 2.4 cm coin is subtended by an angle of 10". I've drawn my diagram and found that $D=\dfrac{d}{\alpha}$ using the small angle approximation. Since $10" = 4.85\times10^{-5}$ rad then $ D = \dfrac{2.4 \times 10^{-2} m}{4.85 \times 10^{-5}rad} = 495$

Are the units of distance in this calculation meters? In other words why would the units of radians be 'dropped'?

The units of radians are 'dropped' because unlike most units, they are dimensionless. Recall that the definition of radian angle measure is the ratio of the length of a circular arc to its radius. Thus, radian angle measures have units of $[\mathrm{Length}]/[\mathrm{Length}] \equiv 1$, i.e. dimensionless.

Radians are units in the sense that they give information about what standard the angle quantity is measured by. Thus, the convention of writing $\mathrm{rad}$ is useful in that it distinguishes it from other ways of measuring angles (e.g., in your question, arc-seconds), but in terms of dimensional analysis, $\mathrm{rad} \equiv 1$, so $\mathrm{m}/\mathrm{rad} = \mathrm{m}$.


If the Sun disappeared, could some planets form a new orbital system? If the sun were to suddenly disappear, the planets would continue to travel tangentially to their former orbits. (This I know from this answer to a somewhat related question here.)
In such a scenario, is it possible (however unlikely) that a new orbital system would form from two or more of the now-rogue planets?  For example, is it physically and mathematically possible that Earth might end up getting captured by Jupiter and thereby become one of its moons?  Or do the size and speed of the planets completely rule out such a scenario, even if the sun disappeared at such a time that the planets' new trajectories would cross?

The issue here is whether pairs of planets can become gravitationally bound to each other.
In the two-body problem the trajectories or orbits are ellipses (bound orbits), parabolas and hyperbolas (unbound). For all practical purposes, an encounter looks like they start out at infinity with some finite speed, approach each other, and then maybe fly away or remain bound. This situation (finite speed when far away) corresponds to a hyperbola, so the generic case will not form a new bound system: some of that energy needs to be shed somehow.
One way this can happen is an inelastic collision, i.e. a merger. A less extreme case is a close encounter where tidal forces dissipate some of the kinetic energy. But the kinetic energy of planets is large, so a single encounter is unlikely to dissipate enough to make them bound.
Another important possibility is a three-body encounter. This is how comets get into short-periodic orbits when falling into the solar system: an encounter with Jupiter (or some other planet) leads to a gravitational "assist" that makes it bound with the sun.
In your scenario where the rogue planets fly off along their tangents it is very unlikely that any start out bound to each other since planetary orbital velocities around the sun are much bigger than the orbital velocities of a bound planet-planet pair at a large distance. Hence they will just fly apart. In theory they might have an encounter before leaving the solar system vicinity (e.g. Mercury by sheer chance sweeping past Jupiter) and that leaves a tiny chance of a three-body interaction with moons or a merger, but the chances are truly tiny. The generic case is just scattering.
Addendum: Some estimates
For a body to be bound to another body of mass $M$ it need to move slower relative to it than escape velocity $v_{esc}=\sqrt{2GM/r}$. If we assume planets move in circles we can then take their orbital velocities for all possible positions along the orbit, calculate the relative velocity, and compare it to the escape velocity.
The result is this diagram, where I have plotted the separation on the x-axis and how many times escape velocity the relative velocity is on the y-axis. The numbers denote which pair of planets I am comparing. The blue horizontal line is $\Delta v=v_{esc}$.

Looking just at the lowest relative velocities shows that the closest to being potentially bound is Saturn to Jupiter were the sun vanish just as they are moving in parallel, but even in this case Saturn has several times Jupiter escape velocity.
One could complicate it with inclinations and elliptic orbits, but it will not change the qualitative picture: the planets are not even close to being bound to each other in the absence of the sun.


Which planets, nebula, and other objects can be seen using Celestron Astromaster 130EQ I have got a Celestron Astromaster 130EQ telescope, which lens should I use to see the various planets and nebula?

When I tried to see Mars it was only a small red dot on scope.

As per the technical description, it's got 6" mirror with 650 mm focal length.
Kindly guide me to which lens of what mm should be used to see Mars, Saturn, Jupiter, Venus, and nebula.

