Can an average person see stars from the bottom of a well or through cardboard tubes? Definitive answer required! A debate about seeing stars or planet during the day below this answer to the Aviation SE question At what altitude might a pilot be able to see at least the brightest stars during the day? seems unresolved.
I'm pretty confident that Venus can be seen during the day through a cardboard tube only because I've seen this mentioned so many times.
But what about stars? And what's this about Aristotle and Sir John Herschel?

Here's a reprint of the debate referenced above just in case some get deleted in the future:

Yes, but the thing with the hole is a hoax, that doesn't work. What does work are binoculars.

Tell it to Aristotle and Sir John Herschel (see edit above).

The link does not confirm the ability to see stars from the bottom of a shaft, but rather refutes it.

Since they failed, even with binoculars, one wonders if sky conditions were less than optimal (a cirrus layer that isn't even readily visible would broaden the point of a star enough to make it invisible against the general scatter). Star visibility from the ground in daylight is well documented and repeatable, and doesn't require visual aids, if only the sky is completely clear.

This article (Hughes 1983, "On Seeing Stars (especially up chimneys") from the Quarterly Journal of the Royal Astronomical Society seems like a pretty good account of the "phenomenon", including both classical and literary claims (e.g., in stories by Dickens and Kipling) and actual experiments. It appears that the Aristotle reference ("people in pits and wells sometimes see stars", from Generation of Animals) is not an eyewitness report by Aristotle himself -- and, given that Aristotle claimed men had more teeth than women, one should always be a bit skeptical about his claims; there is also the possibility that these were sightings of planets, which can sometimes be seen in the daytime. The Herschel reference is to an "optician" Herschel knew who claimed to have seen a star this way once or twice, not something Herschel himself verified.
In Section 2 of the article there are discussions of actual serious (and unsuccessful) attempts to test this, including numerous attempts by Alexander von Humboldt and well-described attempts by astronomers like J. Allen Hynek, who tried (with "several members of my astronomy class") to observe Vega from the bottom of an abandoned smokestack, and some measurements taken at the botton of a tall chimney which verified that the brightness of the sky was unchanged from that seen outside.
There are also some discussions of the sensitivity of the human eye. The conclusion I get from this is that it is barely possible that someone with good eyes and an extremely restricted field of view (much smaller than "from the bottom of a well" or a "cardboard tube") might be able to make out Sirius, but probably nothing fainter.
Edited to add: This article (Können et al. 2015) reports a daytime sighting of Sirius from the La Palma observatory. They note that there were able to do this because they had pointed their telescope at Sirius, and looked along the finder scope (knowing the offset), so they knew exactly where to look. (Even then, it was difficult: "Both of them saw Sirius, but it was difficult; as Wallis put it during his attempt: 'It comes and it goes.'") Knowing exactly where to look is important, because

the limiting magnitude of photopic vision decreases linearly and
rapidly with distance from the center of field of the human eye: at 2°
from the center the limiting magnitude has worsened by 1 magnitude;
at 4° by 2 magnitudes [16]. This means that for an unintentional
sighting, a star should be roughly 2 magnitudes brighter than the
limiting magnitude. Venus meets this requirement, but for Sirius there
is little hope for an unprepared sighting.

(Short answer: Generally speaking, you cannot see stars from the bottom of a well, chimney, mining pit, etc., during the daytime. And you certainly aren't going to see multiple stars from the bottom of a canyon, even if there's a lovely evocation of that idea at one point in The Lord of the Rings.)


Does light from the sun reach the surface of Venus I am aware we have images from the surface of Venus from the Venera missions but I am not sure if those images are taken using the visible spectrum the human eye can see. This is why I was wondering if light reaches the surface of Venus with how dense the atmosphere is or would it be pitch black?

Yes, a surprising amount of sunlight does reach the surface, so you very nearly answered your own question. The Venera spacecraft used visible light to take its pictures. Bear in mind that being nearer to the sun than we are, Venus gets more light from it than we do. What surprised me is that like Earth and Titan, there is a substantial gap of clear air between the ground and the cloud base. The atmospheric pressure is 90 atmospheres, and temperature about 460 C. On Earth, sulphur dioxide from volcanoes and other sources is oxidised in a few weeks to become sulphur trioxide, which forms sulphuric acid on contact with moisture. This process is impossible on Venus, where there is only a minute trace of oxygen in the atmosphere, so the origin of the clouds of sulphuric acid said to be there is a bit of a mystery. Photographs of the surface show no sign of acid erosion.


Is the sun too small to self-ignite? I recently heard in a discussion that the sun is not massive enough to self-ignite via core-pressure. It simply has not enough mass to "generate" the gravitational force needed for that.

The reason why the sun is still burning is that there is a really small chance that two Hydrogen-Atoms fuse without needing the activation-energy/the activation-energy being provided by a random effect like High-Energy-Particles. Even though the chance is really low, there are so many Hydrogen-Atoms that overall the chance is actually high (law of big numbers). Once some fusion started it provides enough energy to fusion more Hydrogen, leading to our sun shining.

The person who presented it only stated that there are sources, but sadly the ones I found from his material didn't support this (nor did they deny it). The sources where about other topic related to stars. I can't ask the person himself as he already changed the university.

Stan has essentially answered this in his comment, which I will attempt to spell out a little more laboriously.

The significant majority of our Sun's energy output comes from the proton-proton chain.  This was advocated by Eddington back in the 1920's, but at that time your basic concern was a very real and major problem.  Objects with like electrical charges repel each other.  In particular, protons will repel other protons since all protons have a positive charge.  What they knew of the sun then indicated that the core was far too cold for protons to overcome this repulsion—at least not even remotely close to the rate that was obviously necessary to produce a brightly shining sun.

With the development of quantum mechanics it was determined that a process known as quantum tunneling would give two protons a non-zero probability of "overcoming" this repulsion.  But not in the sense of they just suddenly gain enough energy (hence why "overcoming" is in quotations).  Instead, in the sense that the "fused together" state has the same energy as the "just about to repel each other before fusion can occur" state, and they just randomly switched from the latter to the former, despite every intermediate stage between the two of them requiring more energy than is available.  That this is possible is one of the many non-intuitive features of quantum mechanics, and I think it would be beyond the scope of this question (and site) to try to get much more precise.

Still, even this failed to explain why our sun was very obviously fusing atoms to the extent that it was.  If you fuse two protons together, you are left with an incredibly unstable state: the diproton.  As soon as one of these forms it pretty much immediately breaks up into two distinct protons.

In the late 30's Hans Bethe (a man who eventually won the Nobel prize and was a part of the Los Alamos team that developed the atom bomb) proposed that yet another random quantum mechanical event would save the day: beta decay, a feature of the recently discovered fundamental force known as the weak force.  In this situation one of the protons in the diproton undergoes a beta decay into a neutron before the diproton separates, at which point you have a stable deuterium nucleus: one proton, and one neutron.

That so many unlikely events have to occur for two protons to produce one deuterium is the reason why the sun's lifetime is approximately 10 billion years.  That the sun shines as bright as it does is, as you suggest, a sheer numbers game: there are truly prodigious numbers of protons in the sun's core, so even though the proton-proton chain is really unlikely, we have so very many chances at it that there is a great many successful fusion events.

Basically every other stellar fusion reaction proceeds much more rapidly than those that produce deuterium.  You may be familiar with the hydrogen bomb, which fuses its fuel very rapidly.  One of the key differences (and there are many) between the hydrogen bomb and a star like our sun is that the bomb is fusing together deuterium (and tritium) into helium, and is not using the proton-proton chain.  Stars have to produce large supplies of deuterium directly, and this is a lengthy process that consumes roughly 90% of a star's life—even for very massive stars, whose lifespans are orders of magnitude shorter than our Sun's.

(Note that very massive stars are rather more complicated beasts, in fact, but this is again beyond the scope of this question)


Evidence of CMB redshift Are there any known spectral lines shifted by ~1100? If not, then how certain is mainstream that the CMB has a redshift of ~1100? All I see is a blackbody radiation curve void of spectral lines.

The redshift of the CMB is not measured, it is calculated.

The brief explanation is that as the universe expands and cools, it becomes energetically favourable to form bound atoms. The reduction in free electrons allows the universe to become transparent and photons escape as a blackbody radiation field and eventually form the CMB.

The temperature at recombination, $T$, can be calculated using well understood physics (see https://en.m.wikipedia.org/wiki/Recombination_(cosmology)) and if the current temperature $T_0$ is known, then the redshift is simply given by $ z = T/T_0 -1$.


