Can a frost line exist around a protoplanet? The Frost line is the border that marks where volatiles such as water, methane, ammonia, etc. (commonly referred to as 'ices') are able to condense into solid grains in the protoplanetary disk around a forming protostar. Beyond this border, these ices form a significant part of any coalesced planetary body (like the many icy moons of the gas giants), whereas inside this border, the bodies that form are primarily rocky (the four terrestrial planets, Mercury, Venus, Earth, Mars).

Can such a frost line exist in the accretion disk around a hot gas giant such as a proto-Jupiter?

Some soft evidence for this: Jupiter's four Gallilean moons are of similar sizes and follow a clear trend from the innermost outwards:

Io: Mostly silicate rock and iron sulfide. Density is 3.5 $g/cm^3$.
Europa: Mostly silicate rock with an estimated ~100 km layer of ice/water. 3 $g/cm^3$.
Ganymede: Silicate with an estimated ~800 km layer of ice/water. 1.9 $g/cm^3$.
Callisto: Approximately 50% ice. 1.83 $g/cm^3$.

Source: Mainly articles in Wikipedia

Now, there are other mechanisms by which ices can be disproportionately lost from Io, mainly through its extreme volcanism. But this doesn't necessarily explain the trend seen in the other moons.

To be specific, I'm not asking if a protoplanetary frost line is responsible for the distribution of elements in Jupiter's moons, merely if such a frost line can exist and contribute to the observed element distribution.

Yes they can exist. The effect of ice lines in circumplanetary discs has been studied, see for example Heller & Pudritz (2015), who propose that the ice lines may act as a barrier preventing moons from being lost as a result of inward type I migration.


Why, in the Solar System, all the mass seems to be concentrated at the centre? Is this a consequence of planet formation in accretion disks ?

It's because of the Sun.

It might be good if I give a quick overview on star formation before I get to the meat of the issue. Here's star formation in a few simple steps:

Giant Molecular Cloud forms. A large region of gas and dust, essentially a dense version of the interstellar medium, coalesces into an interstellar cloud. GMCs can be tens or hundreds of light-years across, enough to give birth to many stars. Within the GMC, some regions will be slightly denser than others.
A portion of the cloud collapses. A certain region of the GMC collapses, generally due to an outside disturbance. The most commonly cited cause is a supernova shockwave that compresses portions of the GMC, although close passes between galaxies have been known to incite star formation. My favorite example is the Cartwheel galaxy.
The region heats up. There is quite a lot of matter pressing in on what has now become a protostar, and so it heats up. Eventually, conditions become such that hydrogen fusion is possible. The protostar, now a pre-main-sequence star, begins to shine.
A protoplanetary disk forms. At this point, the star dominates this region of the GMC. Matter nearby is pulled towards it by the force of gravity, and a circumstellar disk forms. It may be composed of gas and dust. Eventually, small grains of dust collide and form bigger grains. Planetesimals form, then protoplanets, and finally planets.

The reason that there isn't more matter in a given stellar system is that the star dominates the surrounding area. It pulls in nearly everything around it during its early life. Much of the region of the collapsing cloud is made of molecular $H_2$, and so it is pulled in and used for fusion.

Now the question translates to 'Why isn't the protoplanetary disk more massive'? The answer is that when the disk formed, much of the matter that was in its inner reaches spiraled into the Sun. This is partly due to the Poynting-Robertson effect, where photons from the Sun pull dust grains in. Over billions of years, the star can accumulate much of the matter than was originally close to it in the disk.


Why does the moon looks like having different filters sometimes? Why does the moon look more yellow or more LED-like?
Thanks in advance

TL;DR: because of the Earth's atmosphere.
When the Moon is low on the horizon, the light we receive from the Moon has to go through a lot of layers of the atmosphere. Rayleigh scattering makes this light more yellowish/orangish, for the same reason sunsets look orange.
A very thin layer of clouds can also change the appearance of the Moon.
Sometimes, the Moon ends up in the Earth's shadow. In this case, the light that hits the Moon first went through the Earth's atmosphere. Because of Rayleigh scattering (again), the light that went through the atmosphere is red. This gives the Moon a reddish tint.


What is the largest (angular) dark area in the sky? Question

Within our current knowledge, what is the largest (circular, say) area in the sky which is empty/dark? I'm thinking in terms of angular size.

Motivation

I was wondering what was the largest star as seen from the Earth, other than the Sun of course, in terms of milliarcseconds, say. A search quickly turned up R Doradus which is huge (57 mas as seen from the Earth). But I wondered if it was possible to rule out any (angularly) larger dark objects by finding stars whose light would be blocked if there was such an object.

Of course we should eventually be able to rule out each such area -- as Hubble Deep Field showed, if you look long enough there's a lot to find everywhere -- but relative to our current knowledge, what are the largest gaps?

I expect that the answer would be in a region which has been missed by a sky survey and which coincidentally doesn't have anything we can see.

Barnard 68 is the first thing that comes to mind for me, it is a little over 10' across and is opaque in the visible spectrum.

I'm sure there could be something much larger out there, but like I said, this was the first to come to mind.


Infalling observer could never cross Black Hole event horizon? I have an interest in cosmology, but I would no way consider myself an expert. It would be great if the more knowledgeable contributors would be able to answer the question I have posed.

The intense gravitational field near a black hole causes time dilation. This means, to an observer close to its event horizon, time  runs more slowly. It can mathematical be expressed as:

Time dilation = $\frac{\delta t_f}{\delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$.

Where $\delta t_f$ is an interval of time a long distance from the black hole, $\delta t_0$ is the time interval a distance $R$ from the black hole’s centre and $R_s$ is the radius of the event horizon.

This formula means that time dilation becomes infinite at the event horizon so to a distant observer a falling object would take an infinite time to reach the event horizon and would never cross it.  However, in the frame of an observer falling into the black hole it should them take a finite time to reach the event horizon. To me the complexity is that all black holes are believed to ‘evapourate’ and eventually disappear in a finite time, by emitting Hawking radiation. Although this an incredibly slow process. For example, a solar mass hole would have a lifetime of $10^{66}$ years (in a time-frame measured a distance from the black hole).

My Question
Assuming Hawking radiation exists and  black hole evaporation occurs, does this mean that an infalling observer would never actually reach the event horizon, because the black hole has finite lifetime and would evaporate before an infalling observer would reach it?

Further Note
Since posing this question I have had one further thought, which may or may be relevant, since my knowledge of GR and quantum theory is pretty sketchy.

The way that the time dilation formula works is that when the observer is a distance of one Planck length ( ~ $10^{-35}$ m ) away from the event horizon of a solar mass black hole (which has radius ~3km), the time dilation would be still only be  ~ $10^{19}$. However, the lifetime of a solar mass black hole is ~$10^{66}$ years. Therefore, even when they are within 1 Planck length of the event horizon, the time dilation near the black hole is not sufficient for the observer to see the black hole evaporate.

It is often pointed out that the infinite time dilation at the EH is what is known as a "coordinate singularity", rather than an essential singularity that says something is physically singular there (like never reaching the EH for the infaller).  A coordinate singularity is essentially a hiccup in the language we are using, like the way a north-south-east-west coordinate system is of no use at the North pole.  The problem is that if you want to talk about "whether something has crossed the EH yet", you must match up your own concept of "now" with some time at the EH, and that matching up is the job of a coordinate system.  So saying "it hasn't reached the EH yet," and repeating that every day for infinity of time here at Earth, is not the same thing as a physical statement about what is happening at the EH, it's just a reflection of the way we are matching up our "nows" with theirs-- it is purely a matter of the chosen coordinates.

To know what is physically happening to the observer at the EH, the ideal solution  would be to simply ask them, and they would say "yes I am crossing the EH now".  Here it gets tricky, because the normal coordinates we use to talk about the EH (the one you were using also), says that we could never receive that communication.  But those coordinates make physical assumptions that are violated by having something falling into the black hole.  At this point it would be best to have a GR expert, but I believe what happens is that the very act of having mass fall into the black hole in effect moves the EH out to swallow that mass, and we should therefore be able to receive the communication that the observer has crossed the EH in a finite time.  This must be true because, as pointed out, the very existence of the black hole is like information coming to us that stuff has crossed its EH, an EH that formed and moved out as the black hole was created by mass infall.


When will Sirius be closest to the solar system? Sirius is currently approaching the solar system, at a rate of 5.5 km/s, consequently in the future it will be closer and therefore brighter. When will Sirius pass closest to the solar system? How far will it then be from the sun, and what magnitude will it reach?

Sirius is moving towards the sun at 5.5 km/s, and according to Sky and Telescope, will reach its closest distance in 60000 years, at which time it will be 7.8 light years distant, and be at magnitude -1.64, only slightly brighter than it's current magnitude -1.46.

Calculations using the latest Hippacos results give a slightly different value: a closest approach of 8.18 ly in 46000 years, and consequently a slightly dimmer peak. However Sirius is not moving rapidly relative to the sun, and will not change its brightness greatly in the next 100000 years or so

It will be the brightest star in the sky for another 90000 years, and no star will match its current brightness for another million years (when delta Scuti will reach magnitude -1.84.


Time of day of the K-Pg asteroid impact? Is it possible to determine what daytime the Chicxulub impactor striked the Earth 66 mya? Was it day or night, what was the current phase of the Moon etc.? Can sediments, rhytmites or something else hold this information?

Given that the uncertainty in deltaT, the difference between the uniform dynamical time needed to predict the positions of the Sun and the Moon and civil (solar) time was already 4.5 hours by year -4000 (https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html), then I would say by 65 million years ago we would have no idea of which way the Earth was facing at the time. Unless the dinosaurs left eclipse records that would let us calibrate the much faster spinning Earth back then, we would have no means to work out what time of day the impact happened.


