Is the universe older than 13.7 Billion years? I came to know that the cosmic horizon is the part of universe beyond which we cannot see. So if the light from 13.7 billion years ago is what we can observe, does that mean the universe is older than that and we can't see it?

No, the oldest light we can see is as old as the Universe$^\dagger$. This is simply because light started traveling at that time, so after one year you'd se light that had been traveling for one light-year, after 1 million years you'd see light that had been traveling for a million light-years, and so on. There  is (most likely) more Universe beyond the horizon, but light from those regions just haven't had the time to reach us yet.

You might think that the distance to the cosmic horizon today is 13.7 billion light-years, but because the Universe is expanding, the distance is much larger (in fact 43.6 billion light-years).

$^\dagger$Because the Universe was opaque the first 380,000 years, we can't see all the way, but that's a technical detail.


How to convert quasar flux time series from observed frame to rest frame? I have flux time series of quasars in the observed frame and want to convert it in the rest or quasar frame. Any one who can help? Thanks

This is not an answer, but it is too long to put in the comment.

I am not familiar with quasars. In general, converting fluxes from an observed frame to a rest frame requires following consideration: extinction correction, cosmological expansion, k-correction. I think the simplest form is $f_{rest} = (1 + z)^n f_{obs}$ where $z$ is redshift and $n$ is some integer depending on how to express the unit of fluxes. This considers only the cosmological expansion. But, I am not sure if this transformation is valid with your quasars because they might be too far that $(1+z)^n$ is no longer a good approximation.


What would happen if a body were to fall into a neutron star? We know neutron stars as a very massive object with extremely strong gravitational forces that composes mostly of neutrons.

I couldn't help but wonder, what would happen if an object fell into a neutron star, what would become of it? Will it turn it into neutrons as well? And will be any sort of accompanying radiation emission?

No detailed calculations, but a qualitative answer: Depending on the trajectory of the impactor, the results will vary a bit, but it's clear, that the potential energy of the impactor will be transformed into a high amount of kinetic energy before the impact happens. The kinetic energy will then be transformed mainly into heat during the impact, transforming a substantial part of the mass of the impactor into x-rays and gamma rays.

The remnants of the impactor will be transformed into a plasma, with most of the electrons moving independently of their former nuclei, and dispersed mainly into the atmosphere (a thin layer of a few millimeters) of the neutron star. The energies will be high enough to trigger nuclear fusion as well as fission, together with other high-energy particle reactions. Part of the energy will be transformed into magnetic fields, which can also be very strong on neutron stars.

No much intermixing with the interior of the neutron star is to be expected in the first instant for small impactors due to the high inertia and density of the inner parts of the neutron star.

In some cases the impact could trigger the collapse of the neutron star into a black hole, depending of the mass of the neutron star, and the mass of the impactor.

More on the inner structure of neutron stars on Wikipedia.
("Matter falling onto the surface of a neutron star would be accelerated to tremendous speed by the star's gravity. The force of impact would likely destroy the object's component atoms, rendering all its matter identical, in most respects, to the rest of the star.")

More about the Chandrasekhar limit of neutron stars.


Why aren't other kinds of energies considered in the mass-energy of the universe? Mass-Energy of the Universe:

5% ordinary matter,
27% dark matter,
68% dark energy

What about other energies such as thermal energy, potential energy, kinetic energy, chemical energy, and radiant energy?

Various missions such as WMAP and the Planck Satellite have measured the mass-energy content of the universe. You tend to see images like the one below produced by these scientific ventures.

Credit: NASA/WMAP Science Team

This breaks the entire mass-energy content of the universe into three buckets: Dark Energy, Dark Matter, and (the mass of) atoms. But what about the rest of the universe. Surely there's other mass-energy out there such as the energy in the curved space-time metric, or kinetic energy of atoms, or radiation energy in photons, or the mass-energy of neutrinos! Have the astronomers and physicists forgotten about those things? Well the answer is no, they're just not telling you about it because it doesn't matter.

Equation for the universe

The universe's content and evolution is defined by something known as the Friedmann equation (technically there's a set of equations, but that's not relevant here). One important and useful form of this equation is given below.

$$\frac{H(a)^2}{H_0^2} = \Omega_{0,R}a^{-4} + (\Omega_{0,b} + \Omega_{0,c})a^{-3} + \Omega_{0,k}a^{-2} + \Omega_{0,\Lambda} +\ ...$$

This equation may seem complicated, but let me explain the parts. The left hand side more or less is the expansion rate of the universe in the past (or future) compared to the present day expansion rate. The right hand side is the sum of all the various contributions to the mass-energy of the universe. The important parameters here are the various "energy density parameters", denoted by $\Omega$. I've included the parameters for radiation ($\Omega_{0,R}$), baryonic matter ($\Omega_{0,b}$), cold dark matter ($\Omega_{0,c}$), space-time curvature ($\Omega_{0,k}$), and dark energy ($\Omega_{0,\Lambda}$), but you could add more energy density parameters as well. The ultimate point I'm going to show is that there are only a few energy density parameters that are actually important or significant and the rest can be ignored completely.

Thermal/Kinetic Energy

You mention things like thermal and kinetic energies (which are effectively the same thing since thermal energy just comes from the total kinetic energy of a collection of particles). We can't and don't count these because there's no way to absolutely define the thermal or kinetic energy of something. To define kinetic energy of a given mass requirements specifying it's velocity, but that velocity must be defined with respect to a particular reference frame. Special relativity tells us that there is no absolute reference frame though, which means I could define your velocity with respect to me, or the Sun, or the center of the Galaxy and get a different kinetic energy for you with each definition.

Radiation Energy

Radiation energy is just the energy in light. The universe is flooded with photons (a very large portion of which are CMB photons) and these photons have non-zero energy. But that energy is almost completely negligible in the current universe. From WMAP results, it is estimated that $\Omega_{0,R} = 5\times10^{-5}$ which equates to contributing about $0.005\%$ of the Universe's mass-energy budget. It would be a tiny, tiny sliver in the above graph. Interestingly, this wasn't always the case. If you run the universe backwards, you find that very early on, $\Omega_{R}$ was the dominant source of mass-energy in the Universe.

Neutrino Energy

Similar to radiation energy, neutrino energy is negligibly small. In part, this is because the neutrino mass is so tiny. I don't have an estimate for $\Omega_{0,\nu}$ but I can say it's probably as small or smaller than $\Omega_{0,R}$ and equally insignificant.

Gravitational and Chemical Energy

These types of energies are what's known as potential energies. They're a weird sort of energy because they're not actually real. They're potential. It's the energy you would get if you performed some action (e.g., moving a body through a field). If you hold a ball up high, you might say that ball has gravitational potential energy, but the ball doesn't actually have that energy, it just has the potential to gain energy when you drop it. We simply say it does have that energy to make balancing energy more tidy and simple. When you do drop the ball, it gains energy by falling through the gravitational field and taking energy from it. So really, your question boils down to talking about the energy inherent in background gravitational fields. The same is true for chemical energy. Chemical processes such as extreme oxidation, aka fire, release energy, but that energy comes from breaking atomic bonds and that breaking of bonds involves converting electric potential energy to something else.

Ultimately you just want to add in energy from the quantum fields that permeate the universe, $\Omega_{QF}$. I can't estimate this number, nor have I ever even seen someone try, but I can almost assure you that this number would be extremely tiny, even more so that the previous considered energies. Again, we find it's negligible and would by the tiniest mote of a sliver on the above pie chart.

Space-Time Curvature Energy

There's energy in space-time itself. This is known from General Relativity. This space-time energy of the Universe is denoted by $\Omega_{k}$. The interesting thing about this is that it appears the universe is flat! This implies, to decent precision, that $\Omega_{0,k} = 0$. This energy is more than negligible, it is non-existent (to within observational precision).

Conclusion

I'm sure you could think up a lot of different types of other energies you might want to include in this, but hopefully you see the point I'm driving at. They're all completely negligible. The arguably largest contribution comes from radiation and that's been shown to be $0.005\%$ of the Universe's mass-energy budget. What we see is that the Universe is dominated by Dark Energy, has a bit of Dark Matter, a small amount of atoms, and negligible amounts of the rest of everything else. And overtime, the Dark Energy portion of that pie chart will get bigger and bigger until we'll be able to throw out atoms and dark matter from the plot as well since they'll be just as insignificant as neutrinos and radiation are now.


Does the shape of a supernova remnant depend on the progenitor star's magnetic field? The magnetic field activity in a star can be very chaotic and my question is are they somehow related?

The factors that affect the shape of supernova remnants include:

Interstellar medium (the gas and particles between stars)
Planets that surrounded the star
Magnetic field (in the same way magnetic fields cause auroras)
Mass distribution of the star before supernova

There are probably more, but that's all I can think of at the moment


Good textbooks for a descriptive astronomy course Brief tale of woe

My department at a small public university in the US has decided to begin offering a descriptive astronomy general education course for the first time in a couple of decades and we have no institutional memory of how it was done before nor anyone on staff who's taught such a beast.

And I've been tapped to teach it.

Requirements

It's to be a one-semester course with three hours a week of classroom instruction and a three hour lab scheduled late enough that we'll have some dark sky early in the semester and more as the year goes on (schedule has it in Fall Semester). Expected enrollment in the neighborhood of 20 to 30. I won't have a teaching assistant unless I can rope a physics major into doing it somehow (we're an all undergrad department).

As a "general education" course mine will mostly attract students majoring in non-science subjects, biological science or human science. The students will have had enough math to know (in principle) how to isolate any variable in, say, the law of universal gravitation but many won't remember at first and some will resist; many will have chosen this class in an effort to avoid 'hard' classes like our gen ed chemistry, physics, or geology courses.

