Why do certain massive stars leave no remnants? Mass and metallicity are the two main determinants for a star's fate. This is simple enough. What's more complicated is how exactly these determine the star's fate. For example, you can see in this image I got from Wikipedia's supernova article:

Generally, as metallicity decreases and mass increases, black holes become more likely to form. At a specific mass and metallicity range, however, there are no remnants left behind. From that point, increasing mass and decreasing metallicity allows for black holes again.
How does this work? Why would there ever be an area in which no remnants are left behind? In fact, it seems almost random, since the range lies right between direct black hole ranges.

The gap appears because of pair instability supernovae. In short, as one looks at such massive stellar cores at increasing temperatures, an ever-larger fraction of the photons are sufficiently energetic to spontaneously form electron-positron pairs. True, they soon recombine, but there is nevertheless a loss in (radiation) pressure, which causes contraction, and thus a further increase of temperature. If unstable, as appears to be the case in such massive stars, this leads to more high-energy photons, more pairs, and a further loss of pressure.

Models of such objects suggest that the collapse leads to a sudden, rapid ignition of oxygen- and silicon-burning reactions. In a mass range between about 150 and 250 solar masses (corresponding to core masses of very roughly half of that), the thermonuclear explosion is enough to rip the core apart, and there's nothing left to collapse. This is a bit like how the sudden onset of carbon burning destroys white dwarfs in type Ia supernovae. More massive cores are more strongly bound by gravity, and models suggest that the nuclear explosion isn't enough to unbind the core.

So below about 150 solar masses (and at higher metallicities, I think), the core doesn't reach pair-unstable conditions. Above about 250 solar masses, the induced nuclear explosion isn't powerful enough to destroy the core.


Do hypervelocity stars have Oort clouds? If you had a hypervelocity star travelling through space incredibly quickly, would the hypervelocity star's Oort cloud remain intact? Would the high speed of a hypervelocity star shed the comets and dwarf planets in its Oort cloud due to the star's high speeds, or would the Oort cloud get stretched into a teardrop shape? Could the Oort cloud remain unaffected?

There is no problem with orbiting a star that moves very fast. Remember that space is mostly empty, so there is nothing like air resistance or friction that would tend to deform the orbit.

Stars that become hypervelocity stars likely tend to lose their Oort clouds. They gain their velocity by passing very close to a massive body and getting a velocity increment, for example by having a secondary star or planet ripped away by tidal forces. That already doesn't bode well for the stability of any Oort cloud.

Simulations show that planets tend to be ejected from hypervelocity systems unless they are very tightly bound, since they have lower mass and respond more strongly to the velocity changes. This suggests that the lightly bound and small bodies in the Oort cloud will be scattered - likely at even higher velocities.


What is Gravitational Lensing? What is the effect known as gravitational lensing? How does it work? What objects would be able to cause this effect?

Gravitational lensing is the effect that large amounts of gravity have on the path of light. Here's an example of the effect:

The animation shows a black hole passing in front of a galaxy (simulated). In theory any object with a surface mass density larger than the critical surface mass density would be able to produce similar such effects.

What's happening is that light goes in a straight line along the spacetime fabric. When a very massive object is in the fabric, it bends the fabric - since the light follows a straight line on the fabric, the line is also bent. This means that the light goes in a not-straight line, and thus a lensing effect.


What kind of reddish wood is it exactly that was historically used to line the inside of observatory cupolas (domes), in order to absorb moisture? I was told that it was "cedar", but that is an unclear term botanically.

Probably western red cedar, thuja plicata, especially if the observatory is in North America. Western red cedar is easily obtainable in North America. It has a light, red, wood that is particularly resistant to decay, and ideal for use in exposed positions.

  The soft red-brown timber has a tight, straight grain and few knots. It is valued for its distinct appearance, aroma, and its high natural resistance to decay, being extensively used for outdoor construction in the form of posts, decking, shingles, and siding.[28] It is commonly used for the framing and longwood in lightweight sail boats and kayaks. In larger boats it is often used in sandwich construction between two layers of epoxy resin and/or fibreglass or similar products. Due to its light weight—390 to 400 kg/m3 (24 to 25 lb/cu ft) dried—it is about 30% lighter than common boat building woods, such as mahogany. For its weight it is quite strong but can be brittle.

Wikipedia: Thuja plicata

It may be worth pointing out that while "cedar" refers to a group of Cedar woods, trees of the proper Cedrus genus are rare and localized:

  Cedrus (common English name cedar) is a genus of coniferous trees in the plant family Pinaceae. They are native to the mountains of the western Himalayas and the Mediterranean region, occurring at altitudes of 1,500–3,200 m in the Himalayas and 1,000–2,200 m in the Mediterranean.

Wikipedia: Cedrus


If wormholes exist and behave like we think they do, can we cross them without "negatively charged matter "? I heard and read that a wormhole will collapse if they transport any positively charged matter. So, if wormholes exist, they probably close as soon as normal matter enters it, how can we access them without negative matter?

The best current answer is "We can't".  We do not presently know of any way to stabilize wormholes without large negative energy densities. (Some argue that even that isn't enough, but I don't believe that the matter is settled, even on a theoretical level.)

Note that:

(a) No wormhole has ever been observed, nor have we observed anything which is a predicted effect of a wormhole: Wormholes are entirely theoretical.

(b) Having said that, Wormholes are a prediction of General Relativity, which is a very well-tested and well-established theory, so it's unwise to bet against them existing.

(c) Macroscopic wormholes are a purely GR phenomenon and while the fields involved are strong, they are not so strong that they are in the realm where we expect the eventual unified theory of quantum gravity to give markedly different results.

(d) And by the way, we haven't a clue how to create large negative energy densities.


What kind of telescope do I need to see most of the Jupiter's moons? I have a simple Newtonian reflector telescope. Using it, I am able to see the Galilean moons of Jupiter. However, Jupiter has much more moons than that (Wikipedia says 67 have been discovered this far). What kind of telescope I would need to see most or all of them?

Unless you're really rich, unfortunately you won't be able to see all of them.

Jupiter's fifth largest Moon, Amalthea, has an apparent magnitude of $m$ = 14.1. Comparing this to the magnitude of Europa, the dimmest of the Galilean moons, which is 5.3, tells us that Amalthea is roughly 3000 times less bright. Your telescope thus needs to have an area 3000 (or radius ~55) times larger for Amalthea to have the same apparent brightness.

In general, without a camera on your telescope, the dimmest object you can see depends on your vision, but on average, humans are able to see objects of magnitude 6. That means that in order to detect an object of magnitude $m_\mathrm{obj}$, you need a light-collecting area which is larger than you pupil by a factor of
$$f = 10^{(m_\mathrm{obj}-6)/2.5}.$$
Thus, in order to be able to just barely detect Amalthea, you need a telescope which is larger than you pupil (radius = 6 mm) by a factor of ~1738, i.e. has a radius of 25 cm.

It quickly becomes really difficult to see them; for instance, Elara, which is the eighth largest moon, has an apparent magnitude of 16.3 and thus requires a 70 cm telescope. And remember, this is just on the threshold of what you can see.

Of course, if you mount a camera to your telescope and take images with long exposure times, you can get away with smaller telescopes. In fact all moons smaller than Amalthea were discovered this way, some of them by cameras on the Voyager space probes.


How to find out if a galaxy is in a specific cluster I am trying to find out how to find out if a certain galaxy is located inside a cluster. For the famous ones this is quite easy (sometimes even on wikipedia!), but I haven't been able to do it for some less-known galaxies. I have tried to look them up on Simbad, but I couldn't find it written down anywhere (even for the ones I know to be in clusters). Also generic "googling" has not helped me. Do you have any suggestions?

You want to use NED (NASA Extragalactic Database) rather than SIMBAD, since the former has more galaxy- and cluster-specific information.

I'm going to assume you have a reasonably accurate position for your galaxy as well as its redshift (in km/s). The basic idea is to identify candidate galaxy clusters which are close enough to your galaxy on the sky and in redshift so that the galaxy could potentially be a member of one of them.

Determine what angular radius $R$ corresponds to 1 or 2 megaparsecs at the redshift of your galaxy (e.g., using Ned Wright's Cosmology Calculator)
Do a search in NED for Galaxy Clusters within R of the coordinates of your galaxy. If your galaxy has a name, you can skip the coordinate lookup and use the "Near name" search form; if you only have coordinates, there's a "Near position" search form.

A. Put the estimated angular radius $R$ (converted to arc minutes) in the "Search Radius" box;

B. Set the "Selection in Redshift" form to "Between" and give it lower and upper limits of $V_{\rm gal} - 2000$ and $V_{\rm gal} + 2000$ in km/s, where $V_{\rm gal}$ is the redshift (in km/s) of your galaxy;

C. Under "Selection by Object Type", select "Galaxy Clusters" in the "Classified Extragalactic Objects" list.

That should give you a list of galaxy clusters within the specified radius of your galaxy which are at a similar redshift. Note that the same cluster may come up multiple times under different names (i.e., from different cluster catalogs), but it should be clear from the coordinates and the redshifts which are the same clusters.

Now the question is whether your galaxy is plausibly a member of a given candidate cluster or not. Ideally you should find some measurement of each cluster's velocity dispersion $\sigma$; if the galaxy's redshift isn't within $\pm 3 \sigma$ of the cluster's redshift $V_{\rm cluster}$, then you can ignore that cluster, since it's very unlikely that a galaxy with that relative velocity with respect to the cluster is actually gravitationally bound to the cluster.

You can't really do better than a probabilistic assessment: roughly how likely it is that the galaxy is a member of a given cluster. Galaxies which are members of a cluster are more likely to be close to the center (on the sky); member galaxies closer to the center can have a higher range of relative velocities, while member galaxies further away should have velocities closer to the cluster redshift.

