For colonization purposes, what is so good about Titan? I have read that Titan is the best candidate for humans to colonize in our Solar System.
However it has no water, it's atmosphere is not breathable and it is -290° F on the surface. It doesn't sound too hospitable to me. Have they forgotten that humans can't live without water?
If this is the best candidate then I think that humans are doomed. So why is Titan the best option for humans to colonize?

There is nowhere in the solar system except for parts of Earth where humans have any hope of surviving without heavy technological support. There is next to no free oxygen anywhere, nowhere has a surface whose temperature remains in a range we can tolerate (generously let's say 230-320K) and nowhere has liquid water on the surface. So, at a minimum, we will need a sealed habitat and some form of protective suit when we travel outside it.
Terraforming -- transforming the environment of some planet to be close enough to Earth's to allow us to live without protection is much discussed, usually for Mars or Venus. Even the most optimistic detailed proposals I have seen suggest it would take centuries or millenia to complete.
Even more speculative is the idea of transforming humans to be able to live on other planets relatively unchanged. Although popular with SF, at the moment I don't think we really have the understanding to say anything specific about this.
So for short-term colonisation, we are left to decide which planet would present the least difficult technological challenges for hi-tech base.  There are a few somewhat serious contenders:

Mars -- relatively close, has hydrogen, carbon and oxygen accessible on the surface (as ice and CO2), atmosphere protects against micro-meteorites and some radiation, surface temperatures not too far from habitable.
Venus (upper atmosphere) -- pressure and temperature are both reasonable; fairly close; limited hydrogen availability; no easy supply of metals; corrosive atmosphere; good micrometeorite and radiation protection; not much to do
Ceres -- very low gravity, no atmosphere, plentiful water ice, easy access to space, easy access to metals etc (from other asteroids), good for supporting space activity
Moon -- very close, no atmosphere, almost no hydrogen, enough of a gravity well to make landing and taking off non-trivial
Titan -- rather remote, plenty of oxygen, hydrogen, carbon and nitrogen, atmosphere protects against radiation, comfortable pressure -- you'd need a heated dry suit or equivalent, lots of interesting stuff in the neighbourhood, cold.
Europa -- plenty of oxygen and hydrogen, lots of radiation on the surface, deep in Jupiter's gravity well and no atmosphere for aerobraking, so not very appealing on the surface, but a "submarine" colony under the ice might be viable, the ice will provide radiation shielding and some warmth and you might be able to swim outside in just a dry suit and breathing gear.

You pays (a lot of) your money and you takes your choice.


How long after the Big Bang would CMB radiation have been between 273 and 373 K? Cosmic microwave background radiation is 2.7 K today. When would it have been between 0 and 100 C?

The temperature of the cosmic microwave background scales as the inverse of the cosmic scale factor $a$. i.e. When everything was at half the separation it is now, then the CMB was twice the (absolute) temperature. The scale factor in turn is reciprocally related to the redshift by $a/a_0 = (1 +z)^{-1}$, where $z$ is the redshift and $a_0$ is the present-day scale factor, usually taken to be 1.

From this we see that
$$T(z) = T_0 (1+z),$$
where $T(z)$ is the temperature at redshift $z$ at some time in the past and  $T_0 = 2.73$K is the temperature of the CMB now. For the range of temperatures you specify requires redshifts in the range $99< z< 136$.

Getting from a redshift to a time in the universe since the big bang is not so simple. The answer depends on what you take to be the "cosmological parameters" - i.e. the values of the cosmic matter density, the dark energy density and so on.

However, we can avoid looking "under the hood" and use a cosmology calculator. The one I have linked to is for a "flat" universe and has default values for what are currently good estimates for the matter density and present-day Hubble parameter. For the range of $z$ I found above, this corresponds to an age of between 10.8 and 17.3 million years after the big bang.

I guess you are perhaps thinking about these ideas that life could have been around about 15 million years after the big bang.


What is the future of our universe? Is the universe heading towards a Big Freeze, a Big Rip, a Big Crunch or a Big Bounce? Or is it part of an infinitely recurring cyclic model?

What is the future of our universe?

Like StephenG said, nobody knows for sure. But we do have confidence that the universe is expanding, and we also have confidence that the expansion is speeding up. So extrapolating from that, the future looks cold and lonely and bleak. A bit like life for the older generation!

  Is the universe heading towards a Big Freeze, a Big Rip, a Big Crunch or a Big Bounce? Or is it part of an infinitely recurring cyclic model?

I'd say the Big Crunch and the Big Bounce are out. The universe didn't contract when it was small and dense. Instead it expanded, and that expansion is increasing. So it looks like we're in for a Big Freeze. However I wouldn't rule out a Big Rip of sorts. Have a look at page 5 of this paper where Milgrom mentions the strength of space. Then think of the balloon analogy, but make it a bubble-gum balloon, in vacuum.

Image courtesy of the one-minute astronomer.

The skin gets thinner and the balloon expands, then the skin gets even weaker, so the balloon expands even faster, and so on. Bubble-gum bubbles sometimes end badly, and there's something about  this article by Phil Plait that strikes a chord.

  Or is it part of an infinitely recurring cyclic model?

I don't know. I have no evidence to suggest that there's any kind of recycling going on, and I can't think of mechanism by which this might occur. I have of course read about "conformal cyclic cosmology", but I just don't buy it, along with other stuff from Penrose.


Does a gravitational wave loses energy over distance? How does a gravitational wave travel over vast distances without losing its energy? Could they go on forever as long as the Space?

Thank you

tl;dr: Yes. Feynman's beads on a string argument

The other answers skirt what I think is the issue that the OP is asking about. In a lossless medium a spherical wave packet itself, caused by a disturbance, will not "loose energy" itself. If you integrate over a large volume you get a constant energy versus time. Of course the flux per unit area will decrease by the same rate that the area increases, but that's a trivial result.

But, when gravitational waves pass matter, they really do deposit energy in the matter, and thereby become attenuated. The effect is very very tiny for a given amount of matter, but it is non-zero and the universe is pretty big.

In this answer to the question Transfer of energy from gravity back to other “more familiar” forms of energy? I've said:

  This is from the Wikipedia article mentioned.
  I think this is really worth reading, as it not only describes some of the physics associated with the accepted answer and it's citation of Feynman's argument about a bead a stick, but perhaps gives a little insight into the development of the field as well.

    Feynman's argument

    Later in the Chapel Hill conference, Richard Feynman — who had insisted on registering under a pseudonym to express his disdain for the contemporary state of gravitational physics — used Pirani's description to point out that a passing gravitational wave should in principle cause a bead on a stick (oriented transversely to the direction of propagation of the wave) to slide back and forth, thus heating the bead and the stick by friction. This heating, said Feynman, showed that the wave did indeed impart energy to the bead and stick system, so it must indeed transport energy, contrary to the view expressed in 1955 by Rosen.

    In two 1957 papers, Bondi and (separately) Joseph Weber and John Archibald Wheeler used this bead argument to present detailed refutations of Rosen's argument.(5)(6)

    (5) Bondi, Hermann (1957). "Plane gravitational waves in general relativity". Nature 179 (4569): 1072–1073. Bibcode:1957Natur.179.1072B. doi:10.1038/1791072a0.

    (6) Weber, Joseph & Wheeler, John Archibald (1957). "Reality of the cylindrical gravitational waves of Einstein and Rosen". Rev. Mod. Phys. 29 (3): 509–515. Bibcode:1957RvMP...29..509W. doi:10.1103/RevModPhys.29.509.


What is the smallest diameter black hole that can be formed naturally? I was watching the video I Put a 1mm Size Black Hole Next to Earth and This Happened - Universe Sandbox 2 discussing what would happen if a 1 millimeter black hole is orbiting the earth, and I began questioning myself, is it really possible to naturally have such a small black hole?

What is the smallest size that a black hole can form naturally in real life?

The mass at which a star is able to condense into a black hole upon its death is called the Tolman–Oppenheimer–Volkoff Limit, which is roughly 2.17 solar masses. At this mass, the Schwarzschild radius of the black hole (the size of its event horizon) is about 6410 meters by the equation

$$r_s = \frac{2GM}{c^2}.$$

EDIT: I'd also like to add that the theoretical minimum mass for a black hole is the Planck mass, $2.2\times 10^{−8}$ kg, which would yield a radius of $3.267\times 10^{-35}$ meters, though these would not be formed by stellar collapse, though they could have formed as primordial black holes in the very early universe.


Clearing the Neighbourhood: Definition of 'Orbital Region'? I am trying to determine how wide a 'neighbourhood' a planet is expected to clear.

However, the Wikipedia article on the matter provides no clarity, simply stating that a planet is expected to clear it's 'orbital region' with no definition provided.

I suspect that the correct answer is that a planet is expected to clear a region around it's orbit equal to the radius of its Hill Sphere, but I am not certain of it.

Furthermore, for a planet of Earth mass, the Hill Sphere radius is a measly 0.01AU, which doesn't seem like much of a neighbourhood!

Is there a formal way to define the 'neighbourhood'?

The IAU gives no definition, it leaves the term somewhat vague.

Various calculations of the "planetariness" of solar system bodies use slightly different notions of the "neighbourhood"

Soter's $\mu$ considers bodies to be in the neighbourhood if they share a common radial distance (ie the orbits cross each other) and the orbital periods differ by less than an order of magnitude

Margot's $\Pi$ uses $2\sqrt 3$ times the Hill radius.

Stern and Levinson's $\Lambda$ in the original paper consider the neighbourhood to be bodies that cross the orbit of the planet, while not in an orbital resonance, to be in its neighbourhood, in a somewhat complex way that requires knowledge of the relative inclinations and and orbital velocities of bodies that cross orbits.

Changing the details of these values doesn't change the conclusion of the calculations: There is a clear and measurable difference between bodies like Mars and Neptune that are (to use Stern and Levinson's term) uberplanets and Ceres and Pluto that are unterplanets (or dwarf planets as termed by the IAU).


Advancement of perihelion, data I was looking for accurate and up-to-date data regarding the advancement of the perihelion of the planets in the Solar System, the only ones I've found so far are at least 90 years old bout I really doubt that no progress has been made so far in this field... does anyone know where to find newer ones?

In addition to this, can anybody make an estimation about the difference between the Moon-Earth system precession around the Sun and a Moonless Earth system? It should be very small, too small to be detected maybe...