I'm looking of a textbook that touches on things like

Observational coordinate systems, familiarization with the night sky (constellations, I suppose), finding things
The Earth moon system, phases, eclipses, maybe even tides
Kepler's laws
The planets, moons and other bits and pieces in the solar system
The sun in particular and stars in general; including the birth, evolution and death of stars in a low math way
Some observational stuff; parallax, standard candles, HR diagram ...
A unit or integrated material on the planetary missions would be nice
Larger scale structure of the universe; a cosmology unit wouldn't be out of place; how much dark matter (or worse dark energy) can actually be done?
A unit on extra-solar planet hunting would be nice.
A unit on other wavelengths?
What else goes into a course like this?

Please include some detail on what is good about suggested texts and how well it matches with the course I've got to teach. What does it offer that's special?

Random detail

At this point we own a 8" scope (reflector) with a axial mount and motor drive but no automated pointing and two pairs of decent binoculars (I can probably buy more binoculars, but additional scopes are probably out of budget for this).

It's been about 20 years since I pointed a scope up with my own hands, and my only "real" astronomical experience was doing some programming for a CCD backed 14" remote-operated scope in the early 90s (we even managed to track a MACHO light curve and agree with the big boys, woo hoo science!).

The department head thinks that every gen. ed. course in our department should cover the scientific method and some thermodynamics which I think we can fit in.

I sincerely intend to include at least one field trip out to dark sky to just look up with adapted eyes. It's been too long...

The only question I could find on meta suggested that a sufficiently focused recommendation question would be accepted. If this is a no go, perhaps I'll pop into chat at some point.

Kaufman & Freedman's Universe is a very basic astronomy book aimed at non-astronomy science students or astronomy 1st year students, although for the latter it is perhaps a bit too basic. It covers introductions to all the topics you mention, except perhaps exoplanets (but my edition is from the late 90's; my guess is they have it covered too by now). It even gives suggestions for observing projects at the end of each chapter, along with exercises and extra literature. It is also very rich in pictures and explanatory diagrams.


Is there any role today that would justify building a large single dish radio telescope to replace Arecibo? It seems that most of the modern radio astronomy instruments and observation that make the news are interferometers or phased array systems of one kind of another. Is there any application left for which a single huge dish is better (or cheaper) than the same collecting surface in the form of some kind of array of smaller dishes, whether relatively compact, like ALMA, or widely distributed like the Event Horizon telescope or the square kilometer array? In other words is there any point at all in considering replacing Arecibo by anything the same size?

Arecibo wasn't just a radio telescope, it was a radar telescope, bouncing megawatt-level radio signals off various bodies in the Solar System.  A single-dish transmitter is far superior to a phased-array or other composite system, because the beam pattern is a simple Airy disk rather than a complicated pattern formed by tens or hundreds of such disks.


Why do dark objects look white from a distance? (Moon, Ceres, but not Earth!) What makes the Moon shine white in the sky, although it seems to be as dark as asphalt?

Earth's average albedo is 3 and 4 times that of the Moon and Ceres respectively (0.37 vs 0.12 and 0.09). But except for the clouds, Earth doesn't shine bright white in images of Earth taken from cis-lunar space.

Here's an image of Earth seen from the Moon. Although it has 3 times as high albedo, it does not seem to be as bright as the white Moon is as seen on Earth's blue sky.

Here's my take on it. I'm not sure what the mystery is - there appear to be two contributing factors.

Both the Moon and the Earth simply reflect/partially absorb the light that is incident upon them from the Sun. The overall of albedo is of little consequence, since the reflection of any light will result in an object that appears bright against the night sky.  The moon reflects quite uniformly across the whole of the visible range - it actually reflects red a bit better than blue. On the other hand the Earth's albedo is strongly wavelength dependent depending on what is doing the reflecting on the visible hemisphere. See the image below from this site. This injects some perceived colour into reflected earthlight.

The second point is the way the eye works at low illumination levels. Although you might think so on a dark night, the moon is not very bright. It is faint enough that the eye starts to switch over to mesopic or even scotopic vision where the light levels are too low to stimulate the colour-sensitive cone cells and just stimulate the colour-insensitive rod cells.

In the case of the moon, this has the effect of removing the red tinge it would have otherwise and giving the moon a "silvery"/white appearance. This is known as the Purkinje shift.

How the Earth appears to the human eye from the moon I could not say. It is possible that it is bright enough to stimulate the cone cells and thus appear as coloured as the photographs suggest. Either way, you can't tell anything from a photograph other than that the light from the Earth definitely is coloured, whether we perceive it to be or not. Photographs of the moon however do not show any strong colouration. In terms of brightness you also can't tell anything from a photograph showing just the Earth. The Earth is much brighter than the moon in terms of reflected light per unit area and integrated over its disc as seen from the moon.


Why are cirrus clouds more common on Mars? Mars water  ice clouds are often associated with regional-
  to-global circulation and saturation conditions, as opposed to local convective overturning, and so typically exhibit cirrus rather than cumulus forms.

Source: The Atmosphere and Climate of Mars Edited by  Robert M. Haberle et al., 2017 - chapter 5 - subchapter 5.2

Why are cirrus clouds more common than cumulus clouds on mars?

Hmm... here's my guess.
Cumulus clouds form due to warm, moist air rising. With the atmosphere of Mars being so cold, there's little heat to facilitate the formation of cumulus clouds. Cirrus clouds, on the other hand, are composed of sparse ice crystals, which makes them probably more common on Mars than cumulus clouds. After all, supercooled water does become ice.


When was the longest day on record? The Earth's rotation is gradually slowing down. It was significantly shorter in the prehistoric past, but now we need to add a leap second every few years and we've not yet had to take any back. In fifty thousand years, it is thought we'd need a leap second every day.

The slowing down isn't constant, however. The earth speeds up occasionally and we don't need to add leap seconds quite so often.

When was the longest solar day on record?

I suspect the answer is sometime during the 1970s when leap seconds were a yearly occurrence. It is also possible I've made an error of understanding.

(If your answer is the Summer Solstice or the 25 hour DST day, that's not the type of day I'm asking about.)

Maybe I'm reading too much into the question, but there's a few ways to measure a day.   I think 1912 or 1970 is an accounting answer more than the actual length of the day, similar to how 1751 in England was the shortest year, 282 days, because of a calendar adjustment.

Earth takes about 23 hours, 56 minutes and 4 seconds to complete a full rotation and that number is very consistent but gradually increasing because of the Moon and tidal effects.   It's increasing about 23 milliseconds per century or about a second ever 43,000 years or a minute every 2.6 million years.   It also works out to about 230 microseconds per year, less than 1 microsecond per day.

So, in a sense, if you're talking sidereal day, each day becomes the new longest on record, by a tiny bit, unless there's a large Earthquake which can speed up the Earth's rotation a little.  The 8.9 Richter scale Earthquake in Japan compacted Earth slightly and sped up Earth's rotation by about 1.8 microseconds, or slightly more than 2 days worth of tidal slowing.

The longest solar day would be perihelion day, when the Earth moves fastest around the Sun and a single rotation has to move that much further to get the sun back to where it started.   According to this article, obliquity plays a role too, which I'm having a hard time picturing, so I can't explain that.

This raises an interesting problem because Earth's eccentricity is currently decreasing and has been for about 15,000 years.  Earth's eccentricity was considerably higher at times in the past, perhaps a hundred or two hundred thousand years ago, perhaps as high as .04 or .05, much higher than the current .0167, which could have added perhaps 15-20 seconds to a solar day.   The longest day would probably be during a recent peak eccentricity during perihelion.   Unless I'm overlooking something.   Corrections welcome.

Eccentricity Wikipedia article 1

Eccentricity Wikipedia article 2


Why are distant objects observed in the near infrared? I was reading an article that explains why JWST is a successor to Hubble and not a replacement for Hubble. They explained that Hubble's science pushed astronomers to look at longer wavelength. And then they said:

  In particular, more distant objects are more highly redshifted, and their light is pushed from the UV and optical into the near-infrared.

So basically to observe the first galaxies, astronomers have to observe in infrared. My question is why distant objects require observations in the infrared?

Is it because they are at a very large distance from us, so the light has lost a lot of energy on its way so it's detectable in the infrared?

Yes.

When light moves through the expanding space, it loses energy and becomes redshifted. As galaxies form, they a emit a lot of UV light, and are often detected from there emission at the so-called Lyman $\alpha$ light, which has a wavelength of 1216 Å. Galaxies began forming just a few hundred million years after the Big Bang. Since then, the Universe has a expanded by more than a factor of 10 in all direction. The wavelength of the light expands by the same factor, and thus a Lyman $\alpha$ photon emitted in this epoch today has a wavelength of $\sim12,000$ Å, or 1.2 $\mu$m, which is in the infrared.


Is the whole universe is rotating on an axis? Most of the celestial bodies we know are rotating or spinning on its axis.Is there any chance for the universe to rotate on its axis?

From what we observe it is extremely unlikely the universe is rotating, but it is a good question nonetheless.

Perhaps you thought that since we see galaxies moving away from us they could be being pulled outwards as a result of the rotation. All the galaxies we observe (outside the local group) are accelerating away from us, and this includes galaxies in all parts of the universe. If the universe were like a giant spinning cylinder (or whatever object you like) then the galaxies "above" and "below" us in the cylinder would appear stationary since they are rotating at the same radius.

This observation does not support the claim that the universe is spinning on some invisible axis.

As always we should remember that our local universe may not represent the entire cosmos, and we would in fact be spinning around some axis. As more observational evidence is gathered we understand more and more about the universe, and one would be foolish to rule out a theory based on current observational evidence.