Okay, so that's a 130 mm newtonian with an f/5 focal ratio. Per the manufacturer's website, it comes with two eyepieces, 10 mm and 20 mm, giving 66x and 33x magnifications respectively.

First off, make sure your scope is collimated. The procedure is described in their generic Telescope Maintenance document, and also more specifically in the Astromaster Manual. See links on the site. The performance of your scope when observing planets depends a lot on collimation. I made it a habit to check collimation every time before I use the scope; I'm so used to it that it only takes 5 minutes now.

The topic of collimation is huge, just google "telescope collimation" and learn. I'm still learning new things after years in this hobby.

Once the scope is collimated, you could try to push magnification even higher, by using a 2x ... 3x barlow with either of the eyepieces - that would double ... triple magnification. Please be aware that more magnification is not always better; for any observation there's an optimal magnification, not too big, not too small. But I doubt your scope would do well above 200x, even though the theoretical limit is 260x. Probably not worth trying anything beyond the high 1XX values - that is, for your eyepieces, a 3x barlow.

For the Moon, any magnification works. Low magnification gives you broad images with lots of stuff to see. High magnification zooms into particular details. Good collimation is critical.

For planets, it depends.

Saturn looks good at any magnification. If you can see the rings, the shadow of the rings on the surface of the planet, and the main split inside the rings (the Cassini division) then you're doing fine. It doesn't have a lot of detail to justify very high magnification, unless you're looking for the finer ring divisions, but your scope can only see Cassini.

Jupiter is a bit tricky. It has enough fine detail to justify high magnification, but it has low contrast - and then as you increase magnification it gets more washed out, more dark, and hard to see. "Medium-high" magnification works best. Just try different eyepieces, with and without barlow, and see what works best for you. It also helps to keep your eye at the eyepiece, keep tracking the planet, and let your eye adapt - in time your brain/eye system will start pulling more detail out of the image. It takes me at least 2 ... 3 minutes to start seeing the really fine stuff there.

Mars - don't even bother unless it's at opposition. It only looks good when it's closest to Earth. The next Mars opposition is in July 2018, so there's a bit of wait until then (will start to look good a few months before that). Even then, you need several things to see Mars well:

very high magnification
excellent collimation
a good instrument
excellent seeing (the inverse of turbulence)

Otherwise it just looks like a red dot.

Seeing will affect all solar system objects (Moon and planets). When seeing is bad (when atmosphere is turbulent), any object will look bad. It depends on the weather - not just weather at ground level (what you think of as "weather") but weather in the whole thickness of the atmosphere, all the way up (the stuff only airplane pilots usually care about). If you live in North America, the Clear Dark Sky site can give you a prediction of a few parameters including seeing. When seeing is bad, high magnification is worthless (the image is fuzzy and keeps shaking like jello).

Another thing that helps: if there's a large temperature difference between inside and outside, leave the scope outside for a while to catch up with the air temperature. This will reduce internal turbulence and improve image quality on planets. I always leave my scope at "outside" temperature for about 1 hour before observing. When there is no temperature difference between inside and outside, then don't worry about this. This is like seeing (turbulence), but inside your scope.

Please note that light pollution has absolutely no effect on solar system objects (see below re: light pollution). Even when light pollution is very high, the Moon and the planets still look the same.

Dark adaptation does not matter for solar system objects (see below re: dark adaptation). You don't have to be in a dark place when observing them. I've looked at Saturn from a sidewalk under the bright lights downtown (city center), with cars and people passing by - no problem.

Deep sky objects (nebulae, star clusters, and galaxies), a.k.a. DSOs, are different.

These are objects with low brightness. Because of that, your eye cannot see details or colors. Collimation is less important. Seeing also doesn't matter much. Collimation and seeing affect fine details, but you can't see fine detail in DSOs.

What does affect them is light pollution - the light from all sources around you, glowing in the dust, water drops, etc in the atmosphere. This is why it's hard to see DSOs from the city. If you can drive 1 hour away from the city, to a place far from sources of light (cities, factories), then DSOs are easier to see.

Google "light pollution map" and you will find several sites that give you light pollution intensities in your area.

