What exactly are infrared dark clouds and what is the significance of the orientation of their magnetic fields? In the video Why NASA's SOFIA Telescope On A Plane is "Perfectly Balanced" after about 10:04 SOFIA associate director James Jackson says:

We have a spectrometer called GREAT, that’s our workhorse spectrometer that’s probably most popular. We have an imager called HAWC that does far infrared, and we have FORCAST, and we have another spectrometer EXES which is (for) shorter wavelengths.
[…]
Now… we’ve added a new wrinkle to our repertoire, we now have a polarimeter so we can measure the polarization of light, and that tells you how the magnetic fields are threading through these dust clouds or gas clouds. So HAWC does polarization as well, that’s a new capability
[…]
Well Orion for example… the galactic center… we’re looking at infrared… these snaky filamentary infrared clouds called infrared dark clouds. The basic question still is not answered because… does the magnetic field go “this way” with respect to the filament, or go “that way”; perpendicular, or parallel? What SOFIA is finding is that it’s more-or-less perpendicular, which is a surprise.

The text of the Wikipedia article Infrared dark cloud says (in its entirety):

An infrared dark cloud (IRDC) is a cold, dense region of a giant molecular cloud. They can be seen in silhouette against the bright diffuse mid-infrared emission from the galactic plane.
Infrared dark clouds have only been recently discovered in 1996 using the ISO and therefore are in need of further research.
Astronomers believe that they represent the earliest stage in the formation of high-mass stars and are therefore of great importance for understanding the star formation process as a whole.

That's not much to go on. This link is cited in the Wikipedia article and it says a little more in a descriptive manner:

Infrared dark clouds are cold, dense molecular clouds cores seen in silhouette against the bright diffuse mid-infrared emission of the Galactic plane.

and here's one of the images from that page:

8 micron image of G11.11-0.11 from MSX. It is seen in extinction against the bright diffuse mid-infrared emission of the Galactic plane.

The Wikipedia article also links to The early stages of star formation in Infrared Dark Clouds: characterizing the core dust properties but there's no mention of magnetic fields there either.
So I'd like to ask:
Question: What exactly are infrared dark clouds and what is the significance of the orientation of their magnetic fields? I'm also wondering if they produce these fields, or are passively aligned to existing fields.

Molecular clouds are the birthing grounds from which stars form, which occurs when said clouds undergo a collapse.  They contain a significant fraction of molecular hydrogen ($H_2$), and are generally good places for forming other stable molecules.  Molecules and metallicity generally make star formation easier by making it easier to get the non-molecular hydrogen and helium to lose enough energy to collapse into a common core by acting as (kinetic) energy absorbers.  Imagine tossing a ping pong ball at a tennis ball: the ping pong ball bounces off at a slightly lower velocity, and the tennis ball barely moves at all from an even smaller change in its velocity.  And if the cloud is already "cold", then that's even better: there's less energy your particles have to shed.  The presence of molecular hydrogen seems to be particularly important in the collapse of low metallicity clouds, such as would be the case in the early universe. So cold molecular clouds, like infrared dark clouds, should be a great nursery for high mass stars, as you don't have to shed that much energy (relatively speaking) to get things collapsing into a core, and it's got molecules to help it shed what it still needs to.

Exactly how the collapse is predicted to happen depends on the model, and the models depend on the magnetic field of the collapsing cloud.  There are two major models.

Quoting liberally from Magnetic Fields
in Molecular Clouds by Richard M. Crutcher (c. 2012) (all quotes will be from this article, until otherwise stated):

  In the strong magnetic field theory, clouds are formed with subcritical masses, $M < M_\Phi = \Phi/(2\pi \sqrt{G})$ (Nakano & Nakamura 1978), where $M_\Phi$ is the critical mass, $\Phi$ is the magnetic flux, and $G$ is the gravitational constant. Hence, the magnetic pressure is sufficiently strong to counteract gravity and prevent gravitational collapse. Because the magnetic field is frozen only into the ionized gas and dust, neutral gas and dust contract gravitationally through the field and the ions, increasing
  mass in the cloud cores. The magnetic field strength also increases, but more slowly than does mass.

  This process is known as (gravity-driven) ambipolar diffusion. [Ambipolar diffusion, or breaking of flux freezing, may also be driven on small scales by turbulence (e.g., Zweibel 1988.)] When the core mass reaches and exceeds $M_\Phi$, the core becomes supercritical ($M > M_\Phi$), collapses, and forms stars. During the collapse, the magnetic field is dragged inward but cannot become strong enough to halt the collapse.

  The weak-field theory of star formation says molecular clouds are intermittent phenomena, with short (∼$10^6$ years) lifetimes. Magnetic fields are sufficiently weak that the low-density ISM is supercritical ($M > M_\Phi$). Clouds form at the intersection of turbulent supersonic flows. Generally, clouds do not become gravitationally bound, and they dissipate; those that are self-gravitating form stars in essentially a free-fall time (Elmegreen 2000). Supersonic turbulence will dissipate on roughly the free-fall timescale as collapse of gravitationally bound clouds proceeds (MacLow et al.
  1998). Although magnetic pressure cannot stop the collapse, it can dominate turbulent pressure during the late stages of core collapse. An extreme version of weak-field models has the initial field so weak that the medium is super-Alfvenic as well as supercritical (Padoan et al. 2004).

So, the short of the matter is that molecular clouds are important because they are connected to the formation of (high mass) stars. And the magnetic fields of those clouds is important because the exact dynamics of that collapse are dependent upon them.  And all of that is important because it gives us a better idea of how high mass stars form, which is expected to have happened much more frequently in the early universe.

It hasn't, however, yet mentioned why the direction of that field may be interesting (it only mentions the flux).

  Interstellar dust produces thermal emission and extinction of light from background stars. Linear polarization of this radiation provides a probe of the magnetic field morphology in the ISM, including molecular clouds.

So they're curious about the structure of the interstellar medium (ISM) of the galaxy. The radiation of the dust clouds, including the molecular clouds, provides a way to probe the magnetic structure of the ISM.  The paper then recalls how this polarization can arise via radiative torques creating preferential alignments in dust grains, and what this then means.

