How to get ephemeris data for location? How can I obtain ephemeris data which fits to a specific location? Differently phrased: I need ephemeris data of satellites which orbited a specific location. Is there a document where I can look up references?

Right now I am looking in ftp://cddis.gsfc.nasa.gov/gnss/data/daily/2019/ but the subfolders only contain weirdly named broadcast files which don't make much sense to me.

The reason I need these ephemeris files is that I want to test gps-sdr-sim with a USRP I got. I know that those ephemeris broadcast files contain data to every other satellite in orbit (but not in the same file as I understand).

I tried multiple ephemeris files with gps-sdr-sim but none of the data transmitted worked for the location I am in.

Any help would be appreciated.

The merged broadcast ephemeris is in the YYn directory where (YY is the 2 digit year, n means navigation e.g. 19n) and you are looking for the brdcDDD0.YYn.Z where DDD is the day number (of the year) and YY is the 2 digit year as before. These are compressed RINEX files; gunzip etc will be able to uncompress them. The RINEX format is pretty well documented and I've written Python code to read them.

The other files in those directories are the received GPS signals from the various GNSS monitoring sites. You may be able to make progress by using the map on the Archive Explorer to find a site near you, noting its code and then downloading its data file. For me the nearest is VNDP (Vandenberg AFB) and so for 2019-Jan-01 I would download vndp0010.19n.Z from ftp://cddis.gsfc.nasa.gov/gnss/data/daily/2019/001/19n/. Otherwise I think you would be looking at having to do your own visibility code to work out what satellites were visible from your location at the specific time (since the satellites have a ~12 hour orbital period and come and go)

You may want to look at RTKlib (binaries available there at github also) which has a lot of tools for processing these sorts of data and has rtkget which can automate the download of the various types of files from CDDIS etc.


Why is gravity only an attractive force? As per the universal law of attraction, any two bodies (having some mass) experience a force of 'attraction' which is proportionate to ...and ...inverse proportionate ....

Then comes my question: Why it should be force should be of type 'attraction' only ?  Why it should not be repulsion / any other kind of force ?

Because mass is positive

To expand your quote concerning the gravitational force into an equation:

$$F_G = -\frac{Gm_1m_2}{r^2}$$

The force of gravity, $F_G$ is proportional to the product of the masses and inversely proportional to the distance, $r$, squared. Let's break this down and see what might cause $F_G$ to be positive.

In this equation, $r$ cannot be negative because it's a distance between two locations. Two locations cannot be a negative distance apart. And even if they somehow were, the squared would take care of that anyway.

$G$ is the universal constant and always positive. You might argue that it could possibly be negative, but that's not possible. $G$ actually doesn't really exist. It doesn't describe anything fundamental to the physics of the universe. $G$ is simply a bookkeeping constant that allows us to get the right answer for the force based on any choice of units for mass and distance. Technically, if one uses the "correct" units for mass and distance (e.g., the Planck units), then $G=1$ and effectively doesn't exist. Since $G$ is just a scaling factor that depends on the choice of units, it will only be a positive number.

That leaves us with the masses. These are the only things which could possibly be negative. Of course, to get a positive, repulsive force, one mass would have to be positive and the other negative. But what exactly is a negative mass? Mass is the metric which describes "how much" of something there is. How can you have less than nothing of something?

Why can mass not be negative?

If you want to look at this another way, you can show that if mass could be negative, you'd get nonsensical results! Assuming of course, all other aspects of physics were the same. Recall from Newton's second law that

$$F = ma$$

Let's say there are two blocks sitting on a table. One block has a mass $m_1>0$ which is positive and the other has a mass $m_2<0$ which is negative. Ignore all other forces on these two blocks for the moment.

I go up to $m_1$ and I apply a force to push this mass forward. The acceleration that is induced is: $a = F/m_1$. Necessarily, the direction in which $m_1$ moves is the same direction in which I'm pushing. That's all well and good.

Now I go over to $m_2$ and I apply the same force, attempting to push it forward on the table. The acceleration induced on $m_2$ will be: $a = -F/|m_2|$. Note I made $m_2$ positive and pulled out the negative sign. You can see that if my force is forward, the direction the mass moves will be backwards! But here's the problem, my hand is in the way because it's trying to push to mass. As the mass tries to move backwards into my hand, it will be applying a force back on my hand, which by Newton's third law, necessarily mean's my hand is applying more force on the block, which then applies more force on my hand, ... and suddenly infinite forces are being applied or equivalently, these objects are infinitely accelerating. This is described by the concept of Runaway Motion.

If this seems strange to you, that's because it is. If negative masses existed, we'd live in a very weird universe. Fortunately, we live in a universe where physics makes sense, mass is positive, and by extension gravity is always attractive.


What are the advantages and disadvantages of a 2-inch eyepiece versus a 1.25-inch eyepiece? I'm considering picking up some new eyepieces for my telescope and am interested in comparing 2-inch and 1.25-inch eyepieces. Are there any disadvantages to 2-inch eyepieces? Are 2-inch eyepieces only advantageous with longer eyepieces?

When you look into the eyepiece, you've noticed that the image is round, framed by a black ring. That ring is called the field stop, and it's an actual round piece of metal or plastic inside the eyepiece. Its function is to limit the image to the size where the image quality in that eyepiece is acceptable.

If you were to remove the field stop, the image outside of it would start to look bad, and its edges would be fuzzy and barely usable. Also, most people prefer a neat, sharp edge to the field of view, so there's the esthetic element to consider also.

The size of the field stop is given by two parameters:

the focal length of the eyepiece, measured in mm
the apparent field of view of the eyepiece, or how wide the image appears when you look in the eyepiece, measured in degrees of angle

Longer focal length eyepieces will naturally have bigger field stops, keeping everything else in proportion. Eyepieces with a wider field of view will also require a bigger field stop.

IMPORTANT: the field stop, obviously, must be smaller than the diameter of the barrel of the eyepiece, otherwise it would not fit in. This is what dictates the lower limit for the diameter of the barrel. The barrel must be bigger than the field stop. You need to increase the barrel diameter when the field stop grows too big.

So, when you stick to short focal length eyepieces with a narrow field of view, the required field stop is pretty small, and it can easily fit into a 1.25" barrel.

But with longer focal length eyepieces, especially with modern eyepieces with a wide field of view, the field stop can grow pretty large, and eventually becomes bigger than the 1.25" barrel. That's when you need to move up to a 2" barrel. That's what drives the need for larger size barrels.

Long time ago, there used to be eyepieces with 0.965" barrel diameter, because back then opticians didn't know how to make wide field eyepieces. Then the field of view kept increasing, so opticians moved up to 1.25".

Eventually the newer 2" standard was adopted, as the field of view kept growing. At 32mm focal length, a Plossl eyepiece with a 52 degree field of view can fit in a 1.25" barrel, but a more recent 68 degree design requires 2".

Nowadays there's a new standard emerging, with a 3" barrel diameter, imposed by the very wide field eyepiece designs invented recently. At 30mm focal length, an eyepiece with an 82 degree field of view could use a 2" barrel, but if you demand 100 degrees FoV at 30mm, you need to move up to 3".

Of course, larger barrel diameters make heavier and more expensive eyepieces. So you need to mix and match. At smaller focal lengths, the 1.25" barrel works well. For longer focal length eyepieces, as you move up in terms of the size of the field of view, eventually you must switch to 2".

If you only use Plossl eyepieces, you can have all of them in the 1.25" format, because even at the longest focal lengths that make sense in practice, the Plossl field of view is narrow enough to not require a big field stop.

As a practical example, my collection of eyepieces currently is about evenly split between 1.25" and 2", with the 2" barrels being reserved for eyepieces over 18mm focal length. All my eyepieces have an 82 degree field of view.

Of course, with a bigger barrel the eyepiece is larger, there's more glass in it, it's heavier and it can be more expensive. This is why manufacturers try to use the smaller barrels when it makes sense, and switch to bigger barrels only when physics and math force them to take this step.

If you need a certain focal length combined with a certain field of view, and that requires a bigger barrel, then by all means use that size. It's kind of a mix-and-match situation.


When did scientists discover that the Sun has a life cycle and that it is going to die? I have been researching the history of scientific studies about the the Sun. However I have been unable to find out much information about how the scientific consensus started forming around stars' life cycles.
I know that it has been known since Galileo that some star have sudden changes in luminosity during their life. But I was unable to find out when scientists actually discovered that stars are born and die, and especially how this influenced the view on the future of the Solar System.
I suppose this happened somewhere at the start of the 20th century, when more information and research about supernovae was being found, but I couldn't find more precise information about it.
Any more precise information about this is greatly appreciated.

Although various astronomers have speculated that the Sun was a star (some were imprisoned or even burnt alive for such heresy), this was not known definitively until 1838 when Friedrich Bessel used parallax to calculated the distance to 61 Cygni.
In the late 19th century, Lord Kelvin provided rather small (less than 100 million years) for the ages of the Sun and of the Earth. He did not know about fusion. The nature of what powers the Sun was not known until the 1920s based on observations of the mass of a proton versus that of a helium-4 nucleus. The exact mechanism wasn't known until 1938 when Hans Bethe developed models by which hydrogen fuses to form helium.
This model provided the basis for knowing that the Sun will eventually die as it will eventually run out of fuel. Working from this, a number of scientists developed models of what happens inside stars of various masses based on the growing knowledge of stellar nucleosynthesis.


How to specify SM libraries for MOOG I am running MOOG on OS X Yosemite and with $MOOG running the abfind driver I can force fit elemental abundances with an appropriate input line list and model atmosphere. MOOG crashes, however, when I try to use it to generate plots, even though I know I have SuperMONGO (sm) installed on my computer because $sm works fine. In particular, MOOG will crash with the following errors:

*******************************************************************************
                  MOOG IS CONTROLLED BY DRIVER abfind
*******************************************************************************

wav. correl.:  slope =  -7.791E-07  intercept =   7.505  corr. coeff. =  -0.122
Fonts file not found
File graphcap is not defined
Can't get graphcap entry for nodevice
No such device nodevice
File graphcap is not defined
Can't get graphcap entry for x11
No such device x11 -bg black -title MOOGplot -geom 700x800+650+000

How can I get MOOG to find these libraries?

MOOG will search for the libraries through a .sm file. To solve this error, download an example sm file like this one, and specify where your libraries are in this file. Then use mv .sm ~/.sm to place this in your home folder, then MOOG should be able to create plots with SM.


