Why is true and average anomaly of planet important? Why is a true and average anomaly of the planet important? What useful information do they give us?

Given one body orbiting one another, the radial distance between the two and the true anomaly are the polar coordinates of the orbiting body. This wouldn't be all that useful if there was no way to predict where the orbiting body will be at some point in the future.

The mean anomaly along with Kepler's laws of motion do just that. Kepler's first law says that the planets orbits about the Sun are ellipses. His second law says that the rate at which a planet sweeps out area (with the Sun as the central point) is constant. His third law says that the period of a planet's orbit depends on the semi-major axis length but not eccentricity.

This means that a planet in an eccentric orbit and a planet in a circular orbit, both with the same semi-major axis length, will, on average, exhibit the same motion. In particular, their periods will be exactly the same. It's easy to predict where the planet in a circular orbit will be at some point in the future because that planet's true anomaly is a linear function of time. Calling this the mean anomaly, this linear relationship means that
$$M(t) = M(t_0) + (t-t_0)n$$
where $n$ is the mean motion:
$$n = \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$
Here, $T$ is the period, $a$ is the semi-major axis length, and $\mu$ is the constant of proportionality in Kepler's third law.

What's needed is a bridge between the mean anomaly and the true anomaly that makes the true anomaly satisfy Kepler's second law. This bridge is the eccentric anomaly. Mean anomaly and eccentric anomaly are related via Kepler's equation,
$$M = E - e \sin E$$
Here, $E$ is the eccentric anomaly and $e$ is the eccentricity. Note that by this equation, it's trivial to calculate the mean anomaly given the eccentric anomaly and the eccentricity. Simply plug the values in the right hand side and evaluate. What's wanted is a means to calculate the eccentric anomaly given the mean anomaly and the eccentricity. Unfortunately, there is no close-formed inverse of Kepler's equation in the elementary functions. Fortunately, there are many ways to solve Kepler's equation for the eccentric anomaly. One easy way is to use Newton-Raphson iteration.

What's still needed is a relationship between eccentric anomaly and true anomaly. This relationship can be represented in closed form:
$$\tan\frac\theta2 = \sqrt{\frac{1+e}{1-e}}\tan\frac E2$$


Latitude and longitude of Eugene Shoemaker's final resting place? Eugene Shoemaker was an esteemed and celebrated astronomer and planetary geologist. Is it possible to localize their final resting place with a latitude, longitude and feature name?

A small portion of Eugene Shoemaker's ashes were placed on the Lunar Prospector mission. After completion of the mission's primary objectives:

Lunar Prospector was deliberately impacted onto the shadowed Shoemaker crater on the lunar surface at 09:52:02 UT July 31, 1999.
NASA Mission Description

No coordinates are given in that source, however, the Google Earth rendering of the Moon has a placemark for the crash site.  No coordinates are listed.
A source from before the impact occurred lists the planned crash site as 87.70 S 42.0 E, which is very close to the placemark, and gives an estimated crash time within a minute of the actual event.
Thus, the likely resting place of that portion of Shoemaker's ashes may be gives as: 87.70S 42.0E, Shoemaker Crater.
Note: The idea of crashing into the crater was put forth by the University of Texas at Austin, and they had a small site about the crash.  I was able to find a snapshot from the period that has working links on the Internet Archive here.
There was also an entire site devoted to the amateur astronomy aspect of the crash, also on the Internet Archive: lunarimpact.com.
Information from these sites indicates that the reason the crash time was different was due to a stronger-than-expected deorbit burn, and thus the actual crash site might be closer to 87.7S, 42.1E, which is exactly where the Google Earth marker is. The last mission report, however, lists it as 87.7S 42E.


Why wasn't CHEOPS data taken during passage through the South Atlantic Anomaly downlinked in this case, resulting in gaps in photometry? Section 4.1.2. CHEOPS in Six transiting planets and a chain of Laplace resonances in TOI-178 says

Due to the low-Earth orbit of CHEOPS, the spacecraft-target line of sight was interrupted by Earth occultations and passages through the South Atlantic Anomaly (SAA), where no data were downlinked. This resulted in gaps in the photometry on CHEOPS orbit timescales (around 100 min). For our observations of TOI-178, this resulted in light curve efficiencies of 51%, 54%, 65%, and 86%. For all four visits, we used an exposure time of 60 s.

CHEOPS is in a ~700 km dawn-dusk sun-synchronous 92.8° inclination orbit aligned with local sunset/sunrise (06:00, 18:00) and an image of the South Atlantic Anomaly at 560 km in the 1990's is shown below. It is also discussed in Space SE's

What is the South Atlantic Anomaly, and what can I do to protect satellites from it?
Why would Space Cube 1.0 have so many On-Orbit Upsets over South America?

We can probably assume that the enhanced energetic charged particle flux due to Earth's lower magnetic field in that region of space has something to do with it, but what exactly?
Questions:

Why wasn't CHEOPS data taken during passage through the South Atlantic Anomaly downlinked in this case, resulting in gaps in photometry? Is this out of an abundance of caution at each investigator's discretion, or is it standard, or at least recommended to do so?
Is the problem enhanced noise in the focal plane due to charged particle tracks in the CCD itself, or the potential for digital disruptions like flipped bits in registers or solid state memories? In other words, is the problem mostly analog or digital?

note: CHEOPS observations are already constrained by needing to shield its cryogenic focal plane from radiative heating from both the Sun and the Earth in low Earth orbit:

For a given 48 hour observing period, what fraction of the celestial sphere is available to CHEOPS?
Will CHEOPS use DSN or ESA ground stations, or both?
How many flavors do dawn-dusk Sun-synchronous orbits come in?
Wikipedia's CHEOPS Instrument System (CIS):

he required photometric precision will be achieved using a single frame-transfer, back-illuminated CCD detector from Teledyne e2v with 1024 × 1024 pixels and a pixel pitch of 13 µm. The CCD is mounted in the focal plane of the telescope, and will be passively cooled to 233 K (−40 °C), with a thermal stability of 10 mK.

South Atlantic Anomaly at 560 km in the 1990's from source

The high density of energetic charged particles causes problems for any instrument system transiting through the South Atlantic Anomaly. Here, for example, is a description from the Hubble Space Telescope WFC3 Instrument Science Report 2009-40 (Barker et al. 2010, "WFC3 IR SAA Passage Behavior") of its effects on HST:

The South Atlantic Anomaly (SAA) is the region where the Van Allen
belts are closest to the Earth and is distinguished by elevated levels
of high-energy particle radiation (protons and electrons) which
adversely affects the scientific instruments of low-earth orbit
satellites such as HST. Short and long term effects may include: high
cosmic ray (CR) hit rates, high dark and background levels, residual
glow and count rates, damage to the detector lattice, reduced charge
transfer efficiency, flipped memory bits, and Single Event Upsets that
can suspend the operations of HST instruments.

Below is a figure from that report, showing three 159-second exposures taken by the infrared channel of WFC3 with the shutter closed. The left and right exposures were taken before and after transiting the SAA (the left image's brighter background is Earth light leaking into the camera). The middle image was taken deep inside the SAA. They note that the cosmic-ray incidence is about 50 times higher in the middle image.

It's standard to not do observations when passing through the South Atlantic Anomaly. From the HST Cycle 29 Primer's "Orbital Constraints" section:

No astronomical or calibration observations are possible (except for
some specialized and infrequently executed WFC3 observations) during
spacecraft passage through the SAA because of high background induced
in the scientific instruments and FGSs.

(While certain types of observations with the WFC3 instrument may be permitted, the document also notes that some instruments can't be used on an orbit that passes even through the edge of the SAA, so different instruments can evidently be affected more or less strongly.)
My guess (knowing very little about the details of how CHEOPS observes) is that "no data were downlinked" means either:

The telescope kept taking data during the time it passed through the SAA, but the data were not downlinked because it was known they would be too noisy to use;
"No data were downlinked" is a compact (and maybe slightly awkward) way of saying, "No data were downlinked because the telescope was not taking data during that time".

Edited to add: A quick look at the most recent CHEOPS Observers Manual reveals this discussion of the SAA

... a target is typically observable whilst flying through the SAA,
however the data taken will show an increased number of glitches and
noise spikes due to the impact of high energy particles. Data taken
during IOC showed that images acquired during SAA passage are not
amenable to photometric reduction due to the huge number of cosmic ray
hits. Therefore, the images taken during SAA passage are not
downlinked to ground in order to save bandwidth.

(So it's #1 of the two possibilities I suggested.)


How much does the equivalent width of a line change by the introduction of 5% scattered light? How much does the equivalent width of a line change by the introduction of 5% scattered light? We know the equivalent width is defined as $W = \int_{-\infty}^{\infty} \bigg(\frac{1-F_{\nu}}{F_c}\bigg) \, d\nu$; where $F_{\nu}$ represents the flux in the line and $F_c$ represents the flux in the continuum.
The measured equivalent width is $W_m = \int_{-\lambda_o}^{\lambda_o} \frac{I(\lambda)*(F_c - F_\nu)}{D_c} \, d\nu$ in which $\lambda_o$ is the spectral range over which the profile can be traced, $I(\lambda)$ is the instrumental profile, and $D_c$ is the apparent continuum.
If we choose $\lambda$ to be 0 at the center of the line, and the range spans 200 Angstroms, then does the equivalent width of the line change by $200 A \cdot 0.05$ = 10 A? So is the equivalent width of a line changed depend on our range? I.e. width of a line of 140 Angstroms with 5% scattered light would alter it by 0.7 A. Am I correct?

Imagine your line as a rectangle of width $w$ and depth $d$ relative to a normalised continuum.
Without scattered light, the area blocked off by the line is $wd$ and if the continuum level is normalised to 1, then the equivalent width is $wd$.
Now add 5% scattered light. The height of the continuum is 1.05 (but we're going to renormalise it) and the depth and width of the line are still $d$ and $w$.
Now renormalise by dividing by 1.05. The width of the line is the same, but the depth is smaller by a factor 1.05 and thus so is the equivalent width.


Algorithm to fit galaxies Do you know of any other data analysis algorithm that fits 2-D analytic functions to galaxies and point sources directly to digital images, other than GALFIT? Especially to subtract lens light from strong lensing images?