Thank you in advance!

I might be misunderstanding your question, but using HORIZONS (http://ssd.jpl.nasa.gov/horizons.cgi) with these settings should give you the data you need:


Halley's Comet duration of visibility When Halley's Comet is visible in the sky, how long does that last? Is it only visible for a week or is it up there for a month, or several months? What I'm actually wondering is, though it arrived in 1910, could it still have been visible in the sky into 1911?

Typically it was visible a few weeks only, about 3...6 months at most. There is an excellent summary in Broughton (1979). In 1910 the peak visibility was in April, the last recorded sighting of Halley without instrumentation on 13th June.
Yeonan (1986) gives an overview including instrumentation; while it has positional and brightness data for dates in 1911, they are for a magnitude > 10, thus far beyond being visible with the naked eye.
So no, in 1911 Halley was not visible anymore without then bleeding-edge astronomical telescopes.


Strange light in the sky A few weeks ago my daughter phoned me to say she was looking at a very large bright object in the sky in the direction of the Preseli hills. She wondered if it was a planet. I went out to have a look and sure enough this was much brighter and much larger than a star or planet. I looked at it for several minutes and quite suddenly it just, for want of a better phrase, went out. Just completely gone.

Last night I saw something very similar. Again it was much larger than a star but this time appeared yellowish in colour. I kept looking as I was doing a few small outdoor jobs and then, although I didn't see it happen this time, it suddenly disappeared. I am mystified as to what these lights could have been. Looked on internet and hoped someone else may have seen them but no luck.

If you have a date and time you can go to Flightradar24 and recreate all of the planes that flew over the area. An oncoming large body jet can look a lot like a yellow star if it has its "headlights" turned on, and if it is coming in your direction and descending, it can look surprisingly motionless. Since the light is focused forward, it can suddenly dim if the plane changes direction.

The lovely NATS video A Day Over the UK shows a substantial amount of East/West air traffic in that area, including several flights off of the main corridor.

However if you watched it for 20 minutes, this can obviously be ruled out.

Venus and Mars were too close to the horizon at 21:00 GMT in the UK, but I recommend you play with in-the-sky.org a bit. Choose Planetarium. It should know your approximate location based on your internet access so as long as the location is roughly correct the only think you need to do is choose a date and time to recreate, and enjoy! I hear the site works on phones nicely as well.

above: Screen shot fro in-the-sky.org.


false to stable vacuum transition if the universe transitioned from a meta-stable vacuum to a stable vacuum we assume a universal death scenario. However since it cannot propagate faster than the speed of light, from any particular starting point, the universal expansion has crossed an intrinsic "event horizon" so how can this actually happen in any conceivable time scale?

also - cheeky extra question - could energy rich galaxies dislocate in some way from energy poor regions of spacetime

The standard picture of vacuum decay events is very much like nucleation in a phase transition: it starts somewhere, and then expands at a constant speed - lightspeed in this case. So the future lightcone of the nucleation event will be true vacuum.

In an accelerating expanding universe like our current $\Lambda$CDM cosmology future lightcones actually have a finite radius as measured by co-moving coordinates (e.g. from the perspective of matter moving passively with the expansion of the universe). Beyond a certain (current) distance there are galaxies that will never be reached by a light signal we send now - when it gets to the distance they currently have they will have moved much further away, and it never manages to close the gap. Same thing for vacuum decay: the front will always expand outward at lightspeed, but large parts of the observable universe will expand away so fast that it will never catch up with them.

In about 100-200 billion years galactic clusters will become separated by exponentially growing gulfs and quickly lose contact (they stay the same size because they are gravitationally bound). They become separate "island universes". Beyond this point vacuum decay would just be a local issue to one island universe.

Extra question: if a galaxy has too much energy it will implode as a black hole - mass and energy are equivalent in relativity. Whether that counts as dislocation is unclear. But overall regions with more matter density tend to contract while less dense regions expand more, which is why we get the cosmic web of sheets and filiaments of dark matter plus galaxies, and expanding voids. Galaxies in the voids (they do exist) could be described as energy poor (even though their stars are doing just fine) and they would experience everything moving away from them not just because of universal expansion but also gravitational contraction towards remote masses.


How does the center of a black hole can contain so much mass? We know that

  "In the center of a black hole is a gravitational singularity, a
  one-dimensional point which contains a huge mass in an infinitely
  small space, where density and gravity become infinite and space-time
  curves infinitely, and where the laws of physics as we know them cease
  to operate."

So where does this mass go? A measure of how much matter is there in a object is called mass right? so how does that matter stays in a 1-D space?

We know that matter occupy space so how?

Edit: My question is about how can a singularity exists not about singularity itself.

The notion of "centre of the black hole" is misleading. It treats the spacetime in and around the black hole as if it were normal flat spacetime, with no gravitational distortion.  In fact the spacetime is highly curved. So curved, in fact, that "space" behaves more like "time".  Instead of "where is the singularity" you should ask "when is the singularity."

The answer to this question is always "in the future".  There are no singularities in our past.  This is significant, since when we say "there is a pen on the table" we mean that the pen was there a few nano-seconds ago, because of the time light takes to reach us. Everything that we say "exists" actually exists in our past.  Singularities are a "when" that have a "before" but don't have an "after". Nothing can have a singularity in its past, because there is no future after the singularity.

So in this sense, singularities don't exist. We can't study them because they are always in the future (due to the extreme bending of spacetime). If we fall into a black hole we will (rapidly) reach the singularity but there is no way to report back about it, since the singularity is surrounded by an event horizon. Some of us may (if we are unlucky) fall into a black hole. For these people the singularity is in their future. But it is not possible for a singularity to be observed, as it can never be in the past of any timeline.

So a black hole behaves "as if" all the mass was concentrated in one point. But don't think of the singularity as a "thing" in the centre of the black hole, but think of it as a "when", and always remember "this is weird, and analogies and metaphor are never a complete explanation".


How are the three character IAU Minor Planet Center's Observatory Codes assigned? Why are some letters so popular? Answers to Latitude, longitude and altitude of Ckoirama Observatory; where can things like this be looked-up? point to https://www.minorplanetcenter.net/iau/lists/ObsCodesF.html which currently lists 2198 observatories.

The first character of the three character codes has 36 possible alphanumeric values. and the second two are only digits. That allows for 36*10*10 = 3600 possible codes.

I plotted the ones that are currently used and the pattern is interesting. Why are some letter codes fully used (all 100 two digit suffixes are populated) whereas others are completely empty? I haven't noticed an alphabetical associations.

Y = 0 to 9 are the digits and 10 to 36 are A through Z: Shading means the code is used.

For some reason I can't stop comparing the data to the famous Arecibo message, a tiny 1-bit bitmapped image beamed to space containing DNA, amino acids, our solar system and other goodies:

import numpy as np
import matplotlib.pyplot as plt
# blob is the unformatted lines from https://www.minorplanetcenter.net/iau/lists/ObsCodes.html
threes = [line[:3] for line in blob.splitlines()]
key = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
a = [[x] for x in key]
for t in threes:
    a[key.find(t[0])].append(t)
b = np.zeros((36, 100))
for i, thing in enumerate(a):
    c = [int(x[1:]) for x in thing[1:]]
    for d in c:
        b[i, d] = 1
plt.imshow(-b, interpolation='nearest', cmap='gray')
plt.gca().set_aspect(2)
plt.show()
pairs = []
todegs = 180/np.pi
for line in blob.splitlines():
    try:
        cos, sin = [float(x) for x in (line[13:21], line[21:30])]
        lat = todegs * np.arctan2(sin, cos)
        lon = float(line[4:13])
        pairs.append([lon, lat])
    except:
        pass
print(len(pairs), len(blob.splitlines()), float(len(pairs))/len(blob.splitlines()) )
lon, lat = np.array(list(zip(*pairs)))
lon = np.mod(lon+180, 360) - 180
plt.figure()
plt.plot(lon, lat, 'ok', markersize=1)
plt.ylim(-90, 90)
plt.xlim(-180, 180)
plt.gca().set_aspect('equal')
plt.show()

I asked the at the minor planet center how the codes were decided, and the answer was

  Historically, the observatory codes were assigned ascending by longitude toward east (from prime meridian): 360 degrees were divided by numbers. When three digit numerical codes were not sufficient, letters plus two numbers were used again in bands toward the east. Some bands are already full (e.g. H**). New observatory code is assigned in the available letter+number band (based on its longitude).


Is it possible to observe Polaris and the Southern Cross from the equator during an equinox Is it possible for an observer at Earth's equator to simultaneously observe Polaris and the Southern Cross during an equinox?

The March equinox would be a good time to see both; the September equinox would not.
Crux
lies roughly between RA 12h to 13h, Dec -55$^\circ$ to -65$^\circ$.
For best results, be at latitude 10-15$^\circ$ N so that both Crux and Polaris are well above the horizon, and look between 11:00 and 14:00 local
sidereal time
so that Crux is near the meridian.
That would be around midnight in March-April, late evening in May, wee hours in February, or daytime in July-December.


Equations of motion for Earth and Moon One of the simplest versions of Newton's equations of motion for the Sun, Earth and Moon can be obtained by making the approximation that the three bodies are perfect spheres. In this approximation, the pair potential between body i and j is  $$\frac{-G m_i m_j}{ r_{ij}},$$ where $r_{ij}$ is the distance between the bodies: one can solve the equations numerically and get the orbits.

This simple model does not include multiple, additional factors, such as

the fact that the Earth is not perfectly spherical (equatorial bulge),
Earth's tides,
the influence of other celestial bodies,

and others.

Amongst all these factors, what is the dominant one as for predicting the orbit of the Earth? In other words, what is the leading factor that should be included in the simple model above in order make its predictions for the Earth's orbit closer to the observations?

I may have found a partial answer to my question in the Navigational Astronomy chapter of 'American Practical Navigator' by N. Bowditch . On page 228

  [...] nutation, an irregularity in the motion of the Earth due to the
  disturbing effect of other celestial bodies, principally the Moon

and on page 230

  The precession of the Earth’s axis is the result of gravitational
  forces exerted principally by the Sun and Moon on the Earth’s
  equatorial bulge. The spinning Earth responds to these forces in the
  manner of a gyroscope. Regression of the nodes introduces certain
  irregularities known as nutation in the precessional motion.