A quick example: Is NGC 4889 ($V_{\rm gal} = 6495$ km/s) a member of a cluster? At its redshift, 1 Mpc = 33 arc minutes; if we search on NED for galaxy clusters within 33 arcmin of NGC 4889 with redshift limits of 4500-8500 km/s, we get 6 results, 4 of which are just the Coma Cluster in different catalogs, with $V_{\rm cluster} \sim 6900$ km/s. Since the various distances ("Separ. arcmin") translate to $\sim$ 100-300 kpc from the cluster center, and since the Coma cluster has a velocity dispersion of $\sim$ 1000 km/s and the difference between NGC 4889 and the cluster is only $\sim$ 400 km/s, it's a very good bet that this galaxy is a member of the Coma Cluster. (It's actually one of the two cD galaxies in the cluster center.)


Astronomical telescope making I am thinking about making a telescope. I have a 100cm focal length lens which I can use as objective lens in my telescope. So which is the best focal length I can use as eyepiece in my astronomical telescope?

The first question anyone asks about a telescope is "what is the magnification?" It is almost always not the most important thing. Any telescope can magnify a million times, given a short enough eyepiece - the problem is, how good the image is.

For observing planets, the main thing is resolving power - the ability of the telescope to discern tiny details. The resolving power is limited by aperture, or the diameter of the primary lens or mirror. If the primary lens diameter is measured in mm, and the resolving power in arcseconds, then the limit is:

resolving power = 100 / aperture

Your 50 mm diameter lens would have a resolving power limited to 2 arcseconds in ideal conditions. Any details smaller than that would be blurry, no matter how much magnification you put into that thing. For comparison, Jupiter's apparent diameter is between 30 and 50 arcsec, Mars' between 3.5 and 25 arcsec, Saturn's between 14 and 20 (without the rings).

With that aperture you would see the rings of Saturn, and a couple equatorial belts on Jupiter. Venus shows a crescent when it's far from the Sun. Mars is a little orange disk when it's closest to Earth. I've looked at planets through 50 mm of aperture, and you can see quite a few things that way.

The Moon also looks interesting, you can see craters and mountains, plus it's available every month.

Because resolving power is limited by aperture, as you give it more and more magnification, at some point the image is just huge and bloated; big, but blurry. So there's a limit to the useful magnification - again, this depends on aperture. If the aperture is measured in mm, the formula is:

maximum magnification = 2 * aperture

Your 50 mm aperture would support up to 100x magnification before things get too blurry.

So how do you calculate magnification for a given instrument? It is the ratio between the focal length of the primary lens (or mirror) and the focal length of the eyepiece (the lens close to your eye):

M = F / f

For your 1000 mm focal length primary, you achieve 100x magnification with an eyepiece with 10 mm focal length. Anything shorter than that would just bloat the image uselessly. Eyepiece focal lengths that would make sense in your instrument would be between 10 mm and 120 mm.

Words of advice:

Make sure the lenses are perfectly aligned. When you build the instrument, have some way to adjust the direction of the primary lens, so as to align it with the main axis of the instrument. Some tweaking will be required - tilt the primary a fraction of mm, check the results, tilt again, repeat until the image looks best. This is called collimation and it's very important for the overall performance of the instrument.

Have some way for the eyepiece to move back and forth until you achieve perfect focus. This will have to be re-adjusted every time you observe. The adjustment is quite sensitive. A real focuser would help a lot, but this could be achieved also with two tubes sliding into each other, held in place by friction, like in an old pirate captain's telescope.

For astronomy, when the scope is focused at infinity, the distance between primary lens and eyepiece is the sum of their focal lengths. E.g. with a 1000 mm primary and a 10 mm eyepiece, the distance would be 1010 mm. This assumes a convergent eyepiece.

The telescope will not work well hand-held. The image will be jumping around too much. You will need some kind of mount. At the very least put the top of the tube on a fence or something. But astronomical instruments work best when they are on a rock solid mount. Even a broomstick hammered into the ground, with a loop on top for the tube, is better than nothing.

Whichever way you mount the primary lens, don't squeeze it too hard in the lens cell or tube. It will deform the lens and make the image bad. Some pressure on the edge is okay; lots and lots of pressure not okay. You could glue the edge of the lens on some kind of ring. Just use common sense.

This is totally doable. I've built a telescope just like that, many years ago. It's great to see what you can achieve on your own.

Good luck!


Current delay for radio signals from the Mars InSight mission I've been meaning to post it on the day when InSight landed but just got around to it now.

When they broadcasted landing in real-time, it really wasn't in real-time because radio signals from Mars can take from 3 to 24 minutes to reach the Earth (depending on the distance between Mars and Earth) if I am not mistaken.

So what was the delay on the date InSight landed? Each time someone mentioned "real-time" I cringed a little bit because everything already happened some time ago and we were just only learning about it now.

Current delay for radio signals from the Mars InSight mission.

The delay at the time of the landing was 8.07 minutes (8 minutes 4+ seconds), as Mars was at the time at a relative distance from Earth of 0.9703 Au.
Current delay as at 8th December 2018, we're at a distance of 8.85 (8 minutes 51 seconds) light minutes or 1.064 Au.

Using WolframAlpha's tool here, you should be able to satisfy your curiosity without effort.

Times taken are at midnight on those dates and vary minimally throughout the day (eg. the delay on the day following launch at midnight was approx four seconds longer).

This does not take into account the time delay between the signal being sent to the satellite and it being decoded by your box and displayed by your tv or the time it may have taken to be distributed over the net if you watched it that way. I'm not able to determine if any Broadcast-Delay was deliberately introduced, which is a common practice and commonly is about 7 seconds.


if the universe is infinite shouldn't it have experienced an infinite number of universe ending events? or is it that the definition of infinity used here excludes such events like there are an infinite number of odd numbers but 2 isn't in that set?

The universe has a finite age, about 13.7 billion years. So there has not been time enough for infinite events in our past light-cone.
There are rather few events in mainstream physics that are universe ending. Closed universes eventually implode, and if certain forms of dark energy exist they can cause a Big Rip where the universe expands to infinity in a finite time. But obviously observers cannot look back on them in their past. This is also true for other local disasters like false vacuum decay.


Unable to achieve focus I feel like this is going to be a duplicate of this question, Astrophotography - Unable to achieve focus with current setup.

However, the suggestions to go to a straight adapter doesn't apply in my case. I have a Canon Rebel XT, with a 2" 48mm adapter https://www.amazon.com/gp/product/B00T3P5ABC/ref=oh_aui_detailpage_o01_s00?ie=UTF8&psc=1, on my Orion XT10 dob, http://www.telescope.com/Telescopes/Dobsonian-Telescopes/IntelliScope-Dobsonians/Orion-SkyQuest-XT10i-IntelliScope-Dobsonian-Telescope/pc/1/c/12/sc/27/p/102013.uts.

When I attach the adapter to the camera and focus on a planet or star, I can't achieve focus. My focus adjuster on the scope can't get close enough. It's as if the camera is just out of range of the focuser.

I'm not sure what the issue is and would love to hear some suggestions. I'm just a noob getting started and have a lot to learn so I apologize at all the beginner questions here.

This is a common problem with Newtonian type telescopes and DSLRs, the sensor in the camera is too far back to reach focus. As mentioned in the comments a low profile focuser may help or move the main mirror up the tube, I've seen this done on some Skywatcher Newtonians to solve similar problems.


How is the Universe's Expansion Accelerating if the Hubble Constant is Decreasing? I am just getting into the field of cosmology and was wondering what it means practically to say that the Hubble Constant is decreasing while the expansion rate of the Universe is accelerating. I am not sure how this phenomenon is possible. How do we observe this?

It is my understanding that the Hubble Constant is the rate at which the Universe is expanding, and it is decreasing with time. (Although it is asymptotic)... Is this conception flawed?

The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points.
The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator.

In the future, the cosmological constant, $\Lambda$ could result in an exponential expansion with time. A simple piece of maths shows you that the Hubble parameter is a Hubble constant only for an exponential expansion.

For example, suppose $H(t) = (da(t)/dt)/a(t)$, where $a(t)$ is the distance between two arbitrary points in the universe at epoch $t$. Now let's have an expansion rate $da/dt \propto t$ (ie accelerating). But in this case $a(t) \propto t^2$ and $H(t) \propto t ^{-1}$ (ie decreasing with time).

Some extra details:

The solution to the Friedmann equation in a flat universe is
$$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$
where $\rho$ is the matter density (including dark matter) and $\Lambda$ is the cosmological constant.

As the universe expands, $\rho$ of course decreases as $a(t)^{-3}$, but $\Lambda$ remains constant. Thus the first term on the RHS becomes unimportant if indeed $\Lambda$ is a cosmological constant.

Thus the Hubble "constant" actually decreases from its current value $H_0$ and asymptotically tends towards $ H = \sqrt{\Lambda/3}$ as time tends towards infinity.


Surface height distribution for Jupiter's moon Io? The moon Io has around 100 mountains:

These structures average $6 {\rm\, km}$ in height and reach a maximum of ${\rm 17.5 \pm 1.5\, km}$

I am now looking for the overall histogram, i.e. elevation distribution of Io's surface and/or a hypsometric curve for Io.
References

JPL: Topography of Io

This blog by Butcher and Conwey on Geomorphology (or should we say io-morphology?) was a good starter. It seems usual to deal with the distribution of mountains and volcanos on Io. There are height maps (DEM) which give the distribution colour-coded. The closes to a hysometric curve found for Io is in this paper by White et al (2014) which gives a hypsometric curve where they exclude the volcanos in figure 9.
Noteworthy might also be this poster by Wiliams et al with geological map but also a height distribution given single measurement "points" (where the points are areas of various size).


'Geocentric' view of orbits Given our real solar system how would planet orbits plot look like if we fix earth in the middle (put earth in the middle and draw a real orbit of a given planet) ?

What would be the shape of its orbit? Is there any software that allows you to do such plot ? Does anybody have an example of it ?