How to project galaxy data in x y z coordinates? For a data visualization project, I want to extract galaxy data (in x y z coordinates) from galaxy catalogue of within 150 mpc and project each of those galaxies as dots and visualize them in a 2D projection of a spherical volume. Need your help in getting this data set and how to achieve it?

Would be grateful for any help in any way!

You could try the Hyperleda database. You will have to learn to use the database, which possibly means learning a little bit of SQL. For starters you could select all items with redshift $cz < H_0\times 150$ km/s, using redshift as a distance indicator. This should be roughly ok at 150 Mpc (though I don't see any viable alternative).

The returned catalogue should have (at least) RA, Dec, redshift, possibly a size and morphology and, for a small subset of the galaxies, a redshift-independent distance estimate. Given that very nearby galaxies (in the local group or a $d < 50 $Mpc) are not fully part of the "Hubble flow" you might choose to use these in preference to a redshift-distance where available.

X,Y,Z coordinates (based in the equatorial system) can be calculated using this prescription or in Galactic coordinates (with respect to the centre and orientation of our Galaxy can then be found using the operations defined in Johnson & Soderblom (1987).

I have no idea how complete this database will be. Certainly there will be an issue with extinction near the Galactic plane.


What method is used to calculate the 'quality' of a solar/planetary image? What sort of algorithm is used to calculate the relative quality of an image, such as is performed by PIPP, Registax, or Avistack?

Such applications take planetary or solar images and prior to stacking and sharpening, they can optionally select the best images. When looking at images to see which have been selected by the algorithms, I sometimes see images that I wouldn't have selected. This has piqued my curiosity as to what methodology is employed in ranking or grading the images. With a bit of a primer to understand the concepts, I would like to put some thought into how such algorithms might be improved...

A little further research, and I found out that the basic principle, from which several variations derive, is to take the sum of the squares of the difference between adjacent pixels, to get a score.

The principle is that a higher quality picture has a higher probability that there will be significant differences in adjacent pixel values, i.e. that there can be a significant variation from pixel to pixel, whereas a lower quality picture will have similar valued adjacent pixels, as they have "blurred" together. Therefore, when taking the square of the difference of adjacent pixels and summing them together, higher quality pictures will show a significantly higher score, or summation.

I'm going to demonstrate this with a simplified example below, but I'm interpreting the meaning of "adjacent pixels" in a particular way, and I am hoping that if I am incorrect, someone will tell me and I can modify this answer appropriately.

Imagine we are imaging an object that exactly fills the frame of our small 16-pixel camera. In the middle of this object is a very dark square, that coincidentally happens to exactly fill the central 4 pixels of our camera. If the seeing for this object is perfect, then the camera will be exposed with a value of 255 for each of the light pixels, and 0 for each of the dark pixels.

This is how the differences between adjacent pixels would be calculated (I'm assuming only horizontal and vertical differences, not diagonal):

However, movement of the camera or poor seeing resulted in the light being spread more between the pixels, the calculations of the differences between adjacent pixels would be different:

The score, that is the sum of the squares of the differences, for the first (perfect) image is 520200.

The score for the blurred image is 304200. Therefore the sum of the squares of the differences between adjacent pixels suggests that the first picture is the better one.

This strategy isn't perfect, obviously. The presumption is that better images will exhibit a better score due to larger differences between adjacent pixels, but this is an assumption.

The squaring of the difference is intended to provide greater weight to larger differences, but I came across a comment on the PIPP support website that says that PIPP has modified the basic algorithm to give still greater weight to sharper contrast changes over lower contrast changes - so maybe squaring the differences isn't enough sometimes.

Apparently this quality measurement strategy doesn't work when the object is small in size on the sensor, and may be bright or even over-exposed. The above methods may select blurred and smeared frames over frames that are actually better. In this case, a strategy that looks at the histogram of values for the frame and selects the one with the higher peak, may produce a better selection of appropriate 'best quality' frames. It assumes that the histogram will have a higher count of frames sharing a pixel value across a smaller range, rather than spreading the light out across a wider range of pixel values.

An alternative quality measurement strategy simply adds up the values of all the pixels in an image. A higher score will indicate more light got through to the sensor, which implies that less cloud (for instance) got in the way.


How to convert between AB and Vega magnitudes? This might be a bit silly, but I'm really struggling with a conversion.
I have apparent AB magnitudes: $$m_{AB} = -2.5 \log_{10}(f_\nu [\mu Jy]) + 25$$ that I would like to convert to Vega magnitudes, because I am trying to make a figure similar to the one below from Stern et al. 2005. I used the conversions from this site: http://www.astronomy.ohio-state.edu/~martini/usefuldata.html, but it came out all wrong. Should I not use these? Because the axes contain colors, I suppose I don't have to worry about the units of $f_\nu$, because this will create an offset that cancels when you subtract two magnitudes, right? But I find that one color-conversion must be positive, because
$$[3.6]_{AB} - [4.5]_{AB} = [3.6]_{Vega} + 0.79 - [4.5]_{Vega} - (-0.09) \\= [3.6]_{Vega} - [4.5]_{Vega} + 0.88$$
while the other is negative, because
$$[5.8]_{AB} - [8.0]_{AB} = [5.8]_{Vega} +0.02 - [8.0]_{Vega} - 0.45\\
= [5.8]_{Vega} - [8.0]_{Vega} - 0.43$$
but if you look at this figure below, you see that you need two corrections with the same sign. Any thoughts on my attempt at conversion? Thank you.

A few things:

The conversion between magnitudes is dependent on the filters being used, you can see it on the tables from the website you linked. That being said, the conversions available in there will not be useful for you if you are trying to replicate Stern et al 2005, as they use redder filters (in the website with the conversions, the longest wavelength available is 2.19 microns, whereas the shortest in Stern et al 2005 is 3.6 microns). If you can find the proper relationships, you should get the expected results.
The paper states that they used IRAC data, so it is a good idea to start looking over there. In fact, you can find some relations in point 7 here and you could also take a look at point 3, might be necessary (I do not know what data you are working with). If you are not using IRAC data, you might as well look for a different source for the conversions.
Are you perhaps using the UltraVISTA catalog? If not, it might also be a good idea to check it out. Their README contains some details about the photometry and references about how they acquired the data (including IRAC), which could give you some more hints.


When astronomers "look into the past" are they accounting for the expansion of space? Every time astronomers look "13 billion" years into the past, not only is there more space than there normally would be at that instant which increases from observation to observation, but there was also expanding space for eons prior to that which must offset photons over those billions of years.

I haven't seen astronomers ever actually state they've reconciled that difference whenever they talk about their observations, so

1: Do they actually?

2: How do they?

3: How does that change the reality of their public statements?

When astronomers say something is seen "13 billion years in the past", this is (in most cases) a model-dependent statement, but a perfectly valid statement that tells us how long the light from an object has been travelling to us and which does take account of universal expansion.

What has actually been measured is a redshift, and to convert that into a "look-back" time requires an understanding (and a model) of the expansion history of the universe.

In the actual scientific research publication, as opposed to a press release, the details of what assumptions have been made will be stated. The current uncertainties in the cosmological parameters are small - they would not for example lead to uncertainties in the look-back time of even a few percent.

However, there is always a possibility that our understanding of cosmological expansion is wrong. In particular, the assumption that the cosmological constant is constant might be wrong, in which case there will be systematic shifts in estimated look-back times.


Why do we have the cosmological constant? Since the cosmological constant is not required to explain that the universe seem to be expanding, why do we have it?

What other factors cause us to have that constant?

Background:
Without the cosmological constant, distant stars should be affected by a great redshift. The amount of redshift is a function of their distance from us. This is due to gravitational time dilation. We are looking 13 billion years into the past, where the universe was very dense. Those stars should be experiencing extreme gravity, causing Einstein shift.

Since we DO have the cosmological constant, we are now looking for other explanations for the redshift.

Reason 1:

Let's look at the Friedmann equations without the cosmological constant.

$$  \frac{\dot{a}^2 }{a^2} = \frac{8 \pi G \rho}{3}-\frac{kc^2}{a^2}$$

The term on the LHS is just the Hubble constant squared $H^2$ which can be measured the direct measurement of recession velocity of galaxies

The density term can be said to be a combination of $\rho_{matter}+\rho_{dark- matter}$ both of which can be measured directly;$p_{matter}$ by observation of matter in our galaxy and other galaxies while $\rho_{dark- matter}$ by rotation curves of galaxies.

The curvature constant $k$ can be estimated today by the anisotropy measurements in the CMBR.

As it turns out the parameters don't fit and we need more mass-energy in the universe(almost 2-3 times of that we had estimated).

So comes along Dark energy or basically the cosmological constant. Cosmological constant or the dark energy are just two ways of looking at the equation,either as just a constant or a form of mass-energy(though we have solid reasons to believe the latter).

And this is our picture of the universe today:

Reason 2:

Now historically the cosmological constant was necessary for an altogether different reason.

The second Friedmann equation without the cosmological constant looks:

$$   \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) $$

Now this predicts for normal type of matter,the universe must decelerate.($\ddot{a}<0$)

Now,people measured the redshift of the type-1a supernovae and found out the quite paradoxical result that the universe was being accelerated in its expansion.

Since normal matter can't explain this type or behaviour,we again have to look at Dark Energy(or the cosmological constant).And so with the cosmological constant the equation becomes:

$$ \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$

Thus $\ddot{a}>0$ is possible.

Therefore the cosmological constant is necessary to both explain the current rate of expansion and the accelerated expansion.

So finally the accelerated expansion can be explained and today we have the $ΛCDM$ model of the universe.