  [T]he grain extinction cross-section is greatest perpendicular to $\mathbf{B}$, so the maximum of the polarized light from background stars is parallel to $B_{POS}$, the magnetic field direction in the plane of the sky. Conversely, maximum polarized emission is perpendicular to $B_{POS}$. The predicted degree of polarization depends very weakly on magnetic field strength, so dust polarization does not directly give the magnitude of the magnetic field.

This describes how they can measure the direction of the magnetic field of a molecular cloud, a topic from your clip.  It notes that it's a poor measure of the actual field strength, and we need the flux $\Phi$ to check how a given star formation model matches up with observations.

The paper continues on and discusses some measurements of the direction of the magnetic field of molecular clouds.  The results it cites indicated that the field has been observed to be either parallel or perpendicular, and that the direction can change between different parts of the same cloud.  So I'm not sure why your clip mentions a perpendicular measurement as surprising; perhaps it's the "more-or-less" that is key, as if there's enough data to suggest a preferred (perpendicular) alignment then that would be interesting and probably unexpected.

The paper also discusses how it is nevertheless possible to obtain some estimates of $\Phi$, as well as the difficulties in doing so accurately.  This lets us tie everything together: the polarization of the radiation from the molecular cloud allows us to measure the direction of its magnetic field relatively easily; and with that and some various techniques we can then measure the actual flux $\Phi$ of the magnetic field; which we can then use to determine if the cloud is sub-critical ($M<M_\Phi$) or not.  And when we know that, we can now more readily probe how accurately star formation models, such as those mentioned above, reflect the actual reality.


What percentage of the celestial sphere can the Moon cover? I was solving a task that said (paraphrasing):

  What percentage of the night sky can the Moon cover during the entire
  year, when observed from all points on Earth (what percentage of the
  night sky can be studied with the occultation method)?

This is a task I read from a book, and the solution states that the Moon covers a "ribbon" of width $\alpha$ where $\alpha = 0.5 ^{\circ} + 2*1^{\circ} + 2*5.3 ^ {\circ} = $ angular diameter of the Moon + 2*the Moon's daily parallax + 2*the inclination of the Moon's orbit to the ecliptic; and length $360 ^{\circ}$.
Why is this like this, I understand the diameter and parallax part, but I couldn't visualize the problem very well. Why is the third number the inclination to the ecliptic and not to the equator?

If the Moon's orbit were fixed, the area covered by the Moon would just be $360^{\circ}*(0.5^{\circ}+2*1^{\circ})$. The 5.3 degree inclination relative to the ecliptic (or some other inclination relative to the equator) would not be a factor.

What is important to this question about the coverage is that the orbit changes over time. In particular, the position of the ascending node (where the Moon's orbit crosses the ecliptic) moves along the ecliptic with a period of 18.6 years (Wikipedia article on Lunar precession).

In a plot of the sky using ecliptic coordinates (so that the ecliptic is a straight line), the 5.3 degree inclination of the Moon's orbit to the ecliptic is shown as "year X" in my figure. The precession changes the orbit to positions such as year Y and year Z in my figure, and during the entire 18.6 year cycle, the entire band within 5.3 degrees of the ecliptic is covered by the Moon. In other words, the ascending node moves through all 360 degrees of ecliptic longitude. This results in the term 2*5.3 in your equation. (Note: I do not know if the precession is to the "right" or "left", so do not interpret my figure to mean that year Y is later than year X. Year Y could be earlier than X. I'm sure someone will correct me!)

If the same plot were made in equatorial coordinates (so that the equator is a straight line), the band covered by the Moon is of course the same on the sky. However, the precession along the ecliptic limits the ascending node of the Moon's orbit on the equator to a small range or right ascension; the ascending node does not travel around the entire equator. Representing that band using equatorial coordinates and determining the area inside the bands is much more difficult.


German equatorial mount I have a Celestron Astromaster 130 EQ. After aligning  my telescope's polar axis with Polaris, in order to track stars I need information about the coordinates in terms of right ascension (RA) and declination.
But in all the astronomy apps I have found that they have fixed the RA and declination coordinates and are tracking stars with respect to azimuth and altitude.
How am I supposed to use this data? Please help!