How to estimate my eyepiece FOV? As a follow-on to my question about seeing Jupiter and Saturn at the same time in my scope, I'd like to ask how I might determine the Field of View (FOV) of my eyepieces. More specifically, I'd like to determine their aFOV (Apparent Field of View?) because that is a configuration setting in Stellarium when you are using ocular view and would like to specify the specs of your eyepiece.
I have two eyepieces that came with my AstroMaster114 telescope. One is 10mm and the other is 20mm. Beyond that, I have no idea what type they are (Plössl? Ortho? Erfle?).
I'd also like to understand how this FOV/aFOV might be impacted by the use of my Celestron #93640 2x Barlow adapter.
Here are pictures of the eyepieces:
Celestron 20mm Erecting Eyepiece

Celestron 10mm eyepiece

Random references from the internet:

Telescope Field of View - How much can you see?
Determine Your True Field of View
What is a Field of View in Astronomy and Why is it Useful?
Celestron YouTube video: Absolutely Everything You Need to Know about Astronomical Eyepieces & Visual Observing

As long as we need moderate accuracy only, there are several ways to do this.
The stated focal length of your eyepiece is approximate, 10 mm might really be 9.6 mm or 10.4 and they simply didn't bother with the details, each little lens' focal length inside the eyepiece will have some manufacturing tolerance. If your telescope's focal length is stated to be 1000 mm it might be 960 mm or 1040 mm and there'd be no reason for them to measure every mirror and give you it's own focal length to several digits of accuracy. So let's call these methods estimates +/-10%.
Examination of the eyepiece itself.
At the "input" end of the eyepiece there will be a field stop, an aperture that defines the your field of view (FOV) in the sky. It's likely a metal ring with a circular hole in it.
If the ring has a diameger $d$ of 10 mm and your telescope's focal length $f_T$ is 1000 mm, then you can ignore all of the other math and say that your FOV is $d/f_T$ = 10/1000 = 0.01 radians or 0.57°. Technically it is
$$FOV = 2 \arctan \left(\frac{r}{f_T} \right)$$
where $r=d/2$ is the radius of the hole. That doesn't matter much here but it does next.
Since your question asks for the apparent FOV of the eyepiece, you can get it by remembering that the plane of the aperture is what the eyepiece of focal length $f_E$ focuses on with a conjugate at infinity (for your eye to refocus to your retina), so the eyepiece's apparent FOV (aFOV) is given by
$$aFOV = 2 \arctan \left(\frac{r}{f_E} \right).$$
IF for some reason your eyepiece didn't have an aperture stop, you could bring a ruler into focus when looking through it and trying to focus your eye at infinity (perhaps by keeping both eyes open and focusing across the room with the other eye) and count the number of millimeters visible across the diameter. This doesn't sound easy though.
Looking through the telescope
At night you can measure the time it takes for a star to move across the FOV of your eyepiece and get the  real FOV of the system, then divide by the telescope's focal length $f_T$ to get the apparent FOV (aFOV). The Earth rotates 360° degrees on its axis every 23 hours 56 minutes 4 seconds and change (86164 seconds), so at a declination of 0 stars move 0.00418°/sec or 0.251°/minute. If you use something at a different declination, just multiply those speeds by $\cos(dec)$ to slow it down.
This gives you the real FOV of your system, again multiply by the magnification ($f_T/f_E$) to get aFOV.
You can also just look at something of a known or measurable or estimatable size far away, like a house or yardstick or meterstick to estimate the system FOV and multiply by magnification to get eyepiece FOV.


Motion of the sun as observed from mercury I recently found this animation which shows the motion of sun as observed from mercury. It seems as if the sun stops in between, retraces a bit and then continues towards the west. What is the reason for this strange motion? Is it because of variation in the distance between mercury and the sun? Or is it because a day on Mercury is longer than Mercury's year?

Both effects combined.

Day being longer than year would just have retrograde motion of the Sun on sky, but no direction change.

Variation of the distance alone happens on Earth, and we have no such effect.

But the combination of both factors, in the precise amount they have on Mercury, makes this effect happen.


What is driving the expansion of the Universe? Is it just the momentum from the Big Bang? If yes, how can it account for the acceleration of the expansion?
Why isn't the collective gravitational force slowing it down?
What part does dark energy play in the expansion?

As far as I understood it, no one really knows. It is observable that the universe is expanding with increasing speed, therefore there has to be a form of energy driving it. We can meassure (or at least try) the amount of energy and find $\Omega_\lambda = 69.11\%$ (see $\lambda$CDM-Model)

To your questions:

1: I wouldn't imagine it as Big Bang momentum, for it is also changing. After the Big Bang, there was (or at least had to be, to make our comological model work) a period of inflation, meaning rapid expansion, which then slowed down again. I can't think of an explanation which includes some kind of finite momentum.

2: On small scales (speaking of galaxies), where densities are high, expansion doesn't act. Gravity is working against it.

3: It is responsible for the expansion. How exactly is not understood. (Little personal comment form my side. If a physicist calls something 'dark', it mostly likely means it is not understood)


Why do satellites arcs end abruptly when observed from Earth I recently got an app that let me track the ISS. I noticed that during the time it's visible, it disappears before reaching the horizon, and sometimes reappears soon after for a bit.

Can somebody explain to me how the ISS and other satellites orbit? Specifically, what makes their visible period such a narrow one, that doesn't span all the way down to the horizon? (which is what my layman brain expected)

Because satellites are only visible when they are in sunlight, they are not visible when they go into the Earth's shadow. The app most likely predicts where this occurs and ends the arc.

In other words, it does not make sense for an observer to look for a satellite when it is not visible, so there is no need to draw the path when it is in the shadow.


Computationally intensive technique to search for TNO's within the noise of historical photographic data? I remember reading recently about a technique to search through lots of old survey images to try to identify trans-Neptunian objects that were so dim that they were "lost in the noise" in single exposures and so have here-to-fore escaped detection.
This was a few months ago I think, but now I can not remember where or exactly when.
From what I remember it was a computationally intensive technique where images were effectively overlayed with offsets determined by hypothetical test orbits. In other words suppose a field is photographed one a year for 10 years. 1 or 2 sigma noise-like dots of diffraction-limited size will be ignored in any one image, but if many of them show dots that overlap when images are shifted by 0.126 degrees per year, then this may be due to a real TNO at 200 AU. ($\text{1 year} \times (\text{180} / \pi) \times \sqrt{GM/a^3}$)
Question: What was this computationally intensive technique to search for TNO's within the noise of historical photographic data that I'm remembering? Where was it recently published or at least written about?

Dark Energy Survey data have yielded several TNO discoveries.
Bernardinelli et al. 2020 describe their method in detail.
First they compute a trial orbit from images where a transient object was detected. Then

Final validation of the reality of linked orbits uses a new "sub-threshold confirmation" test, wherein we demand the object be detectable in a stack of the exposures in which the orbit indicates an object should be present, but was not individually detected.

Their figure 10 shows a candidate object passing that test; figure 11 shows one failing.
Rice and Laughlin 2021 use a baseline subtraction and shift-stacking technique to search TESS data for TNOs without prior detections.
After verifying that they recover a few known TNOs this way, they list a few unknown candidates for further observation.
If neither of these is what you're thinking of, ADS's References and Citations links may help you find it.


How can white dwarf form Oxygen ? (Temperature problem) I’ve got a question about white dwarfs and oxygen.

I read in a book that a temperature of 100 million degrees is required to fuse Helium in the core of a red giant. The Helium fuses into Carbon by the triple-alpha process.

It’s also written that, after a temperature of 350 million degrees is reached, the core of a red giant ceases to be degenerate. So, the core is able to expand and its temperature is controlled. The core won't reach a temperature of more than 350 million.

But my teacher has said that a temperature of 600 million degrees is needed to form Oxygen from Carbon: Carbon fuses into Neon and by photodisintegration gives Oxygen.

So, how can there be Oxygen in white dwarfs, if we don't reach a temperature higher than 350 million degrees? By what process can oxygen be formed in a red giant?

Some oxygen is produced during CNO cycle processing of hydrogen, starting with carbon nuclei. Oxygen is also produced by alpha capture onto carbon nuclei at temperatures well below 350 million K.

Both of these occur in and around the cores of low mass stars before they become white dwarfs. Neon production is not required.


Do all massive stars explode? I've read a few articles written in $2008$ that some stars which have enough mass just collapse into black holes without a supernova, is this proven?

There are a variety of models of stars collapsing into black holes without proper supernova explosions. These are often called failed supernovae or direct collapse black holes - although the former seems more common. Failed supernovae can happen if the initial shock wave rebounding from a collapse loses enough energy, causing it to fizzle out. Neutrinos produced in the core often supply this energy, but a small enough burst may not be enough to keep the shock going. There might then be a simple collapse.

The minimum mass for the progenitor of a failed supernova is unknown (and as Rob Jeffries pointed out, this is metallicity-dependent). Some early models (e.g. Fryer (1999)) found that stars above $40M_{\odot}$ could produced failed supernovae, while newer ones suggest that this could be as low as $25M_{\odot}$. This lower limit would include a sizable portion of red supergiants - which could be a possible solution to the red supergiant problem.

A number of searches, largely targeting supergiants, have been carried out. Just a few are

Adams et al. (2016), using the Hubble Space Telescope to examine N6946-BH1, a disappeared red supergiant of $25M_{\odot}$.
Reynolds et al. (2015), looking at past Hubble data. They found one possible candidate, a $25\text{-}30M_{\odot}$ red supergiant in NGC 3021.

So far, we have no definite confirmation that these stars were failed supernovae. However, they do seem plausible candidates, and there are many more possibilities out there. One group proposed monitoring one million supergiants in an attempt to catch failed supernovae. Perhaps they'll be successful.


Question regarding the path of sun crossing zenith before the summer solstice I stay in Bangalore, India (12.9716° N, 77.5946° E). For my latitude, the Sun crosses zenith (the highest point in the sky) on 24th of April every Summer, which happens to be about 2 months prior to the June solstice (summer solstice to us here in the Northern Hemisphere).

Shouldn't the Sun be reaching its highest point on the longest day, i.e. on June 21st, instead of April 24th?
Doesn't the Sun take the longest arc across the sky on the day it passes through zenith?

For reference, the angle at which the Sun rises/sets in my location is about 13 degrees, which happens to be almost vertical to the horizon.

Question 2: Doesn't the Sun take the longest arc across the sky on the day it passes through zenith?

Answer: There are two effects. (In the remaining description, I will be referring to an observer in the northern latitude.) As the Sun moves northward from 0 degrees declination, the radius of the circle that it travels is smaller. In other words, the total length of the 20 degree declination circle is smaller than the 10 degree declination circle which is smaller than the 0 degree declination circle (the celestial equator). But at the same time, the fraction of the declination circle that is visible above the horizon is greater. At some combination of declination, these two effects result in the maximum length of the declination circle above the horizon. If the declination is north or south of that optimized solution, the length of the arc is smaller.

Skipping the calculations for now and going right to the solution, here is a graph showing the length of the arc for various latitudes and declinations. This graph plots the results relative to the length of the arc along the celestial equator which is 180 degrees regardless of your latitude (ignoring refraction and the Sun's diameter).