Interfacing FORTRAN programs for model atmospheres with IDL or Python I'm trying to use model atmospheres such as Kurucz or MARCS for the first time personally and I'm wondering if it is better to wrap it with IDL or Python in terms of writing the code, and getting the values I wanted from the model atmospheres into arrays for printing into tables.

Essentially is it better to approach this issue using IDL or with Python, or is it just more of a preferential thing if anything?

After looking into the various wrappers and ways Python and IDL work with FORTRAN I came to the conclusion that it is more of a preference which language you use. However as people continuously add astronomy support to Python this may change, but as of now, it seems to be a question of preference.


How to recognize exoplanet transit I am using Python package lightkurve for exoplanets searching by the transit method. When I download light curve of some star and apply periodogram, I find frequency and power of periodic components in the light curve. However, I noticed that multiples (0.5x, 2x, ...) of the original period are displayed too. Here is example for Kepler-6b.
import lightkurve as lk
import numpy as np

light_curve = lk.search_lightcurvefile("Kepler-6", quarter=1).download().PDCSAP_FLUX
light_curve.scatter()

periodogram = light_curve.to_periodogram(method="bls", period=np.arange(0.5, 10, 0.0001))
periodogram.plot()

I could take only strongest period (3.24 d), but what if there are more exoplanets (1.08 d, 9.71 d, ...)? I thought that if I use the light_curve.fold(period) method, I can tell if it is a transit (there is only drop of flux) or not (there are more drops of flux). However, Kepler-20f has also multiple drops of flux after fold light curve (because of other planets?).

How can I tell if it is a planet transit or not?

Of course you will get multiple peaks in the periodogram. The Fourier series representing a non-sinusoidal signal will contain frequencies at multiples of the fundamental frequency. Similarly, you can have a periodic signal with double or treble the period which will look identical, but where the phenomenon causing the signal repeats either two or three times during each cycle.
The solution is of course to plot the data folded on the proposed period, as you have done.
In the examples you have shown, clearly the period of 3.23d is the only valid period for a claimed exoplanet, since the transit only takes place once per cycle and there is no sign of any other features or excessive scatter in the folded light curve at that period.
For the 9.71d period to be valid you would either (i) have to have three planets, equi-spaced around the same orbital circle, each with a 9.71d period, and of identical size so that the transits are the same depth. That is not a stable situation. Or (ii) you are unfortunate enough to have one planet with a  3.23d period and another with a 9.71d period that transit at exactly the same time. But it is not possible to do that without having unequal transit depths and causing scatter in the transit shape when folded at 3.23d.
I cannot make out what your point is about Kepler 20f. This would produce a single feature in a light curve folded on a period of 19.48d, but the amplitude of that signal is small (and I can't see it). Both Kepler 20c and 20d produce transits that will be $\sim 10$ times as deep, with periods of 10.8d and 77d respectively.


What is the maximum planet mass for human habitation? What is the maximum planet mass for human habitation? How many Gs can humans withstand on a constant basis?

When astronomers discuss the "habitable zone" around a star they're really concerned with someone likely to provide liquid water and the right amount of radiation for basic life (cells) as we understand them.  But this doesn't mean they're suitable for humans.

Maximum mass is not an issue, it's gravity that matters.

We need a few things from gravity to be human habitable.

Firstly we're "designed" (evolved) to operate in a one-gee environment.  As far as I know there have been no studies on long term high gee effects on humans (low gee, yes, but not high gee).  However, considering the "plumbing" and structure involved in a human, it seems unlikely to me we'd adapt long term to more than a small percentage larger than one-gee.  Much more on a sustained basis would seem likely to increase physical stress on the bones, muscles, tendons and so on and lead to more rapid development of related illnesses and disabilities.  The plumbing (fluids are being moved all over your body) can cope with quite high gees for short periods, but it's simply not designed to cope with sustained high gee.  The strain on the circulatory system strikes me as the greatest vulnerability.

Movement in high gee is also an issue.  I'd be surprised if people could adapt to long term requirements without serious health side effects.  We'd also be more vulnerable to injury from falls and slips.

Consider how long you could walk in 1.5g.  It would be the equivalent of walking while carrying something weighing half as much as you do yourself. If you were 100 kilos, how long could you walk while carrying 50 kilos ?   If you had to do that all day, everyday, how long before the strain caused serious injuries and other health problems ?  Not long, I think.

So just on this basis I'd suggest a maximum sustained gee of about 1.1 is ballpark.  Walking around with ten kilos when you are 100 kilos sounds like a lot all day everyday, but it's probably doable.

But gravity does more than provide a way to nail us to the ground.  It provides a way to nail the atmosphere to ground at an atmospheric pressure high enough for us to use, and not too high for us.

Now it would appear from human habitations in the deep sea, that we can sustain quite high atmospheric pressures, so that would not seem likely to be a problem.  What tends to hurt humans is shifting quickly from high to low pressures.  As I've argued that we can't sustain high gees for long and need something close to one gee for health reasons, the atmosphere doesn't seem to be a problem in that case.

So what does 1.1g mean in terms of planet size and density.

The gravitational pull is proportional to planet average density and radius.

So we need a planet that's around the 1.1g mark meaning :

$$1.1 \rho_{earth} r_{earth} \approx \rho_{new-earth} r_{new-earth}$$

where $\rho$ is density and $r$ is radius and this represents the limit.

So this may make it easier to find suitable planets.  We aren't restricted to specific mass ranges, but can look for planets that are less dense and larger or more dense and smaller.

But this does make me wonder if the notion of human colonization of planets ever made any sense.  Building our own large habitats in space is starting to make a lot more sense to me as it restricts us less.


How can I see a nebula? I've recently been seeing a bunch of pictures of nebulae and I'm just fascinated by their beauty and complexity. Is there any kind of telescope that would make it possible to view it from here on Earth?

Yes, indeed! Many nebulae are visible from Earth in a small and cheap telescope, and even to the naked eye (if you are standing in a sufficiently dark place).
In fact, yesterday I was watching the Orion Nebula with my 4.5" telescope (which is worth $200 or so) from my apartment in the middle of Copenhagen.
The term "nebula" is a bit of a… well, nebulous term, as it covers several distint phenonomena, for instance:

Galaxies,           e.g. the Andromeda Galaxy (though this is an old expression)
(Open) stellar clusters,   e.g. the Eagle Nebula
Globular clusters,  e.g. 47 Tucanae
Planetary nebulae,  e.g. the Eskimo Nebula
Supernova remnants, e.g. the Crab Nebula

However, you should be aware that even in a large and expensive telescope, the nebulae do not look anything like the beautiful images you find on the internet. With your eye, you will merely see diffuse, whitish clouds. The beautiful colors arise only in images taken through telescopes with long exposure times. If you want to make such pictures, you will need a telescope with a motor (since the telescope must follow the rotation of the sky) and the possibility of attaching a camera to it.
People are often disappointed when shown a nebula through a telescope. I think the beauty of it lies in knowing what it is that you are seeing, knowing for instance that it lies 600 lightyears away, that you are looking 600 years back in time, and that stars are right now being formed there.
Googling something like "nebulae visible in a small telescope" should get you started. Good luck!


Why isn't the moon dark in a solar eclipse? In the following partial solar eclipse, the moon is "invisible" rather than being dark.  By contrast, if someone stands between my view and the sun, they will appear as a dark figure instead of invisible.

Why doesn't the moon appear as a dark circle instead of being invisible?

Light refracts in the atmosphere, causing the light to scatter in many directions. This light scattering is what causes our day to be blue, our sunsets to be their red-orange-violetish colors and allows us to see the light of the Sun shortly before sunrise and after sunset. As this only begins happening in any significant amount after the light enters Earth's atmosphere, there should not be any notable difference in the color or intensity of light between you and the moon and any other part of the sky at a relative angle from the Sun.

On the other hand, the foreground does not have direct light on the photographer's side (because it is nearly between the sun and the photographer) and there is a very limited amount of atmosphere to scatter light. As a result, less light will be coming to the camera from that direction.


Is the eventual heat death of the universe due to the expansion of the universe? I am reading about the accelerating expansion of the universe and something struck me. The observation that things further away were accelerating away faster and that early observations made it look as though we were situated at the center of the universe.

Specifically, it seems as though the universe itself is expanding and not that the things within the universe were moving away from each other. Instead it seems as though the expansion itself is creating more space between the contents.

Recently I also saw an amazing video on the eventual death of everything in the universe. This seems weird to me because as I understand it energy and matter don't go away they just change.

And it got me thinking, what if this eventual heat death of the universe is simply this expansion eventually ripping everything apart until every particle simply is to far apart to interact.

Basically, as things progress the expansion will affect the contents of the universe even down to the particle level.

Is this the case? Or am I misunderstanding this subject?

As far as I understand it, the heat death of the universe is a consequence of entropy, not expansion. All processes result in the shifting of some energy to higher entropy. Though the observable universe is an open system, the entire universe is an isolated system, so as more and more energy gets shifted to higher entropy over time the universe will eventually reach a state where effectively no work can be done.


How big a dish do I need for radio astronomy? I've recently become interested in the idea of building my own small-scale radio telescope. A quick online search finds a few instructions on how to build this using a satellite dish. These suggest a dish of radius of around 1 metre, but the only objects that you seem to be able to detect are the Sun, the Earth, the Moon (possibly) and communications satellites.

This sounds like it would be become boring fairly quickly, and I would like to observe some deep-sky objects. Some example objects that I fancy observing (depending on how strong their signals are) are Betelgeuse/Sirius, the Crab Nebula and the Andromeda Galaxy.

N.B. I am quite satisfied with pointing the dish at an object and the signal strength increasing. I don't expect to get any kind of image.

Evidently, you cannot pick these up using a metre-wide dish, so what is the minimum size required to pick up these objects (noting that I can't exactly build Arecibo in my back garden)?