The really short answer is ... you don't (use Alt/Az data).  You own a German Equatorial Mount that uses RA/Dec values and have no need of the Alt/Az coordinate system.
A slightly expanded answer is that use of coordinates is meant to help you find objects.  Realistically it would be challenging to do with your telescope mount because the RA/Dec setting circles on the mount are not highly accurate (and sometimes move when they aren't supposed to move).  More on this later.
Right Ascension (& Declination)
The Right Ascension and Declination system for objects in space is somewhat like Longitude and Latitude coordinates on Earth.  But since Earth spins, we can't use actual Longitude / Latitude values.
A object directly above Earth's North Pole has a Declination of +90°.  An object above Earth's South Pole has a declination of -90°.  A declination of 0° means the object lies along a path somewhere directly above Earth's equator.
Right-Ascension works like a clock (which is why it is expressed in hours and minutes) based on 24 hour day.  If you look at the setting circles for your telescope, you'll probably notice that the Right Ascension setting circles have two sets of values... one counting up, the other counting down.  Those are actually to allow for use in the Northern vs. Southern Hemispheres (a telescope in the Southern Hemisphere is aligned to the Southern celestial pole, but the motor drive needs to spin the telescope in the opposite directly as compared to the Northern Hemisphere.
There is a 00 hour 00 minute position in the sky.  It's "name" is the "First point of Aries" (a confusing name because the point is not in Aries ... for reasons I'll get to in a moment).
The point is defined based on the position of the Sun (relative to the background stars ... which of course we cannot see during the daytime) at the Vernal equinox (when the Sun crosses the equator from the Southern Hemisphere into the Northern Hemisphere on or around March 20 each year).
But since Earth's axis precesses very slowly (a cycle that takes nearly 26,000 years), the point moves.  When the point was originally defined, the Sun was in Aries... but a couple of thousand years later, it has migrated to Pisces ... and is actually very close to Aquarius.
So imagine if you could point your telescope at the First Point of Aries (the 00 hour 00 minute 00 second point) and then ... switch off the motor and just watch the sky drift along from East to West.  How long would you have to wait before the imaginary "longitude" value of the object you want to view in the night sky is in front of your telescope?  That time duration... is how the RA coordinate gets it's value.
With the exception of solar system objects (planets, comets, asteroids, etc.) objects in the universe seem as if they don't move.  They do move... but are so incredibly far away that it requires very precise measurements to detect any movement at all.  This means that, for the most part, we can assign them fixed coordinates that don't change (these coordinates are occasionally updated).  On the timescale of a single human lifetime, we don't notice the movement.
So this means, conveniently enough, that we can mostly get away with a simple look-up table to find objects and ignore their proper motion.
This is why the RA/Dec values do not change when you look up objects in those apps... because they don't actually change from hour to hour or day to day.  Many years are required to notice even very subtle changes ... and using very accurate instruments to take those measurements.
Also, because of the way this coordinate system works... the RA/Dec coordinates for these non-solar-system objects are the same for all observers, everywhere on Earth, regardless of date or time (well... regardless of date or time excluding very large amounts of time... such as thousands of years ... at which point their movements would change enough to notice.)
One little tiny nit... Polaris is not located at Earth's celestial North Pole ... but it is close.  It's roughly 2/3rds of a degree away from the pole.  But keep in mind that the angular width of the Moon (as observed from Earth) is just about 1/2° wide in the sky.  This means there is enough angular width between Polaris and the true Celestial North Pole... that you could fit the Moon in that space with a little room to spare.
For visual astronomy ... pointing your mount at Polaris is "close enough".  A computer driven telescope would miss it's "go to" targets ... but probably get close enough that those objects would appear in the Finderscope (and just need to be re-centered). Not a big deal.  (For astro-imaging using long camera exposures ... a more accurate alignment is needed.)
Altitude Azimuth
This coordinate system is much simpler to understand.  But it has a major drawback.
The azimuth is simply the compass heading... in which compass heading do you need to be facing in order to view the object, and... how high above the horizon line do you need to look?  The compass heading is the Azimuth.  The angle above the horizon is the Altitude.  Simple enough.
But suppose you are collaborating with someone about an object you are observing ... so you want to share coordinates with your collaborator.  But your collaborator is not nearby ... they live on another continent.  The Earth is spherical and the curvature of the Earth means that the Altitude and Azimuth you use ... isn't the same as the Altitude and Azimuth your colleague will need to use.  Even if they live along the same latitude and are simply in an adjacent time zone... their coordinates will be different enough to completely miss finding your object.
This is why Altitude/Azimuth values aren't useful in catalogs.  They change ... a lot.
The German Equatorial Mount
The benefit of this mount is that once you orient the mount so that it's major axis (the Right Ascension) is parallel to Earth's axis, then as the Earth rotates from West to East ... the telescope mount can be moved East to West.  If this mount were equipped with a motor and if that motor spins the mount at the same rate that Earth spins... then the telescope will remain fixed on the same position in the sky.
Also, adjusting the mount in the RA direction causes the telescope to move in a perfect east-west direction.  Adjusting the mount in Declination moves the telescope in perfect north-south direction.
This simplifies the process of tracking objects in the sky because once you target and center on an object, only the Right Ascension motor needs to operate to track an object.  With an Alt/Az mounted motorized scope, a computer needs to interpolate the movement of both axes in order to track an object.  The exception is... if that telescope happens to be located directly at Earth's North Pole or South Pole ... then an alt/az telescope works just as well as a German Equatorial Mount (RA/Dec).
Converting from one system to the other requires a bit of math -- but not particularly complicated.
Using RA/Dec to find objects
I previously mentioned RA/Dec (for non-solar-system objects) is mostly just a lookup-table of values that don't change in a very noticeable way within the span of a single human lifetime.
Computer-controlled "go to" telescope mounts have to be "aligned".  The mount is polar aligned (physically).  The computer will typically want to know your geographic coordinates as well as the date & time to have a good idea of what objects should be visible in the sky.  Most "go to" mounts will then ask that you accurately center the telescope on a couple of bright stars ... and use this information to create a "fix" on the entire celestial sphere.  From this ... the computer can now automate moving the telescope to any object you want to observe.
As your telescope mount is not not motorized and controlled by a computer, what point is there to have setting circles with the RA/Dec values?
It turns out you can (theoretically) use the setting circles as a kind of analog computer.
Suppose you want to locate some hard-to-find object.
Start ... by pointing the telescope to a known easy-to-find object.  A really bright star such as Sirius... or Vega... etc.
Look up the RA/Dec values for that really bright star.  Now adjust the setting circle values on your mount to match the values in the table.
Next, find the RA/Dec values for your hard-to-find object.  Then re-point your telescope'S Declination axis to the declination of the hard-to-find object.  Now do the same for the RA value.
Theoretically your telescope is now pointed at the hard-to-find object.  I say "theoretically" because large professional telescope used many years ago to find these objects (before telescopes were controlled by computers) had very large settings circles that were much easier to use.   Consumer telescopes usually have tiny setting circles that are sometimes rather frustrating to use (My personal experience with my own budget mounts were that the bits that were meant to be adjusted to match an object and then remain "fixed" as you move the mount ... often slipped and moved while adjusting the mount.  This made accurate pointing a fairly frustrating experience.)
This means "star hopping" is probably the preferred method to find objects (find the nearest easy-to-find object .. then move the mount only slightly to locate your hard-to-find object).


Open source library to calculate image from signals of a distributed set of radio dishes? The National Radio Astronomy Observatory's Q&A "How To Make Images With a Radio Telescope" says

The most straightforward way to make a radio image with your satellite antenna system, then, is to use an array detector, which I believe would be very difficult to find.  Alternatively, one can make an image by pointing an antenna to an array of nearby positions on the sky then creating a contour map of the measured signal strength.  This is a rather slow process, though, that only produces an image much later, after you have gathered all of your individual pointing measurements.

Assume now that a lot of people all over a continent (e.g. Europe) would have (fixed, non-moveable) radio dishes pointing somewhere into the sky, with their field of view somewhat overlapping. They have different dishes and different receivers. However, given their exact location on the globe, knowing their exact time-stamp of observation, at least theoretically, I should be able to combine the measured time-series of radio-signals to a radio-image of the sky, correct?
The idea already bothers me a while and I am wondering whether there is some open source software I should be aware of to tackle the problem of caculating such an image, ideally in python (but possibly in any other programming language).
Any input is appreciated!
Related

Radio Astronomy and Imaging
In astronomical interferometry, what values do the points in the uv-plane have?
Can one build a distributed radio telescope?
How big a dish do I need for radio astronomy?