So the dominant effects as for predicting the orientation of the Earth's axis seem to be the Earth's ellipticity combined with the attraction of the Sun and Moon.


Why does Jupiter have so many moons? Jupiter has a great many moons - in the hundreds, and they're still being discovered.

What is the current theory for where all these moons came from? Are they rocks flying through space captured by Jupiter's gravity?

Mass.

The more massive a body, the larger the gap between its lowest and highest orbit; the range of speeds at which a random body entering its gravity is likely to remain as its satellite. Sun has millions of satellites if you count all the asteroids; smaller planets tend to have one or two moons at most (Pluto with five being a notable and not fully explained exception)

To a lesser degree there's a matter of shape too. A regularly round body will have more regular and stable orbit than a potato-shaped one. Jupiter, being a gas giant is perfectly round. This doesn't play that much of a role though, especially with higher orbits.

And last but not least, no destabilizing influence of other bodies. It's very hard to maintain a lunar orbit - artificial satellites around our Moon last only a couple years each, because relatively close neighborhood of Earth tends to destabilize orbit of anything orbiting the Moon. Jupiter being a single massive planet with relatively tiny (relatively to its mass) moons doesn't have them influence each other all that much.


What are the color progression of a typical sunrise? "Dawn simulators" are used to simulate a natural sunrise, or at least the physiological effects of one.

What temporal sequence of color values (e.g., hue, saturation, brightness, and timing) corresponds to a typical spring or summer sunrise? Are the methods for calculating changes in these values based on season?

These steps are in there for credit

I used Stellarium to simulate a sunrise, and took screenshots of the progression. Each screen shot was 2 seconds apart, and stellarium was accelerated to 10 minutes a second.

Then, using imagemagik to crop batch those 13 images with this command:

mogrify -crop 1280 1024 *.png

(those numbers are the size of my screen. I cropped to get rid of my second screen)

Next I put the images into Color Thief and got the pallet of each image. This is the result of those 13 images:

Those are the colour pallets. You can pick whatever colours you want from them with a colour picker, such as KColorChooser.


Doesn't the energy conservation law imply, that the universe can't come to a Big Freeze? .. because, how would all the current matter and energy manifest then? Isn't everything coming to complete stop, minimal temperature, zero interaction state impossible? Wouldn't there still be some rare minor events occuring? And if so, given infinite amount of time couldn't these result in events resulting in some kind of chain-reaction?

The current model ($\Lambda$CDM) predicts that space will expand exponentially, and keep on doing it indefinitely far into the future. The end result, after the stars burn out, galaxies disperse, protons decay and black holes evaporate is a very thin soup of stable particles (see Adams & Laughlin 1997 or their popular book The Five Ages of the Universe).

Since each stable particle finds itself separated by exponentially growing distances due to the expanding space they will not be able to interact. Their density scales as $\rho(t)\sim \rho_0/a(t)^3 = \rho_0 e^{-3t/H_0}$. The probability of a collision over time $dt$ for a particle is roughly $\sigma v \rho(t) dt$, which if we integrate to infinity becomes $\Pr[\text{any  collision}]=\sigma v \rho_0 \int_0^\infty e^{-3t/H_0} dt = \sigma v \rho_0 H_0/3 $ which is finite - there will only be a finite number of interactions for each particle, ever.

At this point temperatures are extremely low since all background radiation has been redshifted to undetectable levels, and only the very cold horizon radiation remains ($\approx 10^{-30}$ K). Particles end up in a equilibrium state with this heat bath, and this is the real heat death of the universe.

This low temperature is still enough to rarely excite the particles, or generate new particles... or Boltzmann brains or entire worlds. Maybe. Eternity is a very long time, so low-probability processes will happen if they are allowed. They will however be so rare that the effects of one will typically dissipate long before anything else happens.


Rewriting the equation for the virial mass of a cluster I have a question that I don't know how to solve, any help is appreciated.

Here it is:

Rewrite the equation for the virial mass of a cluster so that if velocities are entered in km/sec and dimensions in Mpc, the resulting mass is in solar mass

Well the equation is: $M=\frac{5(v_r)^2R}{G}$

I think I have to use dimensional analysis but I really don't know how to get there

$$[M]=\mathrm{\frac{(\frac{km}{s})^2 km}{\frac{m^3}{kg \cdot s^2}}}$$

with one $\mathrm{mpc}$ being $3.09 \times 10^{19} \ \mathrm{km}$

Thanks in advance :)

This is something that crops up a lot in Astrophysics and Astronomy, so here is the general approach.

Get your equation into SI units. Thus
$$M = \frac{5v^2R}{G} = 7.5\times10^{10} v^2 R,$$
with $M$ in kg, $v$ in m/s and $R$ in m.

Now if you wish to replace $v^2$ by $(v/1000)^2$ (which is velocity expressed in km/s), you would simultaneously multiply the numerical constant at the front by $1000^2$. Similarly, to replace $R$ by $R/3.09\times 10^{22}$ (which is $R$ expressed in Mpc), then you have to multiply the numerical constant by $3.09\times10^{22}$ to leave everything unchanged. This then leaves your result still expressed in kg. To express it in solar masses, simply divide by a solar mass of $2\times 10^{30}$ kg.

Thus
$$\frac{M}{M_{\odot}} = \frac{7.5\times 10^{10}\times 10^6 \times 3.09\times10^{22}}{2.0\times 10^{30}} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$
$$\frac{M}{M_{\odot}} =  1.16\times10^{9} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$

Finally, check your result makes sense. Galaxy clusters are of order a Mpc in size, have velocity dispersions of hundreds of km/s and consists of hundreds of galaxies with total mass of $\sim 10^{14} M_{\odot}$ or so. This seems to check out - usually if a mistake has been made you will be many, many orders of magnitude out.


How could we search for life in TRAPPIST-1 It is possible to search for life from here, how?
What methods could be used?
It is possible to look for organic matter?
Could we know if there is life according to the components of the planets atmosphere?

As far as I know, and aside from direct communication, there's only one way to "find life" on another planet. That method is to look for molecular oxygen in the atmosphere. Molecular oxygen hates existing by itself and loves to combine with other atoms and molecules through oxidation. The Earth's atmosphere is currently ~30% $\mathrm{O_2}$, but that's only because we have a huge amount of plant life (primarily algae) constantly producing it. If you removed all life from the Earth, the oxygen in the atmosphere would quickly oxidize out and you'd be left with an atmosphere primarily of nitrogen and a few smaller constituents.

What this means is that if we do see a substantial amount of oxygen on a planet, the best bet is that that oxygen was artificially produced by life. It's not ironclad evidence, but its a very strong indicator. We can search for oxygen through a few techniques that already exist. They're still being developed, but the technology is very quickly coming to fruition. Ultimately, you'd need to look at the spectrum of the planet's atmosphere. This can be achieved by either looking at the star's light as it passes through the planet's atmosphere, or else observing at numerous wavelengths and seeing how the transiting planet's radius changes (due to changing atmospheric opacity at different wavelengths).


What are the raisons d'être for the Large Binocular Telescope "binocularity"? edit: The short form of the question is Why are there TWO telescopes?? A longer version follows.

The Large Binocular Telescope (and LBT blog as source of images below) has two 8.4 meter diameter mirrors, allowing for the possibility of interference in one dimension.

I'm aware that a single baseline, two-element stellar interferometer can be used to measure the separation between a pair of stars by looking at the Fourier transform of the intensity vs time of the interference. A single frequency peak is produced by one star, but if there are two close together, there will be "beating" in the time series that shows up as two peaks in the FT.

But I'm wondering if the LBT was designed and built with a plan to do more than that. Just for example, did plans include an ability to use interferometry to help reconstruct 2D images, or enhance the resolution beyond that which is already provided by the adaptive optics used in each half separately? Were there other plans or justifications?

Possibly helpful; Current Status of the Facility Instrumentation Suite at The Large Binocular Telescope Observatory

Open image in new window for full size:

I asked myself the same question and back then, I found the reasoning on the German Wikipedia page pretty comprehensive. I am translating, summarizing and expanding the corresponding section in the following:
By using two mirrors instead of a single one has a couple of advantages:

The double telescope collects the same amount of light as an 11.8-metre mirror, but is from an engineering point of view easier to built than a single large mirror, because the center of mass is outside the telescope(s). An example structural problems of large moving telescope was e.g. the failure of the Green Bank Telescope in November 1988. In radio astronomy, the approach of having multi-antennas on a single moveable structure is used for long, e.g. in the Pluton Complex
The LBT's optical resolution can be increased by interferometry to that of a 22.8-metre mirror.
Interferometry will also be used to virtually block out the light from central stars in order to make their possible planets directly visible, sometimes called nulling interferometer.

Last but not least, I remember hearing that LBT's mirror size is the largest feasible non-segmented mirror. This hypothesis is somewhat supported if you look at the list of largest optical telescopes.


In which two constellations, were the two gravitational wave events of 2016? As best as is known, what is the latest thinking on in the issue of,

in which constellation was GW150914, and in which constellation was GW151226?

(That is to say, within the large uncertainty, as best as is known in what direction is the likely center of that uncertainty?)

The data is indeed very uncertain. As for GW150914, LIGO posted this image of the probable area of the sky:

The most probable banana-shaped region stretches from Crux to the Large Magellanic cloud. Comparing the image with a constellation map, there are several small constellations that may have hosted the event. Those are Mensa, Chamaeleon, Musca and Crux.


What implications does younger volcanism have on the Moon's thermal evolution? In the paper Evidence for basaltic volcanism on the Moon within the past 100 million years (Braden et al. 2014), suggest that features such as Ina (image below) represent a far more recent age of volcanism (relatively speaking, around 100 million years).

Image source: NASA Science News, where they have a different non-volcanic theory.  However, this question is not about whether the features are volcanic or not.

From the Braden et al. article abstract:

  The morphology of the features is also consistent with small basaltic eruptions that occurred significantly after the established cessation of lunar mare basaltic volcanism. Such late-stage eruptions suggest a long decline of lunar volcanism and constrain models of the Moon’s thermal evolution.

Assuming the latest models indicate volcanism as recent as within the 100 million years ago, what are the implications for lunar thermal history of younger volcanism?

A younger lunar thermal history implies that the Moon was still undergoing a contraction and cooling process, which was initiated when it began accreting material together. Evidence has shown that the moon has no tectonic activity, so all geologic features are volcanic.