Are black holes different sizes thanks to the different sized stars that created them? If the point in space that belongs to the collapsed star (that formed the black hole) is infinitely dense, why are black holes different sizes? Is it due to them feeding on stars and everything else that they swallow?

The size of a black hole is not defined by the expanse of its matter content but (usually) its event horizon.  The radius of the event horizon  (of a non-rotating and uncharged black hole) depends on the mass:

$$r_S = \frac{2 G M}{c^2}\\ \small M : \text{black hole mass} \\ \small G: \text{gravitational constant} \\ \small c : \text{speed of light} \\ \small r_S : \text{Schwarzschild radius}$$

Yes, the size of black holes as final stage of stellar evolution stems from the original star's mass, though some mass is ejected/radiated away during the supernova.

Black holes do grow (increasing $M$ and $r_S$) when they accrete matter from disks or companion stars.

Not all black holes we (kind of) observe are from stars: for example supermassive black holes in galactic centers, which can have billions of solar masses.  However, it is not yet fully understood, how they form.

Note: The most common black hole descriptions actually don't include any matter.  They are defined in an empty spacetime except for one singular point that is usually interpreted to hold all the matter.  The mass of the black hole is then assigned by identifying the gravitational pull in the Newtonian approximation at some distant point.

How the interior of real, non-vacuum black holes looks like, is still up to discussion.


What liquids can be found in the void space? Is there any material natural or otherwise a free floating liquid that can exist in space?

https://space.stackexchange.com/questions/32274/what-liquids-last-the-longest-in-space

No liquid can be completely stable in a vacuum, since all liquids have some non-zero vapour pressure, and so will evaporate at some rate. However some liquids have an exceptionally low vapour pressure, and so can be used in a vacuum.

The vapour pressure of silcone fluid DC705, which is used in diffusion pumps is 2.6e-8, and it is designed to function in a high vacuum.

If a location could be found at which it was warmed sufficiently to remain liquid (in deep space it would just freeze, whereas too close to the sun and its vapour pressure would rise) it could remain in a liquid state for some time.  Not indefinitely but it could be stable for a while.


Why are tides high only on a full moon day? As far as I've known, sea tides are higher on a Full moon day. And that is due to the "gravitational pull" of the moon.
But we know that moon is visible because of reflection of sun's rays. So, even on a new moon's day, the moon is still there, but it's not visible.
So, why doesn't it affect the gravity on earth?

Thanks in advance.

There are two main gravitational causes of tides: the Moon, and to a lesser extent the Sun.

When the moon is full or the moon is new, the Earth, Moon and Sun are roughly aligned, and the Lunar tide combines with the Solar tide to give a "Spring tide" that has a larger range.

When the moon is at first quarter, or third quarter (ie a half moon), the solar tide and the lunar tide are acting against each other, and you get a "neap tide" with a lower range than a spring tide.

The size of the tide on a particular day at a particular location is strongly affected by the local shape of the coast and sea floor, Some places only have one tide each day, or tides of different size, due to these effects.

Note the name "Spring tide" is not from the season. You get Spring tides every New and every Full moon. It comes from an old English word meaning "to bulge" The tide "bulges" during New and Full moon.

So you are not correct to say that tides are higher on the day of Full moon. The New Moon's gravity does create tides, and tides at New moon are as large as the tides at Full moon, and larger than the tides at half moon.


How can a low-mass star increase its mass to 1.4 Msun? In my astronomy class I learned that a only low-mass stars (< 0.5 Msun) will contract, and then become degenerate, until it is a white dwarf. However, we also learned about the Chandrasekhar limit, which states white dwarfs have a maximum mass of 1.4 Msun.

If the starting star has less than 0.5 Msun, how does the resulting white-dwarf reach up to 1.4 Msun?

Am I perhaps wrong in my assumption that the mass of the resulting white-dwarf depends on the mass of the original star?

You were taught wrong. Stars of up to about 8 solar masses will end up as white dwarfs. But it is only their cores that become degenerate and end up as the white dwarf. The rest of the envelope is lost during the giant phase due to a dense wind.

There is a non-linear, but probably monotonic, relation between the initial progenitor and final white dwarf masses (see below, from Kalirai 2013) - the Sun will likely end as a 0.5 solar mass white dwarf, but in normal stellar evolution, degenerate white dwarf stars can only be produced up to about 1.25 solar masses by the most massive progenitors. Any more massive than this and it is likely that the core does not become degenerate before igniting and burning through the heavier nuclear fuels. The most massive, probably single, white dwarf known is "WD 33" in the cluster NGC 2099 and has a mass of $1.28^{+0.05}_{-0.08}\ M_{\odot}$, is likely made of an O/Ne mixture, and had an estimated progenitor mass on the main sequence of $>3.5\ M_{\odot}$ (Cummings et al. (2016).

In order to get a more massive white dwarf, up to the Chandrasekhar mass (about 1.38 solar masses for a C/O or O/Ne white dwarf in general relativity), it almost certainly needs to accrete mass from a close binary companion or be the result of some sort of merger.

This is the leading candidate to explain type Ia supernova.


Have parameters been determined for the Fourier components of a Milky Way spiral density-potential model? Spiral arms in galaxies can be modeled by potentials of the form
$$\Phi_{\text{spiral}}(R,\theta,z,t)=\sum_mA_m(R,z)\cos(m\theta-m\theta_0-\phi_m(R)-\Omega_pt)$$
where $A_m(R,z)$ is the $m$th-order coefficient based on some potential - such as a Miyamoto-Nagai potential - $\phi_m(R)$ is a function typically of the form
$$\phi_m(R)=\alpha_m\ln (R/R_{0,m})$$
for some characteristic radius $R$, and $\Omega_p$ is some pattern speed. I've read that typically the $m=0$ component dominates, followed by the $m=2$ term. There may also be an $m=4$ term, though it is typically weaker.

This model describes the disk component of a galaxy with two main spiral arms. The spiral arms of the Milky Way are more complicated; some consider it to have two major arms and two more subordinate arms, while others consider it to have four major arms.

Have any potential or density models of the Milky Way been created using these Fourier components, and if so, what are the parameters used (the $m$s, $A_m$s, $\phi_m$s, and $R_{0,m}$s)?

Yes.
Antoja et al. (2011) created two models: The Tight-Winding Approximation model (TWA) and the PERLAS model. They decided to use just the $m=2$ term for their TWA, based on data from the Spitzer Space Telescope.
The amplitude $A(R)$ was
$$A(R)=A_{sp}Re^{-R/R_{\Sigma}}$$
Their $\phi_2(R)$ (denoted $g(R)$ in the paper) was of the form
$$g(R)=\left(\frac{2}{N\tan i}\right)\ln\left(1+\left(\frac{R}{R_{sp}}\right)^N\right)$$
where $N$ measures, in their words,

how sharply the change from a bar-like to spiral-like occurs in the inner regions.

They approximated using $N=100$, although $g(R)$ should properly use the limit $N\to\infty$. $i$ is the pitch angle of the arms, which they measured to be $15.5$ to $12.8^{\circ}$. $R_{sp}$, the beginning of the spiral, was found to be $2.6$ to $3.6\text{ kpc}$. For the other constants, $A_{sp}$ was taken to be $650$-$1100$ or $850$-$1300[\text{km s}^{-1}]^2\text{ kpc}^{-1}$, depending on the exact arm shape, and the scale length $R_{\Sigma}$ was set to $2.5\text{ kpc}$.
The pattern speed $\Omega_p$ was set in the range $15$-$30\text{ km/s}$, which fits nicely with other values I've seen. $\theta_0$ (denoted $\phi_0$ in the paper) was set to $88$ to $60^{\circ}$ - a little bigger than the spiral arm pitch angle. putting all this together, we get a potential of the form
$$\Phi_\text{spiral}(R,\theta,t)=A_{sp}Re^{-R/(2.5\text{ kpc})}\times\cos\left(2(\theta-88\text{ to }60)+\left(\frac{2}{100\tan (15.5\text{ to }12.8^{\circ})}\right)\ln\left(1+\left(\frac{R}{2.6\text{ to }3.6\text{ kpc}}\right)^{100}\right)-\Omega_{p}t\right)$$
A $z$-component could be multiplied to get a three-dimensional potential, likely exponentially decaying with height.
Griv et al. (2013) also used density-wave theory. Their potential was
$$\Phi(R,\theta,t)=\Phi_{\text{mean}}(R)+\tilde\Phi\sqrt{\frac{R_0}{R}}\cos\phi$$
where
$$\phi=\phi_0+m\left[\Omega_pt-\varphi+(1/\tan p)\ln(R/R_0)\right]$$
which is essentially the same potential as discussed in the question, with
$$A(R)=\tilde{\Phi}\sqrt{\frac{R_0}{R}}$$
The authors chose $m=1,2,3$ components, with different parameters for each. $p$ (in degrees) ranged from $-6$ to $-20$, while $\phi_0$ (also in degrees) ranged from $239$ to $247$. $\Omega_p$ was from $32.4$ to $37.6$ ($\text{ km/s}$), also consistent with other models. $\tilde{\Phi}$ saw the most variation, ranging from $-65$ to $-390$ (in $\text{kpc}^{-2}\text{s}^{-2}$). $R_0$ was set to $8\text{ kpc}$, I believe. Again, an exponentially decaying function should account for the $z$-dependence.


Where does the CO2 in Mars atmosphere come from? As far as I know, Earth and Mars were relatively similar when the solar system formed. So most of the CO2 was used to form carbonic rocks (right?). I found that tectonic movements are the reason that the CO2 on Earth could be released again, which don't exist on Mars.

So how comes that the atmosphere has 95% CO2?

The early atmospheres of Venus, Earth and Mars all had a lot of CO2, Carbon and oxygen are some of the most common elements in molecular clouds, and so it is no surprise that their most common and stable compound should be common in atmospheres.

On the Earth, life happened.

The carbon in the atmosphere was fixed by autotrophs (particularly blue-green algae and the later plants) into the bodies of living things, and then into rocks formed from those bodies: Chalk, Limestone, Coal and Oil. This left an atmosphere dominated by Nitrogen, and later by the toxic waste products from those algae: oxygen.