I have tried searching for such image but most results are connected with historical theory .

@Ingolifs' answer is exactly right. I'll just add to it some more plotting using the Python package Skyfield.

The "inner" plotted bodies (from outer to inner) are Mars, (Earth), Moon, Venus, Mercury, The Sun. The data is for 2000 days.

The "outer" plotted bodies are Uranus, Saturn, Jupiter, Earth, and The Sun. The data is for 2000 months.

Distances are in AU.

Click the plot to see larger, or just run the Python script and make your own plot.

class Body(object):
    def __init__(self, name):
        self.name = name

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader, EarthSatellite

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

AU = 149597870.700  # km

load    = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data421    = load('de421.bsp')
data405    = load('de405.bsp')
data       = data405   # need the longer time range for the outer planets
ts      = load.timescale()

things  = ('sun', 'mercury', 'venus', 'earth', 'moon', 'mars',
           'jupiter barycenter', 'saturn barycenter', 'uranus barycenter', )

bodies = []
for thing in things:
    name = thing.split()[0]
    body = Body(name)
    bodies.append(body)
    body.obj = data[thing]

sun, mercury, venus, earth, moon, mars, jupiter , saturn, uranus = bodies

days_inner = np.arange(2000)
times_inner = ts.utc(2019, 1, days_inner)

months_outer = np.arange(2000)
times_outer = ts.utc(2019, months_outer, 1)

for body in bodies:
    body.posn_barycentric_inner = body.obj.at(times_inner).position.km
    body.posn_barycentric_outer = body.obj.at(times_outer).position.km
for body in bodies:
    body.posn_geocentric_inner = body.posn_barycentric_inner - earth.posn_barycentric_inner
    body.posn_geocentric_outer = body.posn_barycentric_outer - earth.posn_barycentric_outer
    body.posn_heliocentric_inner = body.posn_barycentric_inner - sun.posn_barycentric_inner
    body.posn_heliocentric_outer = body.posn_barycentric_outer - sun.posn_barycentric_outer

print (jupiter.posn_geocentric_inner.shape)

if True:
    plt.figure()
    lw, fs = 0.7, 12

    plt.subplot(2, 2, 1)
    plt.title('geocentric inner', fontsize=fs)
    for body in (sun, mercury, venus, earth, moon, mars):
        x, y, z = body.posn_geocentric_inner / AU
        plt.plot(x, y, linewidth=lw)

    plt.subplot(2, 2, 3)
    plt.title('heliocentric inner', fontsize=fs)
    for body in (sun, mercury, venus, earth, moon, mars):
        x, y, z = body.posn_heliocentric_inner / AU
        plt.plot(x, y, linewidth=lw)

    plt.subplot(2, 2, 2)
    plt.title('geocentric outer', fontsize=fs)
    for body in (earth, mars, saturn, uranus):
        x, y, z = body.posn_geocentric_outer / AU
        plt.plot(x, y, linewidth=lw)

    plt.subplot(2, 2, 4)
    plt.title('heliocentric outer', fontsize=fs)
    for body in (earth, mars, saturn, uranus):
        x, y, z = body.posn_heliocentric_outer / AU
        plt.plot(x, y, linewidth=lw)

    plt.show()


Number of galaxies per redshift Does exist something like a distribution of number of galaxies per redshift interval? I'm studying galaxy formation and evolution, and I'm curious to know how many todays' galaxies (with redshift <0.01 I would say) we see.

Yes, for a given galaxy type…
The term "a distribution of number of galaxies per redshift interval" does make sense, but usally we restrict ourselves to a certain type of galaxies, e.g. Lyman $\alpha$ emitters, Lyman break galaxies, sub-mm galaxies, etc. The reason is that galaxies are detected in a number of different ways, and no single selection technique captures all types of galaxies.
…and comoving volume…
Because the Universe is expanding with time, the "true" (physical) number density of galaxies — and everything else (except dark energy) — decreases with time. To be able to compare densities at different times, it is useful to measure them in comoving coordinates, which is the coordinates that expand along with the Universe. In these coordinates, densities only change if some physical process happen. For instance, the  physical energy density of photons decreases both because the volume of the Universe increases and because the photons redshift. In comoving coordinates it also decreases, but only due to the redshifting since the volume change is factored out.
…and luminosity.
Moreover, since galaxies have no well-defined lower size/mass/luminosity threshold, we have to define a threshold above which we count them.
For this reason, rather than talking about "the number density of galaxies per redshift", we usually use the very useful observable luminosity function (LF) which measures the number of (a certain type of) galaxies per luminosity bin, and per comoving volume, at a given redshift.
LFs have been probed out to very high redshifts, at the times when galaxies were just beginning to form at $\sim$half a gigayear after Big Bang (e.g. Schenker et al. 2013).
You end your question by asking about the number of galaxies at $z\le0.01$, which would be considered very low redshift in the context of galaxies. You can find the answer by considering all kinds of selection criteria, checking for overlaps between surveys (most local galaxies will be detected by multiple techniques), and integrating over volume the LFs down to your chosen threshold (e.g. Small Magellanic Cloud-sized). Doable, but not trivial. An alternative is to integrate numerically calculated dark matter halo mass functions, using e.g. this online tool, and assuming the every halo hosts a galaxy. I get a number density of $n\sim0.4$ halos per $h^{-3}\,\mathrm{Mpc}^{3}$, or $1.2\,\mathrm{Mpc}^{-1}$. The comoving volume out to $z=0.01$ is $V=3.3\times10^5$ Mpc, so the total number of (super-SMC-sized) galaxies is $N = n V \simeq 3.8\times10^5$.


Why the lows of the GRB 2014-04-27 in Andromeda? The Swift gamma ray burst space telescope has measured a GRB according to the link below. But why the oscillations in the measured gamma rate? Why wasn't just an increase detected, but also as big decreases? The average rate doesn't look to have increased during the burst. If averaged out over a minute the line would've been flat. Is there a fundamental astronomical explanation or has it to do with the technicalities of the instrument?
http://www.universetoday.com/112194/possible-gamma-ray-burst-detected-in-andromeda-would-be-closest-ever-observed/

It turned out to be not real GRB event, just a false alarm.

My understanding is that, since the source is actually a steady X-ray source, you have a high count rate level at the base:

I am not expert of GRBs, but this high-oscillation behavior seems a characteristics of GRBs: you can see few examples here.
In fact, the BAT time resolution is very high ($\sim100\mu s$), then you can resolve very tiny details of the source's timing event.

About the source variability, this is something more theoretical (so I can not be of much help), and there is no accepted theory, but qualitatively, in some models the gamma rays are produced by clumpy clouds of relativistically ejected material, which could account for the high variability indeed.

Other source: Wikipedia classification .


Why has Saturn stopped contracting gravitationally? The following text is from here:

The Jovian planets get their heat from the Sun and from their
interiors. Jupiter creates a lot of internal heat and releases this
heat by emitting thermal radiation. In fact, Jupiter creates so much
internal heat that it emits almost twice as much energy as it receives
from the Sun. The only reasonable explanation is that Jupiter is still
slowly contracting, almost as though it has not quite finished
forming.
Saturn and Neptune also appear to be emitting more energy than they
receive from the Sun. While we are certain Saturn is not still
contracting, it seems clear that Neptune is still contracting. Uranus
is the only jovian planet not emitting excess internal energy.

It make sense that the gas giants were formed from the force of gravity i.e. clouds of gas and dust particles contracted under their own weight, eventually becoming planetary bodies with solid core and gaseous atmospheres. For bodies whose bulk is mostly gas, it will contract towards its center and during such process, it will releases stored gravitational energy which will eventually convert into heat energy. Saturn is also a gaseous planet with rich gaseous atmosphere, so in theory, it should also contract under its own weight towards the center. So, why is it considered to have stopped contracting?
It is written here that Saturn has a different process for this internal heat formation because astronomers believe that unlike Jupiter, Saturn's formation process is complete and so, the heat energy is due to the falling of helium from clouds. Now, luminosity of planet (total power emitted by a spherical body) comes from the release of potential energy due to the gravitational contraction.
$${U = -\frac{GM^2}{R}}$$
$${L = \frac{dU}{dt} = \frac{GM^2}{R^2} \frac{dR}{dt}}$$
$${\frac{dR}{dt} = \frac{LR^2}{GM^2}}$$
Since, all the terms has non-zero value, there should be a non-zero value for the differential, even for Saturn implying that Saturn should contract with time. So, why astronomers believe Saturn's formation process to be complete (implying no further gravitational contraction)?
Related on physics.SE: What keeps a gas giant from falling in on itself?

The rate at which a planet contracts is determined by how much heat it can radiate from its surface and how close it is to the minimum radius allowed by electron degeneracy pressure (which is independent of temperature).
The contraction process is that heat radiates from the surface and is supplied by gravitational energy released by contraction. The process stops when the central pressure becomes determined by electron degeneracy at high densities. At this point the interior pressure decouples from the temperature (unlike a perfect gas) and the planet can cool at almost constant radius. The same process occurs in white dwarfs and brown dwarfs.
To first order, the radii of Jupiter and Saturn are similar and they have similar atmospheric compositions, so can radiate heat at a similar rate. But Saturn has far less gravitational potential energy to radiate and its centre reaches degenerate conditions earlier. In your equations you can see that $dR/dt \propto M^{-2}$ so Saturn contracts faster.
The equations you write down are essentially correct, but your statement that the terms are all non-zero is not. Effectively, Saturn is not luminous because of contraction, so $L \sim 0$ in these equations, and so is $dR/dt$ . The measured luminosity of Saturn instead comes from the residual heat in its interior and the chemical differentiation of helium and not through gravitational contraction.