In case it matters, I live in a rural area, ~5 miles from the nearest town.

I originally emailed Jodrell Bank this question but they ignored me :P

I am a member of Astropeiler Stockert e.V., and we are fortunate enough to be able to approach this problem coming from the "large side" :-) We have a 25m, 10m and 3m telescopes as well as an interferometer made from two 1m satellite dishes available. All these dishes can be used to do interesting things, but you'll need to match the instrument to your target (and, in a hobbyist setting, often also the target to your instrument).

First of all, you should think about what frequency band you want to work on. 21cm (1420 MHz) is the classical hydrogen line, which lends itself well for mapping neutral hydrogen in the spiral arms of our galaxy. You can expect comparatively strong signals in a quiet frequency band there. Lower frequencies (and large bandwidths) are interesting for pulsars, higher frequencies give access to more interesting phenomena but will require a lot of work on the high frequency side. So my recommendation would be to start out with 21cm.

Secondly, what to look for? In general, the following areas of observation are easily accessible for amateurs:

continuum emission of the Milky Way, perhaps up to creating maps
spectral measurements of selected, "bright" (as in "intense radio sources") objects
Pulsars

Let's look at them in detail:

Continuum measurements are easily done using dishes of any size, but I'd start from 3m upwards to get more interesting results. Calibration of the backend receiver is not trivial, and you should plan a bit of time for it.

Spectral measurements also benefit from larger collecting areas, but you'll also need to set up a suitable backend. They are, however, possible from 3m upwards if you are happy with a bit of mapping of intra-galactic velocities.

Pulsars require a lot of collecting area and bandwidth, so they are mostly the domain of large instruments. We are currently observing several dozens of the brightest pulsars on our 25m dish. The 10m might be sufficient for some of the very brightest. And while there is a report of pulsar measurements using a 3m dish and an RTLSDR frontend, this is a feat that requires experience and dedication. So, I'd recommend this only for 8m and above dishes.

Interferometry with 2x1m dishes (20 GHz) is rather interesting, but very involved in terms of analysis. Two receiving systems help mitigate local fluctuations, and with that setup, we can observe sources down to 2 Jy (with a looong integration time). This setup will give you access to one or two handful of interesting targets like M1, W51 or Cyg A.

Finally, I'd like to recommend the EUCARA conference series (European Conference on Amateur Radio Astronomy) and the SARA group as great starting points. They have conference presentations available online that show what other amateurs are doing.


How many earths fit in the observable universe? I was pondering our insignificance, when I wondered - how much smaller is our planet then the (observable) universe? And being as I don't know how to do the math, I'm asking it here.

So how many of our planet (in space it occupies - i.e. ignoring space between the space between the spheres) can fit inside the known/observable universe?

Without checking the numbers in detail, according to Wikipedia, the volume of the observable universe is about $3.5\cdot 10^{80} \mbox{ m}^3$, and the volume of Earth is about
$1.08321\cdot 10^{21} \mbox{ m}^3$.

By dividing the two volumes we get a factor of $3.2\cdot 10^{59}$, or written as decimal number: The observable comoving volume of the universe is about
320,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000-times the volume of Earth.


Can we find out whether early Venus was Earth-like or not? It has been speculated that Venus billions of years ago could've had a much different atmosphere with liquid water on its surface and possibly life. Partly thanks to the cooler young Sun. But is it possible for science to find out?

What traces would remain that our probes could measure? AFAIK Venus was completely resurfaced 300-600 million years ago so I suppose there won't be any geological signs of water of the kind found on Mars. And the atmosphere obviously has changed dramatically if Venus was once Earth-like. What should we look for to discover the evolution of Venus? What instruments should we land there, or fly in its atmosphere or orbit?

My ad-hoc opinion: This wouldn't be the first step of Venus exploration. Geologic in-situ investigation of the resurfacing hypotheses would already be a very challenging mission.
Might be, one could find some metamorphic remnants which have survived the last resurfacing, and one could determine the age of rocks. Might be there exist some old layers below the surface, which haven't been molten up completely.

Some kinds of crystals are more resistant to heat than others. They may also contain some records of more ancient epochs. Comparing their isotopic ratios might tells something about the ancient atmosphere, or at least about ancient geology. But conditions on the surface of Venus are very hard for probes.

More feasible would be a detailed isotopic analysis of Venus' atmosphere, as e.g. proposed as VDAP mission concept, or by follow-up balloon missions. A submarine-like ballon diving into the hellish conditions of the lower atmosphere or even to the surface for a short period might be an approach.

A TLS (tunable laser spectrometer) has been suggested to analyse the atmosphere.

May be an additional mass spectrograph would be useful. If short dives close to the surface become feasible, cameras would make sense. A LIBS (laser induced breakdown spectroscope) would make sense only on direct contact with the surface due to the atmospheric absorption. APXS wouldn't probably work due to the heat, unless samples could be taken for analysis higher into the (cooler) atmosphere. IR spectroscopy could possibly work, if cooling can be achieved somehow (e.g. by insulation and adiabatic decompression). If surface samples can be taken during short surface contact, more ways to analyse rocks would become possible. The more instruments, the more weight, the larger the ballon, the larger the probe, the more expensive. But I'm far from elaborating technical or scientific proposals here for space agencies.

Hence VDAP with a small TLS and a QMS (quadrupole mass spectrometer) - as already proposed by hugely experienced scientists - would be a good start.


Does the Milky Way move through space? Does our galaxy moves through space? Or does it stay in a single location? If it does move, what causes it to move?

Does the Milky Way move through space?

Yes it does.

  I'm very fascinated with space, although I don't have a degree or any formal education, I'm still very in love with everything about it and want to learn constantly.

Good man Mike.

  One thing I ask myself is if our galaxy moves through space?

It does. When we look at the Cosmic Microwave Background Radiation we see a "dipole anisotropy" due to the motion of the Earth relative to it:

Image courtesy of William H. Kinney's Cosmology, inflation, and the physics of nothing

See Wikipedia for more:

"From the CMB data it is seen that the Local Group (the galaxy group that includes the Milky Way galaxy) appears to be moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276°±3°, b = 30°±3°.[82][83] This motion results in an anisotropy of the data (CMB appearing slightly warmer in the direction of movement than in the opposite direction).[84]"

627 km/s is quite fast. See this article, which says it's 1.3 million miles an hour. The speed of light is just under 300,000 km/s or 670 million miles per hour, so the Milky Way is moving through the Universe at circa 0.2% of the speed of light. Also see the CMBR physics answer by ghoppe which talks about the CMBR reference frame, which is in effect the reference frame of the universe.

  Or does it stay in a single location? If it does move, what causes it to move?

I'm afraid I don't know why it's moving. Perhaps it's because the Universe is full of things moving in fairly random directions. Like a gas.

  Hopefully the question makes sense, if not I can elaborate.

It certainly makes sense to me!

Edit 13/09/2017 : as Dave points out in the comments, there are other motions, including the motion of the solar system around the galaxy, which is circa 514,000 mph. (See the Wikipedia Galactic Year article). And the motion of galaxies isn't neat and tidy either.


Calculate the true diameter of stars from photographic plate I am reading the book "Fourier Analysis" from T. W. Körner, and in chapter 95, he explains how you can compute the diameter of stars on a photographic plate. He says that "Since observations of the nearest stars at six-monthly intervals (i.e. using a diameter of the earth's orbit as a surveyor's base line) enable astronomers to measure the distance of these stars, knowledge of the apparent diameter (i.e. the diameters of the discs on the photographic plate) would then enable us to calculate the true diameters of the nearest stars".

So how do you calculate the true diameter of a star given its apparent diameter on a photographic plate? Is it related to the "zooming capacity" of the telescope one uses?

And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?

Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.

In principle, this is basic trigonometry: you have a measured angle ($\alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $\sin \alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $\alpha \approx d/D \rightarrow d = \alpha D$ (assuming $\alpha$ is in radians).

As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).

So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $\approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $\sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.

It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.

And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?

Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.


Diameter of any galaxy I've been trying to figure out how to find the diameter in kpc of any galaxy given that one knows the distance to it in Mpc (d) and also the angular diameter/size (A) in arcsec. I had a formula before to find the diameter of a planet: $D=\frac{2\pi dA}{360degrees}$ but this is only if the celestial object is round. Then I asked my friend who gave me another formula $D=d*10^6 *sin (\frac{A}{3600})$

My question is, is the second formula the correct one if I want to find the diameter of any galaxy no matter what shape it is or is there another one which I do not know of?

Given the angular diameter $a$ in radians and the distance $d$ in Mpc, you can get the actual diameter $D$ from:
$$D = d\tan{a}$$

Using the small angle approximation, you get:
$$D = da$$

$a$ is in radians, so to get the distance in Mpc from the angular diameter in arc seconds you'd need to convert the angle in arc seconds to the angle in radians: $a = \frac{2\pi A}{360\times3600}$, where the factor $3600$ is used to convert arc seconds to degrees, and $\frac{2\pi}{360}$ to convert from degrees to radians:

To get the diameter in kpc:
$$D = 1000\times d \frac{2\pi A}{360\times3600}$$
$$D = d \frac{\pi A}{648}$$
$$D \approx \frac{dA}{206} $$

where $d$ is in $\textrm{Mpc}$, $D$ is in $\textrm{kpc}$, and $A$ is in $\textrm{arcsec}$.

Here $D$ is the diameter for round objects (even for disk galaxies seen at an angle). If the object is not round then this would normally be the maximum diameter.

[EDIT (see answer and comments from HDE 226868)] For irregular galaxies you may also need to have more information (viewing angle) to find the real maximum diameter of the galaxy. But that information is (I think) only available for galaxies in the local group.