The two most general, publicly available packages for galaxy image fitting (other than GALFIT) are probably
Imfit and
ProFit. (Note that I am the developer of Imfit.)
There is also Lenstronomy, which is specialized for fitting gravitational lenses; this might be more relevant to your particular needs, though I know very little about it.


Do origin theories imagine each planet to first orbit the Sun very irregularly before stabilizing? What I was wondering is that, for example, toss a few balls into zero gravity space randomly, what would be the steady state motion?   Would not all the balls go on weird directions and undergo irregular orbits due to the interacting and irregular gravitations. I imagine the largest or more dense planets to begin to influence the smaller and eventually it would stabalize.  Is this what origin of the universe theories imagine, or am I totally missing something here? Would the smaller ones eventually all rotate around the largest? All in the same plane, more or less, and all in the same direction? It seems possible but I have no idea if this is mathematically derivable.

What are theories that explain chaotic motions into the present, same plane, same direction planetary orbits around the Sun?

Actually, the sun, planets and other small bodies are all created from the cloud of dust and gas that existed before the solar system. As the gas and dust collect towards the center, the mass rotates and soon an accretion disk forms (which is why the planets are nearly on the same plane).

So your notion of the beginning of the universe is incorrect. It did not start with a bunch of planets (or stars) randomly thrown into space. Also all the stars in a globular cluster are rotating in the same direction; the stars are in orbit in the cluster.

Added:
Yes, they did have irregular movements. It started with a large cloud of gas and dust, some from the big bang, mostly from exploded stars systems. The particles all had different directions and speed.

When gravity influenced enough particles to group and start accumulating is when the mass began to spin (conservation of momentum). The spinning mass flattens out to an accretion disk and planets can start to form (along with the sun).


Question about the formation of elements So, elements such as iron, gold, etc. are formed from different types of supernovas, whether it's type 1A or 2. My question is, is it possible that there are elements that are formed that we never received during the formation of Earth and that we are unaware of? Or have we discovered every non man-made element?

It's very likely, that we don't have discovered every non man-made element. For some elements there exist only very short-lived isotopes.

Plutonium...

  is the heaviest primordial element by virtue of its most stable isotope, plutonium-244, whose half-life of about 80 million years is just long enough for the element to be found in trace quantities in nature

More about naturally occurring nuclides.

This paper mentions the detection of a curium ion in cosmic rays.
The process which forms curium can also form elements beyond curium, hence it's likely that there exist short-lived isotopes of elements in nature, which have not yet been detected. Research in this field is going on.


Bright light in the sky, in the '90s, what could it have been? Sometime in the summer months during the years from 1993 to 1998, my friend and I saw what looked like a gamma-ray burst (or what looked like an artist's impression of it). It looked like a star that got very bright (within a few seconds) and we saw what looked like a line shooting out from the sides of it that spread across our field of vision. All this happened in less than 10 seconds.

I can't confirm when it was any more exactly than that, unfortunately, as we were walking near my friends house at night in the summer, as we did hundreds of times in those years.

According to Wikipedia, the first naked-eye brightness gamma ray burst happened in 2008, so I imagine that's not it.

What could it have been?

Gamma ray bursts are very rarely bright enough in visible light to be seen with the naked eye. This is why they weren't detected until there was a network of satellites looking for atom bombs. You didn't see a GRB.

If what you saw was Bright, and Fast it was likely to be a bright meteor, sometimes called a fireball. These appear as a very bright, fast-moving "shooting star". They are rare, but not so rare that it would be possible to identify a particular event from the information that you give.

A reflection from a satellite can look like a star that brightens and then fades. The Iridium satellites are well known for this though other satellites can also produce "flares". Iridium flares have been visible since the late 90s.

There are other mundane possibilities, but these would seem to be the most likely. Remember memory is a funny thing. After 20+ years thing that you "remember" might be different.


Why does the Solar System have no (natural) satellites of satellites? What are the conditions for a planetary moon to have a satellite of its own? How far do the Solar System's bodies fall from the necessary threshold?

This is due to the mechanics of capture.

For one object (moon) to be captured by a another (planet), some energy has to be removed from the system. If the incoming moon has an existing satellite then it would be ejected, carrying a lot of kinetic energy.

If a small body were to be captured by the planet/moon combination, it would usually be captured by the larger object. Again some energy has to removed from the incoming object by being transferred to one the objects already present. One way that could occur is for the existing moon to be accelerated into a higher orbit.

In summary, it could happen, but with a very low probability.

For much more information, see this answer to the question How do moons get captured?.


What is the temperature of outerspace? I want to know whether the temperature in Outer space is higher than earth's or cold.Is the temperature in outer space vary as it does on earth?

The temperature in the heliosphere is about one million Kelvin, lower (but still very hot) in the outer regions of the heliosphere, higher close to the sun.
But the density of the plasma is very low, so you wouldn't feel the heat (except close to the sun).

Within Earth's exosphere, reaching up to 10,000 km, the temperature is lower, roughly 1,000 to 1,500 C.

In interstellar space there exist hot (e.g. 10,000 K) and cold gas clouds.

Due to the very low density, hot and cooler fractions can penetrate each other, such that there doesn't need to be a uniquely defined temperature.

The large-scale average temperature of outer space should be 2.7 K (about -270.5 C), close to the CMB black body temperature.


When was the last time a great conjunction coincided with a solstice? This December, a great conjunction will occur nine hours after winter solstice, which means that for much of the world the two events will be on the same day. When was the last time a great conjunction occurred within 24 hours of a solstice?
As a rough estimate, consider that there are 96 hours per year that are within 24 hours of a solstice. So you'd expect about 1% of great conjunctions to be within this window. Given that great conjunctions happen once every 20 years, a great conjunction and a solstice should occur within 24 hours about once every 2000 years.

It depends how you define “conjunction.” For example, this year, Jupiter and Saturn will be at the same ecliptic longitude and the same right ascension within hours of each other on December 21, 2020. However, in 2000, they were at the same right ascension on May 30, but at the same ecliptic longitude on May 28, 2000.
I have done calculations (using the VSOP87 planetary theory for positions and Meeus’ Astronomical Algorithms, Second Edition, 1998, for dates of solstices), for conjunctions measured by ecliptic longitude. The result is at https://astronomie.quebec/conjsols.html
This doesn’t give you the precise moment of conjunction; it just lists solstices where Jupiter and Saturn were at less than 5° from each other, between the years −1000 and +3000. As you can see for 2020, the spacing between the two in ecliptic longitude (i.e. neglecting latitude difference) is 0.0387°. The last time these planets were so close in ecliptic longitude at the solstice was in −204 (i.e. 205 BCE).
The only other times in the time span calculated where Jupiter and Saturn were at less than 0.1° in ecliptic longitude (again, neglecting latitude) at the solstice are in −860 and in +2159.
Interestingly, it is in 2020 that the planets are so close in latitude as well at the solstice; their spacing was slightly more that 56′ in 861 BCE and slightly more than 19′ in 205 BCE; and will be more than 1° 12′ in 2159 CE. (Again, these are spacings at the solstices; they may come closer at other dates than solstices, but your question was solstices.)


In zero gravity - I'm floating inside a torus as they spin it up, do I float do the "ground"? Imagine this scenario - I'm in a space station with one of those huge torus shaped constructions, that you spin to simulate gravity through centripetal force. But it's turned off right now, so I'm floating in the air.

When happens when they turn it on and it starts spinning?
I can imagine what would happen if I was grabbing on to something. I'd be pulled around, and the angular velocity would drag me to the outer part of the torus. But if I'm just floating there, with no friction to the ground to catch me, do I stay put, with the thing just whizzing past me?

Yes...

In the scenario you describe, you (and the air) will remain at rest, while the torus speeds up.  One hopes there are no radial partitions, or, if there are, one comes along to pick you up fairly early in the session.

Unfortunately, the rotating torus will impart its velocity to the air in contact with all the irregularities in the wall/floor/ceiling.  This would then be transferred to the rest of the air.  Floating there, you would experience a very slight breeze blowing you in the direction of rotation of the station.  This in turn would cause you to slowly drift radially outward, until you hit the floor and rolled along until you were at rest with respect to the floor.

Even if the station is not running up from zero rotation rate, you can experience some strange effects.

If the station is big enough, you can travel with low energy.  Simply run against the rotation of the station.  When you are running at the rotation speed, lift your feet, and drift in place until your destination comes along.  Put down you feet, start running, and slow to a walk.

Be careful, though.  If you face in the wrong direction and run at that same speed, your apparent weight will quadruple as you crash to the floor...


Sunset on [Mountain] from [Viewing Point] How do I get the days in this year when the sun will set exactly on top of Fuji as viewed from Enoshima.

I shamefully admit that I just want to take a nice picture of Fuji.

My attempt at an answer of my own question. I have no confidence that it is correct, and I'd like to know whether there's an easier way to do this…

https://www.movable-type.co.uk/scripts/latlong.html tells me that the bearing is 90.3°.

A quick application of middle school trigonometry tells me that the visible elevation of Fuji is about 3413 m or 2.90°.

Next, I went to Stellarium, fixed my position to Enoshima, and exported the Ephemeris (F10 menu) of the Sun in a 5 minute intervals with horizontal coordinates from February till May.

Finally, I used a quick python script to find the line in the Ephimeris export where the sun is closest to my desired coordinates (90.3°, 2.9°)

import re
from pprint import pprint

height = "Height"
azi = "Azimut"

# Quick and dirty parser
pre = re.compile("([+-])(\\d+)°(\\d+)'([0-9.]+)\"")
def ph(v):
    m = pre.match(v)
    if m:
        sign, *nums = m.groups()
        return sum([float(s) * 60 ** (-e) for e,s in enumerate(nums)]) * float(sign + "1")
    try:
        return float(v)
    except:
        pass
    return v

res = []
with open('ephemeris2.csv') as f:
    head = next(f).split(", ")
    for line in f:
        php = [ph(v) for v in line.split(", ")]
        res.append(dict(zip(head, php)))

# This is where the magic happens!
filt = [k for k in res if k is not None and height in k and azi in k]
filt.sort(key = lambda v: (v[height] - 2.9) ** 2 + (v[azi] - 270.3) ** 2)
pprint(filt[:5])

This yielded me

{ 'Azimut': 270.13194444444446,
  'Datum and Time': '2019-03-25 17:39:00',
  'Height': 3.0792222222222225 }

I'm unsure whether I can just convert the Bearing to Azimuth like that, but I suspect that a simple mathematical mistake is even more likely…


Planets and Pluto? Neptune? If one of the rules to be a planet is that it needs to clear ALL objects from their orbit, does this also make Neptune a non-planet? Since it has thus far failed to clear Pluto (and based on an answer from the physics board, the kuiper belt) from it's orbit. Or does this rule not really apply to planets and we should welcome Pluto and a few other dwarf-planets into our family of planets?