But even from the city, there are many DSOs that you can see. M31 (the Andromeda galaxy), the Orion nebula, the M13 cluster, the Dumbbell nebula - all these are clearly visible in an instrument like yours even from very light-polluted places. I've watched M13 many times from the middle of the Silicon Valley.

The trick is how to find these objects. Get a book called "Turn left at Orion" - it's a guide for beginners showing how to find many DSOs. Get a good planisphere (star chart) like this one and keep it next to the scope when you're observing DSOs. Start with something easy like the Orion nebula.

Also you could google "how to find the Orion nebula" (or M13, etc) and you'll find online guides which are sometimes helpful.

Usually you don't need high magnification for DSOs. Low and medium magnification work well - but it depends; e.g. I've watched M13 in high-ish magnification and it looks good. What really does help here is eyepieces with a wide field of view, because it makes it easier to find the objects. I'm guessing your eyepieces are of the Plossl design, with a 52 degree FoV, which is fine, but more is better for DSOs. Now don't go ahead and buy some fancy expensive eyepieces - keep them proportional in price with the scope, and the original ones should be fine for now.

Another thing that really helps with DSOs is telescope size - bigger is always better. That's also true for solar system objects, but it's even more important here.

Finally, make sure your eyes are dark adapted before looking for DSOs. Observe from a dark place, and stay in the dark for at least 15 ... 20 minutes before observing. Do not use anything with a screen (smartphone, computer) because that destroys your dark adaptation. Do not use flashlights. There are flashlights for astronomers, with a red LED and a dial button, and you need to dial down the light to the least amount that allows you to see and read the star chart. Very weak red light will not destroy your dark adaptation. You could also improvise a red flashlight, but make it as weak as possible (or put a dial button on it).

Clear skies to you and good luck!


Does a photon need to have EXACTLY the right energy to be absorbed by a gas molecule? From an answer to this question, https://physics.stackexchange.com/questions/281660/how-does-an-electron-absorb-or-emit-light,

Absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and
final states of the system. (the atom or a molecule as a whole)
i.e., by the absorption of a photon, the system could access to some
higher permissible quantum mechanical energy state. If there is no
pair of energy states such that the photon energy can elevate the
system from the lower to the upper energy state, then the matter will
be transparent to that radiation.

Given that the photon's energy is proportional to the photon's electromagnetic frequency, and that the frequency is subject to tiny Doppler shifts due to differences in velocity between the emitting and absorbing molecules, how can a photon have precisely the same energy as the exact quantum state change for a gas molecule?  Does the photon energy only have to be very close to the quantum state change for a molecule to absorb it?  If so, what happens to the extra(or lesser) energy absorbed by the molecule? Does this margin contribute to the width of observed absorption spectra in radio astronomy?

The Physics SE answer (or the part quoted) was incorrect. The photon does not have to have "precisely" the right energy to cause a transition. The reality is that there is a non-zero probability of causing a transition at all photon energies, but the probability distribution is sharply peaked at the energy we calculate to be the energy difference between the two energy eigenstates of the absorbing atom/molecule.
There are a number of effects that cause this broadening of the frequency response of a bunch of atoms/molecules to photon energy.
The transitions have a "natural width" because the transitions take a finite amount of time. This is encapsulated in a form of the famous uncertainty principle, which means that when viewed over a short time period, there is a fuzziness to energy levels in the atom/molecule. A classical analogy is to try and define the frequency of a damped harmonic oscillator. The larger the damping, the quicker the oscillation dies away and the broader is the continuous frequency spectrum.
Related to this is collisional broadening. Transitions can be interrupted/truncated by collisions, again resulting in a broadened frequency response.
A photon of fixed frequency (note that having a group of photons at fixed frequency is equally impossible, for similar reasons at the emitting end) will encounter atoms/molecules travelling with different velocities and as a result their frequency responses will be Doppler shifted by the appropriate amount.
The application of electromagnetic fields can also broaden and shift the frequency response.


Why did Mercury not appear to transit through the middle of the sun? Why did Mercury not appear to transit through the middle of the sun (the equator)? I initially thought that this was because of the ecliptic, but, then I thought that actually we'd see it above the solar equator, wouldn't we? So I'm confused.

