Difference in the baryon acoustic oscillation measurements I understand from this Wikipedia page about baryon acoustic oscillations (BAOs). I can also see that BAO measurements were first detected using luminous red galaxy samples.

Now, many papers use galaxy clusters to measure the BAO signal. What is the advantage/difference between using galaxy samples and galaxy cluster samples to measure the BAO?

EDIT:

Here are some of the papers using galaxy clusters for BAO detection.

Estrada et al. 2008
Hong et al. 2012
Veropalumbo et al. 2014

I think the most important reason is that galaxy clusters more strongly adhere to the BAO shape than individual galaxies do. Galaxies tend to have more dispersion than clusters, making the BAO signal from galaxies smeared out and thus harder to detect. This is alluded to in the Veropalumbo et al. 2014 paper introduction.

  As tracers of the biggest collapsed structures, they are more strongly clustered than galaxies

Another potential reason is that the redshift of a cluster is easier to determine than the redshift of a single galaxy. There are multiple galaxies in a cluster, giving you multiple chances to measure a redshift for that cluster. If you can't measure a redshift for a single galaxy, you can't use it in your analysis. And of course all galaxies in a single cluster are going to have the same redshift to a good approximation. The Estrada et al. 2008 and Hong et al. 2012 both indicate $\sigma_z = 0.01$.

Note also that more often than not, these studies use the photometric redshifts rather than spectroscopic redshifts. Photometeric redshifts are generally less reliable as they rely on correlations (that isn't to say spectroscopic redshifts are perfect though). If you can measure a few photometric redshifts in a cluster or else measure the photometric redshift for the bright, central galaxy with a high signal to noise, you can reduce your redshift errors, compared to measuring it for a single galaxy.

Lastly, in order to use a particular catalogue to measure the BAO signal, the catalogue needs to be both deep (i.e., measures everything down to a specific redshift/magnitude limit) and wide (i.e., measures across a large field of the sky). If your catalogue doesn't meet both of these conditions, it can't be used.

Clusters are easier to spot than individual galaxies. They can be found (as the Estrada et al. 2008 paper describes) by first finding the large, luminous galaxies which generally reside in the center of the cluster and are fairly easy to find. The cluster itself is then identified. Looking at individual galaxies may suffer from magnitude limiting effects or other biases, meaning that your analysis may not include all galaxies that exist at a given redshift. If that is the case, your two-point correlation function will not be accurate and you might miss or inaccurately characterize the BAO signal. I would expect this problem is mitigated by using the easier-to-identify clusters.

Just to point out that cluster samples are likely to be more or less complete, the Estrada et al. 2008 paper is quoted as saying

  Tests on mock catalogs indicate that the MaxBCG sample should be $\gtrsim 90\%$ pure and complete for clusters with $N_{200}\ge 10$.

The requirement of a wide catalogue is also pretty restrictive. The BAO is inconveniently large so not many deep galaxy catalogues are going to be wide enough to even see the BAO. The SDSS is one of the few galaxy catalogues that comes to mind that is wide enough and they have already found the BAO in that. Cluster catalogues are usually built up not from a specific sky survey as galaxy catalogues usually are, but from individual observations or else conglomerated from many galaxy surveys. As such they can be much wider.


damage from an Asteroid If an Asteroid, the size of the one that killed off the dinosaurs , passed by very close to the Earth but still remained outside of Earth's atmosphere, would it cause any damage ?

An asteroid that remains outside Earth's atmosphere, and therefore does not impact Earth, will have no effect on the planet.

Asteroids, even biggish ones like the dinosaur killer, are pretty small compared to the Earth. They do not have enough mass to graviationally affect earth's orbit noticeably. And if they miss earth, the pass quickly.

The concern would be that if an asteroid can pass close to Earth, that it may collide in a future orbit. The asteroid's path will be changed by the earth's gravity. However, a very close encounter will enable astronomers to get a very high quality position on the rock, and so be able to predict its future path with very great accuracy.

Comet tails aren't a worry. We passed through Halley's tail in 1910, the effect on the Earth was precisely nothing. ☄


Is a comet necessarily confined to the solar system? I had the following question:

  A comet enters the solar system. Assume that it is influenced only by the Sun, not by any of the planets. Which statement about its orbit is correct?

And I chose as correct statement:

  Its orbit might be an ellipse, or is might be a parabola or a hyperbola (because a comet is not necessarily confined to the solar system forever).

I want to know what it means by a comet is not necessarily confined to the solar system forever.

Does it mean that eventually the comet will "die" out, with the sun vaporizing all of its ice, so it will basically cease to exist while it is in the solar system?

Or does it mean that the comet can leave the solar system? If the comet enters the solar system it is subject to the mighty force of the suns gravity, then how can it escape?

It means that the sum of the kinetic and gravitational potential energies can be greater than zero. Such objects are not bound to the solar system. They fall towards the Sun on hyperbolic trajectories, picking up speed and kinetic energy as they do so. After rounding the Sun they then have enough speed to escape from the solar system again.

The origin of such comets is not known for certain - some may have come from other star systems. Some comets can enter the solar system on parabolic orbits - these have come from a long way out - perhaps the Oort cloud - and are just bound to the solar system. But interactions with other planets and the Sun can result in them gaining energy and leaving the solar system on unbound hyperbolic orbits.


Have we ever seen the formation of a blackhole? So I know that the formation of a supernova has been observed and that we have taken images of blackholes, however, has the formation of a blackhole ever been observed. How do we observe it?

“We think that ‘The Cow’ is the formation of an accreting black hole or neutron star,” said Northwestern’s Raffaella Margutti, who led the research. “We know from theory that black holes and neutron stars form when a star dies, but we’ve never seen them right after they are born. Never.” ... Another team of astronomers, led by Paul Kuin from, an astrophysicist at University College London, came to a different conclusion. They think that the Cow is a star that’s been ripped apart in what’s called a “tidal disruption event.” Information on that here.

There are only 25 Stellar black holes and candidates.

If 'The Cow' was a dwarf star being consumed by a blackhole then the most likely occurrence that has been observed is N6946-BH1.
"Candidates outside our galaxy come from gravitational wave detections. The disappearance of N6946-BH1 following a failed supernova in NGC 6946 may have resulted in the formation of a black hole." The list of other ones is here and for information on the disappearance of N6946-BH1 here.


To find the mass and radii of hill sphere of a planet after passing through a region of planetary nebula Consider a planet with initial mass m* kg and radii of hill sphere as r*.

The  planet it traveling (assume) perpendicular to the nebula at a speed of 13km/s and the nebula is approaching the planet at 15 km/s.

The nebula has a density of D kg/km^3.

The planet travels through this region for 1.4yrs in which it covers 570.5 million km.

Is is possible to calculate the mass and volume of hill sphere of the planet at the end of 1.4 yrs? If so how?
The problem I face is that with gain in mass the planets radii of hill sphere increases, hence more mass falls in which in turn again causes the hills spheres radii to increase, this process goes on and on till the end . I'm struggling to imply integration!

(In the hill sphere formula you can consider a and M to be constants)

Thank you!

I have come up with a technique to find a solution to this problem as shown in this paper
A Study to Show : Jupiter : Our Next Star
My 11th grade research paper 2018


Stellar systems: what is the difference between virial, dynamical and thermodynamic equilibrium? I'm currently going through Binney & Tremaine (2008) on my own to learn about stellar dynamics. I also have been perusing additional online resources such as this scholarpedia wiki.

Often when distinguishing between collision-less vs collisional stellar systems, the virial theorem is invoked along with the equations for "crossing time" (also known as "dynamical time") and "relaxation time." A large galaxy is said to be collision-less because its relaxation time is many orders of magnitude higher than its age, whereas a dense stellar system (e.g., a globular cluster) is collisional because its relaxation time is less than its age.

But what is the relationship between this so-called "relaxed" state and virial, dynamical, and thermodynamic equilibrium? What do the three different kinds of equilibria intuitively mean?

For example, I have heard that large galaxies are assumed to be in virial equilibrium and then people derive "dynamical masses" (why not "virial masses"?). What would it take and/or mean for a large elliptical galaxy to be not just in virial equilibrium, but also in dynamical or thermodynamic equilibrium?

Thermal equilibrium

Thermal equilibrium relies strongly on the idea of equipartition of (kinetic) energy. In a stellar system, this means that the total kinetic energy is divided evenly amongst all the stars. This doesn't imply that the velocities are all the same; they can't, because not all the masses are the same.

Dynamical equilibrium

Dynamical equilibrium means that over dynamical timescales, the system is stable - basically, it will not succumb to core collapse due to a gravothermal instability. Note that it may not be possible for a system to reach thermal equilibrium, even if it is in dynamical equilibrium. In a system with two main types of stars, it must satisfy the Spitzer stability condition (see Fregeau et al. (2001) and these results):
$$\left(\frac{M_2}{M_1}\right)\left(\frac{m_2}{m_1}\right)^{3/2}<0.16$$
where $M_1$ and $M_2$ are the total masses of types 1 and 2.

Virial equilibrium

Virial equilibrium comes about when the system satisfies the virial theorem (see Meylan (2000)), i.e.
$$2\langle T\rangle+\langle V\rangle=0$$
where $T$ and $V$ are kinetic and potential energies.


Is our universe included inside a black hole? This question repetitively comes to my mind but I've never found any good scientist article to argue about it, only esoteric or speculative web articles without much valuable background.

So let's suppose we're in the middle of a "black hole" ...

To me, here is why this would make sense:

Big bang

The word big bang itself may be inapproriate, indeed it is said to be an "explosion" but an "explosion" without any center.

With the idea of being inside a black-hole, it is easier to represent this as being the matter entering the black-hole. From our position, and as we are attracted by the singularity, everything seems to expand wherever we look at. And as far as we can look at, we can only see the cosmic background radiation.
Dark matter

On the contrary of matter entering the black-hole, dark matter is the matter already in front of us and we'll never be able to access it as its light (i.e. information) cannot go backward within the singularity.
Time/Gravitation

Time is a dimension like any other and as we feel x,y,z dimensions as a field of view, we fell time as an irrepressible attraction toward the singularity. Gravity may be see also as a less irrepressible dimension (depending on mass around).
Other black holes

Other black-hole we "see" may let us envision the shape of the universe not as simple torus, but higher dimensional "toric" shape with "connections" everywhere (maybe shapes like high order atomic orbitals ?).