I answered this same question at physics.SE. I specifically joined this part of the SE network to address this duplicate question at this site.

The astronomy community faced two crises with regard to what constitutes a "planet", first in the mid 19th century, and more recently at the start of the 21st century. The first crisis involved the asteroids. The second involved trans-Neptunian objects. Both crises challenged astronomers to question what a "planet" was.

1 Ceres, 2 Pallas, 3 Juno, and 4 Vesta were discovered in quick succession during the first decade of the 19th century. There was no international astronomical organization at the time of these discoveries; the International Astronomical Union wouldn't be formed for another century. Instead, the designation of what constituted a "planet" fell on the major astronomical almanacs such as the Berliner Astronomisches Jahrbuch (BAJ). Those discoveries at the start of the 19th century were treated as newly discovered "planets". This situation remained static for about 40 years.

That changed in 1845 with the discovery of 5 Astraea. During the 1850s, the list of objects orbiting the Sun grew to 50, and during the 1860s, the list grew to over 100. The response of the BAJ and others was to demote Ceres, Pallas, Juno, and Vesta from planethood status to some lesser status, either minor planet or asteroid. Astronomers didn't have a clear-cut concept of what constituted a planet other than that they should somehow be large. Ceres, the largest of the bunch, is not very large. The end result of all of these discoveries starting in 1845 was that the first four discovered asteroids were demoted from planethood status.

The second crisis started in 1992 with the discovery of (15760) 1992 QB1. By 2006, the number of trans-Neptunian objects had grown significantly. Were these things "planets", or something else? Some astronomers, notably Alan Stern, wanted the term "planet" to be extremely inclusive. Most astronomers balked at this idea.

Paradoxically, it was Alan Stern himself, along with Harold Levison, who provided the key criterion of "clearing the neighborhood" that lies at the heart of what the IAU deems to constitute a "planet." Their paper, Stern and Levison, "Regarding the criteria for planethood and proposed planetary classification schemes," Highlights of Astronomy 12 (2002): 205-213 suggested splitting "planet" into two categories, "überplanet" (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune) and "unterplanet" (Pluto+Charon, Eris, Ceres, Sedna, and a host of others).

Stern is being quite hypocritical when he rants that there is no clear-cut boundary between "planets" and "dwarf planets." The boundary is huge, and Stern knows this. The ratio of the square of an object's mass to its orbital radius about the Sun is key in determining whether an object can clear most of the junk from the vicinity of the object's orbit. There is a five order of magnitude difference between the smallest of the planets and the largest of the dwarf planets in terms of this ratio. This five order of magnitude difference figures predominantly in that paper by Stern and Levison.

The only difference between the proposal by Stern and Levison versus the voted-upon IAU resolution is that while Stern and Levison wanted to designate hundreds (and perhaps thousands) of objects into subcategories of "planet" ("überplanet" and "interplant"). On the other hand, the IAU chose to designate those objects as the mutually exclusive terms "planets" and "dwarf planets". This is consistent with how astronomers dealt with that first crisis. Planets should be "large." Stern and Levison provided the necessary ammunition to distinguish large from not so large.


SIMBAD TAP queries seem to indicate there are no distance measurements for a great many nearby stars, including Alpha Centauri or Barnard's Star I'm attempting to create a map of all objects within 50LY, including main-sequence stars, binaries, planemos, etc. As such, I'm trying to do it all from first principles, building a micro-ETL to scrape SIMBAD data and build a model. However - I'm experiencing some issues with getting distance measurements from ADGL queries. Watch the following:
Using TAP (http://simbad.u-strasbg.fr/simbad/sim-tap)
SELECT * FROM BASIC WHERE main_id like '%alf%Cen%'
Returns 4 rows:

However - when I check out the mesDistance table with this query:
SELECT * FROM mesDistance WHERE oidref IN (3396297, 3396059, 3396054, 7880120)
Returns empty.... So maybe there's a missing parent. Let's check h_link
SELECT * FROM h_link WHERE parent IN (3396297, 3396059, 3396054, 7880120) OR child in (3396297, 3396059, 3396054, 7880120)
Which yields no new objects. 3396059 is the top level object and it has no distance measurement.
However, many other objects show up just fine, such as Proxima Centauri (OID = 3379714 )
It's VERY hit or miss where distance measurements exist and where they don't, but it makes it impossible to query objects based on relative distance. I'm guessing that maybe this has something to do with the age of the catalogs or studies which create a new distance measurement. If a catalog is old enough, it probably wasn't digitized or forced into the SIMBAD ADQL model...
Does anybody have some good suggestions on how to get distance data for nearby stars? Preferably something modern, or with the ability to join it to SIMBAD references. I'd like not to have to do a fuzzy join based on RA/DEC or scraping through alternate IDs. That can get messy or break table grain.

Well - not sure why distance and parallax tables are empty, but I missed something super simple, the PLX_VALUE column on the BASIC table.
SELECT * FROM BASIC  WHERE plx_value IS NOT NULL ORDER BY plx_value DESC
OR convert to distance with D = 1000/PLX_VALUE for dist in parsecs


What are Trojan Asteroids and where are they located? Asides from the main Asteroid Belt, there are Trojan Asteroids - what are these and where are they located?

A Trojan Asteroid is defined by the Swinburne University Astro page "Trojan Asteroids" as:

  Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points.

These are often divided into 'leading' and 'trailing', as per the diagram below (from the source linked above):

Even though the term Trojan asteroid is often associated with the 4800 or so that lead and follow Jupiter, it actually applies to any asteroid that displays this behaviour, associated with any planet.

There are numerous Trojans co-orbiting with Mars and Neptune.  Recently, single Trojans have been found co-orbiting Uranus and even Earth.


What might the surface of Jupiter's rocky core look like? Jupiter is posited, in some models, to have a rocky core about the size of a few Earths.  If we could see through the thick gaseous atmosphere, what topographical features might we find on the rocky core?  I know that no probe has ever sent data from Jupiter's (hypothesized) solid surface, and that orbiters probably can't penetrate the clouds to produce a topographic map, but do we at least have any educated guesses as to what sort of surface features are and are not likely to exist?  For example, given Jupiter's massive gravity, would the surface be extremely smooth?  Or would it not be out of the question to observe things like impact craters or mountain ranges?

There's not going to be any surface features.

First of all, let's assume you could look through a thick soup of metallic hydrogen. This material has densities starting at that of water, going to the densities of solid rock and even above, while still remaining liquid.
On a sidenote: No space probe will ever be able to look into that mess. Only gravitational measurments will help.

In recent years, it has been shown that under such conditions typical minerals are thermodynamically not stable, and thus will dissolve into the hydrogen-helium mix.
Thus, the core will not have any well-defined boundary, but it's going to look more like the following sketch (emphasis on sketch!):

which I took from this author's website, which has also a lot of follow-up in-depth research material linked.

In short: Essentially any topographical features can be ruled out. That thing down there will be a liquefied, high-pressure mess.

It is of course correct to say that Jupiter might not even have a core, but to build a planet with 1 Jupiter mass and one Jupiter radius one usually requires a certain amount of high-Z material in the core, which is why the core-idea is very attractive.
Also, going around on conferences this year, it seems that JUNO does find a 5-25 Earth masses core, but there is no publication on this yet.


Camera reaching the event horizon This question came to my mind after reading this.

I think it is safe to assume that the supposedly thrown camera would not reach the event horizon in one piece. So my question is :

If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon, and if so, why?

If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.

For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at
$$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$
where the the second derivatives of the potential, $\Phi_{,jk}$, form the so-called tidal gravitational field. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.

Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.

To be relativistically correct, the tidal forces on an free-falling object are described by the equation of geodesic deviation, in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor:
$$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$
In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called spaghettification.

Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:

$M\sim 10\,\mathrm{M_\odot}$: $\sim 10^6$ Earth gravities;
$M\sim 10^4\,\mathrm{M_\odot}$: $\sim 1$ Earth gravity;
$M\sim 10^6\,\mathrm{M_\odot}$: $\sim 10^{-4}$ Earth gravities.

The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.


Just how fast is a Fast Radio Burst thought to be? According to Wikipedia;s Fast Radio Bursts; Frequencies and dispersion are recorded they

  The component frequencies of each burst are delayed by different amounts of time depending on the wavelength. This delay is described by a value referred to as a dispersion measure (DM). This results in a received signal that sweeps rapidly down in frequency, as longer wavelengths are delayed more.

The time between the arrival of the pulse at two different frequencies can be used to generate a kind of measure of distance, based on a dispersion constant. The measure does not have units of length, but of integrated electron density over the path from source to observer.

Using some fancy Fourier tricks one could first undo the $1/\nu^2$ delay and then try to reconstruct what the original pulse might have looked like before dispersion.

Has this been done? If so, how fast (narrow in time) might the original disturbance be? A millisecond? Less?

The publication describing the original detection of the first known FRB (Lorimer, D. R., Bailes, M., McLaughlin, M. A., Narkevic, D. J. & Crawford, F.: A Bright Millisecond Radio Burst of Extragalactic Origin. arXiv:0709.4301) has a plot of the measurement that makes the effects of dispersion on this particular FRB nicely visible. Take a look at Fig. 2 in the paper.

The actual signal is less than 10 ms, while dispersion delays the signal by around 200 ms over the 200 MHz frequency range between 1.3 and 1.5 GHz (note that this relationship is nonlinear).