This is an interesting question!
I'll post a partial, preliminary answer and let's see where this takes us.
The NRAO block quote mentions two methods, interferometry and intensity mapping (like how a camera works). There may be hybrid options and other esoteric options, but lets look at these first.
See also this answer to Which techniques are used to convert radio signals received by antenna to images? and since there are so many good answers to Radio telescope targeting those might be worth a good read as well.
Interferometry
This is where all dishes point in the same general direction and you have:

...the exact location(s) on the globe, knowing their exact time-stamp(s) of observation...

and those time stamps must be made to nanosecond relative accuracy; you use the phase information of the radio waves. You either hook them up with coax or fiber optic cable in real time, or write at up to gigabit/second speeds to hard drives and then cart them all to one location and run a very very big and special computer called a correlator to analyze the data.
Intensity mapping
This is just making an intensity map by sampling many spots in the sky and building up a histogram; basically exactly like how a camera works but kilopixels rather than megapixels. Each telescope is an independent sensor, and it contributes to the the big map. You might start with a grid of pixels and as the antennas drift into each box you measure the intensity of radio power, and write that number down in that box. Earth's rotation scans in RA already. so no matter where they are, the dishes should be evenly spaced in declination.
See answers to How did single dish (or single receiver) radio telescopes originally generate images?
Hybrid and other esoteric options
There is intensity interferometry but I don't understand it well enough to write authoritatively on it. It looks at fluctuations in statistical noise, and so you can probably get away with less critical timing; each dish would record the instantaneous power level at perhaps microseconds or even millisecond intervals. I haven't a clue how this works with arrays; I've only heard it used for two receivers, but you could certainly analyze it pair-wise. (see also Narrabri Stellar Intensity Interferometer)
You could also do something fun/interesting with pulsars, but I don't know what exactly.

Intensity plus Doppler frequency can sometimes give you distance if you  have a model; this is an early radio map of neutral hydrogen in our galaxy from  Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)
 click for larger


Is it dark inside a black hole? Assuming you could be standing inside a black hole, would it be light or dark around you?

The only light you would see, would be from the stars outside the black hole. Any light that is generated inside the event horizon would be refracted towards the singularity. You wouldn't see the object in front of you since the light would not reach your eye.


How thick is the ice on Europa? The Wikipedia article about Europa states

It is estimated that Europa has an outer layer of water around 100 km (62 mi) thick; a part frozen as its crust, and a part as a liquid ocean underneath the ice.

But it doesn't say how thick the icy crust and how thick the liquid ocean is. How many kilometers thick is the crust of frozen water on Europa? Is it roughly the same thickness all across the moon?

From this NASA FAQ page about Europa:

Theory and observation indicate that Europa's icy shell is around 15 to 25 kilometers (10 to 15 miles) thick, overlying an ocean approximately 60-150 kilometers (40 to 100 miles) deep. Support for this hypothesis comes from observations of pits, domes, and spots on Europa's surface. The size and spacing of the features suggests that they are due to churning within the ice shell, and theory suggests that such churning (called convection) can occur only if the shell is between greater than about 15 kilometers thick. Measurements of the height of domes on Europa (up to a kilometer or just over half a mile high) also suggest that the ice shell must be fairly thick for the domes to be so tall.

However, it seems like there has been a some disagreement on this matter:

Some scientists have argued that the ice shell might be thinner, only a few kilometers thick.

But, the text doesn't seem to elaborate on any exact figures, so I'll look for some more sources.
Here's a paper that looks at Europa's ice shell:
Constraining the thickness of Europa’s water–ice shell: Insights from tidal dissipation and conductive cooling
From the abstract:

...when considering the present thickness of Europa’s ice shell, if tidal heating is included at a global dissipation rate of ∼1 TW, the shell is found to be, on average, approximately 28 km thick. However, if this dissipative heating is solely restricted to the shell, the local rate of heating may vary significantly due to crustal compositional heterogeneities and it is shown that this process may, in turn, produce thermal maxima in the crust, which could lead to local melting and structural instabilities...

So, with the combined information from these sources, I would stand by a figure of somewhere between 15-28 km. The reason I don't go any higher is because I don't see an error bound or range of values in the paper's abstract - it just says "approximately 28 km thick" - and the full paper is unfortunately behind a paywall. As the paper's abstract says, there are local heating phenomena that could occur and cause the crust to vary in different places, which could explain why the figure isn't that precise. The reason I don't go any lower, on the other hand, is because that NASA FAQ didn't specify any exact values for their mention of "few kilometers thick", and didn't provide much explanation as to why those said scientists supported that idea.


How far above the Earth's surface does interplanetary space begin? The Kármán line is often cited as the beginning of outer space.  The Voyager probes have identified the boundary between the heliosphere and interstellar space.  Is there a agreed upon definition of the boundary of interplanetary space?  Is it the Earth's magnetopause?

It it wherever is convenient for the article you are writing. There is no single consenus definintion of "interplanetary space".  If you are in orbit around a planet then you are not in interplanetary space, if you are between planets, then you are.  But there is no substantial change in the matter filling space as you move from planetary orbit to interplanetary trajectory

So you can use any reasonable point that fits the article you happen now to be writing


Why aren't there blue elliptical galaxies or yellow spiral galaxies? Elliptical galaxies are universally old and yellow; about three-quarters of all ellipticals have no significant star-forming gas or dust left, and even the quarter or so of ellipticals that are still forming stars do so at much lower rates (and with gas and dust reservoirs that are far closer to depletion) than is the case in irregular and spiral galaxies.

In contrast, spirals and irregulars have large amounts of gas and dust, breed stars like rabbits, and, as a result, contain huge quantities of young, bright, blue stars.

Why are elliptical galaxies so universally old, yellow, and star-forming-material-starved, while other types of galaxies are young, blue, and rich in gas and dust?  Shouldn't there be a bunch of young, blue, gas/dust-rich ellipticals complementing their yellowed seniors, and many yellow, geriatric, no-longer-star-forming spirals and irregulars to go along with the young blue star factories?

