the moon - strange moonset around 10 pm (3.2 2014, NZ) - explanation?

during a visit to New Zealand we observed (on 3.2 2014) a strange moonset in Te Anau.

Around 10 pm, shortly after sunset, the moon appeared, but then quickly began to go down, and about 1 hour or less it "touched" the mountains on horizon and shortly after it completely disappeared and then we did not see it anymore.

I did not bring a camera or camcorder, so I only have this bad photo from a mobile phone:

The whole process was pretty fast, so I only have one picture. I've never seen anything like it - usually (in Europe) the moon comes out after sunset rather than going down :-)

There was also one guy living there more than 10 years, but he was apparently surprised too.

Can anybody explain this? Is it something normal in this location?

thanks

cosmology - The largest discrepancy in the history of science

There are roughly two possibilities: either there isn't a large vacuum energy, this would imply that there is something missing from quantum field theory. We don't know what it could be.

Or the zero point energy really is as large as QFT predicts, but there is something else that prevents it from having a large cosmological effect. We have no idea what that could be.

We might get more of an idea if we find out what Dark Matter consists of, or a good quantum theory of gravity we might know more.

the sun - Solar Noon: meridian crossing time versus time of maximum elevation

There are a couple of things you have to think about in order to understand the problem at hand. The first is that there are three coordinate systems at play. The first is the celestial coordinate system, which is a coordinate system based upon the latitude and longitude of the Earth (the celestial equator is the Earth's equator projected out into space). See this post.

The second coordinate system is that of the solar system: The plane of the solar system (which contains the sun and planets, and traces out a line in the sky called the ecliptic). Now because of the tilt of the Earth's axis with respect to the normal vector of the plane of the solar system, the celestial equator is not the same as the ecliptic. What's more, is that the relative position of the ecliptic and the celestial equator change throughout the course of the year.

Lastly, there is the altitude/azimuth coordinate system which locked to the observer. The zenith (the point at which all horizons are equidistant from the point you are looking at) is $+90^{circ}$ in altitude and is a degenerate point in azimuth, and all horizons are $0^{circ}$ altitude but lay on a unique azimuthal arc stemming from the zenith and intersecting the horizon.

Below is a diagram showing the relationship between the first two coordinate systems, the celestial coordinate system and the coordinate system of the solar system.

Why does all of this matter for the question you've asked? It's because as you look due East and due West, you're looking at $0^{circ}$ declination (the celestial equator intersects the horizon at exactly due East and due West) whereas the ecliptic intersects the horizon at different points. These points change throughout the year as the relative position of the ecliptic changes with respect to the celestial equator. Now, the amount the sun travels on the ecliptic over a 24 hour period is $0.9856^{circ}$ ($360^{circ}/365.25 text{ days}$; actually I may want to use a sidereal day here but nonetheless it's about a degree).

Due to this, there is potentially a little bit extra in altitude that the sun may travel throughout the course of a day (since the path of the ecliptic is neither aligned with the celestial coordinate system, nor aligned with the local alt/az coordinate system of the observer; unless you're at the north or south poles in which case your local alt/az coordinate system is the same as the celestial coordinate system, however both are still mis-aligned with the coordinate system of the solar system). I think the word "approximate" needs to be there.

All told, I don't personally know how to quantify the extra bit of altitude the sun may gain from the misalignment of the ecliptic with lines of constant altitude and from the sun's motion along the ecliptic, but I would welcome references and calculations from other people. This little bit extra may not be a really small number. In other words, you might want to see how the rate of change of altitude of the sun at that point changes simply due to the Earth's rotation and how it may compare to the rate of change of altitude due to the motion of the sun along the ecliptic.

EDIT:
I also just looked to see what the ecliptic looks like on the sky (as seen from Philadelphia), and how it compares to the local alt/az coordinate system. Below is a picture with all coordinate systems present to try to illustrate the point I was trying to make.

The green line is the meridian, the orange(ish) lines are alt/az coordiantes, the blue line is the celestial equator, and the red line is the path of the ecliptic. Hopefully you can make this out sufficiently well.

fundamental astronomy - Speed of light through the ISM and Wavelength

You can find a neat description and some examples of the effect here. This is known as the pulsar dispersion measure. As you correctly say, waves with longer wavelength (lower energy photons) are delayed with respect to shorter wavelength radiation from the same phenomenon.

When electromagnetic waves travel through a plasma, they excite currents in the free charged particles. In such cases it can be shown (using Maxwell's equations) that the waves propagate with a relationship between their frequency $omega$ and "wavenumber" $k = 2pi/lambda$ given by
$$omega^2 = omega_p^{2} + c^2k^2,$$
where $omega_p$ is known as the "plasma" frequency and equals $(4pi n_e e^2/m_e)^{1/2}$ for the electrons in the plasma (i.e. it depends on the electron number density $n_e$.).

Now, if you have a bunch of photons emitted as a pulse, the relevant velocity is the group velocity given by $v_g=domega/dk$. So
$$v_g = frac{c^2 k}{(omega_p^{2} +c^2 k^2)^{1/2}} = cleft(1 - frac{omega_p^2}{omega^2}right)^{1/2}$$

This converges to $c$ when the frequencies are high (wavelengths are short), but is slower when frequencies are low (wavelengths are long).