References:

1:http://en.wikipedia.org/wiki/Friedmann_equations

2:http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html

3:http://en.wikipedia.org/wiki/Lambda-CDM_model


Accelerating universe expansion and standard candle Recent stars (and standard candle supernova), afaik, contain more heavy elements than older stars. As consequence, I guess, recent stars and supernova should be slightly dimmer than older stars? If previous sentence is true, slightly dimmer Ia supernova will appear as if it is more distant leading to the conclusion that universe is expanding faster. Does measuring of universe expansion rate and "dark energy" takes such possibility in to the computation?

Most type Ia supernovae are thought to arise from the thermonuclear detonation of white dwarfs that are composed almost entirely out of carbon and oxygen.

These white dwarfs are the cores of relatively low-mass stars that have lived their lives, gone through stages of core hydrogen and helium burning, leaving behind degenerate carbon/oxygen cores that become cooling white dwarfs after the outer envelope has been shed during the asymptotic giant branch and planetary nebula phases. As such, their composition, at least to first order, is almost independent of the initial composition of the star from which they were formed. That is, even if the progenitor star had a very low initial metal content, the white dwarf produced would still be almost exclusively a carbon/oxygen mixture, which had a similar Chandrasekhar mass and a similar explosive potential.

It is well known however that not all type Ia supernovae are the same. It has long been known that their light curves are subtley different and there is a so-called stretch factor that can be applied to get a "corrected" peak magnitude. a.k.a The width-luminosity relation.

More recently there has been a realisation that type Ia supernovae could arise from both accretion or mergers and there is clear evidence that the amount of radioactive Ni varies from explosion to explosion. A very recent paper by Milne et al. (2015) has however challenged the view of metallicity independence. They claim there are two populations of type Ia SNe, connected with progenitor metallicity, and that these populations become more apparent at high redshift when looking at rest-frame ultraviolet emission. The gist of their conclusions is indeed, as your question supposes, that this may go some way to ameliorating (but not eliminating) the need for dark energy.


At Mars' equator how long does twilight last? With Mars' thinner atmosphere I'm assuming twilight would not last as long compared to the duration on Earth.  Has there been any determination of the length of twilight on Mars at the equator or any other latitude?

According to https://mars.jpl.nasa.gov/MPF/science/clouds.html twilight lasts up to two hours because of dust high up in the atmosphere scattering the sunlight. Mars Pathfinder is in Ares Vallis, 3°North.


Why does lithium fuse at lower temperatures than hydrogen? This is a basic question, but it's been bugging me. In the Wikipedia article for lithium burning, it states that:

  Stars, which by definition must achieve the high temperature (2.5 × 10^6 K) necessary for fusing hydrogen, rapidly deplete their lithium. This occurs by a collision of lithium-7 and a proton producing two helium-4 nuclei. The temperature necessary for this reaction is just below the temperature necessary for hydrogen fusion.

I would imagine that with lithium having more protons, it would have a stronger Coulomb repulsion and require higher temperatures to fuse with hydrogen. Now, this article is pretty sketchy since it cites no sources, and I would normally dismiss this. But according to here, lithium burning occurs in protostars, before hydrogen fusion even takes place. How could lithium fuse with hydrogen at temperatures that low?

The slowest reaction rate in the pp chain determines how quickly hydrogen can "burn" in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay.

The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of the PPII series of reactions that convert helium 3 into helium 4. It does not involve weak interactions and therefore the cross-section at a given temperature is much higher than the first step in the pp chain, regardless of the stronger coulomb repulsion between the reactants. It is therefore initiated at lower temperatures - as you say, Li "burning" occurs in low-mass pre main sequence stars well before hydrogen "burning" commences (but after deuterium burning).


Why do sunspots appear dark? Sunspots, such as this one, appear dark:

Why?

Typical sunspots have a dark region (umbra) surrounded by a lighter region, the penumbra. While sunspots have a temperature of about 6300 °F (3482.2 °C), the surface of the sun which surrounds it has a temperature of 10,000 °F (5537.8 °C).
From this NASA resource:

Sunspots are actually regions of the solar surface where the magnetic field of the Sun becomes concentrated over 1000-fold. Scientists do not yet know how this happens. Magnetic fields produce pressure, and this pressure can cause gas inside the sunspot to be in balance with the gas outside the sunspot...but at a lower temperature. Sunspots are actually several thousand degrees cooler than the 5,770 K (5496.8 °C) surface of the Sun, and contain gases at temperature of 3000 to 4000 K (2726.9 - 3726.8 °C). They are dark only by contrast with the much hotter solar surface. If you were to put a sunspot in the night sky, it would glow brighter than the Full Moon with a crimson-orange color!

Sunspots are areas of intense magnetic activity, as is apparent in this image:

You can see the material kind of getting stretched into strands.
As for the reason it is cooler than the rest of the surface:

Although the details of sunspot generation are still a matter of research, it appears that sunspots are the visible counterparts of magnetic flux tubes in the Sun's convective zone that get "wound up" by differential rotation. If the stress on the tubes reaches a certain limit, they curl up like a rubber band and puncture the Sun's surface. Convection is inhibited at the puncture points; the energy flux from the Sun's interior decreases; and with it surface temperature.

All in all, the sunspots appear dark because the are darker than the surrounding surface. They're darker because they are cooler, and they're cooler because of the intense magnetic fields in them.


"Esperance" rock on Mars formed by erosion or weathering? or is there any rock on Mars not formed by erosion? I was wondering if the "Esperance" rock that was found on Mars by the Opportunity rover was formed due erosion or weathering? I have been reading a lot online but I did not seem to find anything. Does anyone know a type of rock on Mars that was not formed by erosion?

Rocks on Mars, just like rocks on Earth, can be igneous (formed from the cooling magma or lava) Sedimentary (formed from sediments that were laid down by water or wind) or Metamorphic (sedimentary rocks that have been changed by exposure to heat and pressure) (there may be other categories, such as meteorites)

After they have formed, rocks are subject to erosion and weathering. These processes don't form the rock, but can shape it or move it. The process of weathering produces sand, silt and clay, that can later form new rocks.

The Esperance rock has a significant amount of "clay" (a sedimentary rock). The clay in Esperance was formed by the weathering of other rocks to very fine particles, these then gathered together in wet environment to form a clay. When the water evaporated, the clay hardened into a rock. The significance of Esperance is that the water under which it formed must have been chemically neutral, not strongly acidic.


How long will star formation endure before the eventual heat death of the observable universe? Lord Kelvin's prediction of heat death indicates the eventual end to star formation in the observable universe. Have any astrophysicists made any predictions for the endurance of star formation in the observable universe?

This is actually very well researched. Wikipedia has quite a good summary, stelliferous era ends after about $10^{14}$ years, assuming an expanding universe.


Comparing info for WISE and GALEX I'm trying to compile a table showing the wavelengths and flux at m=0 for WISE, Spitzer and GALEX, but I can't find the right data for GALEX. Here's what I have so far:

        Band    Wavelength      F_0
                micrometers    Jansky
        ------------------------------
        FUV     0.152
        NUV     0.227
        U       0.365           1810
        B       0.445           4260
        V       0.551           3640
        R       0.658           3080
        I       0.806           2550
        J       1.220           1600
        H       1.630           1080
        K       2.190           670
        W1      3.4             309.54
        W2      4.6             171.787
        SPIT8   8               64.13
        W3      12              37.674
        W4      22              8.363

I got most of the info from here. The information I found for GALEX on page 687 of this paper gives m_0 for NUV and FUV, but no F_0. How can I change m_0 to F_0?

GALEX magnitudes use the AB system, which means that $F_{0} \approx 3631$ janskys for both NUV and FUV, by definition.


Measuring declination in negative hours of arc? I'm running the program found at https://www.projectpluto.com/source.htm in order to find the Right ascensions and declinations of celestial bodies at various points in time. An odd thing I've noticed is that the declination is reported in hours, minutes, seconds as opposed to degrees, minutes, seconds and I'm not sure how to use that. Especially considering the hours seem to lie on [-24,24] which doesn't make sense for the [-90,90] range for declination measured in degrees.

An example:

  Sun RA=10:11:50.1 Dec= 11:08:50.8

This is sampled from Los Angeles, California at 8 PM PDT (3:00 UTC) so I would expect when I ran these numbers to calculate elevation and azimuth that the elevation would be a good bit under the horizon.

I tried scaling the hours by 90/24 so that it would be back in the [-90,90] range but when I tried running those numbers through this tool, it ended up reporting that the sun would actually be just above the horizon which is definitely wrong.

So with that said, how should I be interpreting this data to get the result I expect? I'm pretty new to this, so I may be lacking something incredibly fundamental.

It's not hours, minutes, and seconds, its degrees, arcminutes and arcseconds.

There are 360 degrees in a circle, 60 arcminutes in a degree and 60 arcseconds in an arcminute.

Since the planets lie close to the ecliptic, and the ecliptic tilted at 23 degrees relative to the equator, the declension of the planets rarely exceeds +-23 degrees, giving an appearance that their dec. is like a time.


Will there be any more planets come into our solar system and make it their "home"? Could we have more planets come into our system and get trapped by our suns gravitational field and rotate?

Short answer: Yes, but it is very unlikely.
The probability that a random object drifts into the Sun's sphere of influence (SOI) is low (excluding tiny stuff like space dust and small rocks). But the chance that a full-blown rogue planet enters our Solar System is many orders of magnitude lower. Let's set up a Drake equation:
Let $R_n$ be the number of rogue planets in the galaxy, $R_{ns}$ the fraction of rogue planets that may enter the solar system, $R_e$ be the probability that one enters the Sun's SOI, $R_i$ be the probability that the planet will survive an encounter with a Solar System planet, $R_s$ be the percentage of all possible orbits for the rogue planet after the interaction that are long-term stable, and $R_o$ be the probability that the rogue planet gets captured into an orbit. Therefore, $$R_o=R_nR_cR_iR_s$$


Using a refractor telescope 5" aperture, 900mm focal length for terrestrial viewing Can I use a refractor telescope 5" aperture, 900mm focal length on an equatorial mount for terrestrial viewing.