Galaxies consist of dark matter, stars, and gas. While gas is "collisional", i.e. it may interact hydrodynamically and cool, dark matter and, effectively, stars are collisionless. Hence, it is relative easy for an originally more or less spherical, gas-rich galaxy to collapse along the axis of rotation, while centrifugal forces prevents to collapse in the plane. In contrast, a spherical gas-poor galaxy will tend to keep its shape.
Spiral galaxies are blue
That is, a gas-rich galaxy will tend to settle down into a flat disk. Being gas-rich, such a galaxy will keep forming stars, and a spiral pattern will form.
Stars come in all sizes, but the most massive ones dominate the total luminosity (since $L\propto M^4$), and since massive stars shine with very energetic light, they are bluish/white.
Elliptical galaxies are yellow
Elliptical galaxies are the result of a merger between two or more galaxies. If the merging galaxies are very different in size, the smaller will just be "eaten" by the larger. But if the galaxies are similar-sized (and don't have too similar angular momenta), the event can be quite dramatic:
Tidal forces mess up the shape, gas is stripped from the galaxies, and collisions of gaseous clouds ignite massive starbursts. Such major mergers are quite efficient at suppressing angular momentum, allowing large amounts of gas to reach the galactic centers, feeding the central supermassive black hole leading to vigorous quasar activity (e.g. Hopkins et al. 2008).
Both the gas stripping, the radiation pressure from newborn, massive stars, the feedback from their fast and explosive deaths, and the active galactic nuclei, cause the merged galaxy to lose its gas. As the massive, blue stars die out, and with no gas to form new stars, only the smaller, yellowish/orange/reddish stars are left. The galaxy is said to be "quenched", ending up "red and dead".
Why not the opposite?
So the answer to your question is: The reason spiral galaxies are not red is that they keep forming new stars, the most luminous of which are blue.
And the reason elliptical galaxies are not blue is that they're unable to form new stars, so only the red ones are left.
Addendum thanks to Peter Erwin:
Spiral galaxies may eventually deplete most of their interstellar gas (partially through outflows) and stop forming stars. This cause them to gradually change color toward yellow. However, without the feedback from star formation, such a galaxy will also tend to lose their spiral structure, ending up instead as a lenticular (or "anaemic") galaxy (see e.g. Elmegreen et al. 2002).


What is the final destiny of a neutron star? As I understand, neutron stars are born as extremely bright, extremely fast spinning cores of stars dying in a supernova. However, several websites tell me that within a course of a few years, the surface temperature of a neutron star falls from several trillion kelvin to only a few million kelvins. Furthermore, with the passage of time, the spinning speed of the neutron star also decreases considerably.

This raises the question: what is the final destiny of a neutron star? Does it always stay so horribly magnetic, hot and fast spinning or does it keep degrading into some form of cold, extremely dense star core with a much weaker magnetic field or do some of its features (specially the magnetic field strength and spin) stay at hightened levels forever (or at least several hundreds of billions of years)?

Neutron stars have extremely small heat capacities. That is because they consist largely of degenerate fermions and the heat capacity is further suppressed if, as expected, those fermions are in a superfluid state.

This has (at least) two consequences:

(a) they cool down extremely rapidly - neutrino emission processes are highly effective, in the first $10^5$ years or so of a neutron star's life, at reducing its interior temperature to a few $10^7$ K and the surface temperature to $<10^6$ K. After that, the dominant cooling process is photons emitted from the surface ($\propto T^4$) and neutron stars rapidly fade from view thereafter.

(b) However, the low heat capacity also means that it is easy to keep a neutron star hot if you have any way of adding energy to it - such as viscous dissipation of rotation by friction, accretion from the interstellar medium or ohmic heating by magnetic fields.

No isolated neutron star surfaces have been measured with temperatures much below $10^6$ K - i.e. all observed isolated neutron stars are at young ages. The situation is summarised in section 5.7 of Yakovlev & Pethick (2004). Without any reheating, a neutron star would reach 100K in only a billion years - this is already utterly invisible. The reheating mechanisms must play some role for older neutron stars, but as Yakovlev & Pethick state:  "Unfortunately, no reliable observational data on the thermal states of such stars are available". In conclusion, nobody really knows at the moment what the long-term ($>10^6$ years) fate of neutron stars is in terms of their temperature.

The situation with regard to spin and magnetic field is more secure. There are not the same mechanisms available to spin-up an isolated neutron star or regenerate their magnetic fields. Both are expected to decay with time and indeed the spin-down rate and magnetic field strength are intimately connected, because the spin-down mechanism is the emission of magnetic dipole radiation. The magnetic field decays through the generation of currents that then ohmically dissipate (providing a source of heat) or perhaps more rapidly via currents generated by the Hall effect or through ambipolar diffusion.

For pure magnetic dipole radiation, one predicts $\dot{\Omega} \propto \Omega^3$. For typical surface magnetic field strengths of $10^8$ T, pulsars spin down to periods of around a few seconds in less than a million years, at which point the "pulsar activity" switches off and we can't see them any more, unless they are in binary systems and accreting matter in order to spin them up again. Unfortunately, there is very little observational evidence to pin down how fast magnetic fields decay (because we don't see old, isolated neutron stars!). The decay of B-field cannot be very fast, certainly timescales are longer than $10^5$ years. Theoretical estimates of B-field decay timescales are more like billions of years. If this theory is right then neutron stars would continue to spin down very rapidly even after the pulsar mechanism has ceased.


If nothing travels at the speed of light, except light, how can a black hole also pull light into itself? Is it because it has super massive gravity and energy? Or is it some sort of an anti-light component in it because it used to be a star?

Gravity is a force, and it need not have a "speed"1.

A gravitational body sets up a gravitational field around itself. Note that by the time the light wave approaches the body, this field has already been set up. The gravitational force does not need to "reach out" and "catch up to the light" — it's already there.

In fact, it's not even really a force (though we can treat it as one to an approximation). Gravity bends the fabric of spacetime around it, messing with the meaning of a "straight line". From different reference frames, different lines appear straight. To the light wave, the inspiraling path seems "straight" and thus it follows that path. To an external observer, the light wave is not going straight.

1. It does, but that's the speed at which changes to the field propagate, not the speed at which it "catches up" with other things. Changes in the gravitational field propagate at lightspeed.