In terms of an intuitive physical picture, yes you could think that the refractive index is frequency dependent, but the difference here is that the reason for this dispersion is that waves travelling through a conductive medium are "lossy" - that is the currents induced also encounter resistivity and therefore the waves heat the plasma.

jupiter - Can magnetism escape a black hole?

Nothing "escapes" a BH - in the sense that a signal originating inside the event horizon remains forever inside. If something is observed moving away from the BH, then it was generated outside the event horizon. If it was generated inside, it would never be observed at all, forever and ever.

Gravity itself does not "escape" a BH - and neither does "not escape". Gravity is simply a characteristic of the metric of spacetime. If spacetime is warped in a certain way, gravity can be measured to exist. A BH is simply a very powerful distortion of spacetime, nothing more, nothing less. It is generated by a concentration of mass/energy, which warps spacetime, and then that concentration becomes trapped by this distortion that it has produced.

In that sense, gravity is simply part of the BH, because it's spacetime being warped, and because a BH is essentially just that. Its gravitational field is part of itself, extending to infinity (but getting weaker with distance). It doesn't "escape" because there's nothing escaping.

It's like having a plastic bag tied into a knot to keep water inside, and someone asks "so how does the plastic escape the knot?" The plastic does not "escape" the knot, the knot is part of the plastic.

This all becomes easier to understand when you realize that gravity is not a thing, it's just an effect of spacetime being distorted.

---

EDIT: I think what you were really asking was - can a BH have its own magnetic field? The answer is yes.

A BH can have 3 characteristics: mass, spin (rotation), and electric charge (a.k.a. the no-hair theorem). All other characteristics of the matter falling in it are lost, except these three. If you drop a proton into a neutral BH, then the BH acquires a charge equal to one proton, and that's a measurable electric field.

Now consider a spinning BH with an electric charge, the Kerr-Newman metric. You have a charge, and you have spin. That means you have magnetism. So, yes, a BH can have a magnetic dipole. However, the rotation axis and the magnetic dipole axis must be aligned - a BH cannot be seen as "pulsing". Again, no signal from inside the event horizon can be observed outside.

However, you should not imagine the electric (or magnetic, same thing) field as "escaping" the BH. It does not escape. What happens is, when the charges were swallowed by the BH, the lines of electric field remain "glued" to the BH, which then acquires a charge. Those lines of electric field have existed forever, they don't "escape" anything, and continue to exist after the charge is trapped by the BH.

Note: electric fields and magnetic fields are one and the same. One could appear to be the other, depending on the motion of the observer.

Which kind of properties can we get for cosmic ray particles hitting on an optical ccd?

Cosmic ray "hits" are artifacts caused when high energy particles (cosmic rays, often muons) slam into atoms in the CCD itself and liberate large numbers of electrons, which then show up as bright spots and streaks when the CCD data is read out. They can also occur as a result of radioactive decay processes in the material in the detector and instrument itself and also the immediate surroundings.

They are a tremendous nuisance that can be removed to a certain extent by median stacking images - though this also fails if you get into a situation where cosmic ray hits are found in the same pixels of $>1$ images in your stack. This limits the lengths of exposures that can safely be made with astronomical CCDs. Depending on exact usage, this limit is usually between 30 minutes and an hour.

The flux of muons and the distribution of angles at which they reach the detector is actually very well known already. It is largely unaffected by local shielding (unless your telescope is buried far underground!)

It is possible to distinguish different types of events (muons, locally produced gamma rays etc.) from looking at the pattern made on the detector (e.g. Groom 2004), but I don't think you can tell much about their energies, unless it could be by looking at the flux as a function of how much shielding (lead?) or how far underground the detector was, since often only a small fraction of the cosmic ray energy ends up deposited in the CCD.

star - is pan-starrs' data available to public users?

Looks like you've to be member of the PS1 Science Consortium to be able to log in.
And you'll need some database query knowledge, at least by using the PSPS Query Builder, or more advanced programming capabilities (XML, SOAP, Python, Perl). A short documentation of the database tables can be found here.
If you like to dig deeper into the database, this FAQ page may be a starting point.

The image processing pipeline (IPP) is described in this paper.

planet - The role of gravity during planetesimal accretion

The very first stage of planet formation involves purely inelastic (i.e., sticky) collisions between pairs of small particles of dust to make slightly larger particles of dust, pairwise collisions between these slightly larger particles to make even larger particles, and so on. So far, the only gravitation that comes into play is that of the central protostar and the disk as a whole. The gravitation of these little objects is very small.

Then some magic happens that makes the little rocks that result from this very first stage grow into planetesimals that do gravitate. How do little rocks (a few centimeters in diameter) grow into little planetesimals (a few hundred meters in diameter)? This is the key open problem in the theory of planet formation.

