Calculating longitude from star culmination Here is the problem: I need to calculate longitude of observing place. And I know time (in UTC+1) of star culmination (Sirius) and star altitude. I also know RA (right ascension) and Dec (declination) of the star and I also know RA of the Sun for that day. I already calculated latitude but I can't figure out how to calculate longitude.

Thank you.

This is my first post, and I've accidentally deleted almost all I've written!

a later edit with an answer

The RA of a culminating star gives you the local sidereal time and comparison with Greenwich Sidereal Time gives your local longitude by noting the difference between the two.

If local sidereal time is less than Greenwich, your position is west of the prime meridian at Greenwich; if it is more, then your position is east of Greenwich.

Four minutes of Right Ascension = One degree of longtitude.
One minute of RA = 15 arc minutes of longitude.

A worked example:

Greenwich Sidereal time 17:42
          Local Sidereal time 16:36 hours

      Greenwich is ahead (more) than local, therefore local is West.

The difference = 42 minutes plus 24 minutes

= 1 hour 6 minutes of RA to the west of Greenwich

Thus, 60 minutes = 15° longitude

Add  4 minutes =   1°   longitude

Add   2 minutes =   0°30' longitude

= longitude 16°30'W

Which is accurate to 0°05' of the usually given figure for my location.

http://www.abecedarical.com/javascript/script_clock.html will show GST

I use a Sidereal Clock app on my Android tablet, set to show both GST and local sidereal time at my location.

information for background

This is the historical background for the search for a reliable method of calculating longitude when at sea

The Quest for Longitude and the Rise of Greenwich – a Brief History

I've edited this several times because I'm finding my way around how this works and made several editing mistakes! I've also tried to add another link, and got an error message, but no explanation. Is there a limit to the number of hyperlinks permitted?


What happens if two or three stars collide with each other? After finding out that there is a solar system with three stars, I did extensive research. I couldn't get a definitive answer. What would happen if they collided with each other?

I know there would be total destruction everywhere, but I would like to know what, including more scientific short- and long- term effects, would happen?

If you can't explain three stars, two stars colliding would be okay. I'm still an amateur at astronomy, so please don't make the answer too complicated.

The collision of another star with our sun would be a very effective way of ending all life on Earth. Fortunately stars are so far apart that the the chance of it happening in the life of the sun is very close to zero.

However in the densely packed cores of globular clusters the rate of collision is much higher, and there may be a collision somewhere in the milkyway once every 10000 years or so. Three way collisions don't happen (or are exceptionally unlikely) but would be similar to two two-star collisions.

The result varies depending on the size of the stars, their density, the relative velocity, and whether a collision is head on or not. Generally a denser star is likely to come out of a collision in better shape. So a red giant, could have most of its very rarified outer layers pulled off and the core escaping.

On the other hand, if two main sequence stars, like our sun, collide, they first compress each other. Some of the stellar envelope escapes perpendicular to the collision (think of jam being squeezed out of a sandwich). But within a few hours the two stars have mixed and merged, and   a new, larger star formed.

The mixing adds lots more hydrogen to the core of the star, which causes it to rejuventate. The first evidence of stellar collisions is in globular clusters, in which there are some large blue stars that seem younger than all the other stars in the cluster (Blue stragglers)

A white dwarf and main sequence star collision is bad news for the star. The collision causes the core of the star to heat up and a massive release of thermonuclear energy. Plenty more than is needed to completely disrupt the star. The star is completely destroyed in the process. However the white dwarf is relatively unscathed.

White dwarf-white dwarf collisions can, depending on the size result in gravitational collapse to a neutron star.

The paper Stellar collisions, mergers and their consequences is a very readable summary.


As matter approaches a black hole, does it speed up? If so, how do we know it speeds up?  Doesn't time slow down as gravity increases?  If time slows down around a black hole, is it possible matter doesn't actually speed up?

The answer is neither yes or no or possibly both.

Take a simple example. If something falls freely towards a black hole along a radial path, and is observed by someone who is far from the black hole, its velocity (according to the distant observer) is given by
$$v = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}c\, ,$$
(e.g. see chapter 6 of Exploring Black Holes by Taylor, Wheeler & Bertschinger - freely available) where $r_s$ is the Schwarzschild radius and the negative sign just indicates an inward velocity with $r$ decreasing.

If you plot this function (see Fig.2 in Ch.6 of Taylor et al. - freely available) you will see that initially the magnitude of the velocity increases as $r$ decreases, but as $r\rightarrow r_s$ then $v \rightarrow 0$ and the falling object appears to come to a standstill (actually, because the light from the object is gravitationally redshifted, this may not actually be observed). However, if the velocity first increases and then slows to a standstill, then it must go through a maximum!

The maximum observed speed in this scenario is achieved at $r=3r_s$ and is $0.384c$.

Of course this story is different for different observers. If you are the falling object then your speed just keeps increasing through the event horizon and towards the singularity. On the other hand, an observer who was somehow able to hover just above the event horizon would measure the falling object's speed as just below $c$ as it passed.


What does it mean for a star to go nova or supernova? Can I safely observe these? What does it mean for a star when people say it goes 'nova' or super nova, what are the differences?

More importantly, can I safely observe these with an amateur telescope? I imagine they would be big enough to spot should one happen!

It's a matter of size and stellar evolution.  There are many, many types of stellar explosions.  The University of Arizona has one page that describes these types.  generally, a Novs is not what we think of (i.e. a star exploding). That's actually a Type II Supernova.
According to that site:

Novae are frequently (perhaps always) members of binary systems where the more massive star evolves more quickly and becomes a white dwarf. The less massive star loses some of its hydrogen-rich outer atmosphere to the white dwarf where it collects.  The high gravity and intense heat of the white dwarf make conditions suitable for hydrogen fusion into helium. When enough hydrogen collects, it explodes in a fusion reaction, causing a nova. The gas ejected in the explosion moves outwards at 1000-2000 km/sec.

The Supernova has two distinct types:

The "Type I" supernovae arise when the white dwarf star member of a binary system accretes so much matter from its companion that it is tipped over the 1.4M Chandrasekar limit. The white dwarf collapses very rapidly (just a few seconds) until the infalling layers hit the very hard surface of the neutron star. The resulting shock goes back through the collapsing layers, and they explode by hydrogen fusion.

And the one we are more familiar with from popular media:

The "Type II" supernovae are the result of a massive star consuming all of its nuclear fuel and then exploding. The black line traces the evolution of a massive star schematically, until it reaches the supernova stage.

Portland Community College actually lists a lot of the variations that happen in the different types of supernovas out there.

Type I: supernovae WITHOUT hydrogen absorption lines in their spectrum
Type II: supernovae WITH hydrogen absorption lines in their spectrum.
Type II
The type II supernovae are massive stars whose iron cores collapse and then rebound, shock heating the outer layers of the star, which then explode outward.
Type I
The type I supernovae are subdivided into three subclasses, called (boringly enough), Type Ia, Type Ib, and Type Ic.
Type Ia: no hydrogen lines, no helium lines, strong silicon lines
Type Ib: no hydrogen lines, strong helium lines
Type Ic: no hydrogen lines, no helium lines, no silicon lines
Type Ib and type Ic supernovae are massive stars which lost their outer layers in a stellar wind before core collapse.
Type Ib supernovae lost their hydrogen-rich outer layer, revealing the helium-rich layer immediately below.
Type Ic supernovae suffered more mass loss as supergiants, losing both the hydrogen-rich layer and the helium-rich layer (revealing the carbon-rich layer below).
Type Ib and type Ic supernovae are essentially the same as type II supernovae.

As for observing them, since none of them are expected to be close to us (i.e. no stars near us able to go supernova), it should be safe. The problem is predicting when one will happen.  For instance, Betelgeuse is expected to go supernova "soon", but in stellar terms, that could be well over 100,000 years from now. The list of candidates in the Milky Way doesn't bode well for one in your lifetime. Although astronomers are often seeing them in other galaxies, and some are observable by amateur astronomers.  In fact, just this past August (2013) an amateur discovered a nova.


Universe and black holes I know that the radius of the observable universe is about 46.6 billion light years and is expanding at every instant of time due to dark energy . So is it possible that we our entire universe could be inside a black hole and that the edge of our observable universe represents the event horizon which in this case is expanding faster than the speed of light due to matter being constantly accreted inside it . Where that black hole is a part of another universe and that universe of another black hole and the pattern goes on . Is this possible ?

I know that the radius of the observable universe is about 46.6 billion light years and is expanding at every instant of time due to dark energy.

There's some ambiguity about dark energy. It's said to be responsible for the accelerating expansion of the universe as opposed to the expansion. See the Wikipedia dark energy article where you can read that the term "dark energy" was coined by Michael Turner in 1998. This was in a paper that referred to the type 1a supernovae observations made by the Supernova Cosmology Project and the High-Z Supernova Search Team. Their observations indicated that the expansion of the universe is accelerating, which was contrary to expectations and led to Nobel prizes in 2011.

  So is it possible that we our entire universe could be inside a black hole

No. We aren't in a black hole. A black hole is a place where the vertical light beam can't get out because the "coordinate" speed of light at the event horizon is zero. As Stephen Hawking said, the universe is like a black hole in reverse, but it isn't the same as a black hole.

  and that the edge of our observable universe represents the event horizon which in this case is expanding faster than the speed of light

The edge of our observable universe is a horizon of sorts, see Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Davis and Lineweaver. But it isn't the same sort of horizon as the black hole event horizon.

  due to matter being constantly accreted inside it.

That isn't supposed to be the reason why the universe is expanding.

  Where that black hole is a part of another universe and that universe of another black hole and the pattern goes on. Is this possible ?

Reputable people have talked about black holes as baby universes but there's no supporting evidence whatsoever. My understanding of general relativity and cosmology is such that I think such hypotheses are flat out wrong. So I would say no, it is not possible.


About Radius and Surface Area I am researching about the radius of a star and a its surface area. One question I have is about the effect of changing radii in stars. If for example we have one star with radius $r$ and another one with radius $2r$, we know that the area would mathematically be $A$ for the first star and $4A$ for the second star since $A=4 \pi r^2$. However, I intuitively think that this wouldn't be the case since if I doubled the radius, the surface area will quadruple. But, that means that there will be more molecules and whatever 'materials' the star is made of, meaning that there would be a greater gravitational force towards the core of the star. Would it thus make sense to say that there would be some sort of 'compression' that occurs which pulls all of this extra mass towards the center (effectively making the star denser), and thus the star with $2r$ actually has a surface area of $3\frac{1}{2}A$ rather than $4A$?