Mars was too small to hold onto lighter gases like Nitrogen, and these escaped into space, there is just a trace of the heavier gases: CO2, but at a very low pressure.

Venus retains a dense atmosphere dominated by CO2


Is S2 still the fastest known star in the galaxy? Wikipedia's entry for the star S2 says that it has

the fastest known ballistic orbit, reaching speeds exceeding 5,000
km/s (11,000,000 mph, or 1⁄60 the speed of light) and acceleration of
about 1.5 m/s2 (almost one-sixth of Earth's surface gravity).

But the referenced link to the press release Surfing a Black Hole is dated 2002.  Have any stars with faster orbital speeds at perimelasma (periapsis for a black hole orbit) in the 19 years since this press release?  Is it possible that there is a star on an even faster orbital trajectory than S2 or has any such possibility been ruled out?

S0-2 (I assume you mean the star orbiting the Milky Way central black hole) actually reaches speeds of 7650 km/s at periapsis.
However, if you read further down the wikipedia article you will see that there are stars S0-102 and S62 which have shorter orbital periods than S0-2 (11.5 and 9.2 years vs 16.05 years).
Star S62 is also in a highly elliptical orbit ($e=0.976$, Peißker et al. 2020) and has a periapsis distance that is even closer to the black hole than S0-2, where it has a speed clocking in at about 30,000 km/s$^{1}$. No actual value or error bar is given in the paper.
The plot shows its orbit (in red) compared with other stars in the region.
STOP PRESS
A telegram issued by Peißker et al. 2020 claims detection of a new winner - S4714. They appear to have revised downwards their estimate of the peri-centre speed of S62 to 20,000 km/s and say that this new, very faint object, has a peri-centre speed of 24,000 km/s.

$^1$ Of course that is the speed as seen by a distant observer. If you were to go there and put a speed gun on it as it went past, then it would be a touch larger.


Why do we presume that the universe has always expanded? When I read about the universe having always expanded from a 'singularity', I can't help but wonder if that is perhaps not the case. If a singularity is a point at which everything is infinite and is, theoretically, impossible, why are we quite so convinced it existed? What evidence is there that the universe didn't start out as a single square centimetre or even perhaps half the size it is now, because weve only known it's been expanding for the last century or so because of technological advances? Perhaps it was sizeable to begin with but the pressure of more atoms forming has pushed it outwards? Probably a ridiculous theory but Im intrigued to hear what others think about the whole subject because I just don't think the current presumption is based on any real evidence and is likely to be totally wrong.

We can, after making some approximations, trace the movements and locations of the matter in the universe back in time, using the laws of physics (mainly general relativity).

At the moment we see a certain density of matter which is, on a large enough scale, expanding uniformly -- as if space were being blown up so that every bit of it doubles in size every seven billion or so years.

When we extrapolate that back we find that long ago the matter density was much higher, but it was still expanding in a similar kind of way, and this would seem to hold true back to a state of very high density indeed, at which point our existing understanding of the physical laws stops giving us an extrapolation that makes any kind of sense and we have to stop.  Up to that point though, the extrapolation gives us a picture which is pretty consistent with all the various ways we have of observing it.

So, as someone has said, the universe might have been created moments ago with everything, including our memories set up just so, but this doesn't seem a very useful line of speculation. Instead we say that, based on the physics we understand, and everything we can see and measure in the universe, it looks like its been expanding ever since it was incredibly hot and dense about 13.8 billion years ago.


Undisturbed planetary alignment after galactic collision How do planets still orbit around the empty space where a black hole was after a black hole absorbs the milky way's black hole and keeps moving through space without disturbing surrounding planets that were orbiting that black hole?

The short answer is that they don't. If an even larger SMBH passed through the centre of the Milky Way and "swept up" the existing SMBH, all the stars and planets that are close to it (within a few tens of light-years, maybe more) would experience huge disruption and might end up orbiting the new black hole and being carried away, or flying free, or whatever.

The stars and planets much further away, which make up the bulk of the Milky Way wouldn't be much disturbed, because they are held in their orbits by their gravitational attraction for each other, and for the dark matter halo which suffuses the galaxy and only a small contribution comes from the gravity of the SMBH.


Is there some upper limit in the moon size distribution? The planets range hugely in sizes, Jupiter is more than 300 times as massive as Earth. But the sizes of the large moons seem to be uncorrelated with the sizes of the planets they orbit. Why doesn't Jupiter have a giant moon with a mass hundreds times larger than the Earth's Moon's?

Among gas giants, Saturn at a third of Jupiter's mass has the largest of all moons and Triton way out at 30 AU Neptune is the 7th largest of all moons. The theories of the origin of the moons seem to vary much too, without explaining any upper size limit. The Moon was formed by an impact. Triton and maybe Titan were captured. Other large moons were formed along with their planet. Is this upper size limit on moons imaginary, coincidental or is there some reason for it to be this way? I of course mean a hypothetical expected upper size limit for exomoons in general, not only statistics about the Solar System moons.

The wonderful illustration here is from Emily Lakdawalla at The Planetary Society, it is now my desktop background.

Small point, but Ganymede is slightly larger than Titan as moons are measured by their solid surface, though with atmosphere, Titan appears slightly bigger.   The drawing is probably accurate, but slightly misleading.

It's worth adding that Titan is out-gassing so it may have, quite some time ago, been larger than Ganymede.

Source

According to wikipedia, the larger, close to Jupiter and Saturn moons are unlikely to have been captured moons.

https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Moons

  Jupiter and Saturn have several large moons, such as Io, Europa,
  Ganymede and Titan, which may have originated from discs around each
  giant planet in much the same way that the planets formed from the
  disc around the Sun.[76] This origin is indicated by the large sizes
  of the moons and their proximity to the planet. These attributes are
  impossible to achieve via capture, while the gaseous nature of the
  primaries also make formation from collision debris unlikely. The
  outer moons of the giant planets tend to be small and have eccentric
  orbits with arbitrary inclinations. These are the characteristics
  expected of captured bodies

and the 3 listed methods of moon formation are:

Co-formation
Impact
Capture.

The largest co-formation ratio is Titan to Saturn, about 1 to 4,000 in terms of mass.   Ganymede to Jupiter, less than 1 to 12,000.   The wiki article points out that ratios of 100 to 1 are likely giant impacts and you're unlikely to see that kind of ratio in co formation, so a conservative estimate, the largest co-formation you mgiht see might be around a few hundred to 1.   If you figure the largest planets are about 12-12.5 Jupiters (more than that, they become brown dwarfs, though a cool brown dwarf orbiting a sun might get called a planet too at greater than 13 Jupiter masses), but lets say, 12 Jupiters.  That's about 4,000 earth masses.    If we look at a co-formation ratio of a few hundred to perhaps over a thousand to 1 as about the maxiumum, there's a possible ballpark answer of a handul of earth masses might be about as big as a co-formation moon is likely to get in a perfect formation situation, And those would probably be pretty rare.

Giant impacts are different.  Pluto-Charon could be an impact generated dwarf planet/moon system and Charon is 1/8th the mass of Pluto.    How likely that kind of impact is to happen on a super earth is hard to say, but theoretically a super-earth with a giant impact could have a large moon too - a bigger earth with a bigger impact and bigger moon, but there's problems here.  Giant planets are likely to form close to the sun and the orbital sphere of influence gets smaller closer to the sun.   Also, giant planets might tend to form thick gaseous atmospheres, retaining all their hydrogen and helium, so too big, it might begin to resemble a gas giant, which aren't as good at creating debris when impacted, so there's a few limiting factors here and we might be limited to, certainly several times larger than our moon, but perhaps not as large as co-formation moons, further from the sun.

Finally, moon capture.    While there's no examples of this in our solar system, I see no reason why a large planet couldn't capture another planet, though you then have to determine whether it's a co-planet orbit, which I think a captured planet should be considered, or whether the captured object is a moon. - personally I think a captured planet that orbits a planet would be a co-planet system, so I don't like the capture as a method for getting as large a moon as possible, though I suppose it could be looked at either way.

We know almost nothing about moons in other solar systems (exomoons).  The observations are just barely possible with current technology, but very difficult.  That's likely going to change in the next few years, but for now we know virtually nothing in other solar systems, so, any present predictions on the largest possible moons would have to be modeled or estimated.

A few articles on exomoons:

http://www.astrobio.net/news-exclusive/new-exomoon-hunting-technique-could-find-solar-system-like-moons/

http://discovermagazine.com/2012/jul-aug/06-hunting-moons-outside-the-solar-system

http://www.space.com/25438-exomoon-around-alien-planet-discovery.html

http://phys.org/news/2014-10-exomoons-abundant-sources-habitability.html

http://blogs.scientificamerican.com/life-unbounded/has-an-exomoon-been-found2/


Is the Sun visible from Proxima Centauri to human eyes? I know that the light coming from Proxima Centauri is not bright enough to make it naked-eye visible from the Earth. Is the Sun naked-eye visible from Proxima Centauri?

Well, there's two things we'll need for this: apparent magnitude (the brightness that an object appears to have) and absolute magnitude (the actual brightness an object has). Both of these scales are logarithmic, with brighter objects being lower and dimmer objects being higher. Astronomers have determined that the Sun's absolute magnitude is 4.83. Knowing this, we can find the apparent magnitude of the Sun from Proxima Centauri's location. Apparent and absolute magnitudes are related by the equation:

$$M = m - 5 (\log_{10}{d}-1)$$

Where $M$ is the absolute magnitude, $m$ is the apparent magnitude, and $d$ is the distance, in parsecs. Astronomers have determined that Proxima Centauri is 1.3 parsecs from us. So the apparent magnitude can be determined as:

$$m = 4.83 + 5(\log_{10}{1.3}-1) ≈ +0.4$$

As made clear in this paper, most humans can see objects with apparent magnitudes as dim as $5$ without using tools. So yes, it is certainly visible, and would be quite bright. It is between Procyon and Achernar, the 9th and 10th brightest star on Earth's night sky.

For a comparison, Proxima Centauri has an apparent magnitude of $+11.13$ from Earth's perspective. If we want to compare the two, we could use this formula:

$$v_b = 10^{0.4 x} = 10^{0.4×(11.13-0.4)} \approx 19588$$

So the Sun would appear almost 20,000 times brighter from Proxima Centuari than PC appears from the Sun.