Is there any reasonable scientific background to argue/contradict about this ?

There's a lot to pick apart in everything you try to propose, as it includes a lot of far fetched (or at least rather non-standard) claims.  I am frankly not up to attempting to address every one of them, if for no other reason than that it makes the question as a whole rather too broad for my taste (and perhaps more in the territory of Physics.SE, which has quite a lot of answered questions concerning black holes).

There is, however, the following simple and amusing observation: current estimates of the mass-energy of the observable universe tell us that it is too dense to be a black hole.  That might sound a little weird if you're not familiar with black holes.  In fact, the density of a (non-rotating, Schwarzchild) blackhole is inversely proportional to the square of its mass, and the radius is directly proportional to the mass.  More explicitly:
$$r=\frac{2 G M}{c^2},$$
$$\rho(M) = \frac{3 c^6}{32 \pi G^3}\cdot \frac{1}{M^2},$$
where $r$ is the radius, $\rho$ is the density, $M$ is the mass, $c$ is the speed of light, and $G$ is the gravity constant.

The order of magnitude estimate for $M$ is $10^{54}$ (and $M\geq 10^{54}$ in particular), which makes the universe ~3 times too small at least.

A key fact here is that the universe is not static with respect to itself. See this Physics.SE Q&A in particular.  I'll quote the end of Lubos Motl's answer, in particular:

Our Universe, dominated by the dark energy, is already rather close to an empty de Sitter space which is, from many viewpoints, analogous to a black hole except that the interior of the visible part of the de Sitter space is analogous to the exterior of a normal black hole, and the analogy of the interior of a black hole is everything that is behind the cosmic horizon - where we don't see. It is misleading to create the analogy with the static black holes directly because our Universe is not static in the normal cosmological coordinates.

In other words, there are lots of important and sometimes subtle issues with the whole "the universe is a black hole" concept.


what can I see with this telescope? I'm very new to astronomy and wondering what I can see with these specs, I honestly have no clue what a lot of these specs mean, and need help thanks.

Features:

Maximum magnification: 675x

Large aperture offers clearly illuminated views

Stable construction provides solid support and control

Metal tripod with slow motion control rod

Strong performance in both high and low magnification

Easy sky focusing with finderscope

Specification:

Aperture: 60mm (2.36inch)

Focal Length: 900mm.f/15

Finderscope:  5x24

Diagonal Prisms: 90degree

Slow Motion Control Rod For Easy Vertial Micro Adjustment

Height: 128cm

Standar: 0.965” Acceddoness Include

Eyepiece: SR4mm H12.5mm H20mm

3X Barlow Lens, 1.5X Erector

Telescopic Aluminum Tripod

it also cost $99, and also wondering if it's worth for the price
here's a link to the site

http://www.dshop.com.au/buy/astronomical-telescope-675x-magnification/F90060M?gclid=CjwKCAjw-qbLBRB7EiwAftBCI7h_7lH3E9goJG4Top0N-OHaraxB0cM8IiZgfygRkHD8DZ1ThARCdRoCzA8QAvD_BwE

I know you didn't specifically ask this, but I feel it answers the spirit of your question. I'm just going to go through the list of features and specifications and explain what they mean.

Features

Maximum magnification: 675x

This should hopefully be obvious, but this claims to provide images that are 675 times larger that they would appear to the naked eye. However, don't get too excited about this number. When it comes to telescopes, the magnification power is not the most important number. Telescopes for amateurs will try to play off this ignorance though and very often talk about how good that telescope's magnification to make it sound impressive.

Large aperture offers clearly illuminated views

The aperture is the physical size of the collecting mirror/lens. Usually aperture size is specified as the diameter of the main mirror/lens and larger is better. A larger aperture means the telescope can collect more light and thus you can see fainter objects you couldn't normally see with your naked eye. The phrase "clearly illuminated views" doesn't really mean much though, its just marketing.

Stable construction provides solid support and control

I guess they're just telling you the telescope won't wobble on it's stand. Which is good. It would be hard to look at a fixed object in the sky while your telescope is wobbling all around.

Metal tripod with slow motion control rod

Not sure it matters that the tripod is metal or not. But they do say it has a "slow motion control rod". Basically, these are rods you can turn to move the telescope very slightly rather than swinging it with your hand. This just provides the ability to finely move the telescope to help point it precisely.

Strong performance in both high and low magnification

Pretty obvious statement, albeit a vacuous one. What counts as "strong performance"?

Easy sky focusing with finderscope

The telescope comes with a finderscope. You can see that as the small "mini" telescope attached to the main telescope. When trying to point your telescope at a specific star, it can be hard to know if you're pointing at the right one. The finder scope provides some basic ability to view a larger region of the sky and try to get your telescope generally pointed in the right direction. From there, you can look through the main telescope and use the "slow motion control rod" to get it pointed exactly right.

Specifications

Aperture: 60mm (2.36inch)

This is the aperture I mentioned above. They're telling you the diameter of the main lens. Bigger is better here because it means the telescope can collect more light and see fainter objects. That's why professional astronomical telescopes are constantly getting bigger and bigger. I'll note that they're probably playing mental games by saying its 60 mm when 6 cm would have sufficed. More than likely, 60 sounds more impressive than 6 which is why they chose to use mm.

Focal Length: 900mm.f/15

This tells you two pieces of information. First, that the focal length is 900 mm. This telescope is a Refraction telescope. The light enters the telescope and immediately hits the main lens. The main lens is curved and bends the light, focusing it to a point. This focused light it sent to the eyepiece which will then direct it right to your eye for viewing. They're telling you that the telescope focuses the light to the eyepiece over a distance of 900 mm.

The other quantity is the f-number. It is the ratio of the focal length to the aperture size. You can hopefully easily see that 900/60 = 15 and thus the f-number is written as f/15. The "/" here does not imply division, it's simply how the f-number is written.

To be honest, neither of these two quantities will really matter much for you. I think they're just quoted because they're common "optics" numbers you'd see with higher end cameras and they want this telescope to sound like a serious product. The only thing about this that affects anything for you is the field of view, that is how much of the sky you can see at one time. This will only matter when looking at extended objects like the Moon. A longer focal length means a larger field of view. The f-number is virtually useless for astronomical applications. It only really applies in photography.

Finderscope: 5x24

This is again the finderscope I mentioned above. It's used to get your telescope pointed more or less in the right direction so you can then look through your telescope and do fine pointing. The number 5x24 signifies that it magnifies by 5x and has a 24 mm diameter aperture. This is going to be cheap and may not focus very well so don't expect anything amazing. You'll only use it to point the telescope though so if you're fine with that then so be it.

Diagonal Prisms: 90degree

This is just the prism at the end of your telescope, and before the eyepiece, that let's the eyepiece be 90 degrees turned from the main telescope shaft. This is useful because you can sit next to your telescope and view through the lens on the side rather than having you align your eye directly with the telescope's main barrel. It just provides easier and more comfortable viewing.

Slow Motion Control Rod For Easy Vertial Micro Adjustment

Already described this above.

Height: 128cm

Just giving you the height of the telescope. Although it's unclear precisely what height this is. Certainly the height changes as you move the telescope around on the mount. Is this the maximum height? Is it the height when the telescope is level? Who knows.

Standar: 0.965” Acceddoness Include

I actually have no idea what this means. Clearly they misspelled Standard. Other than that, I don't know what an "Acceddoness" is.

Eyepiece: SR4mm H12.5mm H20mm

Just providing you with descriptions of the three eye pieces that come with this telescope. The measurements are the eyepiece aperture sizes. A larger size generally means higher magnification and higher field of view.

3X Barlow Lens, 1.5X Erector

There are two additional optical pieces provided. The first is a "Barlow lens". This will sit in front of your other eyepieces and serves to provide an additional 3x magnification. The second is the Erector lens. It will provide an additional 1.5 magnification but it also will "erect" your image. The optics of your telescope will flip your image upside down. This is just how optics works (and it's the same reason you appear upside down in a spoon). In truth, when looking at astronomical objects, it doesn't really matter if you're looking at it upside down or not, but cheaper telescopes like to provide the illusion of having fancy gadgetry like erector lenses. There's not much of a reason to flip the image to be "correct".

Telescopic Aluminum Tripod

It's just a basic metal tripod that lets you sit your telescope on a stable mount. The legs can adjust to different heights and are thus "telescopic" which has nothing to do with the fact that it is a tripod for a telescope.


Neptune's Internal Heat Source Unlike its sister planet Uranus,Neptune has a substantial internal heat source which gives rise to a dynamic atmosphere & weather.

Wikipedia says:

  As with Uranus, the source of this heating is unknown, but the discrepancy is larger: Uranus only radiates 1.1 times as much energy as it receives from the Sun; whereas Neptune radiates about 2.61 times as much energy as it receives from the Sun. Neptune is the farthest planet from the Sun, yet its internal energy is sufficient to drive the fastest planetary winds seen in the Solar System. Depending on the thermal properties of its interior, the heat left over from Neptune's formation may be sufficient to explain its current heat flow, though it is more difficult to simultaneously explain Uranus's lack of internal heat while preserving the apparent similarity between the two planets.

Have plausible theories been proposed for the source of the heat necessary to keep Neptune's ocean liquid?

Could there be sufficient methane in the atmosphere to produce a significant greenhouse effect?

Neptune and Uranus have estimated core temperatures of 7273 K and 5255 K respectively. Uranus and Neptune however likely do not have liquid water oceans. The combination of pressure and temperature on both planet results in unsuitable conditions for liquid water oceans. In fact for liquid water oceans to form on Neptune, it's predicted it would have to cool down.