Could I create a staircase to the moon and walk to it? Assuming one could create a staircase and walk to the moon, would escape velocity still be a factor in leaving the Earth's gravitational pull? I'm not a physicist but am just curious to know.

A space elevator has been proposed. With currently available technology it would be very expensive to build one, but not necessarily impossible.

Assuming the staircase or elevator being fixed at some point on Earth near the equator, the Earth's gravitational pull is compensated by centrifugal force at a height of about 36,000 km, in the geostationary orbit.

That's the height you would have to climb to, before you could slide "down" to the Moon along an even more expensive construction stepwise adjusting your velocity to that of the Moon. The total trip would be at least about 400,000 km, that's ten times around the Earth.

Your weight would decrease, the higher you climb from ground to the 36,000 km height. Assuming you make 3 km of height each day, it would take you about 34 years to reach geostationary orbit.

Then sliding down with 10 m/s would take at least one more year to reach the Moon. Hence, it's possible with a one-way trip of about 35 years, assumed the technical infrastructure is provided.

But it looks easier to take an appropriate vehicle.


Ecliptic line plot in a map Is there any url that can give me  data of ecliptic line that i can plot in a map. i.e i need earth's latitude and earth's longitude of ecliptic line for a specific time and date. Right now i only have Sun's declination data and its longitude, with which i can plot only  one point, but ecliptic line is like sine wave as i read on the internet. How can i get the data of complete ecliptic line for a given day and time. I am kind of new to this, kindly apologise if my question is wrong and show me the right way.

The ecliptic is a plane that is inclined 23.43 degrees (approximately, insert your more accurate value as needed). One position on the ecliptic is 0 hours Right Ascension, 0 degrees Declination (0,0). If you plot that on the Earth at the date and time you have, the rest of the ecliptic's path on the Earth can be calculated as follows:
$$\tan(23.43) \sin(longitude-longitude_{equatorcrossing}) = \tan(latitude)$$

where $longitude_{equatorcrossing}$ is the longitude where the ecliptic at 0 hours Right Ascension crosses the equator.

As a check, the latitude should be 23.43 degrees when the longitude is 90 degrees east of the (0,0) longitude point, and the latitude should be -23.43 degrees when the longitude is 90 degrees west of the (0,0) longitude point. In other words, if (0 Right Ascension, 0 Declination) corresponds to 15 degrees east longitude, then 105 east longitude is where the ecliptic is at the maximum northern latitude, and 75 degrees west longitude is where the ecliptic is at the maximum southern latitude.


Changes to Earth's orbit Any time a spacecraft comes in close proximity to a planet and if the spacecraft has the right angle then it is able to use the planet's velocity to move itself further into space.

According to Newton's 3rd law: every action has an equal reaction.

In this case when the spacecraft uses for example the Earth's gravity to speed up, Earth will move towards the spacecraft. Earth's orbital change will be very small because the spacecraft's mass is small compared to Earth's mass but what if a big asteroid comes in close proximity or what if we use Earth's gravity to catapult our spacecrafts and keep doing it for an extended period of time.

What could happen in this case? Could that have a dramatic impact to Earth's orbit?

Gravity assists such as this are a form of elastic collision. There's a bit of number crunching here (hopefully no mistakes!), so you'll want to be familiar with the basics of momentum, kinetic energy, and the conservation thereof.

  Question: If Ceres (the largest known asteroid and nearly 500 km in diameter) used Earth to perform a gravity assist to increase its own velocity, by how much would this slow the Earth down, and how much larger would Earth's orbit become?

The orbital speed of Earth around the sun is $U = 29.8~\mathrm{km~s}^{-1}$. So at a mass of $$M = 5.97\times 10^{24}~\mathrm{kg},$$

it has a kinetic energy of

$$K = 2.65\times 10^{33}~\mathrm{J}$$ and momentum $$P = 1.78\times 10^{29}~\mathrm{kg~m~s^{-1}}.$$

So let's say Ceres is performing a gravitational slingshot as in the simple diagram below. Ceres has a mass  $m = 9.47 \times 10^{20}~\mathrm{kg}$. It approaches Earth at velocity $v$, and after the slingshot its final velocity is (up to, for a low-mass object) a velocity of $2\times U+v$.

The total momentum of the system must be conserved. Ceres has changed direction and thus gained a significant amount of momentum in the leftwards direction: the same momentum that Earth must then lose. Kinetic energy is also conserved. So, we have a system of equations, where the subscripts i and f are initial and final momenta and velocities. M and U are the mass and velocity of Earth, m and v are that of Ceres.

$$MU_i^2 + mv_i^2 = MU_f^2 + mv_f^2$$

which says that the sum of the initial kinetic energies of the two objects must equal the sum of the final kinetic energy. We also have conservation of momentum:

$$MU_i + m\vec{v}_i = MU_f + m\vec{v}_f $$

Solving these equations, the solution is

$$v_f = \frac{(1-m/M)v_i + 2U_i}{1-m/M} $$

If Ceres approached Earth at $v_i = 30~\mathrm{km~s}^{-1}$, I get a solution of $v_f = 89.6~\mathrm{km~s}^{-1}$ - even for such a massive object, the $v_f \approx 2U+v$ approximation is extremely good. This means that Ceres' velocity has nearly been tripled by the gravity assist.

So, the final momentum of Earth is

$$MU_f = MU_i - mv_i - mv_f = 1.78 \times 10^{29}~ \mathrm{kg~m~s^{-1}} $$

  In fact, Earth's linear momentum will only decrease by $mv_i + mv_f =
 1.13 \times 10^{23} ~\mathrm{kg~m~s^{-1}}$. From this change in momentum and Earth's mass, we find its orbital velocity decreases by
  $0.019~\mathrm{m~s}^{-1}$.

  Approximating a circular orbit (using $r=GM_{sun} / v^2$), Earth's
  orbit widens by 190 km. Sounds like a lot, but bear in mind that's 190
  km out of 150 million!

Ceres is many orders of magnitude larger than any satellite that we could launch. So we could never practically use spacecraft to change our orbit significantly, and even an enormous near-miss asteroid would be of little consequence. But, it hasn't stopped some from trying!