If your latitude is between 0 and 23.4 degrees, the longest arc is before the Sun reaches the zenith. For example, an observer at 20 degrees latitude witnesses the longest arc when the Sun is at about 13 degrees declination which is 7 degrees before the Sun reaches the zenith. If you live north of 23.4 degrees, the Sun is never overhead. North of 30 degrees, the longest arc is when the Sun is highest in the sky.

I can go into more details and the math, but I do not know if I should do it here or maybe in a new question.


Adiabatic versus convection I am reading about the interior temperature structure of Jupiter. It says in various texts that there are regions in Jupiter's atmosphere which are well-approximated by a dry adiabatic lapse rate.

Does this mean that in these regions there is convection happening? Is adiabatic just a general synonym for convection?

An adiabatic process in physics is at first a process where a system of limited size does not exchange any heat with the outside world, or where the energy exchange timescale is very long compared to the interesting timescale under which the adiabatic process occurs.

It is also loosely used as a term meaning 'constant in the framework of a certain approximation'. One sees this often used in quantum mechanics where it essentially means 'quasi-static', or in plasma physics where particles can gyrate in a field gradient. If the field gradient is much smaller than the gyration radius, such that every individual particle gyration experiences a near uniform field, then the particle experiences an adiabatic change of the underlying field.

Coming back to planetary/stellar atmospheres, here the two concepts are used interchangeably. Should a piece of atmosphere experience a super-adiabatic temperature gradient, then this situation is usually unstable towards buoyant instabilities. As soon as convective motion sets in, the temperature gradient will be brought down towards the adiabatic one.
This is a very general process and happens in stars as well as planets, if no other physics is present. Thus it is used interchangeably, while people using it are aware that this is a bit sloppy.


Number of photons in universe? I'll take number from a star in unit time and extrapolate I'll take number per average star, multiple by appropriate numbers , star/Galaxy, galaxies/universe, 14 billion years, and try to figure their location.

I was looking for a rough number and a distribution function.

Extending Zephyrs comment as community wiki

See https://physics.stackexchange.com/questions/196366/number-density-of-cmb-photons

In this response, the number of photons per cubic meter is estimated as $n_\lambda = 10^8$. Since the observable universe has a volume of the order $10^{80}\ m^3$, there are about $10^{88}$ photons in the observable universe.


Which stars have been named after astronomers? Most apparent stars have ancient names. But some of the newer (since 200 or so years) discoveries have been given the name of some prominent astronomer. Kapteyn's star is one example, and there are several more in that Wikipedia article. Edward Barnard et cetera. But there aren't too many, are there? It seems hard to get ones name attached to a star, or am I missing a bunch?

Kapteyn and his tie knot matching star

Astronomy was never easy, there's always more to see. Star namer van Biesbroeck in action with his flaring tiny red dwarf.

You probably need someone better versed in the history of astronomy than me, but I'll give it a stab.

As you've already noticed, the Wikipedia article on stars named after people has some entries, and that's probably about it.  In terms of modern naming procedures, we tend to use catalogue names. Unlike, say, asteroids, there's no systematic way of naming stars after people or anything else, so stars usually continue to be referred to only by catalogue number. Those rare stars that are named after people (or something else) are usually because they are now somehow historically associated with that system.  Unless, of course, the catalogue is named after someone!  Another exception are those very bright stars with pre-telescope names, but those aren't named after people (e.g. many scientific papers refer to Betelgeuse as such).

I guess in short, there's just no standard modern naming procedure by which a star can end up with a person's name, so few exist, and probably most are listed in the Wikipedia article.


Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology I was reading Appendix F of Steven Weingberg's book "Cosmology". In this Appendix he works out the perturbations to a cosmological fluid described by non-relativistic hydrodynamics and Newtonian gravity.

It turns out that the first order perturbations satisfy,

$$
\frac{\partial \delta \rho }{\partial t } + 3 H \delta \rho + H \vec{X} \cdot \nabla \delta \rho + \bar{\rho} \nabla \cdot \vec{v} = 0, \qquad \tag{1}
$$

$$
\frac{\partial \delta \vec{v}}{\partial t } +  H \vec{X} \cdot \nabla \delta \vec{v} + H \delta \vec{v} = - \nabla \delta \phi, \qquad \tag{2}
$$

$$
\nabla^2 \delta \phi = 4\pi G \delta \rho. \qquad \tag{3}
$$

Weinberg applies the following Fourier transform to these equations,

$$ f(\vec{X},t)  = \int \exp \left(  \frac{i \vec{q} \cdot \vec{X}}{a} \right)  f_{\vec{q}}(t) \ \mathrm{d}^3\vec{q} $$,

where $f(\vec{X},t)$ is a place holder for $\delta \vec{v}, \delta \rho, $ and $\delta \phi$.

The resulting equations he gets are,

$$
\frac{\mathrm d \delta \rho_{\vec{q}}}{\mathrm d t } + 3 H \delta \rho_{\vec{q}} + \frac{i\bar{\rho}}{a}\ \vec{q} \cdot \delta \vec{v}_{\vec{q}} = 0 \qquad \tag{1'}$$

$$
\frac{\mathrm d \delta \vec{v}_{\vec{q}}}{\mathrm d t } +  H \delta \vec{v}_{\vec{q}} = -\frac{i}{a}\ \vec{q}  \delta \phi_{\vec{q}} \qquad \tag{2'}$$

$$
\vec{q}^2 \delta \phi_{\vec{q}} = -4\pi G a^2 \delta \rho_{\vec{q}} \qquad \tag{3'}$$.

For the most part these new equations can be obtained by making the substitution $\nabla \rightarrow i \vec{q}/a$.

My question : There doesn't seem to be any terms in the transformed equations which correspond to the terms $ H \vec{X} \cdot \nabla \delta \rho$ and $H \vec{X} \cdot \nabla \delta \vec{v}$. Weinberg makes no comment about their absence. Is anyone aware of a legitimate mathematical reason for these terms to disappear in the transformed equations?

The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question.

The $a$ is the occurring in the Fourier transform is the scale factor which has a time dependence. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,

$$
\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q}
$$

$$
 = \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q}
$$

$$
 = \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q}
$$

$$
 = \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q}
$$

The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel.


When examining an exoplanet's atmosphere is the star's emission spectra or planet's light used? My understanding of the main method we use to figure out an exoplanet atmosphere composition is that when a exoplanet transits their sun, visible light passes through the planet's atmosphere, and absorbed by the elements in the atmosphere. We then observe these absorption lines to figure out the chemical composition right? Or am I getting some details wrong here?
My main question is, does the light that passes through the planet's atmosphere is the sun's light or the sun's emission spectra?
Also will different stars starlight cause different absorption lines? As in say some planet X orbiting a M class star as opposed to the same planet X orbiting a G class star. Considering both the stars are different but the planet are same, would the absorption bands of the planet be different in each case?
References

ESA: Starlight yields clues to exoplanets’ atmospheres
Lecture by  Michael Richmond: Spectroscopy of exoplanets/spectra/spectra.html

If you look for a simple answer by easy principles and not technicalities this would be the picture...
When the exoplanet transit between our point of sight and its star, its disk block part of the light. This results in a dimming of the star light we receive.
The latter can be analyse as we do for light of any source. It is well approximated by a black body spectrum with dimmer lines corresponding to the absorption lines of the elements/species present in the star atmosphere.
If the exoplanetary disk that block the light has an atmosphere as well, the constituents of the latter will absorb their typical lines and this results on the appearance of dimmer and/or new lines in the star spectrum as received.
In principle there could be a scattering effect of the exoplanetary atmosphere, too. For instance an observer on the moon during a sun eclipse would observe the sun fainting overall, the absorption lines due to terrestrial air, and a global different spectral distribution toward the red due to scattering, also due to air.
However, in case of far exoplanetary systems, it is already astonishing that we can do the line analysis as for all the above just happen to the little portion of the star flux that is intercepted by the planet and at same time by our telescope.

My main question is, does the light that passes through the planet's atmosphere is the sun's light or the sun's emission spectra?

Emission by the star atmosphere to us or to the exoplanet is mixed with the proper photosphere emission. Practically, it is the star light whatever you like to call it. It has not really an influence on the difficult above measurement. They are difficult right because we must see tiny changes relatively to a much brighter and bigger (angular) sized source.

Considering both the stars are different but the planet are same, would the absorption bands of the planet be different in each case?

No. In principle there could be a difference in the easy of detection, but the absorption lines are typical of the absorber not of the excitation source.
Obviously we can't search for absorption happening in the UV using just visible light. But the very broad emission of stars provides quite the same spectral coverage in term of wavelengths.


Moving light in night sky Today at about 2 am I have spotted a light very similar (perceived speed and apparent brightness, motion due East) to ISS. At very first I thought to have spotted the station but immediately I doubted about as I feel that the observation time is too far from sunrise. ISS is indeed ruled out as there is not email from the spot the station service in my inbox. Can satellites orbiting at higher orbits be see even in deep night?  I feel that answers is yes based on simple geometry. However what could reflect so much light? I think there are not objects with reflecting surface so big to be compared to those of ISS...

If you are in the Northern Hemisphere, it is currently close to midsummer. The Sun is never very far below the horizon in the North, even in the middle of the night. There are plenty of satellites that are visible even late into the night. Heavens Above

Most are not nearly as bright as the ISS can be. However, against a dark sky a satellite may appear brighter. Lots of the objects that are visible then are actually rocket bodies. They often have a higher orbit than functioning satellites.


Metallicity of Celestial Objects: Why "Metal = Non-metal"? Metallicity of objects refers to the amount of chemical elements present in it other than Hydrogen and Helium.

  Note: The other elements may or may not be actual metals in the true sense of their defintion.

But why did astronomers use such a term as metallicity? What is the history and cause behind coining such a term that may be (or rather actually is being) confused with metal-content of that celestial object ?

Is there any scientific objective or explanation to this? I do not believe it is just a random acceptance, but perhaps there is a not-so-solid motive behind doing this. If there is, what is it?