The core of the problem is that objects the size of a spec of dust to an object the size of a small pebble move with the gas that forms the bulk of the disk. The outward pressure from the gas in the disk makes the gas and small objects move at a slightly less than orbital speed. Once an object reaches the centimeter scale or so, the pressure from the gas no longer provides the support needed to keep those objects moving at that slightly suborbital speed. The objects start to become separated from the gas. By the time an object reaches a meter or so in diameter, the gas is equivalent to a hurricane force headwind.

One consequence of this is that collisions become ever more violent as objects grow beyond a centimeter in size, and hence are more likely to break colliding objects apart than to make them coalesce. An even bigger problem is that that headwind makes centimeter- to meter-sized objects spiral into the protostar, and very quickly (a few hundred years from 1 AU). A number of proposals have been put forth to overcome this meter barrier problem. So far, none appears to have stuck. (Sorry for the pun.)

Once objects reach some critical size, the drag from the gas reduces in importance thanks to the square-cube law and the gravitational attraction increases. That critical size varies from hundreds of meters to tens of kilometers, depending on which paper one reads (experts vary). These objects are "planetesimals," and there are lots of them. These once again combine readily to make planetary embryos. From then on, its just a mopping up process as gravitation either makes these planetesimals / planetary embryos combine or ejects them from the star system.

Does Pluto have impact craters?

It was predicted by some, before the images arrived, that Pluto's craters would be much shallower than perhaps on other rocky objects/satellites. The reason is the icy composition and the relatively low impact velocities in the outer solar system (Bray & Schenk 2015).

There are indeed reported quotes in recent days that there are no impact craters on Pluto (e.g. here or here), based on the highest resolution images.

However, I'm with you - the things you have marked look like very large impact craters, or the remains of them, and so what we are seeing is that there are no small ones.

The gist of what I have heard/read is that this indicates geological activity on Pluto, which is effectively removing the evidence of relatively small impacts and that the surface we see is mostly younger than 100 million years old.
Watch this space, a lot of people are just flapping their lips at the moment (including me).

Another possibility - raised by Wayfaring Stranger - could be that the expected number of impacts was (hugely) overestimated. This is discussed by Singer & Stern (2015). There are indeed uncertainties, and it is hoped that cratering on Pluto & Charon would help constrain the number densities and size distribution of small impactors. My impression is that the uncertainties could be at levels of up to factors of 10 - and this seems to be discussed quite carefully in Greenstreet et al. (2015) (behind a paywall, sorry), who also comment in the abstract that surface ages judged from cratering are " entirely dependent on the extrapolation to small sizes [of the impactor distribution]".

Further Edit: LDC3 suggests all the craters could be too small to be seen. The high resolution images that have been released could resolve craters of a diameter of a couple of km or more and cover a few percent of the surface. According to Singer & Stern (2015), the number of craters at this size and above on Pluto they expected in such an area is $sim 1000$. So I think that either the planet has been resurfaced recently, the craters have been eroded in some other way or their estimates for the number of impacts/size of the impactors is wrong.

the sun - Name and language of symbol in astronomy book

I've been working on a programming project involving Peter Duffett-Smith's Practical Astronomy With Your Calculator (3rd Ed.), which has a large number of mathematical/astronomical formulas in it. There's a part at the end of the book for symbols and abbreviations, which contains this:

After doing some Google-fu, I was unable to find the name of the symbol. Would anyone smarter than me (which should be quite a few of you... heck, quite a lot) know what this is? Hopefully there's Unicode for it...

solar system - Pluto's orbit overlaps Neptune's, does this mean Pluto will hit Neptune sometime?

No. From 1979 to 1999, Pluto was the eighth planet from the sun. In 1999, it slipped beyond Neptune to become the ninth. But Pluto's 248-year orbit around the sun takes it 17 degrees above and below the plane in which Neptune and the other planets travel.

So their paths don't actually cross as they swap positions. Imagine you are the sun in the middle of your back yard. The fence is Neptune's orbit. You toss a boomerang way out over the neighbor's houses and it comes back, being on both sides of your fence during its travels without hitting the fence. Of course, activity like that can be frowned upon, and in Pluto's case helped lead to its demotion.

Reference: Will Pluto Neptune Hit

planet - North Pole Right Ascension/Declination to axial tilt conversion

This is wrong, but may help someone find the right answer:

You can convert right ascension and declination to a 3 dimensional
unit vector in the J2000.0 ICRF reference frame using the standard
formula for converting spherical coordinates to rectangular
coordinates and using a radius of 1:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}

The ra and dec of the north ecliptic pole is ra,dec of
{270, 66.5607083333} per Wikipedia, so the unit vector representing
the north ecliptic pole is:

{0, -0.3977771648286046, 0.9174820582119942}

As it turns out, there is conflicting data for Ceres, perhaps because
of the semi-recent DAWN flyby.

Instead, I'll use Vesta, where wikipedia explicitly states (https://en.wikipedia.org/wiki/4_Vesta#Rotation):

north pole pointing in the direction of right ascension 20 h 32 min,
declination +48 [... which] gives an axial tilt of 29 [degrees]

Converting to degrees, Vesta's north pole's ra,dec is
{308,48}. Using the formula above to find the unit vector, we get:

{0.411957936296447, -0.527282113378167, 0.743144825477394}

If we take the dot product of two vectors and divide by the product of
their lengths, we get the cosine of the angle between them. In this
case, both vectors have length 1, and the dot product is 0.891563
whose arc-cosine is right around 27 degrees.