Yes, potentially.  There are a few limitations:

It will be easier to use in alt azimuth mode, which in practice means setting the polar axis to 90 (ie straight up).  Not all equatorial mounts can do this.

With the standard astronomical diagonal, images will be inverted left to right.  You can get diagonals for terrestrial viewing that will fix this.  If you don't use a diagonal, the image will be upside down.

The field of view will be quite restricted if you use a typical eyepiece.  For example a 20mm plossl would give a magnification of 45 and a field of view of about 1 degree.  To get more useful field of view, you would need longer focal length eyepieces, with with a wide angle design.  These are more expensive than ordinary eyepieces.

Unless your scope is an apo, you may see quite noticeable chromatic aberration, particularly at the edges of objects with a bright background (eg birds).


Why do we use FITS format for scientific images especially in astronomy? How is it different from formats such as JPEG, PNG etc? I am new to astronomy and I found that the telescopic dataset can be stored in the FITS format instead of JPEG. Why do we have a special image format just for scientific images?

File formats tend to be industry/field-specific, with the format, tools, and expectations of the field coevolving to become more dependent on each other over time. JPEG co-evolved with amateur digital photography, PNG co-evolved with the web. Likewise, FITS co-evolved with astronomical data processing, and so is naturally more suited for that purpose than formats that had entirely different goals and communities involved in their development.

(Pedantic nit: JPEG is the name of the compression method and the group that designed it. The file format is technically "JFIF", but everybody colloquially calls it JPEG.)

As an image format, FITS has a number of desirable qualities that are lacking in JFIF and PNG, that are crucial for scientific data:

Storage of more bits per pixel (CCDs can records 12, 14, or more bits), and also floating point values.
Storage of arbitrary number of data channels (scientific data may have many, or other, frequency bands than the RGB that JPEG and PNG are limited to).
No lossy compression as is typical (though not strictly required) for JPEG.
Higher resolution (JPEG/JFIF, for example, is limited to 65,535 pixels in each direction), and FITS is also capable of storing 3D data volumes.
Support for unlimited metadata in the header, for example the sky coordinates, information about the telescope, etc. JPEG and PNG don't have the fields and aren't set up to record arbitrary metadata.

Also, the use of FITS for astronomy predates the existence of both JPEG/JFIF and PNG -- FITS was standardized in 1981, JPEG in 1992, PNG in 1996. So even if those formats were suitable (which they are not), by the time they were invented there was already widespread use and existence of astronomical image processing tools that were geared to FITS files (and growing archives of astronomical data in FITS format), so it would have been a major effort to switch formats that would never be undertaken unless a new format not only met but exceeded all the prior requirements of the field.


Will I be able to see both Jupiter and Saturn at the same time in my Astromaster 114? I have a Celestron Astromaster 114 Newtonian reflector telescope. The specs:

aperture: 114mm
focal length: 1000mm
focal ratio: f/8.77

I also have a t-ring so I can connect a Nikon camera to it.
Will I be able to see both Jupiter and Saturn at the same time thru this telescope? I understand from this article that they will come within a tenth of a degree of each other. The article says this is "fifth of the width of the full moon" which, based on my experience, suggests that I will indeed be able to see both, but I'm hardly any expert.
I'm considering borrowing a friend's film camera to try and snap a photo of both in one image, but would like to avoid the trouble (thanks, COVID!) if this won't happen.

This can easily be tested using software such as Stellarium, where you can visualize the field of view with given focal length.
If you have the software installed, click on "ocular view" (the most left button in the upper right corner). The following view is what you culd expect on December 21 with a 10mm eyepiece and a focal length of 1000mm:

And this is the view when attaching a Crop-sensor (APS-C) Camera to a telescope of focal length 1000mm:

So you should be able to see both planets and some moons in your telescope.
Some notes:

These simulated views do not necessarily represent the actual view to 100% accuracy. I did leave the "atmosphere" setting turned on, but actual visibility and brightness will depend on your location (light pollution and other atmospheric effects such as haze might reduce visibility)

Because I was lazy, I used the default telescopes and eyepieces. This ignored aperture (or rather, it was different than on your telescope), but it won't have an effect on magnification. If you wanted, you could add your own telescope in the software's settings.


Is there anything different about the gravitation around a non-spinning black hole and a neutron star of the same mass? If we perform a simple experiment by orbiting identical spacecraft around a non-rotating black hole and a non-rotating neutron star of equal mass at a "safe" distance, and ignore differences other than gravitation (e.g. magnetic fields and charge) would there be any differences in their orbits predicted by general relativity?
"Safe" distance means far from the event horizon of the black hole and where tidal forces don't disrupt the spacecraft.
After writing this answer I am now wondering if it's 100% correct.

Birkhoff's theorem is very useful: in general relativity, if you are in vacuum and there is a spherically symmetric gravitational field, then it will be the Schwarzschild solution. This solution only depends on the mass, not on the size of the object. So the neutron star and the black hole will give rise to exactly the same orbits.


What's to be gained from the New Horizons mission once it's beyond the Kuiper Belt? Now that we have seen what Pluto has to offer and New Horizons is headed to the Kuiper Belt, is there anything we can gain from the mission in interstellar space after it travels beyond the Kuiper Belt?

Possible rendezvous with a Kuiper belt object and Voyager type data on the heliosphere:

  "We should have power until the 2030s, so we can get into the outer part of the heliosphere," says Spencer. "As long as we can continue to get good data—and persuade NASA to pay for it—we will keep getting the data, because we will be in a unique environment that we've never been in before."

Added:
NASA just picked the object for flyby.

  a 30-mile-wide chunk of ice known as 2014 MU69 that orbits nearly a billion miles farther out.

Not much is currently known about the object. We'll no doubt learn more from the January 2019 encounter.

More info from Emily Lakdawalla and Casey Dreier at the Planetary Society:

New Horizons extended mission target selected


Is it true that we see the center of the milky-way for only half of the year? I'm not an astronomer, please excuse my non-formal language.
Since we are located in one of the arms of the Milky way, the center of the galaxy should be in one direction from our location, while in the other direction, we stare into the "other side".
So at nights as we are in the orbit area that is between the sun and the center of the galaxy we are able to see the center (the "milky-way"), while when in the other side of the orbit, at nights, we are not able to see it.
Does this mean we only see the milky way (meaning the milky-way's famous pattern) half of the year?
In the below diagram, the small circle is our year-orbit around our sun. Red dot is the earth location on day 0 of the year and the green dot is the earth location on day 180th of the year.
I know that as we are orbiting our sun, it also orbits the galaxy, but since it is so slow, it is negligible for this questions.

Is this reasonable?
I mean, if we look at the sky, we will see the center of the galaxy only during half of the year?

No, although there are times when it can't be seen, it isn't true that it is visible for 183 days of the year.
The general question could be "If I take an arbitrary location on the sky (say a randomly chosen star) will it be visible for half the year?  The answer is "it depends on the star!" Polaris and many other stars are circumpolar for Northern hemisphere observers and are visible every night,  contrariwise, Polaris is never visible for Southern Hemisphere observers.
Now, the galactic centre happens to be located in Sagittarius, so co-incidentally is quite close to the plane of the solar system.  This means that for about two months of the year, the sun is too close to make observations easy, certainly in visible light from the Earth's surface.  At other times it will be visible, perhaps briefly.  Sagittarius and the galactic centre are also quite a few degrees South of the equator, so it is more easily visible from the southern hemisphere.
At 50 degrees North, The galactic centre sets at twilight at the start of October and doesn't rise until dawn at the middle of February,  between February and October (about 7½ months) the galactic centre is above the horizon at some point during the night. (The exact number of days depends on your definition of "dawn" and "above the horizon") More southerly observers will see the galactic centre for longer. 50 degrees South, the galactic centre is above the horizon at some time during the night between January and November.  In Antarctica, the galactic centre is circumpolar, and so is above the horizon every night. But as the summer sun in Antarctica doesn't set, there are days when you can't "see" it. The optimum location would be 30 degrees South (or in space)


How to isolate H$_\alpha$ from Nitrogen doublet emission lines I am trying to find out line intensity and FWHM of H$_\alpha$6565 emission line from a quasar spectrum obtained from SDSS DR-14Q. Visual inspection of spectrum shows H$_\alpha$ being contaminated with [NII] 6549,6585 emission lines. I tried fitting voigt profile on the emission line to extract FWHM and line intensities. But the problem is voigt profile is not being fitted properly onto the data as shown here
I tried fitting 3 gaussian profiles on the data as there are 3 peaks visible. While fitting multiple gaussians it is clear that the gaussian profile doesn't fit over the left and right wings of the data.

I have already subtracted the underlying continuum and tried to normalize the flux as well.
Can anyone comment if I am doing it correctly because in both cases it seems that there is some sort of data loss? Or can anyone suggest a better way to do that?

I would've posted this as a comment, but I lack sufficient reputation.
I like your idea of fitting three Gaussians, but I think you should try fitting with one large positive Gaussian and two small negative Gaussians.
My physical justification for this is that I would expect a lot of self-absorption from the H-alpha line, which should be well-modeled using the larger negative Gaussian. I'm not sure if the other notch is absorption from another line, but it also looks like a negative Gaussian would fit well there.


stellarium software: selective artificial satellite display The Stellarium satellite plugin by default shows all the artificial satellites' orbits around Earth. I’d like to be able to view only certain satellites so that the view is not cluttered with every satellite in existence. Is there a means to create a custom satellite list, or view only relevant satellites?