The mare are a more recent example of this activity, and have been shown to have fairly young ages, as you indicated. The moon still has some internal heat from its formation that may be leaking out through convection and cooling.

Even though the moon has a relatively small mass, it didn't lose its internal heat as quickly as the earth because there was no tectonic activity to facilitate it fast enough, and convection is a slow process.

As the moon cools, it contracts, creating sharp ridges and cracks in the surface (see features known as 'scarps', or 'garden'). These contractions contribute to the "deposit" of material formed in the images that paper discussed. As the cooling surface shrinks, it breaks the crust through a thrust fault, which pushes some material higher and shoves other material lower.

What may appear to be river channels on the moon are also lava flows. These were initially underground lava channels whose roofs collapsed to expose these subterranean features. More recent incidents of molten material rising to the surface and cooling can contribute to some of the younger ages recorded in the basaltic rocks.

Tldr; Evidence of younger volcanic activity confirms existing theories for the contraction and cooling process of the moon, just that it didn't end as soon as we initially thought


Can I see Harvest moon in India? I just saw articles which say the Harvest moon will be visible in the US.

Apart from the US, in which other country can people see it? Can I see it from India?

"Harvest moon" just means the full moon at around the time of the equinox on September 21.

The full moon is low in the sky during summer (June and July in the Northern Hemisphere) and high during the winter.  So September is a month when the full moon is much higher in the sky than the moon in August. In the past this allowed harvests to continue after dark.

The full moon in September can be seen all over the world, but it may not have the same cultural significance that it had in Europe and North America.


What do "star noise" and "sky noise" refer to for TESS (Transiting Exoplanet Survey Satellite)? The eoPortal article for TESS (Transiting Exoplanet Survey Satellite) gives a fairly in-depth summary of the space telescope design and planned operation. Some further background information on its observation mode and orbit can be found in the video linked below.

Based on the caption (also shown below), I'm guessing that "star noise" is possibly related to shot noise; the counting statistics of the incoming photons or perhaps the photoelectrons thereby produced, and "sky noise" is the cumulative effect of all the other actual light present in the spatially binned area of a given star, including Zodiacal light within the solar system as well as all other sources.

If so, and even if not, I don't really understand the units of $\sigma$ as parts per million per square-root hours, and why sky noise would vary as $I^{-1}$ while star noise expressed in these units varies as $I^{-1/2}$, where $I$ is the intensity of the light from a given star.

below: Figure 12: Top: Expected 1σ photometric precision as a function of stellar apparent magnitude in the IC band. Contributions are from photon-counting noise from the target star and background (zodiacal light and unresolved stars), detector read noise (10 e-), and an assumed 60 ppm of incorrigible noise on hourly timescales. Bottom: The number of pixels in the photometric aperture that optimizes the signal-to-noise ratio (image credit: TESS Team)

Excellent background on TESS: https://youtu.be/mpViVEO-ymc

The key formula for signal-to-noise calculations in photometry can be written as something like
$${\rm SNR} = \frac{S_{\rm star}}{\sqrt{S_{\rm star} + S_{\rm sky} + \sigma_R^{2}}},$$
where $S_{\rm star}$ is the photons counted from your target star, $S_{\rm sky}$ is the amount of unrelated photons in your photometry aperture (due to sky or other stars) and $\sigma_R$ is the readout noise in your detector.

The numerator is of course your (interesting) signal and the denominator is the noise. The reason the noise terms are added in the way they are is that each term inside the square root is a variance. i.e. The variance of the detected signal from the star is equal to the number of detected photons (a property of Poissonian counting statistics).

Now for more complexity we can express the signals as being photons per unit time (I'll use a small $s$ for these), such that $S_{\rm star} = s_{\rm star} t$, where $t$ is the integration time.
$${\rm SNR} = \frac{s_{\rm star} t}{\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}}},$$

Now the thing defined as $\sigma$ in your plots is the inverse of the SNR, plotted on a logarithmic axis.
$$\sigma = (s_{\rm star }t)^{-1}\sqrt{s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2}} $$
$$\log \sigma = -\log s_{\rm star} -\log t +0.5 \log(s_{\rm star}t + s_{\rm sky}t + \sigma_R^{2})\ \ \ (1)$$
This is being plotted against the $I$ magnitude, which is proportional to $-2.5\log s_{\rm star}$.

So if we now look at each of these contributions individually, we see that the noise due to "shot noise" on the star should indeed be a straight line on your plot. From eqn (1), ignoring other sources of noise
$$ \log \sigma = -0.5\log s_{\rm star} -0.5\log t$$
Thus in units of parts per million and scaled for the square root of the exposure time, this will appear as a straight line in a log-log plot with gradient of $-0.5/-2.5=+0.2$.

Now just look at the contribution due to the sky background, from eqn (1), ignoring other sources of noise
$$\log \sigma = -\log s_{\rm star} + 0.5 \log s_{\rm sky} -0.5 \log t$$
So the gradient here is -1 in log space (the sky term is just a constant offset in the log-log plot) and +0.4 in terms of (log) ppm/magnitude. Hence the gradient due to the sky noise term is twice as steep.

Finally, looking at the readout noise contribution
$$\log \sigma = -\log s_{\rm star} + \log \sigma_{R} -\log t$$
Therefore this has the same gradient as the sky contribution (as you can see in the plot), but the offset in the log-log plot should depend on the exposure time. This makes sense because the brightness at which readout noise will dominate does depend on how long the exposure is. So there is some unspoken (at least in the caption you show) assumption about how long the exposure time is in order to place the contribution from readout noise on the same graph, although the gradient is understood.


Is there a theoretical limit on telescope's resolution? Is there a theoretical limit on what we can see with a telescope of fixed size based on Earth's orbit or the Moon, that is in Earth's region of Solar system outside the Earth's atmosphere?

Why not instead of sending probes to Pluto and Ceres, just invest efforts into building better telescopes, so to photograph the surfaces of these bodies in greater detail? I suspect that all necessary information in fact reacheds Earth's region, it is just matter of sensibility and exposure to fix it, am I right?

The absolute limit of a telescope resolution is given by diffraction.
No matter how perfectly built and aligned is a telescope, you cannot resolve angles smaller than
$$\theta \propto \frac{\lambda}{D}$$
where $\lambda$ is the wavelength of interest and $D$ is the diameter of the telescope. This is why a number of millimeters and radio telescope (and also some -antennas) are huge.

The first figure here shows the basic principle. Imagine that you can divide what you observe in tiny squares and consider each one as a point like source. Each one will generate a diffraction pattern when passing into the telescope and if two points are too near, you cannot distinguish between them.

Wisely you put your telescope in space: turbulence in the atmosphere degrade the signal, and the best/biggest telescopes have hard times to go below $0.5 arc seconds$

Interferometry comes to the rescue increasing $D$ from the telescope size to the distance of two (or more) telescopes (called baseline). Interferometry has been used in radio astronomy since decades.
From what I hear optical interferometry is much much more complicated and as far as I know the only large scale attempt is the VLTI project.

So you could imagine to have a constellation of relatively small telescopes spread over hundred thousands or millions of kilometers. But this have the huge problem that you have to know the position and timing of every one of them with an impressive precision (my guess is that the precision in position is of the order of $\lambda$).

The other problem is light collection. If you want to see something very faint you have two options: 1)you observe for a lot of time or 2) you build a bigger telescope (6 to 40 meters in diameter).
And here also the largest baseline interferometers cannot do much, as the amount of light that they collect is just the sum of the light collected by the single telescopes.

To conclude: to observe Pluto or Ceres with high enough accuracy you would need a large number of large space telescopes very far away one from the other with perfect telemetry. It's far easier and cheaper to got there to take pictures.


Why do we need to add a second to 30 June and risk upsetting Internet? We linked a physical constant (the time of one second) to a physical non constant (the rotation of the earth) so there will have to be adjustments - or not? Is it important that the UTC in one line on earth is the same as the astronomical time? As the earth slows down, it means that this line slowly changes position. When the line travels 1 hour away we can bring it back by an hour change if necessary. If currently we add 1 second say every 3 years, it would take thousands of years to change by 1 hour. The last time a second was added, in 2012, problems were caused in some systems. Why do we need to do it?

There's quite a few things here that need comments.

First off, the "linking" you describe was actually the first definition of the second (day, hour, and all the way down to second). Remember, since the dawn of civilization we've used celestial bodies to keep time for us.

The phenomena of the Sun and the Moon had good enough precision for all practical purposes until very recently. It was then that the definition of the time unit was rebased to a different phenomenon. BTW, the second of time is not a physical constant. It is a unit of measure currently defined as a multiple of the frequency coming out of the transitions of a certain atom between two different states.

Of course, the motion of the Earth follows its own clock, different from atomic phenomena, so it ought to drift off after a while. Do we "need" to make adjustments then? That's debatable. We still prefer to keep the Earth's clock and our clock in sync, and therefore we make adjustments when necessary. But is it needed in an absolute sense? No, not really. It's just convenient.

As to the risk of "upsetting the Internet", this is an alarmist rumor the likes of the "Y2K bug". I actually happen to work in a related field - I've done infrastructure at a variety of tech companies in the Silicon Valley; the NTP infrastructure (Network Time Protocol - what keeps Internet services in time sync) is something I'm very familiar with. A jump of 1 second should be and will be tolerated by essentially everything that's out there now.

Will there be stragglers here and there that will hiccup when the time takes a tiny leap? Yes. And that's a bug, the engineers need to look at the software and fix it. Remember, a lot of existing services (like Google Search, or Instagram) have servers and instances right now that are off by more than 1 second (just the way there are cars coming out of the assembly line with missing bolts once in a while), and usually nothing bad happens as long as the drift is not too big); fix NTP, the time base jumps on that instance, and again nothing bad happens in the vast majority of cases.

Relax, nothing important will be "upset". In fact, I doubt that the vast majority of people will be affected in any perceivable way.


Why does the closest approach of star S2 to Sgr A* not appear to be near the focus of its elliptical orbit? Phys.org's Very Large Telescope sees star dance around supermassive black hole, proves Einstein right links to several ESO videos, including The star S2 makes a close approach to the black hole at the centre of the Milky Way.

In the screen shot below, the location of Sgr A* is nowhere near a focus of the elliptical orbit.