So, if Vesta's orbit were the same plane as Earth's orbit, the axial
tilt would be 27 degrees.

However, since Vesta's orbit is inclined 7.14043 degrees to the
ecliptic, this answer is incorrect.

I don't think you can find the correct answer without using Vesta's
inclination and the longitude of its ascending node.

Both of these values are known, but I can't figure out how to use them
to get the correct asnwer.

star - What is the origin of the OBAFGKM classification system?

The Wikipedia article on stellar classification has the full list of letters:

Class | Surface Temperature
---------------------------
  O   | > 33,000 K
  B   | 10,000 - 33,000 K
  A   | 7500 - 10,000 K
  F   | 6000 - 7500 K
  G   | 5200 - 6000 K
  K   | 3700 - 5200 K
  M   | 2000 - 3700 K
  L   | 1300 - 2000 K   -
  T   | 700 - 1300 K    |  These are those new letters
  Y   | < 700 K         -

History

This Cornell doc explains much of the history behind the system:

In the 1890s, many scientists were interested in developing a classification scheme for the stars. Edward C. Pickering at Harvard University, together with his assistant Williamina P. Fleming, assigned stars a letter according to how much Hydrogen could be observed in their spectra: stars labeled A had the most Hydrogen, B the next most, and so on through the alphabet. There were 22 types in all: this scheme was rather cumbersome, and it wasn't clear what its physical significance was.

In 1901, another of Pickering's assistants, Annie Jump Cannon, also begun to work on the classification sequence. Her meticulous observations led her to simplify the 22-type scheme into a simple sequence of temperature: OBAFGKM. Not only was the scheme much simpler than the previous, but it related the amount of hydrogen observed to a physical property of the stars for the first time.

space - Can there be an infinity of humans in the Universe?

IMO, this sounds more like a philosophical question rather than an astronomical question, but since I like both subjects, I'll give it a shot.

Note: This is hypothetical question, so this answer is purely subjective (and hypothetical).

Infinity

• If the number of stars is infinite, so would the numbers of humans.

I believe that in every system where an entity of infinity is introduced, all other entities will (eventually) be infinite. But, (and this is the fun "part") you cannot measure infinity. It would give you infinite results over a time span of infinity.

Take your example for instance. If there were infinite numbers of stars. Picture youself flying around in a spaceship with the goal to find out if this question is true or not. What do you see? "Worse case" scenario; you see only stars. Wherever you go. Forever. The opposite; only humans.

One might think that it's the "in between" that is the most likely scenario. But given the probability theory, we should flatten out at approx. 50%.

Finite

If there is a finite numbers of stars then I think it's "safe" to conclude that there ain't an infinite number of humans as we're all made of starstuff.

Conclusion

Yes: If there's an infinite numbers of stars.

No: If there's a finite numbers of stars.

How do I remove fungus from a telescope mirror?

Several points need to be made.

1. Performance

The performance of the whole optical stack is incredibly resilient w.r.t. small spots on the primary mirror. The mirror might look visually very, very dirty, but the performance of the whole instrument will remain essentially the same. Even if you chip the mirror at the edge, it shouldn't matter. Take a marker and scribble a dozen lines on the mirror - it doesn't make a large difference for performance, not greater anyway than what the 4 vanes of the secondary mirror already do (we can start a separate discussion about diffraction artifacts from spots and lines on the mirror - but the bottom line is, don't worry about it).

With that in mind, you should be conservative when deciding to clean the mirror. Let it accumulate some amount of dirt, the scope will be okay.

2. Permanent damage

That being said, mold or other living things may chemically corrode the mirror. Again, a few spots of corrosion do not impact performance, but you don't want to allow this process to continue.

Dust will generally not corrode the mirror. Salt (from ocean breeze) will usually corrode it in time. With mold it's hard to say, but it's best if it's removed.

3. Prevention

Mold grows because the mirror is wet. The mirror fogs up usually when you bring the scope back into a warm room from outside; the mirror is cold, and water in the warm air condenses on it.

Procure a cheap, small fan, and point it at the rear end of the scope while it's warming up after you bring it back in. The air current will reduce water condensation on the mirror. Turn the fan off after 30 ... 60 minutes.

Also, keep the OTA (the big tube of the telescope) capped off while not in use. This will reduce dust deposits, and therefore mold will find less food to grow. At least cap off the top end and store the scope in a vertical position; that will reduce dust deposits greatly. The bottom end is less important.

Avoid storing the scope in a place with large day/night temperature variations. When the mirror is colder than air, condensation may form. This can happen spontaneously with scopes stored in a shed outside. The mirror fogs up in the morning, when air is getting warmer but the scope is still cold. This is much less of an issue with small scopes, but it's something to keep in mind.

You could cap off both ends while in storage, and place a bag of silica gel inside the OTA (don't drop it on the mirror). That will suck up all humidity inside the OTA, but you have to bake the silica gel in the oven every so often to keep it working effectively. You could buy silica gel online, it's very cheap. Avoid temperature variations and the gel is not necessary.