To first order, the relative abundances of the heavier elements to iron (for instance) are constant. So the metal content of a star is shorthand for the content of any element heavier than He. (NB: we now know this is not true in many circumstances and elements can be grouped by synthesis process - for example we can talk about "alpha elements" - O, Mg, Si etc. produced by alpha particle capture; or s-process elements - Ba, Sr etc. produced by the s-process. We know that the ratio of say O/Fe gets larger in more "metal-poor" stars, but Ba/Fe gets smaller. So talking about a single "metallicity" parameter only gets you so far, and the truth is more complex (and interesting).
The next point is why they are referred to as "metals" rather than another term like "heavies" or something. I would guess this is down to a bit of history and the fact that initial abundance analyses in stars were (and still are on the whole) done in the visible part of the spectrum (e.g. in the early part of the 19th century by Hyde Woolaston and Fraunhofer). The most abundant elements heavier than He are in fact not metals; they are Oxygen, Carbon, Nitrogen and Neon. However, the signatures of these elements are not at all obvious in the visible spectra of (most) stars. Whereas the signatures (absorption lines) of elements like Fe, Na, Mg, Ni etc., which decidedly are metals, are often very prominent.
Thus, there is a reason and some history behind the name "metals". It is that apart from hydrogen and helium, the metallic elements have the most prominent features in the optical spectra of stars, whereas in most stars the signatures of the more abundant non-metals are hard to see.


Singularity at the time of origins According to the big Bang Theory, the universe have started from a point and it describes how the universe expanded from a very high-density and high-temperature state.The initial singularity was a singularity of infinite density thought to have contained all of the mass and space-time of the Universe. If it was so dense,  why didn't it engulf the other parallel universes just like a supermassive black hole?

If there are no other universes,  why the existing black holes didn't suck the entire universe yet?

First thing to mention is all our limited knowledge (best guesses) from back than only goes back to the fraction of time just after the big bang not the creation moment itself. Moreover you have to remember that spacetime itself was also created at that moment as well as all the stuff in it. This means that using the term singularity as we understand from the core of a blackhole, which is to say an infinitely dense and infinitely small point/puncture IN the fabric of space doesn't really marry up. The universe no matter how tightly packed conditions were at this early stage had no space to warp/puncture into.

The universe has no centre, it began everywhere, simultaneously and instantly with all matter distributed uniformly and in equal density. Without a central point all matter was pulling evenly on its self in all directions so wouldn't coalesce at a single point and collapse down as it does in a black hole.

And as to your mention of supermassive black holes sucking in parallel universes "they don't"

Hope this helps


Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as 0 (zero)? Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as, or treated as 0 (zero)?

I think one of the sciences that most precision is required is astrophysics.

I'm aware of my hardware capacity to lowest floating number, but that's not what I'm asking about.

Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as 0 (zero)?

No, there isn't.

Astronomy overlaps with cosmology and there can be some pretty big and small numbers.

However most physical parameters also have units and the units chosen affect the numerical values.

A great example is the Jansky:

  The jansky (symbol Jy, plural janskys) is a non-SI unit of spectral flux density, or spectral irradiance, used especially in radio astronomy. It is equivalent to 10^−26 watts per square metre per hertz.

In some cases one might even imagine going the other way and using L☉/ℓP² myr (solar luminosities per square planck-length per million years) which would require numerical values about 1x10^-110 times smaller.

Instead, astronomers and other scientists usually decide is something is small enough that it doesn't matter within the context of the current discussion. THey use words like negligible and small to indicate they're going to ignore it.


Having some weird issues with my tracking (Advanced VX) I need your help! I am having some weird tracking issues with my tripod (Advanced VX). When I setup and calibrated my scope yesterday I took this picture of NGC6888 (90s exposure):
As you can see this image looks quite good tracking wise.

However due to unknown reasons it seems that 1 in 3 images is always bad and has tracking issues, like this one:
This movement always occours in the same direction, it seems that for some reason one axis of the mount is moving faster periodically, causing these tracking issues.

Funny is that is seems to be occouring periodically. I always get 4 images good, 2 images bad, 4 images good, 2 images bad, ...

I tried fiddeling around with the anti-backlash settings (I never touched them before) but that didnt fix it. My current anti-backlash settings are: RA +- 55, Dec +- 60

Does anyone know where those tracking issues could come from and how to fix them? Note I also have a 5s mirror-lock so it should not come from vibrations of the camera.

Looks like you have a periodic error issue. The worm period on the AVX is about 9mins I think. The issue has a period of 90sec * 6 = 540sec (9min).


How do we know that the intrinsic brightness of Cepheid variables corresponds to their period? As I understand it, beyond parallax the next way of estimating the distance to astronomical objects is by using a relationship between the period and intrinsic luminosity of Cepheid variables.  How do we know that relationship?  Are there some close enough that we used parallax to validate it, or is there a theoretical reason, or something else?

This relationship was discovered empirically by Henrietta Leavitt by comparing the apparent magnitudes of stars in the Magellanic clouds. Since these clouds are far away (200,000 ly) and relatively small (7,000 ly), the difference in distances to the 47 different Cephid variables she observed could account for only a small difference in brightness.  By plotting the brightness against the period she noticed that there was a relationship, cephids with longer periods were brighter.

As you guessed, modern measurements are made via parallax for close stars. The Hipparcos satellite compiled a vast list of parallaxes, and by measuring a large number of periods, we can calculate the relation.


Could the KIC 8462852 (Alien megastructure star) be explained by orbiting gas clouds? I was wondering about the light dimming of the 'alien megastructure' star.
From what I understand, for it to be a planetary body, it would have to be some 20 times the size of Jupiter.
Is it possible that it could be caused by some huge, dense gas cloud, orbiting around the star?

No.

The light curves don't look at all like anything transiting. The ingress and egress profiles and the timing are way beyond that kind of explanation. Not planets, not comets, not clouds, not alien super structures.
No infrared light from dust or gas has been detected, as they should've been if the starlight had heated small particles. And since this is an F star with about 50% larger diameter than the Sun, which contains 99% of the mass in our Solar System, you can imagine how huge any comet/dust/gas cloud has to be to cover 22% of it.
No brightening of the starlight has been detected, as this allegedly huge"cloud" orbits behind the star and should reflect light to us. Shouldn't those huge icy comet clouds reflect some starlight?
Jupiter is about as large as a planet can get, even with 10+ times more mass, its volume would not increase by much. Gas giants differ much more in mass than in volume. An object 20 or 200 times more massive than Jupiter would be a roughly Jupiter sized red dwarf star, and that is not what has been observed at "Tabby's star".

The anomalies correlate strictly with the telescope's orbital period and orientation and are obviously a man made artefact because of some unforeseen malfunction of the instrument, nothing astrophysical is involved.


Might Oort cloud comets be exchanged between solar systems? Considering their distance from their parent stars, might Oort cloud object such as comets be exchanged between passing stars (assuming that other stars have similar Oort clouds)?

You can exclude the 'considering the distance' piece - of course Oort Cloud objects could transfer between different gravitational fields.

However what is it you think will make this transfer? Without some sort of gravitational impetus why would one of these objects leave the solar system? And if you do manage to slingshot one out of the solar system at a high enough speed to exit the Sun's gravity, remember that most directions end up very far away from any other solar systems.

Tl;Dr sure, but not very likely


Measurement uncertainty of the computed center of gravity from pixel image Establishing the position of celestial bodies (stars, and other objects) is based on imaging technology. Depending on the resolution of the imaging sensor used, an object is identified by a group of neighboring (bright) pixels. These pixels surround the center of gravity of the detected celestial body.

In image processing, the central position of a round object (a sphere) is generally determined by computing the center of gravity of the pixels surrounding the detected object. The weights are the intensities of the pixels in the vicinity of the brightest (central) pixel.

My question: how is the variance of this center-of-gravity estimate being calculated, within astronomy?

Every pixel value $S_i$ on the detector at $\vec x_i$ has some error $N_i$: CCDs for example have a background noise $N_\text{bkg}$ from read-out electronics, thermal noise, and sky background, plus a Poissonian photon noise $N_S=\sqrt{S}$.  In many cases this noise follows a Gaussian distribution reasonably well.  After subtracting the background, a position measurement

$$ \vec x = \frac{\sum{S_i \vec x_i}}{\sum{S_i}} $$

has the uncertainty

$$ \operatorname{std} \vec x
= \left[\sum_i \left(\frac{\partial x}{\partial S_i} N_i\right)^2 \right]^{1/2}
= \frac{\left[\sum_i
    \left(
      \sum_j\frac{(x_i - x_j) S_j}{\sum_k S_k} N_i
    \right)^2
  \right]^{1/2}}{\sum_k S_k} $$

$$ N_i^2 = N_\text{bkg}^2 + N_{S_i}^2 $$

using Gaussian error propagation ... I hope I have run the math correctly.  The way I've written it is a bit weird, but shows one important property of this method: If you look at the sum over $j$, you can see how the noise of a pixel is basically weighted by the pixel's distance to the centroid result.  The same error on a pixel value has more impact if the pixel is farther away from the centroid.

Better methods account for the uncertainty $N_i$ on pixel values in the first equation already.  You can do this by introducing additional weights in your centroid or go to fitting models, which is the “usual” method in astrometry, as far as I have seen.

This more involved approach to measuring positions uses the point spread function $P(\vec x_i - \vec x_\text{obj})$ of the instrument under given observing conditions.  The model can be an approximation, e.g. a Moffat function, or an empirical model build from unconfused bright stars in the image, e.g. a spline interpolation.  For point sources, the typical least-squares fit of position and flux of the model to an image easily produces results close to the statistical optimum of parameter estimation in terms of uncertainty.  With our modern computing power the easiest way to get the uncertainty for a given model and data noise is often a bootstrapping algorithm.

Of course extended objects require a bit more work in the model, for example assumptions on their shape, as you have given in your question.