(So technically I'm asking if $A\neq 4\pi r ^2$ in case of stars because more mass (due to the extra radius) would mean that the gravitational force towards the core would be greater, making the star denser in the process, but also have less surface area than what doubling the radius would mathematically correspond to.)

You seem to be confusing the simple mathematical relationship between radius and surface area, and the more complex relationship between mass and size.

If you double the radius of a sphere the surface area quadruples.  This is pure maths, and is not particular to stars. The volume is multiplied by 8.

But in a star, increasing the amount of matter by a factor of 8 will not result in the volume increasing by a factor of 8. The relationship between the mass of a star and it's volume is not simple and depends on factors like age, and the particular combination of elements in the star.

Doubling the radius does quadruple the surface area.

Doubling, or even increasing 8 fold, the mass does not quadruple the surface area.


clarification of the Kozai mechanism As Wikipedia says,

  In celestial mechanics, the Kozai mechanism, or the Lidov–Kozai mechanism, is a perturbation of the orbit of a satellite by the gravity of another body orbiting farther out, causing libration (oscillation about a constant value) of the orbit's argument of pericenter. As the orbit librates, there is a periodic exchange between its inclination and its eccentricity.

My questions are:

Question A
Which is the least massive object? The tertiary object, which is the furthest object, or the satellite in the inner binary? It seems it is not necessary that the tertiary object be the least massive one, which violates what Wikipedia says.

Question B
How does the three-body system evolve?

  There is a periodic exchange between its inclination and its eccentricity.

Whose inclination and whose eccentricity? Please specify them using m0, m1 or m2 in the figure below.

The orbit of the inner binary should become and more circular. Can it become circular, eccentric, circular, eccentric?

Question C
The inner binary will lose energy during the whole process, right?

Which is the least massive object?

Quoting Wikipedia,

  In the hierarchical, restricted three-body problem, it is assumed that the satellite has negligible mass compared with the other two bodies (the "primary" and the "perturber"), . . .

This is the case studied in Kozai (1962), specifically, the case of asteroids being perturbed by Jupiter. While not massless, the difference in mass is large enough that the asteroid's mass is negligible.

  How does the three-body system evolve? . . . Whose inclination and whose eccentricity?

Wikipedia is again rather direct in this, stating that the conserved quantity $L_z$ depends on the satellite's orbit's eccentricity and inclination:
$$L_z=\sqrt{1-e^2}\cos i$$
As the satellite's mass is negligible, it will not have any significant effect on its perturber.

  Can it become circular, eccentric, circular, eccentric?

This is basically asking if the eccentricity (and therefore inclination) can be described by some periodic function. Again, this is given in the Wikipedia article. A slightly different but just as simple xpresison is given in Takeda & Rasio (2005):
$$\text{Kozai Period}=P_{\text{perturbed}}\left(\frac{m_{\text{star}}+m_{\text{perturbed}}}{m_{\text{perturber}}}\right)\left(\frac{a_{\text{perturber}}}{a_{\text{perturbed}}}\right)^3(1-e_{\text{perturber}}^2)^{3/2}$$
In the approximation discussed above, in cases of extreme mass difference, $m_{\text{perturbed}}\to0$.

  The inner binary will lose energy during the whole process, right?

The whole thing is periodic, so no energy is lost.


As days get longer, why isn't the added daylight split evenly between sunrise and sunset? I have a basic understanding of the sinusoidal relationship of the rate of change of daylight as the year goes on. But it seems intuitive that the changes should be equally split between sunrise and sunset.

For my location, Bend Or., todays sunrise/sunset is: 07:40 and 16:43 And for January 31st it is: 07:24 and 17:14

So sunrise will be 16 minutes earlier and sunset will be 31 minutes later. It appears that there is twice the added daylight to sunset than to sunrise. Why aren't the differences equivalent?

It's a consequence of the equation of time, which represents the time difference between apparent solar time (time as measured by a sundial) and mean solar time (a fictitious device; essentially mean solar time is time as measured by a clock). Below are two plots from the referenced website.

Figure 1: Equation of time. Currently (early January), a sundial will be slightly behind a clock and is getting slower.

Figure 2: Time derivative of the equation of time. Apparent solar time changes most rapidly from early December to early January, with the maximum change occurring on December 25.

Days are currently getting longer in the northern hemisphere, but noon is advancing by 20 to 25 seconds per day. Sunset advances by half of the increase in the length of day plus that 20 to 25 second advance in noon. Sunrise advances by half the increase in length of day less that 20 to 25 second advance in noon.

The increase in the length of day (time from sunrise to sunset) depends on latitude. In Bogota Columbia (4°35′53″N latitude), length of day increases by only 3 to 10 seconds per day during January, growing throughout the month. That tiny change in length of day means sunrise continues to advance throughout the month of January in Bogota. On the other hand, in Reykjavík, Iceland (64°08′N latitude), length of day increases drastically from day to day, even at this time of year. Sunrise started becoming earlier in Reykjavík on Christmas day.


Why are all quasars so far away? Why are all quasars so far away?

If the universe is homogeneous, we could expect to have a homogeneous distribution of quasars, but all of then seem to be far away from Earth. Why is that?

There are essentially two reasons.

First, quasars are rare objects, so even though they are homogenously distributed on large scales, the average distance is large. Moreover, the brightest quasars are even rarer, but visible to large distances, so their average distance is even larger.

Second, most quasars were most active at redshift $z\sim2$. This is thought to be because quasar activity is triggered by galaxy mergers, which bring gas into galactic centres so that the supermassive black holes can be fed. Galaxy mergers and star formation peaked at $z\sim2$.

Finally, there is some bias in the sense that astronomers are looking for the most distant (and hence oldest) of objects, because they are more interesting than nearby objects when it comes to learn about the formation history of the universe.


How does the surface gravity on Mars vary between the equator and its poles? A day on Mars is similar in length to the Earth's day. On Earth, due to rotation, the surface gravity varies from 9.78 m/s² (0.997 g) on the equator to 9.832 m/s² (1.003 g) on  the poles. I wonder how the gravity varies on Mars where the day is of similar length. We hear of different values, such as 3.69, 3.711 or 3.72076 m/s². All values around 0.38 g. But I wonder what is the exact value of the equatorial gravity on Mars (including centrifugal forces), what it's on the midlatitudes and on the poles.

The Phoenix probe landed near the north pole; do we know whether it has a gravimeter and measured the surface gravity on its landing place? I know the Curiosity rover which landed close to the equator measured a gravity of around 3.71 m/s² from which I conclude the 3.711 value is for the equator (0.378 g).

When I calculate the gravity of Mars with (G x m)/r² it doesn't include the centrifugal force of Mars' rotation.

The following is excerpted from from this answer to the Space Exploration SE question "Do Curiosity's reported measurements of Mars' surface gravity (~3.717 m/s^2) include centrifugal effects?".

It essentially answers my question. I would appreciate if someone gave the info about the gravimeter on the Phoenix lander too.

So the surface gravity on Mars is:

3.70703 m/s² (0.378 g) on the equator
3.71683 m/s² (0.379 g) on the midlatitudes
3.73493 m/s² (0.381 g) on the poles

  So here's a complete tally using

  $$a_G = -GM \frac{\mathbf{r}}{r^3}$$

  From Geopotential_model; The_deviations of Earth's  gravitational field from that of a homogeneous sphere:

  $$a_{J2x} = J2 \frac{\mathbf{x}}{r^7}(6z^2 - 1.5(x^2+y^2))$$
  $$a_{J2y} = J2 \frac{\mathbf{y}}{r^7}(6z^2 - 1.5(x^2+y^2))$$
  $$a_{J2z} = J2 \frac{\mathbf{z}}{r^7}(3z^2 - 4.5(x^2+y^2))$$

  $$a_C = \mathbf{r_{xy}} \omega^2 $$

            magnitudes shown only (sign indicates generally "up" or "down")
                  at the equator        at the pole
              h = 0     h = -4500 m        h = 0
GM          -3.71317     -3.72303        -3.75729
J2          -0.01092     -0.01088        +0.02236
centri      +0.01706     +0.01697         0.0

vector sum  -3.70703     -3.71683        -3.73493

  Do Curiosity's reported measurements of Mars' surface gravity (~3.717 m/s^2) include centrifugal effects?

  Yes they do! At 5.2 degrees south latitude and -4500 meters altitude (bottom of Gale Crater) the acceleration is -3.7168 m/s taking into account GM, J2, and centrifugal effects.

  So as Cheap Trick tells us and Meatloaf reiterated more elegantly in Roadie:

  Everything works if you let it!

  Here's some Python for double-checking my math:

def accelerations(rr):

    x,   y,   z   = rr
    xsq, ysq, zsq = rr**2

    rsq   = (rr**2).sum()
    rabs  = np.sqrt(rsq)
    nr    = rr / rabs
    rxy   = np.sqrt(xsq + ysq)
    rrxy  = rr * np.array([1.0, 1.0, 0.0])
    nxy   = rrxy/rxy
    rm3   = rsq**-1.5
    rm7   = rsq**-3.5

    acc0  = -GM_mars * rr * rm3

    # https://en.wikipedia.org/wiki/Geopotential_model#The_deviations_of_Earth.27s_gravitational_field_from_that_of_a_homogeneous_sphere
    acc2x = x * rm7 * (6*zsq - 1.5*(xsq + ysq))
    acc2y = y * rm7 * (6*zsq - 1.5*(xsq + ysq))
    acc2z = z * rm7 * (3*zsq - 4.5*(xsq + ysq))

    acc2  = J2_mars * np.hstack((acc2x, acc2y, acc2z))

    accc = nxy * omega**2 * rxy

    return acc0, acc2, accc

import numpy as np

halfpi, pi, twopi = [f*np.pi for f in [0.5, 1, 2]]
degs, rads        = 180./pi, pi/180.