4. Clean up procedure

This is the procedure recommended by Optic Wave Labs:

https://www.youtube.com/watch?v=9Y8xFnXFVGQ

I make mirrors but I don't coat them myself (yet) - however, I've found that procedure to work very well on both coated and non-coated glass.

A few minutes of initial soaking in water makes everything easier later on.

After soaking it, add a few drops of detergent and rub very gently with your clean fingertips or with a soft, clean, lint-free cloth. Don't put pressure on the mirror, just try to detach the dirt from mirror in the horizontal direction and let it wash off. Don't drag dust particles across the glass, just detach them and let them go; this is why fingers are better than cloth - because you can feel large particles coming off.

When done, wash off the detergent with lots and lots and lots of water.

Using tap water for most of the cleaning procedure is absolutely fine. If you can drink the water, it's fine for the mirror too. Don't use hot water if the mirror is cold, there's a very small chance of cracking it, especially with cheap, mass produced glass.

When you're done cleaning, rinse the mirror with distilled water. Again, a few water spots on the mirror don't matter to performance, but the mirror looks better if it's spot-free. Tap water tends to leave water spots if you don't follow up with distilled water rinsing, and most people find the spots not pleasing from an esthetic perspective (but performance is same). So just splash a bottle of distilled water on it at the end and you'll be fine. No touching after rinsing.

Some dirt or mold may not be removed by the usual water-based procedure. In that case, use isopropyl alcohol, the highest concentration you can find (over 90% is fine). Rub gently with your clean, non-greasy fingertips or a lint-free cloth. If it doesn't come off quickly, leave it alone. Alcohol has probably killed the mold anyway.

Acetone might be used for things that resist even alcohol. But again, you have to use your judgement: a few little spots won't affect performance, so is it worth the trouble?

In any case, rinse with distilled water at the end, then let it dry off in an inclined position (carefully to not tip the mirror over on its face). Put a towel on the table in front of the mirror, so if it does topple over, at least it falls on something soft. Do not let it dry in a horizontal position, you'll get water spots and dust.

If there's some permanent damage in a few little spots, don't worry about it. The mirror will work essentially the same.

It is recommended that you do the water-based procedure every 1 or 2 years, depending on how dirty the mirror is. If dust contamination is moderate, leave it alone.

5. Post-cleanup

After everything is done, re-collimate the telescope carefully. But this should go without saying, since collimation is part of the regular maintenance of newtonian telescopes.

exoplanet - Max. speed that a space probe can communite with the Earth?

Question: Which is the average or maximum speed of a telematically sent message that a space probe can send? Is as simple as radio waves travel at the speed of light in a vacuum?

Context: I have been asked to make a first estimation about how much time is needed to send a space probe and then receive the signal.

I am supposing that the space probe is sent at $15 km/s$ but I have not idea of the telematic signal speed.

Thank you in advance!

distances - What is the radius of observable universe- 46 billion LY or 200 yottameters?

I have found two answers for the radius of the observable universe. Wikipedia (and other places, including this site) say approximately 46 billion LY. I am researching this to study for the National Science Bowl, and their question database says 200 yottameters, which is about 23 billion LY (half the size that wikipedia says). There are other places which cite 200 yottameters as the radius. Which is the more accurate answer?

extra terrestrial - How will the Breakthrough Initiatives affect other astronomy observations?

BACKGROUND

The Breakthrough Initiatives is in the news because of a $100 million private donation for a SETI project. It doesn't seem to be for building any new observatory (but maybe for buying computing power for handling the data for SETI purposes?). It will use two of the greatest radio telescopes in the world, Parks and Green Banks, and optical Automated Planet Finder Telescope at Lick, some part of the time during 10 years.

QUESTIONS

• Does this mean a zero sum redistribution of observatory time from other astronomical purposes to SETI?




• How does it work when someone can buy the usage of these telescopes? Isn't there a danger in letting popular scifi ideas decide how the great telescopes should be used? Do recruited big names like Hawkins and Drake decide how they should be used anyway??

earth - Why does this graph for sunlight intensity on land has a steeper slope during sunrise as compared to sunset?

I got this image while checking weather data for a city in North India using Mathematica's Wolfram Alpha query

I noticed one feature in the graph which i could not explain .
Why does the encircled part 'A' which denotes sunrise has slightly higher slope as compared to encircled part 'B' which denotes sunset !

I also checked the graph for summer month and the pattern was exactly opposite

Summer Month

Am i interpreting the data wrong or is it so that sun achieves its highest intensity during the day faster in winter months than in summers ? What could be the reason behind this ?

the sun - Is the Jupiter-Sun system considered a binary system of some type?

Since Jupiter is very massive, it is the only planet (in our solar system) that has a center of mass with the Sun that lies outside the volume of the Sun. (Source)

If Jupiter was a star, they would form a « binary star ».

If the Sun was a planet, they would form a « double planet ».

Since the Sun is a star and Jupiter is a planet, does this have a particular name?

Does Jupiter have a special status or a particular effect in our solar system because of its heavy mass?