This video of the transit from NASA shows exactly what I'm talking about. Mercury is shown relatively low on the solar disk.

I'm looking specifically for why it was visible in the bottom half of the sun, not the top half?

Mercury's orbit is inclined 7 degrees relative to Earth's orbit.

This drawing is exaggerated. When Mercury is "between" us and the Sun it might be below, above, or passing through the Sun's disc from our point of view.

Another post from Universe Today explains this also.


A point of light moving between the stars I noticed you replied to a question before about this topic and i came across your answer. Way back in the 80s when i was in my teens i saw a point of light moving between the stars at a steady rate across the sky. It was about the same size and brightnesz of those visible. It was moving in such a was as it approached a star ir veered off left or right then continued on its original path across the night sky. I showed my sister who was older than me at the time. It kind of reminded me of a whiplash efect using gravity to maintain spèed. Mabie too much Star Trek, but you get the idea. I am not an astronomer so forgive my ignorance, but a plane or a satalite would not move in this manor would it? I am convinced to this day that it was not one of ours. Can you clarify this as it was definately avoiding the stars?

It's almost definitely not alien, if that's what you're asking.

A satellite in low Earth orbit (around 200 km altitude) would have needed to move about 60m in the time you noticed it for the movement to have been visible. So, given that satellites at this altitude generally travel at around 7km/s or 7000 metres per second, it is not inconceivable that it could have been a man-made satellite correcting its orbit.


Why is this Einstein ring pink? I'm having a hard time understanding this image in the CNN News item This glowing ring of light is from a distant galaxy.
The caption (from this version of the article) says:

This  image shows observations taken with the NASA/ ESA Hubble Space Telescope of two magnified galaxies behind massive galaxy clusters: the pink glowing haloes reveal the gas surrounding the distant galaxies and its structure. The gravitational lensing effect from the clusters multiplies the images of the galaxies, producing up to 4 images of the same source.

Trying to parse the caption, I think that the two objects in the center are magnified images of the galaxies in the background, and so should be related to the source of the light in the Einstein ring. But they are yellow and the ring is pink.
Question: So the Einstein ring is picking up light not from the background galaxies, but from a halo of gas around them? Is there a better source to read about this image and everything that's going on here?
The image contains the following label at the top right corner:

click for full size:

The article then says:

The MUSE instrument on the European Southern Observatory's Very Large Telescope has identified halos around distant galaxies. Previously, images of these glowing rings lacked detail, but astronomers were able to use a gravitational lensing effect of galaxy clusters to see detail and structure.
Gravitational lensing occurs as an effect of Albert Einstein's theory of general relativity: mass bends light. So the gravitational fields of galaxy clusters bend light, acting like a magnifier of the galaxies behind them.
The best observations of the halos by MUSE were released Monday, presented by Adélaïde Claeyssens during the annual meeting of the European Astronomical Society in Lyon, France. Claeyssens is a Ph.D. student at the Centre de Recherche Astrophysique de Lyon.

Which further confuses me because the caption mentions Hubble but the article mentions data from MUSE at the VLT. I wonder if the image is a file photo and only peripherally related to the content of the news article?

OK, guessing rather wildly in the absence of solid information....

The press-release page is here; that provides a bit more information
than the news article, and includes two images of (different regions of) the cluster and a hint that the cluster is named MACS0940.

I think the main image is a color composite from HST imaging. It shows a galaxy cluster (the yellow galaxies). Also visible is light from an Einstein ring, the gravitational lensing by the cluster of a (single) distant background galaxy. The thin, white, ringlike streaks within the pink arcs are the stellar light from this galaxy (probably UV light from massive, recently formed stars); the broad pink arcs themselves are (I'm guessing) the Lyman-alpha emission as seen by MUSE. (So the image is a composite of both HST data and MUSE data.) Thus it is plausible to say that the Einstein ring is made up of stellar light from the background galaxy and (in pink) emission from gas in and around that same galaxy; the assumption is that said gas is mostly in a halo around the galaxy.

The second "magnified" galaxy referred to in the news article is probably in a different part of the cluster, and doesn't form a nice Einstein ring; it is presumably the four pink blobs in the second image on the press-release page.