Paper on this at: Cornell University

(Irrelevant, but fun) Diamonds oceans might exist. Uranus and Neptune are made up of 10% carbon some of which is methane. Methane when exposed to the conditions available in Uranus and Neptune can form diamond. Diamond can melt and form liquid diamond under again, the conditions on Neptune and Uranus. Therefore there could potentially be diamond oceans on Neptune and Uranus.

Paper on this at; Nature Materials

Two plausible theories considered for Neptune source of heat is that as you said, the methane concentration in the atmosphere is high enough to retain the formation heat after the formation of the planet. The other one is that the tidal forces generated by Triton's retrograde degrading elliptical orbit, but the paper on it concludes that it doesn't explain the difference between Neptune and Uranus.

Another thing to consider is that Uranus might be cooler than it should be rather than Neptune being warmer. This could be a result of an Earth sized
indirect collision earlier in its lifetime. The collision causes this decrease in emitted temperature because it deposited a significant amount of material around the core which made heat transfer from the core to the outer parts of the planet much slower. Note that if this collision were to occur, it is what would've caused the unusual tilt of 97.7°.

Paper on this collision at: The American Astronomical Society

To the second question, the methane in the atmosphere of Neptune, as covered before, can produce a significant greenhouse effect.

  Neptune emits 2.7 times more energy than it receives from the Sun. This access energy powers the atmosphere to produce the storms that are not seen on its twin planet Uranus. The source of internal energy cannot be due solely to leftover energy from formation (i.e. Jupiter) since Neptune is smaller and would have radiated away the energy long ago. Nor is it due to an unusual chemical change, such as the helium rain for Saturn. Rather, it appears that Neptune is more efficient at trapping leftover formation heat due to the fact that methane is highly abundant in Neptune's atmosphere, and methane is an excellent insulator of heat (i.e. the greenhouse effect). Neptune has a sub-zero type greenhouse effect that is trapping formation heat that should have been radiated billions of years ago like Uranus.

This extract is from:
University of Oregon

In regards to why Uranus atmospheric methane doesn't appear to have as much of an effect as on Neptune, it could be that either Neptune simply has more methane in it's atmosphere or that a combination of factors, some of which mentioned above causes the difference. Beyond that I do not know.

Paper on Tritons Tidal Heating:
University of Texas


What forces expelled these huge clouds, then blocked further progress, yet allowed it to maintain its threads? From Wikipedia and NASA:

Caption: NASA: S74-15583

  (July 1973) --- A huge solar eruption can be seen in this Spectroheliogram obtained during the Skylab 3 mission by the Extreme Ultraviolet Spectrograph/Spectroheliograph SO82A Experiment aboard the Skylab space station in Earth orbit. SO82 is one of the Apollo Telescope Mount experiments. The SO82 A instrument covers the wavelength region from 150-650 angstroms (EUV regions).

The caption may have been written in the 1970's. Is there still a mystery about the forces, or are things somewhat better understood now? The only other (fundamental) force that could be relevant is electric, but I wonder if that was meant to be implicit (as electromagnetic). See statement highlighted in bold below, as the caption continues:

  The magnitude of the eruption can be visualized by comparing it with the small white dot that represents the size of Earth. This photograph reveals for the first time that helium erupting from the sun can stay together to altitudes of up to 500,000 miles. After being ejected from the sun, the gas clouds seem to have come to a standstill, as though blocked by an unseen wall. Some materials appear to have been directed back toward the sun as a rain, distinguished by fine threads. At present it is a challenge to explain this mystery--what forces expelled these huge clouds, then blocked its further progress, yet allowed the cloud to maintain its threads. Both magnetic fields and gravity must play a part, but these curious forms seem to defy explanation based on magnetic and gravitational fields alone. The EUV spectroheliograph was designed and constructed by the U.S. Naval Research Laboratory and the Ball Brothers Research Corporation under the direction of Dr. R. Tousey, the principal investigator for this NASA experiment. On the left may be seen the sun's image in emission from iron atoms which have lost 14 electrons by collision in the sun's million-degree coronal plasma gas.  (emphasis added)

If I understand correctily, this is a telescopic view with a diffraction grating in the optical path - the clear image and its "echoes" are the result of each strong, narrow emission line producing an image at a different angle of the grating's dispersion.

This was a coronal mass ejection.

Those 1973 astronomers weren't looking at the picture correctly. They didn't have the tools at that time to look at the picture correctly. Coronal mass ejections (the term used now) were only discovered a couple of years prior to that picture taken from Skylab, via the Orbiting Solar Observatory 7 satellite. Those early 1970s astronomers called such events "coronal transients". OSO-7 took 45 minutes to transmit a single 256x256 image back to Earth. A proper look at the picture took a few more years to develop. Those coronal mass ejections (the term used now) eject massive amounts of solar material into space.

When looked at via a single still image, it does indeed appear that material is looping up only to fall back to the Sun.

Image credit: SOHO (ESA & NASA)

A better view of a coronal mass ejection became available once those satellites had the ability to send more than an image per hour back to Earth. The image below portrays a coronal mass ejection that occurred in 2001.

Image credit: SOHO (ESA & NASA)


What is the direction of the movement of the solar system in relation to the galaxy's plane I got from this answer that the axis of the solar system is tilted of about 63 degrees in relation to that of our galaxy, so I can assume that we move through space (at least locally) roughly in a direction that is 63° tilted in relation to our solar system's plane, is that correct?

If so, then: is there also some net velocity of our solar system (probably small) in a direction that is perpendicular to the galaxy's plane, as opposed to the main direction that is "parallel" to the galaxy's plane?

Thank you in advance.

EDIT

I think my question is different from this one in that the latter asks about the distance from the solar system to the galactic plane, with no regards as to the speed with which this distance is changing. On the other hand, I'm interested not on the distance, but on the speed with which this distance is changing, which can promptly be approximated with respect to time with a sine function from pela's answer, which was my main intention.

I think your question is answered by the duplicate mentioned: but here are the relevant highlights.

Humphreys & Larsen (1995) suggest, using star count information, a distance of $20.5 \pm 3.5$ pc above the Galactic plane; consistent with, but more precise than the Bahcall paper referred to by Schleis. Joshi (2007) is more guarded, investigating some systematic uncertainties in the estimation techniques and ends up with distances between 13 and 20 pc above the plane.

The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this "peculiar velocity" is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and $W=7.17 \pm 0.38$ km/s (up and out of the plane).  (Dehnen & Binney 1998). Different authors arrive at velocities that differ by $\sim 1-2$ km/s from these values and so this would probably be a more conservative estimate of the uncertainties.

The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from http://www.visioninconsciousness.org/Science_B08.htm to show this oscillatory motion. The oscillations will not be exactly sinusoidal because the restoring force towards the plane does not very linearly with height above the plane.
As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc (Matese et al. 1995), it will be roughly 30 million years before we cross the plane again.

EDIT: Actually I'm glad I revisited this question because I think the picture is not very good at all. As the Sun takes ~230 million years to go around the Gaaxy, it should only execute 3 complete vertical oscillations in our Galactic year, whereas the picture implies many more. Secondly, the Sun executes a radial oscillation with a period of around 160 million years, which is not even indicated!


What would a black hole merger look like? Ship 1 is out side the black hole. Ship 2 is nearing a black hole. To ship 1 it would appear that ship 2 would be moving slower to the point it would seem like ship 2 stopped moving the closer to the black hole ship 2 gets.

I imagine that 2 naked black hole(an infinity dense point with out a surrounding black hole) merging would look like a blur from the speed of them, but and because of the time dilatation is this gif picture below a good rendering of how the merger would appear to ship 1? Then how fast does a black hole move when merging??

I'm pretty far from an expert, but there are a lot of moving parts to this question and it's not easily answered, partially because we don't have a good definition of precisely what a black hole is.   Special relativity tells us that it's a point singularity surrounded by an event horizon from which noting can escape.  Quantum physics tells us that we don't know what it is, but hawking radiation can escape the event horizon and information is conserved on the outside . . . er, I think.   A recent gravity wave discovery (see special relativity), indicates that they lose a pretty fair portion of mass when they merge.

  To ship 1 it would appear that ship 2 would be moving slower to the
  point it would seem like ship 2 stopped moving the closer to the black
  hole ship 2 gets.

This is the common explanation, but I'd like to add some detail to it.   This only works if the ship falls direct into the black hole with zero tangential velocity.

If we imagine a ship that's both reflecting visible light and radiating infrared light as it falls into a black hole, it speeds up very fast as it falls into the black hole, until the time dilation takes over and it appears to slow down, at which point it's very close to the black hole.  It's velocity is a significant fraction of the speed of light - somebody else can do the math if they like, but If memory from another question is correct, about 1/3rd the speed of light when the slowing down takes over - to the observer.  To the ship it continues to accelerate.

The ship continues to red-shift however, because the light that escapes the ship is red shifted both by the SHIP's increasing velocity combined with the gravitational red shift, so as a visual, you'd lose sight of the ship pretty quickly.

Something else happens because of a quirk with relativistic gravity.   An object reflects and emits light in all directions, but as it gets close enough to the black hole, the only light that can escape is almost entirely perpendicular.   Light can only escape the photonsphere of a black hole if it's close to perpendicular, and maybe my wording could use improvement there.  This question has a very nice answer on orbital velocity around a black hole.  Tangential light gets considerably bent or trapped entirely.  What this means is you pretty quickly will lose sight of anything that isn't directly perpendicular between the ship, the center of the black hole and where you are viewing from.   To actually continue to see the ship you'd need numerous cameras spread around the black hole and the view would need to be reconstructed.

And even if not for the photonsphere, the fact that parts of the ship would be closer to the event horizon than other parts, and that light would travel from the ship to your eyes at slightly different directions, you'd likely see some distortion, similar to the high velocity simulations you might have seen on relativity tv shows, or this website gives some examples, or when Randy Johnson throws a baseball.

And lets not forget, any object would likely break apart or spaghettify due to tidal forces, unless you had a really big black hole where spagetification wouldn't be an issue and the warping would be less but the photonsphere would still be a problem.