Since Jupiter-Sun's center of mass lies outside the volume of the Sun, that means that the Sun moves around that center of mass. Does this have an effect on Mercury, Venus, Earth and Mars orbits?

star systems - Sources of Turbulence in the ISM

Turbulence sources:

There are numerous sources of turbulence in the interstellar medium, at all scales:

• at large scales, there is the shear from galactic rotation. One way to sustain turbulence and to couple large and small scales would be the magnetorotational instability (MRI).


• at large scales, gravitational instabilities can also play a significant role, through spiral structures.


• outflows and jets from forming stars play an important role, releasing a lot of energy in the ISM.


• in star forming regions, massive stars are also important. Radiation and stellar winds from massive stars are an important input of energy in the ISM. And eventually, the most massive ones will blow up in supernovæ, releasing even more energy.

Therefore, one could then regards separatly three processes related to massive stars:

• stellar winds


• ionizing radiation


• supernova explosion

Importance for star formation:

They are all relevant to star formation, one way or another. One key property of turbulence is to cascade from large to small scales; therefore, even if you inject turbulence at large scales (galactic scale) you'll get turbulent motions down to the scale of a molecular cloud.

A nice illustration of the tubulent cascade is the Larson's relation (Larson 1981):

The Larson's relation shows the evolution of the velocity dispersion with the size of the structure you're looking at. Velocity dispersion is an indicator of turbulence. Indeed, these dispersions are non-thermal: knowing the typical temperature of the MIS (about 10 K), one can estimate the thermal velocity of, for example, the CO molecule ($v_th = sqrt{2kT/mu m_H}$, with $k$ the Boltzman constant, $T$ the temperature, $mu$ the mean molecular wright and $m_H$ the hydrodgen atom mass) that is about 0.07 km s$^{-1}$. Measured velocity dispersions are of the order of 1 to 10 km s$^{-1}$, and these is interpreted as a turbulence signature (and estimate).

Details:

Energy rates: values are given (roughly) for the Milky Way

• MRI: $ small dot e = 3 times 10^{-29} {rm erg cm}^{-3} {rm s}^{-1} $;


• Gravitational instabilities: $small dot e = 4 times 10^{-29} {rm erg cm}^{-3} {rm s}^{-1}$;


• Outflows: $small dot e = 2 times 10^{-28} {rm erg cm}^{-3} {rm s}^{-1}$;


• Ionizing radiation: $small dot e = 5 times 10^{-29} {rm erg cm}^{-3} {rm s}^{-1}$


• Supernovæ explosions: $small dot e = 3 times 10^{-26} {rm erg cm}^{-3} {rm s}^{-1}$


• Stellar winds: it strongly depends on the type of star: it varies as the power of -6 of the star's luminosity. It therefore ranges from an energy comparable to a supernova explosion (or even more for Wolf-Rayet stars) to almost nothing.

---

Sources:

Can two comets travel together as one?

If the closing velocity is very low, the energy can be dissipated by the comets squishing. You might consider them to merge into one comet in that case. You might be left with a bunch of pieces all travelling together. Whether you call that one comet, the original two, or many is a matter of terminology-all the stuff will be in the same orbit.

galaxy - What will happen to life on Earth when the Andromeda and Milky Way galaxies collide?

First, note that by the time Andromeda is close enough for collisions with wandering stars to become a concern, Earth's average temperature will have changed significantly, and the planet will be unrecognizable.

When Sol is 8.5 billion years old, it will still have hydrogen available for fusion, but as it fuses it contracts and expands differentially. The contraction causes hydrogen fusion to become more favorable, so that Sol will have 50% greater power output ($6 times {10}^{26} mathrm{W}$) and 3% greater effective temperature ($6000 mathrm{K}$). Fusion also causes Sol to lose mass at a prodigious rate (currently $4 times {10}^9 mathrm{kg/s}$); it will release $6 times {10}^{43} mathrm{J}$ from fusion, which corresponds to $7 times {10}^{26} mathrm{kg}$. That is about one hundred Earth masses of sunlight but only $1 over 3000$ the mass of Sol. Gravitation with Earth decreases proportionally, so Earth's orbit might on average expand $3000 mathrm{km}$ per billion years. Other gravitational effects might change Earth's average distance by as much as $6 times {10}^5 mathrm{km}$, 4‰ of an astronomical unit. Expansion of Sol's outer layers due to reduced gravitation will increase its radius by 20%, $3 times {10}^5 mathrm{km}$. Thus Earth will receive nearly 50% more power as well.