What is the significance of the discovery of a pulsar flipping between radio and x-ray emissions? This morning on the radio I heard about the discovery of a pulsar that, due to its proximity to a companion star, was flipping between a state where it emitted radio waves and one where it emitted x-rays. The flipping between states was said to be due to material from the companion star falling onto the pulsar, causing it to rotate more quickly.

What is the significance of this discovery?

This newspaper article from the UK on the discovery contains more information, but doesn't go into detail about why this discovery is important.

The significance of this discovery, according to the European Space Agencies web article "Volatile pulsar reveals millisecond missing link" is the first time a pulsar in a crucial transitional phase between has been observed, and that this explains the origin of mysterious millisecond pulsars.

The significance is as stated by the ESA (bolding mine):

  This bouncing behaviour is caused by a rhythmical interplay between the pulsar's magnetic field and the pressure of accreted matter.

What this tells us about pulsar dynamics is

  When accretion is more intense, the high density of accreted matter inhibits the acceleration of particles that cause radio emission, so the pulsar is not visible in radio waves but only through the X-rays radiated by the accreted matter. When the accretion rate decreases, the magnetosphere expands and pushes matter away from the pulsar: as a consequence, the X-ray emission becomes weaker and weaker, while the radio emission intensifies.

The article also contains contact details of scientists involved


If the sun is a second or third generation star, where is the neutron star? It is widely believed that before the Sun was another star, more pristine in its composition (mostly Hydrogen and perhaps some Helium). And maybe before it was yet another. This blog explains this via the distribution of materials. Now, these previous stars that pre-date the Sun, must have been massive, right? If so, shouldn't they have formed a neutron star or black hole once their lives ended? But there isn't one of those at the center of our solar system. What other stellar evolution paradigm would explain this?

Stars are not like animals, with well-defined parents. They form from interstellar gas clouds that in turn originate from stellar winds and supernova ejecta from other stars in a mixture with primordial gas. It is unlikely that the sun is "descended" to a great degree from any particular past star.
(Even if it had been, the sun formed about 18 galactic rotations ago and we are nowhere near the bodies in that past neighbourhood now, due to mixing)


Are binary stars with only one visible sibling common enough to contribute noticeable to dark matter? Single-lined spectroscopic binaries is a binary system where only one star can be detected.
Are they common enough to contribute noticeable to dark matter?

The luminosity of normal stars is a strongly increasing function of mass. e.g. $L \propto M^3$. If another star is "hidden" in a binary system, then it is of lower mass. So the amount of hidden mass is less than what is seen.
Of course this can (and is) be accounted for when estimating the mass present in luminous matter because we know typical binary frequencies and properties. The luminous matter is only about 1% of that required to explain the properties of the universe as a whole and less than 10% of the mass required to explain the dynamics of ours and other galaxies, so even a factor of 2 is not going to solve the problem.
Further, from big bang nucleoynthesis calculations and estimates of the primordial abundances of helium and deuterium, we know the missing mass is not "baryonic". It is not normal matter made if protons and neutrons.
So hidden binary companions are not an important contributor to the dark matter problem.
You ask for a quantitative answer. Well, a typical binary frequency for low-mass stars is 30-40% and they have a flat mass ratio distribution (i.e. the binary companions have a uniform distribution of mass). Given that stars make up less than 1% of the critical density of the universe then even if one ignored the "binary problem" (which astronomers don't), then the additional mass in hidden binary system is less than 0.15-0.2% of the critical density.
This is to be compared with the ~30% that needs to be there to satisfy the dark matter problem (and as I mentioned, the vast majority of that cannot be in the form of normal matter like binary star companions in any case).
To be even more specific. Hidden binary companions make no contribution to dark matter at all, since they are luminous. What they change is the average mass/luminosity ratios of stars that must be used in order to estimate a mass from a luminosity. The values that are used by astronomers to estimate the amount of luminous matter attempt to take account of a binary population. Are the effects of binarity noticeable? Yes of course we have many means to show that stars have a less massive companion. Are they noticeable on large scales - no. The precision with which the dark matter content of an individual galaxy is estimated (e.g. from a rotation curve) will have uncertainties of at least 10% and the contribution of hidden mass in stellar binary systems will be tiny compared with that.


Space expansion in layman terms So far I got to understand the expansion of space is not to be understood as stars drifting further apart through space. There's something more fundamental - e.g. you can't simply measure speed of it, nor tell where the center of the universe (the midpoint of expansion) is. Well, some smart people got me from knowing the answer that is "Simple, obvious, neat and wrong"  to "I have no clue what it is".

Is there a resource/method that would allow a layman to understand how to think about distortions of space like its expansion; to visualize and understand the concept correctly and without erroneous simplifications; understand it before trying to delve into underlying details of physics?

Basically, if two particles are placed with no other interaction between them, the distance between them will increase.

Imagine living on the surface of a balloon which is being blown up. Your size stays fixed, because you're more or less rigid, but items not attached to you will move further away. Your ruler, another rigid body, stays fixed in size (though it may bend to accommodate the new curvature — this isn't so important). But two rulers (which are not attached to each other) move further away.

  ...also, if everything, including all our measurement devices expands at the same speed, how can we determine the fact it's expanding? :D

On the outset this seems true, however there are other forces at play here. Our measurement devices are held together by electromagnetic interactions, and the strength of these will not change. So the measurement device will hold itself together.

Imagine two faraway atoms. When space expands, the distance between the two atoms increases. However, the size1 of the atom does not — this is determined by electrostatic equilibrium (and quantum mechanical considerations), and this remains unaffected. Even if the atom was stretched, it would rebound.

This scales up to measurement devices, so they don't get distorted either. Indeed, the expansion of space only really makes sense when you look at galaxies — these are pretty far away (when not in the same supercluster) and they don't have any interactions maintaining an equilibrium distance between them.

1. Whatever closest analog we have to "size" for atoms; eg the area which contains 99% of the charge density; or the nth Bohr radius.


The Jupiter Effect Book This evening I have been thinking about the anxiety caused in 1982 by predictions in John Gribbin and Stephen Plagemann 1974 book The Jupiter Effect.
I was a teenager in 1982. The planets aligning somewhat together on one side of the Sun did not cause any problems on Earth.
I am curious to know how frequently the same 5 planetary positions recur, say Mercury in Aries, Venus in Scorpio etc., when we restrict ourselves to the 5 naked eye planets.  (By the way, I do not believe in astrology; I am just trying to find the date of an ancient text that quotes planetary positions).
Say, if the orbital periods of Mercury, Venus, Mars, Jupiter and Saturn are 1/4, 2/3, 2, 12 and 30 years, then the least common multiple (LCM) is 60 years.  So, the same planetary positions will repeat every 60 years.  Is my logic correct, though my orbital periods are not exact?  I know very well that the 5 visible planets do not have the resonant orbital periods that I have mentioned above, so I am sure that the same planetary positions do not recur every 60 years.  If I used the correct orbital periods, will I get the correct answer to my question, whatever the correct answer may be?  The above Wikipedia article says that the 1982-type planetary alignment occured in 1128 AD. One of the article's references says that all planets more or less line up behind each other on the same sector of the sky once every 500 years.
Thus, my real question is: Is there a simple method to calculate (without using astronomy software) how frequently the same combination of planetary positions recur?
The method does not have to be too accurate.

The inner planets are not resonance orbits with each other,  this means that the ratio Earth-year:Martian-year is irrational. And the exact postions of the planets don't recur.
The positions do approximately recur, but the time between approximate recurrences depends on how accurate you want the recurrence to be.  "accurate to 90 degrees?"  "15 degrees?"  "1 degree?"  Are you only counting right-ascension, or are you also looking at the declension too (the planets don't orbit in exactly the same plane)  Are you willing to make Keplarian approximations, or do you want to include perturbations.
The motions of the planets are simple enough physics, but not simple ratios.  So predicting a "recurrence to a particular accuracy" needs physics to forecast.


Why Earth's perihelion occurs 3rd January rather than 1st? Why Earth's perihelion occurs 3rd January rather than 1st January? Is there any effort to correct this discrepancy?

Back in 1246 CE, the perihelion coincided with the December solstice. The Earth was at perihelion at 1 Jan in 2005 and will be on 6 Jan in 2096, GMT time. The drift is not monotonic, but it does progress overall throughout the centuries. Eventually, it will drift towards the March equinox and beyond, to come back to the winter solstice in around twenty thousand more years. This is primarily due to Milanković cycles of Earth's eccentricity and precession index.

But the problem isn't just effectively unfixable; it's just not a problem in the first place. The apsides were seldom a concern for calendars. The practice of 1 Jan as the new year may have begun with the Romans, but it was not continuously practiced throughout Europe, e.g., the early medieval new year began at several different days from early to late March at different times in history, and sometimes at Christmas.


Did a piece of Halley's comet strike the Earth 1,500 years ago? A friend showed me an article saying that there's evidence that a piece of Halley's comet may have struck the earth about 1,500 years ago. Is anyone here aware of such evidence?

Little pieces of Halley's comet hit the Earth every spring. It's called the Eta Aquarid meteor shower, but I think you are thinking of something bigger.

Dallas Abbot's paper notes that at the 533-540 point on the GISP2 ice core there are Sn-rich particles, Ni-rich particles and cosmic spherules. These she identifies as evidence of a minor cometary impact. The resolution of the GISP2 core is sufficient to place this impact in the late spring, consistent with the Eta Aquarid meteor shower, which is known to be produced by Halleys comet.

Dallas notes that an increase in cometary dust in the atmosphere could partly explain the extreme weather events of 535-536, in combination with a minor volcanic eruption. Although there are other, competing hypotheses, and this is not considered to be settled science.


If it only takes 8 minutes for that light to reach the Earth, what took up most of the time? If it only takes 8 minutes for that light to reach the Earth, what took up most of all that time to get to the Earth?