R_mars  = 3396200.0
GM_mars = 4.282837E+13   #  m^3/s^2  https://en.wikipedia.org/wiki/Standard_gravitational_parameter
J2_mars = GM_mars * R_mars**2 * 1960.45E-06     # https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

Req = 3396.2 * 1000. # meters https://en.wikipedia.org/wiki/Mars
R   = Req - 4500.    # https://en.wikipedia.org/wiki/Gale_(crater)
Rpo = 3376.2 * 1000. # meters https://en.wikipedia.org/wiki/Mars

sidereal_day  = 1.025957 # https://en.wikipedia.org/wiki/Mars
T             = sidereal_day * 24 * 3600.
omega         = twopi/T
print "omega:   ", omega
print ''

aaacs = accelerations(np.array([Req, 0, 0]))
print "Req: ", Req
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())

print ''
lat = rads * -5.4
aaacs = accelerations(np.array([R*np.cos(lat), 0, R*np.sin(lat)]))
print "R: ", R
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())

print ''

aaacs = accelerations(np.array([0.0001, 0, Rpo]))  # avoid divide-by-zero (lazy!)
print "Rpo: ", Rpo
for thing in aaacs:
    print thing, np.sqrt((thing**2).sum())
print "total: ", sum(aaacs), np.sqrt((sum(aaacs)**2).sum())


At what distance does Proxima Centauri become visible to the naked eye? The red dwarf star Proxima Centauri is the closest-known star to the Sun (being about 4.2 ly or 1.3 pc away), however it's not visible to the unaided eye for it is too small and dim. I'd like to know how close one should get to see Proxima well, like most other stars. In Celestia the star would become visible to me at about 0.75 ly away. Is Celestia right?

"Back of an envelope" calculation:

Proxima Centauri has an apparent magnitude of about $11$. The faintest objects visible to the unaided eye have magnitudes about $6.5$. So we need to decrease Proxima Centauri's magnitude by about $5$ in round numbers, which corresponds to an increase in brightness of about $100$ times. To achieve this we would have to be about $10$ times closer to Proxima Centuari than we are now, which gives a distance of about $0.4$ ly.

At a distance of $0.75$ ly the apparent magnitude of Proxima Centauri would be about $7.5$, so I think Celestia is being a little optimistic.


Does the Sun belong to a constellation? Each new star we find is generally considered to be part of the constellation it is nearest to.

Our Sun is obviously a star, just much closer. Is our Sun part of any constellation? If so, which constellation is it a part of?

Constellations are human constructs to make sense of the night sky.  When you are trying to find your way around, it helps to "chunk" stars into patterns and assign those groupings names.  When I want to point out a particular object in the sky (say Polaris, the North Star), I start by pointing out a familiar constellation (say Ursa Major, the Big Dipper).  From there, I can tell my friend to follow this or that line to get them to look where I'm looking:

With the advent of computerized telescopes and large data sets, constellations are less important for professional astronomers.  However, many stellar databases use Flamsteed or Bayer designations, which assign stars to constellations.  In order to include all stars, the sky is divided into irregular regions that encompass the familiar constellations.

So, which constellations is the Sun assigned to?  Well, from the perspective of someone on the Earth, the Sun moves through the constellations throughout the course of the year.  Or rather, Sol moves through the region of the sky where some of the constellations would be seen if its light did not drown out distant stars.  Our moon and the rest of the planets move through those same constellations.  (The Greek phrase which gives us the word "planet" means "wandering star".)

The current position of the sun against the background of distant stars changes over the course of the year.  (This is important for astrology.)  It's a little easier to make sense of with a diagram:

So perhaps a better question is:

  What constellation does the Sun belong to today?

Presumably an observer on an exoplanet would assign Sol to some constellation that is convenient from her perspective.  But from our perspective within the Solar system our sun, moon, and planets are not part of any constellation.


absorption line from the chromosphere If I understand correctly, the H-alpha and Ca II K lines are absorption lines of the sun and allow to see the chromosphere. Following the Kirchhoff-Bunsen law, an absorption line is produced by a gas which is cooler than the source of light.

However the chromosphere is hotter than the photosphere and the temperature increases outward (https://www.researchgate.net/figure/The-figure-plots-profiles-of-mass-density-dashed-line-and-temperature-solid-line-in_fig6_312376685). This is in contradiction with the previous hypothesis. Then how can we see the chromosphere in absorption line? What am I missing in the reasoning? Is the Krichhoff-Bunsen law only valid for a thin medium and here we would be in presence of a thick one?

thank you in advance for your help.

Most of the absorption line comes from the photosphere in the region where the temperature is still dropping.  Nevertheless, you can also get an absorption line where the temperature is rising, due to scattering effects.  The Kirchoff-Bunsen law you describe is a simplified rule that assumes the lines form in thermal equilibrium with the local temperature, but sometimes scattering produces non-thermal-equilibrium ("non-LTE") effects.  With scattering, line photons that get absorbed get re-radiated right away without coupling to the local temperature, and their re-emission has a 50% chance of being radiated back downward, where they will be reabsorbed deeper in the atmosphere.  This contributes to absorption, especially near line center (and can cause a feature called "central reversal" there).  But it is true that you can also get emission lines from very strong chromospheres where the Kirchoff-type effect does occur, see the profiles from chromospherically active dwarf stars (called "dMe stars").


What does gravity-wave noise resemble locally? Now that we know that gravity waves exist,  we can consider to design a new unit of measurement,  i.e.  deciGravs (from dB) and we can predict what the backround noise of dG's is locally and in space.

Consider a theoretical high-res LIGO that can graph gravity waves here and far away,  we can ffwd the record to represent the dG's at human timeframes.

What are the loudest local waves? Can we see the moon as the loudest local wave,  followed by other planets and moons,  are local bodies much louder than local bh waves?

There is an expected 3D wave map type locally and in deep space that we can model with graphical animations? Can we draw it easily in 2D/3Dusing 5 lines of code for every body and lots of processing?

Perhaps it makes sense to define a frequency chart of dG's and astroHz and a bit like the electromagnetic waves,  and chart gravity-waves,  amplitude and frequency trends and types?

Perhaps supernovas and solar flares and many background waves appear on graviCharts.  Can we know what graviCharts and backround noise of gravity resembles here and far away?

For orbiting objects, gravitational wave intensity depends on the second time derivative of the quadrupole moment of an isolated system's stress–energy tensor (which for the wimpy objects in our vicinity is proportional to the mass times the line of sight acceleration) and drop off with the distance. So you can easily estimate the GW intensity of any object.  Hint: It's small for nearby objects.

As requested, I have tried to estimate the GW amplitude and I found the numbers somewhat surprising.  I looked at the Moon, Jupiter, Io, the LIGO neutron star and a pair of supermassive BHs if they were at the Galactic center.

I got the mass of each object in Suns, the orbital radius in miles, the orbital period in years and the distance in AU.  The angular acceleration is computed as r * omega2. Multiply by mass to get GW radiation at source. Divide by distance to get GW amplitude observed at Earth. (All units relative and arbitrary.)

Object             Acceleration   Mass*Acceleration  Est. amplitude
Moon               3 x 10**7      1                  4 x 10**2
Jupiter            3 x 10**6      3 x 10**3          7 x 10**2
Io                 1 x 10**10     5 x 10**2          1 x 10**2
LIGO Neutron star  1 x 10**19     3 x 10**16         4 x 10**6
Supermassive BHs   1 s 10**9      2 x 10**15         3 x 10**6

(The supermassive BH numbers are really quite arbitrary, since you can assume any separation you like and pretty much any mass.  I picked 1 AU and 1,000,000 suns mass.  You can easily get much larger numbers, but the frequency will typically be too low to be seen by LIGO.)

The other really important thing to remember is that in spite of Io being about as bright as the sources LIGO has observed, the frequency of Io's radiation is about 10-5 Hz. LIGO is only really sensitive in the 10-1000Hz range and has no chance at all of ever seeing IO or Jupiter or the Moon or even a pair of supermassive BHs that are not in the last milliseconds of merger.  If eLISA gets built, it can probably see at least some supermassivs BH pairs.


Simple experimental evidence that Earth revolves around Sun What are the simplest experiments or calculations that give evidence that the earth revolves around the sun? Can you please explain them and reference the history? Many simple explanations such as this cite observations such as that relative position of two stars are observed from earth vary every night - which would not be true if the stars orbited the earth. But isn't the observation also consistent with a model where the stars orbit the earth but do so at different speeds, while the earth still orbits the sun? Simple explanations would be helpful.

The answer is ironic: Without good instruments, there is no evidence. The people who thought that the Sun went around the Earth were perfectly correct as far as the actual evidence went until the early 1700s and mid-1800s when two lines of evidence opened up that showed that the Earth moved.
Aberration of Starlight
Wikipedia has a correct but over-complicated explanation. The easiest way to think about it is to imagine yourself at a stop sign in a car in the rain, and the rain is falling straight down. When you start moving, the rain's apparent direction of fall changes so that it appears to be falling from ahead of you and slanting down towards you. That's aberration.
In the early 1700s, the stars were discovered to be shifting position, and in 1727, James Bradley correctly identified it as abberation of starlight due to the motion of the Earth around the Sun. (For any star in the ecliptic, the Earth is moving towards it at some time of the year and away from it six months later.)
Parallax
Wikipedia's article on parallax is better, and I refer you to it for details. Basically, if you hold your finger up before you and look at it with your left eye closed, and then with your right eye closed, it appears to jump with respect to the background -- the wall beyond or the trees outside or whatever. Switch back and forth between your eyes quickly to see it clearly.
As the Earth circles around the Sun, nearby stars also appear to shift their position relative to the more distant stars.  A key point here is that there were good scientific reasons to suppose that the stars were much smaller than the Sun. Seen through a telescope, stars showed disks and if they were like the Sun, their distance could be deduced from those disks. And they were close enough that if the Earth really went around the Sun, parallax should have been observed. But it wasn't and the lack of any noticeable parallax was a strong empirical argument against Heliocentric theories.
In reality of course parallax exists, but the parallax of all stars is small, because they are much further away than was estimated from their disks. (The visible disks were actually diffraction disks and not true disks at all -- but it was not until nearly a century later that diffraction began to be understood.)  Friedrich Bessel first measure the real parallax of a star in 1838.


What would happen if we released water into deep space? If we went to deep space by a spacecraft and released water into space, what would happen? Would the water freeze to ice or would it remain as a liquid?

Water can't exist in a vacuum.  Water in space (atmosphere, no gravity) tends to form into a ball, but take away the atmosphere and it would rapidly fly apart.  It actually wouldn't look like boiling.  You get boiling on Earth because water is held in a pot by gravity.  In empty space the water would rapidly and somewhat explosively disburse

This article says it would look something like what happens when you throw hot water into very very cold air and it rapidly turns to snow.  That's probably a pretty good visual image if you imagine no gravity and everything would just continue to fly apart, not slow by air resistance or fall to the earth.

