Finding a location geographically of a celestial object, in this case the moon at one of its quarters

How do I find right ascension 18h23m52.34s on the celestial sphere?

You use a star chart that shows the lines of right ascension and declination. Based on the stars that the line passes by, you can estimate the line in the real sky. It is similar to looking at a map of the Earth and using latitude and longitude to locate a city. (By the way, 0 hours right ascension, 0 degrees declination is the location of the vernal equinox on the celestial sphere. The bright star at that location is the Sun at the moment of the vernal equinox ;-)

Perhaps your question should be "How do I locate the young crescent Moon?". The answer to that question is that it would be better to calculate the azimuth of the Sun at sunset, and the altitude and azimuth of the Moon at 10 minute intervals afterwards. Based on knowing where the Sun was located on the horizon at sunset and the difference between the Moon and Sun, you can estimate where the Moon should be in the sky. There is no need to know where right ascension and declination are located among the stars (other than calculating the position of the Sun and Moon); the sky will be bright, so it is unlikely that there will be any stars visible that will help you locate the line of right ascension.


Could vacuum energy dim standard candles?

Locally, energy and momentum must be conserved. So no, a photon cannot lose energy or change momentum by interacting with virtual particle pairs that have a net energy and momentum of zero.


Calculation of right ascension and declination

The answers to this question are very clear and much better than I could have written.

I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30'W not E.

We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-

diff in RA = Long/15
           = 16° 30'/15
           = 1H 06m

Then:-

RA = RA at Greenwich - diff in RA
   = 22H 20m - 1H 06m
   = 21H 14m

Now for Declination:-

The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked 'H'. Now draw upwards 9 lines and mark the top ,'Z' and downwards 9 lines and mark the bottom 'N'. You should end up with something that looks like this:-

Which is, basically a picture of your meridian as you sit and gaze south at it. 'Z' is your zenith, the point above you in the sky and 'N' is the nadir, it's opposite out on the celestial sphere through the earth beneath your feet.

Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we'll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-

Now we put in the body we are looking at. We know it's elevation, just over 43°, we can just call it x as you didn't specify a body. We can also label some of the distances we know, to end up with:-

We already knew that Lat was H -> Q, that's how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-

dec = Alt - Lat
    = 43° 40' - 28° 18'
    = 15° 18'N

As 'x' is to the north of 'Q', then Declination is north.

So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18'.

This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you'll be doing it in your head. I hope this has helped a bit.

I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.


If quasars are powered by black holes, why are they so bright?

The light emitted isn't actually from the black hole itself but the swirling matter that surrounds it called the accretion disk. The light emitted from the accretion disk is far enough to escape from the pull of the quasar. A link from wikipedia sufficiently explains this:

  Quasars are believed to be powered by accretion of material into supermassive black holes in the nuclei of distant galaxies, making these luminous versions of the general class of objects known as active galaxies. Since light cannot escape the black holes, the escaping energy is actually generated outside the event horizon by gravitational stresses and immense friction on the incoming material.

I am supposing that part of the question, you are asking why quasars are brighter than black holes. That's because of the larger amount of in falling matter into the black hole creates greater friction thus more light.


Information cannot "leave" the black hole. There is no way (by our current framework of physics) that an entity inside the event horizon can send a signal out.

However, entropy can leave. The black hole has entropy proportional to its surface area, and this roughly corresponds to "the number of ways that black hole could have been created". The surface area decreases due to Hawking radiation, so it can release entropy back to the universe. Of course, the growth rate due to infalling mas is usually many orders of magnitude more than the shrinkage.


On a Hertzsprung-Russell diagram, why is the main-sequence line squiggly?

Here is another plot of a Hertzsprung Russell diagram (luminosity versus temperature), but this time based on theoretical models. (The plot is from D. Prialnik 2000, An Introduction to the theory of stellar structure and evolution). Note that the zero age main sequence is well behaved in this plot. Luminosity and temperature are related by smoothly changing power laws that are largely due to the varying temperature dependence of energy generation and modes of energy transport within stars of different mass.

Your diagram is actually an absolute magnitude versus colour diagram. This is not the same thing. The transformation between absolute magnitude and luminosity and especially between colour and temperature are distinctly non-linear and dependent on what particular spectral features happen to be present in the photometric bands that are used to construct the plot. For example, there is a knee for cool M-dwarfs because the B-V colour "saturates" at a value of about 1.6 because there is hardly any flux in the B and V bands and their ratio doesn't change very much with temperature because of the characteristics of the opacity sources in the atmospheres of these stars at those wavelengths.

To demonstrate that the plot is not necessarily "wiggly", the plot below shows the colour-magnitude for the Hyades and Praesepe clusters (basically at the ZAMS) in the Gaia filter system (from Babusiaux et al. 2018). Nice and smooth - though it doesn't cover as big a luminosity range as the Hipparcos plot in your question; the main wiggle at B-V~0.3 is still in the Gaia data (off the top of the plot), but the M-dwarf "knee" is smoothed out by the redder filter response.


Finding the Uncertainty in a Spectral Line

If you are doing chi-squared fitting, then a value for the uncertainty (or an estimate) should have been returned by the software. If not, you could perhaps manually estimate by fixing the line centre at various values, allowing the other parameters to be fitted, and seeing where the minimum chi-squared increases by 1 over the global minimum.

Failing that, output your spectrum as an ASCII file and fit it with something that does provide uncertainties like gnuplot.

Failing that, here's a rule of thumb. The line centre is determined with a precision given by the FWHM divided by the signal to noise ratio per pixel.

Note that none of the above take any account of uncertainties in the wavelength calibration.

EDIT: It appears that lmfit does estimate uncertainties. https://lmfit.github.io/lmfit-py/confidence.html


Why do M dwarfs emit more X-ray radiation than larger, more luminous stars?

A large fraction of M-dwarfs are highly magnetically active because they have deep convection zones (or may be fully convective) and they have longer spindown timescales than higher mass stars, so remain at higher magnetic activity levels for longer.
The magnetic fields are generated by some sort of internal dynamo. The dynamo efficiency appears to increase with rapid rotation and deeper convection.
The magnetic activity manifests itself with a large surface coverage of equipartition magnetic fields (i.e. the magnetic pressure is similar to the gas pressure), with strengths of several kG. These fields protrude above the surface, and turbulence in the photosphere causes magnetic reconnection, acceleration of charged particles, and the evaporation of heated material into the corona, which emits the X-rays. Typical levels of X-ray emission for fast rotating M-dwarfs are $L_x/L_{\rm bol}\simeq 10^{-3}$, about 10,000 times greater than the Sun.
The large scale magnetic field is also responsible for slowing down the star's rotation, via the linkage of the ionised stellar wind to the field lines. At some distant point from the star, the wind decouples from the field, carrying away angular momentum. It is observed, through studies of clusters of stars with known age, that while sun-like stars have spindown timescales of 100 million years or less, lower mass M-dwarfs can remain rapidly rotating for a billion years or more.
Note that young ($\leq 100$ million years) sun-like stars also have $L_x/L_{\rm bol}\sim10^{-3}$; they just maintain these high activity levels for less time.


Not clear on how to calculate sub-radar latitude

In the common case asteroids will rotate around a nearly fixed axis, like the Earth. That axis of rotation defines a north pole and south pole for that asteroid. The north pole is the one where, when a viewer is seeing the asteroid from above it, the asteroid is rotating counter-clockwise. Once you have north and south poles, you can define an equator and, thus, a system of latitude. I have no idea what is done for asteroids that are undergoing chaotic tumbling.

The "sub-radar latitude" is the asteroid's latitude that is directly facing the radar viewer. A good example from the literature that uses the term many times, and includes a short definition, is arXiv:1101.3794. A short excerpt:

  [...] implying significant change in the object’s sub-radar latitude (the angle between the asteroid-Earth line and the object’s equatorial plane).


How did the object CO-0.40-0.22 get its name, and how is it distinct from CO-0.40-0.22*?

An earlier paper on the object (Oka et al. (2015)) explains that CO-0.40-0.22 is a "high-velocity compact cloud". The first discovery of such an object was two decades ago (Oka et al. (1998)), when CO 0.02-0.02 was found. The naming convention used for that object was
"CO" + Galactic longitude + Galactic latitude

For instance, CO 0.02-0.02 has latitude $0.02^{\circ}$ and longitude $-0.02^{\circ}$; CO-0.40-0.22 has latitude $-0.40^{\circ}$ and longitude $-0.22^{\circ}$. Naming based on coordinates is used for many other radio sources.
The * in the name of the continuum point source is, I believe, in continuation of the tradition set by the discovery of Sagittarius A*, the point source in Sagittarius A. It is often necessary to distinguish between the two. Quoting from Goss et al. (2003),

Eight years after the discovery, one of us (Brown) invented the name Sgr A* to distinguish the compact source from the other components in the galactic center and to emphasize the unique nature of this source. Brown (1982) proposed a model of Sgr A* consisting of twin precessing jets with a period of 2300 years. The model has not stood the test of time but the name immediately was accepted. As an example, the VLBI results discussed by Lo et al. (1985) uses the name Sgr A*; the review article by Lo (1987) also uses this nomenclature.
Bob Brown provides the following rationale for the name: “ Scratching on a yellow pad one morning I tried a lot of possible names. When I began thinking of the radio source as the “exciting source” for the cluster of H II regions seen in the VLA maps, the name Sgr A* occurred to me by analogy brought to mind by my Phd dissertation, which is in atomic physics and where the nomenclature for excited state atoms is He*, or Fe* etc.”

The difference between CO-0.40-0.22 and CO-0.40-0.22* is simple: The former is a high-velocity compact cloud, and the latter is a point radio source inside it, possibly an intermediate-mass black hole. The formation, evolution and interactions of the two are likely tied together, but, as with Sagittarius A and Sagittarius A*, they need to be distinguished.


How can I convert Bayer/Flamsteed designations to Hipparchos numbers?

As the names and numbers are essentially arbitrary, the Hipparcos designation cannot be calculated from the Bayer or Flamsteed designation.  Instead you have to look up the star in a database.

There is a HD-DM-GC-HR-HIP-Bayer-Flamsteed Cross Index on line.  (Direct link)You can get a computer to do the look up for you. Or if you only have a small number of stars, you can use SIMBAD and do the look-up by hand.