My point is, the story of the ship remaining visible is a nice simplification, in reality it's oversimplified.  It ignores tangential velocity and it's frequently not mentioned that it would both spread out and quickly become so red shifted that it couldn't be recognized.     The particle nature of light would suggest that in not too much time, the ship would be reduced to individual photons flickering and you'd need an enormous number of telescopes (assuming you could build a telescope to capture light that far red shifted), and an enormous amount of time to capture enough photons, to even track the ship at all.   In practical terms it disappears pretty quickly.   You might see it slow down and red shift and stretch for a while, but not very long.

And if the ship falls into the black hole with tangential velocity, well, that's even worse for keeping a visual track on it, and almost everything in space that falls into something else, has tangential velocity.  The idea that the ship appears to freeze kind of in place and fade is probably not accurate at all but if any experts here want to correct me on that, I invite correction.

I wanted to point out this complexity before giving a layman's answer to your question.

The best way I can personally imagine what it would look like when two black holes fall into each other is to think of the photons outside the event horizon and what happens to them.

This is probably a pretty good estimate of what it looks like from ship 2.    I think images like this are computer generated and not bad.  One thing I disagree with is that these images are always drawn with a kind of ring around the black hole.

The black circle that you see is actually the photonshpere, not the event horizon and it's not completely black because some light with a tangential velocity can escape the photonsphere, but it's probably pretty dark, unless something very energetic is going on inside.

Also, those black holes above are drawn with a kind of light-circle around them.  That's probably not accurate either.   The entire space is warped, so there's no light circle.   We imagine seeing a circle around an object that blocks a light because light can wrap around an object and we think of things as having edges, but that's not accurate.   There would be warping around the photonsphere, extending outwards, but not a circle of light.  I suspect there's a good chance you wouldn't see one black hole pass infront of the other, you'd just see it disappear, like black on black, then reappear as it warped space behind it when it was no longer infront of the other.

I don't object to the liberties taken to make videos like that more visible, but if you're asking what it would look like, I felt those corrections were worth pointing out.

Onto your questions:

  because of the time dilatation is this gif picture below a good
  rendering of how the merger would appear to ship 1?

(give posted above as well).

It's an OK rendering to give amateurs an idea of what it would look like.  Time dilation is largely irrelevant because the black holes would orbit each other very very fast, unless they were very large black holes.   That's another difficulty with your question.  The answer changes somewhat with 2 supermassive black holes, like is expected to happen when Andromeda and the Milky Way's black holes are expected to merge, and with stellar mass black holes.

Stellar mass black holes would orbit each other very quickly and the time dilation would slow it down a little bit, but not enough where it would appear slow.   The interesting stuff happens when they get very close.  Take the collision that was detected by Ligo, when two roughly 30 solar mass black holes merged.    A 30 solar mass black hole has a radius of about 90 km and if the two objects are orbiting each other close enough where you'd want to park a space ship and watch them collide, figure about 5 or 10 times that distance, say 900 km apart.   Objects travelling at relativistic speeds around a (3.14 x 2 x 900), 5,652 km track, at relativistic speeds, you're looking at several orbits per second, give or take, that would likely speed up as they got closer.   Time dilation doesn't catch up fast enough.   You might get some red shifted slowed down bits and pieces, but like the space ship, the idea of watching the black holes kind of freeze on each other is probably not what it would actually look like.

A few other things to keep in mind.   As the black holes spiral towards each other, they lose mass, so their event horizons actually shrink, which would be hard to observe until the very last moment where most of the gravitational wave energy is lost.

And, the black part is actually the photonsphere, not the event horizon.   It could look to the viewer that the black holes are touching when it's just their photonspheres shaking hands, so to speak.   If it's a small black hole and a bigger one, the small one could actually disappear into the larger one's photonsphere before the merger actually happens and the enormous pulse of gravitational wave energy is released.

So, the video probably isn't a very accurate representation of what it looks like, even allowing for the slow-down, but it's not a bad one either.   I don't think it's meant to be accurate as much as it's meant to try to show what happens. It's like when they give Jupiter all these colors that it doesn't really have.  It helps people like us "see" Jupiter, but an actual picture would be less colorful.   It's OK to sacrifice some accuracy in order to make something more visual.

  I imagine that 2 naked black hole(an infinity dense point with out a
  surrounding black hole) merging would look like a blur from the speed
  of them,

OK, here's where I'm in over my head, cause this is hard.   I imagine that you're doing a thought experiment, not actually proposing "naked black holes", that is, if you could see the two singularities spiral towards each other, what would it look like.

First thing is, you can't see a singularity.   Imagine you want to see one, so you sail your ship into a black hole.  Sure it's a suicide mission but you will be the first person to see a "naked singularity" - WRONG.   No light can travel from the theoretical point singularity outwards, so as you're inside the event horizon, not much changes.  You see stuff that fell in with you, with different variations of tangential velocity.  You see things stretched and warped by the curvature of space but you can't see the singularity because it doesn't send any photons in your direction.   You could maybe observe some hawking radiation but even that is iffy, because unlike the border of the event horizon where one particle can escape and one can leave, inside, both particles would be drawn to the singularity and probably annihilate each other like good virtual particles are supposed to do.   You'd see space, star light behind you, and stuff around you but you can never see the singularities (assuming that's what happens anyway, we don't have a good answer for what happens inside a black hole).

While you couldn't see it, you could model it.  Two theoretical point singularities would, in a decaying orbit, spiral into each other.   The gravitational waves released inside the black hole would be enormous, but, somehow, all that energy inside the black hole would likely remain trapped inside the black hole.   The gravitational wave energy that's lost probably all happens before the theoretical singularity crosses the other black hole's event horizon, or before the event horizons touch and merge, so to speak.

But if you imagine there are two point singularities in this scenario, I see no reason why they wouldn't just very quickly spiral into each other.  To get a better picture of this, we'd need an accepted model for quantum gravity, which we don't have.  There are also models of black holes where the matter accrues on the outer edge of the event horizon, where time stops and there is no central singularity.

So the real answer to your question is that nobody knows what it would happen, but if we imagine two point singularities when two black holes merge, the laws of special relativity suggest that they would fly into each other very very fast, perhaps a tiny fraction of a second.  Gravitational energy would be emitted as they spiral into each other, as defined by special relativity, but I don't imagine that energy would escape, it would remain inside the event horizon, but on that, I'm just speculating.   There's many reasons why this question probably can't be neatly answered.

There's also a kerr black hole possibility where the point singularity is more of a rotating hoop or ring in order to retain it's angular momentum, but I've always had a hard time seeing how that meshes with losing energy via gravitational waves . . . but I'm not going to try to explain that one, cause it's above my pay grade.

Apologies if I waxed poetic a little too much, but I think this question almost works as a thought experiment, though it's probably impossible to answer.


How massive does a main sequence star need to be to go type 1 supernova? We know the mass a white dwarf needs to be. That's well defined by the Chandrasekhar limit, but before a main sequence star turns into a white dwarf it tends to lose a fair bit of its matter in a stellar nebula.

According to this site, the white dwarf that remains is about half the mass of the main sequence star, with larger stars losing a bit more.

So, the question: Is it correct to say that a star with a mass of about three solar masses will eventually go supernova, similar to a type 1 supernova, even when it's not part of a binary system? Has that kind of supernova ever been observed?

Or does something else happen like in the final stages of that star? Does it keeps going though collapse and expand cycles, losing enough mass that when it finally becomes a white dwarf it's below the Chandrasekhar limit in mass?

Mostly, what I've read on supernovae says that type 1 supernovae happen when a white dwarf accretes extra matter and reaches the limit and type 2 supernovae are much larger and require about 8-11 solar masses to generate the iron core which triggers the supernova. What happens with the death of the star between three solar masses and eight solar masses?

This is ground that is probably duplicated in a variety of questions here and on Physics SE, so I'll keep it brief. You have also mixed in several different questions.

The Chandrasekhar mass has very little to do with determining what initial mass of object will end up as what particular type of stellar remnant (or black hole).

Whether a star will end up exploding as a supernova depends primarily on its initial mass, but also whether it has a binary companions. There are (basically) two routes to supernovahood.

If the star is more massive than about $8M_{\odot}$ it will progress through several nuclear burning stages. The core of the star does not become degenerate and continues to get more dense and hot through each burning stage. It ends up as iron. Once the core mass of iron exceeds about $1.2M_{\odot}$ (which is the Chandrasekhar mass for an iron composition), then it collapses and we get a type II (core-collapse) supernova.

In this route a $3 M_{\odot}$ star gets nowhere near being able to go supernova. It will burn hydrogen and helium, produce a degenerate core of carbon and oxygen. This degenerate core can cool whilst maintaining the same pressure. The outer layers are shed through thermal pulsations and a dense stellar wind in the asymptotic giant branch phase, leaving behind a white dwarf. The relationship between the initial mass of the progenitor and the final mass of the white dwarf is not a straightforward fraction. It probably is about 50% for a star like the Sun, but the fraction is more like 15% for a $7M_{\odot}$ initial mass. The maximum mass of a white dwarf formed in this way is probably about $1.1-1.2M_{\odot}$ and some way below the Chandrasekhar limit for a C/O white dwarf ($\simeq 1.39M_{\odot}$).

The preceding paragraph is more-or-less what should happen for all stars between about $0.6 M_{\odot}$ (except they haven't had time to do so yet) and $8M_{\odot}$, except that there is a small "grey area" at the upper mass end ($7-9M_{\odot}$) where you might produce slightly more massive O/Ne white dwarfs.

Once a white dwarf has formed and if it is in a binary system, then the white dwarf could merge or accrete more mass. At some point close to the Chandrasekhar limit, it ignites. This causes a type Ia (detonation or deflagration) supernova explosion (or at least this is the leading model for how this works). This is really the only route whereby a star with initial mass $<8M_{\odot}$ could end up going supernova.


Does the holographic principle contradict eternal inflation? As it is for example explained here, Andrei Linde's chaotic inflation is among the potential winners or survivors of the spring cleaning induced by the potential detection of primoridal gravitational waves. As this type of inflation is eternal, the new data might also lend some updraft to the multiverse business etc ...