In cold space the water would freeze, but some of the liquid would also effectively also boil away, even in very cold temperature because liquid isn't stable in a vacuum.  As I said, some of it would boil, but it wouldn't look like boiling, It would fly apart faster than it would freeze but you'd also see it freeze fairly rapidly.   If the water was enclosed, freezing would take some time, but in a vacuum and spreading out, it would freeze into snow and ice crystals rapidly.

A percentage of the water would become gas and that would help carry some heat away from the remaining water, speeding up the freezing process.  That gas would fly away at high velocity and not be visible.   If you collected and weighed the ice-crystals after the water was tossed into space you'd probably find that a percentage of the water escaped as gas.

You can get an idea of water turning into ice by heat transfer and escape from forming into gas molecules in a video that show what happens to water in a vacuum chamber.   A vacuum chamber can't replicate the rapid spreading out due zero gravity but it demonstrates the water turning to ice, even at room temperature.   Here's one video of that.


How to obtain distance modulus and Total extinction from Isochrone fitting to CMD? This is my first post here.

I am studying the following article: https://arxiv.org/abs/1901.03574.

On Page 5, Sec 3.1, the authors are presenting their results they obtained via Isochrone fitting to the CMD of the two GCs being discussed. My questions are:

How can I find out the distance modulus and the Total extinction $A_{F606W}$ from just the fitting? I know that Isochrone fitting is done when we need to estimate the age. In order to fit, we need to move the Isochrone both vertically (Distance modulus?) and horizontally (Color excess/Reddening/Extinction?). Is my understanding correct?
How do they convert the Reddening in filters to reddening in B-V? What is the technical details/physics of it?
What is the role of the Extinction law in getting these numbers?
In the end, they calculate the distance, which is simple once we get the distance modulus. Why is it the distance from the Sun? Did we assume a heliocentric coordinate system somewhere?

Thank You.

The isochrones are not straight lines in the CMD. Extinction moves the stars both redward (right) as well as fainter (down) in the CMD. The vertical shift (the extinction $A$) is calculated from the horizontal shift (the reddening) using an extinction law. (e.g. $A_V = 3.1 E(B-V)$). The effects of extinction and distance (which only moves the stars vertically) can therefore be separated.
In the case of the paper referred to, the globular clusters have a horizontal branch and a giant branch. The horizontal branch is almost ...horizontal and the giant branch is almost vertical. This enables the authors to be independently sensitive to a vertical and horizontal shift of the model isochrones, as described above.
+3. There are simple conversion factors published in several papers. Extinction is wavelength dependent, so the spectrum of a star is attenuated in a wavelength dependent way  The reddening in any colour (e.g. E(B-V)) is just the difference in extinction at the two wavelengths (well it needs to be integrated over the photometric bands) i.e. $A_B - A_V$. These differences in extinction coefficients can be calculated for any pair of bands. They depend on the wavelength dependence of extinction, aka the extinction law, but also on the intrinsic spectrum of the star (which is frequently ignored). Extinction laws  can be empirically determined by observing stars with a known intrinsic spectrum.

'4. Does it matter? Have you compared a parsec with an astronomical unit?


Were radio spectroscopic observations ever used to first measure a distance successfully? Usually, especially for an extragalactic object, its redshift is determined by optical spectroscopic redshift. But the angular resolution of early radio observations is poor and an optical counterpart needs time to be found. Is there any celestial object whose distance was determined firstly by the redshift of radio spectral lines?

Yes, radio spectra have been used extensively to find the distances and locations of HI regions and molecular clouds inside the Milky Way. Observations of the 21 cm hydrogen line and/or several carbon monoxide lines (in particular, $\text{CO}(1\to0)$) enable us to make radial velocity measurements of clouds within the galaxy. From there, some geometry (see Roman-Duval et al. 2009) allows us to determine the distance to the cloud, if the cloud lies farther from the galactic center than the Sun does.
Unfortunately, if the cloud is closer to the galactic center, there are two possible solutions for the distance between the cloud and the Sun, something called the kinematic distance ambiguity. It occurs because at two possible points along the line of sight - if the cloud's orbit is smaller than the Sun's - the projected cloud velocities are identical. There are a few ways we can try to do this, including HI absorption (Kolpak et al. 2003).
Here's a diagram to make the distinction clearer, from Roman Duval et al.:

Figure 1. Sketch of the HISA method to resolve the KDA. In the inner Galaxy, a single Galactocentric radius (determined by the radial velocity of the cloud)
corresponds to two distances along the line of sight, a near (in blue) and a far (in red) kinematic distance. The near and far kinematic distances correspond to the same
radial velocity $V_r$, which is the projection of the orbital velocity $V_0$ of a cloud around the GC onto the line of sight. At the tangent point, the orbital velocity of a cloud is parallel to the line of sight. In this case, the radial velocity is maximal and the near and far kinematic distances are identical. The cold HI embedded in a cloud
located at the near kinematic distance absorbs the 21 cm radiation emitted by a warm HI background located at the far distance. Consequently, the HI 21 cm spectrum
toward a near cloud exhibits an absorption line that is coincident with a 13CO emission line from the cloud. A cloud located at the far distance does not lie in front of
a warm HI background emitting at the same velocity as that of the cloud. Therefore, there is no absorption feature in the HI 21 cm spectrum toward a cloud located at
the far kinematic distance.

The technique has also been applied to extragalactic sources (though often lines in the optical region are used, even for radio galaxies). One example is the Leo Ring discovered by Schneider et al. in 1983. It's mostly neutral hydrogen, so the 21 cm line was initially used to derive a redshift, recessional velocity (a system velocity of $V=960\;\text{km s}^{_1}$) and distance. That velocity measurement corresponds to $d\simeq14\;\text{Mpc}$, fairly close to that of the Leo Group. I wouldn't be surprised if other extragalactic blobs of neutral hydrogen have also had their distances measured via the 21 cm line.


Would the electron cyclotron-maser emission mechanism affect Proxima b's ability to retain an atmosphere? In a recent arXiv preprint, Pérez-Torres et al. "Monitoring the radio emission of Proxima Centauri" claim the detection of radio emissions synchronised with the orbit of the planet Proxima b. They explain this as the result of electron cyclotron-maser (ECM) emissions, making the Proxima–Proxima b system a scaled-up analogue of the Jupiter–Io or Jupiter–Ganymede systems.
Would this mechanism make it more difficult for Proxima b to retain an atmosphere?

This mechanism, being actively emitting in the radio-wavelengths, is certainly negligible for the overall atmospheric energetics at Proxima b. One can conclude this by taking the band luminosities from the cited paper ($\rm 2.51\times10^{20} erg/s$, p.3, first paragraph) and compare them to the solar constant at the planets orbit, which should amount to $\rm9\times10^{23}erg/s$.
The latter is however only the thermal energy contribution given by the stellar effective temperature, there might be other significant chunks of energy contributed in the UV and X-bands.
The luminosities thus considered would be important for driving both bulk escape and jeans escape. Being negligible in the radio-range, shows that this process is unimportant for the retention of a bulk atmosphere via thermal mechanisms.
In terms of non-thermal mechanisms it's possible to imagine that there would be some resonant absorption of the ECM emission at the planetary radius, driving scattering of heavier species. Considering however the scalings of the cyclotron frequency $f=qB/{2\pi m}$, and the fact that the magnetic field is weaker at the planet compared to the emission region at the star, the scattered particles would have to be lighter than electrons. Those particles do not exist in atmospheres in significant numbers, only in particle accelerators.
Hence we can conclude that non-thermal processes driven by ECM heating will also not play a role for atmospheric escape.


What would a gas giant consisting of only earths atmosphere look like? Putting aside how such a planet could be formed and just lets assume it exists.

What would an gas giant "look" like if it just consisted of the gas compound earths atmosphere consists of.

Could we just watch through it?
Would we see a burning core and therefore see something like a dim star?
Or would there be still causes that would block vision, even if there were no other components included?

And would such a planet be able to hold it self together?

Jupiter's atmosphere, if you could see through it, the inside of the planet would begin to glow visibly red from heat perhaps a couple/few thousand miles below it's upper clouds.

Source

The problem is, some two thousand miles of atmosphere above that visible glow is almost certainly well past the point of transparency.   Add to that, mixing and cloud formation would make seeing through all that impossible - no matter what the atmosphere was made of.

Hot gas rises and as it rises, some gas condenses from gas into liquid droplets (clouds), but these clouds form in updrafts so even though they start falling back towards the planet, the thick atmosphere and the updraft movement maintains the clouds for some time.   There's different layers of clouds too.   Jupiter's upper most clouds are Ammonia, but below that there's water clouds (and perhaps others).   go deep enough you might get sulfur clouds (I'm not sure) or more dense hydrocarbon (gasoline like) clouds.

An atmosphere that thick maintaining transparency is impossible.   Hydrogen and Helium are (to my knowledge) just as transparent as Oxygen and Nitrogen, maybe more, but the mixing and condensation gives Jupiter a constant cloud cover, different colors mostly represent different temperatures and depths from which that gas came from.

If you gave Jupiter an Oxygen-Nitrogen atmosphere but maintained it's inner core, the inner hydrogen would rise and mix with the Oxygen-Nitrogen.   Chemistry would inevitably happen and the Oxygen would bind with the hydrogen and make water.   Some of the Nitrogen would probably form Ammonia and in the end, Jupiter might not look that different.   Initially it might have a bluish tint, but only temporarily.  It would likely always have clouds and mixing some heavier/darker gases from below.   You'd never see it's core, no matter what.

Now if you had a planet with only one gas in it's atmosphere and entire makeup - avoiding any chemistry, say a planet made entirely of hydrogen or entirely of Argon or Oxygen - then you might get specific colors relating to the specific element.   But such planets are pure fantasy.

My best guess is that Jupiter's appearance wouldn't change much if you gave it an Earth-like atmosphere, but Jupiter is basically made of gas, so if you did that, it would just mix and it's inner hydrogen would rise to the top and Oxygen-nitrogen would sink.   It's the trace gases that give Jupiter it's color anyway, not it's hydrogen.