Your idea about algorithmically removing the effects of dispersion on the signal is regularly done in practice, search for "dedispersion". At our (hobbyist) observatory, we are using D. Lorimer's own sigproc package to do this, and it seems to be in widespread use amongst professional observers as well. The basic idea is to simulate a classical filterbank arrangement and shift each filter channel according to the DM.

From Lorimer et al. (cited above):

  Figure 2: Frequency evolution and integrated pulse shape of the radio burst. The survey data, collected on 2001 August 24, are shown here as a two-dimensional ‘waterfall plot’ of intensity as a function of radio frequency versus time. The dispersion is clearly seen as a quadratic sweep across the frequency band, with broadening towards lower frequencies. From a measurement of the pulse delay across the receiver band using standard pulsar timing techniques, we determine the DM to be 375±1 cm−3 pc. The two white lines separated by 15 ms that bound the pulse show the expected behavior for the cold-plasma dispersion law assuming a DM of 375 cm−3 pc. The horizontal line at ∼ 1.34 GHz is an artifact in the data caused by a malfunctioning frequency channel. This plot is for one of the offset beams in which the digitizers were not saturated. By splitting the data into four frequency sub-bands we have measured both the half-power pulse width and flux density spectrum over the observing bandwidth. Accounting for pulse broadening due to known instrumental effects, we determine a frequency scaling relationship for the observed width W = 4.6 ms (f/1.4 GHz)−4.8±0.4 , where f is the observing frequency.
  A power-law fit to the mean flux densities obtained in each sub-band yields a spectral index of −4 ± 1. Inset: the total-power signal after a dispersive delay correction assuming a DM of 375 cm−3 pc and a reference frequency of 1.5165 GHz. The time axis on the inner figure also spans the range 0–500 ms.


How to calculate the moon's illuminated fraction tilt? I use "Astronomical algorithms" book by Jean Meeus for programming algorithms for the moon data calculations. I've already created a lot of methods but stuck on the calculation of the moon's illuminated fraction tilt.
Here is the example of how the moon tilt looks like (when you switch date the moon disk turns and so does the illuminated fraction).
I've already calculated: illumination (48.1), phase angle (48.4), position angle of the illuminated bright limb (48.5), parallactic angle and etc. So I think I'm pretty close but still can't get the sufficient result.
I think that the tilt depends on the position angle of the illuminated bright limb but the tilt looks almost the same during the day and the position angle changes its value rather drastically.
I would appreciate any assistance.
UPD
I think I've found the right formula for the tilt calculation (page 347, and in the image attached):
ZenithAngle = MoonPositionAngle - ParallacticAngle.

But I'm not sure if I understand where ZenithAngle exactly is. I made a picture where I marked ZenithAngle (ZOC) and AlphaAngle which is actually the angle I need to calculate.
So if ZenithAngle is ZOC then AlphaAngle = ZenithAngle - 90.

Also I created a sample page with angles values, moon illumination and limb turned by the angle.

Your update is correct.

PA is the position angle of the bright limb (measured eastward from celestial north).
q is the parallactic angle between the zenith and celestial north.

Then the angle of the bright limb relative to a horizontal line ($\alpha$) is PA-q-90.
Keep in mind that PA is independent of the observer's location (as long as they are on the Earth), and it is well defined for any given date and time. The parallactic angle is dependent on the observer's location, date and time. I would say that it is well defined only when the Moon is above the horizon. (Of course it can be calculated when the Moon is well below the horizon, but what does that mean?)
For example, imagine two observers separated by 60 degrees of longitude. The Moon is above the horizon for each observer. The PA of the bright limb is the same for each observer at all times. But the parallactic angle is different for each observer since it depends on where the Moon is located relative to the meridian. As the Moon crosses the sky, the parallactic angle also changes significantly from hour to hour, and therefore the tilt you are calculating is different based on date, time, and location. (The PA changes with time also, but it is a slower change except around the time of New and Full Moon.)


Can this mirror be cleaned? I dug up my old 6" Dobsonian telescope from storage and the mirror looks like the following:

The picture is down the tube, so you can see the spider holding the secondary. To my eye, the spots look like corrosion and not cleanable, but does anyone actually know what they are?

It was stored relatively carefully - not totally sealed but in a house, with the tube covered with a plastic bag. So huge temperature changes are not likely, but it does quite humid here. It was last in action > 10 years ago.

The only other thing I know about the scope is that it was made by Criterion, sometime before the 1980s.

The spots could be mold or fungus.
If a gentle cleaning with distilled water, a little detergent, and sterile cotton (no abrasives!) fails to remove them, there are several vendors who can strip the old aluminum coating and apply a new one.
If you don't want to go that far, you can probably find a buyer who would.
Criterion scopes have a considerable fan base.


Theoretical sudden appearance of a black hole in BBC drama show *spoiler warning* I asked this on the ‘movies’ stackexchange site but they suggested I ask here, it’s a purely hypothetical query based on a BBC drama show called ‘Hard Sun’, feel free to kill the question if it’s inappropriate here.

In the TV show’s series finale an expected ‘extinction level event’ happens and appears to be a black hole close to our sun, it already has developed a spiral of hot gas around it drawn from the sun.

Clearly this is fiction but I’m keen to understand just how badly-considered this is. Firstly the show is set in London and everyone seems amazed when the image below rises from the dusk yet clearly it's been in existence for the time taken to create the spiral (hours at least?) so nations east of the UK would have seen it for that time and reported on it globally surely?

My real question is more around is there any potential scenario where a black hole of that size could develop suddenly and get to that size quickly enough for that image to be a surprise? My understanding, very much as a scientific-layman, is that a black hole of that size would have to have the mass of at least one of our suns to get to that size (probably more) - is that a valid assumption? If so where has the mass come from? Also wouldn't the sudden existence of something else with at least the same mass as our sun suddenly appearing at the centre of our solar system have enormous gravitational impact on everything in the system, including Earth? We'd have massive volcanic/tectonic impact wouldn't we?

So again sorry if this is inappropriate, I was just incredulous at how badly-considered the scientific aspects of this fiction were and wanted to check that my understandings were correct.

Screengrab:

Firstly the show is set in London and everyone seems amazed when the image below rises from the dusk yet clearly it's been in existence for the time taken to create the spiral (hours at least?) so nations east of the UK would have seen it for that time and reported on it globally surely?

And we've kind of a large number of space based satellites that do nothing but monitor the Sun and all the people monitoring them would certainly spot this minor event.

  My real question is more around is there any potential scenario where a black hole of that size could develop suddenly and get to that size quickly enough for that image to be a surprise ?

About the only way I can think of is someone creating a Kugelblitz, although it's pretty far fetched.

Maybe a very freak collision of two high mass objects head-on, but that would create far worse problems - essentially an absolutely huge explosion and we would not be looking at the result quietly afterward.

  My understanding, very much as a scientific-layman, is that a black hole of that size would have to have the mass of at least one of our suns to get to that size (probably more) - is that a valid assumption ?

There's a minimum size for the formation of a black hole by gravitational collapse of a star and it's larger than our Sun's mass.

There is, however, no theoretical minimum for the size of a black hole if you can make one some other way.

  If so where has the mass come from ?

Ask the writers. :-)

  Also wouldn't the sudden existence of something else with at least the same mass as our sun suddenly appearing at the centre of our solar system have enormous gravitational impact on everything in the system, including Earth?

Yes that's a major issue.

Any black hole large enough to start siphoning off the Sun's outer layers would need a significant mass and it would have a significant impact on our orbit - and do so more or less immediately.

Our orbital speed is enough to maintain our almost circular orbit around the Sun.

With something as massive as a small star added close to the Sun, our orbital speed would no longer be enough for a circular orbit and we'd be in an elliptical orbit about the Sun-BH pair.  It's more complicated than than that, but that's the broad strokes.

But that ellipse would be with our current distance as an aphelion, whereas we'd have a smaller perihelion distance.  So we'd get warmer.  And it would not take much to raise temperatures here uncomfortably.  Very roughly if we were to have a closest approach of 0.72 AU (Venus's current orbit) we'd find our temperatures rise by perhaps 24 degrees Celsius, which would be enough to make the planet unlivable for us - ( maybe 60 C at the equator and enough to melt significant parts of the polar ice and permafrost ).  And as this change would happen every six months as we orbited moving from close in to far out, we'd be permanently shifting climate.  And that ignores greenhouse effects - we could end up with a runaway greenhouse effect that raised temperatures to lethal levels everywhere.

  We'd have massive volcanic/tectonic impact wouldn't we?

That's a hard one to call.  This would depend on how far apart they Sun and the BH were and how massive the BH was.

But we'd have lots to worry about.


Primordial supermassive black holes. How old are they? I know that astronomers have detected supermassive black holes in the early universe. What was the age of the universe at which these black holes were detected?  In other words how long is it thought to take for these supermassive black holes to form and what are the mechanisms by which they formed, other than from supermassive stars?

The first supermassive black holes must have been formed earlier than about 750 million years after the big bang, since there is evidence for luminous quasars at redshifts of up to 7 and with likely black holes of a billion solar masses or more at their centres (e.g. Momjian et al. (2013).

The candidate mechanisms to produce such objects in a short period of time are hyper-Eddington accretion onto stellar black hole seeds; the direct collapse of large, primordial gas clouds into black holes; or the merger of stellar-sized black holes in dense clusters, followed by gas accretion. These possibilities are accessibly reviewed by Smith, Bromm & Loeb (2017).


Is it possible to view the Andromeda galaxy with the naked eye in Greece? I am living in the northern hemisphere in Greece and in class 4 area (light pollution scale). Is it possible to see the Andromeda galaxy with a naked eye?

Assuming you mean Bortle class 4, then the Andromeda galaxy (M31) should be a fairly easy object with the naked eye. It won’t be particularly impressive, but rather will look like a hazy blob. On the other hand it is more than 2 million light years distant, so it’s impressive to be able to see it at all.


What's the name for [the other kind of planet] in a binary star system? This XKCD what-if talks about rainbows on planets in a binary star system.   It points out that there are two types:

circumbinary planets, where the planet orbits far from and around both stars
[the other kind of planet], where the planet orbits one of the stars and the other star is distant.