But I always thought that because of the holographic principle, which means when applied to the exponentially expanding universe that stuff moving away can be thought of as being attracted to the cosmic horizon which takes to role of a black hole horizon, it does not make sense to talk about regions or any other universes beyond the cosmic horizon anyway.

So, does the holographic principle contradict eternal inflation as it arises for example from Linde's chaotic inflation? What does the detaction of primordial gravitational waves then mean when taking the holographic principle into account?

There is no obvious implication to re-assess the large-scale topology of the universe.
You can still stereographically project infinity to a point on a sphere, if you need.
Implications to the holographic principle aren't immediate, at least.

When talking about objects beyond the cosmic horizon, it's a difference, whether we talk about observation or mathematical models. When talking about mathemetical models of the universe, it makes sense to talk about space and objects beyond the cosmic horizon, because we can get a higher symmetry of the theory. We can use mathematically simple structures like de Sitter spaces as simplified models of the universe, with no need to add an observer to the structure.

For an evidence-based science it's of course an extrapolation far beyond immediate observation, hence questionable. But there is also no evidence-based argument to assume, that the universe looks fundamentally different from a distant observer with a different cosmic horizon overlapping with our cosmic horizon, but reaching beyond.


What observations will be done during the New Horizons flyby of Ultima Thule? On New Year's Eve 2018, the New Horizons spacecraft will fly past Ultima Thule (2014 MU69) at about a third of the distance it flew by Pluto some years ago. So it should be able to get some detailed photographs, but the closing speed is rather high. What scientific observations will be done during the flyby?

Is it a case of a couple of snaps, assuming the camera is pointed correctly? Or is there more that can be done? Any other instruments that will get useful data?

Emily Lakdawalla in her Planetary Society blog post of 17 December describes fairly well what we can expect:

A few precious data downlinks in the days before and after the flyby will contain some small photos from New Horizons’ Long Range Reconnaissance Imager (LORRI), which takes grayscale snapshots. Lower-resolution color comes from the Multicolor Visible Imaging Camera (MVIC), a component of the Ralph instrument. The first images that we get back from New Horizons will not be the best images that New Horizons has taken.

Because of the uncertainty in the position of MU69...

... the real position of the body may be completely outside the field of view of a photo focused on the predicted position. In order to make sure it doesn’t miss the target, the spacecraft will have to scan its instruments across a region of space in which it is most likely to find the tiny world, taking a lot of data on empty space to make sure to get data on the object.

She shows what to expect in the days around closest approach, using comet 67P images as an example:

The Core phase will last 10 days, from 24 December to 2 January. Nearly all of the science will happen in just a 2-day period around closest approach, with intense imaging and spectroscopy to map MU69, study its composition, figure out its rotation rate and pole, and search for dust, rings, and satellites
Several things will make the MU69 flyby significantly more difficult than the Pluto flyby. MU69 is small, dark, and faint. New Horizons won’t see it as more than a dot until 3 days before closest approach. MU69 is darker, and the sunlight dimmer, than the LORRI camera was designed for. Also, because MU69 is so small, New Horizons has to pass much closer to it than it did at Pluto in order to see interesting detail on the surface, resulting in a rapidly changing viewpoint near closest approach.
As a result of all of these factors, LORRI’s images may be a little smeared and will be noisier than we saw at Pluto. MVIC is less affected by these challenges; the way that MVIC works, it’s possible for New Horizons to compensate for the low light and high speed by sweeping more slowly across MU69 to build up a stronger signal.

Other measurements to be made are:

Toward the end of 29 December, science data gathering will really begin. A dense set of observations will gather a lightcurve on MU69, a measure of how its surface appears to change with rotation, over a period of about 6 hours.
30 December will see another 6-hour block of science, a mix of color imaging, deep satellite searches, and lightcurve data. The same happens early on 31 December, after which New Horizons enters the most intense, Inner Core phase of the flyby.

You should read the blog post for all information, because she gives much more detail than can reasonably be quoted here.
She also quotes more resources.
Here is another simulated image of Ultima Thule, showing how the primary camera aboard NASA’s New Horizons spacecraft will “see” the Kuiper Belt object as it approaches for a 1 January flyby:
Source: Johns Hopkins University Applied Physics Laboratory
Here is the operations time line overview for the approach:

(source)
That flyby distance decision has now been made: the spacecraft will remain on course for a 3500 km flyby.


How Can I interpret this particular light curve without reading data on the side There is only one dip in flux recorded. How can I interpret this particular light curve and find planetary radius
https://exo.mast.stsci.edu/exomast_planet.html?planet=55Cnce

The radius of the planet is 0.17% of Jupiter's radius.  Under the "Planet" info on the left, it says $R_P$ which is the radius of planet, and since the units are shown as $[R_J]$ this is the value is in comparison to Jupiter. It also gives the $M_P$ (mass of planet) in relation to $M_J$ (mass of Jupiter) as well.
The interesting thing here is the size of the star - it's smaller than our Sun.
To get the radius of the planet, find the radius of Jupiter and multiply it by 0.17.
A "year" on that planet is shorter than one of our days and it's orbiting extremely close to its host star, as well.  The distance is given as 0.015 AU (astronomical unit which is the distance from the Earth to the Sun).


How many times can iron be detected for the first time in an exoplanetary atmosphere? I see today that there's a story about the first detection of iron in an exoplanetary atmosphere, in this case in the atmosphere of the ultra-hot Jupiter KELT-9b.

I remember there being a story a couple of months ago about the detection of iron rain in an exoplanetary atmosphere, the example being another ultra-hot Jupiter, WASP-76b.

It doesn't seem plausible for both of these to be the first detection of iron in an exoplanetary atmosphere. So which one is it?

Neither of these, interestingly enough, is the first time iron has been detected in an exoplanetary atmosphere. Other groups (Hoeijmakers et al. 2018, cited by both papers) have detected absorption lines of iron and other metals in the day-night transition zone of exoplanets, during transit. Ehrenreich et al. 2020, the new ESO paper, did something similar.

Now, Pino et al. 2020 are referred to in the spaceref.com article as having "directly demonstrated" the presence of iron in an exoplanet atmosphere for the first time. What I believe "directly" in this case refers to is the detection of iron emission lines, rather than iron absorption lines. This is what makes Pino et al.'s observations different: they observed KELT-9b shortly before it passed behind the star, allowing them to view the hot, day side of the planet and thus iron emission.

In other words, I think the spaceref.com article is playing fast and loose with its wording - Pino et al.'s results don't constitute the first time iron has been detected in an exoplanet atmosphere, just the first time iron emission lines have been detected. They got around that by characterizing emission lines as "direct"$^{\dagger}$, implying absorption lines are "indirect". Whether you agree with that is up to you, I think.

If you had to write an article labeling one study as the first detection of iron in an exoplanetary atmosphere, you could make a good argument in favor of Hoeijmakers et al., but I think the better option might be to simply not try to describe any of these in that manner. Some nuance here wouldn't be a bad thing.

$^{\dagger}$Hoeijmakers et al. refer to their observations as the first "direct detection" of iron, adding a layer of complexity and irony.


Is there a difference between the terms 'elliptical' and 'elongated' for galaxies? While studying the Solar System, I found that some galaxies are either elliptical or elongated. What's the difference?

tl;dr: "Elliptical" refers to a special kind of galaxies. "Elongated" is a loose term meaning that a galaxy is stretched out in one direction.

Etymologically, "elliptical" means having a shape that can be described by a certain mathematical function — the ellipse. For a three-dimensional object, the proper term is "ellipsoidal", but still it refers to being described by a certain equation. In contrast, "elongated" is a more loose term, meaning something like "significantly deviating from a spherical shape in one direction".

In the context of galaxies, the term "elliptical" refers to a certain type of galaxies that tends to be not actively star-forming, have no or very little structure (as opposed to the beautiful spiral arms of spiral galaxies), be quite massive, as well as other characteristica. Having no ongoing star formation, the massive, short-lived stars, which have blue colors, have died long ago, leaving behind the less massive, red stars. In addition, elliptical are often quite dusty, further reddening the light. Thus, ellipticals appear red in color.

The shape of an elliptical galaxy is described by the relative size of its three axes, $a$, $b$, and $c$. If two axes, say $a$ and $b$, are roughly the same size and are larger than $c$, the galaxy is said to be "oblate", whereas if they are less than $c$, it is said to be "prolate".

As in the etymological meaning, an elongated galaxy is not a specific type, but rather refers to a galaxy departing from spherical shape in one direction. This could for instance be a prolate elliptical, but it could also by any other galaxy that has been distorted by merging with another galaxy.

Usually, though (I think), it will refer to ellipticals. The "elongation" of an elliptical is defined as $10\times(1 - b/a)$, where $a$ and $b$ is now the observed axes, i.e. of the two-dimensional projection on the sky. Ellipticals are classified according to this number as E0 (being spherical, such that $a=b$) to E7 (being very elongated). In principle, you could have even higher E-numbers, but that's not observed.


Observations of a perigee coincident with a full Moon Hopefully this will be a quick and easy question, with a quick and easy answer, but why is it that we observe a perigee Moon coincident with a full Moon approximately every 13 months? Is this related to the Earth-Moon dynamics, perhaps the eccentricity of the lunar orbit?

The time between full moons is called the the synodic month, and the time between perigees is called the anomalistic month. The anomalistic month is shorter than the synodic month by about two days, (27.6 days and 29.5 days)

Since the anomalistic month is two days shorter, and there are nearly 28 days in a month, it takes about 14 months for the two cycles to come back into alignment. Thus there is a perigee at full moon every 14 (not 13) months.

More accurately the beat period between the two cycles is 411.78443 days. See wikipedia: https://en.wikipedia.org/wiki/Full_moon_cycle


When the universe expands does it create new space, matter, or something else? I am wondering what exactly is meant when it is said the universe expands. Does it simply create new space for matter to fill, does it also create new matter/dark matter to fill that space, or am I way off? Thank you for any help!