Layman's answer.  :-)


The formula in the wikipedia article explicitly uses integer division with round toward zero. Python's integer division uses round toward negative infinity (i.e., floor).
The wikipedia article formula repeatedly uses (m-14)/12. This evaluates to -1 for months 1 and 2 (January and February), zero otherwise. You can use the wikipedia article formula in python3 if you change every instance of (m-14)/12 in the wikipedia article formula to any of the following:

int((m-14)/12)The single / performs floating point division. Python's built-in int function rounds toward zero.
((m+9)//12-1)The inner (m+9)//12) evaluates to 0 for January and February (m=1 and 2), 1 for all later months.
(-1 if m<3 else 0)While more verbose, this does a better job of conveying the intent of the wikipedia article's (m-14)/12.


What would happen to the Sun's evolution if it collided with another yellow dwarf? How would the Sun evolve if it collided with another G-type main sequence star, perhaps a more massive one like Alpha Cen A? Since its mass would then be above the Chandrasekhar limit, could it eventually become a neutron star instead of a white dwarf? Or would it lose enough mass until then so that it would still become a white dwarf, just like Sirius A (whose mass is above the Chandrasekhar limit too but will develop into a white dwarf anyway due to mass loss)? Would the Sun's yellow-dwarf-to-red-giant phase be prolonged due to the additional mass? When largest, would the red giant be bigger than the currently estimated 256 solar radii?

If the Sun collided with another star about the same mass, then its mass would be slightly less than 2 solar masses, as some material would be ejected away from the merger. This would result in an A-type star, as the merger's mass is about 2 solar masses. A good example of a 2 solar mass star is Fomalhaut A, which is an A3V  star. Therefore, this merger would be somewhere around Fomalhaut's position on the HR diagram.
If the mass of a star is doubled, then its lifespan is decreased by eightfold, meaning that the merger would live for about $\dfrac{10}{8}= 1.25 \text{ billion years}.$ With that said, the transition to the red giant phase would not be prolonged, but shortened.
When this merger becomes a red giant, its core will have less mass than the Chandrasekhar limit, and become a white dwarf. At its largest, it will exceed 256 solar radii, up to about 400 radii, according to Wikipedia. χ Cygni is a red giant that is about 2 solar masses, which is a good reference to the aging merger. This star will become a white dwarf, so that is further proof that the Sun-merger will have a shorter lifespan as well as not have enough mass to exceed the Chandrasekhar limit.


Validity of ephemeris time In the wikipedia article for ephemeris time as first adopted in 1952, it does not explain clearly how or why it was superseded. For example, it reads:

In 1976 the IAU resolved that the theoretical basis for its current
(1952) standard of Ephemeris Time was non-relativistic, and that
therefore, beginning in 1984, Ephemeris Time would be replaced by two
relativistic timescales intended to constitute dynamical timescales:
Terrestrial Dynamical Time (TDT) and Barycentric Dynamical Time
(TDB).[28] Difficulties were recognized, which led to these being in
turn superseded in the 1990s by time scales Terrestrial Time (TT),
Geocentric Coordinate Time GCT(TCG) and Barycentric Coordinate Time
BCT(TCB). [16]

So, from this it appears that ephemeris time was never superseded and it continues to be evolved with a different theoretical basis?
From reading Robert Newton's books such as "Ancient planetary observations and the validity of ephemeris time", it would seem that ephemeris time is invalid because of non-secular accelerations in the motion of Mercury and Venus.
So, I guess my question is: does current astronomical theory consider ephemeris time to be invalid or valid, and if it considers it to be valid, then what is current evaluation of Newton's calculations and their implications for the validity of ephemeris time?

It depends on what one means by "Ephemeris Time." Officially, the Newtonian-based Ephemeris Time has been superseded by the relativistically-correct Barycentric Dynamic Time (TDB) and Barycentric Coordinate Time (TCB). However, the concept of what was meant by Ephemeris Time carries on to this day. The intent of Ephemeris Time was to use the solar system as a timekeeping device. This is what TDB and TCB do, but in a relativistic rather than Newtonian setting.

TCB is a theoretical time scale whose clocks are infinitely removed from the solar system, are co-moving with the solar system barycenter, and that tick at the rate of one tick per 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom (at a temperature of 0 K). (Note well: This is the same definition currently used for an SI second.) That TCB clocks are well removed from the solar system means that a second measured by an atomic clock on the surface of the Earth ticks a tiny bit slower than does a TCB clock.

These slightly different tick rates was perceived as a problem by some. The solution was eventually to redefine TDB as a scaled version of TCB such that, on average, TCB ticks at the same rate as atomic clocks on the surface of the Earth. Atomic clocks on the surface of the Earth tick faster / slower than would TCB clocks based on whether the Earth is further from / closer to the Sun than average. But averaged over the course of years, TCB and TAI remain more or less in sync.

Regarding the more or less: TDB is updated every year. As technology improves, these subtle updates change the timing of past events. TAI does not do this. As technology improves, older atomic clocks are replaced with newer ones. Atomic clocks have a shelf life of a decade or so. This can result in discontinuities in atomic time. This is not a problem with TDB as it is re-issued annually. This also means that TDB is independent of atomic time.


Error propagation methods for orbit parameters Recently I've encountered an article in our local astronomical popularization magazine. It is about a well known task of Sgr A*-black hole's mass estimation. The article is written as a step by step guide for young researchers. For example, the orbital parameters of the S2 star should be found using a drawing on a sheet of paper.
That is why I believed there is no need for very precise calculations
for this task.
However, in addition to the $ x $, $ y $ coordinates of the S2 star, I was also given the corresponding uncertainty values $ \Delta x $, $ \Delta y $ (19  coordinate measurements for the time period from 1992 to 2003):
\begin{array}{|c|c|c|c|}
\hline
time    & x      &   \Delta x    & y      & \Delta y \\ \hline
1992.226 & 0.104 & 0.003 & -0.166 & 0.004 \\
1994.321 & 0.097  & 0.003 & -0.189 & 0.004 \\
...     & ..     & ..    & ..     &       \\
2003.353 & 0.077  & 0.002 & -0.030 & 0.002 \\
2003.454 & 0.081  & 0.002 & -0.036 & 0.002\\
\hline
\end{array}
Using direct least squares method for S2 orbital fitting by an ellipse I've found parameters
$a,\ b,\ c,\ d,\ e,\ f$ of the conic equation: $$ax^2+bxy+cy^2+ dx+ey+f=0$$  I've
also decided to simulate the error propagation in a Monte Carlo fashion to find
the uncertainty in the estimates of $a,\ b,\ c,\ d,\ e,\ f$. And I succeeded in that.
But, here is the question: is there any other appropriate,
"analytical" way (not a Monte Carlo method) to find the uncertainty in
the estimates of $a,\ b,\ c,\ d,\ e,\ f$ which may be used in usual astronomers' practice?

ProfRob's comment is actually already the answer:

Unless you can linearise the model, there is no analytic error estimate.

I consider error estimation very important, so I would like to show a bit more details here: A way of estimating the error of a model is by Gaussian error propagation, see in particular the part of the Wikipedia entry on non-linear combinations. Let me briefly summarize the idea behind as I find the Wikipedia article not so easy to follow. I will also use different variable names so that they do not overlap with those in your question.
We start with a function $\varphi$ which depends on different variables $x_1, x_2, \ldots$, meaning $\varphi = \varphi(x_1, x_2, \ldots)$. We assume that we can linearize around a certain point $\tilde{\bf x} = (\tilde{x}_1, \tilde{x}_2, \ldots)$, which mainly means writing the function at that point as a Taylor series.
Then, we can determine the Gaussian (maximal) error in the vicinity of ${\bf \tilde{x}}$.
$$\left. \Delta \varphi \right|_{\bf \tilde{x}} = \left. \frac{\partial \varphi}{\partial x_1} \right|_{(\tilde{x}_1,\, \tilde{x}_2,\, \ldots )} \!\!\!\!\cdot \Delta x_1 + \left. \frac{\partial \varphi}{\partial x_2} \right|_{(\tilde{x}_1,\, \tilde{x}_2,\, \ldots )} \!\!\!\!\cdot \Delta x_2+\cdots$$
In this formula you calculate partial derivatives $\frac{\partial \varphi}{\partial x_i}$ with respect to each variable $x_i$ and plug in the values of the variables at the central point ${\bf \tilde{x}}$. The $\Delta x_i$ is the error estimate for each individual variable $x_i$.
How to we apply that to your problem where you have fitted an implicit equation? Since you excluded the Monte-Carlo approach (which is the usual method for this case), I suggest the following recipe:

Formulate explicit functions $a(x,y), b(x,y), \ldots f(x,y)$, e.g.
$$ a = - \frac{bxy + cy^2+ dx+ ey + f}{x^2}$$
Linearize the functions around $x_0,y_0$. The $x_0$ is called $x$ in your table, and respectively $y_0$ is in the $y$-column.
Determine $\Delta a, \ldots, \Delta f$ by partial differentiation.
Plugin in $\Delta x$ and $\Delta y$ as also given in your table.


I'm having trouble achieving sharp telescope focus I've been having trouble getting really sharp focus on my telescope. Using Saturn as a reference, with a 20mm eye piece it is very sharp. But when I go to 10mm or 6mm, it gets bigger and blurrier. I re-adjust focus each time I change eye pieces, but the best I can get at the stronger magnifications is still very blurry compared to 20mm. The 20mm will have nice sharp edges on the planet and rings, whereas 10mm and 6mm have fuzzy edges on the rings, and the planet itself.

I'm using the Celestron NexStar Evolution 8 Schmidt-Cassegrain Telescope. I'm in a suburban area so I'm sure there is light pollution, could that be a factor?

Last night I was observing for about an hour, which I believe would be enough time for the temperature of the telescope to match the temperature outside.

I also tried combining the 20mm with a Celestron 2x Barlow lens, just with an idea that maybe it would make a difference, but it was pretty much the same.

Is there something I'm doing wrong?

Your telescope has a 2000 mm focal length, and 200 mm aperture. With the 20 mm eyepiece, you get 100x magnification. With the 10 mm lens, you get 200x. Several things could cause what you describe:

Seeing (turbulence) might be bad. It is common in many places that less than great seeing means everything at 200x and over might start looking blurry. But seeing changes with the seasons, or day to day, hour to hour, or indeed from one second to another. It's random. If you get the exact same results all the time, and the image never improves even after many weeks of trying, then perhaps it's not seeing that's the culprit.