What is the proper name for [the other kind of planet] ?

I don't know about names for the planets, specifically, but the orbits are called S-type and P-type:

S-type: The planet orbits around one star, and the host star has a binary companion (i.e., "the other kind" in the XKCD comic)
P-type: The planet orbits around both stars of the binary


Luminosity of black hole accretion disc The luminosity of a black hole accretion disc gaining mass at a rate $\mathrm{d}M\over \mathrm{d}t$ can be estimated as

$${1\over 12}{\mathrm{d}M\over \mathrm{d}t}c^2$$

That is a substantial proportion of the rest mass of the in-falling matter. The linked document explains that the factor 1/12 is because at less than 3 times the event horizon radius the matter "spirals in without
radiating more energy"

What causes matter in the accretion disk not to radiate beyond this limit? It is twice the radius of the photon sphere, is the reason general relativity?

Three times the Schwarzschild radius corresponds to the closest stable circular orbit around a black hole. The general idea is that as matters moves in towards the black hole it gets stuck in an accretion disc where angular momentum has to be moved outwards in order to allow the matter to move inwards. The generic mechanism is some sort of viscosity, which heats the gas and hence you get radiation.

However, once the matter gets inside $3r_s$, that problem disappears. There are no stable orbits, no angular momentum loss or viscosity is needed and the material is able to flow (rapidly) straight into the black hole.

Thus when we observe black hole accretion discs we expect them to be truncated at $3r_s$.

So I think the argument then is along the lines of - the gravitational potential energy of unit mass falling to $3r_s$ is converted into an orbital kinetic energy of $0.5v^2 = GM/6r_s$ per unit mass and the rest is converted to radiation. Thus
$$L = \left[\frac{GM}{3r_s} - \frac{GM}{6r_s}\right] \frac{dM}{dt}$$
$$ L = \frac{GM}{6r_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$


How would rainbows appear on other planets? Are other planets capable of producing rainbows? How would those rainbows appear? Can rain, clouds or ice from elements other than water produce rainbows?

Related: https://space.stackexchange.com/questions/34357/rainbow-space-probe

Rainbows occur when sunlight shines through rain. This is rare in the solar system. Rain (of sulphuric acid) might be common enough under Venus's clouds, but there is no sun. Conversely, there is plenty of sun in Mars, but no rain, and only very rare clouds.

It rains on Titan: methane rain. Methane has a lower refractive index than water (1.27 instead of 1.33), which would make the rainbows slightly larger (though not by much 42->52). However the atmosphere of Titan is hazy, and while there is some light on the surface, the sun's disc is not visible.

There is rain in some layers of the gas giants, but again not on the outer layers where the sun is visible.

It is likely that the Earth is the only place in the solar system where rainbows are a common phenomenon.


Could the unseen companion of HR 6819 be a close binary system? Recently there was an announcement of a black hole candidate in the HR 6819 system (ESO Instrument Finds Closest Black Hole to Earth). In the proposed system, the black hole forms a close binary with the B3III primary of the system, and a Be star is located on a wide outer orbit forming a hierarchical triple:

 ┌ B3III star
┌┤
│└ black hole
┤
│
└ Be star

Reading through the research paper Rivinius et al., 2020, it is mentioned that there is no X-ray emission detected from the system indicating that it is either not accreting significant quantities of material, or that it may not be a black hole. The paper notes that the companion cannot be a single main-sequence star (which they estimate would be of spectral type B7 or earlier) because this would appear in the spectral lines, which given the mass leads to the inference that the companion is a black hole as it is too massive to be a neutron star or white dwarf.

As far as I can see, I don't see any mention of the possibility that the companion could itself be a close stellar binary, i.e. the following architecture:

 ┌ B3III star
 │
┌┤
││┌ close binary, star 1
│└┤
│ └ close binary, star 2
┤
│
│
│
└ Be star

A binary of near-equal-mass stars would have lower luminosity than a single star of the same total mass, which might help it avoid detection.

I'm wondering if the configuration would be possible given the short period (40 days) of the companion's orbit, but maybe the data is sufficient to rule this out.

The absolute minimum mass of the unseen companion is $4.2 M_{\odot}$ (and my reading is that this really is an unlikely minimum, requiring an abnormally faint primary for its spectral type and an orbital inclination of 90 degrees, but let's go with it) and the argument is that a main sequence star of this mass would make its presence known in the spectrum. But let's suppose that instead of a $4.2M_{\odot}$ main sequence star, we had  a pair of A stars with $2.1M_{\odot}$ each, in a tight binary. The luminosity of ths combined binary would be less than 25% that of a $4.2M_{\odot}$ single star, so might have gone undetected.

The orbital period of the B3 giant primary with the putative black hole is 40 days. Assuming the minimum mass of the system as a whole is $9.2M_{\odot}$ (for an inclination of 90 degrees), then Kepler's third law yields a binary separation of 0.48 AU.

The radius of a $2.1M_{\odot}$ A star (like Sirius A) is about $1.7R_{\odot}$, so the minimum separation of a pair of such objects in a binary system is about $3.4R_{\odot}$, or 0.016 AU.

As a rule of thumb, as long as the separation of the closely orbiting stars is less than about ten times that of the wider binary then that ought to be stable to tidal disruption (though it also depends a bit on the masses too).

There are examples of systems where there is a wide binary, where one of the components is itself a triple featuring another close binary, in the catalog of binary stars by Tokovinin (1997). These would be multiples with a level 112 in his notation and there are a dozen examples in this catalogue. Perhaps the most famous is Tau CMa, which consists of an O supergiant in a wide, just resolved, 250 year orbit with another bright B-star. However, the O-type star is in a 154 d binary system with a pair of eclipsing B-stars that orbit each other every 1.28 d.

So I think you have a legitimate case that this may not be what the authors claim.

Edit:
I was interested enough by this to email Rivinius and put this scenario to him. The reply raised one or two interesting objections.

Firstly, he suggests that the reduction in luminosity of the secondary (compared with a main sequence single star of double the mass) might not render it invisible in the spectrum. The reason is that the A-type stars have narrower (metallic) lines that could be observed than the B-type giant. He cites the star 66 Oph as an example where this has been done. I am not totally convinced by this, or at least not convinced that it has been demonstrated.

The second point is better. If the inclination is anywhere near edge-on (as required for the unseen companion to be at its minimum mass), then in a close binary you would expect eclipses or at least significant ellipsoidal modulation due to distortion from spherical symmetry. Using Spica as an example (a 4d binary) Rivinius argues that even with dilution that this modulation would have been picked up in the SMEI and TESS light curves discussed in the paper.

If the binary were more face-on to reduce this ellipsoidal modulation, then of course the companion mass would increase dramatically from the 4.2 solar mass minimum, making even a binary companion impossible to hide.

So my conclusion changes. I'm reasonably convinced by the second argument - that a close binary companion would be revealed through ellipsoidal modulations.


working with stellar spectra in fits format in python Hey I am new to working with astronomical data in Python. I wanted to start working with stellar spectra and I am having trouble with the data. To get a first look I just wanted to plot a spectra (flux over wavelength). I downloaded the Miles library (http://research.iac.es/proyecto/miles/pages/stellar-libraries/miles-library.php) and started working with a single fits file. My first steps where as follows :
import numpy as np
import matplotlib.pyplot as plt
from astropy.io import fits

hdul = fits.open('s0013.fits')
data = hdul[0].data
h1 = hdul[0].header

I read the header using print(repr(h1)) and know I wanted to plot the spectrum. The data has shape (1, 4367) and I am not sure how to proceed to achieve 2 arrays one with the wavelength and one with the flux to plot the data. I am sorry if this question is stupid but I can not figure it out.
Cheers

The information you need to recreate the wavelength array is in the World Coordinate System (WCS) of the header, specifically:
CRPIX1  =                 1.00
CRVAL1  =            3500.0000 / central wavelength of first pixel
CDELT1  =             0.900000 / linear dispersion (Angstrom/pixel)

which lists the starting/reference pixel of the wavelength array (1.0), the wavelength value at the start point (3500 angstroms (assumed)) and the step per pixel (0.9 Angstrom/pixel). To read this information, it is best to use a WCS library rather than trying to interpret them directly as they can be more complicated and there are many subformats of FITS WCS.
Fortunately astropy has a module to make this easy (starting from your code above and extending it):
import numpy as np
import matplotlib.pyplot as plt
from astropy.io import fits
from astropy.wcs import WCS

hdul = fits.open('s0013.fits')
data = hdul[0].data
h1 = hdul[0].header
obj_name = h1.get('OBJECT', 'Unknown')

flux = data[0]
w = WCS(h1, naxis=1, relax=False, fix=False)
lam = w.wcs_pix2world(np.arange(len(flux)), 0)[0]

plt.plot(lam, flux)
plt.ylim(0, )
plt.xlabel('Wavelength (Angstrom)')
plt.ylabel('Normalized flux')
plt.savefig(obj_name + '.png')

This will produce the following plot:
If you want to do more extended manipulation of spectra, particularly with the bewildering variety of wavelength and flux units, it might be worth looking at synphot and specutils which build on Astropy and add more direct support for spectra beyond simple numpy arrays. For example, you could make a synphot SourceSpectrum from the above by doing:
from astropy import units as u
from synphot import units, SourceSpectrum
from synphot.spectrum import Empirical1D

source_spec = SourceSpectrum(Empirical1D, points=lam*u.AA, lookup_table=flux,
                   keep_neg=True, meta={'header': h1})


So, what, exactly, is a tropical period? From timezone.com's On the Units of Time Part III: The Year February 19, 2001, by Edward Hahn:

Uh, Oh…He’s Going to Ask That Question Again…
So, what, exactly, is a year?