Satellites are provided by a plug-in (that is enabled by default)
Configuration (the spanner-and-star button) -> plug-ins -> Satellites -> configure.  You can configure which satellites to display from here.
It might be worth noting that meteosat-8, a geostationary satellite has a magnitude of about 19, and so is not a naked eye object, and is in fact beyond the capabilities of most amateur equipment.


Earth's precession and the position of the Sun in the sky To my understanding, the Earth's precession causes an approximately 20-minute difference between the sidereal year and the tropical year. Also to my understanding we use the tropical year for our calendar which means we ignore those 20-minutes. These 20-minutes will cause the calendar to "shift" one day every 72 years so if we don't do anything about it in a few centuries winter and summer will switch places. So every century they add an extra day to the calendar to counter this.
This I understand. What I don't understand is that this 20-minute difference means that the Sun's position in the sky will vary by 20 minutes every year. Meaning if the Sun rises at the right ascension of 0 this year, next year it should rise at the right ascension of 20 minutes, and so on. But each year the Sun will actually rise at the same spot in a particular day.
How is this possible? Where do these 20 minutes go?

Yes, the Gregorian calendar is based on a good approximation of the mean tropical year, so the dates of the equinoxes are stabilised.
Geometrically, the equinoxes occur when the Sun's apparent path on the ecliptic crosses the celestial equator. The March equinox point (aka the First Point of Aries) is the 0° point for ecliptic longitude and also the 0 hour point for Right Ascension.
Due to the precession of the equinoxes, the First Point of Aries moves relative to the stars by about 50 arc-seconds per year, which gives rise to the 72 year value you mentioned.
The upshot of this is that a table of RA and declinations of stars must specify the epoch that it uses, so that the given positions can be adjusted to give the actual RA on a given date. A very popular epoch for the last few decades is J2000 which is 12:00:00 noon (Terrestrial Time) on the 1st of January, 2000, which is Julian date 2451545.0 TT. In UTC, that's 11:58:55.816.
There's a brief explanation on Wikipedia of the effects of precession on Right Ascension.


Can gravitational waves pass through a black hole? As the title says, what happens when a gravitational wave approaches a black hole?  I would presume that something interesting happens because of the way spacetime works near black holes but I have no knowledge to back it up.

No, gravitational waves cannot pass through a black hole.

A gravitational wave follows a path through spacetime called a null geodesic. This is the same path that would be followed by a light ray travelling in the same direction, and gravitational waves are affected by black holes in the same way that light rays are. So for example gravitational waves can be refracted by gravitational lenses just as light waves are. And just like light waves, if a gravitational wave crosses the event horizon surrounding a black hole it is then doomed to travel inwards to the singularity and can never escape.

There is one caveat to this. When we talk about a gravitational wave we generally mean a ripple in spacetime that is relatively small. Specifically it is small enough that the energy of the gravitational wave does not significantly affect the spacetime curvature. So when we calculate the trajectory of a gravitational wave near a black hole we take the black hole geometry as fixed, i.e. unaffected by the wave, and we compute the trajectory of the wave in this fixed background.

This is exactly the same approach as we use for calculating the trajectories of light rays. Since light rays carry energy and momentum then, at least in principle, they have their own gravitational fields. But for both the light rays and gravitational waves likely to exist in the universe the energy carried is too small to make a significant contribution to the spacetime curvature.

When you say in your question:

  I would presume that something interesting happens because of the way spacetime works near black holes

I would guess you are thinking that the gravitational wave could change the geometry near a black hole, but as described above typical gravitational waves don't have enough energy to do this. It would be reasonable to ask what happens if we give the wave enough energy, but the answer turns out to be that it no longer behaves like a simple wave.

Gravitational waves exist in a regime called linearised gravity where they obey a wave equation that is basically similar to the wave equation light obeys. If we increase the energy so much that gravity becomes non-linear (as if the case for black holes) then the oscillations in the spacetime curvature no longer obey a wave equation and need to be described by the full Einstein equations. For example it has been suggested, but not proven, that really high energy gravitational (or light) waves could interact with each other to form a bound state called a geon. I confess that I'm unsure how much work has been done studying oscillations in this regime.


Do comet tails increase as comets move away from the Sun? Do the tails of comets increase on their return trip from the Sun?

The only data I could find indicates that the brightness is similar approaching and receding:

Source
This is for the unusual almost "tailless" Comet Wirtanen .


Photographing Venus with a phone through a telescope I took a picture of Venus through my telescope (Celestron Astromaster 130eq) using a 10mm lens with a Barlow lens.

The picture isn't really good since I took it with my phone and I still don't have a camera.

Also, it isn't a matter of focus since I fiddled a lot with it.

I don't know what to do to clear the image and reduce the brightness. Should I use a filter? and which one?

Phone cameras just aren't designed to take good photographs of astronomical objects. The sensor is too small and the configuration options (like exposure time) are minimal.

For astronomical photography, a better option is to use (the body of) a DSLR (or the analog version); the lens is provided by your telescope. Another option is a specialized CCD camera.


Why do medium stars collapse to form supernovae while big stars collapse to form black holes? I understand that a star runs out of fuel slowly and gradually by fusing heavier and heavier particles (because of pressure by gravity) until the last heaviest particle is not able to fuse at the core of the star. This is where gravity takes over, and mass collapses (emitting gases out) to become a smaller dense star (with no energy left to give, just the gravity to absorb).

What I don't understand is why medium stars collapse to become a supernova which emits light while big stars collapse to form black holes which absorb all light.

A supernova is caused by the sudden (less than 1 second) collapse of the core of a massive (initially more than 8 solar masses in total) star. Less massive stars do not undergo core collapse and do not produce core-collapse supernovae.

The thing that determines whether a supernova remnant is a neutron star or black hole is largely the mass of the core at the onset of core collapse, but may also depend on how effectively the rest of the star is blown into space by the supernova explosion and how much material falls back onto the collapsed core.

If the collapsed core has a mass of less than about 2 solar masses (and perhaps up to 3, though there is no observational evidence for this), then the collapsed core, consisting mainly of neutrons, can be supported against its gravitational weight and stabilised as a "neutron star". The support is provided by neutron degeneracy pressure and the extreme repulsion between neutrons when they are squeezed together, which is caused by the strong nuclear force.

If the collapsed core is more massive, or if more material from the exploding star falls back onto the collapsed core, then even this strong nuclear repulsion will be unable to support its weight. In fact, above about 3 solar masses, General Relativity dictates that no force can possibly support the weight of the compact remnant and it will collapse further to become a black hole. No light (or anything else) can emerge from inside a black hole; so-called Hawking radiation will be negligible; so the only light received from such objects (after any supernova has faded), will be due to material being accreted into the black hole that is heated as it spirals towards it.

A third possibility exists, which is direct collapse to a black hole. It is possible that for very massive stars (at least 25-30 solar masses), that the collapsing core is never stabilised even as a proto neutron star, that no bright supernova is produced and that much of the mass of the star directly collapses into a black hole. This is a very active research topic since the discovery of 20-30 solar mass black holes by the LIGO gravitational wave experiment.

To summarise, I can do no better than quote from an excellent review paper by Fryer (1999):
"Thus for core collapse models we can define three regimes of compact object formation: (1) low-mass, core-collapse stars drive strong explosions with little fallback and produce neutron stars; (2) moderate-mass stars produce explosions, but the fallback is sufficient to form black holes; and (3) high-mass stars are unable to launch shocks and collapse directly to black holes. The question for core-collapse theorists, then, is to determine the limits for these regimes.",
where the separation between these regimes is (1) $\geq 8$ up to perhaps 20 solar masses; (2) 20-40 solar masses and (3) more than about 30-40 solar masses, but these boundaries depend on metallicity, rotation and a correct understanding of supernova and neutrino physics.


Is there any official star gazing location list? I recall when I was a child (many years ago) my father taking us out to gaze at the starts. I remember being able to see the Milky Way with our naked eyes (North Central New Jersey). Nowadays living down here in Maryland I can count the stars on my hand because of all of the light pollution.

So I ask, is there some sort of listing that shows where the best places are to gaze without a telescope up at the stars?

I'd like to see the Milky Way again and not the white glow of the city lights.

You may be out of luck. If you live pretty much anywhere in the North East, there are very few locations where you can see a truly dark sky. You might be able to get somewhere where you can see the faint glow of the milky way band, but to see a really dark sky, you pretty much have to be out west in the middle of no where.

One resource to help you out is to look up the Bortle Scale. This is a metric used to describe how dark the sky is in a given area and thus what you're likely able to see with the unaided eye. There are numerous sites which attempt to predict or inform what your Bortle scale number is at your specific location.

In general though, you can get a feel of the light pollution in a particular area by using light pollution maps like this one or this one. Attached below is an image from the first link showing the general light pollution level across the US.


Orbits of Trojan Asteroids I understand that the Trojan points are located 60 degrees ahead and behind a planet in its orbit.  However, since there are quite a number of Trojans in Jupiter's orbit, they cannot all be exactly at that point. Presumably they orbit around it.
So what sort of orbits do they have? Are they circular or anywhere near so, or are they elongated along the orbit of Jupiter? If the latter how far can they get from the 60 degree point?