Why does this large Newtonian telescope's front cover have two or three holes in it? The Michael Bernardo video How to use an Equatorial Mount for Beginners shows a large Newtonian telescope on an equatorial mount.
The cover of the telescope's large aperture shows what looks like three holes each with its own removable sub-cover, but I might be misunderstanding what I'm looking at, perhaps the "central object" is just a handle to rotate the cover when the other two holes are open.
But that raises the question "why two holes" rather than just one. I'm pretty sure that these are not used for this kind of thing but it would be great to know how they are used!

What are these?
These are used to reduce the aperture of the instrument. While you see three in this image, likely only one is removable.
The instrument appears to be a Newtonian reflector. Which means it would have a secondary mirror near the front (and center). So the thing that looks like a knob in the center of the telescope cover is probably just for grip (and not likely removable).
One of the off-axis caps is probably actually a cap. That can be removed.  The other thing that looks like a cap is typically non-removable. It is a holder for the cap on the opposite side (the one that can be removed).
Purpose
If you observe a bright object. Such as the Moon. You can use this to reduce the total amount of light.
Solar filters are large aperture telescopes will often employ the same idea. The Sun is plenty bright. Collecting "enough" light isn't the problem. By using a solid cap with an off-axis (off-axis only because it is necessary to avoid the central obstruction in the telescope) can help limit the total energy entering the telescope to avoid over-heating internal components and damaging the instrument.


Is something wrong with my luminosity calculation? A few million years after a white dwarf forms, its surface temperature reaches $100000\text{K}$, while its radius is $0.01R_\odot$. Would this mean that its luminosity is $\Big(\dfrac{100000}{5778}\Big)^4 \cdot 0.01^2 = 8.972 L_\odot$? If so, why is it that way (I do understand the Stefan-Boltzmann law), and does that mean that all neutron stars are less luminous than white dwarfs?

Yes it would.
It is that way because the effective temperature is defined to be $(L/4\pi \sigma R^2)^{0.25}$.
The radius of a neutron star is about 10 km $(1.4\times 10^{-5}R_\odot)$. They are born with surface temperatures of around $10^8$ K. The coldest white dwarfs have effective temperatures of about 3000 K.
The luminosity ratio is
$$ \frac{L_{\rm NS}}{L_{\rm WD}}  = \left(\frac{10^{8}}{3000}\right)^4 \left(\frac{1.4\times 10^{-5}}{0.01}\right)^2 = 2.4\times 10^{12}\ .$$
Thus the most luminous neutron stars are (briefly) much more luminous than the coldest white dwarfs. They would remain more luminous until they cooled to $\sim 10^5$K, which could take as long as $10^7$ years (Lattimer & Prakash 2004).


How do astronomers distinguish the spectral lines? As a black body a star emits a continuous spectrum of light and other radiation. I've learned they are capable of knowing of what elements the stars are made of because of the spectral lines. But how do they distinguish those lines from other radiations?

The lines show as either peaks above the black body emission (for emission lines) or troughs below the black body emission (for absorption lines). Though stars more often have absorption lines and nebulae emission.


Beta Andromedae (Mirach) and distances mentioned in original Cosmos series from 1980 In one of the episodes from Carl Sagan's show Cosmos he explains that Beta Andromedae is the second brightest star in the constellation Andromeda, and is 75 light years away. The link to the video is here.

I've looked up this star on Wikipedia and it says it's the brightest star in the constellation, not the second brightest, and instead of 75 light years away it says it's about 197 light years away.

I know that this show Cosmos is quite old. It aired in 1980. I understand our measurements are more accurate nowadays, but I was just wondering if anyone knew why the discrepancy with the brightest and second brightest star claims, and also that the Wikipedia article says it's 197 light years away, a difference of about 260% from the 75 light years mentioned by Sagan. That's quite a significant difference.

His information of the distance from us to the center of our galaxy is quite close: he says 30 thousand light years; Wikipedia says 27 thousand light years. As to our distance to Andromeda, he says: 2 million light years; Wikipedia says 2.5 million light years.

Have I completely got the wrong star that I'm looking up?

Looking at the SIMBAD data page for Beta Andromeda shows the source of both the parallax (distance) and the magnitudes. In this case, as it will be for many bright stars, the source of the parallax is the reprocessed data from the Hipparcos satellite, described in this paper. Prior to the launch of the Hipparcos satellite by ESA in 1989, parallaxes were very difficult to obtain and only low precision values were available for the closest stars.

The parallax given in SIMBAD for Beta Andromeda is $16.52\pm0.56$ milliarcseconds which translates to a distance of $\frac{1}{0.01652"} = 60.5\pm2.1$ parsecs or $60.5\times3.26=197$ light years. This small parallax would have been extremely challenging or impossible to measure accurately with the pre-CCD technology before 1980. Errors of several hundred percent were not uncommon. The prior parallax measurement which probably was an improvement on what was available in 1980 at the time of Cosmos is from van Altena et al. 1995. This lists a parallax of $47.7\pm7.9$ millarcsec, a nearly 5x larger error and which gives a distance of 68 light years.

Similarly we can see that the $V$-band magnitude comes from this collection and the difference between Beta Andromeda ($V=2.05$) and Alpha Andromeda ($V=2.06$), the in-theory brightest star in the constellation is only 0.01 magnitude. Measuring a star that bright to that precision was, and in fact still is, quite difficult as most detectors will saturate. So it's not particularly surprising given how close Alpha and Beta Andromeda are in brightness, that the early measurements got them reversed when Beta is in fact (very slightly) brighter.

This high brightness probably means that we will not see a more accurate from the Gaia satellite, the successor to Hipparcos. The Gaia DR2 data release lists a brightness limit of $G\sim3$ which may be improved slightly in later data releases with more sophisticated data processing and treatment of saturated stars.


In which countries are there organised Messier Marathons? Just out of curiosity...

In which countries have there been or are current, organised Messier Marathons?

Thank you for all answers and effort for this!

As far as I have researched, there are really many countries that organize Messier Marathon, but also many astronomers organize it locally only for them and don't publish it on web, so I could write here nearly every country.


Chances of life on other planets, how watertight is the idea that 'As soon as life could occur, it did occur’? I’ve heard it reasoned that ‘Almost as soon as life could occur on Earth, it did. Therefore it seems likely that life develops easily, and thus that there’s probably lots of life out there’.
Is this notion undermined by any of these things?:
– If life develops easily, it’s likely to have developed independently multiple times on Earth (any evidence of this?).
– The fact that life developed soon after particular conditions emerged indicates only that that’s what happened on that occasion. It doesn’t indicate frequency. It might occur once every trillion years under those circumstances.
– Life could happen under many circumstances that life as we know it doesn’t exist. If there’s no evidence it did on Earth under different conditions, how does the reasoning hold?

Yes, early life on Earth does not imply easy life.
Full disclosure: I have a paper on this, Snyder-Beattie, A. E., Sandberg, A., Drexler, K. E., & Bonsall, M. B. (2021). The Timing of Evolutionary Transitions Suggests Intelligent Life is Rare. Astrobiology, 21(3), 265-278.
The mental model most people have is that there is some constant rate $\lambda$ of life emergence on lifeless worlds, so we should expect the probability of life at time $t$ to grow like $1-\exp(-\lambda t)$, with the expected emergence time at $t=1/\lambda$. Earliest life was conservatively 3.77 Gya and oceans 4.5 Gya, so that gives estimates like $\lambda \sim 1.37$ per Gy. One can of course quibble about whether the rate is constant, how long the window of emergence is, and many factors.
But as the question asks, that we observe life early in Earth's history may not be good evidence. Suppose $\lambda = 10^{-12}$ per year, making the probability of getting life over a 10 billion year sun-like star lifespan 1%. Any intelligent observers that evolve will be on one of these rare 1% planets. If they try to guess how likely life is on other worlds they will estimate a $\lambda$ that is much bigger, typically $10^{-9}$ or so based on when life emerged on their world.
But this bias is likely to be much worse, since intelligent observers come about after a long evolutionary chain that may contain several fairly unlikely steps. For most of Earth's history life was very simple, and becoming eukaryotes, land-living or intelligent may have required very unlikely steps with their own low probabilities. Observers will always find themselves on worlds where these steps have been passed, even if they have their own super-low probabilities and the observers are extremely rare lucky cases.
When you run the math on this, you will find that the hard steps get roughly evenly distributed across the habitable time of a planet. If you need 5 hard steps (say: life emerging, photosynthesis, eukaryotes, multicellularity, intelligence) the observers will typically see life show up around 1/6th of the habitable time, and they show up 5/6th of the way to the end.

Random samples of histories where rare transitions happen but a full set of transitions happens before the end of the time interval. Colour indicates the number of the transition.

Probability density of the transitions over the habitable period of the biosphere for 5 very hard transitions.
Getting back to the questions: if we were to discover several independently evolved kinds of life on Earth (or elsewhere in the solar system, even if extinct), then we would have good reason to think that at least the life emergence step was easy and there will be lots of life in the universe. Same for discovering life that can thrive in very different environments.
It might also be that true life emergence is very rare, but then spreads by panspermia across the galaxy. In that case it would seem as if life was easy, but it is just received from elsewhere.
However, later hard steps will still tend to push appearance of life early since there is a need for a lot of time to have the other unlikely things to happen so that the very rare observers can emerge. What we did in our paper is to show how the pattern of steps can be used to constrain estimates of life emergence, but there is plenty of room to argue what steps were truly hard or even when they happened. But I would agree with the original post: early life does not have to imply that life is easy. But the way to know the difficulty of life is of course to look for it, not just do statistics.


Finding Jupiter, Saturn I have a Dobson Omegon with a 640mm focal length and a 102mm aperture.
My finderscope is broken but I can do without.
I have located the planets with the naked eye but I do  I do not see them through my eyepiece.

I can focus perfectly on the moon, using a 20mm eyepiece. (I have a 6mm eyepiece as well.)

How should I proceed?