@PM2Ring's comment

FWIW, that's a mean tropical year, which is slightly different to a Vernal Equinox year (~365.242374 days). See https://www.timezone.com/2002/09/25/on-the-units-of-time-part-iii-the-year/

and that includes the following:

(A side note: Some people out there might be wondering why I didn’t use the Tropical Year of 365.24219 days length? In Duncan Steel’s book Marking Time, he points out that both the US Naval Observatory and the Royal Greenwich Observatory make a mistake when saying that, “the tropical year is defined as the mean interval between vernal equinoxes.” This is not astronomically correct – the tropical year is the average time between consecutive occurrences of any date on the calendar. Since the earth on average orbits the sun in the same amount of time regardless of when you start your measurement, shouldn’t they be the same? Well…they aren’t – the vernal equinox year is an average over some number of orbits measured on a single calendar day, while the tropical year averages once more across all the days of the year. The difference is about 16 seconds a year – not much, but even so the people who designed the Gregorian calendar knew enough to aim for the vernal equinox year since they were trying to stabilize the date of Easter, and not the entire year.)

Wikipedia's Tropical year does indeed refer to the Astronomical Almanac Online Glossary (2015) and while the link does not work for me sitting on a bus where I am Wikipedia quotes it:

the period of time for the ecliptic longitude of the Sun to increase 360 degrees. Since the Sun's ecliptic longitude is measured with respect to the equinox, the tropical year comprises a complete cycle of seasons, and its length is approximated in the long term by the civil (Gregorian) calendar. The mean tropical year is approximately 365 days, 5 hours, 48 minutes, 45 seconds.

Question(s):

Is the US Naval Observatory wrong somehow here? Have mistakes been made? Or are there misquotes?
Is it possible to explain in everyday English the geometrical difference between a "Vernal Equinox year" and a tropical year or a mean tropical year?

The basis of an answer may be within the links I've included or it may not be; while "both the US Naval Observatory and the Royal Greenwich Observatory make a mistake" could be true, it's also possible that the blogpost may not have it right, and I can't access the Astronomical Almanac Online Glossary (2015) to resolve the difference.

Is the US Naval Observatory wrong somehow here? Have mistakes been made? Or are there misquotes?

That's a matter of opinion. In Duncan Steel’s opinion, yes, they made a mistake. In the US Naval Observatory's opinion (and also that of Her Majesty's Nautical Almanac Office), there is no mistake. The Nautical Almanac defines a tropical year as the time from one mean vernal equinox to the next. Steel is using a more pedantic definition. Increased pedantry does not always equate with increased correctness.

  Is it possible to explain in everyday English the geometrical difference between a "Vernal Equinox year" and a tropical year or a mean tropical year?

Right now, Northern Hemisphere spring and summer (Southern Hemisphere autumn and winter) last longer than do Northern Hemisphere autumn and winter (Southern Hemisphere spring and summer). This situation will be reversed in about 13000 years. These changes are due to the Earth's axial precession. That the Earth undergoes axial precession has multiple ramifications, three of which are that:

The length of a tropical year (no matter how one measures it) is less than is the length of a sidereal year.
The times from one March equinox to the next, one June solstice to the next, one September equinox to the next, and one December solstice to the next are not constant.
The times from one March equinox to the next, one June solstice to the next, one September equinox to the next, and one December solstice to the next are not equal to one another.

Elaborating on the third point, the calculated length of one mean December solstice to the next was 365.242740 days (ephemeris days) on 1 Jan 2000 while one mean June solstice to the next was 365.241626 days, or about 1.6 minutes shorter than the duration from one mean December solstice to the next.

So what is a "tropical year"? The US Naval Observatory, which has a much longer history than does Duncan Steel, defines the tropical year as the passage from one March equinox to the next. The US Naval Observatory inherited this definition from Her Majesty's Nautical Almanac Office, which in turn inherited it from the Church of England, which in turn inherited it from the Catholic Church. One of the key drivers of the Gregorian Calendar was getting the date of Easter correct. (Getting the date of Easter correct was very important to many Renaissance minds.)

Duncan Steel defines the tropical year otherwise, as an average value of the times it takes for the Sun's longitude as perceived from the Earth to advance by 360°. That's fine, but to argue that people who use a different definition is not just wrong. It's "wronger than wrong". The fact is that there are multiple definitions of what constitutes a "tropical year", all of which are valid.


Does "spectral type" and "stellar classification" refer to the same thing? Should the tags stellar-classification and spectral-type be merged? has been asked in meta two weeks ago. It requires some careful consideration but so far no response has been forthcoming, so I'm asking this question here to get a definitive answer that can then be applied there.
Wikipedia's Stellar classification begins

In astronomy, stellar classification is the classification of stars based on their spectral characteristics.

and while there is one subsection labeled simply Spectral types to my untrained eye it looks like there may be other classification schemes besides O, B, A, F, G, K, M, L, T.
Question: Do the astronomical terms "spectral type" and "stellar classification" refer to the same thing? Or at least always when applied to stars? Might "spectral types" also refer to a way to classify asteroids?

Spectral type is only based on the star's color. You need additional the mass or luminosity to make a stellar classification form it.
Example: Giants may belong to the same spectral class as main sequence stars.
Of course, if you don't simply look at the color but measure the spectra you see the differences. Spectroscopy is one of the most important tools for astronomers.


What are the advantages of refractor telescopes over reflector telescopes? I've been thinking about getting a new telescope. From what I've read online, reflector telescopes are newer and better then refractor telescopes. Is there any reason to get a refractor telescope instead of a reflector?

reflector telescopes are newer and better then refractor telescopes

It's true that the first examples of reflectors we know of were made shortly after the first refractors we know of. But it's not correct to say that they're (always) better. First, each telescope model you might buy has its own specifications and build quality, which have to be carefully considered in terms of what you're trying to view. Second, certain refractors can outperform certain reflectors for certain tasks: for instance, many refractors can offer better views of planetary/lunar/cluster details than similarly priced reflectors could. Reflectors designed such that eyepieces have to be used at a standing height have the disadvantage of being difficult to use for things (e.g. planets) near the horizon. Refractors could also have the advantage of being potentially smaller and more portable than, for instance, something like a 10" dob. Reflectors have their advantages too: lack of chromatic aberration, for instance (although they do suffer from other forms of aberration). I hope this gives you a better idea of the pros and cons of different kinds of telescopes.


How many light years away is Earth from the closest outer edge of the black hole at the center of the Milky Way? How many light years is Earth away from the closest outer edge of the supermassive black hole in Sagittarius A* at the center of the Milky Way?  I don't mean the distance to the center of the galaxy, but to the point in the black hole's circumference where it is closest to us.

As @Py-ser already said, there is the very clear information about that, although his link correctly is here.

As we can read there, the distance of this is around 25000 light years, with a precision of 1400 light years.

Normally, the size of most galaxies is around some ten thousands of light years. Compared to the black holes, their size (the diameter of their event horizon) is some kilometers (stellar black holes) or some astronomical units (some millions - hundred millions of kilometers).

Even the size of the event horizon of the greatest known black hole is around the size of the solar system, which is around some thousandth of light years.

You don't need to ask, "how far we are from the nearest edge of Sagittarius A*", you can simply ask "how far we are from Sagittarius A".


Plotting The Kepler Field Full View I have downloaded the calibrated ffi(full field image) for the Kepler field, aside from the primary HDU which contains the metadata, they contain 84 channels, each in the form of an image HDU, representing a small part of the field - I want to plot the full field in astropy, how would I go about it?
The Kepler ffi file can be found in the link below
https://archive.stsci.edu/missions/kepler/ffi/kplr2013098115308_ffi-cal.fits

Yes! I can do it for K2 and I'm guessing it should be similar.
You can see with fits.info(fits_file) that it has the 84 extensions (21 modules * 4 channels per module) you mention.
Then you can select a channel, e.g. 65, and prepare also the wcs coordinates so you get the RA and dec associated with the image for the axis.
n = 65

with fits.open(fits_file, mode = "readonly") as hdulist:
    wcs_info = WCS(hdulist[n].header)
    cal_image = hdulist[n].data
    header = hdulist[n].header

Then you can just plot:
plt.figure(figsize = (15,15))

fig = plt.figure(figsize=(12, 8))
ax = fig.add_subplot(1, 1, 1, projection=wcs_info)
im = ax.imshow(cal_image, vmin = np.percentile(cal_image,1),vmax = np.percentile(cal_image, 98), origin='lower')

ax.coords[0].set_ticklabel_position('l')
ax.coords[0].grid(color='white', ls='dotted')
ax.set_ylabel('Dec')
ax.coords[1].set_ticklabel_position('b')
ax.coords[1].grid(color='white', ls='dotted')
ax.set_xlabel('RA');


Is having a surname which begins with the letter “A” a big advantage for a researcher in astronomy/astrophysics? Is changing our surname a good idea? Is having a surname which begins with the letter “A” a big advantage for a researcher in astronomy/astrophysics?

If I have a surname which begins with the letter “M”, and I haven’t published any paper yet, would changing my surname to a name which begins with the letter “A” be a good idea? Or would the advantage gained be too small to make it worth the effort?

Approximately what percentages of research papers in astronomy/astrophysics order their authors …

purely alphabetically?
half by contribution and half alphabetically?
purely by contribution?

I would say that it matters not-at-all whether your name is Aardvark or Zygote. Astrophysicists are not stupid, we can see when a large author list has been alphabetised. If anything I would say it is a disadvantage, since a casual glance at an author list might lead someone to think you were the lead author of any paper, even one with just two authors, simply because of your surname.

In practice, the tiny minority of author lists are alphabetised in Astrophysics papers. Off the top of my head I can think of the gravitational wave papers by Abbott et al. and the MACHO papers by Alcock et al. Most many-author papers have a nucleus of authors at the start of the work and then a longer list of names, possibly in alphabetical order, of people who contributed to the overall project or mission. Even these papers are a (growing) minority and it is still true that the majority of Astrophysics papers have a relatively small author list ordered by contribution.

Finally, I will add another personal opinion. The presence of someone's name on many long-author list papers does not necessarily impress me unless they have lead some of them. It is more impressive for a young scientist to have forged their own path and perhaps written a few lead-authored papers with small author lists or even a sole author paper. Unfortunately, much "big science" does involve big consortia and though it can be exciting for an early-career scientist to be part of such an enterprise, it can be much harder to make your mark outside of that consortium.