As an example, look at Earth's only confirmed Trojan

By Phoenix7777 - Own work Data source: HORIZONS System, JPL, NASA, CC BY-SA 4.0, source
Now, to understand what is happening here. The yellow dot is the sun. The blue dot is the Earth.  Although the Earth is orbiting the sun, the "camera" is turning so that it appears that the Earth is roughly stationary (it wobbles slightly due to the eccentricty of the earth's orbit)
The pink dot is the asteroid 2010 TK7.  It has an orbit that is eccentric, and so sometimes it is much closer to the sun than the Earth, and at other times it is further, but its orbit takes nearly exactly one year.  However the exact shape of the orbit changes, sometimes it moves around the sun in slightly less than a year, so it starts to catch up with the Earth, but as it nears the Earth, the Earth's gravity tends to pull it forward, and out, causing it to slow down and move away from the earth.  The whole cycle takes hundreds of years.
Note, the actual shape of the orbits are elliptical about the sun, the odd shapes is a result of the camera turning at one revolution per year.
This kind of orbit is said to librate about the L4 point. It is called a "tadpole orbit" Trojans don't have to remain exactly at the L4 point, they can stably orbit in a tadpole orbit around the L4 point.


Why do galaxies have arms? A couple of years ago I started playing a game called Kerbal Space Program (KSP) and my understanding about how orbits actually work increased dramatically. Because of KSP I was looking at a picture of a galaxy with its spiral arms and something struck me as very odd. The stars closer in should orbit the galaxy center faster than the stars farther out and that would smear out these lanes of stars until they were no longer recognizable. Something weird is going on here!

So why do galaxies have arms? What's bunching these stars together like that?

I noticed that the center of galaxies are all smeared out and bulbous like I think that the rest should look like. So what if all that mass rotating around a common center is warping space around it, like the water draining down your sink, and this warping is what catching matter/stars into the lanes.

The usual explanation of spiral arms involves density waves.
Nonuniform motion leads to matter alternately bunching up (boosting star formation) and spreading out.
Like cars on a congested road, stars move through regions of greater or lesser density over time.
The rotational smearing you anticipate is known as the "winding problem."
This Scientific American article
offers three explanations by professional astronomers.


What if our galaxy didn't have a SMBH? From my understanding, it is believed that almost every big galaxy and especially spiral galaxies have supermassive black holes (SMBH's) at the center. Also, from what I've read, a SMBH isn't required for a galaxy to exist since and in layman's terms, the matter inside a galaxy such it gas clouds and formation of stars will keep it gravitational-ly in check.

If I am right or even wrong, then would there be significant difference if our galaxy did not have a SMBH at the center? Significant enough for noticeable differences even here on Earth?

The supermassive black hole (SMBH) in the center of the Milky Way (MW) — called Sgr A* [Sagittarius A-star] — has no direct impact on our galaxy. Its mass is only a few million Solar masses, and if you remove it$^\dagger$, it will only affect the most central stars, which would suddenly continue in straight paths out through the MW. These stars would almost surely not hit any other stars or something like that (since stars are really, really far apart), but some of them have velocities high enough that they may escape the MW.

If Sgr A* weren't there to begin with, things might look a little different. There seems to be a relation between the mass of a galaxy's SMBH and the velocity dispersion of the stars in its central bulge; the so-called M-sigma relation. so MW without Sgr A* would mean a more ordered center. Our Solar system is located in the disk, far from the center, and their is evidence that SMBHs have little impact on the disk (Gebhardt et al. 2001). However, in their early phase (as an active galactic nucleus), their extreme luminosities cause galactic superwinds which blow out gas and may quench star formation (Tombesi et al. 2015).

$^{^\dagger}$Removing Milky Way's SMBH is left as an exercise for the reader.


Is there a photo that shows a iron meteor or asteroid in space in raw form having no layers of fusion crust? A raw iron meteor or asteroid in outer space should be in its raw form having no signs of fusion crust from the friction of atmospheric entry and should therefore show distinctive signs like the posted image.

In short, no.

The reason is that while finding meteorites on Earth is hard, finding them in space is a lot harder. A small asteroid, weighing only a few kg can't be spotted while it is still in space. The first we know of them is when they make their fiery descent.

On the other hand, there are some large asteroids that are metallic, such as 16 Psyche, and we have direct images of other large asteroids. Metallic asteroids are relatively rare, and have a higher albedo than stony asteroids, and much more than comets. However they are not shiny balls of metal. After a few billion years in space they absorb 80% of the light that falls on them.


Why Trojans haven't merged into moons Jupiter's trojans are located in the L4 and L5 Lagrange points.
These two points are stable, so why don't all the trojans have already merged into small moons?
And since it's not the case, what prevents them to do so?

The Jupiter Trojans are not at the Sun-Jupiter L4 and L5 Lagrange points. They are instead in pseudo orbits about those points. These pseudo orbits appear from the perspective of a frame of reference that rotates at Jupiter's orbital rate (i.e., a frame in which the Sun and Jupiter are more or less fixed). Unlike central body motion, which results in planar orbits, these pseudo orbits in general are not planar, and they librate between in front and behind the L4 or L5 point over a period of about 150 years.
The very large volume means interactions are rare, and the crazy pseudo orbits means collisions are more likely to be violent than not. That said, at least one Trojan asteroid, 624 Hektor, is most likely to be a peanut-shaped contact binary.


Is solar wind a source of water? Solar wind may contain some hydrogen and oxygen atoms which is the composition of water, some of them may move to the poles of earth, my question is, is solar wind also a source of water in earth?

Perhaps I have always been misunderstanding the concept of space weather, rain from space now that's a cool idea.

Seriously though, the solar wind particles are at such a a high temperature that molecules cannot form, instead it is an ionised gas containing mainly Hydrogen and Helium and only very tiny traces of other elements.


How can we tell that the sun is moving with theories such as the theory of relativity? In a nutshell, the theory of relativity says that we cannot know that the ether is moving because if we were on the sun, the earth would seem to be moving and visa-versa. Apparently, we can't find absolute motion; we can only find relative motion.

How do we know that indeed the earth moves?
What is relative motion?
What is absolute motion?

We don't know. All that we know is that the Earth moves relative to the Sun and relative to other things in the wider universe. That is what we mean by relative motion.

A measurement of absolute motion requires you to define an absolute reference frame. We don't have one of those - you can choose your reference frame to suit your purposes. The beauty of relativity is that one of its axioms is that the laws of physics remain unchanged whichever reference frame you pick.

Conrad suggests using the cosmic microwave background as an absolute frame of reference. Whilst I agree that this is a useful frame of reference - it demonstrates with what velocity we (or any other object) are moving with respect to the local co-moving volume; there is no sense in which this is absolute or has any special properties beyond its convenience. The laws of physics remain unchanged whatever your velocity with respect to the cosmic microwave background.


Relativistic Mercury orbit around the Sun My question is the following,would the scalar of precession of Mercury orbit be the same if Mercury orbited Sun in the oposit direction?

General relativity is only one cause of the precession of Mercury's orbit, and not the largest factor. Gravitational perturbation by the other planets and the non-spherical sun also cause precession. Newtonian gravity predicts precession of 532 arcseconds per century, but 575 arcseconds is observed. The discrepancy can be explained if one uses General Relativity as the model of gravity.

The extra precession of 43 arcseconds caused by GR would be the same, the perturbations due to the other planets would not be the same, unless they were also orbiting the sun in the opposite direction.


How do galaxies move in space? Kind of like a shower thought just occurred to me. We've all seen lots of animation about spiral galaxies, how they crash... in virtually all of them, I recall they move in a frisbee way, that is their velocity vector is on the same plane as their disc. That sounds wrong... right? What keeps them from tumbling in a random direction in space? Why can't they move like an American football? Or a tossed coin? You know, all the stars within can still go around merrily, but nothing is keeping the whole spiral from moving in a chaotic way.

Before you say that I'm completely wrong to even speculate that, think of our neighbor Andromeda. Fact: its angle to us is not completely edge-on, perhaps 25deg. Fact: many scientists @NASA & other great institutions made high-confident predictions that we're on a head-on crash course with it. Just those 2 facts mean Andromeda OR Milky Way can't be moving along its own plane (or both, lol).

EDIT: when I said galaxies I meant spirals in particular. Also, I think the question above can be rephrased as: in the case of us & Andromeda, what will we see during the next 4.5Gy? Will we see our sister galaxy tumbling while coming toward us, i.e. its angle changes, its appearance dramatically transforms, not just getting bigger in apparent size?

Galaxies move though space independently of the orientation of their axis of rotation. That this is true can be appreciated from the fact that their direction through space is relative; that is, in the reference frame of an observer that is passing the galaxy in its plane, it is moving like a frisbee, whereas in the reference frame of an observer who is moving along its axis of rotation, it is moving like an American football:

Galaxies form by accreting gas which often lies in filaments of gas, and this is thought to have a tendency to align their spins with the filaments. Higher-mass galaxies grow by mergers along filaments, which should align their spins perpendicular to the filament. I'm unsure if it has actually been confirmed observationally, but it is seen in simulations (e.g. Krolewski et al. 2019). Thus, a merger between a massive and a low-mass galaxy will have a tendency to make them meet with their spins at right angles of each other. But this alignment is only in a broad, statistical sense, and mergers at all angles do indeed occur.


Would the solar corona be visible from the distance of the outer planets? If the photosphere was blocked, before the sun rises on an airless moon in the outer solar system, would either the inner or outer corona be visible to the naked eye?

If the sun is ever perfectly eclipsed as it is on Earth, then yes, the corona would be visible. The only reason we can't usually see the corona is because it is way dimmer than the rest of the sun. If the entire sun other than the corona is ever covered fully by another object, then the corona will be visible.


Massive Nearby Stars Is there a possibility that I will find records of stars with masses around 25-30 solar masses within 30 parsecs of solar system. If not, how far do I need to look?

There aren't any.

A 25-30 $M_{\odot}$ main sequence star would have a spectral type of $\sim$ O7 and an absolute $V$ magnitude $M_V \simeq -5$ (see Zombeck 1982). A giant star with this mass would be even more luminous.