Is there an advantage to the equatorial region of the far side of the moon for a radio telescope or would any crater on the far side work? If nearly any crater on the far side of the moon will work, and not just equatorial craters, then we could pick craters near the possible future Artemis base at the South pole which would allow easier access.  My assumption is as long as the crater is outside the range of Earth and the satellites in orbit it would be essentially the same as an equatorial crater.

tl;dr: The less complicated you want to make your dish, the closer to the equator you want to put it.

Since you are asking about single craters I assume this question is about using a crater as a natural pre-form for a single, large dish antenna like Arecibo Observatory and FAST.
On the Moon with no atmosphere or ionosphere nearly the whole sky is available for good signal, but as you move away from the current zenith position (i.e. "straight up") there are several problems. One is that the effective area of the telescope may be reduced for geometrical reasons at steep angles, and the other is that the optics of getting a good focus is increase.
Below you can see that Arecibo can't look much beyond 18° away from the zenith, and while the much newer and more complex FAST dish with it's changing dish shape articulated by moving cables and actuators can go as far as 60° away, it loses useful collection area beyond 26° not even counting $\cos(\theta)$ losses.
You will also notice that each telescope's maximum zenith angle attempts to be larger than the geographical latitude of its location, so that it can see the ecliptic.
Moving your lunar dish near the poles means you will lose access to a lot of the sky and be restricted to a "cap" of declinations either near +90° or -90°.
So the less complicated you want to make your dish, the closer to the equator you want to put it.
Arecibo
From Overview of the Arecibo Observatory; ALFALFA, Undergraduate Workshop , Gregory Hallenbeck January 14, 2013

Latitude: 18° 20′ 39″ N
Can move dome to zenith angle position of 19.7°, But only to ~18° with good performance

FAST
From Wikipedia:

Latitude: 25° 39′ 11″ N
The maximum zenith angle is 60-degree when the effective illuminated aperture is reduced to 200 m, while it is 26.4-degree when the effective illuminated aperture is 300 m without loss.33,3
Although the reflector diameter is 500 metres (1,600 ft), only a circle of 300 m diameter is used (held in the correct parabolic shape and "illuminated" by the receiver) at any one time.22 Thus, the name is a misnomer: the aperture is not 500 m, nor is it spherical.
3Project FAST
22The Five-Hundred-Meter Aperture Spherical Radio Telescope (FAST) Project
33The optics of the Five-hundred-meter Aperture
Spherical radio Telescope


Radio Astronomy and Imaging So, I've been starting to investigate radio astronomy, and am wondering about if certain things are possible from an amateur standpoint. I was looking at this powerpoint(that discusses building a tiny radio telescope), http://nightsky.jpl.nasa.gov/download-redirect.cfm?Doc_ID=347&Doc_Filename=IBTcom1%2Eppt .

On Slides 11, 12, and 13, it shows various radio emissions overlayed on the orion constellation. As well as Slide 17 and 18.

I'm wondering, is the data acquired in these images, doable from an amateur standpoint? From previous research I think I can build some detection devices for the Hydrogen Line emission(1420Mhz/21cm line), and that could lead to data(per my initial/basic understanding)  that's shown in image 1. Am I right in that assumption, and are the other emissions or data doable from an amateur standpoint?

There are tons a amateur radio astronomy clubs and groups.  Radio Astronomy was basically started by an amateur named Grote Reber, where he basically did the experiment that you are pondering to do in 1937 in his backyard in Wheaton, Illinois.  You can visit that exact instrument at the NRAO facility in Green Bank, West Virginia.  It was named a National Historic Landmark in 1989.

Just Google Amateur Radio Astronomy

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

http://www.nrao.edu/whatisra/hist_reber.shtml


How often do meteorites hit the Earth? I heard of one that occurred in 2008 in the Nubian Desert in Sudan, how often does it happen? Is there a catalogue of all hits or are they very rare?

Related link (This is what inspired my interest not defining what I'm looking for)
https://en.wikipedia.org/wiki/2008_TC3

At the time of asking I had not thought about the distinction between hitting the atmosphere and hitting the Earth.

There is definitely not a catalogue of all meteorites hitting Earth. For instance, the ones falling in desert areas and in the ocean aren't found, and even the ones falling in more populous regions are easily mistaken for normal rocks.

Meteorite rate

Meteorites come in all sizes, from sand grains to dinosaur-annihilating rocks.
The size distribution of these rocks (as well as most other things in space) follow an approximate power law, meaning that the smaller the meteorites you consider, the more often they hit Earth.

Estimating the total number of meteorites hitting Earth is difficult, but one study (Bland et al. 1996), looking at the composite of the earth on three arid site and considering the weathering over time, finds that between 36 and 116 meteorites over 10 g hit Earth every year per $10^6\,\mathrm{km}^2$. Multiplying by the surface area of Earth $A_\oplus$, that's roughly 20,000 to 60,000 meteorites hitting Earth per year.

Meteor rate

UPDATE: You want to know the rate of incoming meteors, i.e. all rocks that hit Earth's atmosphere. By definition, a meteorite is a meteor that lands on the surface of Earth. Before it enters the atmosphere and becomes a meteor, you can call it asteroid, meteoroid, rock, pebble, or dust grain, depending on its size and your mood.

Taylor et al. (1996) have monitored incoming meteors with the AMOR radar in New Zealand and detected in one year 350,000 trails from incoming meteors of size 10–100 $\mu$m vaporizing in ~100 km's height. I don't know exactly how large an area AMOR monitors, but from simple trigonometry, and as an order-of-magnitude estimate, if the events happen in $h \sim 100\,\mathrm{km}$'s height, the radar can "see" a distance $d=\sqrt{h^2+2R_\oplus h} \sim 1100\,\mathrm{km}$, and thus covers an area $A\sim4\times10^6\,\mathrm{km}^2$. Thus, again multiplying by $A_\oplus$, this amounts to the order of 40 to 50 million meteors per year.

Killer meteor rate and the Sudan asteroid

The meteors discussed above is probably not what you had in mind. As mentioned, the bulk of these "rocks" are actually more like sand grains. So what is the rate of devastating killer-rocks from space? Brown et al. (2002) fit a power law to several different data sets (LINEAR, Spacewatch, NEAT and others), and find that the number $N$ of meteors above a diameter $D$ in meters is roughly
$$
N(>D) \simeq 37 D^{-2.7}.
$$
The following plot is from their paper, with the long black line showing their fit, and the short black lines indicating the uncertainty:

You mention the asteroid 2008 TC3, which had a diameter of 4.1 m.
According to the fit, such an event happens roughly $1/(37[4.1\,\mathrm{m}]^{-2.7})\sim$ once per year.

You can also see that the largest meteor hitting Earth in a century would be $(37\times100\,\mathrm{years})^{1/2.7}\sim20\,\mathrm{m}$, equivalent to $\sim500\,\mathrm{kiloton\,TNT}$.
Similarly, hundred-meter-sized objects ($\sim1\,\mathrm{megaton\,TNT}$) should hit us roughly once per 10,000 years. By extrapolation of their fit beyond observational data, objects of $D>1\,\mathrm{km}$ — or 100 gigaton TNT — should hit Earth on average once every ~3.4 million years.


Can I build a telescope with a lens from a CD-ROM as eyepiece? I saw some article on the web about making an eyepiece of a telescope with the lens from a CD-ROM. Will this really work? I have an objective lens of 100 cm focal length. What will be the magnification if I could make a refractive telescope? Can I see the planets and moon with that? Can anyone suggest the arrangement of the lenses?

This article gives a construction guide for how to make an eyepiece using a CD ROM player lens: How to make an eyepiece for a telescope.

As to what the focal length of your  eyepiece will be, you can refer to this paper, in particular Figure 1 from which I derived the following relation:

$f=0.73r$

where $f$ is the focal length of your lens, and $r$ its radius (this is a rough approximation, of course). So given the fact that a typical CD drive lens is about 5mm in radius, its focal length will be only a few millimeters.

Now, assuming the focal length of your telescope is 1000mm, the magnification will be a few hundred times. What you will be able to see with that depends on the width of your objective lens and how well you will have built your eyepiece, but you should probably be able to see the Moon with your custom eyepiece. As for the planets, it will be hard to get something good given how much you are magnifying—not impossible, but be aware that magnifying several hundred times requires a good weather, a good telescope and a good eyepiece.


Keplerian Elements for Approximate Positions of the Major Planets Simple question here. I am working on something that needs to compute an orbital path using Keplerian Elements. When doing some research I came across this JPL page: Click Here. Within that link is a PDF titled 'Keplerian Elements for Approximate Positions of the Major Planets'.
From what I can tell, it looks as though this PDF is an excerpt from a larger document. I'm making this assumption based off of the equation numbering. They start at 8-30. From my experience, this is usually due to the fact that there were more equations/sections preceding these equations in the overall doc.
I would love to explore this larger document in order to learn more and find some answers to questions that have developed after reading through the doc. Does anyone know what this document is taken from?
Thanks!
Full Answer
Just for those who are looking for a complete answer to this question, I wanted to combine the two answers below into one complete answer (without taking credit and making my own answer).
As JamesK mentions, this excerpt is actually pulled from the book, Explanatory Supplement to the Astronomical Almanac. I now own this book and it is chock full of useful information, including the excerpt I was looking for. It also dives into a deeper explanation of everything in the excerpt I posted. If you don't want to buy the book, Huy Pham provided a link to the full chapter that includes the excerpt I linked.

I believe you are right, this is extracted from the "Explantory supplement to the Astronomical Almanac (2006?)"  (although it might date back to the 1960s...)
For example a similar PDF document gives that citation.
The supplement is published in a separate binding from the main astronomical almanc and a new version is published each year.  Finding a copy may be challenging (its not going to be on the shelves of your local Waterstones).


Is it possible to steer the sight of a Liquid Mirror Telescope using a plain mirror scheme? If yes, why hasn't it been done? I have read about Liquid Mirror Telescopes in a number of places, and according to these sources, a major disadvantage of these telescopes is that they can only "see" straight up.