DIY telescope to observe nebulae How can I make a good telescope that is powerful enough show nebulae and other deep sky objects ? (consider low budget build)

Absurdly cheap and (potentially) very fun!
The absolute cheapest telescope to make would be extremely hard to make, you'd be doing something for the first time, but it would at least be fun!
Make your objective lens out of ice, and find a way to polish it smooth. Don't make it as big as these though. You can then use an eyepiece out of a pair of old binoculars or a toy telescope or use a dime-store jeweler's loupe for an eyepiece.
From Has anyone ever tried to make a simple telescope using ice?
These are solar collector projects, a telescope objective would have to be polished to a much better shape.

Here is a functioning camera lens made from ice for example, from this answer there:

click images for larger
Source:  Mathieu Stern
Chromatic abberation is a problem, so use a filter that removes blue and green (I guess you'd call it a red filter) and look at things that are mostly hydrogen.
From Alex Kozik's Orion Nebula spectral composition we can infer that the Orion Nebula would still have the shape of the Orion Nebula in the presence of chromatic aberration if got rid of green and blue light.

From James Kaler's1 The Spectrum of the Orion Nebula:

1Professor Emeritus of Astronomy, University of Illinois

But seriously folks
If we throw out any hope of an electronically driven and computer controlled equatorial mount and just want to look at things as they slowly drift then spend most of your budget on an f/8 spherical mirror, either a second hand one from a discarded telescope or one out of a catalog like an Edmund Scientific from the 1970's. Seriously I don't know if there are catalogs where you can order premade mirrors any more, but it was fun to dream!
You may have to invest in a telescope mirror making kit. There are an abundance of resources on how to make home-made mirrors, and as long as you are f/8 or longer you don't need to worry about figure correction (making it a parabola instead of the natural spherical shape it will want to have)
Next, put your mirror in a Dobsonian mount made from plywood or scrap metal, depending on your materials and particular skills and tastes.
You'll need a diagonal mirror or prism to send the light out the side and a stick or spider to hold it near the center of the aperture. If you are only going to use one eyepiece then you can find a simple way to hold it in place, but you've got to do some reading about how to align the whole thing.
That would be an excellent follow-up question here!
You can also take a cue from Sir William Herschel and instead of using a diagonal in the center of the aperture, just peer over one edge with the eyepiece and (since your telescope will be a lot smaller) use a diagonal mirror or prism attached to the edge. You'll have more distortion, but if you increase your f/ number (the one below was f/10) and stick with low magnification, you'll see some nebula, at least as opposed to not seeing them. :-)
I used a small front-surface mirror, some alligator clips, an eyepiece from an old pair of binoculars, a cardboard tube (a la Clyde Tombaugh) and a weak concave magnifying mirror for my first 10 cm Herschellian reflector telescope and had a truly thrilling time!
From here:

click for larger

left: Source, right: "Two Dobsonian telescopes on display at Stellafane in the early 1980s." Source


Why is Earth's rotation considered fast? I'm currently reading Atom (Krauss book), and in chapter 11 there is this text:

  This collision can explain many things, including the high rate of spin of the Earth (the grazing collision would have twisted it around like two football players who collide while running in opposite directions).

This is in reference to a proto-planet hitting the early Earth, which led to the formation of the Moon.

Why is Earth's spin considered high rate, given that Mars' spin is pretty similar without having a big moon? Or have I misunderstood what is being referred to?

The interactions of the Earth and its moon have slowed the Earth's rotation significantly over the billions of years since their formation.  Wikipedia cites a paper that gives a figure of approximately five hours for rotational period in the past.  We don't know of a similar method to slow down Mars.

This implies that the Earth must have had a much faster rotation than Mars in the past.


Why is the "green" comet C/2014 Q2 (Lovejoy) red in this picture? In this nice NYTimes 'survey article' "Our Vast Solar System
and Its Many Explorers" there is a NEOWISE image of Comet C/2014 Q2 (Lovejoy). See here also.

The NASA description of the image says:

  The red color is caused by the strong signal in the NEOWISE 4.6-micron wavelength detector, owing to a combination of gas and dust in the comet's coma.

So I understand that this is a false color image, and the image at 4.6-micron is overlaid using red. So I should really ask: why is Comet C/2014 Q2 (Lovejoy) (and probably many) comet actually bright at 4.6 microns "...owing to a combination of gas and dust in the comet's coma." Is the gas fluorescing and the dust scattering? Or is this thermal radiation from the dust? The explanation in quotes doesn't really explain why "the red color is caused by the strong signal (around) 4.6-micron wavelength (is) owing to a combination of gas and dust in the comet's coma."

below: The more familiar green glow of this (and many) comet from here.

Mainzer et al. 2014 characterize the performance of the reactivated NEOWISE.
Having run out of cryogenic coolant for the original WISE's 12 and 22 μm bands, it only detects in the 3.4 and 4.6 μm bands.
The comet looks red in the false-color infrared image because, relative to the 4.6 μm "W2" band, it emits less in the 3.4 μm "W1" band than stars do.
Figure 3 of the paper shows a similar image in which an asteroid also looks red.

Section 4.1 discusses comets, attributing the 4.6 μm band detection to a combination of reflected sunlight and thermal radiation from dust, 4.26 μm emission from CO2, and a little 4.67 μm emission from CO.
They estimate how much is due to gas vs. dust by comparison with ground-based observations near the same times, as in Figure 17.


Finding a location using the direction of shadow, UT and date approximately Does anyone know how to find a location by using :

the direction of shadow relative to the normal (for example azimuth
angle)
the Universal time (UT) and date

I have a satellite image and I know north, time and date. One of the cues that I can use is, for example, in the northern hemisphere the direction of shadow is north.

Would anyone know if it were possible, and how, to calculate the location based on these parameters?

Planetarium software, such as Stellarium, or astronomical computation software such as pyephem can compute the exact position of the sun at any date, time and location. It is then simple trigonometry to calculate the length of a shadow:

if the sun is $\theta$ degrees above the horizon, a vertical building of height $h$ will cast a shadow on horizontal ground of length $\frac{h}{\tan(\theta)}$.


Is the Drake equation based on a logical fallacy? I often wonder why the Drake equation gets taken so seriously and why popular scientists in general say "it's probable' life exists elsewhere in the cosmos.
My problem is that the mechanism for abiogenesis is as yet unknown.

We don't know the likelihood that RNA like molecules can assemble from chance or for that matter that this is the source of life on this planet (as opposed to coming via a comet). We do know that we have no record of it recurring and as yet Miller-uri style experiments have not yielded results.

The argument I keep hearing from Dawkins, Sagan and many others is that the universe is vast and life occurred relatively quickly in our planet's history which suggests it is common.
I detect a logical fallacy based on the inductive reasoning of using one occurrence in order to establish a pattern.

Think of it this way. If I've never played cards and get dealt my first hand which is a royal flush I might think its quite likely based on my past hand that I will get another in the next deal. This doesn't change if the deck contains a hundred suites and the odds of a royal flush is astronomical - if my first hand is a royal flush.

Similarly the vastness of the cosmos tells you nothing about the likelihood of other life if you don't the probability of its occurrence.It could be very likely or so rare as to only occur once.
Does anybody have an opinion on this?

To directly answer the question, the Drake Equation is not based on a "logical fallacy" (which means a different thing, philosophically). The logic is fine, the real problem is we don't have any meaningful data.

Regarding abiogenesis (the first formation of some simple life): as you correctly state, we know only that it happened once (here, presumably but that doesn't matter). However the argument that it happened quite quickly on our planet is meaningless statistically, because that's just a sample of one.

(If anyone thinks this answer is too waffly or off topic just say so and I'll delete it by the way; I think the question is a fair one but we just can't present any data, just reason.)


How often do they move most/all of ALMA's dishes? The Phys.org article Unknown treasure trove of planets found hiding in dust reports on recently published radio measurements (~1.3mm) of planetary systems forming around stars in Taurus:

Scientists base this scenario of how our solar system came to be on observations of protoplanetary disks around other stars that are young enough to currently be in the process of birthing planets. Using the Atacama Large Millimeter Array, or ALMA, comprising 45 radio antennas in Chile's Atacama Desert, the team performed a survey of young stars in the Taurus star-forming region, a vast cloud of gas and dust located a modest 450 light-years from Earth. When the researchers imaged 32 stars surrounded by protoplanetary disks, they found that 12 of them—40 percent—have rings and gaps, structures that according to the team's measurements and calculations can be best explained by the presence of nascent planets.

The article ends by saying:

Going forward, the research group plans to move ALMA's antennas farther apart, which should increase the array's resolution to around five astronomical units (one AU equals the average distance between the Earth and the sun), and to make the antennas sensitive to other frequencies that are sensitive to other types of dust.
"Our results are an exciting step in understanding this key phase of planet formation," Long said, "and by making these adjustments, we are hoping to better understand the origins of the rings and gaps."

Changing from 15AU to 5 AU suggests a roughly factor of 3 increase in the overall scale of the pattern of antennas. The linked paper in ApJ Feng Long et al. has a preprint in ArXiv as well, which says:

All observations were obtained from late August to early September 2017 using 45-47 12-m antennas on baselines of 21∼3697 m (15∼2780 kλ), with slight differences in each group (see Table 1).

Several years ago I remember reading about the design of ALMA and there are a lot more places to put dishes than there are dishes. I think there were five optimized configurations that all looked to have a similar spiral shape, but each one was maybe a factor of 2 or 3 larger than the previous.
Since the dishes take time to move and setup and calibrate, this probably doesn't happen that often.
Question: How often does it happen? How often do they move most of ALMA's dishes to change resolution?
If there's a site that shows how ALMA is set up now, and what the next setup will be and when, that would be great to know about as well.

above: "Until recently, protoplanetary disks were believed to be smooth, like pancake-like objects. The results from this study show that some disks are more like doughnuts with holes, but even more often appear as a series of rings. The rings are likely carved by planets that are otherwise invisible to us." Source Credit: Feng Long/ALMA

The list of array configurations is given in the Proposers Guide and ten configurations have been defined for the 43 12-meter antennas. They have the form C43-x where x goes from 1 (most compact; 15-161 meter separation) to 10 (least compact; 244-16200 meter separation). The changes between configurations are shown in the Configuration Schedule; over the ~12 months of Cycle 6, it looks like they change configuration 14 times. There is a month long "Antennae Relocation Shutdown" in May 2019 as they transition from fairly compact to least compact - presumably because it takes a longer time to move the 43 dishes out multiple kilometers

As an example, here is a snapshot of the planned schedule for a year starting 01-Oct-2018 as it stands now:


Why is Enceladus's albedo greater than 1? Enceladus, one of Saturn's moons, has a geometric albedo of 1.38 and a bond albedo of 0.81. How can the geometric albedo of Enceladus be greater than one? What are the similarities or differences between a geometric albedo and a bond albedo?