EDIT: As a brief, empirical check on my answer, I looked at arXiv for 22nd November 2016. There were 66 new astrophysics papers listed. Of these, 29 were written by three or fewer authors. Only five were written by $>10$ authors and only one paper, from the Pierre Auger collaboration, appeared to have an alphabetised author list. Try hard not to laugh - the paper is by Aab, A. et al. Good luck in your efforts to beat that.


Flux vs. Velocity representation of spectra? I'm looking for a basic explanation of the flux vs. velocity representation of a spectrum and how it's obtained from the regular flux vs. wavelength representation. A good example of this is in Arav et al 2001 (see here, Fig. 2 and 3).

More specific questions I can ask are:
• How do you convert between wavelength and velocity in these figures (i.e. what phenomenon are we using to make the conversion & what's the equation)?
• What thing has the velocity in question/what does the velocity represent? For example, is it the astrophysical object moving in this direction?
... But any further elaboration is much appreciated. As much as I've seen these (I'm trying to brush up, it's been a while since I've done astro stuff), I can't find any explanation for this; perhaps I don't know the right jargon, but yeah it's difficult to find a simple answer. Thanks in advance for the help.

I think the velocity scale is calculated something like this:
$$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$
where $\lambda$ is the observed wavelength and $\lambda_0$ is wavelength of the Mg II line, corrected for the redshift of the quasar ($z=0.868)$ it is observed towards.
The confusing thing is, for Fig.~3 in the cited paper, that the Mg II feature is actually doublet with lines at rest wavelengths of $\lambda_r =2796.35$ and $2803.53$ Angstroms. Therefore
$$\lambda_0 = \lambda_r (1+z) = 5223.58\ \ {\rm or} \ \ 5236.99\ \ {\rm Angstroms}$$
corresponding to the thick and thin solid lines on Fig.3 respectively.
In other words, zero on the velocity scale would correspond to a Mg II absorber at the redshift of the quasar (an intrinsic absorber), whereas negative velocities correspond to absorbers at lower redshift, by the velocities indicated.
For a check on this - calculate the velocity of feature "b", which appears to be at about 5221 Angstroms in the observed spectrum for the redder doublet component (the thin solid line in Fig.3). Using the above definition, this would be a velocity of -915 km/s.
Looking at Fig.3, there seems to be a major problem. Feature "b" is at -680 km/s.  However, if we look further on at Figs. 4 and 5, we see feature "b" plotted at -850 km/s in the Mg II spectrum, which is much closer, and would perhaps be in exact agreement if feature "b" was actually at 5222 Angstroms (which looks unlikely) or that a 4th, unspecified decimal place had been used in the redshift. e.g. if $z=0.8676$ then $\lambda_0 = 5235.87$ Angstroms, and the velocity of a feature measured at 5221 Angstroms would then be -852 km/s in perfect agreement with Figs. 4 and 5.
Either way, there is major disagreement between Fig. 3 (which I suspect is in error) and Figs. 4 and 5 (which could be reconciled with the above analysis if the redshift was 0.8676).
I'm sorry this is not very conclusive and I certainly wouldn't accept it without someone else verifying that the x-axis of Fig.3 appears to be problematic.


Is our central black hole actually at the CG of the galaxy? It seems fairly certain there is a large black hole, about 25,000 ly from Earth roughly in the center of our galaxy, in the direction of the Sag. A radio source.

• In fact, is it thought to be literally at the center of gravity of our galaxy?

Does we know, or is this uncertain?

• Indeed, do we actually know where the CG of the galaxy is?

I observe that the term "Galactic Center" seems to be an astronomical term. (Example.) Can an astronomer tell me: does that term mean what we believe to be the CG, or rather, is it a nominal mapping point?

• If the black hole IS exactly at the CG of the galaxy ... why?

It would seem to me a big BH could form any old place, and perhaps still be orbiting or oscillating the CG - right?  If the SBH is literally the CG of there galaxy, is there a reason for that?  (Like, "SBH can only form at the CG" .. or "stellar collisions can only happen at the CG" .. or whatever, those are only wild guesses.)

Or indeed, is the SBH the "main player" in the galaxy - just as the solar system formed around the sun (so it would be silly to ask "why is the sun at the center"), did the SBH sort of co-form with the galaxy (or something?)

In any event, is the SBH at the actual CG of our galaxy? Do we know?

In simple terms: yes, the Milky Way's supermassive black hole (SMBH) is at the center of the galaxy, we know approximately where the center is (but not terribly precisely), and we should expect the SMBH to be there.

You can define the Galactic Center using the orbits of stars and gas: what is the (average) center of their orbits? You can also use the spatial distribution of the same (which is probably a bit closer to what you're thinking of as a "center of gravity" or "center of mass"). This is never going to give you an exact answer, and since the galaxy is not perfectly symmetric or unchanging, there will always be some uncertainty and variation.

There's a 2D coordinate definition, which is the point on the sky where Galactic latitude and longitude both = 0. This is intended to approximate the direction toward the GC, but since it was defined back in the 1950s, no one is worried if modern attempts to measure the (3D) center of the galaxy might be slightly different.

As far as we can tell, yes, the SMBH is at the Galactic Center. If a SMBH did form outside the central region of a galaxy, dynamical friction would make it spiral into the center of the galaxy (and damp any radial orbital oscillations as well). This would also cause a SMBH from another (smaller) galaxy that merged with the Milky Way to spiral into the GC and merge with the SMBH there; see my answer to this question.

The SMBH is not the "main player" in our galaxy: it has a mass of about 4 million solar masses, which you can compare with the total stellar mass of about 50 billion solar masses. (And a dark-matter mass of roughly one trillion solar masses.) Nonetheless, there is evidence that SMBHs "co-form", in some sense, with their host galaxies. In particular, the mass of SMBHs is fairly tightly correlated with several properties of the host galaxy (e.g, the mass of the "bulge", or the velocity dispersion of stars in the inner part of the galaxy).

[Edited to fix a typo.]


How do we know how fast the surface of a neutron star is rotating? I understand that we can count pulses in time, but how do we know that the magnetosphere is rotating along with the surface of a neutron star?

Can we measure the rotation of a neutron star in any other way like red-blue shifts? It should be pretty difficult due to the small radius and big distance but is it possible? Is it already done?

How do we know how fast surface of neutron star is rotating?

I suppose we don't know the answer with total confidence, but I'd say we think we know with good confidence. Because we have robust evidence along with a model that matches that evidence.

  I understand that we can count pulses in time, but how do we know that magnetosphere is rotating along with surface of neutron star?

I suppose we don't actually know for sure. But I can't think of any process whereby a stationary spherical body can emit regular pulses of radiation.

  Can we measure rotation of a neutron star in any other way like red-blue shifts? It should be pretty difficult due to small radius and big distance but is it possible? Is it already done?

No, and no. At least as far as I know. But most of what I know came from Frozen Star by George Greenstein. The first half of the book is all about neutron stars. It dates from 1984 and it could be a little out of date. But as far as I know there's not much wrong with it, and it's a good read regardless. Note that you can buy a second hand copy fairly cheaply.


Extent of knowledge of potential future gamma ray bursts What do we know about the locations and directions of potential Gamma ray bursts that may intersect with Earth in the future, or are we mostly blind to when such an event directed our way may occur & cause a mass extinction event?
In other words, are we able to predict to any degree at all when one might occur close enough that there may be some potential impact to life or life on Earth or infrastructure?

Although not being an expert on the field, I found a multitude of studies dealing with events which may be used as indicator that a GBR is about to happen somewhere. Research on GBR precusor events range from smaller gamma ray precursors, over gravitiational waves, to thermal signatures, and neutrinos:

Sarp Akcay: Forecasting Gamma-Ray Bursts using Gravitational Waves (2018)
Thomas Koshut et. al: Gamma ray precursor activity as observed with BATSE (1995)
L. Piro et al.: Probing the Environment in Gamma-Ray Bursts: The Case of an X-Ray Precursor, ... (2005)
E. Troja,  S. Rosswog, and N. Gehrels: Precursors of short gamma-ray bursts (2010)
Frédéric Daigne, Robert Mochkovitch: The expected thermal precursors of gamma-ray bursts in the internal shock model (2002)
Coppin, Nick van Eijndhoven for the IceCube Collaboration: IceCube search for high-energy neutrinos produced in the precursor stages of gamma-ray bursts (2019)


If we pumped an large EMP into High Earth orbit and it detinated would it act as a shield against solar radiation? Could this pulse potentially shield spacecraft from harmful solar radiation.

I was assuming an EMP would have little effect as a radiation shield because the pulse would not be rapid enough to counter act the suns energy

f we pumped an EMP into high earth orbit and it detinated would it create a pocket of radiation-free space

I can think of 2 ways to create an EMP and while both involve explosives in both cases the EMP is an effect of the blast and not a detonation per se. Would it create a radiation free "bubble "? Yes but only briefly and hardly worth the trouble. In fact the first way an EMP is created is as a side effect of a nuclear explosion. Hardly worth the trouble then if the goal is to clear out a patch of space to then fill it full of fallout. Rapidly collapsing a strong magnetic field essentially by crushing the magnet in an explosion may also produce an EMP although less strong and your space cleared of radiation would be but for a fleeting moment and only potentially cleared of radiation susceptible to electromagnetic influence. If you think of the charged particles that you are trying to deflect as bullets and the electromagnetic lines of force as walls then your walls have to be strong enough to withstand the bullets or the particles  will blithely rip through the EMP like it wasn't even there.


Is it "weird" that all disc galaxies rotate once every billion years? In a recent paper (Cosmic clocks: A Tight Radius - Velocity Relationship for
HI-Selected Galaxies by Meurer, et al.), it was noted in the conclusion that:

  [This] implies a constant orbital time of ∼1 Gyr at [the outermost] radius [of disc galaxies].

Given what we knew about disc galaxies and dark matter up to this point, is this an unexpected conclusion? My sense is that it is strange that all disc galaxies have a constant rotation regardless of their size, but my intuition may be wrong. I know that dark matter speeds up the rotation of the outer parts of disc galaxies, but isn't it strange that they converge on this value?

It is not weird.

The comment is about a certain class of galaxy, and not just a class but a stage of galaxy formation and development within that class.