In a nutshell, the usually very expensive primary mirror is replaced with a rotating mercury container. The mercury surface will become parabolic with the rotation and earth's gravity, and a camera is placed in the focal point of the parabola (straight above the container). Since the parabolic shape depends on gravity, the primary cannot be tilted and the telescope can only point straight up.

My question is: couldn't one use a mirror scheme to steer the sight of this telescope? For example a pair of big plain mirrors a bit above the focal point, working similarly to a periscope. Then one could change the angles of these two plain mirrors and point to different directions in the sky.

I am assuming a large plain mirror is much cheaper than a parabolic mirror of the same size.

Has this been done? If not, why?

Designs where the curved primary mirror is fixed, and steering is achieved via a moving flat mirror (siderostat) in front of it (or a set of flats) have been done before. One example is the Pfund telescope. The main design constraint that led to the invention of the Pfund telescope was the need to have a focal point which is fixed in space and easily accessible.

Here's a variation of that design; the need for the fixed focal plane is driven by accessibility concerns here (the user cannot follow a moving focal plane):

Also see the related concept of heliostat - same idea, just applied to sunlight. A moving flat reflects sunlight in a fixed direction. Some solar power plants use heliostats.

So, in principle, fixed curved primary + moving flat is doable, and has been implemented a number of times. However, there are substantial issues with this design.

First off, the size and the quality of the flat mirror are not less than the size and quality of the parabolic primary. Let's say you build a liquid mirror telescope because it's easy. Then you have to make a flat mirror of the same diameter, with the same precision, and all the ease of construction is now lost. Flat mirrors are not easier to make than parabolic. In fact, as a mirror maker, I prefer to work on curved catopters - in some ways they are more predictable during manufacturing than large, high precision flats.

There are further issues down the road for the liquid mirror telescope. You would actually need 2 flats, both very large, both very precise. Both would have to be suspended in the air, above the liquid primary, and supported by complex mirror cells just like curved primaries in any telescope. Overall, it will be more difficult to make such a telescope, not less.

So why is the Pfund design used? Not for ease of manufacturing; as I said above, it's actually more difficult to make it than a regular scope. But in some situations you just need that fixed focal plane for some reason. In that case, you don't have much of a choice. Of course, it's not a design that scales well.

  I am assuming a large plain mirror is much cheaper than a parabolic
  mirror of the same size.

That was the incorrect assumption. Once that's taken care of, everything else should be clear.


How was the galactic plane established? As pointed out in this answer the axes of galactic coordinates are defined such that galactic longitude and latitude (0, 0) point toward the galactic center. But then the plane defined by galactic latitude equal to zero would be parallel to the galactic plane only if the Sun were in the galactic plane as well.

Is that the case?

More generally, by what procedure was the galactic plane actually defined, considering a substantial part of it is obscured by dust (at least in the visible) and most of the mass is dark matter and therefore not observed at all (directly at least)?

Question: How was the galactic plane established? Primarily what kinds of observations were used, and how were they analyzed to help decide how this plane should be chosen?

The zero of Galactic latitude (i.e. the Galactic plane) was defined by a working group of the International Astronomical Union. The observations used to do this were 21 cm radio observations of atomic hydrogen, which are unaffected by dust absorption near the plane. The details of the observations are found in Gum, Kerr & Westerhout (1960). This paper uses least squares fits to the 21cm data assuming that the greatest concentration of neutral hydrogen is found in the Galactic plane itself. Whether this is the dynamical plane was considered by Gum & Pawsey (1960), as part of the same IAU study. They concluded that it was (bar the location of the Sun with respect to the plane), citing the flatness and relative narrowness of the neutral hydrogen and the coincidence of the source Sgr A*, which even then was thought to be at the Galactic centre, with zero latitude.

Dark matter was not well explored or widely considered in the dynamics of individual galaxies in the late 1950s, however it is not unexpected that neutral hydrogen should lie in the Galactic plane, as defined by the symmetry of the Galactic potential. A rotating mass of gas, capable of dissipative interactions, should settle to a plane perpendicular to its angular momentum vector.

The 21cm definition of Galactic coordinates cannot itself define the position of the Galactic plane in 3d space unless the Sun is exactly in the plane. Detailed observations of star counts and the dynamics of stars around the Sun and perpendicular to the plane have revealed this isn't quite the case. The details are given in How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? ; the Sun is currently around 17 pc above the Galactic plane and heading upwards. This is close enough that it makes very little difference to the apparent flatness and narrowness of the 21cm data.


Singularities/ringularities and their properties A black hole contains a singularity at its center. It is a zero-dimensional point, and  it's where all its mass is located. So, my first question is:

If a singularity contains all a black hole's mass, when a black hole absorbs matter, the matter should collect at the singularity. Then why does the black hole increase in size? The matter it absorbs is all in the singularity, not in the space around it...

Second, we know that spinning black holes can't have a singularity at the center as a point, because points can't spin, and when they do, they aren't a point. So there is such thing as a ringularity where it is a one-dimensional line bent in a circle that spins, and contains all the black hole's matter.

However, is a ringularity really a bent one-dimensional line in a circle, or an infinite collection of points that form a circle-looking figure?

(My information on ringularities is from a paper published by Cornell University that compared the one-dimensional line to a bunch of zero-dimensional points, and videos by Kurgesagt.)

You need to be a bit cautious about statements like: A black hole contains a singularity at its center. It is a zero-dimensional point, and it's where all its mass is located. What you are referring to is a mathematical structure called the Schwarzschild metric. This is indeed a solution to the Einstein equations for a static black hole, but this particular solution takes an infinite time to form. So given that the universe is only a shade under 14 billion years old we can be confident that as I type this there are no Schwarzschild black holes in the universe.

Indeed, it is far from clear that a Schwarzschild black hole could ever form, even given infinite time, because black holes evaporate. This point is pursed in the question Why does Stephen Hawking say black holes don't exist?

Incidentally, the Schwarzschild black hole doesn't have all its mass at the centre. In fact it is a vacuum solution and contains no matter at all. The mass of the Schwarzschild black hole is actually a geometrical property called the ADM mass.

We run into the same problems with the spinning black hole. The structure with the ring singularity is the Kerr metric, and again while it's a perfectly valid solution to Einstein's equations, like the Schwarzschild metric it also takes an infinite time to form. Actually the Kerr metric is more troublesome still, because it is uncertain whether the structure is stable with respect to perturbations. It seems very likely that real collapsing stars would not form a structure resembling the Kerr metric. More precisely the region outside the event horizon would be well described by a Kerr metric, the region near the singularity would be different.

But if we put these doubts aside and return to what you asked, the answer is that in the Kerr metric the ring singularity really is a ring and not a collection of points.


Where can I find data on Mercury's orbit to fit a model to? I have a project in which a Monte-Carlo method is used to find the optimal parameters for some model that is fit to a data set. I want to fit a model for the orbit of Mercury using this methods, but I can't find any data to fit it to. I have found a lot of papers on the orbit of Mercury that mention observational data, but no actual observational data. Where could I find data indicating the position of Mercury around the sun over a time period, or even a relatively accurate mathematical model from which I can generate my own data by defining it as a function that is stepped through some time interval?

The available astrometric data for Mercury come from meridian transit timings (right ascension and declination) and radar or spacecraft ranging (distance).
NASA JPL has some
here,
and the French IMCCE has some
here.
To use those, you'll have to work out some geometry including the position of Earth in its own orbit.

For your purpose it may be easier to use numerical ephemerides via
JPL HORIZONS or SPICE,
or IMCCE Miriade or Multisat,
which can output Cartesian coordinates if requested.
Of course these are fit to observational data themselves.
The underlying data files (e.g. DE430, INPOP17a) come with papers detailing sources, methods, uncertainties, etc.


Has SETI data been used for astrophysics? How, or why not? SETI collects alot of data from space, mostly radio. I wonder if this data has any astrophysics value other than SETI?

Well, that might depend on exactly what definitions you use.  SETI itself is a broad program.  In this answer I'll focus solely on what I consider the most iconic part of modern SETI, which is SETI@home.

We have the following quote from SETI@home's webpage (emphasis their own):

  SETI (Search for Extraterrestrial Intelligence) is a scientific area whose goal is to detect intelligent life outside Earth. One approach, known as radio SETI, uses radio telescopes to listen for narrow-bandwidth radio signals from space. Such signals are not known to occur naturally, so a detection would provide evidence of extraterrestrial technology.

SETI@home's specific purpose deals with the radio SETI data.  So they are implicitly asserting that the data they collect is expected to be useless outside of their namesake: they're specifically looking for things that are not known to occur naturally.

That being said, it turns out that the SETI@home data collection is almost never active (see here on their site).  It is done passively, as a piggyback collection when the Arecibo radio telescope is being used for other science projects.  So if your question is "is any data useful outside of SETI actually collected while SETI(@home) is gathering data?", then the answer is "yes".

Now, we can also note that one of the original goals of the SETI@home project was to prove the viability and practicality of distributed computing for scientific projects.  This we can say was a resounding success, as it spawned BOINC, which serves as a distributed computing platform for a wide range of non-SETI science projects: Folding@home and Rosetta@home deal with protein folding, Einstein@home is for gravity wave detection, the now-defunct Mesenne@home was for finding Mersenne primes, and many others (my copy of BOINC specifically states that there are over 30 currently active).

So if you are willing to consider the advances they spurred in distributed computing in scientific projects, and how this has impacted scientific projects inside astrophysics (or outside), then the answer is again "yes".

EDIT:  Einstein@Home recently made the following announcement about a gamma ray pulsar that was detected via the program; it includes links to a published article on the detection, including an arxiv version link if you find the official publication is behind a paywall.  So that's a concrete, published, and current discovery in astrophysics.  If you're willing to give SETI some of the credit for having spawned the @Home style of distributed computing projects in science, as suggested above, then there you go.