To answer this, one really has to understand how the geometric and bond albedos are defined. Let's start with the bond albedo since its simpler.

Bond Albedo

The Bond Albedo is just the fraction of energy hitting a surface that gets reflected. To simplify the process, let's say I shoot 100 photons of all the same energy at Enceladus. Of those 100 photons that hit the surface, 81 of them will be reflected back and 19 of them will be absorbed. That means the bond albedo will be $81/100 = 0.81$.

Simple right? As such, the bond albedo can never be greater than 1 because you obviously can't reflect back more photons than you received. This means that a bond albedo of zero implies no light was reflected and all of it was absorbed, whereas a bond albedo of one implies all light was reflected and no light was absorbed.

Geometric Albedo

The definition of the geometric albedo is a bit more complicated. The first thing you have to realize though is that it is not a straight zero to one scale. A geometric albedo of one doesn't mean all the light got reflected like it did for the bond albedo. So let's try to define the geometric albedo.

There are two important points about the geometric albedo that make it differ from the bond albedo.

For the bond albedo, we talked solely about the total amount of incident light and the total amount of reflected light. This says nothing about an observer of said light. We just have some magical way of knowing about every photon that hits and is reflected by your surface and that allows us to calculate the albedo according to the bond albedo definition. The geometric albedo on the other hand specifically takes into account your ability to observe the light from a specific vantage point.

The second major point is that the geometric albedo is a measure of how well your particular surface reflected the light back to you compared to a reference surface. Imagine now you have two surfaces, one is the surface of Enceladus, the other is your "reference surface". This reference surface is an "idealized" reflector which means that it reflects every photon that hits it (making it have a bond albedo of 1). The key here though, is that it reflects light isotropically. What I mean by that is that the light doesn't get reflected in any preferred direction, but rather gets reflected in all directions equally. So now you have your 100 photons hitting your reference surface and all 100 photons get reflected, but because they're all reflected in random directions, only 10 photons actually get to your camera/eye/detector. What can happen with Enceladus though, is that the surface has just the right properties such that 100 photons hit the surface, and 14 photons get reflected to your detector because the surface preferentially reflects light in a specific direction. It's sending more photons to your detector than the reference surface. The geometric albedo is the ratio of how much light gets reflected to your detector by Enceladus over how much light gets reflected to your detector by the reference surface. In this case, $14/10 = 1.4$.

One small, additional point, is that this definition relies on your detector being in the same direction as your light source. In other words, the geometric albedo is a measure of how much your surface can retro-reflect (i.e., reflect back to the source of the photons) compared to the reference surface. Technically, the geometric albedo attains a maximum value when your surface is made up of retroreflectors.


"Up" and "down" in Space From my understanding, "up" and "down" in space is going towards (down) an object's gravitational pull or (up) which is going away from it. I get confused with this explanation, and I believe simply because of how we perceive up and down here on earth.

My question(s) is this: Can we go in all directions in space with the "Earth" interpretation of up, down, left, right, north west, south east, etc if we can get past gravitational pulls of objects in space?

Looking at my Gimp version of the universe, I have the Earth and stars. So going up would essentially be go towards any of the stars in the picture that has the strongest gravitational pull, although each object are on different planes?

In relation to my question above, I have imagined Earth in space as to like the 10 floor on an infinite elevator that can go in all directions. Say we can somehow go through star A and escape it's gravitational pull, we would be essentially "going down" right?

Now my other question: Are there objects in every single direction is space? How about in the Bootes void? Since it's such a massive void and there isn't much objects around to pull you in "down" or go away from "up", essentially can't we go in any direction?

If it may seems that my question about this is kind of confusing, I apologize because I'm confused about this myself!

I think that one thing confusing you, is that up/down is a 2 dimensions concept (just like your image by the way), but the Universe is in 3 physical dimensions (let's not talk about time here).

So when you say « going up from Earth is extracting from it's gravity », it can be in many direction. If two persons are in North Pole and South Pole of Earth, and if they are both going up. They are in fact going in opposite directions.

The second point is that you need a frame of reference. You go up/down « to » or « from » something.
For example, you go up (going away) from the North Pole, but from Sun's or Moon's point of view, it's not going up at all.

The up/down/left/right concepts always need a reference. See this very simple example: If we are facing each other, what is on my left, is on your right.
Your question here is exactly like asking : « What is on my left in the universe ? ». There is no real answer to that.

Hope I could help.


How does the Milky Way look like above 66° North and below 66° South? I read on Wikipedia:

  depending on the time of night and the year, the arc of Milky Way can
  appear relatively low or relatively high in the sky. For observers
  from about 65 degrees north to 65 degrees south on the Earth's surface
  the Milky Way passes directly overhead twice a day

What happens and how does the Milky Way look like out of that range?
If it matters, consider also the all-summer light and the dark winter.

On midnight, right around this time of year, the Milky Way will be in the zenith. Here is an XEphem rendering for the north of finland (65th latitude) for yesterday midnight (the brown outline marks the Milky Way):

You can also see this on photographs by finnish photographer Tommy Eliassen. He has many examples of this on his website. I won't put any in here, because I guess they are copyrighted, and not free to use.

This is an animated 24 hour version of the above sky map, which shows the rotation of the milkyway around the zenith:


How does solar activity (e.g. flares, coronal mass ejections) change over the life cycle of a Sun-like star? As a Sun-like star is born from a collapsed nebula and generates energy by nuclear fusion some of that energy is converted to magnetic fields which fuel solar activity. How does that activity evolve as the star itself evolves?

As in the Earth and more than in the the giant gaseous planets the Sun's magnetic field is generated like in a dynamo. A few facts about it:

The rotating plasma inside the Sun acts like a dynamo creating it's
magnetic field.
The plasma is ionized so all interact with the magnetic field and
vice-versa.
The sun as a fluid body also has Differential rotation, that
means the equatorial part of it rotates faster than the poles. This provokes what is known as the Solar cycle every 11 years.

At the begining of the Solar cycle the Sun's magnetic field is pointed North-South as in the Earth. During the cycle the differential rotation makes the plasma interact with the magnetic field causing it to be displaced. This makes the first the Sun's spots appear near the poles. As the cycle advances the magnetic field get more and more scrambled, it's orientation changes to west-east and the spots appear closet to the equator. Finally the magnetic field can't hold the "pressure" and "breaks" like a rubber band creating the biggest spots and flares. The magnetic field stabilizes again but with a south-north polarity and the cycles begins again.

Take a look at this gif image that shows how the cycle go.

And in this image you can see what is know as the butterfly diagram that shows how the Sun's spots are located closer to the equator as the cycle advances. And it lasts 11 years.


Finding a Kerr Black Holes I understand that Kerr Black Holes are purely theoretical and have not been found anywhere in the Universe. However, they are theorized for possible means of time travel.

I am attempting to prove in a research paper that controlled time travel is statistically improbable for the human species. Therefore if scientists have evidence that there does not exist a black hole within a distance of X from Earth, and it is statistically improbable for the human species to ever be able to travel a distance of X, I can rule out time travel by means of kerr black holes with a significant degree of certainty.

However, I cannot find this "X" threshold, which is why I am asking this question. Have scientists verified that there are no kerr black holes within a distance of X from earth for some X?

1) Nearly all black holes are Kerr black holes.

2) There are roughly 100 million black holes estimated in our galaxy.

3) There is a confirmed black hole about 7800 light years away.

4) There is a potential black hole about 1500 light years away.

5) It is very likely that there are black holes even closer than that, but they are difficult to detect.

Some of your assumptions are wrong, but for now the closest known potential black hole is still a pretty good distance away. Traveling that far with our current technology is impossible, and overcoming that hurdle is not trivial. Regardless, using a Kerr black hole for time travel may not be feasible anyway.


What specific parts of the Milky Way do we see when we look up? I know we can't see the center.  But do we see any of the central bulge?  Are we just seeing nearby stars in the Orion Spur where we are?  In the Sagittarius Arm?  In the other direction, are we seeing any of the arms further out from ours?  Where are the dust clouds we see, in relation to these arms?

But do we see any of the central bulge?

With the naked eye and a dark sky, the galactic center is brighter than other areas of the galactic plane.  Some of the light from that region does make it to earth to make that direction identifiable.

If you look at the ESO night sky composite from Brunier and Tapissier (which is approximately what someone could see with the naked eye given dark skies and good viewing), you can see that while the galactic center is brighter, other directions along the plane seem fairly similar.  That makes it very difficult to say what fraction of scattered glow is coming from one arm or another.  It all just blends together.

https://www.eso.org/public/news/eso0932/

  If all we can see are stars in our galaxy that are relatively close to us, then the Milky Way should look relatively uniform, right?

Not necessarily.  The visible part of the galaxy is relatively thin.  It doesn't have a fixed distance, but a commonly quoted one is 1000ly.  That means that even if light were magically cut off at 3000ly or so, it could appear non-uniform.


What leads to increase in opacity in kappa mechanism? I understand that the kappa mechanism (that leads to star variability) causes an increase of opacity with increasing temperature in partial ionization zones.  However, I'm not sure I understand what causes the increasing opacity with temperature.

Until recently, I thought that the increasing opacity was due to increasing Thompson opacity due to the increasing ionization fraction with temperature in partial ionization zones.  However, some things I read suggested that that was not the cause.  So I wanted to ask, what is the cause of increasing opacity with temperature in partial ionization zones?

It's a bit subtle, the key thing the partial ionization does is to keep the temperature from changing much.  What you really want is an increase in density, not temperature.  The reason the kappa mechanism is important is that it allows heat to be added to the gas when it is compressed, and removed when the gas is expanded, that's what allows for the pulsation to have energy added to it.  Heat is added when the opacity is high, and removed when low, so you need the opacity to go up when the gas is compressed, and go down when the gas is expanded.