Does a photon need to have EXACTLY the right energy to be absorbed by a gas molecule?

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


There are two main datasets available from the PanSTARRS survey which are available from the archive at MAST/STScI. These are the Object Catalog, a (large) list of parameters such as position, brightness, shape etc (full list of catalog fields) and the Image Cutouts service which, well, cuts out sections of the images. (The survey went over the same patch of sky several times in 5 filters so there are deeper stacks available)

If you are looking for moving objects then you will need to use the "warp" single epoch images as a moving object would be averaged away in the stack (unless it was done on the object's motion). You will also need a source of positions in order to cutout the suitable postage stamps in the right places as the PanSTARRS archive doesn't support querying by moving objects or orbital elements. You would need to use something like the Minor Planet Center's Ephemeris Service or JPL's HORIZONS service to generate these positions (use 'F51' as the Observatory Code/Observers Location as this is the MPC site code for PanSTARRS1). Once you have a source of positions, you will probably want to investigate querying the Image Cutout Service via script/code; the Image Cutout Service documentation has instructions and a Python Jupyter notebook showing how to do this.

I should also caution you that one of the goals of the PanSTARRS survey, which my institution was a partner in, was to find moving objects. A lot of effort was put into the Moving Object Processing System (MOPS; paper, which I'm on, available (for free) here at IOPscience) and PanSTARRS was one of the two major sources of new Near Earth Objects (NEOs), see purpley bars in this plot at the Center for NEO Studies. I'm not saying there isn't new moving objects in there, we go back and search through the PanSTARRS data for precoveries of newly discovered NEOs that the automatic software missed on a roughly monthly basis, but bear this in mind before launching into a big project.


What is the best database for identification of spectral lines?

There is the NIST Atomic Spectra Database where you could browse by elements. This the reverse approach, meaning that you have to first query element by element and then see which of the lines you find in the spectrum - that's how we used to do it in our labs during studies:

Then, there is another NIST site where you can enter up to 4 spectral lines in units of electron Volt to be identified:

I am aware that there is also at least one old commercial software (some name starting with Win) to do that, running under Windows 3.1.
There is also custom written C-code used by research groups working on the topic, but I have to dig to figure out whether these software is still available. All I found is a software called Spectrum Analyzer which I have not tested.


Clarification on Greenwich Sidereal Time (GST)

The flaw in your logic is that "RA and L are constants". The right ascension (RA) of the object on the meridian is not constant. It changes continuously, otherwise we would not say that the sky is rising in the east and setting in the west.

Also, the Sun moves about 1 degree per day along the ecliptic, so the RA of the Sun is constantly changing. If the Sun is on the meridian at 12:00:00 every day, and if the RA of the Sun is changing every day, then the same RA is not on the meridian at the same time on every day. (Note: Because of the equation of time, the sun does not cross the meridian at the same time each day. For purposes of this thought experiment, pretend that it is true.)

These two things taken together lead to the fact that 24 hours of right ascension cross the meridian every 23 hours 56 minutes 4 seconds. And this is why the question asks for a date when the LMC culminates (crosses the meridian) at a specific time (9 pm) and a specific location.


Five years have passed, and several papers have been published about this. The latest, Neesemann et al. (2019), compiles previous results and propose a new age estimate. Or rather estimates, because dating such object is tricky! It depends on which model you use (lunar-derived chronology model versus asteroid-derived chronology model, or ADM), and on which part of the crater you use the model on (ejecta blanket versus interior lobate deposits). Hence the large interval in the result: they found an age of 1.6 to 63.7 Ma (ADM on ejecta blanket), which they translated to "relatively young", geologically speaking.


If a planet gained too many moons could the tidal forces of those moons rip the planet apart?

No.
There are two simple arguments, each on its own alone enought to refute such thought:
First way: Tidal force is the gradient of the gravitational attraction. Gravitational force goes like $F_g \propto \frac{M}{r^2}$ and the tidal force thus as $F_t \propto \frac{M}{r^3}$. For a moon and its planet the mutual distance is the same. And as the planet has the higher mass, the tidal forces of the planet would be higher on the moon than the one of the moon onto the planet. Materials are not that different that it plays a major role here. So before a moon destroys a planet, the planet destroys the moon (or you mis-named the two in the first place, calling the planet the moon and vice versa). This may happen for moons within the Roche limit of a planet.
Second way to argue:
The extreme case of 'many moons' is actually a ring of moons, thus a complete ring around the planet. However a spherical symetric mass assembly around a center of mass (here the planet) does not excert any tidal force anymore. Thus a circular mass assembly will only create a small gradient in the normal direction to the equatorial plane and increase the equatorial radius a bit.


Do pulsar beams interact with the interstellar medium?

Yes. the emissions from a pulsar do interact with interstellar medium - this effect is called dispersion. When the emission from the pulsar hits free electrons (and dust) in the interstellar medium, different frequencies in the pulse become "delayed" by different amounts corresponding to their frequencies - specifically, lower frequencies are delayed more than higher frequencies. So, when we receive the signal from a pulsar in a radio telescope, the pulse is "dispersed", or curved. Software is used to de-disperse these pulses by testing out many different "dispersion measure" values to line up the emissions correctly at different frequencies. You can find that here.

This is the main way we can be sure the pulses interact with the interstellar medium. Based on how dispersed those pulses are, and how much "stuff" we think the interstellar medium contains, we can get a very basic idea of how far away the pulsar is when we accurately calculate its dispersion measure.

I don't, however, know of any effects like what you referred to with the laser, resulting in an ability to "indirectly" observe pulsars... the pulses themselves are already quite weak and "seeing" a pulsar directly takes a decent amount of time on a telescope along with large amounts of data processing. From my personal experience in the field along with taking many observations on radio telescopes designed to observe pulsars, I highly doubt that the laser-pointer effect you described is observable.

For more details on pulsar dispersion, take a look here.


If the density of the universe is high enough then the mass of the universe would be enough to eventually stop expansion and cause the universe to contract (the critical density is about $10^{-26}\, \mathrm{kg\, m^{-3}}$).

If this is the case (it's probably not, the universe is probably flat, with dark energy causing expansion to speed up) then the universe is finite in time and always has been. The universe doesn't stop being "infinite".


Large objects can be very faint if they are far enough away. So large objects wouldn't necessarily be discovered a long time ago.  The object you mentioned is very faint and required a long view time to acquire enough photons to "see" it. As stated in a comment by @Pierre Paquette, the object was viewed for over 59 hours.


How do they keep lights from guided lasers in adaptive optics from interfering with the image?

The laser guide stars are indeed outside the science field of view. In the setup for the Wide Field Mode of the MUSE instrument, there is a mirror with a large hole in the middle: the light for the instrument goes through the hole, while the light from the adjacent part of the sky, where the laser guide stars appear (and also the natural star that's used as the "tip-tilt" reference in the Wide Field Mode) is reflected off to the side to the adaptive-optics system ("GALACSI").
Here's a figure (from the MUSE Instrument Description web page) showing the possible positions of laser-guide stars (LGS), and the tip-tilt star (TTS) for MUSE's Wide Field Mode (the TTS can be any reasonably bright star within the GALACSI field; it doesn't have to be in the upper right). The green square is the science field of view; the gray annulus is the part that gets reflected to the adaptive-optics system:

The alternate Narrow Field Mode allows for proper AO correction down to (in principle) the diffraction limit of the telescope. In this case, all the light goes through the central aperture, and then a special dichroic mirror reflects the laser wavelengths to part of GALACSI, while another dichroic reflects all the near-infrared light to a different part of GALACSI, where the TTS star's light (here called the on-axis guide star = OGS) is analyzed; the parts of the optical light not reflected by the sodium-wavelengths dichroic go to the actual science instrument.

(See here for more details.)
But, yes, scattered light from the laser guide stars is a concern. I believe that the Narrow Field Mode deals with thus using the aforementioned sodium-wavelength dichroic; for the Wide Field Mode, there is a narrow-band filter in the science instrument to block light from the laser. Both approaches do exclude light from astronomical sources with the same wavelengths, so you have to plan your observing accordingly. (E.g., if you really want to observe those wavelengths, then you can't use the laser guide stars.)
The figure below (from the MUSE user manual) shows the instrument throughput in the Wide Field Mode as a function of wavelength. The blue and red curves show sub-modes using the laser guide stars ("AO" = adaptive optics), and the fact that they go to zero for wavelengths of $\sim 5700-6000$ Angstroms is due to the filter being used for those modes.


Is there any way to find out what sky surveys are *currently active?*

If you really just want a list, Wikipedia has a pretty good page on astronomical surveys.

For some comprehensive information, including some tabulated values (such as sky coverage for each survey), try this paper from S. G. Djorgovski et al.


What limits the usable focal length of telescopes currently?

To answer the question reworded as: "What limits the quality of telescopes currently?" The answer is mostly: money.

It used to be: atmosphere. But with advances in adaptive optics, ground-based telescopes are achieving what used to require a space telescope. Plus, we have the technology for space telescopes if we want, like the pending JWST.

So pretty much, it boils down to funding. Who is going to spend the money on the expensive technology to see better and further, when science is being squeezed for funding from all quarters, and astronomers can't put forward a business case for a return on investment like a biotech company can for developing a grass that will result in cows burping less methane.


Does Mars' "Northern ocean" have another name? What about the "Inland Sea"?

According to the Wikipedia article on the Mars ocean hypothesis, the ocean is sometimes called the Paleo-Ocean, and sometimes called "Oceanus Borealis" (a.k.a. "Northern Ocean" for those who feel that quidquid Latine dictum sit, altum videtur.)

The inland sea in the article is located in the Simud Interior Basin (part of the Simud Valles). The Nature article describing the findings (Rodriguez et al. 2019) doesn't appear to give the sea a name, although it does use the term "SIB sea" at one point.


How can comets have tails if there's no air resistance in space?

There are two forces that can cause the formation of a tail: the solar wind and radiation pressure.

The first misconception in your question is "the dust [travels] slower than the nucleus".  The tail is not left trailing behind the comet, it is pushed away from the comet by the sun. When the comet is moving away from the sun, the tail is in front of the comet.

Now radiation pressure is small but real. When light shines on something there is a small force. This pushes dust back from the comet in the direction opposite to the sun. The dust is still affected by gravity and a curved dust tail results.