Yes, space is constantly being created. The new space does not hold any matter (like atoms) or dark matter. This means that the density of normal and dark matter decreases at the same rate as the volume increases. However, dark energy, which is something completely different and thought to be a property of vacuum itself, is being created with the new space, so the density of dark energy stays constant.

This in turns mean that while the early Universe (i.e. from it was 70,000 years old and until it was almost 10 billion years) was dominated by matter, the Universe is now dominated by dark energy.

And it will only get worse.


What is a "fundamental location", and why is Mosting A defined as one for the moon? The crater Mosting A was chosen to be the "fundamental location" for the Moon's coordinate system (latitude and longitude). But what exactly is a "fundamental location"? What does it have to do with the establishment of a coordinate system? Is it just so that observers have a way of easily projecting a spherical grid on the moon? And why was Mosting A chosen for the job?

Every coordinate system needs one point defined as exact and unambiguous - the fundamental location which all other location measurements in the vicinity are relative to.
This can be the origin of the coordinate system (like 0° longitude and 0° latitude) - but it can also be any other point instead. You choose that (or one) point which you can revisit the easiest in a reproducable manner.
The given point in the center of the Mösting A crater previously was one of the fundamental points for the Moons reference frame as it was/is easy to identify: it is very round, has a sharp edge, thus the mid-point is easy to identify. There seems to be some original correspondance on the topic available from the late 19th century on ADS and discussions on improving the map accuracy prior to the moon landing which all refer to these properties as being the reason of choice.
Currently it's defined more accurately via the retroreflectors placed on the moon by the Apollo missions. As such the coordinate for that were calculated such that 0°/0° is at the center of the on average visible disk of the Moon and then the location for that retroreflector was defined to be the resulting coordinates. Without such fundamental point, different measurements might use (slightly) differently-defined coordinate systems and comparison of data would become difficult. The retroreflector on that location is the largest one, so the easiest to find.
It's basically similarily arbitrary that the 0° latitude on Earth pass through Greenwich observatory.


Can a star have a ring system? I often hear about planetary ring systems, and even some moons might have them, but how about stars? Can a star also have rings?

They certainly can. A ring is often formed around a celestial body when its gravity rips apart another smaller celestial body. The Sun is really massive, so it could destroy any object that is not dense enough. Just Google about the Roche Limit for more informations (and better explanations).

Now, take a look at our solar system : You have the asteroid belt between Mars and Jupiter, and you also have the Kuiper belt beyond Neptune. Would you consider these being rings ? They sure are not as smooth as Saturn's, but to my mind, they still are rings.


What if the atmosphere of Titan were like the one on Venus? The current mean temperature in Titan is −180 ºC (93 K), the same as in Saturn.

But I was wondering what would be the case if the atmosphere of Titan would be like the Venus atmosphere, whose composition is mainly CO2 and has a pressure of 90 bar, instead of approximately 1.5 bar.

What temperature could Titan reach in that case? Would the greenhouse effect allow the moon to be heated enough to maintain liquid water or is it too far from the Sun to be affected by any kind of greenhouse effect? Maybe the latest question allows to ask another one: How far from the Sun could Venus be so it can have temperatures above 0 ºC?

The question may be answered given two possible alternatives:

Same composition as Venus, but same pressure as in Titan.
Same composition and pressure as in Venus.

If the pressure were the same as the current pressure on Titan, the planet would be slightly warmer for a while, but not for long. The CO2 would soon be frozen out. It is probable that 4 billion years ago the atmospheres of Venus,Earth,Masrs and Titan were much more similar. A hotter Saturn may have warmed Titan in those days. It probably,like the others, had a lot of CO2 in its atmosphere,but that has long ago been frozen out. There is plenty of frozen water on Titan, and if gelogists ever dig down far enough,they are likely to find a layer of frozen CO2 there as well.


How would water-ammonia oceans behave? Scientists strongly suspect that several moons in our solar system have frozen-over oceans of water-ammonia mixture. I've also read speculations on the possibility of surface water-ammonia oceans on exoplanets and exomoons. What I haven't seen a lot of is a discussion on how those oceans might behave differently than the water oceans of Earth.

More specifically, how would the surface water-ammonia ocean behavior differ from our water ocean behavior in terms of:

Ice(s) formation
Evaporation
Surface and underwater currents
Varying ratios of water-ammonia based on temperature or other
factors

I'd like to know if this has been studied and if these behaviors have been described in papers, books, or in-depth articles on this topic. Especially anyting that goes into the specifics of the water/ammonia ratios.

I'll add some personal speculation here. While it is not part of the primary question stated above, it helps to explain the type of problem I'm trying to understand, so may be helpful to narrow down which resources may be most useful.

When subject to increasing levels of heat would the ammonia molecules
in the ocean evaporate before the water molecules, causing warmer
areas of the ocean to have an increased amount of H2O relative to
NH3? (And the cooler areas like the poles having an increasing ratio
of NH3 to H2O).
If the above was true, might the equatorial waters that have less NH3
facilitate the formation of water-ice? If so I could imagine
exoplanets having water-ice-covered equators because they were
warmer, and the cooler poles might be ice-free thanks to the greater
amount of NH3 in the ocean.
Would the ice formed on the ocean be composed of a molecular mix
matching that of the ocean (with H2O and NH3 in the same ratio as the
ocean), or would some molecules freeze-out first, possibly causing
different layers or areas of varying types of ices?
Might the poles of some water-ammonia planets have "continents" of
water-ammonia ices that are frozen from ocean bed up to the surface
(thanks to the fact that frozen ammonia is more dense than liquid,
allowing for entire sections of ocean to freeze solid)?

If my speculation is correct (and I could be wildly wrong since I don't yet understand how a mix of water-ammonia actually behaves) then an exoplanet with an equatorial band of water-ice, mid-latitudes of liquid ocean, and solid polar-caps/continents of water/ammonia ices could be possible. The ammonia in the ocean would evaporate more quickly near the warmer equator, causing that area of a global ocean to have a greater ratio of H2O, and the ammonia precipitation and ocean might freeze at the poles, causing the build up of vast polar continents of ammonia/water ices. I would imagine the currents would function very differently than on earth, with surface and deep-level currents of varying ammonia-water ratios and temperatures. I I also see the possibility of frozen ocean beds in the more ammonia-heavy parts of an ocean, with the surface unfrozen.

Of course the ratio of ammonia to water will be very important. I'd love to see specific information on that!

Any informed clarification appreciated!

It's a big question, but kind of a favorite subject of mine, thinking about exoplanets, so I can give a ballpark answer, and I invite anyone to give correction or give a more technical answer if they like.

Ice(s) formation

An ammonia-water ocean wouldn't be friendly towards ice formation because water ice would sink in the ammonia-water solution and Ammonia ice would sink in liquid ammonia.  There's no range where the ice floats unless you remove nearly all of the ammonia and had it a high percentage of water.

That said, oceans are likely to have dissolved salts in them, especially on rocky worlds that have a very high rock to water ratio.    A water world might have much less salty oceans, but, lets not get too sidetracked.

If there's enough dissolved salts or iron in the water-ammonia mix, then the density might be sufficient that water-ice could remain on top.  Ice tends to form out of nearly pure water, with very little ocean salt, which increases it's buoyancy.   It's possible, with enough saltiness, that ice could form in an ammonia-water ocean.  It's also possible that ice would sink.

Density

I want to point this out now, because it's important.    Using our oceans as a model, higher salt concentration sinks.

Salt concentration by density

That's kind of common sense and is likely to happen with a water-ammonia ocean as well.   Higher ammonia concentration near the surface, higher water (and higher salt) concentration as you go lower.  That's speculation on my part, but it seems reasonable.

Varying ratios of water-ammonia based on temperature or other factors

Higher water concentrations would sink and higher ammonia concentrations would rise and flow towards the lower.  That's simple physics.   Like on Earth, the variation of salinity of surface water isn't all that much.  The same thing would likely happen with an ammonia-water ocean, the surface currents would try to equalize the different ratios.  There would be some variation, but likely not as pronounced as the variation with depth.   You're not likely to get a 90% water ocean in one region and a 40-60 ratio in another region because gravity abhors an imbalance like that.   The 90% water region would be heavier and as a result, lower.  You can have that kind of density imbalance with a solid crust, but not a flowing ocean.

The varying density of the ocean is important for another reason.   On Earth, the salinity being relatively consistent across the ocean surface, the cold polar ocean is more dense and it sinks, and this is what's thought to drive the oceanic conveyer.

There's a range of uncertainty on whether an ammonia-water ocean would conveyer or not.   You'd need a sufficient temperature and/or density variation, keeping in mind that if you begin with a ratio imbalance the ocean would work to correct it, so the planet needs an engine to maintain the conveyer.   On Earth, that engine is the warm equator and cold poles.  Different planets would have different temperature variations.  Venus, for example, has practically no temperature variation at it's surface.

I looked, but couldn't find a table on ammonia-water solution by temperature.  That would be a factor if anyone can find it.

Evaporation

Ammonia in water has a saturation point, based on temperature and pressure.   If the ammonia-water solution exceeds that saturation point, say, on a hot day, then ammonia would bubble out of the ocean fairly quickly.  Perhaps quickly enough to generate a weather system and trigger a density change, and in this case, warmer water sinking and a kind of reverse conveyer, driven by sinking water at the planet's warmest regions.

The phase transition, liquid to gas would remove heat from the water as well.

A planet with an ammonia-water ocean and a surface temperature that exceeds the saturation point could be quite dynamic.

If the planet is slightly colder and/or the ammonia percentage stays below the saturation point, then it's more of an equilibrium system.   Water is a good solvent and it takes in many gases, such as CO2, in small amounts.    Ammonia and Water, unlike CO2 and Water, mix very happily, so evaporation of an ammonia-water ocean below saturation wouldn't be that different to water on Earth, where evaporation is driven primarily by wind and sunlight.  (contrary to logic temperature isn't as big factor in evaporation as direct sunlight).  Wind is also a big player and on Earth, running water from rain or snow melt and transpiration from plants all return water to the atmosphere, which quickly rains or snows back to the surface.   At any given time, Earth's percentage of water vapor in the atmosphere is about 1%, but it varies locally based on air temperature and relative humidity.