Always start with the low magnification (20 mm) eyepiece.
If the target is not in the eyepiece field at first, methodically sweep the surrounding area.
Any sweep pattern that makes sense to you is fine.
When the telescope is pointed near a bright planet, stray light may brighten the background slightly.
After you find and center the target, switch to the high magnification (6 mm) eyepiece if needed.

Of course it's much easier to aim a telescope with a working finder.
If your finder is not repairable, most modern consumer telescopes have a dovetail fitting which makes it easy to install a replacement.


Why does the Moon's terminator look "wrong" in this image? There is a picture of the moon in National Public Radio's on-line article Get Ready For Halloween By Watching The Moon's 'Occultation' Tonight. It looks wrong to me - specifically, the brightness gradation near the terminator - or lack thereof.

The image is credited "A waning gibbous moon occultation will be visible Tuesday night in parts of the United States. JPL/NASA" but I wonder if this is true. Something just looks "wrong" here.

In the original article the JPL/NASA contains a link to this page, which currently contains this image also shown below. It looks more like an actual photograph.

And here is a NASA/JPL image from https://svs.gsfc.nasa.gov/4404 for 2016-10-19 04:00 UT, which if I uderstand correctly is not actually a photograph, but simulated from LRO data. It also illustrates that the terminator is expected to be graded from light to dark and contain contrast from shadowing.

above: terminator from all three images. Realistic terminator on the right show graded intensity and strong shadowing from the highly oblique incident light.

The reason that moon image looks wrong is because it is wrong. It is not a real image of the moon $-$ at least the terminator is not real.

The original article you cite has a link just below their image indicating the source of their image of the moon. That source is the night sky planner, hosted by JPL. You'll find the same image on that website, albeit slightly darker (it seems the NPR people lightened up the image a bit).

If you do some more digging, you'll see that the night sky planner got its image of that moon from someone else. Within the html code, they have the image defined as:

<img alt="the moon" src="http://api.usno.navy.mil/imagery/moon.png">

Clearly you can see that the image was taken from the United States National Observatory.

After a bit of digging, I found out what exactly is going on here. The purpose of this site is to show you the current phase of the moon. To do this, they take a single image of the full moon and artificially shade out a region to make it appear as the current phase of the moon. You can see their process here and how it was done by some guy named R. Schmidt. They've broken down the moon phases into 181 images which you can download here, if you're interested.

As you can see, the terminator on that image is wrong because it is not a real image of the current phase of the moon, but rather a computer generated "shading" of the full moon to indicate the current moon phase.


Does weight influence Earth's spin? If put enough weight on a particular point on Earth's surface disturbing the balance between hemispheres, is it possible that the Earth's spin could change like an unbalanced spinning top?

The Earth does spin like an unbalanced top. The Earth's rotation axis is not fixed. It instead moves in a complex manner due to a combination of external torques exerted by the Moon and Sun, a torque-free nutation due to the oblate shape of the Earth, and also due to changes on and in the Earth.
The torque-induced motions are called precession and nutation, distinguished by period. The largest and slowest of these motions is the axial precession. This causes the Earth's rotation axis to trace out a cone over the course of 26000 years.

(source: nasa.gov)
The torque-induced nutations are also cyclical motions induced by the Moon and the Sun. These are much smaller in magnitude and have a much shorter period. The largest of these has a magnitude of about 20 arc seconds and a period of 18.6 years. All other nutation terms have much smaller magnitude and have shorter period.
The torque-free nutation would have a period of about 305 days if the Earth was solid. The oceans, the atmosphere, and the outer core alter this. The Chandler wobble has a period of about 433 days and a magnitude of less than an arc second. Because the Chandler wobble isn't as predictable as are precession and nutation, it's lumped into a catch-all category called "polar motion." The redistribution of water over the course of a year (e.g., snow on Siberia in the winter but not in the summer) results in a yearly component of the polar motion.
There are lots and lots of other factors, all small. Polar motion is observed after the fact.


What happens to the shrinking universe in the presence of the pressure of Hawking radiation? Black holes release Hawking radiation. Now assume that the universe after all collapses back again, bringing all matter to a single black hole. I assume that this would bend the whole space-time of the universe as well, wrapping it around the black hole.

Now what happens to the shrinking universe in the presence of the pressure of Hawking radiation?

Could there be a stable point where pressure and gravity are balanced?

In a shrinking, roughly 3-spherical universe with only a black hole, Hawking radiation should follow a geodesic line and return to the black hole, without excerting radiation pressure to the universe as a whole.

Therefore it's hard to see, how Hawking radiation should establish an equilibrium with gravity.

More feasible seems, that the shrinking universe can prevent the black hole from further evaporation at some point, since evaporated particles and radiation are directed back into the black hole by the overall curvature of spacetime.


Are lunar occultations visible to the naked eye? The Moon occults a lot of stars. But are these events visible to the naked eye? Won't it be blinded out by the crescent even at its thinnest just before sunrise? Did ancient astronomers actually observe them, or did they deduce them theoretically according to orbital calculations? (I suppose occultation during an eclipse is both invisible and unlikely, right?)

Venus, the brightest planet, like the new Moon always stays near the horizon. But could one actually see even Venus disappear behind the lunar anti-crust? A popular medieval islamic symbol suggests to me that they were well aware of this phenomena. Both the Moon and Venus are still symbolically important in their culture. But was occultation directly visible or inferred?

Yes - especially as the occultation does not have to occur on the bright side. Planets may also be occulted (see http://www.lunar-occultations.com/iota/planets/planets.htm)

Incidentally the symbol is not originally Islamic. It was the flag of Constantinople when conquered by the Ottomans and then adopted by them. The Ottomans later became the Caliphs - the nominal leaders of the world's Sunni Muslims and so the flag was widely adopted across Islam.


How to calculate how much a galaxy moves from its coordinate at redshift 0 to redshift 1? I have two simulation snapshots in my hand at redshift 0 and 1. I know x, y, z coordinates of the galaxies in both redshifts 0 and 1, however there is no way for me to identify a single galaxy in both redshifts, i.e., there is no tracker/tracer for the galaxies that will trace the position of galaxies from redshift 1 to 0.

Now I have halo catalogue for both $z =0$ and $z=1$ and I am interested to trace a certain galaxy from z=0 to 1. I know that galaxy's exact x,y,z at $z =0$. My initial plan is to create a search with $\sqrt{x^2+y^2+z^2}\le r$ in kpc. So any galaxy at redshift 1 with coordinates less than r maybe a potential candidate.

My question is, is there any intelligible guess of what a potential value of r would be in kpc? In other words, I am interested to know with reduced Hubble parameter $h = 0.7$, how much far a galaxy might move from $z = 1$ to $z =0$?

The galaxies don't "move" (unless you have given them a peculiar velocity); space expands, such that the meaning of $x,y,z$ will change.

The Hubble parameter is simply defined as the rate of change of the scale parameter divided by the scale parameter $\dot{a}/a$. But $z$ is also related to the Hubble parameter, so if your epochs are defined by $z$, the value of the Hubble parameter at any particular epoch doesn't matter.

I am assuming that you have a coordinate system based on our Galaxy being at the "origin". In which case, the "initial" position of a galaxy (at $z=1$) is related to where it is "now" at $z=0$.

Redshift and scale factor are related by $ a = (1+z)^{-1}$. So taking your example of a galaxy at $x_0, y_0, z_0$ at $z=0$, then at $z=1$ all galaxies were closer together by a factor of two, and (ignoring any peculiar velocity) $x_1 = x_0/2$, $y_1 = y_0/2$ and $z_1 = z_0/2$.

I don't know what you mean by "halo catalogue". The simple relationship above completely breaks down on scales smaller than tens of Mpc, because galaxies then become influenced by their local gravitational potential. The relationship between galaxy separation and time then has little to do with cosmology and more to do with the dynamics of their local groups and clusters and no general answer can be given, other than to point out that $z=1$ corresponds to more than half the age of the universe - trying to predict the position of a galaxy within a cluster or group based on a snapshot 7 billion years ago is not possible!


What am I looking at here at Google Sky? I was looking through the Google Sky map, when an object caught my eye:

I am really curious as to what this is (just a big star? then why does it look so peculiar?)

It's hard to search for online because I don't know which words to use. If anyone can point me in the right directon please?

It's a star with pretty strong diffraction spikes.
To find out which one it is, you can just look up its coordinates in databases like SIMBAD.
Then you find that it is the star Xi Cygni.
You can also find some information on it on Wikipedia.


Suns Orbital speed and escape velocity question This is a homework question and I don't want an answer but just want to understand what its asking for.

We are given that a star with a mass of x solar masses in our galaxy turns to a supernova and releases y amount of energy. Using this we are asked to calculate the average speed of the ejected mass.

v = sqrt(2*E_k/M) (Kinetic energy equation) gives the average speed.

Finally, the questions asks the following:

  Compare  your  result with our Sun’s orbital speed around the centre of the Milky Way galaxy. Where would you expect this material to end up?

Is it asking to solve and check that the average speed > escape velocity if the star was at the same distance from the sun? Or how far from the center of the Galaxy would it end up?

It's asking exactly what it says- where would you expect this material to end up? Giving a distance from the center of the galaxy may be relevant, or it may not, depending on exactly what speed you end up with, and how it compares to galactic orbital / escape velocity. If the velocity is well above galactic escape velocity, giving a specific distance from the galactic core doesn't make much sense- there will be no fixed distance, the stuff will just keep going, achieving greater and greater distances.

So, the first step is indeed to solve and check if average speed > orbital velocity or not (because the question says to do so), and a good second step would indeed be to check if it's less or greater than escape velocity. And from that information, you can guess generally where the stuff will go- stay reasonably close to the original orbit of the star? or stay in galactic orbit, but spread out in some wildly different range of orbits than what the original star had? or completely ejected from the galaxy?


How was the Earth-Sun distance originally calculated? AFAIK it was possible long before the first interplanetary probes.

Who did it?