TL;DR: Because light doesn't travel at an infinite speed.
Light travels at exactly 299,792,458 meters per second, and one AU is equal to 149,597,870,700 meters. Therefore, it takes about 500 seconds, or 8 minutes and 20 seconds, for light to travel from the Sun to Earth.


What is the orbital path of the newly discovered star-less planet PSO J318.5-22? Recent results from Institute for Astronomy at the University of Hawaii at Manoa claim that there is a Jupiter-sized gas giant planet that is independent of a star about 80 light-years from Earth.

The press release refers to this planet as "free-floating" but I assume that it must be in some kind of predictable orbital path.

What is the orbital path of this newly discovered planet?

There is no "orbital path" detected, that's why it is a "free-floating planet". There is no radial velocity mesured, but the informations given by its kinematic location show that it belongs to the beta Pictoris group, that is a stellar group.

For more dirty details, have a look at the submitted paper on PSO J318.5-22:
http://arxiv.org/abs/1310.0457

Comment:
Apart from that (the following reflects my personal opinion) the term "free-floating planet" is ill-chosen; it is clearly a very low-mass object, but since it is not orbiting around another larger object, it seems to me that the term "planet" is not pertinent. I think that it should be consider more than a "very low-mass brown-dwarf". The problem arises from IAU definition of a brown-dwarf and a exoplanet, that is probably not well-fitted for this kind of objects (and that is not really physical). You will notice, by the way, that this object is moving in a stellar group, which kind of reinforce my point.


Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength? In Tonry & Davis (1979), they describe spectroscopic redshift measurement via correlating with templates at known redshift. In Section IIIa, they say "Because the spectra are binned linearly with ln$\lambda$, a velocity redshift is a uniform linear shift.  Why is this the case?

Here is the relevant paragraph from the paper:

All this means is that you need to bin your spectra in equal intervals of log wavelength for each pixel to be a constant interval in velocity.

First consider the case whereeachpixel is worth a constant interval in linear wavelength. Here we have
$$ \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c},$$
and the $\Delta v$ represented by each pixel depends on $\lambda$, which changes across the spectrum.

But now if you bin in equal increments of
$$\Delta \log \lambda = \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c}$$
and each pixel has the same fixed velocity interval, $\Delta v = c \Delta \log\lambda$, independent of $\lambda$.

This log wavelength binning is a prerequisite for cross-correlation procedures that yield velocities.


Streaks on a stacked star field Stacking by Sequator 100 dng photos of night sky taken with a smartphone resulted in a (beautiful, I would say) populated star field where undesired arc streaks are seen all over the field. It is like as the program treated the vast majority of stars as wanted, while (perhaps the less brighter) others are just rendered as they moved along the 100 frames period.
Align stars was chosen then accumulation.
Question: Is this a common problem using Sequator, or at least happened to some of you? Is there any trivial setting or way to redo the stacking? Thanks.
Attached is a compressed copy of the final photo.

the above is cross posted in Photography SE
https://photo.stackexchange.com/questions/124748/arcs-in-stacked-star-field

Those are probably just "hot pixels" that don't move on the sensor like the real stars do, so that the stacking spread them out in an arc.
I don't think the software processes and treats each star separately, they can't be "skipped stars" mingled with "recognized stars". Notice that they are all 1 pixel wide, whereas stars are much wider.


How often are new astronomical objects (variable stars, supernovae, comets, etc) discovered by amateurs? How often are new astronomical objects (variable stars, supernovae, comets, etc) discovered by amateurs? Where could one report new findings?

The answer is: frequently. There are many amateur astronomers that make it their ambition to discover new supernovae or to observe and report on new variable stars.

As an example, let me cite amateurs Robert Evans, who apparently holds the record for most supernovae found by visual observation, or Tom Boles, who holds the record for supernova discoveries by an individual.

Observations of variable stars can be reported to the Information Bulletin of Variable Stars (IBVS). Supernovae discovery or new comets would normally be reported to the International Astronomical Union Circulars.


Alfven Radius of mainsequence stars Is there any average value for the Alfven radius of main-sequence stars? if there are no such values, then how may I calculate it??

There is no average value. The large scale magnetic fields of main sequence stars are both a function of spectral type and rotation. In turn, rotation is age-dependent. In addition, the gas pressure at a distance from the star depends on conditions in the stellar wind and this too will depend on spectral type, rotation and age.
There are no universally agreed models for the behaviour of these quantities.


Is the Earth-Sun distance 1.012 AU? Wolfram Alpha's query: Sun Earth distance in AU gives:

  1.012 au  (astronomical units)

They definitely use this value in other places too - e.g. Earth-Sun $L_1$ distances from Earth and from Sun don't add up to 1 AU - they are 0.01009 AU and 1.001 AU respectively.

Contrarily, Google gives Earth Sun distance in AU = 1.000 AU.

The top search results for Sun Earth distance 1.012 AU yield a page with a big heading of "Paranormal UFO Aliens Wow!".

Is the 1.012 AU value an error in Wolfram Alpha, or is it some little-known adjustment, redefinition or other quirk of the unit or the Solar System?

As of this moment (2016 May 18, 13:15 UTC), the Earth is 1.0116 astronomical units from the Sun. WA is smart enough to know that "Sun Earth distance in AU" is a time-dependent question.

Update: The stricken text that follows from my original answer is incorrect. I am leaving it present (but stricken) for the sake of humility.

It interpreted your query to mean Sun Earth distance today in AU, and because today is 2016 May 18 and because you didn't specify a time of day, it picked noon (UTC), it in turn interpreted your query to mean Sun Earth distance on 2016 May 18 at noon UTC in AU.

The correct answer: I happened to ask WA the very question posed in the OP a couple of minutes apart and got two different answers (1.014 and 1.015 AU), and this did not occur across a day boundary. WA apparently interpreted your query to mean the current distance between the Sun and the Earth.

You have to be very careful of what you ask WA.

A better query is to ask WA What is the mean distance between the earth and the sun in AU? That query will give you an answer of 1.0000010178 au. An even better question is What is the semimajor axis of Earth's orbit? That will give you an answer of 1.00000011 au. Note the extra zero, but also note that for some reason, the precision is reduced.


Deducing the hour based on RA I'm making a technological project. The goal is to write a program that gets an image of the nightsky and predicts the hour in which the photo was taken.

I implemented a well-known method using the big dipper, but it's not enough. I don't want to limit my program to only work when the big dipper is in the image.

When I read about right ascension, my gut told me seeing a star in the sky and knowing its right ascension (RA) would help me deduce how many hours ago the sunset happened, though I couldn't explain how.

I'm not comfortable enough with this coordinate system, so I need help answering this question:
Say I konw the location a photo was taken, and I can identify the celestial objects in the sky and get their RA, can I make some calculations to konw/approximate the hour? If so, how?

Thanks in advance!

Will you know the date and location from where the image is taken?
If you are sucessfully using the big dipper already you probably using this info.
There is a "degeneracy" between date and time.  The sky rotates once a siderial day (23hrs 56 min) and once a siderial year (which is 20min longer than tropical year: sun to same position). So the sky tonight at say 20:00 is the same as the sky at 20:04 the previous day. In two weeks that's 1 hour. So you need the date.

At the same time (say Universal Time == London time) at each longitude the observer is facing a different direction. So 15 deg long is equivalent to 1 hour in time. So you need longitude. Others have alluded to latitude. You see stars diffrently at different latitudes.   For example at the equator the pole star is on the northern horizon, and invisible in the sothern hemisphere.

The core relationships is between right ascension (like "longitude" for stars), siderial time and hour angle. The hour angle is the angle between the meridial and position of star. Or the siderial time is the right ascension of stars crossing the meridian. You could work this out from an wide field image using a service like https://nova.astrometry.net/ and then convert "siderial time to local time". There's a stack overflow answer plus others.


Which stars have I been seeing? In the past mornings I have used to get out before sunrise for a stroll, to get some exercise while not meeting other people.

I live in Eindhoven (NL), and I get out of home at around 5:30 AM, when the Sun is not yet risen, the sky is just about to start getting brighter.

At about 30-40 degrees to the right from the rising Sun point, about 20 degrees above the horizon, I see a bright white star.

About 5 degrees to the left of this star, I see two reddish objects, about 2 degree apart from each other, much more faint than the other star.

Which stars are they? I have tried to find them in a sky map, but haven't been able to find anything resembling them.

My interpretation would be that these are the planets Jupiter, Saturn and Mars.

The bright star is Jupiter. It is rather lower than 20 degrees above the horizon, it is at about 10 degrees, but rising.  The two other stars, one brighter than the other are Saturn (brighter) and Mars.  Mars is really red, Saturn may be reddened by being low in the sky.  Saturn is about 5 degrees East of Jupiter and Mars is a couple of degrees more.

To get your bearings, locate the "summer triangle" of Vega, Deneb and Altair. This triangle (currently) points towards the area containing the three planets. (Draw a line between Deneb and Vega, and through Altair and continue on a great circle in the sky)

I used Stellarium to view a simulated sky from Eindhoven at this time.


Differences between alt/az in Stellarium and Astropy I've just started to learn Astropy. It's a nifty tool. Using some tutorials, I wrote and then made modifications to a little script that finds the Alt/Az of an object at a certain place and time. I then checked Stellarium's calculation for the Alt/Az of the same object (at exactly the same place and time) and found it to be slightly different. I suspected this was due to slight differences in the underlying coordinates, so I tried manually hard coding the coordinate data for the star from Stellarium; it still came out different. Here is the data without the hard coding:

Object: Rasalhague (alf Oph)
Time: 2018-08-28 23:00 (converted to universal in the script)
Latitude: 35
Longitude: -79
Stellarium reported Az/Alt: +247 35 12.3 / +48 00 57.8
Astropy reported Az/Alt: +247 35 37.4939s / +48 00 08.7586

Manually coding:

RA/Dec on date from Stellarium: 17 35 48.82, +12 32 55.9
Astropy reported Az/Alt: +247 22 42.0379 / +48 09 43.0755

I'm guessing that the difference stems from the conversion algorithm to alt/az.