The ultraviolet light from the Sun ionises the gas and gives it an electric charge. The solar wind carries magnetic fields and the gas (or more properly plasma) follows these fields in a straight line back from the sun.

So space around the Sun is not empty. There is powerful light and magnetic fields that are strong enough to push the dust and gas released by the comet away from the coma, and form the tail.


Why are galaxies distributed as walls and voids, or spider webs?

Sorry I have not enough reputation to comment. I am not going to give a full answer as I am on my phone.

I guess there is a (at least seemingly) neat mathematical explanation for the geometry. But it is simply a result of gravitational collapse. The degree of 'formy-ness' depends on cosmology (for example, need more / less time to form structures depending on cosmology). If you have random distribution of matter in infinite 3d with slight density fluctuation, some patch of the space become denser faster than other patches and become even denser and so on. Every matter pulling each other while clustering themselves results in 3d to 2d, 2d to 1d, 1d to 0d structure of overdensity. So most dense regions are point-like structures (super clusters), and next most dense regions are filamentary, then sheets. The rest are voids. Cosmic expansion changes the strength of gravitational pulling but does not really alter the 'shape' or direction of force, hence little effect on the 'formy-ness' of voids. As long as it is a cosmology with collisionless darkmatter, the same formy structure will form given enough time.
 Cosmological N body simulations use fixed(comoving) volume where only the magnitude of gravitation changes due to the cosmic expansion. And in those simulations you can see (and that's where astronomers learned about) the cosmic structures. There are many movies of cosmological simulations you can easily find.


What is the largest possible size for a brown dwarf?

I think what you are asking is what is the radius of an object that has the maximum mass to be classed as a brown dwarf?

The mass threshold for this is about $0.075 M_{\odot}$. The boundary is a bit fuzzy - there is still some hydrogen fusion, but not enough to produce the luminosity of the object), it also depends slightly on the chemical composition of the object.

However, the fact that you can still see (with a telescope!) brown dwarfs raises a more fundamental problem with your question. Their luminosity comes from gravitational potential energy; they are contracting all the time.

So you can only ask what is the radius of the most massive brown dwarf at a given age?

The oldest (about 10 billion years, though there is no good way to estimate the age of a brown dwarf in the field, or measure its mass), most massive brown dwarf's should be about the size of Jupiter. A similar brown dwarf in a star forming region, born a few million years ago, should be several times bigger.

The plot below (from Nelson, Rappaport & Joss 1986 - newer models are available, but show the same behaviour) shows the radius of substellar objects as a function of time. The plateau at early times is caused by Deuterium fusion in object more massive than about $0.012M_{\odot}$.


How can I calculate the apparent position of stars as seen from a location on Earth at a specific time?

There is doubtless a way to do it in AstroPy, but in case you’re interested, the Skyfield library both supports the operations of loading the Hipparcos catalog and computing apparent coordinates. Here is its overall guide to positions, with running examples in the text:
https://rhodesmill.org/skyfield/stars.html
And here's a more elaborate and realistic example star plot, using the stereographic projection, which is my favorite when doing astronomical work:
https://rhodesmill.org/skyfield/example-plots.html#drawing-a-finder-chart-for-comet-neowise
In Skyfield you will account for location and time of year by using a specific time and location when generating your apparent positions, a maneuver that's included in this example:
https://rhodesmill.org/skyfield/examples.html#at-what-angle-in-the-sky-is-the-crescent-moon


What happens with the black hole's mass when a black hole disappears?

The mass (equivalently energy) is radiated away as Hawking radiation.


The metaphysical aspects of your question can't be answered. Our bodies aren't eternal, and astronomy can't answer questions about the soal.

Observations of the universe suggest that it will continue expanding.

As you note, eventually all the stars will run out of hydrogen, collapse to white dwarfs, which then slowly cool. What happens next is rather speculative. At this time there will be nowhere that is warm in the universe. Everywhere will cool to very close to absolute zero. Complex chemistry will no longer be possible, and there will be no life in the universe.

Some stars will fall into black holes. Others will be ejected from the galaxy. If protons decay, then very very very slowly, the matter will break down, leaving only black holes.

If black holes emit Hawking radiation, then they too will break down.

In this model, the spacetime of the universe is eternal, but nothing within it survives.

Wikipedia has a list of some of the events of the far future


Have we detected galaxies which have red-shifted beyond the visible light range?

Yes, of course. Many, many examples. Telescopes work in the infrared, far-infrared and there are even samples of galaxies that are selected on the basis of their mm emission.

The most distant galaxies detected now have redshifts of 7 or more. This means the wavelength of their light has been stretched by a factor $1+z$ - i.e. by a factor of 8. Thus light in the visible range, say 500nm, now appears at wavelength of 4 microns, in the infrared.

Telescopes that work in this range include the Spitzer space telescope; and many ground-based telescopes. Observations of highly redshifted galaxies are routinely made at infrared wavelengths on telescopses all around the world.

Galaxies are also detected in the far infrared by the Herschel satellite or at mm (getting on for microwave) wavelengths by JCMT or the ALMA telecope.


There are two primary dust populations near 1 AU, interplanetary dust (IPD) and interstellar dust (ISD) [Mann, 2010].  I also discussed dust observations in detail at https://physics.stackexchange.com/a/160627/59023.

Interplanetary Dust

IPD of ~1 $\mu$m size drift sunward due to Poynting-Robertson drag while following roughly Keplerian orbits [e.g., Malaspina et al., 2014].  Closer to the sun, these particles break up due to collisions, sublimation/ablation, and/or sputtering.

Dust grains of ~0.1 $\mu$m size are the so-called "$\beta$ meteorites", which travel away from the sun due to the imbalance of radiation pressure over gravity [Mann, 2010].

The smallest dust grains with $\ll$ 0.1 $\mu$m size, the so-called nanograins or nanodust, act like large pickup ions, which are carried anti-sunward by the frame-dependent convective electric field (i.e., just the Lorentz force) produced when the dust grain moves relative to the solar wind flow (i.e., $\mathbf{E}_{conv} = - \mathbf{V}_{sw} \times \mathbf{B}_{sw}$, where the subscript $conv$($sw$) stands for convective(solar wind), and $\mathbf{V}$ and $\mathbf{B}$ are the bulk flow velocity and quasi-static magnetic field).  These particles can reach speeds in excess of 100 km/s relative to the sun [Meyer-Vernet et al., 2009].

Interstellar Dust

ISD was first discovered by the Ulysses spacecraft, which is ~1 $\mu$m size and moves at ~26 km/s relative to the solar system barycenter.  More recent work [Malaspina et al., 2014] has found a relationship between dust impact count rates and ecliptic longitude.

The reason can be seen from the following.  The Earth's transverse speed about the sun is ~29 km/s.  Thus, when the Earth moves anti-parallel(parallel) to the ISD flow direction the relative dust-spacecraft speed is ~55(~3) km/s, which produced an enhanced(depressed) dust count rate.  This occurs because there is a threshold impact speed necessary to produce a large enough plasma cloud (i.e., $\gtrsim$5-10 km/s depending on dust size) [Meyer-Vernet et al., 2009; 2014].

  What is the origin of the dust near the sun?

The primary sources of ~1 $\mu$m size near 1 AU are cometary debris trails, asteroids, planets, moons, and ISD [Mann, 2010; Zaslavsky, 2015].

  Simply put - wouldn't the gravity pull it in or the solar wind blow it away?

Some are attracted by a combination of gravity and Poynting-Robertson drag while the smaller grains are either "pushed" out by radiation pressure (i.e., $\beta$ meteorites) or "picked-up" by the solar wind Lorentz force (i.e., nanodust).

References

D.M. Malaspina et al., "Interplanetary and interstellar dust observed by the Wind/WAVES electric field instrument," Geophys. Res. Lett. 41, pp. 266-272, doi:10.1002/2013GL058786, 2014.
Mann, I. "Interstellar Dust in the Solar System," Annual Review of Astronomy and Astrophysics 48, pp. 173-203, doi:10.1146/annurev-astro-081309-130846, 2010.
Meyer-Vernet, N., et al. "Dust detection by the Wave instrument on STEREO: Nanoparticles picked up by the solar wind?," Sol. Phys. 256, pp. 463-474, doi:10.1007/s11207-009-9349-2, 2009.
Meyer-Vernet, N., et al. "The importance of monopole antennas for dust observations: Why Wind/WAVES does not detect nanodust," Geophys. Res. Lett. 41, pp. 2716-2720, doi:10.1002/2014GL059988, 2014.
Zaslavsky, A. "Floating potential perturbations due to micrometeoroid impacts: Theory and application to S/WAVES data," J. Geophys. Res. 120, pp. 855-867, doi:10.1002/2014JA020635, 2015.


Which stellar properties can we describe as "first principles" in which we can derive the rest?

Starting from a protostar, one would hope to be able to predict everything about its future development if we knew its initial mass, chemical composition and angular momentum.
Mass is fundamental because it determines how much fuel the star will have and the pressure at its core. Composition is key because among other things it determines the opacity of the stellar material. Angular momentum is important because rotation can alter the structure of the star, if fast enough, and can also enhance internal chemical mixing.
Of course extrinsic factors like interactions are also possible, so the properties of any companions are also fundamental, as might be the birth environment that can do things like hasten the dispersal of circumstellar material from a protostar.
If you are talking about figuring out the past and future of a star observed midway through its life, then in addition to the above, we may need to know how the chemical composition varies with depth inside the star, since it will not be uniform.


What is the distance that the Moon travels during one orbit around the Earth?

The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth's frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example second animation explaining Lunar librations in this answer.

But its orbit can be said, in an oversimplified manner, to be periodic, with no significant apsidal precession (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.

So, plugging our numbers in a calculator, $l = v * t$, we get the Moon's orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is Wikipedia on Moon.


Selecting a Telescope for Viewing Planets

Go with Option 1 (135mm) as it has bigger light collecting area, i.e. diameter of primary mirror. It will enable you to see faint objects and it will also help you to see the dim objects under a greater zoom eyepiece, e.g. 4mm or 10mm.

Also, don't fool yourself with the bigger numbers advertized by telescope vendors like 238X zoom or 300X zoom.

First, these are zooming power using Barlow lenses also the more you zoom in on any object, the less light you gather on the mirror and hence objects will look much fainter in appearance. I will suggest the 10mm eyepiece with a 135mm telescope (Option 1) will be sufficient for you to observe planets like Jupiter, its Galilean satellites, Saturn's rings and also many Messier objects like Andromeda Galaxy M31, the Honey Bee Cluster, and open star clusters in the Auriga constellation.