Your telescope might not be collimated. Look in the user's manual. Is there a collimation procedure you need to follow? (EDIT: yes, there is - page 20) A miscollimated scope indeed behaves as you described - seems "sharp" at low-ish magnification, but quickly becomes blurry if you push magnification up.

Point the scope at a bright star and defocus. Are the rings perfectly circular? If not, it might be miscollimated.

http://www.astrophoto.fr/collim.html

Another factor is optics quality. Optics with lots of aberrations cause "mushy" images and a difficulty to find the true focus. High quality optics produce a scope that is very "snappy" when achieving perfect focus, and make very sharp images when seeing is good. There's nothing you can do about this, but the good news is - this problem is not very common nowadays; optics quality, while not always great, tends to be at least passable in many cases. Quality could be evaluated either on the test bench (which requires equipment), or via direct observation of stars (which requires a very experienced observer and takes time).

Finally, have you compared views with another instrument of similar aperture? The image at higher magnification will always seem a bit more soft, compared to lower magnification - even in perfect seeing with a collimated instrument. Also, the contrast always decreases as you increase magnification. This is why there is always an optimal magnification for each view - sometimes higher, sometimes lower, depending on many factors. The best scope in the world will look mushy and blurry if you push magnification up too much.

Can you easily see the Cassini division in Saturn's rings? If yes, then you're probably in okay shape (collimation and seeing), you're just not used to the softness at higher magnification. I have a very high quality reflector (self-made), similar aperture class to yours (a bit smaller), and Cassini is immediately visible at 180x and looks sharp; it's also visible at 255x but the whole thing starts to look mushy and faded. It looks super-sharp and high-contrast at 136x but it's kind of small. This is in perfect collimation, perfect thermal equilibrium, high quality optics throughout the stack (primary and secondary mirrors, eyepieces), and good seeing typical of N. California.

When seeing is bad, Cassini starts to fade no matter what I do.

Does Saturn look equally soft in the center of the image, as well as near the edge? Good optics (mirrors, eyepieces) make a good image everywhere, cheap optics make an okay image in the center and a blurry image at the edge.

My bet is that it's either bad collimation, or you haven't compared the view at 200x with another scope, or both. But it's hard to diagnose things over the Internet.

You should probably get a 15mm eyepiece as well. You'll use it a lot, probably more than the 10mm.

Light pollution is never a factor for: planets, the Moon, the Sun, most double stars. These things are just too bright to care about it, so observing them from the city is fine. What does matter is seeing (turbulence). Also, it's important that the scope is in perfect collimation.

Light pollution only matters for the "faint fuzzies": nebulae and galaxies. These are low-brightness, low-contrast objects, that are difficult to see from the city. Being low contrast, your eye will not perceive a lot of detail anyway, so for these objects seeing doesn't matter.

Congratulations for being aware of, and following, the rule that says you need to acclimate the scope to ambient temperature. A lot of people are not aware of it. At least you're removing one unknown from the equation. It's always a good idea to let the scope "breathe" outside for an hour before you even begin to observe.

One way to be 100% sure that your scope is focused perfectly is to use a Bahtinov mask. Just make (or buy) one to match the scope's diameter, put it over the scope's mouth, point scope at a bright star, and tweak the focuser until all spikes intersect exactly in the center of the star. Then remove the mask - the scope is in perfect focus. This will probably not help in your case, but it's just one way to remove another uncertainty. It also helps with regular observations - I use it almost every time I observe, it makes focusing so much easier.

The mask could be made from a piece of cardboard and a sharp knife in like 20 minutes. It doesn't need to be a marvel of engineering precision to work well.

http://www.deepskywatch.com/Articles/make-bahtinov-mask.html

One thing that is certainly NOT a factor is the central obstruction. You'll often hear on the Internet how Cassegrain instruments, or other scopes with a large central obstruction (secondary mirror) are supposedly less "sharp" than scopes without a central obstruction, or ones with a very small obstruction.

Yes, the obstruction does matter a little, but not to the extent that Internet mythology would have you believe. A Cassegrain instrument, with good optics, in perfect collimation, can be a superb telescope even at a 40% obstruction. It simply performs like a somewhat smaller instrument, that's all (the physics of the central obstruction are very complex, and would require an entirely different discussion on this forum to fully elucidate).

(Added in reply to a comment below)

Miscollimation is quite likely the #1 performance killer of all amateur telescopes. Some well-made refractors are more or less immune to it, but most reflectors need periodic collimation.

It's a good idea to google around for various collimation techniques, there's so many of them. Some require the observation of stars, some require special devices known as collimators. The technique is also different for different types of telescopes (newtonian, cass, etc).

If you own a reflector, it's a good idea to collimate it once in a while, just in case, like changing the oil on a car. Dobsonian telescopes probably require most frequent collimation.

Check Howie Glatter's site - he makes some of the most precise collimators currently available:

http://collimator.com/

Regardless, just learn any method you can and apply it. If the telescope is massively miscollimated, you'll get an improvement immediately.

Not being able to see Cassini is not a good sign. Maybe seeing is persistently bad (it can happen), or maybe the scope is off-collimation. Can anyone else in your area see Cassini these days in a similar aperture, in a scope known to be collimated well?

Surface detail on Saturn is not easy to see, but should be doable with apertures over 100mm. It's like zones of latitude similar to Jupiter's equatorial belts, but very, very pale and faded. You have to keep looking for a while to see it. Don't worry if you don't see that stuff for now.

Jupiter's Great Red Spot is harder to see nowadays. It's been shrinking and fading for decades now. Go on Wolfram Alpha and type in Great Red Spot. It will tell you if it's visible at the moment. Usually it looks more like a dent in the equatorial belts, rather than an actual visible spot.

http://www.wolframalpha.com/input/?i=great+red+spot

Jupiter works best at not so high magnification, otherwise the contrast gets too low. Around 140x ... 180x should give you decent contrast in your aperture.


Taking a picture of Starlink satellites I really want to take pictures of Starlink satellites, but I am fairly new to this kind of photography so I have a few questions.

What is the best lense to use and do I need any additional filters?
How do I point my camera at the spot where the satellite will pass through, do I need some kind of tracker for it?

Thank you in advance for your answers, I hope my questions made sence.

I'm going to write a short answer to get you going.

Here is a random link that claims to track StarLink satellites after launch, it might be helpful to predict when a newly launched train of satellites passes over your location, I don't know.

http://www.satflare.com/track.asp?q=StarLinkLaunch#TOP

It's not secure and it asks you for personal information. There are instructions in URGENT – Time change for tonight’s Starlink Satellite Train visible over Washington County where there is also this video that can give you some idea what kind of pointing and field of view might work.

Okay on with the question:

  What is the best lense to use and do I need any additional filters?

You don't want a long focal length lens, use a normal or somewhat wide-angle lens. Your best strategy is to search the internet or other Stack Exchange sites for examples that you like and then just see what they use. Satellites move across a big chunk of sky and their visibility is limited by what fraction of their path is in sunlight.

Photography SE has over 300 questions tagged astrophotography and it seems 17 posts also contain "satellite'.

  How do I point my camera at the spot where the satellite will pass through, do I need some kind of tracker for it?

If you know what satellite you want to photo, then there are many websites and programs available to predict when they will pass and where in the sky to look. I use a python program called Skyfield but in-the-sky.org and heavens-above are two websites that are very useful. This has been covered in more detail in other answers here and in other sites.

Even if the trail goes all the way across the sky, it won't always be sunlit, so look west for a few hours after sunset or east a few hours before sunrise. During the middle of the night only satellites in very high orbits will be in sunlight and therefore visible. For more on that see answers to Why do satellites appear to move faster when overhead and slower closer to the horizon?

Satellites move very roughly 1 degree per second, but it varies a lot depending on where they are in the sky (overhead vs near the horizon) and their altitude (see below). So if you tracked them everything else would be a complete blur and your satellite would just be a dot. For more on that again see answers to Why do satellites appear to move faster when overhead and slower closer to the horizon?

If you track the motion of the stars which is only 0.004 degrees per second, then you could expose for a long time and get many stars to show up in your photos. The satellite track will then be a roughly straight line across your photo. That's what most satellite photos look like.

But if you have a short focal length lens and your exposures are only a few minutes, I don't think tracking is necessary at all, just use a tripod. Sometimes people program their cameras to take a new photo every say 10 or 30 seconds and they stack them together later with small shifts so that the stars don't appear to move or blur. In that case they have to process the background objects on the horizon separately or else they'll be blurred.

Have fun!

 from this answer


How many actual stars are Polaris? I am getting conflicting accounts everywhere. I heard astronomers saying Polaris is a 3-star system. Some say they are binary but are optical-doubled with a 3rd.

Phil Plait a renowned astronomer wrote in his book Bad Astronomy that Polaris is sextuple star system.

The above text reads

  ... but due to its proximity to the NCP it has taken on the popular name of Polaris. The star itself is actually rather interesting; it's really a multiple star consisting of at least six stars in orbit around each other.

While in his Crash Course Astronomy Series He said Polaris was a Pentuple star system.

Now, I am not saying Phil is doing anything wrong. I am actually a fan of him, enjoyed reading the books he authored. He may have said one or the other by mistake. I tried asking him this question, it does not look like there is an effective way to submit him an Errata or ask him questions like these.

So my question is, how many stars are Polaris and are they all gravitationally bound to each other or some of them are just Optically grouped?

Polaris consists of multiple stars,

α UMi Aa is the main star. It is a supergiant, and a cepheid variable. α UMi Ab and α UMi B are smaller (but still larger than the sun) and both are in orbit with α UMi Aa, (the former is close in, the latter is further out).

There are two further stars: α UMi C and D. These stars are thought to have formed before α UMi Aa, and so are not considered to be physically associated with α UMi A. There are other stars that appear nearby, one might be physically associated: J021006.2+891626 J021454.3+890852 are their catalogue numbuers. They are faint, but appear to be the same age as Polaris.

In conclusion, Polaris is known to have 3 stars: Aa, Ab and B. The stars C and D are not physically related to Polaris. There may be one or two more stars that are distant dwarfs and may share Polaris's orbit.

Reference http://iopscience.iop.org/article/10.1088/0004-6256/139/5/1968


Why does gravity make time go slower and warp light? I know that objects with lots of gravity are capable of warping light or slowing time (an example being a black hole). I also know this is because of the theory of relativity, but I need someone to expand on that and explain why/how that works.