Now, normally opacity goes down when gas is compressed, because the rise in temperature causes ionization, which actually hurts metal opacity by reducing the number of bound electrons (metal electrons have higher opacity in general when bound, due to resonances, than when free, though the hydrogen and helium we had in the Big Bang are an exception that works the other way-- in a star, it is usually metal opacity that is important and variable).  But increases in density also tend to produce more bound metal electrons, so there's a competition there that temperature usually wins and stars are pulsationally stable.  But in partial ionization zones, temperature changes are reduced by the phase changes eating up the excess energy, and that gives the density increase the edge that allows opacity to rise when the gas is compressed.


Natural satellites for Mercury Would it be possible that Mercury had a natural satellite but the gravitational pull from the Sun became so overwhelming from it growing over the years that it just sucked in Mercury's moon?

The larger an object is and the further away it is from the Sun, the greater its Hill Sphere is. Mercury's is very small (only about $175000km$ in radius from what @userLTK said) and its "moon" would therefore be yanked away by the Sun relatively quickly. See this for more details.  The orbit of this moon would be highly unstable and therefore unpredictable, so it may hit the Sun.


What is behind the Great Red Spot's longevity? On Earth, storms can last a few days. The Great Red Spot on Jupiter is a storm that has been going on for more than 400 years now. What is different on Jupiter that makes it possible for storms to last so long?

Standard answers, as found  here, (partial quote)

Unlike Jupiter, the Earth has land masses that cause major storms to
lose energy due to friction with a solid surface. Without this
feature, Jupiter’s storms are more long-lasting. However, the Great
Red Spot is long-lived, even by Jupiter standards. Researchers don’t
quite understand why, but we do know that Jupiter’s storms that are
located in cloud bands with the same direction of rotation tend to be
longer lasting.

So, we know some of the probable reasons, and don't know the full story yet.  Keep in mind that the current RedSpot might be a 3-sigma event, even for Jupiter.


Are planets more likely to have dense materials at their cores? When planets are forming, does the densest material sink the core and the less dense material "float" on top? What I'm asking is that are the more dense materials more likely to appear in the core rather than the surface. I'm also wondering, can a planet have an iron surface and still have rock beneath the surface in its crust and core?

Essentially, a planet can be considered as a self-gravitating fluid. As was already known to Archimedes, under gravity low-density fluids are buoyant and rise, while high-density fluids sink. So, with some local exceptions, planets are always well structured from the densest material in the core to lower-density material further up.


Satellite/Planetary Orbits All planetary orbits contain 5 unusually stable points. These points are particularly important because they allow man-made satellites to orbit the Sun with a period equal to that of Earth’s. 3 of these points are collinear. Suppose that   is the distance between the centers of mass of Earth and the Sun. Find the distance from Earth’s center of mass to either one of the other stable points in the Earth-Sun system in terms of...

(I'm not looking for a full solution; I just want to know what these points are called)

What are these points it talks about, and what's their mathematical relation to Earth's orbit?

This still isn't homework; I'm just looking for the name of these points.

These are the Langrangian Points http://en.wikipedia.org/wiki/Lagrangian_point -you'll find an explanation of the maths at a variety of sources if you search using that.


Why would the Earth's solar analemma would be still a figue eight even if Earth's orbit was circular? The Earth's solar analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. ... For [a planet] with a circular orbit but significant axial tilt, the analemma would be a figure of eight with northern and southern lobes equal in size.

This statement seems to be true, but it doesn't make any sense to me. If the eight is not due to the equation of time, why is there an eight? Thanks.

It is correct that the Earth's axial tilt alone would result in an analemma of a figure of eight with equally sized lobes.  The axial tilt contributes to the equation of time.

Circular Orbit, no Axial Tilt

Let's look at an example case with a perfectly circular orbit and no axial tilt, to observe what happens with no equation of time.  In this example, it's noon at the equator and prime meridian.  The Sun is directly overhead.  Next, the Earth rotates exactly once.  This isn't a normal solar day, but a sidereal day, about 23 hours, 56 minutes.  The Earth has rotated exactly once with respect to the stars.  Where is the Sun directly overhead now?  The Earth has moved a little bit in it's orbit around the Sun, so that our observer at the equator and 0 degrees longitude has observed that the Sun has moved in the sky with respect to the stars.  The Sun is now directly overhead at a location to the east, a little less than 1 degree longitude away (360 degrees in a circle, usually 365 days in a year).  The Earth must rotate a little more so that the Sun is directly overhead again (which completes the 24 hours in a day).

After each rotation, the Earth must rotate a little more each day to bring the Sun back overhead at 0 degrees latitude, 0 degrees longitude.  After a quarter of a year, the Earth must rotate another quarter of a rotation (90 degrees) for the Sun to be overhead at that spot.  Without that extra quarter of a rotation, the Sun is overhead at 90 degrees east longitude.  That makes sense; the Earth has moved 90 degrees in its orbit.  With a perfectly circular orbit, the rate at which this point moves eastward per day is constant, a little under a degree per day.

These points define a great circle around the Earth that coincides with the equator.  A great circle is a circle on the surface of a sphere whose center is also the center of the sphere; it's the biggest possible circle that can exist on the surface of a sphere.

A Large Axial Tilt

Let's keep Earth's perfectly circular orbit, but let's give it a very large axial tilt, 80 degrees, for the purpose of this explanation.  This Earth now begins at noon on the northward equinox at 0 degrees latitude, 0 degrees longitude.  After one Earth rotation, where on Earth is the Sun overhead now?  It's still a little less than one degree of angle away, but the direction of the displacement has changed.  Instead of this point moving due east, it is mostly northward and only a little bit east.  The Earth needs to rotate far less than 4 more minutes for the Sun to reach local noon for our observer at 0 degrees latitude, 0 degrees longitude.  These points are west of where they would be without axial tilt.  Solar noon has occurred before 24 hours have passed.  As days continue to pass after the equinox, solar noons continue to occur earlier each solar day, as these points continue mostly northward and only a little bit eastward.  The analemma shape as seen in the Northern hemisphere sky is moving up and to the right (west is to the right when looking south towards the Sun in the Northern hemisphere).

A quarter of a year later, it's the northern solstice, so let's see what has happened to our sun-overhead plots as the sidereal days have piled up.  These points have continued to move mostly northward and a little eastward, but now they are moving purely eastward, because at the solstice the sun has stopped moving northward in the sky.  It's still moving at a little less than 1 degree per day along that great circle, but because longitude lines are spaced much more closely together at 80 degrees latitude, one degree along a great circle covers many degrees of longitude.  In other words, the eastward movement of these points has been "catching up" with the longitude of the corresponding no-axial tilt points.  At the solstice, both the no-axial tilt point and the axial tilt point have the same longitude, 90 degrees east.  The analemma shape as seen in the Northern hemisphere sky is moving up and to the left (eastward), where it reaches its highest point.

These points will continue to move through longitude lines rapidly while the Sun's overhead point remains far in the north.  Now, the sun's overhead point is further east than it would be with no axial tilt.  The analemma shape as seen in the Northern hemisphere sky is moving down and to the left (eastward).

As the southward equinox approaches, the sun-overhead points are moving mostly southward and only a little eastward.  The analemma shape as seen in the Northern hemisphere sky is moving down and to the right (westward).  This allows the non-axial tilt points to "catch up", and the longitude differences begin to decrease again, until the time of the southward equinox comes, when the two points coincide again.

Here is a crude ASCII drawing of these points so far, with longitude along the horizontal direction, and latitude along the vertical direction.

                                   (2)
                   K       L        M        N      O
            J                                              P
         I                                                    Q
      H                                                          R
    G                                                               S
(1)F                                                                 T(3)
  E                                                                   U
 D                                                                     V
 C                                                                     W
B                                                                       X
A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y

A = northward equinox
M = northern solstice
Y = southward equinox

The points around (1) are further west than their corresponding points on the equator. At (2), the points have caught up in longitude.  At (3), they are further east than their corresponding points on the equator, but the equator points are catching up.

After the southward equinox, the longitudes of the sun-overhead point diverge once again, with the same mechanism as described above for the northward equinox.  Just reverse north and south, and the eastward movements are the same.  The analemma shape as seen in the Northern hemisphere sky first moves down and to the right, then down and to the left where it reaches its lowest point at the southern solstice, then up and to the left, then up and to the right where it reaches its starting point at the northward equinox.

Back to Reality

With an axial tilt of 80 degrees, the equation of time would show some extreme values, approaching almost 6 hours of divergence from the no-axial tilt case.  With the true axial tilt of about 23.5 degrees, our equation of time difference values are far less substantial, but the effect is real.

The true analemma shape we see on Earth is the combined effect of the axial tilt that we have seen plus the fact that Earth's orbit is elliptical and it slows down during the parts of its orbit when it's farther from the Sun and it speeds up during the parts of its orbit when it's closer to the Sun.

The site Analemma has good explanations about the individual effects of the elliptical orbit and the axial tilt and how they combine to create our analemma.


When will the next summer solstice and a full moon coincide? What day? What day will the summer solstice and full moon coincide? As far as I know the next time will be in 2016. I would like to know what day.

The Summer Solstice in 2016 will occur at 22:34 UT on the 20th June, and the full moon will occur at 11:02 UT on that day (but since they are 12 hours apart that won't be on the same day everywhere).

Solstice data from here (GMT- Grenwich Mean Time =UT)

New and full moon data from here (UTC- Coordinated Universal Time =GMT=UT)


How to convert VSOP's xyz coordinates to heliocentric spherical lbr coordinates? I ran VSOP2013 in Fortran, and I now have the following position $x, y, z$ as well as velocity $v_x, v_y, v_z$ data.

How do I convert the position to heliocentric ecliptic spherical coordinates $l, b, r$?

PLANETARY EPHEMERIS VSOP2013  EMB
Ecliptic   Heliocentric Coordinates:  X,Y,Z (au)  X',Y',Z' (au/d)  - Dynamical Frame J2000

Julian Day 2415021.0
-0.2054467990    0.9615310525    0.0002141233   -0.0171063024   -0.0036576954   -0.0000011815

Looking at the numbers and comparing to JPL Horizons, the best match is to the position of the Earth-Moon Barycenter in heliocentric coordinates. That's of course consistent with the data being labeled as:  EMB, Ecliptic Heliocentric Coordinates.