Ammonia would enter the atmosphere similar to water, but how long it would stay in the atmosphere is a factor.    Air temperature and 100% relative humidity are the upper limits for how much water vapor can exist in the atmosphere.  There are no such upper limits for Ammonia at Earthlike temperature and pressure because Ammonia, unlike water, is a gas at standard temperature and pressure.

As a result, a theoretical Earth-like planet with an ammonia-water ocean is also, very likely, going to have a lot of ammonia in the atmosphere, not the trace 1% of water that's in our atmosphere, but a lot more.   You can't have an ammonia ocean without ammonia in the atmosphere at well, in high concentrations.

For example, to have ammonia raindrops at 1 ATM, the air temperature needs to be colder than -33 C because that's ammonia's boiling/liquification temperature.   Unless the planet was really cold, a lot of the ammonia would just stay in the atmosphere, though, presumably the planet would have water-rain and that water rain would pull some of the ammonia back out and return it to the oceans.   That's the equilibrium, and it's a complicated thing to try to calculate, but the question isn't just how fast would it evaporate, but also, how fast would it return and the return rate would depend on the method of return.   Ammonia, being a gas at what we think of as normal planetary temperatures, means it would likely remain in the atmosphere as well, it wouldn't all go into the ocean.  It would be some of both.

Perhaps over time, the ocean ratio and the rain-water/ammonia ratio would equalize, but again, that's just speculation.

If the planet has very cold poles, where the ammonia could rain out of the atmosphere, or perhaps tidal locking, and a colder night-side of the planet, then the ammonia could return from the atmosphere more quickly.

I'd like to know if this has been studied and if these behaviors have been described in papers, books, or in-depth articles on this topic.

I don't know of any papers.  Theoretical planetary formation is fun to think about, but to do it right, requires some pretty serious computer modeling.   What I've read on the subject is much less ambitious than your question.   Models are done on water worlds and on tidally locked worlds and on plate tectonics for different mass planets.   I've never seen a study on ammonia-water oceans, but I think it's a fun subject.

Earth had to have fairly abundant ammonia at some point because it's one of the primary "ices" in comets.   Comets are mostly ammonia, water, CO2 and CH4, with smaller amounts of other elements.  It's worth asking where Earth's ammonia went.  It may be more chemically reactive than the other elements.   If Earth lost it's ammonia, other planets may lose their ammonia as well in the process of planet formation and the chemistry that goes with it.  It's also possible that ammonia-water oceans are common, perhaps more common than the pure water oceans like the ones we have on Earth.

Perhaps time and better telescopes will tell us more about this subject.

Take my answer with a grain of salt as I'm a hobbyist.


Converting ångström spectral dimension to galaxy speed (km/s) I have a spectral cube (in FITS format) whose spectral dimension is in ångströms. The sampling along the spectral dimension is 0.28A (CDELT=0.28). The observation in the cube is Ha emission of a galaxy at redshift 0.1.

How can I convert the sampling to km/s?

If you want bins that have an equal step in km/s then you will have to rebin in steps of equal log wavelength. On this scale the rest wavelength of H alpha is at zero km/s and then each bin will represent a increment in velocity with respect to that.

See How do I apply a velocity shift to a wavelength array with uniform logarithmic spacing?


What's the closest Venus gets to the moon? I recently observed the beautiful sight of Venus quite close to the crescent moon, and was wondering, what the closest (angular distance as seen from earth) they come.

Are there on-line tables or calculators for finding when such nearest events occur and are visible?

Venus is, on occasion, occulted by the Moon (that is to say, the Moon passes in front of Venus as seen from our point of view).  Exactly when that happens is dependent on your location on the Earth, since the location of the Moon in the skies varies by about a degree or so depending on your location on the Earth, and of course, it might be on the other side of the Earth when it happens (it only lasts about an hour or less).  From North America, it happened in December 2015; it will happen again in September 2017, but only from Australia and neighboring countries.

If you google "Venus occultation" you should get something like what you want.  ETA: Here's one link: http://www.lunar-occultations.com/iota/planets/planets.htm


Are stars really "burning balls of gas"? Even the NASA website refers to stars as "big ball of gas". It's a common statement that's popularized by movies and popular culture. However, stars are technically not in a gaseous state - they're in a state of plasma.

So do stars qualify as being made of "gas" as the proverb would suggest, or is that considered a misconception?

While I agree that it boils down to semantics, I actually disagree on the scientific use of the term in the comments. In astronomy, we know there's a difference between gas and plasma, but we almost always use the term "gas" when talking about what's in stars. E.g. "the fraction of gas locked up in stars" (as opposed to in the interstellar medium). We also talk about the metallicity of gas, meaning both gas and plasma. I think we only use the term plasma when specifically discussing properties that are unique to plasma, like being "frozen" in magnetic fields. And googling ["ionized gas" -plasma] (the '-plasma' is to get rid of articles explaining what plasma is) returns 243,000 hits, of which by far the most seem to be scientific papers and websites.

So I think the answer is: "No, stars aren't made of gas, they're made of plasma, but go ahead and call it gas, you won't offend any astronomer."

However, I agree with Rob Jeffries that we don't like when you call it burning gas. It's nuclear power! It's a million$^\dagger$ times more powerful than fire!

$^\dagger$Why "a million"? Because nuclear reactions are measured in MeV, while chemical reactions are measured in eV.


what is the current explanation for the formation of cosmic voids? Why aren't galaxies distributed evenly in space, but instead form those sponge-like structures, with huge voids between filaments?

Void formation is due to baryon acoustic oscillations. These can be thought of as fluctuations in the dense matter in the early stages of the universe; they became anisotropies, which can be seen in the cosmic microwave background. The fluctuations grew, until they hit the 150 Mpc scale, which correlates nicely to the sizes of voids in the universe today.

There's an interesting section (4.2) in this paper by Robert Brandenberger, which gives a mathematical treatment of how fluctuations of all kinds grew in the early universe. There is a better version of the explanation in Zeldovich (1972), which in fact puts forth the theory. The number density (if you'll pardon the pun) is lesser in that latter, but the former has a good overview, as it is meant to be a broad summary.

The simple explanation is that tiny fluctuations grew over time to become enormous structures.


Different sizes of bow shocks In this image of LL Ori bow shock,

there is a smaller bow shock at the upper right.  Is that second bow shock substantially smaller than the LL Ori bow shock, or is it just much further away?  If the stars are reasonably close, is the upper star significantly smaller that it is producing a much smaller bow shock?  The LL Ori bow shock is measured about half a light-year across.  Any estimates of the size of the smaller one?

After a bit of searching in Aladin. I found that the other star is V* V2170 Ori.
By comparing their respective Simbad pages one can see the following:

LL Ori is actually slightly further away.
V2170 Ori has a GAIA parallax of 2.84 while LL Ori has one of 2.55. Simply using $D=1/p$, this translates to a distance of 0.35 kpc and 0.39 kpc
LL Ori and V2170 Ori have about the same proper velocity of 1.65 and 1.69 mas/yr. So, that should affect the shock size.
LL Ori is a lot brighter.
It has an apparent V-Band magnitude of 11.5, while  V2170 only has 18.7.
LL Ori is hotter.
Its spectral typ is K0-K3e, while V2170 is of spectral type M4.7e.

It is therefore not unexpected that LL Ori has a significantly stronger stellar wind and thus a larger bow shock.

Since the distance is basically the same, you can just have a look at this image with scale to estimate the size of the other bow shock.


How often are supernovas visible by the naked eye from Earth? I heard about the 1054 supernova that was visible with the naked eye.

Of course, these events happen at random. But just like with radioactive decay, it must be possible to determine statistics on these kinds of events.
Wikipedia says

  The total supernova rate in our galaxy is estimated to be between 2 and 12 per century

but this doesn't indicate anything about their visibility from Earth. In fact, quite a few supernovas in the Milky Way probably go unnoticed, so the rate of visible supernovas is likely lower.

In addition, type Ia supernovas can outshine the galaxy they are in, so if one occurred in Andromeda or a Magellanic Cloud, it might be visible. So there possibly could be more visible supernovas than there are supernovas in just our Galaxy.

What is the average time between two supernovas that are visible with the naked eye?

One way to answer this is to look at how often naked-eye supernovae were seen in the past. In the last 2000 years, there have been (at least) eight such supernovae recorded: in 185, 393, 1006, 1054, 1181, 1572, 1604, and 1987. Seven of these were in our galaxy, while the last (SN 1987A) was in the Large Magellanic Cloud.
It's possible that supernovae in the far southern sky would not have been visible to Chinese astronomers (the most reliable and comprehensive records we have), so some additional supernovae visible from the far south may have been missed. We could round the total up to ten, which gives us a rate of about one every 200 years. If you want to worry about some being visible but not recorded (e.g., because they happened during a particularly chaotic time in China), then perhaps the rate would be a little higher -- maybe one every 150 years?


Why does Nancy Grace Roman = 100 × Hubble? Why is the new space telescopes wide field camera so much wider than the old one's? The title of the WFIRST project description (before it was named the Nancy Grace Roman Space Telescope) is The Wide Field Infrared Survey Telescope: 100 Hubbles for the 2020s.
Question: Why does Nancy Grace Roman = 100 × Hubble? Why does the new space telescopes wide field camera so much wider than the old one's?
The telescopes have very roughly the same wavelengths and apertures, so does the new one have one hundred times more pixels at the same resolution, or 100 times larger pixels? Is there better correction optics that allows for a flatter field for a flat array, or a curved focal plane array, or other optical correction?