We start with a fairly constant distribution of matter within the universe.

Stars for when this matter clumps together. This clumping is going to have a fairly even distribution.

Stars then clump together to form structures that we identify as galaxies.

The distribution of this clumping is also going to have a fairly constant distribution.

This carries on up the chain to clusters of galaxies, and clusters of cluster and so on.

Going back to stars, we identify main sequence stars. These are stars that start of with a mass and composition that is very common and then have common life cycles. They take so long to burn up certain percentages of their hydrogen, so long burning combinations of hydrogen and helium and then so long going up the periodic table forming lighter elements and in the process becoming the different types of stars we define as being on the main sequence.

For galaxies it is similar.

Spiral galaxies need to be of a certain size and angular momentum to generate their signature spiral structure. They did not start out as spiral galaxies. And they will not remain spiral galaxies. They are simply how galaxies of a certain size and angular moments appear at certain stages in their development.

This means that the outer edge angular velocities of these types of galaxies (for a specific definition of their outer edge) are going to be similar.

For some information on the development of spiral galaxies and other galaxy types:

https://astronomy.swin.edu.au/cosmos/S/Spiral+Galaxy+Formation


How did DSCOVR capture the Moon's shadow on Earth during the August 21 2017 solar eclipse without the Moon itself also being visible? In this gif you can clearly see the Moon's shadow moving from the West Coast of the U.S. towards the East Coast during the August 21 2017 eclipse.
DSCOVR (the satellite that imaged this) is located at the L1 point, which is exactly between the Earth and the Sun, meaning it always sees the sun-lit part of Earth.
But for a Solar eclipse to happen, the Moon also has to be exactly between the Earth and the Sun, and thus it should be visible in this gif.
At first I thought it was because DSCOVR was closer to Earth than the Moon was, but it's actually more than 3 times farther away.
Geometry dictates the Moon should be between the Earth and DSCOVR since DSCOVR itself is between the Earth and the Sun, so why isn't the Moon visible?

DSCOVR is in a large Lissajous orbit around the Sun-Earth L1 point, about 1,500,000 kilometers from the Earth. At this point it is far from the line drawn between the Sun and Earth; if the Sun were behind you and you were looking at the Earth-Moon-DSCOVR system, you would say that the spacecraft is down and to the right by several hundred thousand kilometers.
I've used the Python package Skyfield plus a location for DSCOVR from JPL's Horizons to plot their positions. Earth is large and blue, the Moon is green (cheese) and smaller, and the very tiny black spot is DSCOVR. In each case I've annotated DSCOVR's position.
I've done some rotations so that my X-axis passes through the centers of the Sun and the Earth during the eclipse.
If I get a chance I'll make a 3D plot of DSCOVR's orbit. The script I used can be found in https://pastebin.com/AXsJM8J1


How is the Milky Way on a collision course with Andromeda when the universe is expanding? At some time in the future it is said our galaxy will collide with Andromeda. Yet at the same time the universe is expanding at a ever-accelerating pace. Using the old raisin-in-the-baking-loaf-of-bread analogy, the raisins grow farther apart and don’t “collide” with each other during the baking process. It seems to be the same way with galaxies and the expanding universe.

How are these two seemingly conflicting postulates reconciled? What am I missing here?

I'm not an expert, but I looked into this recently when toying with the idea of a scale model of the observable universe. To the best of my understanding, the following is approximately correct.

TLDR: The expansion happens on the very large scale of the universe. At the scale of the observable universe the Andromeda galaxy is a next-door neighbour, and at the local scale gravity is strong enough to override any effect of expansion.

The Milky Way is 100,000 light years across, 2.5 million light years brings us to the Andromeda galaxy.

Galaxies cluster in galaxy groups, and 10 million light years span about 50 galaxies and brings us to the edge of what's called the Local Group.

Galaxy groups cluster into superclusters. We're part of the Laniakea Supercluster, which spans about 100,000 galaxies and 500 million light years.

For any appreciable expansion (redshift 0.1) you need at least a billion light years. It's something that happens in the voids between superclusters, and doesn't become really pronounced until you're halfway across the observable universe (redshift 1, 7-8 billion light years as measured in light travel time).

For the "raisins in a loaf" analogy, I think it's not too wrong if you think of the raisins as superclusters rather than individual galaxies. Note that although the dough expands in a baking loaf, individual raisins do not, because there are other forces at work that hold them together.


Orbital elements: Difference between longitude of perihelion and argument of perifocus I am learning orbital elements and one thing isn't clear to me. When I look at the Horizon's ephemeris data for Mars, this gives me argument of perifocus equal to roughly 286.53 degrees on epoch J2000. When I look up its longitude of perihelion for the same epoch here, I get roughly 336.04 degrees. I thought these two parameters are the same thing. What is the difference between them? Why are the two numbers not the same?

I know the answer now. Turns out that longitude of perihelion ($\overline\omega$) is defined as the sum of longitude of ascending node ($\Omega$) on the ecliptic plane and argument of perifocus ($\omega$) on the orbital plane: ($\Omega$ + $\omega$). It's strange to add up two non-coplanar angles, even when the inclination between the two planes is very small, but that's what it is.


Sensitivity of calculated orbital elements to observational errors These days, we have some very precise ways of making measurements, but I'm sure it wasn't so in Kepler's day. So I am wondering how astronomers of that time could make such accurate determinations of planetary orbits, given the likely limitations of the available instrumentation. To put the question another way: How does each added (or lost) digit of precision in the measurement affect the accuracy of the calculated orbital elements? Is there significant compounding of errors due to the nature of the calculations, or is it essentially one-for-one? How much accuracy (to how many digits of precision) must the "three observations" be in order to determine the orbital elements to a given precision? How does that precision degrade over time? If, for example, we wanted to prepare an almanac for Neptune that will be accurate to say, one second of arc in our sky after 100 years?

Kepler used data from catalogs made by Tycho Brahe which had a varying level of precision and accuracy down to about 0.5 arcminutes for a measurement. However, since Kepler thought all systems should operate in the same way, he could use a larger sample verify his laws. Once you know what the relation should be, you can then map the observations to what you would expect. For an individual case, the precision of measuring orbital parameters depeneds on how much of the orbit you have actually observed.


Why aren't ground-based observatories using adaptive optics for visible wavelengths? Adaptive Optics (AO) techniques allow ground based observatories to dramatically improve resolution by actively compensating for the effects of Astronomical Seeing.

The atmospheric effects are quite variable in both time and location. A parameter called Isoplanatic Angle (IPA) is used to express the angular extent over-which a given wavefront correction optimized for one point (usually a guide star, artificial or natural) will be effective. As an example, Table 9.1 in this Giant Magellan Telescope resource shows values for IPA scaling almost linearly (actually: $\sim\lambda^{6/5}$)  from 176 arcseconds at a wavelength of 20 microns to only 4.2 arcseconds at 0.9 microns.

This suggests an IPA of 2 to 3 arcseconds for visible wavelengths, which taken by itself is not a killer limitation.

However, it seems almost all currently active AO work is done exclusively in various infrared wavelengths, apparently down to 0.9 microns but no further. (AO is also implemented computationally to array data in radioastronomy.)

Is this because the observed wavelength needs to be longer than the guide star monitoring wavelength? Because it is simply much harder and there is always Hubble above the atmosphere for visible work so it's not worth the extra effort, or is there another more fundamental reason?

I'm not looking for speculation or opinion, I'd like a quantitative explanation (if that applies) - hopefully with a link for further reading - thanks!

There's a pretty good discussion at this page.

There are several factors at work:

The smaller isoplanatic angle, as you note. This limits how much of the sky you can observe with AO, since your target needs to be within the isoplanatic angle of a bright enough references star. (Even with laser guide stars, there is still a need for a reference star for "tip/tilt" correction.) The difference in angular area on the sky means that the area of the sky that can theoretically be observed with AO will be about 20 times larger in the near-IR than in the optical, just from the difference in isoplanatic angle.
The effects of turbulence are stronger and have shorter timescales in the optical. This has three effects:

A. The corrective optics (e.g., deformable mirror) need to have more movable parts ("a near-perfect correction for an observation done in visible light (0.6 microns) with an 8-m telescope would require ~ 6400 actuators, whereas a similar performance at 2 microns needs only 250 actuators.") and need to operate on a faster timescale.

B. In addition to the electromechanical complexity, you'll have to do far more in the way of calculations to drive all those actuators, and on a shorter timescale. So the computing power required goes up.

C. In order to provide the inputs for the corrective computations, you need to observe the reference star on a much finer angular scale ("A large number of actuators requires a similarly large number of subapertures in the wavefront sensor, which means that for correction in the visible, the reference star should be ~ 25 times brighter than to correct in the infrared."). This limits how much of the sky you can do AO for even more: a star that might be bright enough in the near-IR to correct a region 20-30 arcsec wide isoplanatic patch won't be bright enough to correct the corresponding 5-arcsec-wide isoplanatic patch in the visible.
In order to make corrections, you need to observe the reference object in the optical. This is easy to do with a near-IR setup using an optical/IR beamsplitter: send the optical light to the AO equipment and send the near-IR light to the near-IR instrument. In the optical, you use an optical beamsplitter to send half the light to the instrument and the the other half to the AO equipment. This means that the AO equipment only gets half the light it would if it were used with a near-IR instrument, which makes it (even) harder to do the corrections.

Finally, there is an issue unrelated to the AO itself, which is that you need different science instruments depending on whether you're working in the optical or the near-IR. Optical instruments use silicon CCDs for detection; these are only sensitive out to about 0.9-1 microns. Near-IR instruments use different detectors (usually HgCdTe-based), which are good from about 1-3 microns. (Near-IR instrument also need a different design to reduce contamination from thermal emission from the telescope and optics for observations at wavelengths longer than 2 microns.) So in practice the choice has been: combine AO with a near-IR instrument and get good performance with affordable/feasible technology, or combine AO with an optical instrument and get very limited performance with more expensive (or even, until recently, unattainable) technology.

Nonetheless, there are some optical AO systems starting to appear, such as MagAO on the Magellan telescope (which has both an optical instrument and a near-IR instrument, and can correct for both simultaneously).