At a distance of 30pc, the apparent magnitude of such a star would be $V=-2.6$. Closer examples would of course be brighter. Sirius is the brightest star in the sky with $V=-1.46$.

The nearest O-star may be zeta Ophiuchus which is at about 450 light years (or about 140 parsecs). This might be a bit lower than $25M_{\odot}$.  Betelgeuse, a red supergiant at a Hipparcos-measured distance of $152\pm 20$ pc, probably has  a mass of about $15 M_{\odot}$, but I think this is quite uncertain.

The secondary of $\gamma^{2}$ Velorum has an O7.5V spectral type and reasonably well-determined mass of about $30M_{\odot}$ (Eldridge 2009) and a distance of $336 \pm 8$ pc (North et al. 2007).


Astronomy Olympiad Books I'd like to one day take the Astronomy Olympiad, however I'm at a loss to know where to begin. As it currently stands I have no experience whatsoever with Astronomy. If there are people here with some expertise on the matter, I'd like to know what books you would recommend to study? The syllabus is at the following link: http://www.ioaa.info/page2/page7/page7.html

I happen to have represented my nation at IOAA

Start with "Roy and Clarke's" book : "Astronomy: Principles and Practice"

Move on to the references mentioned in the book.


How often do stars pass close (~1ly) to the Sun? Interstellar exchange of massive objects is difficult across several light years. But as the stars orbit the galaxy the distances between them change. I don't find data for neighbor star distances which covers more than a couple of a hundred thousand years. I can imagine that it is difficult to trace their orbits further given todays uncertainties in distance measurements. But I'm interested in a purely statistical/geometrical estimation of how frequent nearby Sun-star passages have been.

How many times has a star come closer than, for example, 1 light year from the Sun since it formed?
How well mixed have the nearest stars become over time? What fraction of them have followed the Sun since formation and made multiple passages?

I suppose that 1 light year's distance is enough to cause disturbances in the Oort cloud and likely exhange comets.

Here is the very study you are looking for by Bailer-Jones (2014).

Using the re-reduction of the Hipparcos astrometry, he has integrated orbits for 50,000 stars to look for objects that might come or might have come close to the Sun.

The K-dwarf Hip 85605 is the winner on that timescale, with a "90% probability of coming between 0.04 and 0.20pc between 240,000 and 470,000 years from now".

The next best is GL710 a K-dwarf that will come within about 0.1-0.44pc in 1.3 million years.

On a statistical basis, some work has been done by Garcia Sanchez et al. (2001). They estimate, using the Hipparcos data, that encounters within 1pc occur every 2.3 million years. However, the Hipparcos data isn't complete for low-mass, faint objects. Making a correction for this, the authors estimate a $<1pc$ encounter every 100,000 years. The probability for closer approaches scales as the inverse of the square of the separation. i.e. the timescales for approaches closer than 0.1pc is 100 times longer.

There are plenty of big error bars in all this work - recall that an error of 1km/s in velocity leads to a 1pc error in position after 1 million years. All this should be nailed down really well by Gaia results in the next 2-3 years.


Spotting the booster rocket of the Chinese space station? SpaceWeather.com's newsletter states

Sky watchers are reporting two new objects in the night sky: China's new space station and an out-of-control booster rocket that helped launch it. The tumbling booster is flashing brighter than some of the brightest stars in the sky.

My usual star-gazing app does not have the rocket booster in its database (yet), and I did not manage to spot it yet (from cloudy central Europe). Is there an official name for the object? Could somebody help me how to find it in the remaining 2 days while it is still in orbit around Earth? Or maybe somebody could even share a picture or two?
Bonus question: Maybe somebody could also enlighten me in how far any single-use booster-rocket could be in-control after it ejected its payload? Wasn't that the default for many years? Also: Now that we can observe the rocket we know the trajectory and could also forecast when it will crash and where, aren't we?
References

U.S. Space Command tracks Chinese rocket for uncontrolled re-entry from orbit

The name of the module is Tianhe (天和), and the name of the rocket is Long March 5B, but it got a special designation of 2021-035B. It is not really official, but I think it is good enough to get some data. I found this website for current position (live on globe). Your userpage says you live in Europe, so the rocket will fly over Greece at 00:30 CEST (22:30 UTC) tomorrow (in 1.5 hour if you have time to be awake for that long). I also calculated the maximum angular speed (on May 6) of around 1.85 °/s (on May 7, it is 2.7 °/s).
There are already some images; I found only that one. The booster's visual magnitude is around –1, so pretty bright to spot.
So, if you have time, try to look to the South and spot the booster at 22:30 UTC today. Maybe you are going to have luck tonight : )

There are some equations I derived with basic trigonometry about the rocket's visibility:
Distance $D$ at which the rocket can be seen right on horizon:
$$D = R \ \arccos\left(\frac{R}{R+h}\right)$$
Where the calculator calculates radians, $R$ is Earth's radius and $h$ is the altitude of the booster (you can get that on previously given website; at the moment it is 170 km).
The angular height of the booster which is $D$ kilometers away from us (on Earth) is given by:
$$\alpha = \frac{\pi}{2}-\arcsin \, \left(\frac{(R+h)\sin{\frac{D}{R}}}{\sqrt{2R^2+2Rh+h^2-2R(R+h)\cos{\frac{D}{R}}}}\right)$$
Again, the calculator must be set in radians.

Update: The rocket fell on May 9 at 02:14 UTC into the Indian Ocean.


Calculation of Eccentricity of orbit from velocity and radius For a school project I am making a gravitational computer model. As I want to Milankovitch cycles, I need to calculate the eccentricity of an orbit after the model has completed its simulation.

I have thought about calculating the eccentricity using the aphelion and parohelion height, but these are not available as it is a simulation and as it therefore stores the data for one point on the ellipse.

The data I have after the model has completed:
X,Y,Z coordinates of points that it has saved
X,Y,Z velocities of points that it has saved

I have been able to calculate the semi-major axis using the vis-viva equation. I have tried to get the value using the vis-viva equation but I got stuck as I think I need at least the Aphelion or Parohelion to calculate the eccentricity.

My question: Does anyone know how  to calculate the eccentricity at any given point in the orbit using the velocity and coordinates (without simulating all points around it)?

If not, does anyone maybe have a better idea/suggestion that I can try to gain the eccentricity values?

If you have the relative position $\mathbf{r}$ and velocity $\mathbf{v}$ at the same point in time, compute the eccentricity vector:

$\mathbf{e} = \frac{\mathbf{v} \times \mathbf{h}}{\mu} - \frac{\mathbf{r}}{\|\mathbf{r}\|}$

Where $\mathbf{h} = \mathbf{r} \times \mathbf{v}$ is the specific relative angular momentum and $\mu = GM$ is the standard gravitational parameter.

The eccentricity is then given by the norm of the eccentricity vector $\left\| \mathbf{e} \right\|$.


Has the microwave background radiation ever been blocked in order to prove it's really from the sky? A rather stupid question but I was just curious if the radiation could be virtual.  From the instrument....sort of like virtual particles. In other words if you just build a sensitive microwave detector and stick it in a South African Durban Deep gold mine 5 miles under the earth would it do the same thing?

This is kind of old, but I've been reading a bit about CMB and stumbled on this question. If you go back to the original discovery of the CMB, they basically pointed their antenna at the sky and measured the microwave signal. They found one that was "isotropic, unpolarized, and free from seasonal variation" having run the antenna from July 1964 to April 1965. To calibrate the instrument they had a liquid helium cooled reference that they could switch into the receiving circuit. Blocking the CMB signal would be difficult because it's at 2.7 K, and most anything you would block it with would be a larger source (i.e. warmer). Instead, they used a liquid He reference source as a calibration, and then accounted for all other sources of microwave signal. The biggest "other" source was from looking through the atmosphere. You can read about this here:
http://adsbit.harvard.edu/full/seri/ApJ../0142//0000420.000.html
https://www.wmrc.edu/resources/docs/ch1.pdf

To really see what they did, you'd have to read more about that antenna, and there are references in the paper. The original publication is just a letter and so it is terse and refers to previous results instead of offering a detailed description.

...but if you're in orbit, it's easier to make something cold. For FIRAS on COBE, they actually had a device called XCAL (External Calibrator, I guess) that fit into the CMB receiving antenna like a trumpet mute. It looks like they held it at 2.7K and used it to calibrate by moving it into and out of the microwave receiving horn. That's pretty close to blocking the measurement signal. Here's the reference:
https://www.researchgate.net/publication/2669262_Calibration_of_the_COBE_FIRAS_instrument


How many sun-like stars are there in the universe? After yesterday's announcement of the Kepler telescope finding a huge amount of newly observed exoplanets, i saw a headline claiming that as much as 22% of sun like stars in the universe have planets in their habitable zone. There are loads of stars in the universe, so the number of planets in the habitable zone has to be enourmous. But how many of the stars in the universe are about the same size as our sun?

This is a question that concerns the initial mass function (IMF) - an empirical (that is, defined by observations rather than theory) function that describes the statistical distribution of stellar masses.

Edwin Salpeter (1955) was the first to describe the IMF, though if you read Chabrier (2003) there are some reasonably comprehensive explanations of the theory and history. However, these lecture notes are a fair bit more accessible.

From the approximations in the UCSC lecture notes I linked above, I get that around 4% of stars are between 0.7 and 1.3 solar masses (92% are between 0.1 and 0.7 solar masses!).

There are perhaps 100 billion stars in a galaxy and 100 billion galaxies in the observable universe, giving something on the order of $4\times 10^{20}$ (400 billion billion) stars that are about ($\pm 30\%$) one solar mass.