Perhaps the difference stems from the conversion algorithm?

Almost certainly this is the answer.  It's not clear to me exactly what Stellarium assumes for is refraction calculation, but I know Astropy's algorithm.  Given that the altitude you've listed here is below the horizon (i.e., alt is negative), it's not really clear what the correct interpretation is: in particular, atmospheric refraction is not really well-defined below the horizon (since it should be invisible anyway!).

Additionally, even well above the horizon where refraction isn't that important, there are several finicky details that you have to account for to properly do the Az/Alt conversions.  In addition to the precession that the other answers mention, if you want to get exactly the same answer on two tools, you have to worry about things like how much the axis of the Earth shifts due to, say, major earthquakes ("true polar wander"), the effects of General Relativity on the apparent direction of the incoming light ,etc.  Most of these are pretty small, but together they mean it's very hard to ensure all of the software gives exactly the same answer unless they carefully follow the definitions set by the International Astronomical Union.

That said, the differences you see are only ~20 arcsec.  The human eye isn't even capable of telling that small of a difference, and almost all telescopes that people would use Stellarium with probably wouldn't either (at least in an absolute sense).  So for this particular application they essentially do match.


How much of a difference do good lenses make? I recently purchased a Celestron PowerSeeker 114EQ and with it I got three lenses (which I assume are low quality), a 4mm, a 20mm and a 3 x Barlow.

I can get pretty decent views of the moon but a very blurry, small view of Jupiter with the 4mm and Barlow.  I can see the color of the storm rings for example, but barely.  I've not had a chance to have a go at Saturn yet.  A view similar to this...

I've been reading up on this and was looking at purchasing a 4-8mm Plössl lens but I'm not sure if the lens is my weak spot or some other aspect of the telescope.
Another thing to note is that I've been using it in a pretty bright street, looking up very close to street lights.  I've not had a chance to take out somewhere with less light pollution.

So what I'm asking is, how much difference would a lens upgrade make, specifically to my image quality?

...a very blurry, small view of Jupiter with the 4mm and Barlow...

Be aware that a 4mm eyepiece and a 3x barlow at the same time will give you a very high magnification - too high!  For regular Jupiter viewing I would suggest you stick to 100x or 200x at most, unless the air is exceptionally still. (After a few sessions you'll find out what "Still" air means.)

A lot of beginner scopes like this are sold with barlows that tend to give too much power, my advice is to keep it packed away for most sessions.

So I'd stick to using that 4mm on its own, jupiter will look quite small but with practice you can usually tease a bit more detail out of the image.

I don't have one of these scopes, but my experience of comparing cheap modern eyepieces with expensive ones is that the lower cost ones are generally OK these days. 4mm is quite a high power so the extra magnification will cause a lot of blurriness and it will take patience to get the focusing at its sharpest, so don't panic if everything is looking fuzzy at the moment.

For a first stage I would suggest a few things that are nothing to do with the eyepieces at all...

Collimation - look this up, it just means adjusting the two mirrors
so your eye is looking right down the tube in a straight line. For a
long focal length scope like yours, it's not likely to be a problem
unless one of them is wildly out of line.
Tube currents/thermal behaviour - on almost any cold night, when
the tube and main mirror are still warm from being indoors, rising
air currents in the tube will mess up your image and make it
shimmery, at high power. Low power will look OK. It might take an hour or so for the image to improve (a guess)

Anyway, in summary my guess is changing the eyepieces straight away won't make a dramatic difference. Eyepiece makers will say otherwise of course :)


Are any supergiants translucent? Are any supergiants translucent? Some have volumes thousands of times more than the Sun's while having maybe twenty times the mass of the Sun which makes them sound rather diffuse. If there was a very bright distant star behind the supergiant would we be able to see it?

No mass blob of stellar mass is transparent at any wavelength of interest. Opacities $\kappa_{\nu}$(inverse transparency) as function of wavelength becomes really high and broad band at pressures above > 0.1 bars, for all wavelengths.
This leads to the optical depths $\tau_{\nu}$ being enormous and as transmission is $T=1-\exp(-\tau)$, you won't be able to see through any star, ever.


Is the Drake Equation an accurate way of finding the probability of life on planets? I was interested in finding out if there was life on any planets or moons. I saw ReNiSh A R's question, and I saw the answer about there being life on Europa, but it didn't accurately answer what I wanted to find out.

The Drake Equation is on: http://en.wikipedia.org/wiki/Drake_equation.

I have also heard that there could possibly be life in the upper atmosphere of Jupiter, which could have floating beasts, and also on Titan, which does have surface liquids, but I am pretty sure most of it is in solid form.

If the Drake Equation is equal to a low number, then why is it pretty common if there is a possibility of life on 2 moons and 1 planet? Does that make the Drake Equation a non-reliable source?

I'm also looking for an accurate answer, so can you please give it to me in detail!

The Drake equation is little more than a randomly assembled guess of relevant factors for intelligent life, and any use of it requires a number of hard to justify guesses for the various probabilities.

It is otherwise an attempt to give a foundation for making Fermi estimations on the number of civilizations, and the associated Fermi paradox. A Fermi estimation is an order of magnitude estimate. It doesn't try to get a particularly accurate estimate, it only tries to tell you how many digits you can expect the answer to have. And the Fermi paradox is simply stated as "There is an absurd number of stars. So absurd that finding extraterrestrial life should be easy. But in fact it is hard. Why?"

The resolution to the paradox is the second sentence: it assumes we have reasonable basis to suppose life is easy enough for the sheer number of stars to give us lots of observable life. We do not. The Drake equation is in one sense a way of singling out as many factors as possible so we have a concrete goal to figure those factors out, and so the entire problem.

But mostly it is used by pop science and sensationalists to rile us up with lines like "there should be thousands of civilizations more advanced than ours in our own galaxy!" and "there could be life in Jupiter's atmosphere." These things have no solid science behind them, and are just attention grabbers dressed in pseudoscience.


Telescope light gathering power and resolution I'm a little confused. Could you help me?

The question is;
If we increase the radius of a telescope twice, how much does the light gathering power change? What about resolution?

Is it increase 2^2?
For second part i know As the diameter of the telescope's objective increases, the resolving power increases.

Thanks

The collecting area of a telescope is roughly proportional to the square of its radius (for a circular aperture). I say roughly because you also have to factor in the small fraction of light blocked by the secondary if we are talking about reflecting telescopes.

In principle the angular resolution decreases as the reciprocal of the radius (that is, the smallest resolvable angle gets smaller). In practice, this may not be the case for ground-based telescopes without adaptive optics capabilities, where the smallest resolvable angle could be determined (for large telescopes at least) by turbulence in the atmosphere (a.k.a. the seeing).


Does the Sun rotate? The planets rotate as an after effect of their creation, the dust clouds that compressed span as they did so and the inertia has kept it rotating ever since.

It's fairly easy to prove that planetary bodies are rotating just by watching their features move across their respective horizons.

This seems less easy to discern for amateur astronomers  in the case of the sun though.

Does the Sun also rotate as a by-product of it's creation? What evidence is there to support this? Does the sun have any discerning features that make it evident it is rotating?

Yes.

It does not rotate uniformly though, different portions have a different angular velocity (as a body made of plasma, it can get away with this).

Measuring this in theory is pretty easy, we just need to track the motion of the sunspots. This isn't as simple as calculating the changes in relative positions of the sunspots, though, as the Earth is rotating and revolving, which makes the calculations harder. This measurement can be done using the celestial sphere (the field of stars that we see) as a "fixed" reference point and seeing how the Earth and the sunspots move relative to that.

Almost everything in the universe rotates/revolves, at least a little bit, because angular momentum is hard to get rid of. It can be transferred from body to body, but for a body to end up with zero angular momentum, it needs to meet another body with the exact same angular momentum and collide with it in a particular way. Given that this is pretty rare, all celestial bodies rotate.

In addition to that, a non rotating body that is revolving will eventually start spinning due to tidal forces.


How would I measure that I'm at a pole? How would a person measure that he is at a planetary pole? My first inclination is to use a sextant to ensure that the Sun remains at a constant inclination. However, due to the orbit around the Sun and a possible planetary axial tilt, the Sun's inclination may vary over the course of the year and that could affect measurements if the length of the day is a significant portion of the length of the year.

A second inclination is to try to identify a pole star or at least the point about which distant stars rotate during the course of a planetary rotation. The observer could then (at least in theory) measure the inclination of this point assuming that planetary precession takes a long enough time (as on Earth). However, in the most likely chance that there is no bright star near the pole, I don't see how a sextant could measure reliably the inclination of a blank point.

Perhaps the first method could be used, but with other objects in the sky (stars) in place of the Sun. Thus, if a given star's inclination does not change as it (apparently) revolves around the horizon, we are at the pole.

Are there any better methods, assuming a completely self-contained system (i.e. no external satellites)?

You could confirm with taking an image of the star trails.  They would form a circle with the apparent center at the zenith of the location.  You do not need a pole star at all.  Just a night of viewing.

You would also be able to tell based on the height above the horizon that the sun is.  On the equinox, the sun would be on the horizon at noon (when it is at its maximum height).  On the summer solstice, its angle will be above the horizon equal to the angle of precession.  And below by the same amount on the winter solstice.

Measuring latitude using the sun

For any other day in the year, you will need to determine the max angle based where you are between the equinox and solstice.

Compensating for the Earth's tilt

Basically, you would determine if you were at a pole the same way you determine any other latitude.  You are just looking for a result of 90 degrees.


What is the size of the image sensor in the largest optical telescopes? What image sensors (imaging electronics) are used in telescopes? Like CCD, is that the best option?
What is the typical physical size and resolution of the photo-sensitive surface of the currently active largest ones?