In addition to the excellent accepted answer I have written this answer to Why does Earth not appear to be at the focus of TESS' elliptical orbit in this video? and so I know that looks can be deceiving depending on one's perspective.

But I still don't really understand why the offset can look so bad here. For the video did they put the observer (the "camera") at a fairly close distance and thereby distort the perspective, or even if you choose the distance to be infinite can elliptical orbits appear to not have their central body at their focus, or at least along their axis?

Is it possible to state the orbital elements of S2 in such a way and in some coordinate system so that I can plot them in 3D and then try to reproduce this effect?

The orbital elements are on wikipedia:

$$e=0.884\ a=0.125'',\ i=134^\circ,\, \Omega=228^\circ$$

(At an assumed distance of 8kpc, $0.125'' = 1000au$)

It is the inclination that means that the black hole is not at the focus of the projected ellipse.

Imagine a circular orbit, with a central body.  If viewed from a distant but high inclination, the orbit is projected as an ellipse, with the central body at the centre of the ellipse (not the focus)

A distant projection of an ellipse is another ellipse, but the projection of the focus is not the focus of the projection.

Below you can see another example.  The ellipse is in the grey plane, with foci at A and B.  But the apparent ellipse has a completely different axis, and the focus is nowhere near the axis.  These image have been generate with geogebra and no perspective has been used.


What must/do astronomers reveal beyond their academic papers? Academic papers often make claims about observational data and results from computations. The papers only refer to names of data sources (such as an observatory) and of mathematical methods without any specifics, other than some of it illustrated in a few graphs and tables. References lead almost only to other papers of similar format.

In order to replicate (or even peer review) a paper, obviously one must have access to specifics such as the actual calculations, the raw data, the software, the blue prints of the measurement equipment and logs of how it has been operated.

What data do peer reviewers and replicating scientists (and the public) have access to beyond a paper making some claims?

Two examples: The results of the infamous cold fusion press conference of 1990 could never be satisfactorly replicated. But were the attempts based only on the paper they then published, or did they publish on the side also a wealth of data about their procedures? And the more recent "neutrinos travel faster than light" media debacle was, as far as I know, all about the details of equipment, was all relevant information about it made public (for the science community)?

Most catalogs have a public release somewhere on the web. Here is the best listing of astronomical catalogs that I know of:
http://vizier.u-strasbg.fr/viz-bin/VizieR

There is also a NASA Extragalactic Database (NED) that can be helpful: http://ned.ipac.caltech.edu/

In general, astronomical observations are available to the public to be analyzed. Different telescopes/space missions maintain their own archives. Some give an observer a proprietary period (6-7 months is the time frame I have heard given for most observatories) to give the person who requested the observations time to analyze the data before it is released to the general public.

In publications the author(s) will list they catalogs the used to come to their conclusions so replicating scientist can use the same dataset.


How Does Cosmology Constrain the Number of Neutrino Species? I've asked this question already on the Physics SE, but I felt it would be worth posting here as well.

I know that based upon theories of structure formation cosmologists can constrain the sum of the masses of neutrinos - if neutrinos were too light or too heavy it would simply change the power spectrum of the universe.

My question is how exactly is the number of neutrino species constrained by cosmological measurements (i.e. - $N_{eff}$ as measured by WMAP or Planck)? The recent results from the Planck Satellite give us $3.3\pm0.3$, which is consistent with three neutrino species. How is this measured?

The original research paper, p.43, section 6.3.2.

I leave it to someone else to find comprehensible words to describe the connection between CMB, redshift and effective neutrino number.


Was Earth closer to Europa on 1983-11-25 or 1985-07-22? xkcd #2003:

gives a humorous line of succession to the United States presidency. The comic's title text, found by hovering over the comic, states that a tie is broken by whoever was closest to the surface of Europa when they were born.

The only potential tie on the list occurs in item #19. It states the "champion" of the Nathan's Hot Dog Eating contest, but does not specify the men's or women's champion. At the time of the comic's publication, the men's champion was Joey Chestnut (born 1983-11-25) and the women's champion was Miki Sudo (Wikipedia gives only 1985 as her year of birth, but famousbirthdays.com claims July 22).

Both were ineligible to hold the presidency at the time of publication due to not meeting the minimum age requirement of 35. But supposing they both were, who would be first in line by the tiebreaker of distance from the surface of Europa at birth?

tl;dr Miki Sudo

Using JPL's SPICE toolkit, I computed the positions of Earth and Europa for the times in question.

On 1983-Nov-25, Earth and Europa are 935.2 million km apart, while 1985-Jul-22, they are 612.5 million km apart. Miki Sudo wins by 323 million km, given the assumed date for her birthday. If we don't trust famousbirthdays.com for Sudo's birthday, but assume it was still sometime in 1985, she still wins.

In fact, it turns out Joey Chesnut was born at a time when Earth was close to its farthest point from Europa, so he's fairly unlikely to ever win in this tiebreaker:


How are exoplanets with 10,000 day periods discovered with the radial velocity method? The radial velocity technique has apparently been used to discover exoplanets with extremely long orbital periods of ~10,000 days. For example, see this link:

https://exoplanetarchive.ipac.caltech.edu/cgi-bin/TblView/nph-tblView?app=ExoTbls&config=planets

If the period of a planet is ~10,000 days, then the period of the star must be the same. However, this would mean that observations would need to be made over ~30 years to get a complete picture of the radial velocity variation of the star, a time which is far too long. How then are radial velocity measurements used to discover such planets?

You don't need to collect data for one cycle in order to determine the period.

From Newtonian's physics, we know precisely how a trajectory of one particle orbitting around another one would be. Since the instantaneous velocity (v_inst) at any point on the trajectory is tangential to the point on the trajectory, we can decompose the v_inst into the radial velocity which we can match with the observation from, e.g., redshift. With assumption of constant v_inst and trajectory, we can determine the period from just a couple of observations (depending on how accurate the redshift can be measured, and the number of variables in the equation).


The Solar Motion and the peculiar velocities of stars Local standard Of Rest is the hypothetical origin of a reference system with respect to which the motions of all stars in some neighborhood of the sun average out to be zero

which type of motions of stars average out to be zero and how can I justify that their average will be zero?

What is Peculiar Motion? How does it look like?

How we measure the proper motions of stars as seen from the sun as we can only observe from earth?

When you talk about the motion of anything, you need to first define what "not moving" means.  For example I may be sitting in a chair (and not moving relative to the chair), but I know that the Earth is rotating, and orbiting the sun, and the whole solar system is orbiting in the Milky Way.

When I look at the motion of stars, I know that (nearly?) all the nearby stars are also orbiting the Milky Way. But I want to ignore that overall motion, and only consider the motion of the stars relative to each other.

It is possible to measure the motion of a star relative to the observer. We also know the motion of the observer relative to the sun (since we know how fast the Earth spins and how fast it orbits the sun) So by simply subtracting one motion from the other we can find the motion of the star relative to the sun.

But instead of using the sun as a fixed frame of reference, I can look at the local stars, find their average motion relative to the sun, and subtract that motion from each individual velocity. This will give a motion relative to the motion of the stars around the Milky Way. I know the average will be zero because I have chosen to subtract a certain amount from each star's velocity to make the average zero. It is zero "by construction"

(simple example Consider the numbers "3,7,2,3,5". The average is 4. If I subtract 4 from each number I get (-1, 3, -2 -1, 1) with average = 0)

The motion of a star, when you have subtracted the average motion of all the local stars, is called the peculiar motion


Can the diminishing energy of the CMBR be the source of dark energy? I was just told the CMBR energy diminishes with time. Can it be that the lost energy is somehow transferred to spacetime, causing the expansion to accelerate?

No - the decreasing energy in the CMB is already well modeled in the Friedmann equations. The term in the density parameter that is proportional to $a^{-4}$ is the contribution of radiation energy density to the evolution of the universe, the term proportional to $a^{-3}$ is matter density (mostly dark, but includes ordinary matter), $a^{-2}$ is the contribution of the curvature of space-time itself, and the term without any factors of $a$ is the contribution of dark energy. The size of the radiation density, today, is already a small fraction of the matter density (about 0.03% of the matter density, 0.01% of the density of the universe overall - ordinary matter is about 5% overall).

The last time the energy density in the radiation fields was the same size as what's in the matter fields was around $z=3,300$.

I also disagree with @J.Chomel's answer - the energy stored in the radiation field is decreasing. Then energy density in the radiation field scales like $a^{-4}$, and the volume scales like $a^3$. Since the total energy is the energy density times the volume, the total energy scales like $a^{-1}$, just as you would expect with the number of photons being fixed, but the energy in each photon scaling as $a^{-1}$ as the wavelength increases.


Liberator 3 for fits format How to set in Liberator 3 the reverse color scale? I would like to have stars black and the surroundings white. Or could you recommend me another program that has this ability?

I would suggest you check out SAOimage DS9, which is one of the most commonly used FITS viewers (probably the most commonly used one) among professional astronomers. It allows you to display images with a variety of different scalings and color maps (and, yes, there is an "Invert Colormap" setting under the "Color" menu).


Do contracting objects show redshift? If we consider a point on the central part of a contracting object as observed from earth, the point is continuously moving away from us as the object contracts. Moreover, light emerges from a deeper gravitational well as the object contracts. So, shouldn't contracting objects show a redshift? And if so it should be less towards the edges than towards the centre, right?

shouldn't contracting objects show a redshift?

Yes, they should. But do they, actually? That's the better question.

To have any appreciable redshift, the speed of the object needs to be huge. That will be a very short implosion. Also, there aren't many mechanisms that can accelerate implosions that much.

What I'm saying is - it's theoretically possible, but in practice you won't see it that often, if ever.

  And if so it should be less towards the edges than towards the centre,
  right?

Sure. If the object is transparent to its own radiation, then you'd see blueshift from the other side, too. Again, this is very theoretical.


How is phase defined in a binary orbit with eccentricity? Are orbital phases of the black spot in the figure measured from some angle or time/period?
Usually, how is phase=0 defined? It is relative to the line of apses perhaps with phase=0 at periastron?
Any of the anomalies below is defined as orbital phase? Please specify reference papers or books.
E: eccentric anomaly
M: mean anomaly
u: true anomaly

Source

Usually, how is phase=0 defined? It is relative to the line of apses perhaps with phase=0 at periastron?