Hubble was designed in the late 1980s, when electronic detectors were small (the original Wide Field/Planetary Camera had two different 2x2 arrays of 800x800-pixel CCDs); it also had to carry multiple different instruments, each taking up nominal space in the focal plane. So it was optimized to use small fields of view to go with the small detectors.
WFIRST/Nancy Grace Roman is much more tightly focused on large-field surveys, so it can devote most of the field of view to the IR camera. The reason it is now possible to do this is probably a combination of three things:

Large, sensitive, low-noise IR detectors are now available; the current plan is to use a total of 18 individual 4k x 4k detectors -- a huge advance over the original mid-1990s 256 x 256 NICMOS detector installed in HST in 1997, and over the single 1k x 1k detector in the infrared channel of WFC3 (installed in 2009);
We have lots of experience in building array mosaics with lots of individual detectors (the planned $\sim 300$ megapixel mosaic is relatively modest compared with the gigapixel arrays in Pan-STARRS1 and LSST/Vera Rubin);
Computer technology has advanced to the point where we can now store all the data such an instrument would take and transmit it to the ground in a reasonable time. Hubble can currently take and transmit between 2 and 3 gigabytes of data per day; Roman is forecast to take and transmit about 500 times as much data.


Why do satellites with greater altitude travel at a slower speed? For example, a geostationary satellite travels slower than the ISS. What is going on?

Kepler's third law in this situation tells us that
$$ \frac{R^3}{P^2} = \frac{GM}{4\pi^2}, $$
where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P \propto R^{3/2}$.

To calculate the orbital speed, we divide the orbital circumference by the orbital period
$$ v = \frac{2\pi R}{P}. $$

Then substituting for $P$ from Kepler's third law, we get
$$ v = \sqrt{\frac{GM}{R}}$$

Thus, the larger the orbital radius, the slower the orbital speed.

Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force:
$$ m \frac{v^2}{R} = G\frac{Mm}{R^2},$$
where $m$ is the mass of the satellite, which leads directly to the same result.


Apparent size of M31 I am trying to understand something I read on wikipedia about M31.

Wikipedia says that M31 appears more than six times as wide as the full Moon.

But I remember that when I watched it naked eye it seamed a lot smaller, even considering that I was able to see only the central region. For this reason I also used the software Stellarium to check it and setting the same magnification for the two objects (see pictures), the wikipedia information still seems wrong ... or most likely I am wrong!

Where is that I fail?

"At an apparent magnitude of 3.4, the Andromeda Galaxy is one of the brightest Messier objects,[15] making it visible to the naked eye on moonless nights even when viewed from areas with moderate light pollution. Although it appears more than six times as wide as the full Moon when photographed through a larger telescope, only the brighter central region is visible to the naked eye or when viewed using binoculars or a small telescope"

The answer is right there in your screenshots. The 'dimensione' or angular diameter on the sky is 3 degrees for M31, whereas the moon has an angular diameter of only 0.5 degrees. In other words, M31 is about 6 times as wide as the moon.

As wikipedia correctly states, you need a large telescope to see the full extend of M31. With the naked eye of even a small telescope you simply cannot see the spiral arms.

I think this image shows it very neatly:
http://apod.nasa.gov/apod/ap061228.html


Why is there no time in the distance equation using Hubble constant, red shift and speed of light? So I had some problem and I think I solved it, then I began to dig out things around it and I found some equations using red shift etc. to get speed of the object in Universe relative to us and then I found this one:

distance=speed of light/Hubble constant*red shift

It took some time before I found the thing which seems to be wrong with this thing.
How can I get the distance traveled if there is no time mentioned?
Thanks in advance!

Are you familiar with Hubble's Law? Because this is just it, in a slightly less familiar form.

Note that the distance in this formula is the proper distance, the distance between us and some source, not the distance travelled.

Other than that, the speed of light in a sense defines our conception of time and space, so it is mentioned in the formula. ;)


Why don't Neutron Stars form event horizon? Trying to compare density of Black Holes and Neutron Stars I came up with the following:

A typical neutron star has a mass between about 1.4 and 3.2 solar masses1[3] (see Chandrasekhar Limit), with a corresponding radius of about 12 km. (...) Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m^3 [1]

and

You can use the Schwarzschild radius to calculate the "density" of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x10^16 g/cm^3) x (Msun / M)^2 (...)
The value of the Schwarzschild radius works out to be about (3x10^5 cm) x (M / Msun) [2]

Let's take a neutron star from the top of the spectrum (3.2 Msun) and same mass black hole.
Converting units:

Neutron star: 5.9×10^17 kg/m^3 = 5.9 × 10^14 g/cm^3
Black hole: 1.8x10^16 g/cm^3 x (1/5.9)^2 = 5.2 x10^14 g/cm^3

The radius of the black hole would be (3x10^5 cm) x ( 5.2 ) = 15.6km
The 3.2Msun Neutron Star of this density would have volume of 1.08 x 10^13 m^3 which gives radius of 13.7 kilometers
According to Shell Theorem, spherical objects' gravity field strength at given distance is the same for spheres as for point masses so at the same distance from center of same mass (point - black hole, sphere - neutron star) the gravity will be the same.
That would put the surface of the neutron star below the surface of event horizon of equivalent black hole. Yet I never heard about even horizon of neutron stars.
Either I made a mistake in my calculations (and if I did, could you point it out?) or... well, why?

As Francesco Montesano points out, using the wrong mass leads to the wrong answer.  Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.

Since the density scales as ρ ~ M/R^3 and the Schwarzschild radius as Rs~M, the density of BHs scales as ρ~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.


How is IR spectroscopy used to determine the composition of asteroids? I am trying to understand how spectrometers or spectroscopy can be used to calculate the surface composition of asteroids for the purposes of asteroid mining.

It uses the same fundamental conept of mapping between emission-line wavelengths to certain transitions. Similar as we know that 21cm line is the transition of spin-flip of a neutral hydrogen.

Infrared (IR) regions is typically where the molecular lines are seen. Wikipedia has good and brief discussion.

Asteriods are cold so we expect molecules to form. Therefore, we expect to see lines in IR region the most. To be more specific, surface temperature of asteriod is ~346 K, which is equivalent to ~83714 A at max emission (assuming blackbody radiation and following Wien's law); this is IR.

Also note that, what we see is the surface of the asteriod, not the inside.


What exactly does outer space contain? Does outer space contain any gases? Recently i watched a video on youtube,in which NASA installs a camera on a spacecraft.The footage shows some dust particles or some matter floating over the space craft's surface and as well from the thrust.What exactly is that? If there is nothing in vacuum,how come newton third law works out to make it move in turn? Here is the link

http://www.youtube.com/watch?v=HSnOTaCeSuk

Any gas or particles you see in the video are not matter of outer space, but part of the propulsion system or the spacecraft. The high velocity of the gas is a consequence of the propulsion. The exhausted gas (water vapor, aluminium, carbon dioxide, soot, many other compounds, depending on the propellant) continues to move roughly with constant velocity relative to the rocket as soon as it has left the thrusters (Newton's first law); the rocket accelerates as long as the thrusters are active (Newton's third law), but that's not noticible in the video relative to the exhaust fumes. While the rocket is moving through Earth's atmosphere there may be some interaction of the exhaust fumes with Earth's atmosphere.

There is no interaction of gas or fragments with matter of outer space visible  in the video.

Although outer space isn't quite empty. It contains e.g.

interplanetary and interstellar dust, mainly consisting of silicates, but also of other compounds like oxides, water ice, traces of organic compounds,
hot plasma, mainly consisting of protons and electrons emitted by the sun (solar wind),
galactic cosmic rays, mainly consisting of fast-moving atomic nuclei,
electromagnetic radiation across the electromagnetic spectrum, from radio waves to gamma rays,
magnetic fields.


Sky view from Stellarium software vs. Sky view with naked eye I'm slowly starting to interest in astronomy. Currently I'm enjoying in stargazing but unfortunately place where I live is very light polluted.
I'm planing my vacation in a couple of months and one of the attractive places on my list is Tenerife because I can combine both star-gazing and going to the beaches. As per my research, base of Mountain El Teide is perfect for star-gazing so I tried to simulate El Teide location and date in the Stellarium software to see what I can expect. Based on the light pollution index (2) which Stellarium picked for me based on the location I got following picture:

Full resolution image
My question is: How this rendered picture will match my naked eye in the reality at the same location and time (we can assume that there will not be any clouds and that I'll be stargazing when moon is couple of days from New Moon phase). Can I expect something like this or even something better?
I'm aware of all the tips which can maximize my experience but I'm trying to just be sure that El Teide will not disappoint me. I really want have great star gazing-experience and try to see Milky way and much stars as possible because I don't have that luxury in the country where I live.
Any personal star gazing experience from Tenerife location is more than welcome. For my first time I'll plan to not use any equipment (except maybe binoculars)
Btw, there are stellarium settings used for picture that I attached (I assume that this is closest what you can see with naked eye).

From personal experience - yes your Stellarium picture looks like a good dark sky in the Canary Islands. Certainly, the Milky Way is plainly visible, with lots of detail available to the dark-adapted eye. I more frequently view from the top of La Palma, which suffers a little less light pollution than Teide to the East.

The Canarian sky law was largely setup to protect the night sky quality at the observatory on La Palma, but covers all lighting on Tenerife in a straightline view to La Palma. So I would be guessing that the western side of Teide will be the darkest.

I found this picture which illustrates the controlled lighting areas (in green). It is not to scale, La Palma is further away from Tenerife (on the right) than illustrated.


What is the closest candidate for Dark Matter currently? I've both fascinated and frustrated after learning about the existence of Dark Matter.

Can we find evidence of Dark Matter within our Solar system? What is currently the closest candidate for Dark Matter?

And please let me know what are good online resources where I can stay up to date with information about this?

This arxiv post says if there is dark matter in the solar systems, its density should be below $\approx 1.4\cdot 10^{-20} \frac{g}{cm^3}$.

They tried to find the dark matter in the solar systems by measuring its possible gravitational effects.

Note, it is a negative result, i.e. they didn't find anything, but they are suspecting, if the dark matter density would be at least $\approx 1.4\cdot 10^{-20} \frac{g}{cm^3}$, they had found it.

There are various candidates what could it be, the currently considered most probable are the WIMPs (particles interacting only gravitationally and maybe weakly).