Also, please write the answer in somewhat simple terms, as I am not (yet) an expert on this stuff :)

Sometimes things we think of as different are actually just different forms of the same thing. Electric fields and magnetic fields are forms of each other. That is why we now call it the electromagnetic force.

Likewise, energy and mass are also just different forms of the same thing. We can even convert it too.
E = mc^2 ...

And from what I understand, space and time are connected together as well. So it gets called spacetime. If you want to think about it that way, gravity works by warping space, like a heavy marble sitting on a fluffy blanket. The fluffy blanket is spacetime. The heavy marble is a black hole or other super massive object. So now light is still going "straight" along space (the fluffy blanket) but it is bent because the blanket is bent. (Caveat: this is where i am conceptualizing light as a wave, not photon packets. It is actually both.) And if gravity bends space, of course it bends time too, because space-time is connected somehow as the same substance.


Are 3D coordinate data from Sloan DSS-III available & easily accessible to non-pros? Of course the newly announced SDSS-III data and maps from the Sloan Digital Sky Survey with 1.2 million objects, along with all previous data sets are openly available, and I'm sure there are many tools to access, work with, and view the data.

What would be the simplest way to extract a list of coordinates of galaxies in order to just try to visualize it on my own? There may be visualization tools also, if you want to add a link that would be great, but this question is about getting a list of coordinates so I can look at density or even try to plot one dot per galaxy in some slice.

I'm guessing the coordinates might be available in RA, dec, and redshift, and possibly also some calculated/inferred x, y, z.

I use python, but I'm not familiar with AstroPy yet, so if it can be done, albeit inefficiently and/or inaccurately by my writing a straightforward python script on my laptop, that would be the most helpful answer.

edit: if doing this in AstroPy is failry easy, I do have an Anaconda installation, and therefore already have at least a basic AstroPy installation.

Here is a graphic (below), from here in Phys.org attributed there to Daniel Eisenstein and SDSS-III.

Another graphic (below), from here in Phys.org attributed there to Jeremy Tinker and SDSS-III.

SDSS DR12 Catalog Data
looks like a good starting point,
apparently pretty open to those willing and able to figure it out.
Their
SciServer Compute
site hosts Jupyter notebooks to query CasJobs in SQL.

The Large Scale Structure galaxy catalog under
BOSS value added catalogs
may also be relevant.


Solar Eclipses for dummies: Step 1 - moon equatorial orbit around a planet without axial tilt Note 1: I've studied downvoted questions and decided that I either asked too many questions or I sound like I'm setting a fantasy world. So I'm editing my question accordingly.

Note 2: I'm a language teacher. I learnt equations with variables but that was over 20 years ago and I remember little of it; I never learnt trigonometry and other more advanced maths, that's a fact, but if there's no way around it I'll do my best to tackle it.

Note 3: My approach to understanding eclipses, due to the above stated handicap, is removing all the complicating factors (namely, the axial tilt and the moon's 'ondulating' orbit). I assume that, if I can fully understand such a simplistic scenario, I can then follow the mechanics of the real Earth scenario.

Therefore, I have written below the points I am confident are true. The ones I don't know are asked as questions (Q) and formatted in bold.

Every new moon would be a total solar eclipse.
The path of totality would always cover the same area: a 250km  wide corridor with its centre at the equator (so 125km over the northern hemisphere and 125km over the southern hemisphere).
I think the path of totality wouldn't always cover the same area, e.g. always over Africa, so Q1: Is there a simple formula to predict which areas would be in the path of totality?
In real life, the moment of totality can vary from seconds to seven minutes. But in my study example, Q2: the period of totality should always be the same, right? Because the crossing path of moon and sun is always the same (while in real life the Moon can overlap the Sun at different angles and going in different directions)
Q3: How can I calculate this period of time?
What about the latitudes for whom the solar eclipse would only be seen as partial? I understand that the nearest to the path of totality, the more covered the sun will be, and vice-versa. Q4: Is there a set value that says e.g. for each km, another degree is visible?

If this question is still not ideally adjusted, please leave a comment so I can further improve it.

May, or may not.
No.
Q1: A formula: yes, a simple formula: no.
Q2: Not for each point on Earth. This is e.g. due to different distance quotients.
Q3: Depends on distances, sizes, angular velocities, position on Earth.
Q4: Yes, but it's different for each point, not just by latitude. Even at the equator, there may exist points with, and points without totality. That's since the shadow of the Moon isn't cylindrical, but a cone. And because the Earth isn't a plane perpendicular to the Sun - Moon axis. And the distance Moon - Sun is varying due to its orbit around Earth.


Identification of Lunar mare (maria) from a taken picture I'm in Brisbane right now. I took this picture 15 minutes ago from my balcony.

Could you help me with identification of marked Lunar maria? I cannot compare any of these to this map.

I think 1 is Mare Serenitatis and 2 is Mare Tranquillitatis. Am I right?

So your image looks more like this, with the red line being the terminator. The illuminated side in your image is the left side of the reference image, turned upside down.  I think you've gotten it backwards, the mare that you've labeled "1" is the Sea of Tranquility, and the one you've labeled "2" is the Sea of Serenity.


Why N$_2$ is a non-absorbing species in the spectrum of the Earth? The transmission spectrum of the Earth atmosphere is like that (Kaltenegger & Traub 2009):

As you can see, you can find many absorption lines related to some components of Earth's atmosphere: H$_2$O, O$_2$, CO$_2$, O$_3$, CH$_4$... But the nitrogen, N$_2$, is not detected anywhere in the spectrum. Why N$_2$ is not detected, but O$_2$, for example, is?

As your question is based on the plot you posted, I suggest you to look for a lower wavelength range of the atmospheric electromagnetic absorption. A quick search in google gave me this paper, which says:

The importance of molecular nitrogen as the most abundant species in the Earth's atmosphere is evident. The strong absorption bands in the range 80–100 nm shield the Earth's surface from theextreme ultraviolet (XUV) part of the solar radiation.

As you can see, the absorption bands of diatomic nitrogen resides in the <1000 nm, which is in the extreme left of your plot, shadowed by its scale (1 micrometer = 1000 nanometers).


How do I find the signs that planets are in during any given date? Say for example I wanted to know where the planets were on December 19th, 1664, at 1am; how do I figure that out?

You can use the software Stellarium, which is available on most plateforms, and in-browser. With this software, you can show the constellations and the planets at any time and date, and see where the planets are in the sky.

For instance, here you can see that Mars is in Libra and Mercury is in Scorpius.

At the dates you specified, the positions of the planets were the following


What was the size of the universe around 400 million years after the Big Bang? I was trying to find out how large the universe was around 400 million years after the Big Bang as this is when some of the first galaxies were formed but I am having trouble finding out the answer or working it out.
I found some information on the size of the universe after 380,000 years when the CMB was created and by calculating the red shift from the time of creation a size of roughly 80 million light years is given by reducing the observable universes size by a factor of 1100 but I am not sure how to use this to calculate the universes size after 400 million years and I am not sure if as the above calculation is for the observable universe how that related to the actual size.
Is the size of the universe in terms of its length or diameter known or can be calculated after 400 million years from the Big Bang event?

There are two issues: what counts as "the size of the universe", and how to rescale this. The second issue is easier.
If you have something at distance $d_1$ at time $t_1$ it will be at distance $d_2 = d_1 a(t_2)/a(t_1)$ at time $t_2$, where $a(t)$ is the scale factor at the relevant time. To further make things nice, $a(t)=1/(1+z)$ where $z$ is the redshift at the time.
For $\approx 400$ myr and a flat universe, $z=11.35$ and $a=0.081$ (from this calculator).
Note the word "flat" above: flat and open universes are by default unbounded. So in a literal sense they are always infinitely large. However, it is entirely possible that they have a nontrivial topology (imagine a universe that repeats like a hall of mirrors; various cases are possible), except that there is at present no evidence for this possibility - if there is something like going on it is likely larger than our current cosmological horizon.
The observable universe on the other hand is finite, and grows both because light from more and more distant objects arrive and because of the expansion makes their distance larger. This is why the size is not just $ct$, it depends on the expansion. The calculator gives 367.92 Mpc as the distance to the horizon at $z=11.35$, or 1.2 giga-lightyear.


How strong is the force between the Sun and the centre of the Milky Way? I know the Sun is orbiting around the Milky Way, but how strong is the attractive force between them (e.g. what is the order of magnitude in terms of newtons)?

Galactic orbits are not Keplerian: There is not a single massive centre, whose gravity attracts the sun, rather the whole disc, and dark matter halo which surrounds the galaxy. We can't use the inverse square law to calculate the force of gravity without knowing the distribution of mass in the galaxy.

Nevertheless, the Sun's orbit is roughly circular, so we can use kinematics to get some idea of the forces involved:  For circular motion $a=\frac{v^2}{r}$. The velocity of the sun is about 225000 m/s, and we are at a radius of about 2.5e20 m from the centre. The formula above gives a very small centripetal acceleration of 2e-10 m/s²

However the sun is quite massive, 2e30 kg, so using $F=ma$, the force on the sun is of the order 4e20 N.  This is about 0.01 of the force acting on the Earth by the Sun. (3.6e22 N)


Which 16 antenna locations were used this MeerKAT radio image? The final configuration of the MeerKAT radio array will be 64 dishes. In the press release of July 16, 2016 a radio image is shown using a completed sub-set of 16 dishes.

Which 16 are these? I can find documents on-line with coordinates for 64 location, (sometimes numbered 1 through 64, but not here), but I can't figure out which 16 are used to produce this image.

I found images of KAT-7 and MeerKAT in google maps - maybe someday we can get daily ground truth updates from Planet Labs, but for right now I'll stick to stackexchange.

Image from the press release:

Above: MeerKAT array core from Google Maps at (30.7136109S, 21.4399576E). The remaining sites are farther away and can be seen if one zooms out.

Above: KAT-7 array from Google Maps at (30.7217011S, 21.4070518E)

Lorenzo Raynard, spokesperson for SKA Africa was gracious enough to send me the map with the sixteen sites below, and I've duplicated the dots on a larger format map.

It's been quite a while, probably the array has grown since this "first-light" image.

Next step (strictly for fun) will be to try to make a UV map for this configuration.