If you can live with some small uncertainties due to the details of how coordinate transforms may evolve over time, I believe you can just apply some simple trigonometry. You can confirm here.

$$r = \sqrt{x^2 + y^2 + z^2}$$

$$l = tan^{-1}(y/x)$$

$$b = sin^{-1}(z/r)$$

Data in Question: for JD 2415021.0
position (AU):     [-0.2054467990,  0.9615310525,  0.0002141233]
velocity (AU/day): [-0.0171063024, -0.0036576954, -0.0000011815]

JPL Horizions Earth-Moon barycenter J2000 Heliocentric for JD 2415021.0
position (AU):     [-0.2054465857,  0.9615310983,  0.00021393097]
velocity (AU/day): [-0.0171063032, -0.0036576916, -0.00000118253]

JPL Horizions Earth J2000 Heliocentric for JD 2415021.0
position (AU):     [-0.2054522688,  0.9615603123,  0.00021303115]
velocity (AU/day): [-0.0171136285, -0.0036594066, -0.00000182603]

JPL Horizions Earth J2000 Solar System barycenter for JD 2415021.0
position (AU):     [-0.2022720210,  0.9679285296,  0.00010939456]
velocity (AU/day): [-0.0171209826, -0.0036556277, -0.00000165111]


Celestron Astromaster EQ130 I recently purchased a Celestron Astromaster EQ130 telescope and have absolutely no clue about telescopes but would love to learn.
I have set this up but cannot even get a decent view of the moon, any advice of help would be very much appreciated.
Thanks in advance.

Whatever difficulty you're having, the surest way to get past it is probably an in-person tutorial with an experienced telescope user.
Find a local astronomy club via
the Astronomical League,
the NASA Night Sky Network,
Sky and Telescope,
or Go Astronomy.
Then bring your scope to a public observing session, and most likely someone there will be willing and able to help.


What factors influence the Widmanstätten pattern in iron meteorites? The cores of iron meteorites exhibit beautiful Widmanstätten patterns, as shown below in the Alvord meteorite (photo by Jim H):

I understand these patterns form when the molten nickel-iron core of its host body cools. As with normal crystal formation, slower cooling rates lead to longer crystal-formation times, which leads to larger crystals.
Beyond that, what factors influence the formation of the crystal, and in what ways?
I imagine composition and cooling time are the dominant factors....

It’s been a while since I studied this, and meteorites are very much not my field, so I’ll share my current understanding but would welcome corrections from those more informed :-)

These meteorites are mostly iron with a small amount of nickel. For a given composition of iron and nickel, there are different crystal structures that can be formed. Of relevance here are two crystal structures called “taenite” and “kamacite” (in the iron-carbon system where very similar patterns can be formed, these would be called “austenite” and “ferrite”, but they refer to the same crystal structures).

Whether taenite or kamacite is more stable depends both on the temperature and the amount of nickel present. At high temperatures, when the meteorite is young, it will be all taenite. However, as it cools, we eventually reach a temperature where, energetically speaking, the universe would like a mix of both taenite and kamacite.

However, just because it’s now energetically favorable for kamacite to form, there still needs to be a mechanism by which some of the taenite crystal can transform into kamacite.

The Widmanstätten patterns only form at very low undercoolings, in other words when the temperature is only just in this range where kamacite is starting to become stable.

What you’re actually seeing in this pattern are the plates of kamacite that have formed. The exact pattern you see will depend which slice of the meteorite you decide to polish, and is related to the underlying symmetry of the crystal structure.

The reason you don’t see these patterns be so large on earth is for two reasons: one is simply that the grain sizes themselves are much smaller, and so the size of the microstructure is limited by that. The other reason is that the cooling rate is much faster, so the patterns can’t grow as large.

When these plates nucleate, the extent (length/width) of the Widmanstätten plates can grow rapidly, but they will only get thicker at a rate that's limited by the diffusion of nickel atoms between the iron atoms. This is because the sides of the plates have a coherent interface with the parent taenite phase and diffusion is slower, whereas the tips/edges of the plate have an incoherent interface where diffusion is much more rapid.

I believe the rate of the plate growing longer is related to the degree of undercooling, so at very slow cooling rates like those in space, the plates grow longer much more slowly. This then gives time for atoms diffuse on the plate sides for the plate to grow thicker. And therefore, eventually, we have this beautiful microstructure large enough to see with the human eye.

That’s my understanding, however I think there isn’t yet consensus on exactly how Widmanstätten patterns form, the role that diffusion plays, and also the role of other additions in mediating this (e.g. the presence of phosphorous in the case of iron-nickel meteorites).

I hope this helps!


Questions on apparent motion of superior planets Question
Why is it that superior planets seem to oscillate up and down of the ecliptic, from the view of an observer on Earth, with a period equal to the planet's orbital period?
Additionally, why do these superior planets seem to come closer to the ecliptic when they are nearing conjunction and drift away from the ecliptic when they are near opposition?
Attempt
Does it have to do with the Earth's tilt?

The planets do not orbit in the same plane as the Earth, they are inclined by a few degrees. This means that as they orbit the sun, they are sometimes above the plane of the Earth's orbit and sometimes below.  And it takes one orbital period to go the full cycle from above, to below and back to the start.
The reason that they appear further from the ecliptic at opposition is a matter of perspective.  At opposition the planet is close to the Earth. And nearby things appear bigger. So the planet will appear (in general) to be further from the ecliptic at opposition.


"If earth had rings like Saturn": A few further thoughts on the consequences I came across this video recently where artists depicted what it would look like if Earth had rings like Saturn. http://www.wimp.com/earthrings/

Please note: The video contains already some information on what I ask here, but I can't assess its reliability, therefore I'd appreciate any scientific thoughts on it :)

1) Earth has roughly a distance of 1 AU to the sun, Saturn between 9 and 10. Sun's gravity would therefore have a much stronger influence on these rings. Would it already be enough to distort the rings (e.g. even leading to the sun "stealing" parts of it) or would the rings still have a ring shape?

2) How well would the objects in such rings be visible from earth? The Saturn rings are mostly dust and ice as far as I know, but there are surely some larger objects. Depending on material and size some object would probably be visible to the naked eye, right? Is there (probably yes) a formula to calculate what materials and sizes are required to achieve this?

3) Would those rings not be constantly raining particles down on earth? Would there be a much higher risk of dangerous impacts for the earth's population?

4) Are the colours used in the video precise? Don't they depend on the angle? What influence would our atmosphere have onto them?

5) Would those rings have the same orbital inclination or would theirs differ from earth's to an extent where the difference becomes visible?

6) What shadow would these rings cause on earth? Would they be dense enough to "shade out" complete stretchs of land?

I'm aware that these are hypothetical questions. But I find it, especially as an author of sci-fi and fantasy, nonetheless very interesting.

Thank you for your interest!

Pat

This is a big one, so I'll split it into parts based on the questions you asked:

  1) Earth has roughly a distance of 1 AU to the sun, Saturn between 9 and 10. Sun's gravity would therefore have a much stronger influence on these rings. Would it already be enough to distort the rings (e.g. even leading to the sun "stealing" parts of it) or would the rings still have a ring shape?

Sun would not have too much of an influence because the rings are all individually gravitationally bound to the earth (similar to the moon, and you don't see moon being affected by the sun, gravitationally; the rings will be much closer to the earth and hence even less affected).

The moon, on the other hand, would distort a few rings due to orbital resonances (similar to Titan Ringlets: https://en.wikipedia.org/wiki/Rings_of_Saturn#Colombo_Gap_and_Titan_Ringlet and resonant gaps in asteroid belts due to Jovian planets: https://en.wikipedia.org/wiki/Kirkwood_gap), and its effect would be much more pronounced because it is closer to the earth than Titan is to Saturn, though limited to a few resonant regions. There will also be ripples if the plane of the moon is different from the plane of the rings (http://blogs.discovermagazine.com/badastronomy/2009/06/11/saturns-rings-do-the-wave/#.U7VjpvmSw08).

  2) How well would the objects in such rings be visible from earth? The
  Saturn rings are mostly dust and ice as far as I know, but there are
  surely some larger objects. Depending on material and size some object
  would probably be visible to the naked eye, right? Is there (probably
  yes) a formula to calculate what materials and sizes are required to
  achieve this?

Most matter in Saturn's ring is particles up to a few metres in size (http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q6). Resolution of the human eye is about 1' (Is the moon only 60 pixels?), which means at a distance of around 1 to 1.5 Earth radii (given by Roche limit of the earth; this is where the rings will be), the eye can resolve things above a couple of kilometres in size. However, Saturn also has Shepherd moons, which are a few tens of kilometers in size. Such moons, if present, would be visible to naked eye. Also, the gaps would be visible clearly, since they are often big enough.

  3) Would those rings not be constantly raining particles down on
  earth? Would there be a much higher risk of dangerous impacts for the
  earth's population?

There would be collisions/interactions in the rings (I'm guessing, correct me if I'm wrong) which would send particles earthwards. However, the particles which will get thrown will be the lighter, smaller ones (conservation of momentum etc.), and usually small meteoroids are fairly harmless. They will also be slower than the usual meteoroids that enter the earth at above 20 km/s, since the orbital motion of the earth doesn't add up in this case. I am a little unsure about the consequences of them being slower, but anyway, smaller meteoroids should technically be fairly harmless. (Also I couldn't find decent references other than: http://science.howstuffworks.com/question486.htm)

  4) Are the colours used in the video precise? Don't they depend on the
  angle? What influence would our atmosphere have onto them?

I believe so. This section shows true colour images of the rings from two different angles, and there doesn't seem to be a significant change in the colouration: https://en.wikipedia.org/wiki/Rings_of_Saturn#Subdivisions_and_structures_within_the_rings. The atmosphere will probably affect the rings closer to the horizon, with redder colouration similar to sunrise/sunset and moonrise/moonset.

  5) Would those rings have the same orbital inclination or would theirs
  differ from earth's to an extent where the difference becomes visible?

Depends largely on the formation history. This FAQ question here discusses this fairly well: http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q14

  6) What shadow would these rings cause on earth? Would they be dense
  enough to "shade out" complete stretchs of land?

There will be shadow similar to that due to clouds, since the rings are not too thick or dense. In many of the Cassini photos, you can see through the rings, so it is likely that the sunlight would still pass through, but would be dimmed (it would be interesting to see by what factor; I cannot say for sure with my knowledge). Also, remember that the earth will still be rotating but the ring system will be dynamically static, so there will be extended periods of darkness during the day for some regions at different times of the year.