The true anomaly, $\theta$, is the angle between the current location of the orbiting particle and its location in the orbit at which it is closest to the central body (called the periapsis/periastron). The word "phase" is used variously in physics and astronomy, but based on your question, you may mean $\theta$ since it is defined as being zero at periastron.

Are orbital phases of the black spot in the figure measured from some angle or time/period?

It ultimately depends on how you want to compute the trajectory/orbit of the particle (i.e., the black "spot" in the OP image) that is moving about a primary body (i.e., the star in the OP image). Alternate values can be used instead of $\theta$, for example $\displaystyle M$, the mean anomaly.


Is Sahara desert a good place to build radio/optical telescopes? At least Atacama Desert of northern Chile is a good place to build telescopes.

VLT and ALMA locate there.

What about Sahara desert?

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

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

No, the Sahara isn't a good place to build telescopes.
The Atacama desert is used because it is at high altitude, which means that there is less atmosphere to get in the way. Other telescopes are located on mountaintops for the same reason.
The Sahara is mostly at sea level. It's also very hot, so you get lots of turbulence due to rising air, which distorts the image.
The Sahara is also undesirable for other reasons: no infrastructure, unstable regimes, lots of erosion due to sand storms.


How likely and severe is the threat of a gamma ray burst to earth? In the National Geographic article: http://news.nationalgeographic.com/news/2009/04/090403-gamma-ray-extinction_2.html it is suggested that a gamma ray burst likely caused a mass extinction in earth's history.  How severe is the threat of a gamma ray burst to life on earth?  It would be nice to see a chart that considers the threat within the next hundred years as follows (I have no idea what the actual numbers are, hence I am asking):

Severity - Likelihood in 100 years:

Global extinction - 0.0000001%
Major impact - 0.001%
Moderate impact - 0.1%
Minor impact (e.g. deplete ozone layer by 0.01%) - 10%
Observable - 100% ("minor" gamma ray bursts are frequently detected)

Follow up questions are:

Would there be any warning of an impending severe gamma ray burst?
Could we survive in gamma ray burst shelters? (it would appear being underground may be sufficient according to: http://en.wikipedia.org/wiki/Radiation_protection#Shielding_design and http://2013.org/showthread.php/188-How-does-one-prepare-sufficiently-for-gamma-radiation-in-a-shelter-etc)

Something else I found to be shocking is that the Bible appears to predict such an event.  I would rate the predicted event on the major impact level.  I asked about this in the Christian Stack Exchange here:
https://hermeneutics.stackexchange.com/questions/9324/interpretations-of-revelation-168-11/9335#9335

I'll address WR104 first. The National Geographic article calls it a "potential threat." Yet that potential may be low. There are a slew of articles quoting astronomer Grant Hill on the subject. Hill studied the star and found that it looks like it isn't pointing straight at us. Its axis might be up to 45 degrees in another direction, meaning that we'd be fine if it underwent a burst.
From here:

It would appear the original Keck imagry may not have been as straight-forward as it seemed. Spectroscopic emission lines from the binary pair strongly suggest the system is in fact inclined 30°-40° (possibly as much as 45°) away from us.
So, Earth doesn’t appear to be in the firing line of WR 104 after all…

Here and here Hill is quoted as saying that the star looks like it could be pointed at us - which goes against the evidence he found! Finally, this is another example of Hill's cautious attitude: The evidence says we're good, but he's not making any assumptions.

Sabre Tooth, I know you said that you didn't want Wikipedia, but my primary goal here isn't to collect the bounty, so I'm going to reference it in this section. I hope you're okay with that. I may withdraw it, though, if Jonathan indicates that he doesn't want it.
Okay, back to the topic at hand. Wikipedia, of course, has a short tidbit on the frequency of gamma-ray bursts effecting Earth:

Estimating the exact rate at which GRBs occur is difficult, but for a galaxy of approximately the same size as the Milky Way, the expected rate (for long-duration GRBs) is about one burst every 100,000 to 1,000,000 years. Only a small percentage of these would be beamed towards Earth. Estimates of rate of occurrence of short-duration GRBs are even more uncertain because of the unknown degree of collimation, but are probably comparable.

'A small percentage' isn't too specific. The BBC article that is cited in this passage says the following:

Observations of deep space suggest that gamma ray-bursts are rare. They are thought to happen at the most every 10,000 years per galaxy, and at the least every million years per galaxy.

However, the 'small percentage' is never elaborated on.

That's all I have for this edit; more to come later. By the way, this Physics.SE question may interest anyone reading this. . .


antimatter annihilation in stars Electrons annihilate with positrons produced through fusion processes in stars. Which particle interaction produces new electrons so that the sun isn't deplete of electrons? Or is something else happening altogether?

Regular fusion cycles in stars produce neutrinos and positrons as by products. Those positrons annihilate with electrons that are already there in the plasma of the star to produce the light that we eventually see. How are those electrons replaced?

The proton-proton chain ultimately converts four protons into one helium nucleus. The charge of the 4 protons was balanced by 4 electrons, but helium contains 2 protons (and 2 neutrons), so it only needs 2 electrons to be balanced.

As you point out, the process of converting a proton to a neutron releases a positron (and an electron neutrino), and that positron quickly annihilates with an electron.

Here's the diagram from that Wikipedia page of the main p-p chain.

So the process actually consumes 6 protons, and emits 2 protons, a helium nucleus, and 2 positrons (plus a couple of neutrinos), and a couple of gamma photons. The positrons annihilate with 2 electrons, releasing more gamma photons (usually 2 or 3 apiece, depending on the spin alignments of the positron & electron).

If you add everything up, you'll see that the electromagnetic charge balance is unchanged.

We started with 4 protons, which are balanced by 4 electrons nearby in the stellar core plasma. (We can ignore the intermediate pair of hydrogens that are eventually re-emitted). We end up with a helium nucleus that only needs 2 electrons to be electrically balanced, so if those other 2 electrons weren't annihilated then the star would build up an excess of negative charge.


Why can't gravity repel things? Gravity is the result of the curvature of space, so could the curvature actually send objects away from the source?

Apart from the field-theoretical standpoint presented by Stan, one can repel objects in a sense, when taking orbital mechanics into account.

The slingshot maneuver extracts angular momentum and energy of an orbiting mass by the use of gravity. The trick here is that the probe's velocity just gets redirected in the planet's reference frame. But this results in an acceleration in the solar rest-frame that the probe essentially tries to leave.

As you were asking about curvature, I don't know how and if centrifugal potentials that play a role here are described by general relativity.


Can the period of the planet transit across the sun be calculated in the same way the eclipse year is? I read from Wikipedia$-$Eclipse cycle that the eclipse year, which is the period of Earth's hitting a certain node of its orbit around the sun, that is, the ecliptic, and the moon's orbit around Earth, can be calculated as the beat period of the synodic month and draconic month. The recent occurrence of planet transits across the sun lets me wonder if this principle can be applied to the case of a planet's transit across the sun as seen by an observer on another planet in an outer orbit.

When Venus transited across the sun on June 6, 2012, I read it takes place in pattern of the 8-year-interval pairs with each pair at the interval of 105.5 years and 121.5 years alternately, with one in each pair being when Venus crosses the ecliptic plane at its ascending node and the other being when Venus crosses the ecliptic plane at its descending node. This made me contemplate whether the period of the Venus transit across the sun when it crosses the ecliptic plane at its certain node can be calculated in the way the eclipse year is, that is, whether the following holds:

  The period of the Venus transit across the sun as it crosses the
  ecliptic plane at its certain node = The beat period of the period of Venus'
  being in inferior conjunction with the sun and the period of Venus'
  hitting that node.

Early this July I read from EarthSky an observer on Pluto may, through some optical aid, be able to see Earth transiting the sun on July 12, when Pluto was at opposition, which, together with the upcoming event, that on October 24, 2018 Pluto will cross the ecliptic plane at its descending node, is responsible for the Earth transit as seen from Pluto. Last time this kind of transit occurred on January 11, 1931 on account of both that Pluto was at opposition that day and that Pluto crossed the ecliptic plane at its ascending node on September 9, 1930. And next time this kind of transit will occur when Pluto is at opposition 161 years later, around when Pluto will cross the ecliptic plane at the same node it crossed in 1930. This makes me again contemplate whether what follows holds:

  The period of the Earth transit across the sun as seen from Pluto as
  Pluto crosses the ecliptic plane at its certain node = The beat period of the
  period of Pluto's being at opposition and the period of Earth's
  hitting that node.

I think the principal difference between the eclipse year and planet transit across the sun is that in the former case both the moon and the sun are in a closed orbit around Earth while in the latter case only the sun is in a closed orbit around the observer-sited planet (the transiting planet is in a tortuous path from the perspective of the observer). I don't know if this feature would affect the applicability of the beat-period principle. If the principle of beat period as it used to calculate the eclipse year can be applied to the above two examples, I think it should be able to be generally applied to calculate the period of any planet's transit across the sun as seen by an observer on an outer planet, right? That is, should what follows generally hold?