The energy balance of Earth wrt Sol gives the expected surface temperature:

$$
begin{align}
bar{a} = & 0.7 & smalltext{(Average absorption)} \
P_p = & 1366 mathrm{W/m^2} & smalltext{(Average solar flux incident on Earth at present)} \
P_f = & P_p cdot 1.5 approx 2000 mathrm{W/m^2} & smalltext{(In future)} \
sigma = & 5.670373 times {10}^{-8} mathrm{W/m^2/K^4} & smalltext{(Stefan-Boltzmann constant)} \
\
T_p^4 = & frac{bar{a} P_p}{4 sigma} \
approx & frac{0.7 cdot 1366 mathrm{W/m^2}}{2.268149 times {10}^{-7} mathrm{W/m^2/K^4}} \
approx & 4.2 times {10}^9 mathrm{K^4} \
T_p approx & 250 mathrm{K} \
\
T_f approx & T_p cdot {1.5}^{1/4} approx T_p cdot 1.11 \
approx & 280 mathrm{K}
end{align}
$$

Since the average surface temperature on Earth is not $-20 mathrm{°C}$ — it is $+15 mathrm{°C}$ and already around $8 mathrm{K}$ warmer than in an airless future — we can see the atmosphere has a significant role in retaining heat. Assuming increasing cooling needs do not lead to the atmosphere retaining more heat, the average surface temperature can be expected to rise to $+50 mathrm{°C}$.

The average temperature of Antarctica is now $240 mathrm{K}$ in winter and $270 mathrm{K}$ in summer. These can be expected to rise to $270 mathrm{K}$ (just below freezing) and $300 mathrm{K}$ (well above freezing) respectively, and this is a best-case scenario. Antarctica will melt. That will produce the largest component (60%) of sea level increase, in total around $100 mathrm{m}$.

---

If Earth were still inhabited four billion years from now, it is extremely unlikely that Earth would fall into a star from Andromeda.

Space is big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is.

— Douglas Adams, The Hitchhiker's Guide to the Galaxy

The Milky Way is about 100,000 light years in diameter and contains about 400 billion stars. Andromeda is bigger and denser; it may have one trillion stars and a diameter of 140,000 light years. It is 2.5 million light years away but appears six times larger than Sol.

$$
begin{align}
d_M approx & frac{4 times {10}^{11} mathrm{stars}}{{10}^{10} pi/4 mathrm{{ly}^2}} \
approx & 50 mathrm{stars/{ly}^2} \
\
d_A approx & frac{{10}^{12} mathrm{stars}}{2 times {10}^{10} pi/4 mathrm{{ly}^2}} \
approx & 60 mathrm{stars/{ly}^2} \
end{align}
$$

If the two galaxies were simply superposed, there would be about one hundred stars per square light year, viewed from infinitely far along the rotation axis. However, the Milky Way is a 2:1 ellipse as seen from Andromeda, while we see Andromeda as a 3:1 ellipse. Projecting both onto a plane between them, perpendicular to a line between their central black holes, would give a region of overlap with dimensions between $50 times 50 mathrm{{kly}^2}$ and $50 times 100 mathrm{{kly}^2}$, with at most half the Milky Way outside it. Sol is likely to be involved in the collision, since it is about 27,200 light years from the galactic center.

That doesn't mean, though, that Earth will come close to another star, that Sol might collide, or that the solar system will be disrupted.

Considering probabilistically the worst-case scenario (the entire Milky Way falls through Andromeda on their first pass), there is a mean free path for stars. The actual stellar density of the colliding galaxies is:

$$
rho approx 1.4 times {10}^{12} mathrm{stars} / V_{A cup M}
$$

where the union of the two galaxies' volumes would be a very complicated expression. Very roughly, their volumes can be described as joined cones, ignoring their spheroidal dark matter halos (which are mostly harmless).

$$
begin{align}
rho approx & frac{1.4 times {10}^{12} mathrm{stars}}
{left(
frac{1}{2} cdot
left(
{10}^3 mathrm{ly} cdot {10}^{10} pi/4 mathrm{{ly}^2} +
1.4 times {10}^3 mathrm{ly} cdot 2 times {10}^{10} pi/4 mathrm{{ly}^2}
right)
cdot frac{1}{3}
right)} \
approx & 0.28 mathrm{stars/{ly}^3} \
\
V_star approx & 3.6 mathrm{{ly}^3} \
\
r_star approx & {left( V_star cdot frac{3}{4 pi} right)}^{1/3} \
approx & 0.95 mathrm{ly}
end{align}
$$

At a distance of 1.9 light years, Betelgeuse would look a lot like Mars. If we assume disaster results from a star closer than the diameter of the heliosphere (about 200 AU), then:

$$
begin{align}
m = & frac{1 mathrm{star}}{rho cdot pi cdot 4 times {10}^4 mathrm{{AU}^2}} \
approx & 1.1 times {10}^{21} mathrm{m} \
approx & 7.2 times {10}^9 mathrm{AU} \
approx & 1.1 times {10}^5 mathrm{ly}
end{align}
$$

On average, a star can travel 110 thousand light years before it grazes past another, slightly less than the diameter of Andromeda. The proportion of stars from the Milky Way that do not approach within 200 AU of stars in Andromeda is at least $1/e^{1.4 / 1.1} approx 100/400 mathrm{billion stars}$. For Earth to approach within 4 AU of another star (one Betelgeuse radius), it can be expected to travel at least 2500 times farther, which at a relative velocity of 300 km/s would take $9 times {10}^{18} mathrm{s} approx 300 mathrm{billion years}$.

core - Could we fly/drive through Jupiter?

http://spaceplace.nasa.gov/review/dr-marc-solar-system/gas-giants.html

We think of a gas as something very . . . well, airy. After all, air is the gas we all know and love. We breathe it and fly planes right through it with no trouble. So it makes sense to think that a gas planet must be like a big, airy cloud floating out in space.
Saturn in true color.