The book The Transits of Venus, by Sheehan and Westfall, describes how Aristarchus used Hipparchus' calculation of the Earth-Moon distance, who in turn used Eratosthenes' calculation of the Earth's circumference, to calculate the Earth-Sun distance.
Aristarchus of Samos was the first to seriously calculate the distance to the Sun, using geometry.  When the Moon is exactly half illuminated when seen from the Earth (first or last quarter phase), then there is a right triangle between the Earth, Moon, and Sun, with the Moon at the right angle.  Then he could measure the angular distance in the sky between the Sun and the Moon, plus the Earth-Moon distance and geometry, to get the Earth-Sun distance.

The most famous ancient estimate of the earth's circumference as made by Eratosthenes of Cyrene (c. 276-196 BCE), the librarian at the great library at Alexandria.  By using a simple gnomon, he found that at Syene, ... the sun at the summer solstice cast no shadow at all: it was exactly overhead.  ... At the same moment, at Alexandria, the shadow cast by the sun shows that it stood 7.2 degrees from vertical.  This difference is equal to 1/50 of a circle.

Using the distance between the cities, Earth's circumference could be calculated.

Once the earth's radius is known, the earth itself can be used as a baseline for determining still greater distances -- the distance to the moon.
[I]t becomes possible to work out the earth-moon distance indirectly from the geometry of [lunar] eclipses.  Using this method, Hipparchus of Rhodes (fl. 140 BCE) worked out that the distance of the moon was 59 earth radii.  It's a good approximation - with 1 1/2 or 2 earth radii of the modern value.

Using the Earth-Moon distance and the separation of the Moon from the Sun in the sky when the Moon was at exactly half-phase, Aristarchus calculated the Earth-Sun distance.

Aristarchus put forward a geometrical argument, based on determining the sun-earth-moon angle at the time the moon's phase is exactly half.  For this angle, which is actually 89.86 degrees, Aristarchus used 87 degrees; the disagreement is more significant that it might appear because the critical quantity is the difference between the angle and 90 degrees.

Because of this Aristarchus only got a value of the equivalent of "5 million miles", much too small.
Phil Plait has, on his old Bad Astronomy site, an article answering a question about how astronomers originally calculated the distance from the Earth to the Sun (the AU, or astronomical unit).
Huygens was the first to calculate this distance with any kind of accuracy at all.

So how did Huygens do it? He knew that Venus showed phases when viewed through a telescope, just like our own Moon does. He also knew that the actual phase of Venus depended on the angle it made with the Sun as seen from the Earth. When Venus is between the Earth and Sun, the far side is lit, and so we see Venus as being dark. When Venus is on the far side of the Sun from the Earth, we can see the entire half facing us as lit, and Venus looks like a full Moon. When Venus, the Sun and Earth form a right angle, Venus looks half lit, like a half Moon.
Now, if you can measure any two internal angles in a triangle, and know the length of one of its sides, you can determine the length of another side. Since Huygens knew the Sun-Venus-Earth angle (from the phases), and he could directly measure the Sun-Earth-Venus angle (simply by measuring Venus' apparent distance from the Sun on the sky) all he needed was to know the distance from Earth to Venus. Then he could use some simple trigonometry to get the Earth-Sun distance.
This is where Huygens tripped up. He knew that if you measured the apparent size of an object, and knew its true size, you could find the distance to that object. Huygens thought he knew the actual size of Venus using such unscientific techniques as numerology and mysticism. Using these methods he thought that Venus was the same size as the Earth. As it turn out, that is correct! Venus is indeed very close to being the same size as the Earth, but in this case he got it right by pure chance. But since he had the right number, he wound up getting the about the correct number for the AU.

Basically, Huygens used good methods, except for using "numerology and mysticism" to determine the size of Venus.  He was lucky that Venus was almost the size of Earth; that made his estimate for the AU pretty close.
Not long after, Cassini used the parallax of Mars to determine the AU.  (Same article as linked above.)

In 1672, Cassini used a method involving parallax on Mars to get the AU, and his method was correct.

Parallax is the apparent difference in angle observed due to the differing observing positions.  The smaller the parallax, the larger the distance.
However, the precision of the resultant calculation depends on the precision of the observations, and measurements of the parallax aren't that precise.
In 1716, Edmond Halley published a way to use a transit of Venus to accurately measure the solar parallax, i.e. the difference in the Sun's position in the sky due to observers at different latitudes.

Because of the latitude difference of the observers, Venus would appear to move along chords of different length over the disk of the sun.  The motion of Venus being nearly uniform, the length of each chord would be proportional with the duration of the transit.  Thus, observers would not actually have to measure anything; they would only have to time the transit.  Fortunately, existing pendulum clocks were more than sufficiently accurate for this purpose.

They could time the transit, which would last hours, with great precision.  But they had to wait until the next transit of Venus in 1761.  Then, observers observed the black drop effect, which made it very hard to time the event from start to finish precisely.

The black drop effect cannot be eliminated altogether, but it is much more ever in observations made with telescopes of imperfect optical quality (as many of those used at the 1761 transit were) and in boiling or unsteady air.  Confusion about the times of the internal contacts ... yielded contact times that differed among observers, because of the black drop, by as much as 52 seconds.
In the end, there as a wide range of published values, from 8.28 arc-seconds to 10.60 arc-seconds.

But then there was the transit of 1769.  Observations in Norway and in the Hudson Bay were made for the northern observations, and Captain James Cook was sent to what is now Tahiti to make a southerly observation.  Jérôme Lalande compiled the figures and calculated a solar parallax of 8.6 arc-seconds, close to the modern figure of about 8.794 arc-seconds.  That calculation yielded the first fairly accurate calculation of the Earth-Sun distance, of 24,000 Earth radii, which given the Earth's radius of 6,371 km, about 153,000,000 km, the accepted value being about 149,600,000 km.


Using a telescope to look at the Earth History in Some Detail The following paragraph is a scientific fact.

When we look distantly into the universe (with telescopes) we see galaxies
(or whatever else you see) as they were millions years
ago- in other words it is how they looked in the past (this is because of the speed of light).

So my question is could we build a technology (using current science or more speculative/theoretical science) using this scientific fact too see the earth history or past to such a level of detail that we could see historical events (e.g. individual dinosaurs) and the famous people's faces (in history books) - as if we were standing face to face with them- of the long ago past? What's the best level of detail could we theoretically see?

No.

Let's say we want to see Earth as it was 1000 years ago.  Assume that someone has set up a perfect mirror 500 light-years away, so that we can actually see the light that left Earth 1000 years ago. (That's a really big assumption.)

The best telescopes in the world can't see the Apollo landing sites from Earth. We didn't get decent images of the descent stages, which were left on the surface, until the Lunar Reconnaissance Orbiter sent back photos it took from Lunar orbit. See http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html

1000 light-years is about 20 billion times as far away as the Moon. There's no way we could see people's faces at that distance with current technology -- or with any reasonable future technology. (There are physical limits on the resolution of an optical telescope of a given size.)

And all this assumes that we have that perfect mirror out there. As far as I know, nobody has set up such a mirror for us, and all the light that left Earth 1000 years ago is now 1000 light-years away, badly faded, and beyond our reach.

It's conceivable that we could develop faster-than-light travel (which may or may not be physically possible), go out there, build a telescope with a really big aperture, and point it back at Earth. But that's not likely to happen any time soon, and I wouldn't know how to determine how good an image we might be able to get.

And going back millions of years just makes the problem worse.


Is there an example of a star that is already dead, but can still be seen on Earth? While looking up at the stars I wondered, what are the odds that some of them are already dead. I did some research and found a very interesting article here. It states that the odds are rather small, so it is not like about 50% of the stars we see are already dead.

But I still wonder if there is an example of a star we know to be already gone, while it is still visible on earth.

As far as I know, the only way to find out is to watch the stars light. So if the light "goes out", we know the star has been gone for quite some time. Assuming this to be fact, I would have to conclude that such an example does not (and cannot) exist.

Is this true? Or is there in fact a star that is already dead that we know of, which can be seen in the sky?

PS: By "seeing the star" I am talking about a star that looks just like any other star in the night sky (not, for instance, a visible supernova).

Edit: If there is such a star, how did we find out it is already gone?

There is an accepted answer already, but there is a couple known cases of a star we know has gone supernova, and yet we can still see it.  This source describes one such unique circumstance.  The star that exploded happened to be in a galaxy that was behind another massive one from our point of view.  The alignment was just right such that the light formed an Einstein Cross.  The light from each point on the cross takes a different path to get here, and each of those paths are different lengths.  Thus, the different points on the cross show the star at 4 different times in its final years.  The Space.com article I linked was written in 2015. Scientists first noticed the supernova in 2014, and each image of the supernova arrived within a year of when it was first noticed.  If I'm understanding that article correctly, as of now (2017) all of the images of that supernova show the supernova aftermath, and not the star beforehand.  However, there was a period of time where we could see the star both pre and post-supernova at the same time.

You imply in your question that you are focusing on stars within our own galaxy.  I don't know of any such situations of Einstein Cross events closer to home that let us do this.


How do I reduce planetary photometry to standard viewing geometry using phase angle? I'm trying to make light curves for Pluto using old data. Most papers give magnitudes "reduced to standard viewing geometry" of r = 39.5 AU, Δ = 38.5 AU, and α = 1 degree, where r is heliocentric distance, Δ is geocentric distance, and α is phase angle.

When papers give apparent magnitudes, I'm left to perform the reductions myself. I'm fine with the adjustment for r and Δ, but how do I reduce to phase angle of 1 degree?

I would also really appreciate a good source if you have it on hand.

Phase functions are nonlinear in general, but Pluto presents such a small range of phases to Earth that a linear approximation will do:
about 0.03 mag/deg for Pluto and 0.08 mag/deg for Charon, or 0.04 mag/deg for the two together.
The Stern and Tholen book summarizes the estimates up to 1997.
A search for Pluto phase coefficient
yields some of the papers behind those and more recent estimates.