  The period of a planet's transit across the sun as seen from an outer
  planet as it crosses the outer planet's orbit plane at its certain node =
  The beat period of the period of the transiting planet's being in
  inferior conjunction with the sun from the outer planet's perspective
  and the period of the transiting planet’s hitting that node

I apologize for the length of this in advance! Before getting to the answer it should be said from an orbital mechanics point of view all transits are eclipses and all eclipses are transits.  There are a lot of different types of both and the words are used differently but they mean the same thing - something is passing in front of something else from the observers point of view.  Not many would say "Venus eclipsed the Sun" during a Venus transit nor would they say, "the Moon transited the Sun" during a solar eclipse.  But the way cycles are determined is the same for eclipses or transits.
In short, the answer to this is almost yes:

The period of a planet's transit across the sun as seen from an outer
planet as it crosses the outer planet's orbit plane at its certain
node = The beat period of the period of the transiting planet's being
in inferior conjunction with the sun from the outer planet's
perspective and the period of the transiting planet’s hitting that
node

For us to see Venus or Mercury transit/eclipse the Sun, three things have to happen.  The inner planet's nodes must be at the same ecliptic longitude as the Sun, the planet must be crossing the node when this occurs, and it needs to be on the same side of the Sun as us.  (Oh, and the observer needs to be on the Sun side of the Earth with clear skies which is NOT what happened to this observer during the last Venus transit and I'm still bitter about it.)
Various cycles of different lengths have to come together to make this happen which is what I think you mean by "beat period" though I've never heard it phrased this way in astronomy.  It's as if I ate dinner on Friday every two weeks with one friend and every three weeks with another - then the "beat period" of six weeks would mean a scheduling conflict or a dinner party.
The reason I said your conclusion was almost correct is explained by looking at the transits of Mercury on this Wiki page.  Sometimes a transit doesn't happen even though it is on the "beat period" (which astronomers just call cycles).  Lots of smaller cycles line up, or come together, every so often on a larger cycle.  You got it right when you said the two things have to happen - the nodes lining up with the Sun during an inferior conjunction but as Mercury shows the transit can be missed.  If a larger "beat period" was used to leave out those missed transits, there would be extra transits happening mot on the beat if that makes any sense.
Determining these cycles is fairly easy for us on Earth due to the frame of reference of the ecliptic.  Neither the Earth nor the Sun have "nodes" to the plane of the ecliptic because it's the Earth and the Sun which define the ecliptic.  Nodes are where tilted (inclined) planets cross this plane.  It's a handy reference since we live on Earth, but for the observer freezing on Pluto wondering about all of the transits they will get to see with their telescope, they would want the ecliptic be the plane defined by Pluto and the Sun.  Then the Earth and all of the planets would have nodes as they crossed above and below the plane of the Pluto - Sun ecliptic.  From their perspective, there would be inner planet cycles which would repeat with a "beat" when certain cycles lined up again so that Jupiter, for example, transited the Sun.
On the Wiki page you linked talking about eclipses, I found it awkward the way they kept referring to the "Sun being near/at the node" which technically isn't wrong because we know they mean "from the observes viewpoint, the planet's node lines up with the Sun (same ecliptic longitude)" but I think it could give people the wrong impression as if the Sun was at or near that location in 3D space. That's just my opinion, though.
Here is a good Sky & Telescope article on Venus transits and how they will become more rare in the future. And here is one of the most enjoyable websites on the internet - Orbit Simulator. Look through the list until you see something that interests you and then click it to see fantastic browser based orbit details.  They also have a Windows program you can download to have a lot more control over the simulations.
I apologize again for the lengthy reply - orbital mechanics is a subject I love.  And good on you for putting so much thought into the orbits and cycles, not many want to dive that deep into the subject.


How does the Eddington limit make the gravitational field of stars asymmetric? This answer to Is the gravitational field of the sun uniform? says:

Very large stars are asymmetric due to being close to the Eddington limit. The Sun is too small of a star to have the kind of gravitational anomalies that very large stars and rocky planets exhibit.

which is really interesting but leaves me unsatisfied.
Question: How exactly does the Eddington limit make the gravitational field of stars asymmetric?

What does Eddington Limit (Eddington Luminosity) Show?

From wikipedia:

Eddington luminosity
Humphreys–Davidson limit

Hot, massive stars tend to be (a) asymmetric, due to their rapid rotation (at least on the main sequence); and (b) very luminous, and thus nearer to the Eddington limit, but the two aren’t causally connected to each other.  So perhaps that answer was using correlation in place of causation.
The rapid rotation is due to the fact that hot stars lack the dynamo-driven magnetic fields that let lower-mass stars shed angular momentum and spin down over time.  The high luminosity is just a natural consequence of maintaining the higher temperature and pressure needed to support the higher mass, i.e. to maintain hydrostatic equilibrium.
It is closer to being true the other way around: because of rapid rotation, the effective gravitational field at the star’s equator is lower, and thus that part of the star will be closer to the Eddington limit at a given luminosity than a non-rotating star would be.


Do the Bullet Cluster remnants prove that dark matter consists out of particles? Some time ago the remnants of a collision between two collections of stars were discovered. These remnants are called the Bullet Cluster:

The color blue (non-visible in reality) represents the presence of dark matter. Pink represents the normal matter (I'm not sure if the color pink in this picture is actually visible, contrary to the blue).
There are some theories around to explain the nature of dark matter. They can roughly be divided into two groups:
1)Theories that state, dark matter consists of particles
2)Theories that modify gravity (apart from MOND) itself, for example, the theory propagated by the Dutch physicist Erik Verlinde (who's "contraption" here in the Netherlands was announced on TV as a huge breakthrough and for he received the Spinoza Prize. i.e. a lot of money, in 2011; much too much, in my opinion) which see gravity as entropic.
Theories of a modified gravity make the prediction that dark matter is tied to normal matter. Now in pictures of the bullet cluster, it seems like matter and dark matter are clearly separated.
Does this prove that dark matter has to consist out of particles?

TLDR: Not as such, but it does give us some constraints on its properties.
Background:
A galaxy cluster consists of three major components, from smallest to largest proportion of mass: galaxies (visible with optical telescopes), intracluster gas (visible with X-ray and radio telescopes) and dark matter (not directly observable). Dark matter makes up about 80% of the total mass; of the remaining mass, 90% is the hot gas between the galaxies (not exact numbers, but order of magnitude is sufficient here).
The bullet cluster image:
In the image, we see two accumulations of galaxies centred roughly where the blue colouring is. Overlaid in red is the X-ray emission from hot gas. Note the conical shock front to the right. The blue colour, finally, shows the mass distribution as measured through gravitational lensing.
What does that mean?
We see that gas and galaxies are separated, in contrast to what we see in regular galaxy clusters. The implication is that two clusters have collided. While stars and hence galaxies are essentially collisionless and pass right through each other (there is a lot of space between stars within a galaxy so collisions happen rarely if at all), the same cannot be said for the hot cluster gas. So the gas remained in the centre of the collision while the galaxies passed through.
Now we know that the mass ratio of gas to galaxies is around 10:1. But gravitational lensing (which doesn't care what kind of mass is present) shows us that the majority of the mass in this system is where the galaxies are.
And the dark matter?
We see that the dominant component of the cluster mass is collisionless (it is where the galaxies are). It also does not interact with normal matter except through gravity (else we'd see it).
An exotic elementary particle would fit this description well, but so might primordial black holes (hypothetical low-mass black holes formed shortly after the big bang).
From observing the bullet cluster alone, the dark component might also be made up of quite massive black holes or other massvie dark compact objects -- these can be ruled out by observing microlensing events around our own galaxy.
What about modified gravity?
Mond and similar theories would have to explain why gas of a given mass would affect space-time differently than the same mass clumped together into a star.
Note, however, that there is a known component of dark matter: shortly after the big bang, neutrons that were not bound into helium (or deuterium etc.) decayed into a proton, an electron and an electron neutrino. This cosmic neutrino background is still around, but measurements constraining the upper limit of neutrino rest mass make it highly unlikely that modified gravity theories could use them to explain the bullet cluster. Using our standard theory of gravity, its influence on cluster physics is utterly negligible.
Conclusion
The observation of the bullet cluster provides a big problem for theories that try to explain how our universe works without resorting to some form of collisionless, weakly interacting matter as the major component of the matter in our universe.
It puts some constraints on the nature of dark matter but does not by itself rule out a form of dark matter that is much more massive than an elementary particle.
Further reading
Galaxy cluster composition
Neutrino background
Gravitational lensing
Observing intracluster gas with radio telescopes


How can we use hypervelocity stars to determine the origins of the Universe? I was reading this article finding evidence of Universe's origin, which describes that in 1 trillion years we may lose the ability to determine how the universe was created.

The answer seems to be using hypervelocity stars.  These are stars that have been ejected from their galaxy.  Using the expansion of the universe, we can use the velocity of the star to then determine evidence of the Big Bang. But how does this work?  Is there calculations we can do now that agree with our understanding of the universe and its age when working with hypervelocity stars?

I understand finding a hypervelocity star is very hard to do. I guess my thought here is that astronomers have modeled this scenario and found that if we were to find a hypervelocity star, we could then use the scenario above to calculate the age.  I'm assuming the age calculated agrees with current estimates.

How can we use hypervelocity stars to determine the origins of the Universe?

First and foremost, I should clear something up: our knowledge of the Big Bang is incredibly limited, and so we don't even know the origins of the Universe today. General relativity and quantum mechanics both break down as you get closer and closer to $t_0=0$. All we really know is that something happened 13.8 billion years ago. However, if we look at the expanding universe, we can extrapolate that, 13.8 billion years ago, space was much less dilated than it is now.

General relativity suggests that the Universe would be packed into a singularity. This is not a good approach, since GR can't properly explain such conditions (QM is more fit to do so, but that alone can't explain how gravity acted), so we really haven't the slightest clue as to what happened 13.8 billion years ago. That said, we still know when the Big Bang (whatever it may have been) occurred, and that's what future astronomers may be able to find out.

The idea the article brings up is this: by one trillion years from now, the expansion of space will have driven away any other galaxies from our line of view, and CMB radiation will have become too redshifted to be detectable.

However, hypervelocity stars will likely be detected outside of the Milky Way (or rather, Milkomeda). Once these stars get far enough from the Milkomeda's gravitational influence, we can observe their acceleration due to the expansion of space, just as we do for many galaxies today. The 4-page paper the article is based on, Loeb (2011), actually makes this clear in a graph:

As the stars get farther and farther from Milkomeda, the gravity's pull on the stars will weaken and they will be accelerated by the expansion of space. This will really become noticeable once they are about 2.3 megaparsecs from the galaxy. Of course, these stars will not be easily found, but Loeb mentions that future astronomers will likely have advanced their tools and instruments by that time.

Loeb explains that, once we observe their redshifts as their distances increase, we can confirm Hubble's law. After this, we just need to measure the necessary parameters, and finally we can calculate the age of the Universe using the Friedmann equation:
$$t_0 = \frac{1}{H_0} F(\Omega_r,\Omega_m,\Omega_\Lambda,\dots)$$

That's the gist of it. The hypervelocity stars will be the only extragalactic objects detectable by then, and thus we will rely on them to confirm our measurements.

Interestingly, Loeb mentions that observing and analyzing the expansion of the Universe at that time would answer some of the uncertainties we have today:

  The working assumption so far was that the vacuum energy density remains constant and does not decay to a lower energy state over hundreds of e-folding times. Measuring the dynamics of hypervelocity stars at late cosmic
  times provides the added benefit of testing whether the cosmological constant is truely [sic] constant. Indications to the contrary might reveal new physics, such as that considered by evolving dark energy models.