The current largest digital CCD camera is that of the Vera C. Rubin Observatory1 which has a whopping 3.2 gigapixels. The previous largest features on the Panoramic Survey Telescope & Rapid Response System (Pan-STARRS), and has a resolution of ~1.4 gigapixels.[1]
Based on the spec sheet provided by the Vera Rubin Observatory, the LSST camera has a resolution of roughly 0.2 arcseconds per 10 $\mu$m pixel. It is about 5 feet (1.52 meters) wide and weighs over 6000 lbs (2721 kilos, $2.6\times10^9$ dyn).[2] The actual photosensitive portion of the camera is ~64 cm ($4\times10^{34}$ Planck lengths) across.
Cameras of this size often have pointing, calibration and recording issues. These sources of error are usually extremely well understood, but it's still interesting to see. Pan-STARRS has a detailed list on their data site[3]. These issues include:

Randomly missing data that gets filled in later
Pointing errors related to astrometric positions of their targets
Registration issues near the celestial pole resulting in poor photometry

1formerly known as the Large Synoptic Sky Survey or LSST. The acronym has now been repurposed: What is the LSST now? Where does LSST end and Vera C. Rubin Observatory begin?


List of galaxies with their dark matter halo Where can one find the list of galaxies with their dark halo masses, I've searched several sites but I found no such proper list where the masses of their dark matter halo is compiled?
Can anybody help?

This paper by Wechsler & Tinker (2018) should give you a better direction. Depending on the model, different estimates of DM halos come up. You just need to go through the sections on observational evidence and the underlying papers.
https://arxiv.org/pdf/1804.03097.pdf


What kills us in the event of two galaxies combining? I find the joining of galaxies interesting as the two eventually become ONE. On series I saw on a discovery channel stated we die.  Everyone dies. What is it that causes us to die when the galaxies combine? Is it something colliding with the Earth from the other galaxy such as a planet? Is it the drastic element changes on the Earth such as extreme temperatures or flooding, etc. Do we have any percent of surviving on Earth?

There are two reasons why nobody will die when the Milky Way merges with Andromeda:

The merger won't take place for another 4.5 billion years, by which time solar evolution will have long since made Earth uninhabitable (and even if humans did colonize the outer planets, we'd have long since gone extinct or evolved beyond recognition).
The merger itself will be a slow and not very dramatic process - stars aren't going to collide, planets aren't going to be ejected from their solar systems, and while it will look dramatic from a distant galaxy, other than Andromeda looking larger in the night sky not much will change for any intelligent life that's around at the time.

However, there's one potential wrinkle. Both the Milky Way and Andromeda host supermassive black holes at their cores. We've never observed a merger between two supermassive black holes, and modeling the process is theoretically challenging (look up the "final parsec problem" if you're interested). When this does happen, though - long after the merger of the stellar portions of the galaxies - there would be a massive release of energy. Most of this will be in the form of gravitational waves, but if enough of it comes as high-energy radiation it could be bad news for any remaining life.


When was the term "orbit" (in the modern sense) first used and by whom? Wikipedia coyly suggests that some notions and views kept circulating (apparently since Adam and Eve) until finally Kepler, of all people,  ... etc. I find this hard to believe. Please elucidate.

Searching for the definition in dictionary.com you can find this under the word origin and history section:

  Late 14c., "the eye socket," from Old French orbite or directly from
  Medieval Latin orbita, transferred use of Latin orbita "wheel track,
  beaten path, rut, course, orbit Astronomical sense first
  recorded 1690s in English; it was in classical Latin, revived in
  Gerard of Cremona's translation of Avicenna.

I believe that this is what you are looking for.


Is every single large solid body orbiting entirely outside of the frost line covered in ice? Not sure where I heard that but I want to make sure if it's true. Forget about the asteroid belt for a minute: Are all the moons of the outer planets and kuiper belt objects covered in water ice?

No, they aren't all covered in water ice.
As an example, Io, a moon of Jupiter is covered in volcanically deposited sulfur and sulfur dioxide frost.

Solar system mechanisms are pretty complex and varied, so we wouldn't expect similar quantities of water as exist on Earth.  Also, the "frost line" for water in the Solar System isn't the same as the "frost line" for many other compounds like Ethane and Methane.  Saturn's moon Titan has lakes of the stuff:

Instead of water snow, Titan has benzene snow that dissolves as it falls in the lakes!  Delightful...


Why is the magnetic axis of Uranus and Neptune off center? It(Uranus) rotates on its side, tilted almost 98 degrees from the plane of its
orbit around the sun. The axis of its magnetic field is tilted too, at
a 59-degree angle from the rotational axis. The magnetic field is also
off-center, with the field lines emerging about a third of the way
toward the south pole. (source)

Neptune's magnetic field is tilted 47 from the planet's rotation axis
and is offset at least 0.55 radii, about 8,500 mi (13,500 km) from the
physical center. The dynamo electric currents produced within the
planet, therefore, must be relatively closer to the surface than for
Earth, Jupiter, or Saturn. Because of its unusual orientation, and the
tilt of the planet's rotation axis, Neptune's magnetic field goes
through dramatic changes as the planet rotates in the solar wind. (source)

So, it is well established that the magnetic axis of Uranus and Neptune is off-center and due to this, the planets faces the following consequences:

the magnetosphere  is irregular
the strength of magnetic field varies, almost opening and closing periodically as the magnetic field lines disconnect and reconnect
the orientation of magnetic field changes constantly.

But, why is the magnetic axis off-center in the first place? I found no source answering that. Is it due to alignment/orbit/internal heating mechanism of the planets?

Disclaimer: This answer is inspired by Magnetic fields of Uranus and Neptune: Metallic fluid hydrogen by Nellis [2017], who explains the center offset of the ice-giants' magnetic fields by contrasting them with Earth's magnetic field.
Earth's axisymmetric magnetic field
The most researched magnetic field in our Solar System is, of course, Earth. Nellis 2017 explains how coupling between the rotational motion and convective dynamo motions in the Earth's outer core cause the magnetic axis to drive towards alignment to the spin axis:

Because rotational motion (RM) of Earth is strongly coupled into
convective dynamo (CD) motions of its fluid-Fe outer core, planetary
RM stabilizes convective motions that generate a dipolar magnetic
field. If a convective fluctuation occurs which tends to destabilize a
given dipolar axis, then strong RM-CD coupling either drives
convective motions that essentially restore the initial orientation or
CD fluctuations that drive the initial magnetic axes out of
orientational equilibrium are so strong that RM-CD coupling eventually
drives the dipolar axis into an alignment anti-parallel to its initial
one.

Uranus/Neptune non-axisymmetric magnetic fields
Voyager 2 measured both gravitational moments as well as the magnetic fields of both Uranus and Neptune.  These measurements helped develop models of planetary composition as well as models of magnetic field generators. For Uranus/Neptune, the fields are generated near the surface at radii as large as $\approx0.9R_U/R_N$, while the Earth's magnetic field is generated much deeper at $\approx0.5R_E$.
The magnetic generation near the surface, as well as the H-HE envelope (again according to Nellis) means that:

local convective dynamo motions of fluids that produce the magnetic
fields are essentially decoupled from global rotational motions [...].
The dynamos of U/N would then be relatively free to wander as local
convective fluctuations dictate.  Thus, tilt angles and center-offsets
of their fields would vary slowly over the age of the Solar System.

Summary
The Earth's composition and convective liquid outer-core currents cause a coupling between the rotational motion of the Earth and the convective dynamo motion.  This drives the magnetic axis to align with the spin axis, which passes through the center-of-mass.  No similar strong coupling exists for Neptune and Uranus, which means the magnetic axis on each of these planets is free to wander both in orientation and offset from the spin axis and center of mass due to local surface convection currents.


Has Hubble ever been used to try to image a near Earth asteroid? This answer points out that the near Earth asteroid Apophis will likely be close to 2 arcseconds in diameter as seen from Earth during its close approach in 2029. I speculate that if the Hubble Space Telescope were still operational then, it could potentially image the asteroid in visible light at a few dozen pixels in diameter.

This leads me to wonder if the Hubble has ever been used to image† or at least spatially resolve in some way an asteroid during a close pass to the Earth before.

†Here the verb "image" should be taken to mean the act of producing a resolved image of an object so that different pixels correspond to intensity from different parts of the body being imaged. For the purposes of this question please don't consider telescope images in which an asteroid happens to appear but is too far away to be resolved. Thanks!

[rewritten to address the revised question]

Maybe, depending on how fussy you want to be about "resolved".

This is a study from 1995, using observations of asteroid 4179 Toutatis made in 1992 with HST. They reported marginal resolution of the asteroid, as suggested by this figure comparing a deconvolved image of a star (observed with the same filter and imager location) and a similarly deconvolved image of the asteroid itself (each pixel corresponds to about 450 m at the distance of the asteroid):

The appearance of the asteroid is pretty clearly not a point source, but it's also fair to say it's only partly resolved, and mostly just in one direction.

(My admittedly vague impression is that this is one of the best, if not the best, case of HST "resolving" a near-Earth asteroid.)

Most observations of near-Earth objects with HST are, I think, aimed at getting optical information on compositions not possible from other wavelengths, and sometimes refining estimates of rotation rates, as was done this year (using data from 2012) for the asteroid Bennu, currently being visited by OSIRIS-REx.

In practice, you get much better spatial resolution using radar (including line-of-sight distance variations due to the structure of the asteroids from time-of-return measurements, allowing you to construct 3D models of them), so there's not much point in trying to resolve them with HST.


Sedna, VP113 and the likelihood of the PX/Tyche/Thelistos hypotheses Crosslisted question from https://physics.stackexchange.com/questions/131876/sedna-vp113-and-the-likelihood-of-the-px-tyche-thelistos-hypotheses

The recent discoveries in exoplanetary science (specially those findings of planets orbiting far far away from its parent star/stars) raise questions about how much we know about the (true, AIU definition) number of planets in our solar system.