Is there a theoretical minimum mass main sequence star that can become a red giant? Wikipedia's Red Giant page says they range in mass from 0.3 to 0.8 solar masses.   Which corresponds to an initial mass of a bit less than 1 solar mass, about 0.8 or so, if I was to estimate and a few solar masses.   Given that over half the mass is lost prior to the Red Giant stage.

Any stars with less than 0.8 solar mass to begin with wouldn't have had enough time to go red giant, so it's possible that significantly smaller stars could eventually go red giant.   Is there an estimate for how light the main sequence mass can be where the star still becomes a red giant late in it's life?

Perhaps you are interested in the lower mass limit (given in the comments as 0.3 solar masses), and why that lower limit exists.  This is similar to the question of why stars become red giants in the first place.  You are right that given the age of the universe stars less massive than 0.8 solar masses won't have had time, so the lower mass limit is theoretical.  But what sets it?

The answer is, red giants happen because stars can exhaust all the hydrogen in their core, turning it all into helium.  Then there is no nuclear fuel to keep the star in equilibrium, so no balance between the light that leaks out (which sets the luminosity) and the fusion rate (which mostly just responds to that luminosity).  As a result, the core shrinks and heats, and fusion initiates in a shell around the core-- a shell that used to be too cool to fuse.

The temperature in that shell, and key aspects of its fusion rate, are determined by how small and massive the core becomes (the core gets smaller with time because it is losing heat, and it gets more massive because helium "ash" keeps being added to it from the shell).  Eventually the free electrons in the core become what is called "degenerate", meaning they approach their quantum mechanical ground state, and that implies the core gets about as small as the Earth. This also means the shell is extremely hot, and the fusion rate goes essentially berserk.  This requires the envelope to puff out into a red giant.

Hence, to have a red giant, you must have a degenerate core, and you must have shell fusion surrounding it.  But stars less massive than 0.3 solar masses have an internal structure that is purely convective.  Thus they don't run out of hydrogen in their cores until they run out of it everywhere  (and like you say, none have yet done this in our universe, it's a theoretical expectation).  That is why they cannot have shell fusion, and cannot become red giants.


What should I consider in order to accurately calculate the moon phase? Although this question apparently addresses the topic, it does not tell appropriate factors involving the moon phase.

I found two distinct algorithms. this one and also this one. I am less than a newbie, less than a layman, to astronomy - but I am not to programming, but I would like to skip testing the algorithm for a couple of years before choosing either.

After unsuccessfully finding a comparison of different algorithms, I'd like to know of either of these algorithms is less error-prone, or if there is someone less error prone-to these (I know there is no stuff like a perfect algorithm, but I'd like to have a good one), or having appropriate ways to judge an algorithm.

So my question comes like this:

Do these methods have a small error, enough to accurately calculate the moon phase?

Or...

Can you give me the most precise algorithm -or concepts / underlying theory I should know- astronomers would use as of today to calculate the moon phase?

The moon's phase isn't easy to calculate accurately. Fortunately, the 4 main phases (new moon, first quarter, full moon, last quarter) are a bit more stable than the intermediate phases. Good main phase algorithms use a dozen or so sine calculations for the new or full moon, and a similar number for the first or last quarter. Those two algorithms you linked use very simple approximations so their results are rather crude.
The period of the moon's phase cycle (the synodic month) is stable over the long term, but it drifts considerably over the short term. So algorithms like those in your links are usable over a rather long time span, and they can be useful first approximations to feed to better algorithms, but by themselves they only yield precision results by accident. ;)
The guru of astronomical calculations for bodies in the solar system is Jean Meeus, so if you want quality algorithms make sure they reference Meuss. :)
For seriously accurate ephemerides, especially if you need to cover a wide time span, consider using the work of JPL. You can access their ephemerides at the Horizons Web page; a telnet and an email-based interface are also provided.
If you want to do precise astronomical calculations (eg for eclipse prediction) you'll need to learn about the various time systems used in astronomy, in particular Terrestrial Time and its forerunner, Ephemeris Time.
I see from your SO profile that you're familiar with Python. Good stand-alone Python moon phase code isn't easy to find online, but eventually I did find keturn's py-moon-phase, which is a Python port of the core code of John Walker's rather ancient Moontool, which uses algorithms from Meeus and also from Peter Duffett-Smith. I haven't actually tested John Walker / keturn's code, but it does look rather good (although it uses UTC instead of Ephemeris Time or Terrestrial Time). I have very similar (but rather messy) C code that's generally accurate to half a minute or better over the 20th century.
However, there's a package called PyEphem that can do all sorts of astronomical calculations. James K notes in the comments that PyEphem is now discouraged for new projects. The PyEphem developers recommend Skyfield instead. Astropy can also do a lot of what PyEphem does, and much more.


Is ours the first universe? I see here that after a very long time into the future, "Estimated time for random quantum fluctuations to generate a new Big Bang".

This begs the question, do we speculate that our big bang was the first?  Or are we one of many in a never ending chain?  Or do I miss the point entirely (being ignorant in the subject)?

It is very hard to tell whether we are in a unique Universe. The difficulty is the closer we attempt to probe in the fractions of a nano second after the Big Bang our current forumlations of Gravity and the three our fundamental forces do not work. So it is very hard to tell whether our Universe is unique.

However, with all that said, my own personal belief is that our Universe is not unique. The estimated time for random quantum fluctuations to generate a new Big Bang as quoted in the page that you have referred to is an UNIMAGINELY large time scale.

My question I will pose back to you is, how could we ever tell we are a unique Universe, or part of a collection of Universes?


How long would it take a Boeing 747 to go around Jupiter? Assuming a 747 could fly around Jupiter at its top Earth speed (I know this assumption is unrealistic), how long would it take to fly around Jupiter once?

It will depend on the speed of the jet, however:

Jupiter has an equatorial circumference of 449,200 km, a current-generation 747 has a top speed of 988 km/h so it will take about 455 hours or nearly 19 days (18.9) (this doesn't allow for re-fueling).


How many earth-like planets do we know at this moment? Well, the title says everything...
How many earth-like planets have we discovered out there?
Discard too big (made of gas), too cold, too hot, etc.
Consider earth-like sun distance, size, temperature, the rocky ones...

Twelve plus Mars are considered as potentially habitable (as of February 10, 2014).


Why doesn't Titan have a carbon dioxide atmosphere? Earth's primitive atmosphere had large amounts of carbon dioxide, as did the ancient Martian atmosphere. Venus's current atmosphere is no exception either. So why does Titan have next to none of this compound? Is it due to its location with respect to the Sun?

The chemistry of Titan's atmosphere is complex, with reactions occurring between carbon dioxide, oxygen, carbon monoxide, hydroxl, and other compounds. This means that carbon dioxide production and destruction takes place through a variety of reactions (Samuelson et al. 1983), some spurred by ultraviolet light from the Sun (and hence photodissociation). In other words, comparing Titan to Earth and Mars ignores a variety of rich chemical processes; right away, the comparison shouldn't hold.

All that said, the primary source of carbon dioxide loss on Titan is condensation, with photolysis playing a secondary role (Horst et al. 2008). With a surface temperature of 94 Kelvin, Titan is much colder than either Earth or Mars - well below the freezing point of carbon dioxide. Even if you were to add a large influx of carbon dioxide to Titan, much of it would very quickly (from an astronomical point of view) condense into solid form and fall to the surface. Indeed, much of Titan's surface contains significant amounts of dry ice.


Number of planets in a habitable zone with large moons Excluding gas giants, roughly what percentage of planets in the habitable zones of their stars would have large moons like Earth does, as opposed to Mars and its small moons/captured asteroids?

It's not quite the drake equation in terms of your inputs determining your estimate but it's not too far off.   Our telescopes can't get a good look at existing or forming solar systems.  They've spotted several thousand planets but only a tiny number of possible moons and I don't think any of them are confirmed.    It's also not always possible to tell is some of the "earth-like" planets that have been observed are more like sub-neptunes or perhaps water worlds.   Any kind of gaseous/water or ice world planet would have a hard time making an impact moon.  Two rocky bodies colliding are best for that.

Young solar systems are full of collisions during their planet formation period so there are probably enough collisions with the approximate size/ratio of Theia-Earth.  At least, it's probably not uncommon, but size isn't the only factor.   The impact has to be at the right angle to generate sufficient angular momentum, because the main body ideally rotates faster than its Moon orbits.  This faster rotation and the tidal force of the moon on the planet pushes the moon outward over time and stabilizes the Moon's orbit (provided the hill sphere is larger than the Moon's distance over time).   If the main body rotates slower than the moon orbits, the moon is drawn towards the planet and would likely destabilize before long.

An impact too straight on is bad for moon formation and too glancing is also bad.  The impact needs to be within a proper range in size, ratio between the two objects and angle of impact and without a surface with lots of ice and/or water.   The requirements are pretty precise, so stable large collision moon formation should at least be somewhat rare because of all the necessary conditions.   It's also not known with accuracy how many of the "habitable" planets started out rocky.

Migrating jupiters and other planets are another problem.   It's fairly common for Jupiters to migrate which should destabilize a percentage of inner planets with stable collision moons.

The biggest issue is the size of the star.   The habitable zone of red dwarf stars is very close to the star and moons wouldn't be stable around planets in those systems.    Those planets would likely be tidally locked, which means they'd rotate slower than their moon orbits and that would likely lead to destabilizing the moons orbit and the moon crashing into the planet, that is, if the planet had a moon at all, it would need to form in a very tight orbit.

So there's a few estimates to make a guess at an input.   I don't think an accurate estimate is possible but if I was to guess, I think large collision moons will be found.  I don't think they're super rare, and I base that guess on a reasonable abundance of rocky material in solar system formation and a very large number of collisions, so even if the Earth-Theia collision was one in a million, there's enough collisions happening that other two body systems like ours probably create long term stable moons from time to time.  So, rare but I think, not super-rare.

We probably won't have an answer for this until we get a better look with the James Webb Telescope.