Thus, even though much of the Universe will not be observable by that time, we will ironically have more answers about cosmology than we do today, simply by observing the few extragalactic objects that will be visible.

This is the general explanation of how we can find the time since the Big Bang. Remember that even now, we really have no idea about the origins of the Universe — it's a major unsolved problem in physics. All we know is something happened 13.8 billion years ago. Using hypervelocity stars, future humans would be able to determine the time since the Big Bang as well. Don't get your hopes up on humans surviving for 1 trillion years, though.


Spectra and star identification Are stellar spectra unique enough that they could be used to identify particular stars? If so, has any attempt been made to use them for navigation or equipment alignment?

No, you can't use them to identify your star. Indeed, you have too many stars in the universe. Even in our galaxy you can find two stars which have pretty much the same spectrum. You can classify the star though.

There is another problem with redshift and blueshift. It change the spectrum of the star depending its movement.


Why does the Earth have a z-component at the start of the J2000 epoch? When I go to the NASA HORIZONS webpage

https://ssd.jpl.nasa.gov/horizons.cgi

and ask for the VECTOR coordinates of the Earth on 1 January 2000, it tells me that Earth has the following coordinates in the J2000 epoch:

X =-1.756637922977122E-01 Y = 9.659912850526895E-01 Z = 2.020629118443605E-04

These are measured in astronomical units. Quoting from the webpage:

Reference epoch: J2000.0

XY-plane: plane of the Earth's orbit at the reference epoch

X-axis  : out along ascending node of instantaneous plane of the Earth's             orbit and the Earth's mean equator at the reference epoch

Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense              of Earth's north pole at the reference epoch.

But these figures don't make sense! The z value should be EXACTLY zero, since we are measuring at 1 January 2000. Also the x should be exactly 1, and the y should be zero.

What is going on?

You made the same fundamental mistake that Anton Gromov made in his question on the sister Space Exploration StackExchange network site: You used the solar system barycenter rather than the Sun as the frame origin. Had you used the Sun, the apparent discrepancy would have dropped by almost two orders of magnitude.

A much lesser flaw is that you apparently used midnight rather than noon. The J2000.0 epoch is 12 noon Terrestrial Time on 1 January 2000. This doesn't change things by much, but it is nonetheless important. That half day offset has bitten me more than once.

The z coordinate of the Earth's position would not have dropped to zero had you made the above corrections (Sun-centered rather than solar system barycenter, and 12 noon rather than midnight). It would not have dropped to zero even if you had used the Earth-Moon barycenter rather than the center of the Earth.

One reason it would not have dropped to zero is that the Earth's orbit about the Sun is not an ellipse. The Earth's orbit doesn't even lie on a plane! The key cause of this non-planar motion is that the Moon's orbit about the Earth is inclined by about 5.15° with respect to the ecliptic, making the Earth bob up and down with respect to the ecliptic1. Because the Earth's orbit is not truly elliptical or even planar, the ecliptic plane at some epoch is a time-averaged plane centered about the epoch time that makes the Earth's position with respect to the Sun (or the Sun's position with respect to the Earth) average to zero.

A lesser reason is that the J2000.0 mean ecliptic and mean equinox frame was defined over 36 years ago. HORIZONS uses DE431, which was released 6 years ago. JPL's ability to model the solar system improved significantly in the intervening three decades.

1 Strictly speaking, it makes the Sun appear to bob up and down with respect to the ecliptic. The concept of the ecliptic plane dates back to the ancient Greeks, and it retains a vestige of this definition to this day.


Why Only Saturn Has Visible Rings It may be surprising to know that all four giants in our solar system have rings orbiting around them.  Yet only Saturn has a system of rings visible to the naked eye.  Why is that?  What's so different about Saturn in comparison to Jupiter, Uranus and Neptune that makes it the only planet to have a visible ring system?

By "visible to the naked eye", I take it you mean "visible from Earth with a small telescope".

Saturn's rings are largely water ice, and so they reflect more sunlight back to us.

Jupiter's rings, have lower proportions of ice, and lots of smaller dust particles that tend to scatter light forward rather than back to us.

The ring systems of Uranus and Neptune are made of really dark material, so also don't reflect much light in our direction.


Finding Latitude and Longitude of a person or of where a picture was taken this is my first post on the Astronomy stack exchange site. Please let me know how this question is based on y'all's generally accepted criteria for a good question as well as possible improvements.

In my AP comp sci class we were given pretty much total freedom to come up with a project, I remembered hearing about it being possible to calculate a person's position or the position a picture was taken based on time of day and the position of the Sun, that is, longitudinal and latitudinal position. For the picture, I heard Sun position was determined through some implementation of shadows in the picture. I thought this was cool and decided to look into it. I found an article written by a prominent UCLA physicist and a well-researched Wikipedia page. I understand what the Wikipedia page requires in regards to values for calculations and what the equations calculate, but I do not understand how to implement them to find longitudinal and latitudinal location, nor do I understand why it has rectangular equatorial coordinates as that is not included in the three-step explanation in the Wikipedia page intro at the top.

For the UCLA report, I understand none of it except for the matrix mathematics and manipulations, and I'm sketchy on those at best, but I do not understand all the coordinate systems that it mentions nor how to implement them to achieve my objective.

So my question comes down to this: Can anyone clarify the equations, their purpose, their implementation, how to convert an answer to longitude and latitude for either the Wikipedia page or the UCLA report?

OR

If you can explain how to find the latitude and longitude of a picture or person. If you choose to explain, I do need equations or an explanation that I can turn into equations.

Please suggest question revisions and whether this should be posted on a different Stackexchange forum. Thanks.

EDIT: It turns out I answered this question earlier (in a slightly different format):

Cancelling out earth rotation speed, Altazimuth mount

Imagine you are at 35N,106W at 5h local sidereal time.

The celestial north pole is at (0,90) in the celestial frame and
(0,35) in your frame.

The point (5h,0) is (75,0) in the celestial sphere (1h = 15 degrees),
and (180,55) in your frame

The point (11h,0) is (165,0) in the celestial sphere and (90,0)
(setting in the east) in your frame.

Converting to rectangular coordinates, we need the matrix that does this:

{0,0,1} -> {Cos[35 Degree], 0, Sin[35 Degree]}

{Cos[75*Degree],Sin[75*Degree],0} -> {-Sin[35 Degree], 0, Cos[35 Degree]}

{Sin[-75*Degree],Cos[75*Degree],0} -> {0,1,0}

or, generalizing a bit (lst = local sidereal time as an angle, lat =
latitude, notice that lon is irrelevant since we're using local
sidereal time)

{0,0,1} -> {Cos[lat], 0, Sin[lat]}

{Cos[lst],Sin[lst],0} -> {-Sin[lat], 0, Cos[lat]}

{-Sin[lst],Cos[lst],0} -> {0,1,0}

Solving (Mathematica):

m0 = Table[a[i,j],{i,1,3},{j,1,3}]

(* Some simplifying assumptions, intentionally avoiding corner cases *)

conds = {-Pi/2 < dec < Pi/2, -Pi/2 < lat < Pi/2, 0 < ra < 2*Pi, -Pi < lst < Pi}

m = Simplify[m0 /. Solve[{
 m0.{0,0,1} == {Cos[lat], 0, Sin[lat]},
 m0.{Cos[lst],Sin[lst],0} == {-Sin[lat], 0, Cos[lat]},
 m0.{-Sin[lst],Cos[lst],0} == {0,1,0}
},Flatten[m0]],conds]

m = m[[1]]

This gives us

$
   \left(
   \begin{array}{ccc}
-\cos (\text{lst}) \sin (\text{lat}) & -\sin (\text{lat}) \sin (\text{lst})
  & \cos (\text{lat}) \\
-\sin (\text{lst}) & \cos (\text{lst}) & 0 \\
\cos (\text{lat}) \cos (\text{lst}) & \cos (\text{lat}) \sin (\text{lst}) &
  \sin (\text{lat}) \\
   \end{array}
   \right)
$

The inverse of this matrix:

$
   \left(
   \begin{array}{ccc}
-\cos (\text{lst}) \sin (\text{lat}) & -\sin (\text{lst}) & \cos
  (\text{lat}) \cos (\text{lst}) \\
-\sin (\text{lat}) \sin (\text{lst}) & \cos (\text{lst}) & \cos (\text{lat})
  \sin (\text{lst}) \\
\cos (\text{lat}) & 0 & \sin (\text{lat}) \\
   \end{array}
   \right)
$

will transform rectangular azimuth and elevation back to right
ascension and declination and right ascension. This would be an
interesting alternative approach to this problem, but I won't be using
it.

OK, if we convert right ascension and declination to rectangular
coordinates, multiply by the first matrix above and convert the result
back to spherical coordinates, we will have azimuth and
elevation. Let's do that...

The spherical coordinates for (ra,dec) are:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}

Multiplying by the matrix, we get

$
   \{\sin (\text{dec}) \cos (\text{lat})-\cos (\text{dec}) \sin (\text{lat})
\cos (\text{lst}-\text{ra}),-\cos (\text{dec}) \sin
(\text{lst}-\text{ra}),\cos (\text{dec}) \cos (\text{lat}) \cos
(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat})\}
$

(note that lst-ra is sometimes called the "hour angle", which would
simplify the above)

Converting back to spherical coordinates and simplifying:

{x,y,z} = Simplify[m.{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]},conds]

(* NOTE: the use of single argument arctangent here introduces
ambiguity, and the final answer must compensate for this *)

az = Simplify[ArcTan[y/x],conds]

el = Simplify[ArcSin[z],conds]

yielding:

$
   \text{azimuth}=-\tan ^{-1}\left(\frac{\cos (\text{dec}) \sin
(\text{lst}-\text{ra})}{\sin (\text{dec}) \cos (\text{lat})-\cos
(\text{dec}) \sin (\text{lat}) \cos (\text{lst}-\text{ra})}\right)
$

$
   \text{elevation}=\sin ^{-1}(\cos (\text{dec}) \cos (\text{lat}) \cos
(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat}))
$

Assuming I haven't made any mistakes, the next steps would be:

Converting local time to local sidereal time
Inverting these equations to get latitude and longitude from
azimuth, elevation, and time.

I may or may not edit this answer to add those steps, but this
hopefully provides a start.