Which measurements have shown that 2014 MU69 is red? (New Horizons' next target) The recent NASA feature article New Horizons: Possible Clouds on Pluto, Next Target is Reddish

MU69 is actually the smallest KBO to have its color measured – and scientists have used that data to confirm the object is part of the so-called cold classical region of the Kuiper Belt, which is believed to contain some of the oldest, most prehistoric material in the solar system.
“The reddish color tells us the type of Kuiper Belt object 2014 MU69 is,” said Amanda Zangari, a New Horizons post-doctoral researcher from Southwest Research Institute. “The data confirms that on New Year’s Day 2019, New Horizons will be looking at one of the ancient building blocks of the planets.”

Gizmodo elaborates on the significance of the red color:

Speaking today at the the American Astronomical Society Division for Planetary Sciences (DPS), mission scientists talked about the Pluto flyby, but they also looked ahead to the Kuiper Belt encounter. Recent observations made by the Hubble Space Telescope suggests MU69 is red—if not redder—than Pluto. MU69 is now the smallest KBO to have its color measured, and it’s our first sense of what this mysterious object actually looks like.
The reddish color of MU69, like the iconic red splotches on Pluto and its moon Charon, suggest the presence of tholin, a class of molecules that’s formed through the ultraviolet irradiation of simple organic compounds such as methane and ethane. Tholin doesn’t form naturally on Earth, but it’s abundant on the surface of icy bodies in the outer solar system. Using Hubble, scientists have confirmed that MU69 is part of the “cold classical” region of the Kuiper belt, which contains some of the oldest objects in the Solar System.

How was the "redness" of 2014 MU69 detected? The second link mentions that it is Hubble data - is it just photometric (images through different filters) or is there also spectroscopy?
I'm curious, just how red is it? What exactly are the numerical data that are being interpreted as "redness"? Is this exclusively the assumed visual appearance, or is there excess infrared as well?
Also, I wasn't sure if 2014 MU69 is considered a comet or an asteroid. So far everything I've read just calls it a Kuiper Belt Object (KBO). So I've used both tags.

above: graphic from NASA blogs. It looks like they are all red! (joke).

This was a hard one to answer, primarily because of the difficulty in tracking down information.
The Observations of MU69
The extended mission for New Horizons involved adjusting its orbit to do a close fly-by of a Kuiper Belt Object (KBO). The annoying part about this was that they needed to find one first! Ground based observations were unsuccessful due to the faint magnitudes of any potential KBOs and the crowded star field (being that they were looking towards the galactic core). Finally, HST was used and over the course of 194 orbits, between June to September 2014, the entire region of the potentially reachable sky, by New Horizon's designated $130\:\mathrm{m/s}$ delta-v, was scanned using WFC3 instrument and the F350LP filter. Image processing revealed 5 potential candidates. This is all laid out in a paper by the lead of the KBO search, John Spencer. You can choose to view any and all images taken during these observations (since it is in the public domain) by downloading them from Hubble's Legacy Archive. Of course you'll have to do some digging in the 970 total images to find the actual images which contain MU69.
Once the entire region was scanned and a set of candidates were found, follow-up observations were made in late October 2014 of each individual candidate to precisely determine orbital parameters and observational characteristics so as to narrow down which candidate was reachable and presented the most scientific advantage. In this case, the same observing group from New Horizons took the images and once again used Hubble's WFC3 and the F350LP filter. Once again, these images can all be found on Hubble's Legacy Archive. An example of a single image which from this run, containing MU39 (although you'd be hard-pressed to pick it out) can be seen here. Most of this information can be gleaned from the Hubble Legacy Archive and is also described in this paper.
The Minor Planet Catalog (props to AtmosphericPrisonEscape for pointing this out) indicates that further observations of MU69 were taken with the HST in May and July of 2015. I had a bit of trouble tracking these images down, but knowing that John Spencer was the lead in these observations, I went and looked up his 2015 HST proposal which you can read here. With the proper proposal number, I found the set of 15 images during May and July which were taken of MU69. Again, the WFC3 instrument was used with the F350LP filter.
The Color of MU69
In total, we have 38 images of MU69, all taken by John Spencer and his team using the HST's WFC3 instrument and the F350LP filter (that peaks at 5845.871 angstroms). The important point here, and the reason I detailed the observations so much in the previous section, was to determine the exact type of images taken of this KBO, so as to ascertain what science data could have been used to determine color.
Based on everything I've compiled, it appears to me that the statement of MU69 being "red" is a educated guess. Anyone who knows about telescopes will tell you that imaging an object in a single filter is not sufficient to reliably measure aspects such as color. You need at least two filters to determine that because you need to understand how the brightness of said object varies with wavelength.
However, it could be possible to make a guess at the color of this body from a single filter by making a lot of assumptions. This is purely a guess as to what they did, but you can first measure the apparent magnitude of this object in your filter, then, assuming things like size, albedo, distance, etc., you can create a model of this object's spectrum that you'd expect it to have such that it reproduces your observed image. The first paper I linked details how they did inject false KBOs into their images to ensure their KBO finding algorithms worked which indicates they had some basic ability to model KBO images in their data. If your model can reproduce the observations, you can use that model as a basis for what your KBO is likely like and from that you can say that it appears "red". Based on experience with KBOs and in partiuclar pluto, we expect they might be red and that the redness corresponds to tholin, a common, red chemical found on KBOs, so the fact that they believe it to be read is not surprising.


Is the Bondi Radius the Same as the Schwarzschild radius? $$2MG/c^2$$

This is the expression for both, but some authors use one name and it has confused me. Is there any distinction?

The Bondi radius is $2MG/s^2$ where M is the mass, G is the gravitational constant and $s$ is the speed of sound in the medium of the accretion disk.  It represents the boundary between sub-sonic and super-sonic infall of matter into the black hole.

The Schwartzchild radius is $2MG/c^2$ where $c$ is the speed of light and is the radius of the event horizon.


How to find which part of the sky a photo contains? I was taking some photo stacks yesterday, and I also took a plenty of random ones (pointed not at a specific object) too at 200mm.

However, I've ended with a shot but I have no idea where I pointed my camera at, and because of the narrow angle of 200mm, I'm lost without any guiding star/object.

This happened to me before too, so I was wondering whether there is a software or a website that I can upload a photo, and it will analyze the objects, cross match them with known objects and detect which part of the sky it is? Or is there a method that I can use to figure out by myself (remember that there are no well-known bright stars in the image, at least not that I can identify by myself)?

Have you tried nova.astrometry.net?

They set up a web service for doing more or less what you're talking about.


How dense are Saturn's rings? I assume that the A ring is the densest, but I might be wrong. Nevertheless, I failed to find any explicit information over the net about the minimum and maximum of densities for the different rings. As a less practical question: what would one see upon entering one of the denser rings? Would it be obvious that it is a dense but thin (few metres - 1 km) cloud of particles, continuously bombarding the spacecraft irregardless whether it orbits along within the ring, or one would not see more than just a distant haze (if lit by the Sun) that spans over the horizon?

Elements of answer:

It is not an easy question, as we lack of data to constrain strongly the density and mass of Saturn's rings.

However, a first clue of their density is their optical depth (that is a measure of the transparency of a medium). The densest rings are the one of the main ring system (rings A, Cassini Division, B and C) plus ring F; they are characterized by optical depth larger than 0.1 (up to about 5 for the densest part of ring B). Rings in the faint ring system (E and G) have very small optical depth (lower than $10^{-5}$).

With this regard, B is indeed the densest ring of Saturn's rings. (edit) Looking around ([1] and [2]) you can find numbers that help to figure out the density of this ring. Knowing its mass (about $2.8\times10^9$ kg) and its extend (from 1.527 to 1.951 Saturn radii, with a typical thickness of 100 m), you can deduce a mean density of about 0.0167 g.m$^{-3}$ (about $10^5$ times less dense than the terrestrial atmosphere at sea level).

Just for fun:

This gives you an idea of what means the optical depth. In the densest parts of ring B, optical depth is between 2.5 and 5, so you'll definitely notice you're in the ring. It is probably more complicated than that (with ring particles ranging from dust size to house size), but you get the idea.

Side note:

Optical depth is a measure of transparency of one medium; it measures how radiation is neither scattered nor absorbed by the medium. It expression is given by

$$\frac{I}{I_0} = {\rm e}^{-\tau}$$

where $I_0$ is the source intensity, $I$ the observed intensity, and $\tau$ the optical depth.

Sources:

Charnoz 2009


How is the path of an asteroid calculated? How do scientists and astronomers recognize the perfect path or orbit (if it's orbiting an object) of an asteroid or a comet?

How do scientists and astronomers recognize the perfect path or orbit (if it's orbiting an object) of an asteroid or a comet?

They don't, and they can't. The best they can do is estimate the object's trajectory based on observations and based on techniques used to propagate the object (and other objects in the solar system) over time. These estimates are always imperfect:

The measurements upon which the estimates are based are limited in number. This is particular true for comets and asteroids. The historical record on observations of the Sun and the planets go back thousands of years. The number of observations for a newly observed comet or asteroid might go back a few months.
The measurements upon which the estimates are based inherently have some error associated with them. These measurement errors limit the ability to determine the object's true trajectory.
The physical models used to propagate trajectories are simplified and incomplete versions of reality.

General relativity, our best model of how gravitation works, is exceeding complex and highly non-linear. Even the very best precision orbit determination programs use a linear parameterized post-Newtonian expansion of general relativity. The non-linear effects? They're ignored.
Nearby stars and the galaxy as a whole perturb the trajectories of the Sun, the planets, etc. These effects are ignored.
Comets vent gas when they approach the Sun. When you blow up a balloon and let it go, the balloon veers this way and that. There is no predicting where the balloon will go. Comets do the same thing. After time, there is no predicting where a comet will be or where it is going.
Asteroids (and comets) are subject to radiation pressure, the Yarkovsky effect, and the YORP effect. These are very hard to estimate. One of the key purposes of the upcoming Osiris-REx mission is to get a better handle on the Yarkovsky effect.

The numerical techniques used to propagate objects' trajectories are inevitably flawed to some extent. They introduce errors that grow with time into the propagated trajectories. There's always a tradeoff between precision, accuracy, fidelity to the physics, and computational expense.
The solar system is ultimately chaotic. This is particularly the case for asteroids and comets on highly elliptical trajectories. Consider a comet whose orbit takes it from beyond Neptune's orbit to inside Venus's orbit. Over the centuries, that comet has ample opportunities to make a close pass with a planet. There are but slight differences between the planet making a minor change in the comet's orbit, the comet colliding into the planet, the planet sending the comet toward the Sun or another planet, or the planet ejecting the comet from the solar system. This is chaos at its worst.