I will add more to this if something comes to mind. I hope this was helpful.


How Long Will Earth's Year be When Our Sun Goes Red? We live in a planet that orbits 93 million miles from a G-type main-sequence star, or "yellow dwarf".  That is far enough for a revolution of 365 days.  Such is the case of Kepler's Third Law of Motion--the farther a body is from its parent, the longer it'd take to revolve around it.

But in the far-distant future, our sun will balloon into a red giant, a star big enough to swallow Mercury and Venus out of existence.  Earth and Mars will take their place, and the habitable zone (where liquid water is possible) will be moved from between Mars and Jupiter to between Saturn and Uranus.

So, in the future, Earth will no longer be one AU from the sun, which means a far shorter year.  But has anyone made any calculations as to how much shorter a Terran year will be in this future scenario?

The year length depends on the distance between the planet's centre & the Sun's centre, not the Sun's surface. So if the Sun merely expands, Earth at 1 AU will still take a standard year to perform 1 orbit.

However, when the Sun becomes a red giant, it won't just expand. As Wikipedia mentions, red giants shed a considerable amount of mass in the form of gas and dust. It's estimated that the Sun will lose around a third of its mass over the first billion years of its red giant phase, and eventually have a mass around half of its current mass; see here for details.

It's still not clear exactly what will happen to the Earth while all this is going on. It may become engulfed, like Mercury & Venus certainly will be. But even if it's not engulfed, interacting with all that gas & dust is likely to affect its orbit.

If we ignore that complication, we can easily calculate the period of a 1 AU orbit around a Sun of reduced mass. The Newtonian form of Kepler's 3rd law says that
$$T^2= \left(\frac{4\pi^2}{GM}\right)r^3$$
where $T$ is the period, $G$ is the universal gravitational constant, $M$ is the Sun's mass, and $r$ is the mean orbital radius.

If we keep $r$ at 1 AU but halve the Sun's mass, then the period is multiplied by $\sqrt{2}$, giving us a year length around 516 days. That's using the current day length, by then the rotational period of Earth will be considerably longer.

I seriously doubt that the Earth will still be at 1 AU from the Sun, though, so please don't take that figure of 516 days too seriously!


Demo to get Matter Power Spectrum in cosmology I would like please to demonstrate the expression of Power spectrum in Cosmology :
First, I have the relative contrast:
$$\delta_{i}(\vec{x}, z) \equiv \rho_{i}(\vec{x}, z) / \bar{\rho}_{i}(z)-1\quad(1)$$
After, we decompose this relative contrast on Fourrier basis :
$$\delta_{i}(\vec{x}, z)=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \tilde{\delta}_{i}(\vec{k}, z) \exp (\mathrm{i} \vec{k} \cdot \vec{x})\quad(2)$$
and finally, how to find the following expression (3) from (1) and (2) :
$$\left\langle\tilde{\delta}_{i}(\vec{k}, z) \tilde{\delta}_{i}\left(\vec{k}^{\prime},z\right)\right\rangle=(2 \pi)^{3} \delta_{\mathrm{D}}\left(\vec{k}+\vec{k}^{\prime}\right) P_{i}(\vec{k}, z)\quad(3)$$
?
Any help is welcome.

This is just a Fourier transform: (let $\boldsymbol{x}=\boldsymbol{r}_2-\boldsymbol{r}_1$)
$$
\begin{aligned}
\langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle  \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\
&=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3r_2\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2)\rangle e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\
&=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3x\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_1+\boldsymbol{x})\rangle e^{-i\boldsymbol{k}_2\cdot(\boldsymbol{r}_1+\boldsymbol{x})}\\
&=\int e^{-i(\boldsymbol{k}_1+\boldsymbol{k}_2)\cdot\boldsymbol{r}_1}d^3r_1\int\xi(\boldsymbol{x})e^{-i\boldsymbol{k}_2\cdot\boldsymbol{x}}d^3x\\
&=(2\pi)^3\delta_D(\boldsymbol{k}_1+\boldsymbol{k}_2)P(\boldsymbol{k}_2)
\end{aligned}
$$
Here, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ is two-point correlation function (2pcf) in real space. If we assume our universe is statistically homogeneous, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ should have the form $\xi(\boldsymbol{r}_1-\boldsymbol{r}_2)$. So power spectrum is the Fourier transform of 2pcf.
In addition, if we assume our universe is statistically isotropic (not true in redshift space), 2pcf can be $\xi(|\boldsymbol{r}_1-\boldsymbol{r}_2|)$ and power spectrum can be $P(k)$.


Why are planets round? I was studying our solar system and all celestial objects are depicted as round objects, including black holes.. worm holes as hollow cylinders as a pathway across two universes.. even the galaxy is show circular.. Why does this happen?

Gravity makes objects compress to their centers of mass, since gravity extends in all directions.

For planets, gravity continues compressing the object until the rocks cannot be compressed anymore, since the pressure will fight against gravity. At this point (called hydrostatic equilibrium), the object has become a spheroid. One of the requirements for an object to be called a "planet" is that it has reached hydrostatic equilibrium.

For black holes, the shape of their event horizons really depend on whether the black hole is spinning or not. For non-spinning black holes, gravity extends in all directions, so the event horizon will become spherical in shape. For spinning black holes, it will likely be an oblate spheroid.

Most galaxies are only spherical when they are forming. Their angular momentum usually causes them to flatten out over time. However, one part of galaxies, called the "halo", is spherical.

Wormholes are purely hypothetical, so I won't really entertain that.


Where asteroids disappear Scientists claim that objects left on the Moon can lie hundreds of years in the same condition like astronaut Charles Duke family photo. Why we don't see remains of asteroids inside or around craters? Are asteroids going destroyed on impact or go under the surface even without any residue? Why do we not see asteroids that hit  Moon with a small force and remained on the surface for hundreds of years like astronaut's photo? Can craters form for other reasons? For example, from underground collapse?

Asteroids hit the surface of the moon at speeds of about 25 km per sec, so they are mostly vaporised on impact. If a geologist inspected a crater on foot and looked hard enough, he probably would find a few relatively small fragments, particularly if the meteorite was of the iron/nickel variety. A few such fragments were recovered from the Barringer Crater in the United States. It is unlikely an impact would leave no residue at all. but the residue might be difficult to identify. When trying to visualise the impact of, say, a 100-ton meteorite, it is helpful to imagine that it turned into 100 tons of high explosive on impact.

Yes, craters can form for other reasons. Most craters on the moon are impact craters, but a small number are volcanic. Underground collapse sometimes occurs in the vicinity of volcanoes; the resulting holes are usually elongated, and are called graben. Lava tubes sometimes collapse too, leaving a  long, winding  valley like a river bed. These indentations are called rills. Objects left on the moon tend to survive for centuries in good condition, because there is no weather worth talking about, just solar wind (charged particles), cosmic rays and electromagnetic radiation.


Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958) Jan Oort was a pioneer in radio astronomy. Wikipedia says:

  It has been written that “Oort was probably the first astronomer to realize the importance” of radio astronomy. “In the days before radio telescopes,” one source notes, “Oort was one of the few scientists to realise the potential significance of using radio waves to search the heavens. His theoretical research suggested that vast clouds of hydrogen lingered in the spiral arms of the Galaxy. These molecular clouds, he predicted, were the birthplaces of stars.”

The article includes the image shown below, which is a 21 cm radio map of the Milky Way galaxy. (for more on the transparency of dust at 21 cm see this excellent answer to How was the galactic plane established?

You can also read more about Oort's work during this time in this 1976 AIP Oral History Interview interview.

I'm guessing that the center of the plot is the galactic center and the point at 8 kpc above it is the Earth. What is the reason for the two blank wedges, projecting downward and upward from the Earth? Are they geometrical, blind spots from the few early radio telescopes in the 1950's perhaps, or do they reflect real phenomenon in the galaxy?

The original source is The galactic system as a spiral nebula Oort, J. H.; Kerr, F. J.; Westerhout, G. MNRAS 118, (1958) p. 379

TLDR: these wedges are bits where things are moving around the centre of the galaxy at about the same speed as us, so we can't understand what is there.

As it states on page 4 of the paper you have linked,

  the great gap between 315 and 340, where, except at small R, the differential rotation is too small to separate the various arms

The method used to deduce this structure is described on page 2:

  Observations in the decimetre continuum can only give the integrated radiation over the line of sight. The 21-cm line gives discrimination in distance. But although the 21-cm observations give discrimination in distance they cannot by themselves provide actual distances. The distance distribution in a given direction can only be inferred from radial velocities. For this we have to suppose that in each part of the Galactic System the average motion of the gas coincides with the circular velocity at the corresponding distance from the centre.

The map is given by combining several observations at different wavelengths - the 'decimetre' gives the total amount of stuff along the line of sight (because it isn't absorbed much along the way), the 21-cm line gives a measure of distance (because it is absorbed so predictably wakened by the amount of stuff it has passed through), and then to we have to assume a model of Keplerian rotation as a guess at how things are moving around the centre of the galaxy. Combining these three things allows you to map out the material even though you are seeing different bits stacked together along your line of sight.

This doesn't work in the blank wedges because the speed of rotation is too similar to ours, so we can't use the rotation velocity to disentangle distances.


Type of telescope on Star Trek In the first episode of Star Trek: Discovery, we learn that Captain Georgiou has an old-fashioned telescope in her ready room. Michael Burnham inherits it a bit later in the season. I wonder if anyone can identify exactly which telescope it is. What's the model and manufacturer? It seems to be an equatorial-mounted refractor with a finder scope.

I'm not sure this fits this site, maybe better for TV & Movies, but it doesn't appear to be a historic or current telescope.

This website advertises that they sell that model Based on the look and it is similar.   It may be a  nice telescope but it's nothing special.  Certainly not worthy of being a family heirloom passed down for generations, so that doesn't really make sense and it's not a perfect match either, though it's similar.

Larger image here.

It not being a perfect match and there being virtually no references to that telescope representing something real, it's likely not a historical telescope, which is fine.   It could be a future telescope and still be in Captain Georgiou's family for generations.  That would work because they used it in the show to look at things the ship couldn't sense due to some kind of obscuring.  A historic telescope probably wouldn't be able to do that.

Ultimately, I think Signourney Weaver as Lt. Madison said it best.