This is the abstract for an HST proposal on which Claeyssens was a co-I; the data were taken on May 1, 2019, so it's plausible the image is from those observations:

  Recent deep integral-field spectroscopy observations using VLT/MUSE
  revealed two     high-redshift (z>4) lensed arcs in the galaxy cluster
  MACS0940. Notably, both lensed galaxies show highly extended
  Lyman-alpha emission (> 5 arcsec)... one of these galaxies forms an
  almost complete Einstein ring, with the Lyman-alpha arc reaching an
  unrivaled extent of 30 arcsec.


Confused about rubber sheet analogy! How to resolve confusions on the rubber sheet analogy of the spacetime curvatures?
I am a newbie to spacetime curvature. I have watched several youtube videos on Einsteins GR and spacetime curvature where most people used the analogy of the rubber sheet and ball on the sheet. However, I cannot realize that rubber sheet analogy with my reality. For example,

The rubber sheet is a two dimensional sheet on the other hand I see our universe is a three dimensional.
The balls on the rubber sheet are placed from one direction to other (top to bottom) but in reality I see out planets and stars are flying on the space.
Light flows over the rubber sheets and curves around the balls in its world line, but in reality we know everything themselves are travelling with the speed of light in their world line.

How can I resolve these confusions? I am really interested to know detail about them.

The rubber sheet only is not meant to be a qualitative model, it gives one concept and one concept only:  Mass causes curvature of spacetime.
You can't get any more than that from the rubber sheet.  If you have that idea in your head already then you are ready to drop the image because:

The sheet is 2d but spacetime is 4d
The 2d sheet is embedded in 3d space. Space-time isn't embedded in a 5d or higher dimension (or at least, if it is, it is irrelevant)
The sheet has two space-like directions with no time dimension. Space-time has a time-like dimension.
The basic way of finding distances on a flat sheet is $(x^2+y^2)^\frac12$. The basic way of finding distances in 4d spacetime is $(x^2+y^2+z^2-t^2)^\frac12$  (where units are chosen to make the speed of light = 1, eg time in seconds, distance in light-seconds)

You can't create a simple image of curved 4-d space-time. The rubber sheet analogy will hinder further understanding if you try to use it to understand why an object will travel in an apparently curved path in space when in a gravitational field. It's only purpose is to embed the notion that "space-time is curved" it can't tell you anything else about what that implies.
So to address your three points

In fact spacetime is 4 dimensional.
Yes.  The placing of balls is strictly metaphorical. There is no actual sheet.
Object don't move at the speed of light.  In their own frame of reference they are still (by defintion).  If you have two objects, they can move relative to each other, but always at less than the speed of light.


calculating lunar phases Given a date, I'd like to return the lunar phase.

Jean Meeus's Astronomical Algorithms book and the many packages like PyEphem that make use its algorithms provide ways to calculate when a primary phase (New, First Quarter, Full, and Last Quarter).  The secondary phases (Waxing Crescent, First Quarter, Waxing Gibbous, Waning Gibbous, and Waning Crescent) can be easily calculated as falling in between the primary ones.

The percent illuminated the moon (or any other body between the observer and the Sun) can also be calculated for a given point in time.

We can also calculate the subsolar point.

What isn't clear is how long the primary phases should be considered to be in play.  How are labels generally determined for these phases?

The next new moon occurs 2017/7/23 09:45:34 UTC.  When should it no longer be considered new but waxing crescent?

When it is 1.0% illuminated?
When the subsolar point moves 5 degrees west of 180E?  10 degrees?
At 2017/7/24 00:00:01 local time?
At 2017/7/24 00:00:01 UTC?

I'm inclined to use the last.  The moon is "new" for the entire UTC day 0.0% illumination falls on.  The following day it is waxing crescent and remains so until the day the first quarter falls.

Anyone know of a standard here or rule of thumb?

A new moon occurs 2017/7/23 09:45:34 UTC. Instantly after that it is waxing.

A crescent won't be visible for a few days. It reaches first quarter about 7 days later. The moon will appear to be about half full for maybe a couple of days. Nevertheless, first quarter, full moon, third quarter and new moon are moments in time, not intervals.