What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere? Overview:

Taking a planet Ganymede for example (moon, technically I suppose). Ganymede is huge (well, in contrast to other bodies) and I've noticed a pattern between larger bodies and atmospheres. Usually it seems that the larger a body is, the more likely it will have an atmosphere (I know this isn't true, but it definitely seems true, with obvious exceptions). Even when playing Kerbal space program I came up on Vall expecting to be slowed due to its huge size. Big mistake, slammed into the surface. So, if size isn't the end-all-be-all of why a planet would have an atmosphere, what really causes it? Why wouldn't something like Ganymede, doubly protected by it's own magnetic field and Jupiter's field not have an atmosphere? What causes the planet to lose it once an atmosphere is created?

You can ignore my overview rambling, I suppose my one-liner focused question is:

What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere?

Technically speaking, every planet has an atmosphere (though most drop Mercury from the list because it's so thin). Let's start with the big guys, which are pretty straightforward. Gas giants just keep their atmosphere from when they gathered the remnants of gas from the nebula that their parent star formed from. If you're like the gas giants in our solar system, you usually keep this atmosphere for a very long time because you have a strong magnetosphere and you also are fairly far from your star. Your star also isn't particularly powerful. Some exoplanets are categorized as hot Jupiters because they are similar to jupiter in composition but are orbiting very closely to their star. If this is the case, despite the strong magnetosphere, their star may quite literally boil their atmosphere off, either adding it to the star's mass, or blowing it away, sort of like a comet. Eventually they may become what we call chthonian planets, which are hypothetical planets that are the rocky cores left over from boiled gas giants.

How about rocky planets like Earth? They get their atmosphere from a process called outgassing, which basically means that volcanoes burp out gases. Earth is unique in that we believe our atmosphere was at least partially created from comets and wet asteroids. Venus burped out too much greenhouse gases and got a hellhole for an atmosphere. At least it rains there...rains sulfuric acid...and even that boils before it reaches the surface! Mars probably had a thicker atmosphere than today, as evidenced by big volcanoes like Olympus Mons. What happened to it? Well first, Mars doesn't have a very strong magnetosphere because it is simply smaller than the Earth. Because it's smaller its core cooled down faster, and its mantle no longer supported a good magnetic field. More charged particles means a higher chance of atmospheric stripping, which Mars suffered from at least partially. However, we think that a large impact probably knocked a lot of Mars's atmosphere off. Finally, that leaves us with Mercury. Mercury is pretty small and very close to the Sun. It just can't retain an atmosphere because its gravity is weak. The same can be said for most moons in our solar system, as they are typically smaller than Mercury. They are simply too small to retain an atmosphere.

Wait, then how come some moons have atmospheres and others don't? Well the ones that don't, as I said before, are too small to retain one, and they also have to deal with interactions with the solar wind, atmospheric stripping, and interactions with their planet. Some moons have volcanic activity, so their thin atmosphere can be replenished with burped up gases. In fact, some may have weather! I believe Io and Triton both have weather systems, despite having very thin atmospheres. Titan has the thickest atmosphere of the bunch, with one that is about 50% denser than Earth's. Thus, it also has complex weather systems. But why? Titan is similar to Ganymede in structure. Why does Titan get to have an atmosphere? Short answer: we don't fully know why. We think it might've been comets, but we really aren't 100% sure on anything at this point. Go check out the wikipedia article for more.

So to answer your question, atmospheres are usually formed from volcanic activity or, for gas giants, retained gas from nebula remnants. Atmospheres are lost due to low gravity, high temperatures, solar wind, atmospheric stripping, large impacts, and planet-moon interactions.


Gamow Energy, Peak, Width and Window? I am trying to wrap my head around Gamow energy and it’s various terms.
To put it simply I understand higher Gamow energy reduces the chance of penetration relating to the Coulomb barrier.
For the width/window would it be fair to say that a higher value indicates a bigger window so therefore more chance of fusion occurring?
To be clear i am not asking for equations or help with any specific problem sets in nuclear fusion but I hoped some more knowledgeable people than myself could guide me on some simple understanding of the process.
I hope that makes sense?
Thank you in advance.

The likelihood of a reaction occuring at a given energy is a product of the number of particles with that energy (the Maxwell Boltzmann distribution), which decreases with energy, and the tunneling probability, which increases with energy. This product forms the Gamow window.
The total reaction rate (for a non-resonant reaction) is proportional to the area under the Gamow window - i.e. the product of its width and height.
For resonant reactions, that occur over a narrow energy range, all that really matters is how close to the peak of the Gamow window that energy is.


What Makes Stars Hot? Simple question, but can't seem to find the answer anywhere. It can't be nuclear fusion because nuclear fusion occurs as a result of the heat. And it can't be because of gravity because it's believed that the core of black holes are nearly absolute zero. So what makes stars hot?

Stars do not get hot because of nuclear fusion, they become hot enough to sustain nuclear fusion and this process maintains their temperatures. Nuclear fusion actually stops a star getting hotter.

Protostars (before nuclear fusion) get hot because of a well known statistical relationship between the gravitational potential energy of a gas and the internal kinetic energy of the particles that make up the gas. [In an ideal gas, the kinetic energy of the particles is directly proportional to the temperature of the gas.] This is known as the virial theorem, which says that twice the summed kinetic energy of particles ($K$) plus the gravitational potential energy ($\Omega$, which is a negative quantity for a bound object) equals zero.
$$ 2K + \Omega = 0$$

Now you can write down the total energy of the system as
$$ E_{tot} = K + \Omega$$
and hence from the virial theorem that
$$E_{tot} = \frac{\Omega}{2},$$
which is also negative.

If we now remove energy from the system, for instance by allowing the gas to radiate away energy, such that $\Delta E_{tot}$ is negative, then we see that
$$\Delta E_{tot} = \frac{1}{2} \Delta \Omega$$

So $\Omega$ becomes more negative - which is another way of saying that the protostar attains a more collapsed configuration.

Oddly, at the same time, we can use the virial theorem to see that
$$ \Delta K = -\frac{1}{2} \Delta \Omega = -\Delta E_{tot}$$
is positive. i.e. the kinetic energies of particles in the gas (and hence their temperatures) actually become hotter. In other words, the gas has a negative heat capacity. But a hotter temperature usually means more radiation is produced and if the energy losses continue, then so does the collapse.