Question: does the discovery of VP113 and Sedna increase the odds of some kind of planetary object in the edge of the known solar system (Oort's cloud)? How could we find it if it is very cold and dark? I mean, how many Sedna or VP113 objects we would need in order to guess its position and location?

Reading about this topic, I don't know how to quantify the odds of finding

PX(Planet X): An Earth sized body perturbing the orbits of Sedna, VP113 and other similar objects.
Thelistos: A super-Earth (about 5-10 Earth masses times a O(1) factor)
Tyche: (Now disfavoured by WISE data, I think it can not be excluded if it is extremely cold and "dark") Gas giant (Neptune-like, Saturn-like, etc)
Nemesis: A "dark star"/brown dwarf object.

... raise questions about how much we know about the true (IAU definition) number of planets in our solar system.

Eight.

Question: Does the discovery of 2012 VP$_{113}$ and 90377 . increase the odds of some kind of planetary object on the edge of the known solar system (the Oort cloud)?

2012 VP$_{113}$ is a minor planet, that does not imply that it is a "planet".
90377 Sedna is a large minor minor planet, also not a "planet".
The Oort cloud has so many tiny objects that they upset each other's orbit, occasionally sending comets into our solar system. Part of being a planet is having an orbit and clearing your orbit; these objects do neither.

How could we find out if it is very cold and dark?

Every object in space affects the movement of every other object.
We can detect tiny black holes at a distance, they are cold and dark; detecting a new planet in our solar system isn't beyond our ability, there simply isn't one there.
It's not impossible for a large object to begin orbit of the sun and establish that it will orbit a reasonably fixed path (some wobble is OK, drastic changes in direction or too close to something large enough isn't going to maintain it's orbit (and clearing ability)).
Pluto takes 248.00 years (90,560 days) to complete an orbit. Discovered in 1930 it lost its status as a planet in 2006. If something appears out of nowhere it will take time for it to be approved, so no surprises in the next few years.

I mean, how many 90377 Sedna or 2012 VP$_{113}$ objects would we need in order to guess its position and location?

Objects very distant to each other affect each other to a much smaller extent that objects which are closer to each other. As time passes, observations and new calculations are made, results are confirmed, and it passes IAU committee approval something might be discovered and become a new planet, instead scientists are looking outside our solar system for exoplanets.
Beta Pictoris (β Pic) is the second brightest star in the constellation Pictor. It is located 63.4 light years from the Solar System. Scientists believe there are a few planets in orbit around the star.
"The planet companion around β Pictoris" (Oct 2010), by Anne-Marie Lagrange, Mickaël Bonnefoy, Gael Chauvin, Daniel Apai, Et al.:

Abstract
The β Pic disk of dust and gas has been regarded as the prototype of young planetary systems since the 1980s and has revealed over the years an impressive amount of indirect signs pointing toward the presence of at least one giant planet. We present here the recently detected first giant planet around this star. We show how this planet could explain some very peculiar features of the star environment (disk, spectroscopic variability), and how it constrains the scenarios of planetary system formation (timescales, mechanisms).

Scientists can see far and do look for planets, it's unlikely that there's a 11th planet that's been unobserved. They have nicknamed β Pic's planet "b". Here's its photo:

Reading about this topic, I don't know how to quantify the odds of finding:

PX (Planet X): An Earth sized body perturbing the orbits of 90377 Sedna, 2012 VP$_{113}$ and other similar objects.

There is support for the existence of the object, but under the IAU definition imposed by your question it hasn't yet met the requirements to become a planet. The announcement does not mean there is a new planet in our solar system. The existence of this distant world is only theoretical at this point and no direct observation of the object nicknamed have been made. The mathematical prediction of a planet could explain the unique orbits of some smaller objects in the Kuiper Belt, a distant region of icy debris that extends far beyond the orbit of Neptune. Astronomers are now searching for the predicted planet.

Thelistos: A super-Earth (about 5-10 Earth masses times a O(1) factor)

Misnomer. Deleted from Wikipedia, redirects to planet X.

Tyche: (Now disfavoured by WISE data, I think it can not be excluded if it is extremely cold and "dark") Gas giant (Neptune-like, Saturn-like, etc.)

Disproven. Does not exist.

Nemesis: A "dark star"/brown dwarf object.

128 Nemesis is an asteroid, the other Nemesis is a hypothetical red dwarf or brown dwarf, originally postulated in 1984 to be orbiting the Sun at a distance of about 95,000 AU (1.5 light-years). Thought to be somewhat beyond the Oort cloud, it's postulated to explain a perceived cycle of mass extinctions in the geological record, which seem to occur more often at intervals of 26 million years.

Which objects are defined as planets is precisely defined, there's only some overlap for non-planets:

Here is the text of the IAU’s Resolution B5: Definition of a Planet in the Solar System (.PDF):

The IAU therefore resolves that planets and other bodies, except satellites, in our Solar System be defined into three distinct categories in the following way:
(1) A planet$^1$ is a celestial body that
$\quad$ (a) is in orbit around the Sun,
$\quad$ (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and
$\quad$ (c) has cleared the neighbourhood around its orbit.
(2) A "dwarf planet" is a celestial body that
$\quad$ (a) is in orbit around the Sun,
$\quad$ (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape$^2$,
$\quad$ (c) has not cleared the neighbourhood around its orbit, and
$\quad$ (d) is not a satellite.
(3) All other objects$^3$, except satellites, orbiting the Sun shall be referred to  collectively as "Small Solar System Bodies".

1 The eight planets are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.
2 An IAU process will be established to assign borderline objects to the dwarf planet or to another category.
3 These currently include most of the Solar System asteroids, most Trans-Neptunian Objects (TNOs), comets, and other small bodies.


Do some comets spin? If so, how fast? The GIF below is copied from my question earlier What might a CN filter be in the context of comet watching? Is it showing dust, or gas, or something else? where I'd said:

In this post on the website for the Comet Wirtanen Observing Campaign at wirtanen.astro.umd.edu, there's mention of a CN filter being used to observe the comet, and this cool GIF.

It certainly looks like the comet is spinning; it reminds me of images of the density of the solar wind as active areas on the sun "spray" out matter as the Sun rotates, or the patterns discussed in Why is one of these two concurrent fuel-dump spirals blue?.
Question: This leeds me to wonder if I'm just imagining things, or if some comets really do spin? If so, how fast have they been observed to rotate?

Yes, comets spin although measuring it can be tricky due to the coma and outgassing from the nucleus. It's easiest to measure the rotation period when the comet is inactive near aphelion although this is more difficult as the comet is fainter, necessitating a large amount of large (>4m) telescope time which is difficult to obtain.
Searching the JPL Small Body Database (select Object Kind=Comets, set  Limit by object characteristics using rot_per (h) greater than 0, add rotation period (h) to the Output Fields) should produce a list of 21 comets with measured rotation periods:

The initial rotation is likely primordial since the pre-solar nebula the comets formed from would have some initial rotation from the galactic rotation which can be increased as the proto-comets contracted during formation. The spin rate of the nucleus can be changed by the torques caused by the activity-driven outgassing from the nucleus; this was observed for Comet 67P from in-situ measurements during the ESA Rosetta mission (Keller et al. 2015) and also have been observed to change in rotation from remote observations (abstract from Bodewits et al 2018 for Comet 41P)


What is the furthest object in the observable universe? My question is about furthest object in the observable universe except Сosmic microwave background.

About 13.1 billion light years.

The z8 GND 5296 Galaxy mentioned in the article linked by TildalWave, mere 700 millions years old (after Big Bang) would be the mentioned distance from us for the light to reach us at indicated speed, considering the 13.798 bln years old universe. As we push technological barriers, we keep finding more distant objects.

The age of the universe is the hard limit in light years of distance of any observable object - there was nothing to emit light more than 13.798 bln years ago, or respectively, visible from more than 13.798 bln light years away. visible from distance that what was 13.798 bln light years 13.798 bln years ago - or, after adjusting for universe expansion over that time, currently 46.6 billion light years away.

Of course, as RhysW mentions, this limit changes over time as the universe ages and grows. The exact rate of growth is 74.2km/sec per megaparsec, or about 0.000000007% per year. I'd say, comparing to our life and current technological progress, an insignificant factor.


Can radiation affect gravity? I was wondering if the sun's radiation affects the gravity when it passes through space? Like can radiation strengthen or weaken the force of gravity?

Radiation has energy, and energy exerts gravity. You could say that it "strengthens the force of gravity" (it definitely doesn't weaken it), but it would be more appropriate to say that it contributes to the total gravitational field.

The effect it immensely small, however, and is in almost all circumstances completely insignificant compared to the gravity of other forms of mass/energy. But most of the radiation in the Universe is the so-called cosmic microwave background, and because this radiation is gradually redshifted with time, it used to be more energetic, and hence exert more gravity. It is fairly easy to calculate that until the Universe was roughly 50,000 years old, radiation actually dominated the gravitational field.

Prologue:

As Zephyr points out below, the term "exerting gravity" is not really the correct way of describing gravity. In general relativity, objects are assumed follow straight lines in space, unless a force (e.g. electromagnetic) acts upon it. What we interpret as "gravity" is the deviation from what we think is "straight ahead", caused by a geometrical change in the space (and time) — and thus in the mathematical rules we use to calculate e.g. how parallel lines behave, how many degrees there are in a triangle, etc — around a massive object. The deformation of space is given by the Einstein field equations, in which the so-called stress–energy tensor appears. This is a mathematical object consisting of 16 numbers, of which one number (the "$T^{00}$'th component") holds the energy density of everything, i.e. radiation, dark matter, dark energy, stars, planets, bicycles, and — in particular — your mom.