Calculating the date and year when a lunar eclipse will fall on the Winter Solstice I am a novelist working on a new book, titled Werewolves In The Christmas City, which happens to be Bethlehem, Pa. In 2010 when I was working on a book titled The Christmas City Vampire, a lunar eclipse on the Winter Solstice occurred. As you know it was the first one since 1638. On the 2010 occurrence, I added the resurrection of two werewolves-twins-turned during that eclipse of 1638. My question is: How would my professor character figure out that the event will occur in 2010. The time frame I am in right now is 1947. I would appreciate any help you can give me. Math is certainly not one of my strong suites.

A university library in 1947 might have a copy of T. von Oppolzer,
Canon der Finsternisse (1887),
which lists those lunar eclipses on pages 368 and 374.
A reader with a working knowledge of astronomy could make sense of the tables even if not proficient in mathematics or German.

If that reference were not available, a text such as R. Buchanan,
The Mathematical Theory of Eclipses (1904),
would equip the professor to work it out from scratch.
As lunar eclipses are visible from the entire night hemisphere,
they are simpler to compute than solar eclipses,
which are total only along a narrow path.

Eclipses recur in
saros cycles
of 18 years and 11.3 days.
Given dates of some contemporary eclipses,
one could trace
saros 119
back: 1945-06-25, 1927-06-15, ..., 1638-12-21; or trace
saros 125
forward: 1938-11-07, 1956-11-18, ..., 2010-12-21.
The saros numbers were not assigned until 1955,
but the cycles were known to ancient astronomers.


Do any particles in AGN jets escape the galaxy? I have read, at http://www.thestargarden.co.uk/Black-holes.html for example, that whole stars can be ejected from certain galaxies. "These are thought to have been part of a binary star system that broke apart as it approached a supermassive black hole. As one star was captured, and the other was pushed away at a velocity exceeding the escape velocity of the Galaxy."
My question is about the jets that can form at the poles of the rotating supermassive black hole: "At the rotation axis of the supermassive black hole, matter from the accretion disc can be pushed away at the speed of light, creating jets that can extend for thousands of light-years."
At thousands of light-years away, is this matter still gravitationally bound to the galaxy?  (If not, maybe some particles can go around in a big loop that takes a while.)

Yes, definitely.
While some matter returns to the galaxy as a so-called "galactic fountain" (e.g. Biernacki & Teyssier 2018), some material is ejected at super-escape velocity, becoming part of the intergalactic medium.
This is one of the mechanisms responsible for polluting the intergalactic medium (IGM) with metal-rich gas (i.e. elements heavier than helium). A recent observation of this is presented by Fujimoto et al. (2020).
The IGM itself is too dilute to form stars, and hence metals, itself, but is nevertheless observed to contain quite a lot of metals, usually seen as absorption lines in the spectra of background quasars (e.g. Songaila & Cowie 1996; Aguirre et al. 2008). These must be blown out from the galaxies, either by stellar feedback (through radiation pressure, cosmic rays, and supernova feedback), or by AGN activity (see also Germain et al. 2009).
Note however that although material can reach relativistic velocities, "pushed away at the speed of light" is just a tad too fast. Only massless particles can travel at the speed of light, but if you're massless it isn't really a big achievement — photons do that all the time, which is why we see the galaxies in the first place.


How to compensate the effect of tectonic activity in devices like LIGO? In late 2015, the LIGO project team announced the detection of gravitational waves. The detector is (very basically) a laser measuring the distance between two mirrors in vacuum, over a large-enough distance to detect very small changes.

The two mirrors are installed on Earth, and a couple kilometers apart.

I believe, perhaps mistakenly, that tectonic activities (if only vibrations) may change the mirror distance over time, so the LIGO project has probably used some model and technology to nullify their effects.

How can that be done? If there is any mistake there, I wonder about how to validate the experiment results.

There are four aspects to this:

You suspend the mirrors and other components using the best suspension mechanism that has ever been built. This is very effective at isolating the experiment.
There are two detectors thousands of km apart. To claim a genuine signal it should be seen identically in both in less than the light travel time between them.
Many of the astrophysical GW sources (e.g. merging black holes) produce characteristic signals that look nothing like random seismic activity - for instance a correlated increase in frequency and amplitude terminating in a "chirp".
Lastly, LIGO is not immune to seismic noise - it is this that partly limits its sensitivity below frequencies of around 30 cycles per second. The seismic noise component at higher frequencies is much smaller.

The way that LIGO isolates the astrophysical signal from the various noise components that are present is described in this paper by the LIGO and VIRGO collaborations. A crude characterisation of the process is that you cross-correlate your signal against a library of expected signatures from astrophysical GW sources and look for a high degree of correlation (which you wouldn't expect in random noise).

The characterisation of the noise (including seismic noise), a description of how it is assessed and monitored (including how data is rejected when deemed too noisy) is found in a further LIGO publication.


Why do (most of) the planets rotate counterclockwise, i.e. the same way the Sun does? Referring to the mechanisms explaining the solar system formation and to the initial rotation of the gaseous cloud that collapsed, I understand easily why the planets orbit the Sun the same way this one rotate (say counterclockwise) but I can't figure out why this apply to planets rotation too. Thinking about that from Kepler's laws and angular momentum conservation point of view, I might conclude that the planets should rotate clockwise because the velocity of the particles that aggregated during the planets formation was higher closer to the Sun...

Apart from a short explanation, I would like to have a good reference from the literature if possible.

Edit, to make my reasoning more explicit: following Kepler's laws, the particles that aggregate on the "day side" of the proto-planets in the east-west direction relative to the ground are faster than the ones hitting on the "night side" in the west-east direction. If we add all of these contributions, the planets should rotate in the opposite direction relative to the initial cloud (i.e. relative to the actual Sun rotation). I guess something is wrong or missing there (to counterbalance the phenomenon I just described) but I can't see what it is...

New edit: References I found some published articles dealing with this kind of question but I don't have the time right now to read them carefully. If someone is motivated to do so, do not hesitate ;-) If I find the answer to my question amongst these papers, I will post it there later. Of course, you may need to use the network of an institution with a subscription to these editors to access them:

R.T. Giuli (1968a) in Icarus: http://www.sciencedirect.com/science/article/pii/0019103568900821

R.T. Giuli (1968b) in Icarus: http://www.sciencedirect.com/science/article/pii/0019103568900122

A.W. Harris (1977) in Icarus: http://www.sciencedirect.com/science/article/pii/0019103577900793

J.J. Lissauer, D.M. Kary (1991) in Icarus: http://www.sciencedirect.com/science/article/pii/001910359190145J

Even though I'm a professional astronomer (but not an expert on this field), I don't have a good answer. AFAIK, this question is still open. Actually, the fact that most planets in the Solar system rotate in the same sense as they orbit the Sun is an important constraint on possible planet formation scenarios (of which still several are presently discussed by scientists).

For example, the idea that planets form out of dust collected at the centre of vortices in the proto-planetary disc can be almost ruled out, because retrograde vortices are much more stable and long-lived than prograde ones (Jupiter's red spot is a retrograde vortex), and hence planets should be retrograde, but aren't.


Why do X-ray binaries such as GRS-1915 have active and quiet states? X-ray binaries such as GRS-1915 feature a black hole and a regular star. The x-ray luminosity is likely related to the rate of mass flow into the black hole from the star so when there is a lot of mass flow the x-ray luminosity is high but this explanation seems to just beg the question as to why the mass flow would change over time.

Such objects are surrounded by an accretion disk, that is a disk of gas, the gas being accreted onto the central object. However, at this stage of its life, the accretion disk around the black hole of a binary system like GRS-1915+105 is strongly depleted. Therefore, the accretion rate is not sufficient to support a coutinuous burst. Matter fills the outer disk until a critical density is reached, that triggered an outburst. Mass from the companion star is lost by a Roche lobe overflow mechanism.

References:

X-ray binaries in general: Remillard & McClintock 2006
For the outburst X-ray transients in particular: Dubus et al. 2001


What's the distribution curve for meteor sightings? I know that meteor showers have a "peak" observation time.

What does the distribution in time look like? Is one side of the peak steeper/shallower or longer/shorter than the other side, and if so, why?

The meteor sightings in a meteor shower can be approximated to a normal distribution. In fact, research papers use a normal distribution to model meteor showers.

The exact characteristics of the shower will depend on the location, local weather conditions etc. The number of sightings and the number of actual meteors impacting the earth's atmospheric conditions are different as it is the local conditions that determine the number of sightings.

The International Meteor Organisation maintains data on the number of meteor sightings which shows that the sightings can be approximated to a normal distribution.

However, this data has to be used carefully because these refer to the reported sightings and the errors are significant.


Outside of a Universe and its origin Per the Big Bang Theory, the universe was formed from a dense singular point which existed 13 billion years ago and has been expanding ever since. This suggests that the point should have existed somewhere in space implying an outer universe. If we call this point the "Universe", then what do we call the space that held that point? Also how did this single point come into existence in the first place?

You are making the mistake of thinking that there was a single spot in space that everything expanded from. Rather, the expansion was not of things in space, it was space that expanded. It was an expansion OF space, not IN space. There wasn't "space" around the singularity.


Long term temperature effects of an eclipse I was surprised to see a very measurable difference in temperature at my home during the recent solar eclipse.  The temperature was about 93 degrees fahrenheit but dropped to around 86 degrees.  It made me wonder how an unusual eclipse might affect a planetary body.

If there were a situation where someone had put an object into a stable synchronous orbit that would block a percentage of solar radiation, how big would it have to be for an appreciable effect?  An example would be from Futurama's Crimes of the Hot episode Wernstom's giant parabolic mirror that reflected 40% of all sunlight.
If an effect similar to a solar eclipse would be prolonged over a specific area, how would that effect the local atmosphere?  I would assume it would create a stable "cold front" in that area.  Is there any information on how that would affect other passing cold or warm fronts?