Also bear in mind that you live in a country area closer to sea like here in India and the weather will be hot and humid.
So ask the vendor about special coating which protects the mirror from rusting (spots on mirror due to moisture) during the humid monsoon season as at least for 4 months it will be humid.


Who coined the phrase "martian dichotomy"?

It is always difficult to say who is "the first" at something, as there are often cases where you can really dig and find something earlier.
I started searching based on my answer to your similar question, but Hartmann (1973) did not use the word "dichotomy" in his paper.  I did a search through all of the journal articles on my computer for the word "dichotomy," including all articles published in the planetary science journal Icarus since its inception, and I came across Soderblom et al. (1974) "Martian Planetwide Crater Distributions: Implications for Geologic History and Surface Processes."  In it, they use the actual phrase, "The Martian global dichotomy" which is pretty close to the exact term you're looking for.
Given the response to your other question in that the earliest identification of it was from Mariner 9 images, which came in 1971–1972, I suspect that if this Soderblom et al. paper was not the first, it was among the first to use that phrase.  Of course, it also takes time for any phrase or term to catch on, so when the term became common is a completely different question and one that I'm not sure could be answered other than by asking planetary scientists from that era (I know some, I could try to ask if you wanted).


Before we begin: how far a planet is seen from the Sun is called elongation, and it's measured in degrees. 0° elongation means it's right on top of the Sun (or behind); 180° elongation means it's opposite to the Sun (it's highest in the sky at midnight, when the Sun is on the other side of the Earth).

https://en.wikipedia.org/wiki/Elongation_(astronomy)

See the image below, where the Sun and the Earth are shown, along with an inferior planet (such as Mercury) and a superior planet (such as Mars). Elongation is the angle marked ε.

  How can you see it at night?

Both Venus and Mercury are always seen close to the Sun. You can never see them at midnight, true. But you can see them during twilight, or, in Venus' case, shortly after nightfall (and then it sets quickly), or shortly before dawn in the morning (and then it's masked by the light of day). This is not a paradox or an impossibility. Here's why:

Mercury in particular can only be seen during twilight, very close to sunset or sunrise. Its orbit is so close to the Sun in space (only 39% the size of Earth's orbit), it does appear very close to the Sun even in the sky as seen from Earth. You could not see it during the day, of course, because the sky is so bright. But during twilight you can sometimes see it peeking just to the side of the Sun, very briefly - then it either sets in the West, or is swallowed by the light of day.

The greatest elongation that Mercury can achieve is 28°. Hold your arm straight out in front of you and make a fist; twice the size of your fist is nearly 28°. But even that's quite exceptional; most of the time Mercury stays closer to the Sun, closer than half of that angle (less than one fist).

So I should make it very clear: it's very rare that you can actually see Mercury with your naked eyes. It pretty much has to be right at maximum elongation. I've observed its phases (just like the Moon, or Venus) in a small dobsonian telescope a couple times; at max elongation it's close to the first quarter phase, or last quarter, and it looks like a very tiny replica of the Moon seen at that same phase. Kind of like this (except in my small telescope no craters or mountains were visible, because the image was just too tiny):

https://en.wikipedia.org/wiki/Planetary_phase

Venus' orbit is quite a bit wider, about 72% the size of Earth's own orbit. So it can swing ahead or behind the Sun quite widely. But that's all it can do - swing ahead or behind. It can never go "all the way around" the Earth. You can never see it at midnight, when the Sun is on the other side of Earth.

Venus' maximum elongation is 47°. Stretch your arm out and make a fist; now stick your thumb and your pinky sideways as far as they go (like the "hang loose" gesture, but stick them out all the way). The span from tip to tip is about 25°; twice that span is the maximum elongation of Venus.

Now imagine Venus at maximum elongation - two "hang loose" hand spans. The Sun has already set in the West about 1 hour and 40 minutes ago, so this is during the night, it's dark already, and now the Sun is one "hang loose" hand span below horizon. But Venus is at maximum elongation, which means it's one handspan above horizon. That's quite hight in the sky actually, you can see it pretty well.

So, there you have it. You can actually see Venus during the night. Not for long, because it still can't wander off too far from the Sun. It must remain within two "hang loose" hand spans from the Sun, or less than that distance, at all times. So Venus itself will set in the West soon after you see it. It cannot hang around until midnight.

Venus also shows phases, just like the Moon or Mercury. I've watched those too in my telescopes.

You can get more precise angle measurements if you can use an old-school marine sextant, like captains on ships back in the day. The fist and "hang loose" gesture techniques are just approximations. Or just get a protractor and tape two straws to it at the desired angle, and sight along the straws to estimate the angle in the sky.

The superior planets (Mars, Jupiter, etc) are not like that. Their orbits are larger than Earth's own orbit. So they can actually be seen, once in a while, high in the sky at midnight, when the Sun is on the opposite side of Earth.

Please refer to the diagram at the top of this reply, it's quite self-explanatory.

So, as you can see, the standard solar system model in astronomy is in excellent agreement with what you can see with your own eyes. There is no disagreement between observation and the model - if there was, the model would be fixed. That's how science works.


Most stars are of a solar-mass or below. The average number of companions that each stars has (in the sense of being part of binary or higher multiple systems) systems ranges from 0.75 for stars of a solar mass to approximately 0.35 (not a well-established number) for the more numerous M-dwarfs. Let's take a compromise value, say 0.5.
The separation distribution of these multiples peaks at around 50 AU for solar-type stars, reducing to about 5 AU for low-mass M dwarfs. Again, lets take a compromise value of 20 AU. See Duchene & Kraus (2013) for all the details.

So if we take 1000 stars, then 333 of them (roughly speaking) are companions to another 333 stars, while 333 are isolated single stars. (NB This does not mean the frequency of multiple systems is 50%, because some of the companions will be in higher order multiple systems)

Thus, taking your calculation of the separation between stellar systems of 5 light years ($= 3.2\times10^{5}$ AU), then the mean separation is:
$$\bar{D} = 0.667\times 20 + 0.333 \times 3.2\times10^{5} \simeq 10^{5} AU, $$
but the median separation is 20 AU!

This is of course sophistry, because I'm sure your question is really, why are stellar systems so far apart?

Stars (and stellar systems) are born in much denser environments. The number density of stars in the Orion Nebula Cluster (ONC - the nearest very large stellar nursery) is about 1 per cubic light year. The equivalent number for the solar neighbourhood is 0.004 stars per cubic light year. Thus the average interstellar separation in the ONC is 1 light year, but in the solar neighbourhood it is about 6.3 light years.

The reason for this separation at birth is the Jeans length - the critical radius at which a clouds self-gravity will overcome its thermal energy and cause it to collapse. It can be expressed as
$$\lambda_J = c_s \left( \frac{\pi}{G\rho } \right)^{1/2},$$
where $c_s$ is the sound speed in a molecular cloud and $\rho$ its density. For star forming giant molecular clouds $c_s = 0.2$ km/s and $\rho=10^{-23}$ kg/m$^3$. So clouds of scale hundreds of light years could collapse. As they do, the density increases and the Jeans length becomes smaller and allows the cloud to fragment. Exactly how far the fragmentation goes and the distribution of stellar masses it produces is an area of intense research, but we know observationally that it can produce things like the ONC or sometimes even more massive and dense clusters.

From there we know that a new born cluster of stars tends not to survive very long. For various reasons - outflows, winds and ionising radiation from newborn stars are able to heat and expel the remaining gas; star formation appears to have an average efficiency of a few to perhaps 20-30%. Expelling the gas, plus the tidal field of the galaxy breaks up the cluster and disperses it into the field, which gives us the lower field star (or system) density that we see around us.

Once the stars are part of the field they essentially don't interact with each other; they are too far apart to feel the influence of individual objects and move subject to the overall gravitational potential of the Galaxy.
 so your consideration of the force between stars is not really relevant.


What are the most recent minerals modified by water found to date on Mars?

If you have access, I would recommend you to read Carter et al. (2013). It is a compilation of all occurrences of hydrous minerals that have been detected on Mars by spectrometers OMEGA (Mars Express/ESA) and CRISM (Mars Reconnaissance Orbiter/NASA). This data was used to investigate the spatial distribution, composition, and age of hydrous minerals on Mars.
Regarding age, the most recent hydrous minerals are opal and zeolites of Amazonian age (< 3 Ga):

The exposure with the youngest unit age is a series of small cones and
dykes or fractures which have been identified thanks to a dust‐free
window in the Utopia Rupes region, in a unit of Amazonian age. Both
opaline silica and zeolites (or alternatively sulfates) are found
there as shown in Figure 7. Zeolites are found mainly on the butte
that could constitute a small volcano, as suggested by the presence of
radial patterns resembling lava flows, to the southeast especially.
Silica is found on an elongated structure which is either a dyke or a
structural horst. Of interest is the presence of two small cones
exposed in the NE corner displaying opaline silica. Such small cones
on Mars have been interpreted either as volcanic vents or mud
volcanoes [e.g., Allen, 1979; Farrand et al., 2005]. In both cases,
the observation of opaline silica pleads for a local fluid circulation
inside a hydrothermal context consistent with volcano‐ice interactions
as predicted by previous studies [Allen, 1979; McGill, 2002; Farrand
et al., 2005].

However, it does not really answer your question as their presence is likely not due to "liquid water on the surface", but rather to impact-driven hydrothermal systems inside the crust. Exposures of their "horizontal deposits" (i.e. sedimentary) and "alluvial fans and deltas" categories, most likely linked to surface liquid water, are of Hesperian and Noachian age, although I couldn't find which is the youngest.
A limitation to keep in mind: some exposures of hydrous minerals may be older than they seem, as the terrain could have been excavated by an impact long after the alteration process (Noachian terrain exposed by Hesperian or Amazonian impacts).


Why are most lunar maria on the visible side?

The two hemispheres have distinctly different appearances, with the near side covered in multiple, large maria (Latin for 'seas,' since the earliest astronomers incorrectly thought that these plains were seas of lunar water). The far side has a battered, densely cratered appearance with few maria. Only 1% of the surface of the far side is covered by maria, compared to 31.2% on the near side.