It may, sometimes be convenient to refer to "the day that new moon occurs", and in casual use people will say the moon is full while it appears as a circle to the naked eye (usually lasting several nights) These are casually defined for casual use. For precise terminiology say "it is now 3 days 4 hours after New Moon" or similar.


Why don't planets and asteroids have (cometary) tails? Comets have tails. We all know that, and a lucky few have seen the better known. So why don't the planets (e.g. Earth, Mars) have tails? Why don't asteroids don't have tails?

See Jeans Escape. If the average velocity of the volatile molecules is above escape velocity, volatiles will escape. And the with the shallow gravity wells of comets, escape velocity is very low.

The earth and Mars have lots volatile gases and ice. But with their deeper gravity wells, sublimated volatile ices aren't hurled into space as they are with comets.

A lot of asteroids have high enough temperatures that any volatiles would exceed their tiny escape velocity. Which is why a lot of asteroids in our neighborhood have little or no volatile ices. If they ever had ice, it would have long since boiled off. For example, after Comet Wilson Harrington outgassed much of its volatiles, it'd no longer have a tail when coming near the sun. They thought they had lost the comet when they stopped sighting it. Then later they discovered asteroid 1979 VA. Then they noticed they were the same object.

It's speculated there are more "dead" comets like Asteroid 1979 VA. Former comets who've outgassed most their volatiles.


What would happen to black holes during the Big Rip? If the universe expands fast enough to rip atoms apart, what will happen to black holes? Would they rip apart too? What would that be like? Would they expand along with the universe's expansion, or would their gravitational force be able to resist it?

The universe won't expand fast enough to rip atoms apart, but if dark energy turns out to be a fact it will expand fast enough to avoid gravitational collapse. Black holes won't be ripped apart, they will just carry on as normal, as will the galaxies they belong to. But as space expands the galaxies will move further and further apart, going their own separate ways with not enough gravity to keep them together, eventually fading away into the black nothingness of space. Further research may mean that dark energy will have to be revised or reinterpreted, so at the present time a collapsing, blue shifted universe can't be completely ruled out.


A few questions regarding the transit of planets Here we will be concerning ourselves with the transit of a planet with it's parent sun.

I have a few questions that I am not sure about:

1) Do all planets transit their parent sun? It would seem to me that if all planets "cross" their parent star in some form, then depending on our point of view, we should be able to see the transits. For example, we never see a Mars transit, but if we alter our point of view to be farther then mars, then we would.

2) If all the planets are in the same orbital plane, would the transit of one, imply the transit of all the others? For example, our solar system is relevatily in the same plane, and it seems like Mercury/Venus transit, would also imply earth and beyond transits. We just have to be in right perspective.

3) If a planet transits its star, does the probability of a transit increase/decrease/same if planet semi major axis changes? I am not sure about this one. I still think that it would be 100% the fact that a transit will occur, but its more unlikely that we would ever see it, because we need our angle to be just right.

4) If a planetary system has multiple planets, would we detect transits from several planets if they are all in the same orbital plane, or if they have different orbital planes? This kind of builds off of #2, and I think the answer is the same orbital plane, because the "different" orbital planes could be such that are 90 degrees from our point of view, meaning we would never see the transit.

I want to know if my understanding of transits is correct by these questions. Thanks!

When we talk about a transit, we are normally talking about one that is visible from Earth, or at least from some planet in the solar system.

But if you are willing to go to a point is space, it is always the case that every planet is transiting the sun from somewhere. So the Earth is right now transiting the  sun from any position that is directly opposite the sun and more than about 1600000km from Earth (otherwise the sun won't be transited, it will be eclipsed.

The transits of all the planets can be seen from exterior planets, though the transits of (say) Saturn from Neptune are exceedingly rare

If you are in the orbital plane of the planet, you will eventually see a transit.
But you may be waiting a long time. The semi major axis doesn't have any effect.

If the planet in a planetary system are in the same plane, then you may see more than one transit. But you are less likely to see the outer planets as they are further from the star.


What is the angle between the planes of Oumuamua's hyperbola and the Milky Way Galaxy? Viewing the animations of Oumuamua passing through the solar system, it is apparent that the plane of its hyperbola departs from the plane of the ecliptic by a large amount.  The plane of the ecliptic departs from that of the Milky Way Galaxy by a large amount.  Now that there is so much speculation about a potential artificial origin, one might be forgiven for speculating that there might be a coincidence between the Milky Way Galaxy's plane and that of Oumuama.