It was suggested that we could cool down Venus by putting a powder cloud in the Lagrange point between it and the Sun. This will cool Venus in a few months to the point where its atmosphere will start liquify, creating a global CO2 ocean. The surface of an ocean is much easier to colonize than creating flying cities.


How to name planets outside our galaxy? Recently, planets outside our galaxy have been discovered.

How would be the convention to naming these planets? As far as I know, it seems that there is no standard for naming exoplanets even in our galaxy.

But if I'm publishing a scientific paper and I wish to name some of these recently discovered planets, what would be some guildelines on doing so?

The original paper ( and a version on arXiv ) which this relates to does not seem to identify any actual planets.  It's clear from the abstract that it is simply suggesting there is evidence that what they describe as unbound planets seem to exist as an inference of the spectral data they have gathered.

They are not identifying any individual planets at all.  They have not actually discovered any planets and there is nothing to name.

This is the abstract :

  Previously, planets have been detected only in the Milky Way galaxy. Here, we show that quasar microlensing provides a means to probe extragalactic planets in the lens galaxy, by studying the microlensing properties of emission close to the event horizon of the supermassive black hole of the background quasar, using the current generation telescopes. We show that a population of unbound planets between stars with masses ranging from Moon to Jupiter masses is needed to explain the frequent Fe Kα line energy shifts observed in the gravitationally lensed quasar RXJ 1131–1231 at a lens redshift of z = 0.295 or 3.8 billion lt-yr away. We constrain the planet mass-fraction to be larger than 0.0001 of the halo mass, which is equivalent to 2000 objects ranging from Moon to Jupiter mass per main-sequence star.

What they're saying is that, if you accept the hypothesis of their theory and data, there must exist a minimum of about 2000 objects of planet mass per star.  There's no way to identify any individual object and no means to verify their claims that I know of.


Which star did Uranus occult in 1977 when the planet's rings were discovered? In 1977, Uranus occulted a star, and the starlight flickered 40 minutes before and after the actual occultation, revealing Uranus' rings.This is common knowledge to astronomers, but what I can't find out is which star Uranus occulted. Can someone give me the name of this star?

From wikipedia, the rings of uranus

  The definitive discovery of the Uranian Rings [...] use[d] the occultation of the star SAO 158687[...]

The star SAO 158687, also cataloged as HD 128598 is a magnitude 8.7, orange dwarf star in Libra. There seems to be nothing special about the star, except that one day, a planet passed in front of it.


Has the Kepler data been examined for light curves due to phase changes of non-transitting hot Jupiters and other? I found this article http://www.epj-conferences.org/articles/epjconf/pdf/2011/01/epjconf_ohp2010_03005.pdf from 2011 announcing that two dozen secondary eclipses of "hot Jupters" have been seen in Kepler data. I even think I have seen a light curve showing phase changes of a transiting planet. This makes me think that there should be evidence in the Kepler data for non-transiting planets since the amplitude of the light curve due to planetary phase changes shouldn't be much different from that of a secondary eclipse as long as there is a fairly high inclination (say over 60 degrees).

Is anybody aware of such an examination of the data?

Here's an example paper looking for phase curves of non-transiting planets in the Kepler data: Millholland & Laughlin (2017) "Supervised Learning Detection of Sixty Non-transiting Hot Jupiter Candidates".

The candidates themselves can be found on this webpage.

Unfortunately I'm not aware of any RV followup on them to determine which of these candidates are actually planets.


Central temperature and pressure for constant density vs linear density model of stars I just derived the pressure and temperature profile for a linear density model of a star. If I compare the profiles of pressure and temperature for a constant density model of stars, I see that the linear density model central pressure is higher, but the central temperature is lower.

What is the physical reason behind this?

Because the total mass of the star has to be conserved.

If you consider a linear density model, for a star of fixed mass and radius,
you have to make the central density higher than the average density in a constant density model.

Furthermore, because the mass of each shell goes as $\rho r^2$, there needs to more shells with a. density higher than average. This means the average density of a mass shell is higher than in the constant density model and more of the mass is at smaller radii.

Hydrostatic equilibrium demands that the pressure gradient is proportional to $-\rho M(r)/r^2$, where $M(r)$ is the mass inside radius $r$. At any radius, this will be larger on average in the linear density model because more of the mass is at smaller radii.

If the pressure gradient is higher, then the central pressure will be higher. The value of the central temperature then just comes from gas laws.


Are blue and red shift visible? When looking into the sky at night using my bare eyes, I see that stars appear in different colors. From my understanding this is caused by different chemical compositions of those stars which show up in their spectral lines and therefore defining their color.

Does blue and red shift visibly change the color perception when looked at the stars with your naked eye?

No, the blue and red shift of stars is not possible to detect with the naked eye. There are a couple of reasons for this. First the effect is slight: even for an object moving at thousands of km/s (way faster than stars in the milky way move), it is only a tiny change in hue. Next, since the light of stars is a mixture of many different colours (a black body spectrum) if the light from a star is red shifted, then ultra-violet is shifted down and takes the place of blue, the blue moves towards green and so on. The end result is a mix of all the colours, and so very little change in the perceived colour.

The colour of stars has relatively little to do with their composition, and a lot to do with their temperature. Cool stars (either small, or large, old and with puffed up cool outer layers) look red or orange. Hot stars (large and powerful stars) are white or blue-white. Stars have a spectrum that approximates a blackbody, and so the actual composition has relatively little effect on the colour.

The composition of the atmosphere of a star will absorb particular wavelengths of light, and this results in a dark line in the spectrum (not enough is absorbed to change the apparent colour). It is by measuring the position of these lines in the spectrum that red- and blueshift is observed.


Are all models of ocean planets theoretically cloud covered? Take for example an ocean planet slightly smaller than the Earth. Let's say it has a planet-wide ocean that is on the average 10 miles deep. Say for argument sake, that there is a molten, rotating outer core producing a magnetic field and that there are plate tectonics (although not to the same degree as Earth). Also, imagine that there are volcanoes that reach as high as 5 miles below sea level. The planet's effective temperature is similar to Earth (not counting greenhouse). Would such a world without land always be covered with clouds? Would it necessarily have a runaway greenhouse? How about such a world with no vulcanism?

Having given this some thought, it's a bit of a monster question with a few variables that you don't mention.  Planetary tilt, which causes seasons is one.   Length of day, for example, a rotation like Venus' 116 days would be quite different than an earth rotation of 24 hours and a tidally locked planet would also be different.   Atmospheric pressure is another.   I'm assuming Earth like tilt, 24 hour day since you didn't mention, and 1 ATM, but I think, if and when we finally get a good look at earth-like exo-planets, which, granted, might be several decades away, I think we'll see a few unexpected things.

OK, moving on.   Earth likely got much of it's atmosphere from comets which suggests healthy amounts of CO2, CH4 and NH3, maybe N2, some from the NO family, maybe some Helium, Argon and Neon, those 3 I think we can mostly ignore, some hydrogen perhaps from solar flares, which might get stripped along with the helium from the planet over time.

The total amount of comet and asteroid impacts and ratio of gasses add some variability and a really important factor is whether or not the planet has life and undergoes photosynthesis.   Photosynthesis would put O2 into the atmosphere and reduce CO2 and would also have the effect of, over time, reacting with and eliminating the virtually all of the CH4 from the air and any dissolved Iron in the oceans, turning the oceans from a muddy brownish/red to clear blue.  Photosynthesis likely played a key role in turning the Earth from a hot planet to a snowball.

So, lots of variables, but it's a fun speculative question, so I'll give it a shot.

How much smaller than Earth?   I'm going to go with Venus size but if you have different in mind, let me know.   .815 Earth masses, 6,052 km in Radius and gravity of .905 of Earth.  Source.

10 miles deep oceans (16.09 km), surface area, 4 Pi R^2 about 7.4 billion cubic km of water.  (Earth, by comparison, about 1.37 billion cubic km)  Source.   Your planet has about 5.4 times as much water as earth (not counting water in the crust, but you didn't mention that, so lets not go there) - and planet size doesn't affect the predictions much.

  I wonder how much outgassed CO2 etc. would stay dissolved in the ocean
  and how much would accumulate in the atmosphere? Nitrogen?

I'm going to assume that the theory is correct that Earth (and your water world) get most of it's atmosphere and water from asteroids and comets Source and Source, though the 2nd source suggests some could have come from the primordial material of the Earth.   If your water-world has a lot of gas trapped below it's crust in it's primordial material, then out-gassing from volcanism becomes a bigger factor and the volcanoes, not the atmosphere above would keep the oceans saturated with gas.   That's possible, but I'm going to assume that most of the gas is already in the atmosphere following a late heavy bombardment period.   I'll touch a bit more on tectonics later.

Titan Probably got it's thick atmosphere from outgassing Source and perhaps Venus from a recent large outgassing event too - just throwing that out there.

I'm also assuming that your planet wouldn't have permanent ice over it's oceanic poles due to the warmish temperature you implied and likely oceanic circulation not blocked by land-masses would prevent ice formation.

  Would it necessarily have a runaway greenhouse?

Greenhouse Gases, CO2, CH4, NO family, (H20, more indirectly)

If your planet has a lot of CO2/CH4, comparable amounts to the amount of H20, then it's unlikely the oceans could begin to dissolve enough gas and this would lead to a run-away Greenhouse, almost without question.

If your planet has more earth like CO2/CH4 levels, perhaps driven by photosynthesis capturing carbon and released oxygen chemically reacting with the CH4, then the runaway greenhouse can be avoided and CH4 be very low concentration.   CH4 on Earth is currently about 1.8 PPM and prior to farming and livestock and oil drilling & Fracking which can release some CH4, it was probably less than half that much, less than 1 PPM.  In a water world CH4 would probably be even less as there's no biodegrading or digestion of plant-mass releasing CH4.