One commonly accepted explanation for this difference is related to a higher concentration of heat-producing elements on the near-side hemisphere, as has been demonstrated by geochemical maps obtained from the Lunar Prospector gamma-ray spectrometer. While other factors such as surface elevation and crustal thickness could also affect where basalts erupt, these do not explain why the farside South Pole-Aitken basin (which contains the lowest elevations of the Moon and possesses a thin crust) was not as volcanically active as Oceanus Procellarum on the near side.

It has also been proposed that the differences between the two hemispheres may have been caused by a collision with a smaller companion moon that also originated from the Theia collision. In this model the impact led to an accretionary pile rather than a crater, contributing a hemispheric layer of extent and thickness that may be consistent with the dimensions of the farside highlands.

Though the far side has more visible craters, this is due to the effects of lunar lava flows, which cover and obscure craters, rather than a shielding effect from the Earth. NASA calculates that the Earth obscures only about 4 square degrees out of 41,000 square degrees of the sky as seen from the moon. "This makes the Earth negligible as a shield for the Moon. [...] It is likely that each side of the Moon has received equal numbers of impacts, but the resurfacing by lava results in fewer craters visible on the near side than the far side, even though the both sides have received the same number of impacts."

From wikipedia page and an other source.

EDIT: to complete the answer.

From space craft orbiting the moon, it has been found that the centre of mass of the moon is displaced toward Earth. An explanation of seas distribution states that lunar core, aesthenosphere and the mantle is off-centre, whereas the crust is not — so the crust is thicker on the far side (see figure). Thus the basalt would reach the surface more easily on the near side due to thinner crust (to 60 km near side, 100 km far) than on the far side.

From this source.


This question is someone open-ended since you don't provide many constraints, so I'll impose the following two conditions of my own to help me give an answer:

I'll assume that after you're randomly placed in space and time, that you can make any observation one could make here on Earth - that is, you have whatever equipment is necessary to observe anything about the Universe we can currently observe.
I'll assume that you have access to all current astronomical knowledge (in a digital library or some such thing) such as star catalogs, physics books, published papers, etc.

How do we know our location?

In determining location, because we exist in three dimensional space, we have to specify position with three coordinates. The simplest coordinate system in my mind, in terms of what we can measure, are the three coordinates below:

our radial distance from the center of the galaxy,
our height above or below the galactic plane,
and our azimuthal angle from some reference point.

Yes I know these coordinates aren't orthogonal, but they relate well to measurements one could make and simple coordinate transformations could make them orthogonal.

The first one is pretty easy. We can currently measure the distance to Sag A*, the black hole at the center of the galaxy. It's not particularly easy and measurements have fluctuated over the years with respect to our position on Earth (see e.g., Eisenhauer et al. 2003 or Boehle et al. 2016), but it is a measurement which could allow you to define #1.

The second coordinate could be achieved by simply mapping the local star field. A survey akin to SDSS could accomplish this. This is precisely what was done in this Humphreys & Larson 1995. It would take time and a lot of work, but you could make this measurement pretty easily.

So far so good. Now we come to coordinate #3 and suddenly we hit a problem. We need to know our azimuthal position with respect to some reference point. That is, if you look down on the galaxy and consider our angle from some line, what would that line be and what would that angle be? Since the clear intent is to relate our new position to where the Earth is, the most logical reference line is the line connecting the Earth to the center of the galaxy. But how do we know where that line is with respect to our new position? I'm afraid I don't have a good answer here. Others suggested using pulsars but that isn't foolproof. First, you don't know when you're being deposited so it could be billions of years in the past or future when all current knowledge of pulsars is no longer relevant. Second, you might be on the other side of the milky way where we can't actually see anything currently and thus have no knowledge of pulsars over there.

Your best bet might be to map out the star distribution of as wide a swath of the galaxy as you can see to try and understand what the spiral arms look like near you. We've already done this for our current position in the galaxy. Galactic arms move and change slowly, taking hundreds of millions if not billions of years to change significantly. If you can recognize the general structure of the arms, that would give you a decent reference to your azimuthal position.

How do we know the time?

I'm going to break this down into two possible answers. The first assumes you've been placed in a time not too distant from our current time (think less than a million years) and you're still pretty near to our current position.

In the case above, pulsars are the way to go. Pulsars pulse with extreme regularity, so that they're great time keepers. But if your goal is to know how much time has passed, you need to look at the time rate of change of the pulses. As it turns out, the pulsars don't pulse with perfect regularity. The time between pulses decays very slowly over time due to the pulsars losing angular momentum very slowly. The general step to calculating a reasonably precise time is as follows.

Find a pulsar that is currently known and has current measurements of the pulse rate and decay in pulse rate. To make sure you've identified a given pulsar as the correct one, you'll have to map its relation to other stars via a wide survey (the star fields will have changed, but if you're not too far away from our time you can predict with reasonable accuracy where most stars will be or have been in the past).
Measure the pulse rate (and decay rate if you wish, but that may take years or even decades).
Compare your currently measured pulse rate to the present day measured rate and decay rate and do some calculations. You should be able to figure out your current time with respect to the time now fairly precisely. I can't say how precise, but I wouldn't be surprised if you could get within a decade or so of the current time (depending on how accurate your pulsar models are).

The only caveat to the above approach is that pulsars can "glitch". Normally the pulse rate decays smoothly, but every once in a while a pulsar might just glitch and quickly change pulse rates. It is unclear exactly why these glitches occur and they're unpredictable so you have to make the assumption that your pulsar hasn't glitched between now and whatever time you're in. If it has, your time may be off wildly. To mitigate this, you should of course use as many pulsars as you can find. If one pulsar gives a different number than the rest, you can assume it glitched and throw it out.

The second case is more tricky. Now I'm going to assume you're in a time really distant from our own. Millions or possibly billions of years away from now. In that case, any hope in finding a precise time (or even a precise year) is wildly presumptuous. At best, you can figure out your general time within the universe by measuring the current age of the universe. We currently know the age to be $13.799\pm 0.021\ \mathrm{billion\ years}$ based on the latest and greatest measurements. You can see our uncertainty is down to 21 million years. If you make similar measurements of the Universe's age in your time, then the best you'll be able to do is define your time with respect ours with an accuracy of ~20 million years. But if you're a billion years in the future, that's a pretty good measurement in my opinion.


(Thought experiment) if we put a huge mirror into space, could we see back in time to the big bang?

Imagine there is a supernova a hundred years ago. Lots of people see it but nobody has a decent telescope. A hundred years later we point a powerful telescope in the opposite direction, and find a huge alien mirror a hundred light years away, that allows us to watch the supernova.

Light travels at a fixed speed so a hundred years later when we build our telescope we would see nothing in the opposite direction, because the light reflected from your alien mirror would only have reached that mirror, and has another hundred years to reach our telescope.

  My question is, what is the limit of such a technique ?

It's a thought experiment, not a technique.

  Could we see right back to the big bang if the mirror was large enough and far enough away?

Light takes time to reach us, so when we look at very distant objects they're already telling us about the distant past.  This is how we have built up an understanding of what happened in the distant past.

You might want to explore the subject of the Cosmic Microwave Background here.

In principle it would be easier to detect any radiation from the big bang directly (without a mirror) in a similar way.  But the universe is not a quiet place and radiation from the big bang has literally been through everything and won't be as easily detected (if it's possible at all).  We don't even have a proper theory to know what to expect to be able to detect, if anything.

All the mirror does is complicate things.


Would we be able to feel the acceleration of a planet with a highly eccentric orbit?

No.

While you're on that planet, you are also in the same orbit around the star as the planet is. As long as you're in the same orbit as your vehicle (planet, spacecraft), you don't experience any acceleration relative to the vehicle. Whether you're on the planet, or you're just tagging along on the same orbit, is the same thing.

Similar question: in a satellite in a very eccentric orbit around the Earth, is any break in microgravity experienced in various parts of the orbit? Same answer - no. The satellite and everything in it follow the same orbit, so microgravity conditions are maintained.

Now things would be different if you had a very eccentric orbit around a source of much more intense gravity, like a neutron star or a black hole. At the lowest point in the orbit, you could experience tidal forces: the things closer to center experience stronger gravity compared to things further away from center.

But for this to be experienced within a small spacecraft, or within the limits of one human body, the gradient of the field would have to be very strong indeed; the orbit would have to reach very low, very close to the neutron star or black hole.

However, what you would experience would not be orbital "acceleration", but just tidal force. Even so, in extreme cases the tidal force could be strong enough to destroy the ship or kill the passengers.

See the short story "Neutron Star" by Larry Niven for a literary depiction (fairly accurate physically) of this phenomenon. Unfortunately, the story is spoiled by the explanation above, but hopefully learning about gravity compensates for spoiling the story. :)


What is a black hole's "shadow" and the best angle to view it to measure the spin?

The black hole shadow is basically the image of the event horizon. As you know the event horizon is the geometric locus of the points from which a light ray pointing in the opposite side of the singularity (the center of the black hole) can't no longer escape from it. Any other light ray emmited in any other direction from those points would never reach an external observer since the one pointing directly outside the black hole was the one with the best chances to do so.

So we might think that because of the absence of light coming from these region inside the event horizon an external observer might see a black ball of 1 Schwarzschild radius (which is the physical size of the even horizon), but this is not correct. Even if the event horizon marks the physical boundary between both regions the fact is that due to the extreme light bending the actual image of the event horizon (its "shadow") is a distorted view of this surface. For a classical non-rotating black hole this "shadow" (we call a shadow the image created by the absence of light rays but we can trace it with hypothetical "dark rays" that behave in the same way) looks like a ball with 2.6 Schwarzschild radii in size. Much larger than the actual event horizon! To visualize this I can only point to this fantastic explanation by Derek Muller from whom I took this animation.

As you can see we are throwing light rays from infinity into the black hole (that's why they are all parallel in the beginning). Our light rays fall reach the event horizon even if they were not directly pointed there, because they curve. Since light paths can be reversed and the physics still hold we can talk about light rays coming from different parts of the event horizon and reaching the outside observer in the reversed paths. So as you can see not only light rays emmited just barely outside the event horizon pointing towards us will reach us but also rays coming from "the back" of the event horizon can reach us. And as you can see those rays would create an enlarged image of the event horizon since they seem to come from a region located farther than the event horizon itself. So when you look to a real black hole from outside you will see this "black shadow", which is a map reprojection of the surface of the true event horizon where you can see even the 100% of the surface of it from a single vantage point.