The bigger a planet becomes, the heavier is the material weighing down on its center. Think of how it feels to dive under water. If you are wearing a face mask, you notice that as you dive deeper, the mask presses harder and harder on your face. Also, your ears start feeling the pressure even at 2 or 3 meters (5 or 10 feet) below the surface. The pressure you feel on your body is due to the weight of the water above you. The deeper you go, the heavier the water above you and so the greater the pressure on your body. Even on Earth's surface, each square inch of your body experiences 14.7 pounds of pressure due to the weight of the atmosphere above you. If you could dive down to the center of Earth, the pressure on your body would be about 3.5 million times as great! The center of Jupiter is more than 11 times deeper than Earth's center and the pressure may be 50 million to 100 million times that on Earth's surface!

The tremendous pressure at the center of planets causes the temperatures there to be surprisingly high. At their cores, Jupiter and Saturn are much hotter than the surface of the Sun!

Strange things happen to matter under these extraordinary temperatures and pressures. Hydrogen, along with helium, is the main ingredient of Jupiter's and Saturn's atmospheres. Deep in their atmospheres, the hydrogen turns into a liquid. Deeper still, the liquid hydrogen turns into a metal!

But what's at the very center of these planets? The material becomes stranger and stranger the deeper you go. Scientists do not understand the properties of matter under the extreme environments inside Jupiter and Saturn. Many different forces and laws of nature are at work, and the conditions inside these planets are very difficult to create in a laboratory here on Earth. But you can be sure that you wouldn't be able to fly through these bizarre materials! As we now know, the gas giants are much more than just gas.

Is the whole universe is rotating on an axis?

From what we observe it is extremely unlikely the universe is rotating, but it is a good question nonetheless.

Perhaps you thought that since we see galaxies moving away from us they could be being pulled outwards as a result of the rotation. All the galaxies we observe (outside the local group) are accelerating away from us, and this includes galaxies in all parts of the universe. If the universe were like a giant spinning cylinder (or whatever object you like) then the galaxies "above" and "below" us in the cylinder would appear stationary since they are rotating at the same radius.

This observation does not support the claim that the universe is spinning on some invisible axis.

As always we should remember that our local universe may not represent the entire cosmos, and we would in fact be spinning around some axis. As more observational evidence is gathered we understand more and more about the universe, and one would be foolish to rule out a theory based on current observational evidence.

amateur observing - Can I see comet ISON from Saudi Arabia?

I'm not far from you (Israel) and I am also waiting to see comet ISON.

Comets are very unpredictable, and comet ISON has not yet reached the critical part of its journey which will determine visibility for us. Comet visibility is usually due to reflection of the Sun's rays on the coma and tails of the comet, as the nucleus is to small to see directly. So ISON's visibility (and all comets for that matter) depend on how much gas, water, and other materials that it outgasses. This in turn depends on how much volatiles it has, how much the Sun melts them, how much the slingshot around the sun and tidal forces fracture the nucleus, and other factors. So the actual visibility that will occur is very difficult to determine in advance. That said, ISON has all the markings of this being its first trip around the Sun. Thus, we assume that it has many volatiles to release and we assume that ISON will be exceptionally bright.

In any case, ISON will be most visible when it is closest to Earth, near the end of December. It should be most visible in the early morning sky, I'm guessing at about 4:00 AM or so. This is because ISON will be close to the Sun (thus visible close to the time that the Sun becomes visible). Like all celestial bodies, it will rise in the East. Each day the comet will move farther from the Sun and thus will be visible earlier and earlier in the night, until it disappears from view probably sometime in mid January.

You are welcome to visit Beersheba and have a peak through my binoculars!

If the expansion of the universe is speeding up, how long until everything nears the speed of light, and what would happen?

This is a tricky question, because dark energy does not always affect the distances between any two given objects in the universe. Consider this question: Galaxies are moving away from each other then how Milkyway and Andromeda galaxy coming towards each other?. The Milky Way and Andromeda are incredibly far apart, yet they are moving towards each other because of gravity. Likewise, the other galaxies in the Local Group are gravitationally bound to the Group, and won't be leaving it. The gist of this is that it's difficult to determine just where dark energy becomes the dominant player in the universe.

We can calculate the speed (recessional velocity) at which objects move away from each other by using Hubble's law, $v=H_od$, where $v$ is recessional velocity, $H_o$ is the Hubble constant, and $d$ is the distance between the objects. We can use the Hubble constant, along with the Friedmann equations, to derive the Hubble parameter, which can then be used to calculate the expansion of the universe over time. The Hubble parameter is a function of time, while the Hubble constant is constant throughout space-time; it is the current value of the Hubble parameter.

So while we can use the Hubble constant to calculate how fast objects are moving away from each other now, we can use the Hubble parameter to calculate at what time two objects will be moving away from each other at a given speed. I invite you to use the following pages to get you started on some calculations, but I will warn you to not use them for any objects inside the Local Group.

Here are the aforementioned pages:

Deceleration parameter

Scale factor

I hope this helps.