CO2 is more complicated and the amount dissolved in the ocean depends on the total amount available.  If there's not too much CO2 to saturate the oceans, then you have an ocean/air equilibrium which is also affected by temperature.    For a more detailed answer on this, look into Henry's law, but I'm going to cheat and do a quick and dirty calculation rather than use Henry's.

on Earth, the oceans contain about 50 times the CO2 as the Atmosphere:   Source.  And the oceans weigh about 265-270 times more than the atmosphere.   (ocean mass given above, atmosphere mass here.)

So if we estimate this as parts per million (ppm), the CO2 concentration in Earth's atmosphere is about 5 times the concentration in Earth's oceans, and at higher temperatures, that ratio goes up, lower temperatures it goes down, but it never gets to the point that there's no CO2 in the air, cause not possible in an equilibrium.   With 10 mile deep oceans, we can assume a more CO2 in the oceans, maybe 99% vs about 98% ocean to air ratio currently on earth but the 1% in the air would depend on the total mass of CO2 in the water world's ecosystem, ignoring anything permanently trapped deep on the ocean floor, such as sea-shells.

  Would such a world without land always be covered with clouds?

Lets say you have an Earth like Oxygen/Nitrogen atmosphere, 1 bar, a planet very similar to earth but just oceans.  Similar weather (as a starting point), this planet should have more clouds as clouds are more common over the oceans, source, and you would have no low humidity/dry pockets of land, but I would think, probably not 100% clouds, just more clouds.   The cloud effect on temperature has some uncertainty to it, as clouds both both reflect sunlight making the planet colder during the day and they trap heat at night, so they play for both the warming team and the cooling team, but what I've read, the overall effect is pretty small.   Days could be colder, nights warmer.

Also, evaporation tends to cool the air at the surface, which is why islands surrounded by ocean don't ever get the scorching hot temperatures you get in death valley for example, even if they're closer to the equator, so you wouldn't get heatwaves, but you wouldn't get freezing cold either.  (I would think).

Finally, oceans have lower albedo and a water-world earth would likely have no ice-caps, so, Overall, I think it's likely a water-world earth would be warmer primarily due to lower albedo and this would increase atmospheric water-vapor (not clouds but transparent water vapor, which is a greenhouse gas), so a water-earth would probably several degrees warmer on average with no ice-caps.    How much warmer - I have no idea but I don't think a water-earth would be run-away greenhouse.

A water world (Earth) could also have less CO2 in the atmosphere due to more dissolved in water and probably less CH4 from decaying matter on land, so, it might actually be colder, but this would more likely depend on the life cycle's O2 to CO2 ratio rather than oceanic absorption of CO2.   Too many unknowns to say for sure.

Another curious effect to a water world earth is you might get some hurricanes that last for weeks, maybe even months or years in the right conditions, kind of like mini versions of Jupiter's great red spot.   All you need for a hurricane to gain strength is warm water and cold air and with no land for the Hurricane to lose strength on, You could get some real jumbos.   Maybe category 7s, perhaps 8s.  That would be awesome.

Now if (and I think this is the gist of your question), what happens if you decrease the atmospheric pressure by significantly lowering the amount of O2/N2 in the atmosphere (or raising it).  Lower atmospheric pressure lowers the boiling point of water and if you lower the atmospheric pressure enough, then water starts to boil and you get a partial water vapor atmosphere as a result of low atmospheric pressure and lots of water.   I suspect this is unlikely to actually happen as a planet should always have enough other types of gas to prevent this unlikely outcome but a water vapor atmosphere can be approached logically, even if it's not going to really happen.  Now, this isn't just high humidity, this is an actual water vapor atmosphere.

I imagine a water-vapor rich atmosphere would form droplets and rain fairly regularly and be permanent or nearly permanent cloud cover but that's just a guess, though it would probably appear more like low to the ground smog than cloud cover, the clouds might hover much closer to the earth.

If you increase atmospheric pressure, then water has a higher boiling point and takes more energy to vaporize.   A denser atmosphere might hold heat better and be a bit warmer.  It might also have fewer clouds due to more air and perhaps less variation in temperature and circulation, as it's circulation and warm air cooling off that's the primary driver in cloud formation.

Water vapor would still form under a higher pressure atmosphere by wind and by photons.  Sunlight plays an important role in evaporation, perhaps even more than temperature based on pan evaporation studies.

If you have a very thin atmosphere made almost entirely of water (getting back to our improbable example).  You'd drown if you tried to breath it, but as far as temperature of such a planet, at least with earth temperatures, water stays liquid even at quite quite low atmospheric pressure.

At 1/2 PSI (1/29th of 1 atm) the boiling temperature of water is 79.6 degrees C.  Source.   I'm not sure how much heat 1/29th of an atmosphere could trap even if it was mostly water which is a greenhouse gas, so this one is hard to predict.  You'd likely see wild night to daytime temperature swings, frequent boiling of the oceans during the day and very heavy rainfall at night, much more rain than we ever see on earth, as 1/29th of an ATM of water vapor is several times more water than is in the earth's atmosphere at any given time.

If a water-vapor atmosphere planet (unlikely), got enough heat from it's sun, it could certainly turn into a run-away greenhouse, but I'm not sure if that happens at the radiance earth gets.   But a planet like this would probably be subject to bigger temperature swings than we see on Earth and very fast wind.

  How about such a world with no vulcanism?

(Vulcanism - as in, spock, . . . sorry).

We think of plate tectonics and volcanism as a process by which gas is given to the atmosphere, and that's true to an extent.  Certainly any gas trapped inside a planet at formation can be outgassed through Volcanism, but perhaps a more important aspect of plate tectonics isn't the release of gas but the absorption of gas.   Oxygen is very reactive and it combines with basaltic rock to make lighter/stronger rock like granite and bedrock which leads to continents and mountain ranges and all that good stuff.  Nitrogen can bind with basalt too, but Oxygen I think, binds more readily.   Without photosynthesis and the creation of oxygen, not just life on earth but the continents would look very different and they'd probably be much less permanent.

So if all the volcanoes are under water, that limits the atmospheric absorption of oxygen and other gases to just what's disolved in the oceans which could be a lot of NH3, but relatively lower amounts of other gasses.

Another effect of all tectonic activity and volcanoes being under water is that volcanic gas like sulfates and tiny dust particles which can cool the planet after a large eruption, are unlikely to reach the atmosphere in any significant volume, so you're likely to avoid any volcanic cooling like you get on occasion on earth.

Finally, you might think that volcanoes under oceans would warm the oceans and they would locally, but not much overall.   A significant percentage of Earth's volcanoes are under water, but because ocean water circulates, most of the deep oceans stay at a chilly 4 degrees C, as opposed to land which gets warmer as you dig into the ground.  Convection circulates heat much faster than conduction.

Undersea volcanoes in your water world could be very helpful in the continuation and maintenance of extremophile life due to repleneshment of nutrients but not much in the way of temperature change or cloud formation.

Finally - a fun bit worth mentioning.   NH3, which is a very common gas/ice in the solar system.  It's in all the outer planets and comets and outer moons.   The neat thing about NH3 is it's water soluble and it's an energy source to primitive life, so simply by having oceans, stinky but useful NH3 is fairly quickly, mostly removed from the atmosphere and dissolved in liquid oceans and if there's life, it can be consumed, some released as N2 or NO-family and some can be built into amino acids.  Some nitrogen can bind with basaltic rock and magma from undersea volcanoes forming different types of rock, but I'm no geologist, so I'm not sure how much that would happen.

There's no one simple answer to this, but that's my best guess as a layman.   I enjoy thinking about stuff like this.


Can I see the star of Bethlehem at Christmas? In the Christian tradition the first Christians are believed to be three wise men who followed the star of Bethlehem to the stable where Jesus was supposed to be born. Is it with some accuracy known if this is an actual star and if so will I be able to see that star during Christmas in North-western Europe?

The most recent suggestions are that it was a nova or supernova, however no one really knows what the object was. The verbatim records describe it as a star, though it is not certain whether or not it was a "true" star (in those times, a star being any object that had a fixed right ascension and declination). Perhaps it was just a planet.

In any case I suspect it is unlikely you will be able to see it with your naked eye today. Even if you could, its true identity has certainly been lost in history.

Some references for further reading (click on the "Full Printable Article (PDF/Postscript)" link for a PDF of the articles):

What was the star of Bethlehem?, S. Hagar (1918)
Interpretations of the Star of Bethlehem, E. Little (1968)
Can we find the Star of Bethlehem in Far Eastern Records?, C. Cullen (1979)
The Star of Bethlehem: a type Ia/Ic supernova in the Andromeda galaxy?, F. Tipler (2005)


Have we ever observed a body, such as a large asteroid, "hitting" the Sun? Some other SE questions about launching ICBM's into the sun got me wondering whether we have ever observed an object on a path that intersected with the Sun? How close did it get?

Yes

and here's a video of "a Giant Comet Hitting the Sun":

https://www.youtube.com/watch?v=Mat4dWpszoQ

The impact occurred sometime during May 10-11, 2011.  The comet was not named but believed to be a member of the Kreutz family of comets

Many close calls

Before this spectacular plunge we had witnessed several other comets graze (come close without hitting) the Sun.

I've included a nice picture of comet Lovejoy after its close approach of the Sun brought it through the Sun's corona around December 15, 2011:

What happens when a comet hits the sun?

It acts like a supersonic snowball in Hell

  If a comet is big enough and passes close enough, the steep fall into
  the sun’s gravity would accelerate it to more than 600 kilometres per
  second. At that speed, drag from the sun’s lower atmosphere would
  flatten the comet into a pancake right before it exploded in an
  airburst, releasing ultraviolet radiation and X-rays that we could see
  with modern instruments.

  The crash would unleash as much energy as a magnetic flare or coronal
  mass ejection, but over a much smaller area. “It’s like a bomb being
  released in the sun’s atmosphere,” Brown says. The momentum propelled
  by the comet could even make the sun ring like a bell with subsequent
  sun-quakes echoing through the solar atmosphere.