This phenomenon (which is called relativistic light deflection) is also noticiable in other compact objects like Neutron stars. The intense gravitational distortion around them allows for light rays coming from its bright surface to deflect when going outwards and reach your eye even if they where emmited in regions close to "the back of the neutron star". Even if that region of the star shouldn't be accesible to an observer if light rays moved in straight lines (since those parts of the surface lie behind the curve of the star) you still can see them (which is something that can mess with the calculations of their true brightness).

You can make a square grid on top of the surface of the neutron star and see how much of it you can see from far away in this representation:

As you can see we are able to see more than an hemisphere (more than 50% the surface of the star). In fact you can see both polar regions and their surroundings. Well this is the same thing happening in a black hole but in that case the reprojected map is all black (since the event horizon is a uniform featureless surface all around) and you can see the 100% of that surface not only a small extra percent.

Now, this all changes if the black hole is rotating. Why? Because of relativistic frame-dragging. According to Einstein field equations mass-energy does not only curve spacetime but is also able to "twist it" if the object is rotating. We have measured this "twisting" of the surrounding spacetime in our own planet using exquisite instruments onboard the GRACE satellites.

In our case the important thing is that a non-rotating black hole (a Schwarzschild black hole) has a region outside the event horizon where orbiting the black hole in a stable manner is possible, we call it the innermost stable circular orbit (or ISCO). Getting closer to ISCO makes your orbit unstable and you end up falling into the event horizon. But if the black hole is rotating (a Kerr black hole), then ISCO is different if your orbit is prograde (orbit in the same direction as the black hole rotation) or retrograde (goes in the opposite direction around the black hole) because frame-dragging alters the solution. If you orbit prograde the fact that spacetime is been dragged in the same direction allows you to have some push by the black hole and your orbit can be mantained even much closer to the event horizon in a stable manner. On the contrary if you orbit the black hole in the opposite direction you are fighting against the drag of spacetime and thus you will decay more easily, making the ISCO for retrograde orbits a lot higher than the ISCO for prograde orbits.

If you apply this reasoning to photons you can start to notice something interesting. Light coming from far away stars behind the black hole as viewed from an outside observed gets bent in different ways if it is coming from one side or the other due to this frame-dragging effect. If the light ray is coming parallel over the rotating surface of the black hole it is going to be helped by the black hole itself, and some of the angular momentum is going to be transfered to that light ray from the Kerr black hole. If instead the light ray comes anti-parallel to the rotation (which is going to happen in the other side of the black hole), then it might never reach the observer. This reasoning can be applied to the "dark rays" (which do not exist but is a way to trace the shadow which is the absence of light rays) coming from the event horizon and thus the shadow of the black hole is no longer a perfect black disk but an asymetric D-shaped black region, that tells you if the black hole is rotating clockwise or counter-clockwise.

In this animation you can see the appearance of the shadow of the black hole when we increase its rotational speed, as you can see it goes of-center and asymetric as we increase it.

Since frame-dragging goes as the mass rotates you can't notice it so strongly if you watch the black hole from another inclination. In fact the projected rotational speed if you watch a black hole from the poles ($i = 0^\circ$) is zero, and thus the black hole would look just like a non-rotating one. Here you can see the dependance of the shape of the shadow (in red) of a Kerr black hole with a fixed rotational speed when you see it from different inclinations (from the equator to the poles). The event horizon is represented in blue (but remember, you don't see that, you only see the shadow)

As you can see the effect is the same in both cases (changing view angle for a fixed rotational speed vs. changing rotational speed for a fixed inclination), which means that you can't tell the real rotational speed of a black hole just by measuring the shape of the shadow (a disk like shadow could mean a non-rotating black hole or a Kerr black hole as viewed from above for example), but at least it gives you a minumim estimate for the rotation. To study the exact rotational speed we need some independent measurements like for example the inclination of a disk of material around it. In those cases you would have complete information about the angular momentum of the black hole.

Finally here you have a beautifull simulation on what you would see from orbit around a Kerr black hole (I don't know how to embed YouTube videos so...).


Is this picture of a digital camera being oversaturated by bright light source i.e. the Sun?

It's an artifact from the digital sensor and/or the image processing in some cameras.

There have been a number of digital cameras with this issue. E.g. here is a Sony camera where the sun turned black with a specific setting for the picture profile. (In the example video, it's the S-log picture profile at 0:49.)

Sony called it a bug, and released a fix for it.

This artifact only affects a relatively small number of cameras, and it makes sense to think of it as a bug in the camera image processing: Yes, the sensor is oversaturated, but oversaturation is better represented as all white rather than all black.

If you have an affected camera, it's not just the sun that turns black, it's all oversaturated highlights, like this example with car headlights.


You're experiencing a few issues all of which have one solution. Take the Scope outside. Let it acclimate to the ambient air tempurature, and then try looking.

The 1st issue is the mirror needs to be the same temp as the air you're observing in. Since you are indoors you have a very warm optical path this causes distortion and trouble focusing.

2: Because you're looking through a piece of window glass (or more if it's a multipane window) you're introducing distortions that way. It's no wonder you can't reach focus.

Light pollution won't really affect Mars or the other planets too much, so that's not your worry.

Why are you observing from inside anyway?


What was the largest telescope ever fitted with an eyepiece

From this answer, it seems that the Harlan J Smith telescope located at the McDonald Observatory, in Texas, with a 107-inch (2.7 m) aperture is the largest aperture telescope that was ever fitted with an eyepiece.

Some independent evidence is there to indicate that this indeed had an eyepiece.


That looks like a lens effect. Polaris is the brightest object and there is an internal reflection in the lens or the camera.

It's a nice photo under difficult conditions. Can you see NGC 188, an odd open cluster?


What is the longitude of the winter solstice for the various planets?

You might find in-the-sky.org's Custom astronomical graph plotter by Dominic Ford helpful. For instance, for Jupiter, a start date of today, you can obtain the declination graph of the next 10 years for a given location (which you can change as well):

This does not solve your problem that you want to determine the longitude of maximal declination, but still.


Why aren't new stars in Earth's relative proximity constantly discovered?

To the naked eye, the answer is almost certainly no because of the enormously slow movement of stars across the sky and because 5,000 stars may be a lot but only a tiny percentage of the sky is covered by visible stars.

To Hubble, which can see perhaps tens of millions of stars, maybe more, the link here, has a picture of two stars that are approaching crossing each other's path from our point of view.  With a big enough telescope it probably happens from time to time, though I wouldn't want to try to calculate how often, but to the naked eye, I'm comfortable saying no, in fact, it was often assumed that stars didn't move and were fixed in the sky (contrary to what Macrobius said).  That was the popular point of view prior to Halley's observation.

There was also Tycho Brahe's "De Nova Stella" or "new star" which we now know to be a super-nova, and that was quite the surprise at the time.   Nobody thought a new star could appear because they thought the stars were fixed and permanent, but that appearance wasn't by the method you suggest.

Consider how small stars are from our point of view.    Alpha Centauri A, the larger one, it's about 1.7 million KM across and it's about 4.3 light years away, or, 41 trillion KM.   It's diameter is 23 million times smaller than it's distance from us.  That's the equivalent of looking at a golf ball from nearly 200 miles away.    Now if you scatter 5000 golf balls each 200 miles away across the sky and you let them move around very very slowly, how often do you think one golf ball passes in-front of another?   Not very often.   Granted, that's not quite right as the atmosphere spreads stars out a bit so each golf ball is smudged to maybe the size of a basketball, but they almost never pass infront of one another, at least, not if we only take into account 5,000 visible stars.

Now, binary stars, it happens more often if they are lined up right, then they can pass infront of each other and this has certainly been observed by telescope but not to the Naked Eye, we can't visibly tell that Alpha Centauri is 2 stars (3 with the more distant Proxima but that can't be seen by the eye).  They are on average about a billion miles apart but that can't be seen by the naked eye.   It was observed by telescope in 1689.

There simply aren't enough visible stars (and taking HDE's point that most of the 5,000 visible stars weren't cataloged until recently), there's essentially zero chance that it was ever observed that a star appeared "new" by passing from behind another star.

Using Hubble, it can happen, but not to human sight.


Expression for length of sunrise/sunset as function of latitude and day-of-year

The equations necessary for a correct answer are somewhat lengthy, for a web interface that accepts the location and prints a calendar try: https://sunrise-sunset.org/.

For a lengthy explanation of the mathematics see Wikipedia's Calculating the equation of time and Daytime webpages for information about the duration of rising and setting, along with information about the differences in solar noon and the three types of twilight.

This is a QBasic implementation by Keith Burnett for twilight times:

'********************************************************************
'   Sun rise/set /twilight
'   From Explanatory Supplement
'   definitions and functions
DEFDBL A-Z
pi = 4 * ATN(1)
tpi = 2 * pi
twopi = tpi
degs = 180 / pi
rads = pi / 180
'
'   the function below returns the true integer part,
'   even for negative numbers
'
DEF FNipart (x) = SGN(x) * INT(ABS(x))
'
'   FNday only works between 1901 to 2099 - see Meeus chapter 7
'
DEF FNday (y, m, d, h) = 367 * y - 7 * (y + (m + 9) \ 12) \ 4 + 275 * m \ 9 + d - 730531.5 + h / 24
'
'   define some arc cos and arc sin functions
'
DEF FNacos (x)
    s = SQR(1 - x * x)
    FNacos = ATN(s / x)
END DEF
DEF FNasin (x)
    c = SQR(1 - x * x)
    FNasin = ATN(x / c)
END DEF
'
'   the function below returns an angle in the range
'   0 to two pi
'
DEF FNrange (x)
    b = x / tpi
    a = tpi * (b - FNipart(b))
    IF a < 0 THEN a = tpi + a
    FNrange = a
END DEF
'
'       the function below returns the time in 24 hour
'   notation - hhmm - given a decimal hours figure.
DEF FNdegmin$ (d)
c = ABS(d)
a = INT(c)
b = 60 * (c - a)
d$ = STR$(INT(a * 100 + b + .5))
d$ = RIGHT$(d$, LEN(d$) - 1)
WHILE LEN(d$) < 4
    d$ = "0" + d$
WEND
FNdegmin$ = d$
END DEF
CLS
'
'    get the date and geographical position from user
'
INPUT "     lat : ", glat
INPUT "    long : ", glong
INPUT "    zone : ", zone
INPUT "rise/set : ", riset
INPUT " Sun alt : ", altitude
INPUT "   year  : ", y
INPUT "   month : ", m
INPUT "   day   : ", d
day = FNday(y, m, d, 0)
PRINT "  day no : "; day
utold = pi
utnew = 0
sinalt = SIN(altitude * rads) 'go for the sunrise/sunset altitude first
sinphi = SIN(glat * rads)
cosphi = COS(glat * rads)
glong = glong * rads
DO WHILE ABS(utold - utnew) > .001
  utold = utnew
  days = day + utold / tpi
  t = days / 36525
  '     get arguments of Sun's orbit
  L = FNrange(4.8949504201433# + 628.331969753199# * t)
  G = FNrange(6.2400408# + 628.3019501# * t)
  ec = .033423 * SIN(G) + .00034907# * SIN(2 * G)
  lambda = L + ec
  E = -1 * ec + .0430398# * SIN(2 * lambda) - .00092502# * SIN(4 * lambda)
  obl = .409093 - .0002269# * t
  delta = FNasin(SIN(obl) * SIN(lambda))
  GHA = utold - pi + E
  cosc = (sinalt - sinphi * SIN(delta)) / (cosphi * COS(delta))
  SELECT CASE cosc
  CASE cosc > 1
    correction = 0
  CASE cosc < -1
    correction = pi
  CASE ELSE
    correction = FNacos(cosc)
  END SELECT
  utnew = FNrange(utold - (GHA + glong + riset * correction))
LOOP
PRINT "    UT   : "; FNdegmin$(utnew * degs / 15)
PRINT "   zone  : "; FNdegmin$(utnew * degs / 15 + zone)
END
'******************************************************************

This is an implementation by Barry Carter of rising/setting duration (see his GitHub for associated files):

// http://astronomy.stackexchange.com/questions/12824/how-long-does-a-sunrise-or-sunset-take

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "SpiceUsr.h"
#include "SpiceZfc.h"
#include "/home/barrycarter/BCGIT/ASTRO/bclib.h"

// since the sun travels no more than 362 degrees per day, and has an
// angular diameter of no less than 31 minutes, the fastest possible
// sunrise is 31 minutes/362 degrees or 2 minutes; thus, 120 below in
// bc_between below is valid

int main(int argc, char **argv) {

  SpiceDouble radii[3];
  SpiceInt n;
  char direction[10];

  furnsh_c("/home/barrycarter/BCGIT/ASTRO/standard.tm");

  // the year 2015
  double stime = 1419984000, etime = 1451692800;

  // test
  // double stime =  1388559600, etime = 1419984000;

  // lat/lon from argv
  double lat = atof(argv[1]), lon = atof(argv[2]);

  // the sun's radii
  bodvrd_c("SUN", "RADII", 3, &n, radii);

  double *results = bc_between(9, lat*rpd_c(), lon*rpd_c(), 0., stime, etime,
                   "Sun", -34/60.*rpd_c(), 120., radii[0]);

  // NOTE: can NOT ignore result 0 and 1, but must check if they are borders
  for (int i=0; i<1000; i++) {

    double rstart = results[2*i], rend = results[2*i+1];

    // if we start seeing 0s, we are out of true answers
    // TODO: this may break near 1970
    if (abs(rstart) < .001) {break;}

    // if the end result is too close to etime, result is inaccurate
    // same if start result is too close to stime
    if (abs(rend-etime)<1 || abs(rstart-stime)<1) {continue;}

    // TODO: this really inefficient (and not even always correct?)
    if (bc_sky_elev(6, lat, lon, 0., rstart, "Sun", 0.) >
    bc_sky_elev(6, lat, lon, 0., rend, "Sun", 0.)) {
      strcpy(direction,"SET");
    } else {
      strcpy(direction,"RISE");
    }

    // the "day" and length of sunrise/sunset
    printf("%f %f %s %f %f %f\n", lat, lon, direction, rstart, rend, rend-rstart);
  }
  return 0;
}

For an explanation see Wikipedia's Sunrise Equation webpage.

This is the algorithm for rising/setting times:

Sunrise/Sunset Algorithm

Source:
    Almanac for Computers, 1990
    published by Nautical Almanac Office
    United States Naval Observatory
    Washington, DC 20392

Inputs:
    day, month, year:      date of sunrise/sunset
    latitude, longitude:   location for sunrise/sunset
    zenith:                Sun's zenith for sunrise/sunset
      offical      = 90 degrees 50'
      civil        = 96 degrees
      nautical     = 102 degrees
      astronomical = 108 degrees

    NOTE: longitude is positive for East and negative for West
        NOTE: the algorithm assumes the use of a calculator with the
        trig functions in "degree" (rather than "radian") mode. Most
        programming languages assume radian arguments, requiring back
        and forth convertions. The factor is 180/pi. So, for instance,
        the equation RA = atan(0.91764 * tan(L)) would be coded as RA
        = (180/pi)*atan(0.91764 * tan((pi/180)*L)) to give a degree
        answer with a degree input for L.

1. first calculate the day of the year

    N1 = floor(275 * month / 9)
    N2 = floor((month + 9) / 12)
    N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))
    N = N1 - (N2 * N3) + day - 30

2. convert the longitude to hour value and calculate an approximate time

    lngHour = longitude / 15

    if rising time is desired:
      t = N + ((6 - lngHour) / 24)
    if setting time is desired:
      t = N + ((18 - lngHour) / 24)

3. calculate the Sun's mean anomaly

    M = (0.9856 * t) - 3.289

4. calculate the Sun's true longitude

    L = M + (1.916 * sin(M)) + (0.020 * sin(2 * M)) + 282.634
    NOTE: L potentially needs to be adjusted into the range [0,360) by adding/subtracting 360

5a. calculate the Sun's right ascension

    RA = atan(0.91764 * tan(L))
    NOTE: RA potentially needs to be adjusted into the range [0,360) by adding/subtracting 360

5b. right ascension value needs to be in the same quadrant as L

    Lquadrant  = (floor( L/90)) * 90
    RAquadrant = (floor(RA/90)) * 90
    RA = RA + (Lquadrant - RAquadrant)

5c. right ascension value needs to be converted into hours

    RA = RA / 15

6. calculate the Sun's declination

    sinDec = 0.39782 * sin(L)
    cosDec = cos(asin(sinDec))

7a. calculate the Sun's local hour angle

    cosH = (cos(zenith) - (sinDec * sin(latitude))) / (cosDec * cos(latitude))

    if (cosH >  1)
      the sun never rises on this location (on the specified date)
    if (cosH < -1)
      the sun never sets on this location (on the specified date)

7b. finish calculating H and convert into hours

    if if rising time is desired:
      H = 360 - acos(cosH)
    if setting time is desired:
      H = acos(cosH)

    H = H / 15

8. calculate local mean time of rising/setting

    T = H + RA - (0.06571 * t) - 6.622

9. adjust back to UTC

    UT = T - lngHour
    NOTE: UT potentially needs to be adjusted into the range [0,24) by adding/subtracting 24

10. convert UT value to local time zone of latitude/longitude

    localT = UT + localOffset

Here is an implementation of rising/setting times in Basic:

10 '         Sunrise-Sunset
20 GOSUB 300
30 INPUT "Lat, Long (deg)";B5,L5
40 INPUT "Time zone (hrs)";H
50 L5=L5/360: Z0=H/24
60 GOSUB 1170: T=(J-2451545)+F
70 TT=T/36525+1: ' TT = centuries
80 '               from 1900.0
90 GOSUB 410: T=T+Z0
100 '
110 '       Get Sun's Position
120 GOSUB 910: A(1)=A5: D(1)=D5
130 T=T+1
140 GOSUB 910: A(2)=A5: D(2)=D5
150 IF A(2)<A(1) THEN A(2)=A(2)+P2
160 Z1=DR*90.833: ' Zenith dist.
170 S=SIN(B5*DR): C=COS(B5*DR)
180 Z=COS(Z1): M8=0: W8=0: PRINT
190 A0=A(1): D0=D(1)
200 DA=A(2)-A(1): DD=D(2)-D(1)
210 FOR C0=0 TO 23
220 P=(C0+1)/24
230 A2=A(1)+P*DA: D2=D(1)+P*DD
240 GOSUB 490
250 A0=A2: D0=D2: V0=V2
260 NEXT
270 GOSUB 820: '  Special msg?
280 END
290 '
300 '        Constants
310 DIM A(2),D(2)
320 P1=3.14159265: P2=2*P1
330 DR=P1/180: K1=15*DR*1.0027379
340 S$="Sunset at  "
350 R$="Sunrise at "
360 M1$="No sunrise this date"
370 M2$="No sunset this date"
380 M3$="Sun down all day"
390 M4$="Sun up all day"
400 RETURN
410 '     LST at 0h zone time
420 T0=T/36525
430 S=24110.5+8640184.813*T0
440 S=S+86636.6*Z0+86400*L5
450 S=S/86400: S=S-INT(S)
460 T0=S*360*DR
470 RETURN
480 '
490 '  Test an hour for an event
500 L0=T0+C0*K1: L2=L0+K1
510 H0=L0-A0: H2=L2-A2
520 H1=(H2+H0)/2: '  Hour angle,
530 D1=(D2+D0)/2: '  declination,
540 '                at half hour
550 IF C0>0 THEN 570
560 V0=S*SIN(D0)+C*COS(D0)*COS(H0)-Z
570 V2=S*SIN(D2)+C*COS(D2)*COS(H2)-Z
580 IF SGN(V0)=SGN(V2) THEN 800
590 V1=S*SIN(D1)+C*COS(D1)*COS(H1)-Z
600 A=2*V2-4*V1+2*V0: B=4*V1-3*V0-V2
610 D=B*B-4*A*V0: IF D<0 THEN 800
620 D=SQR(D)
630 IF V0<0 AND V2>0 THEN PRINT R$;
640 IF V0<0 AND V2>0 THEN M8=1
650 IF V0>0 AND V2<0 THEN PRINT S$;
660 IF V0>0 AND V2<0 THEN W8=1
670 E=(-B+D)/(2*A)
680 IF E>1 OR E<0 THEN E=(-B-D)/(2*A)
690 T3=C0+E+1/120: ' Round off
700 H3=INT(T3): M3=INT((T3-H3)*60)
710 PRINT USING "##:##";H3;M3;
720 H7=H0+E*(H2-H0)
730 N7=-COS(D1)*SIN(H7)
740 D7=C*SIN(D1)-S*COS(D1)*COS(H7)
750 AZ=ATN(N7/D7)/DR
760 IF D7<0 THEN AZ=AZ+180
770 IF AZ<0 THEN AZ=AZ+360
780 IF AZ>360 THEN AZ=AZ-360
790 PRINT USING ",  azimuth ###.#";AZ
800 RETURN
810 '
820 '   Special-message routine
830 IF M8=0 AND W8=0 THEN 870
840 IF M8=0 THEN PRINT M1$
850 IF W8=0 THEN PRINT M2$
860 GOTO 890
870 IF V2<0 THEN PRINT M3$
880 IF V2>0 THEN PRINT M4$
890 RETURN
900 '
910 '   Fundamental arguments
920 '     (Van Flandern &
930 '     Pulkkinen, 1979)
940 L=.779072+.00273790931*T
950 G=.993126+.0027377785*T
960 L=L-INT(L): G=G-INT(G)
970 L=L*P2: G=G*P2
980 V=.39785*SIN(L)
990 V=V-.01000*SIN(L-G)
1000 V=V+.00333*SIN(L+G)
1010 V=V-.00021*TT*SIN(L)
1020 U=1-.03349*COS(G)
1030 U=U-.00014*COS(2*L)
1040 U=U+.00008*COS(L)
1050 W=-.00010-.04129*SIN(2*L)
1060 W=W+.03211*SIN(G)
1070 W=W+.00104*SIN(2*L-G)
1080 W=W-.00035*SIN(2*L+G)
1090 W=W-.00008*TT*SIN(G)
1100 '
1110 '    Compute Sun's RA and Dec
1120 S=W/SQR(U-V*V)
1130 A5=L+ATN(S/SQR(1-S*S))
1140 S=V/SQR(U):D5=ATN(S/SQR(1-S*S))
1150 R5=1.00021*SQR(U)
1160 RETURN
1165 '
1170 '     Calendar --> JD
1180 INPUT "Year, Month, Day";Y,M,D
1190 G=1: IF Y<1583 THEN G=0
1200 D1=INT(D): F=D-D1-.5
1210 J=-INT(7*(INT((M+9)/12)+Y)/4)
1220 IF G=0 THEN 1260
1230 S=SGN(M-9): A=ABS(M-9)
1240 J3=INT(Y+S*INT(A/7))
1250 J3=-INT((INT(J3/100)+1)*3/4)
1260 J=J+INT(275*M/9)+D1+G*J3
1270 J=J+1721027+2*G+367*Y
1280 IF F>=0 THEN 1300
1290 F=F+1: J=J-1
1300 RETURN
1310 '
1320 '   This program by Roger W. Sinnott calculates the times of sunrise
1330 '   and sunset on any date, accurate to the minute within several
1340 '   centuries of the present.  It correctly describes what happens in the
1350 '   arctic and antarctic regions, where the Sun may not rise or set on
1360 '   a given date.  Enter north latitudes positive, west longitudes
1370 '   negative.  For the time zone, enter the number of hours west of
1380 '   Greenwich (e.g., 5 for EST, 4 for EDT).  The calculation is
1390 '   discussed in Sky & Telescope for August 1994, page 84.


How is Cosmic Microwave Background's temperature measured?

That is what is done.  This is shown in an old xkcd comic https://xkcd.com/54/

The curve shows the distribution of frequencies in the CMB, and by using the marked value of the maximum you can determine the value of T, the apparent (red-shifted) temperature of the CMB


Approximate conversion of redshift 'z' to a time and/or distance, when reading papers?

Use one of the "cosmology calculators". The conversions depend on what you assume for the cosmological parameters.

Here's one that will do what you want. http://home.fnal.gov/~gnedin/cc/

e.g. For the default parameters, $z=7$ corresponds to a look back time of 13.01 billion years, whilst $z=10.5$ corresponds to a look back time of 13.33 Gyr, with the current age of the universe being 13.78 Gyr. Choosing the "luminosity distance" option, we that the object has a luminosity distance of 112 Gpc.

If you want the proper distance you could use http://www.astro.ucla.edu/~wright/CosmoCalc.html
According to this calculator the $z=10.5$ object is currently at a proper distance of 9.76 Gpc (31.8 billion light years).


What is the formula to predict lunar and solar eclipses accurately?

The NASA sites have some very useful resources for this I will list them below:

Lunar Eclipses

This Link has an index for all lunar eclipses from -1999 to +3000, predominantly a statistics page but also has this page that contains how to calculate when lunar eclipses are.

There is more than one formula depending on which time frame you are trying to look in.

This is the formula for eclipses between the year 2005 and 2050:

  $$\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2$$

  Where:   $$y = year + (month - 0.5)/12$$  $$t = y - 2000$$

Solar Eclipses

This Link has an index like above but for all of the solar eclipses from -1999 to +3000.

This link has the formula for calculating solar eclipses. This is the formula for between 2005 and 2050:

  $$\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2$$

  Where:  $$y = year + (month - 0.5)/12$$  $$t = y - 2000$$


Knowing the RA/DEC of a star how do I locate a star from ground?

Calculating local sidereal time exactly is not trivial as shown by the calculation for local sidereal time given at Approximate Sidereal Time:

  The Greenwich mean sidereal time in hours is: $$GMST = 6.697374558 + 0.06570982441908 D_0 + 1.00273790935 H + 0.000026 T^2$$
  where $T = D/36525$ is the number of centuries since the year 2000.

Fortunately thereare a number of online calculators that will do that for you.

You can, however ballpark it knowing that at the equinox in March (~21st, depending on the year, because you also need to take leap years into account), local sidereal time is 12 hours different than local time (without daylight saving). i.e. 6 pm (18:00) is 06:00 LST. This means that RA 06h00m is overhead at 18:00 local time.

Then add 1 hour for each $\frac{1}{2}$ month or 2 hours for each month since then. So, mid-April is about 1 month after the March equinox which means that RA 08h00m is overhead at 18:00 local time, and mid-May will be 2 months so RA 10h00m is overhead at 18:00.

From there, it is just like a clock - midnight is 6 hours later than 18:00 so you add 6 hours to the local sidereal time calculated earlier meaning that in mid-April RA 14h00m is overhead at midnight.

What this tells us is that around the March equinox (~21st), Sirius (RA 06h45m) is overhead at about 18:45, which we can round off to around 19:00 (7 pm) given the roughness of the calculation so far. We can guess then that it will set then around 6 hours later, at about 1 am.

In mid-April, RA 21h00m will be overhead at around 19:00 (7 pm), which means Sirius will have been overhead about 2 hours earlier - around 17:00 (5 pm), which means it will set around 11 pm, and in mid-May will set around 9 pm.

Melbourne is located at ~145° E, ~38° S. From this you can work out the relative angle from the declination (Sirius dec. is -16°43'). Melbourne is at 38° south, so declination -38 is directly overhead. Sirius is at dec. -16°43' (so about -17°) so it will be about 20° north of directly overhead.

So, you would face towards north and look upwards almost to directly overhead. Knowing the LST, you know the RA that is directly overhead and can count towards the west (or east, depending on the month), and knowing that in mid-April Sirius was overhead at around 5 pm and will set around 11 pm, you can find it.

NB. James K's answer is really worth reading.


Why did supernova 1987A's "String of 'Cosmic Pearls'" appear to be so lumpy 20 years later? Is it still?

The "lumpiness" is (mostly) inherent to the ring, and is probably the result of instabilities in the outflowing gas that gave rise to it in the first place. E.g., Akashi et al. (2015):

We suggest that clumpy dense outflowing equatorial rings around
evolved giant stars, such as in supernova 1987A and the Necklace
planetary nebula, are formed by bipolar jets that compress gas towards
the equatorial plane. The jets are launched from an accretion disc
around a stellar companion.... we perform 3D numerical simulations,
and show that bipolar jets expanding into a dense spherical shell can
compress gas towards the equatorial plane and lead to the formation of
an expanding equatorial ring. Rayleigh-Taylor instabilities in the
interaction region break the ring to clumps.

What you're seeing in the image is emission due to heating from the impact of the expanding supernova shock wave. Some of the clumpiness may be exaggerated by the fact that denser clumps of gas can radiatively cool more efficiently than diffuse gas, as explained by Fransson et al. (2015):

If the density of a clump is sufficiently high, radiative cooling will
cause the clump to collapse to an even higher density. The result will
be an optically emitting hotspot (Pun et al. 2002). If instead, the
density of a clump falls below the threshold density for radiative
cooling, the gas behind the transmitted shock will radiate X-rays and
very little optical radiation. This threshold density is a sensitive
function of the velocity, ${{V}_{{\rm clump}}}$, of the shock entering
the clump. Thus, at any given time, the requirement for radiative
cooling acts as a high contrast filter, limiting optical radiation to
only the denser clumps.

The brightness of the ring has faded somewhat since that image was made (more on the east side than on the west), as shown in this composite of HST images (Figure 3 of Larsson et al. 2019):

The figure in the paper itself includes an animation of the evolution.


Would Adaptive Optics be Useful in Radio Astronomy?

In fact, the techniques of adaptive optics are already being used in radio astronomy.  They are implicit in the basic imaging algorithms (e.g., CLEAN) used to produce maps from radio interferometers.  In those cases, they are usually being used to correct for the artificial structure introduced by the way the interferometer samples the sky, rather than for structure imposed by the intervening material.  But at low frequencies (1 GHz and below, certainly) they are also used to correct for the artificial structure imposed on the incoming radio wavefronts as they pass through the ionosphere.  Current large low-frequency instruments (such as the LWA and LOFAR) rely heavily on these methods.


Earth is already pretty bright due to cloud cover, with a typical albedo of .3-.35 -- that is, it reflects about a third of the visible light that hits it.

That means it couldn't get more than about three times as bright even if it were perfectly reflective (albedo 1.0), which means about 1.2 magnitudes brighter.

Spotting Earth from a great distance with the naked eye is less an issue of brightness, more an issue of separation from the Sun. I can't run numbers at the moment to get actual magnitudes or angular separations.