What is the angle between the planes of Oumuamua's hyperbola and the Milky Way Galaxy?

The inclination of the orbit is 122.7 degrees (with respect to the ecliptic), whilst the Galactic plane is inclined at 60 degrees to the ecliptic. So there is 63 degrees between the two.

Such a high inclination by no means rules out that the object is orbiting within the Galactic disk. A study by Mamajek (2017) concluded that it's approach trajectory was entirely consistent with an object travelling at the median velocity of stars around us. It had insignificant vertical or radial motion with respect to the local standard of rest, but is travelling about 10 km/s too slowly for a circular orbit at the Sun's Galactocentric radius (whereas the Sun moves about 10 km/s too fast).

This appears to be discrepant with the inclination angles discussed above until you recall that the sun is moving up and out of the Galactic plane at around 7 km/s.

Have a look at the animation at https://www.cosmos.esa.int/web/gaia/iow_20180925 , it says a thousand words. It shows time running backwards from the time of encounter and shows how Oumuamua comes in from inside the Sun's Galactic orbit and from "above", with respect to the Galactic plane.


Is there a maximum Julian day? I have read various discussions of the Julian day, including that on the Wikipedia, but none of these appear to define the maximum Julian day or proleptic Gregorian year in which that maximum day would occur.

Is there a maximum Julian day? If so, what is it?

Since the Julian day is defined as a continuous count of seqential days starting at Jan 1 4013 BC, and there is no limit to the number of integers available for  the count, there is no specified maximum.

If you are asking when the Julian Day Number will no longer be useful, possible answers are when there is no one left to count it, or when it is replaced by another concept or definition. Note that there are already several shortened versions, such as Modified JD and Truncated JD, used when JDN became too large to fit in smaller computer word sizes with the desired precision.


does the orientation of orbital plane of our solar system rotate with the galaxy As I understand it, our solar system travels through the milky way with the orbital plane tilted about 60 degrees so that we are sort of traveling in the direction of earth's north. My question is if our orbital plane rotates with our galactic orbit so that it maintains its orientation in relation to the center of the galaxy as I've show in this first picture.

Or is the orientation unaffected by the galactic orbit as I've shown in this second picture.

The orbits in our solar system add up to a certain total angular momentum, which is a vector so points in some direction.   To get that vector to point some other way requires torque, it won't happen just because the solar system is orbiting the galaxy. So the "precession" rate of the plane of our solar system is controlled by very different things than the orbital rate, so there would never be any reason for picture #1.

Picture #2 is what you get if there is no torque, so at least there's a reason for it.  The kind of torque you'd need would be a force that points up from the plane on one side of the orbit and down from the plane on the other side, but a force like that would depend on distance from the Sun so it would be different for different planets.  That would tend to make their orbital planes undergo nodal precession at different rates, so they would not be in the same plane. So either such torques have had no effect over the lifetime of the solar system, or else something else shepherds the planets into the same plane.  But it can't be the latter, because Neptune is in the plane but Pluto isn't.  So I think it must be picture #2.


Star names and the HYG Database Related to Convert coordinates between RA/dec and WGS 84 (SRID=4326).

WARNING: I'm a noob in astronomy.

After import all stars from HYG Database (http://www.astronexus.com/node/34) I noticed some missings:

1) NGC catalog seems to have other names ( maybe Gliese )

2) Some stars have catalog names when could have simple (commom) names ('Mimosa' is 'Bet Cru').

In fact, I think the HYG Database is too professional (
select count(*) from hygn where proper_name is not null gives me only 87 stars) and keep the stars names more scientific.

Is there another database like HYG Database I can try?

What about BSC5p for stars brighter than magnitude 6.5? The "alt_name" "is usually the Flamsteed and/or Bayer constellation-based name for the star".

Here a list of more advanced catalogs.

Here a list of some common star names for Hipparcos numbers. Probably not much different from your query result.

On Wikipedia a probably more complete list of more or less wide-spread common star names.

Those common names often are not unique, and not all common names are widely accepted. That's probably why they aren't provided in the star catalogs.

The only way I could imagine to get a solution is to convert the Wikipedia table to a csv file, and to import it to a database table used as a relation from the various catalog names to the proper names. That will probably need quite a bit of manual work.

None of the naming systems is complete, not even necessarily unique. Stars frequently turn out to be not one, but several stars, or no star at all. At some point it's easier to use 'scientifics'.


Understanding energy loss in a telescope aperture due to atmospheric dispersion I'm trying to understand energy losses due to atmospheric dispersion, given the plot below from the Keck telescope.
For a typical seeing I'm told that there's 80% encircled energy in 1". The aperture of the instrument is 2". The plot shows the dispersion as a function of zenith distance for different wavelengths.
What is the maximum zenith distance with < 20% loss at 3200 angstrom due to atmospheric dispersion?
I'm not used to thinking about aperture in arcseconds, but I suppose it's bigger than the seeing, usually? Also, the fact that 80% of the energy is encircled in 1", does that imply that you catch 100% with an aperture of 2" or am I overlooking things?
I don't really know how to use this plot. At a certain fixed zenith distance, if for example the y-value is -3, does that mean the seeing increases with 3" (e.g. if seeing was 1" without dispersion, would it be 4" with dispersion)? Or does it not make sense to think about seeing "without" atmospheric dispersion?

I'm going to make the assumption that the "aperture" of the instrument refers to the width of a fibre that feeds a spectrograph (I can't think of any other plausible scenario).
What you want is to enclose the most power you can within a 2 arcsec diameter in the telescope focal plane at the wavelengths of interest. One confounding effect is the seeing. If 80% of the energy is enclosed within 1 arcsec diameter, then it is a reasonable assumption that you almost get all of it ($>95$%) within a 2 arcsec fibre.
The above consideration applies to light of any single, monochromatic wavelength. If you wish to observe a spectrum over a range of wavelengths, then you must take account of atmospheric dispersion. Light of different wavelengths is refracted by the atmosphere and thus light of different wavelengths is brought to a different focal point depending on its wavelength. The amplitude of this differential displacement in the focal plane depends on how much atmosphere is in the way and so increases with zenith distance.
The graph in your question shows you what that displacement is (in arcseconds) with respect to light at 5000A. What this means is that the range of wavelengths you can get successfully into your 2 arcsec fibre will depend on the central wavelength and the zenith distance.
For example, trying to get a spectrum between the atmospheric cut-off at 3200A and 5000A in a single observation will not be possible above a zenith distance if 50 degrees because the centroid of the images at these two wavelengths differ by more than 2 arcsec. And in fact you will start to lose signal before that because the displacement in the graph is just of the image centroid and obviously if that centroid gets near the edge of the aperture then the blurring effects of seeing will lead to light loss.
If instead the "aperture" refers to a slit width on a more traditional spectrograph, then you can get around this dispersion problem by rotating the slit to be at the parallactic angle, so that the slit lies along the dispersed image of the star and so collects light at all wavelengths (though they will appear in different pixels on the camera).
If you are doing imaging the effects of atmospheric dispersion will depend on the bandwidth of your filter and the zenith distance. With a narrow band filter the size of your images will be limited by the seeing. If you are using broadband filters, especially U and B, then your images can become vertically elongated at large zenith distances due to the differential dispersion.
A solution is to introduce an atmospheric dispersion corrector prior to the focal plane, which reverses the calculated effects of atmospheric dispersion (at the expense of some loss of signal and image quality).


Is the Large Magellanic Cloud contained in the Milky Way? From a distance of 160,000 light years, the Large Magellanic Cloud looks at first glance like an isolated section of the Milky Way, running across southern Dorado into neighbouring Mensa.

Is the Large Magellanic Cloud contained in the Milky Way? I don't think it is, but I'm not entirely sure.

The short answer is that the Magellanic clouds are not part of the Milky way; they are satellite galaxies of the Milky way.

The image below shows the Andromeda galaxy. Note the small extra galaxies, marked Messier 32 and Messier 110. They are not part of the Andromeda galaxy, but are satellites of it. They would appear like the Magellanic clouds of the Andromeda galaxy.