This process is ultimately arrested in a star by the onset of nuclear fusion. This replaces the radiative losses with nuclear energy and the star attains a quasi-equilibrium that lasts as long as it has nuclear fuel to burn.


Will LSST make a significant increase in the rate of astronomical event alerts? The NPR news article and podcast New Telescope Promises To Revolutionize Astronomy updates the status of the "Large Synoptic Survey Telescope under construction on Cerro Pachón in Chile".

NPR's Joe Palca's piece includes sound bites from astronomers, including Caltech astronomer Mansi Kasliwal:

  PALCA: Kasliwal says although LSST will detect these events, other telescopes are better suited to study them in detail. So the plan is to send out an alert to other telescopes when LSST sees something interesting. Of course, that means the other telescope has to drop what it was doing, but Kasliwal says it will be worth it.

Isn't the LSST expected to "see something interesting" quite often?

Due to the size and scope of LSST's surveying capability which is due to it's huge corrected field of view and huge focal plane array and huge image processing and event detection capacity, there is a potential for a significant increase in the rate of alerts generated and sent to observatories and to astronomers' cell phones (one notable example of such things).

Question: Will LSST make a significant increase in the rate of astronomical event alerts? Has there been an estimate of how the overall rate will change when it comes online?

I'm wondering if astronomers will get woken up more often by their phones, or if some observatories will have to make the decision to promptly change observing schedules or not much more frequently.

Yes. The estimates are that LSST will produce about 10 million alerts per night (LSST Alert Distribution presentation) which will be at least a factor of 5x greater than the amount coming from ZTF currently. ZTF is an approximately 10% scale model of what the LSST alert stream will look as there is about 5x fewer alerts and the alert packets contain about 50-60% of the information that will be in the LSST alerts. The ZTF alert stream (Patterson et al. 2019) also uses the same software (Apache Avro, Spark and Kafka) as LSST will use, just running on less hardware.

These alerts (in Avro serialized binary data format) will not be sent directly to astronomers (and certainly not to cell phones), the plan is to send the alerts to a few "brokers" that will do additional filtering, classification and cross-matching to other catalogs to give additional information and context to the alert (which is basically a "this source on the sky changed in brightness by X amount or appeared"). There are several brokers in various stages of development of which the leading examples are ANTARES and Lasair.

Astronomers, or more likely their software systems, will subscribe to these brokers and add filters for the alerts to subset the stream down to the particular types of objects and science that interest them. These will most likely go into their own databases of interesting targets and astronomers or their software will decide on the small number of most interesting ones that will be sent out to other telescopes, some robotic for rapid response science case, some traditionally scheduled. The large number of alerts and triggers for follow-up has the potential to be very disruptive to the prior way of operating observatories, which is one of the reasons why we are working on system like the Astronomical Event Observatory Network (AEON) to handle and coordinate this.


What ever happened to the "Small Comet Theory" of water influx on present-day Earth? This comment (below the question Is there any new water formed on, or brought to, Earth?) links to the 2002 Article in the New Scientist Was all the Earth’s water brought here by comets? which states:

Evidence that there is a constant rain of small comets, each a loose snowball held together by a carbon shell, first came in the 1980s from satellites that looked down on Earth and observed “atmospheric holes”- dark spots 50 to 100 kilometres across- that were appearing in the dayglow of the atmosphere. The dayglow is ultraviolet light emitted when sunlight interacts with the oxygen in the upper atmosphere. Water is the only common substance that could create such holes, and each could be made by a small comet containing between 20 and 40 tonnes of water that broke up into a cloud of vapour as it hit the atmosphere. The case for a cometary origin for the atmospheric holes grew in the 1990s when the trail of a cosmic snowball was picked up as it vaporised over the Atlantic by the visible imaging cameras on polar satellites.
From the number of atmospheric holes, it’s possible to calculate that the atmosphere is being hit by a 20 to 40-tonne snowball comet every 3 seconds. According to Frank, this is enough to add an inch of water to the Earth’s surface every 20,000 years. Over the 4.5-billion-year lifetime of the Earth they could have provided all the water in the oceans. (emphasis added)

The article also links to this University of Iowa page smallcomets.physics.uiowa.edu/.
Is this still believed to be happening? Have there been additional observations? Is the earth receiving water in the form of 20 to 40 ton comets ariving every three seconds?
Question: What ever happened to the "Small Comet Theory" of water influx on present-day Earth? Has it been discounted, modified, verified?

The "small comets" buzz was temporary and has been completely superceded by subsequent evidence that it was all noise in the detectors.  A careful investigation that produced a null result looking for the comets is published in http://adsabs.harvard.edu/abs/2000JGR...10524907M.  The moral is, beware of the "1 sigma effect," where a researcher sees a hint of a signal and is beguiled into seeing it as something real.


The compatibility of the Grand Tack hypothesis with the "core-warping impact" theory of Jupiter's diffuse core In recent years, the Juno mission revealed that Jupiter's core was much more diffuse than astronomers had expected.
One theory is that "within a few million years" of its formation, Jupiter experienced a head-on collision with a planetesimal of about $10M_{⊕}$, adding a lot more mass to its core from the silicate planetesimal, but also causing the core's contents to be broken up and mixed with the inner envelope.
The models used in this theory placed Jupiter at a distance of 5.2 AU from the Sun, which is approximately the same as its semi-major axis in the present day. The planetesimal in question would be at the higher end of the valid mass range for a Super-Earth.
Now, according to the Grand Tack theory, Jupiter originally formed at a distance of $\approx 3.5$ AU and migrated inwards toward the Sun, before gravitational interactions with Saturn caused the two planets to move outward and brought Jupiter to its present-day orbit.
Saturn itself would have formed at a distance of $\approx 4.5$ AU, increasing in mass from $30M_{⊕}$ to $60M_{⊕}$ during the first $10^5$ years of Jupiter's inward migration, before starting its own inward migration. This would have been much faster than Jupiter's, enabling Saturn to "catch up" and the gravitational interactions as described to then occur.
The original formations would all have occurred over a timescale of $\lessapprox 6Myr$, maybe closer to $\approx 3Myr$. (The original paper describing the Grand Tack refers only to "a few Myr"; the figures here are based on "Disc Frequencies and Lifetimes in Young Clusters", which it cites in support of this.) The inward and outward migrations would then have occurred in an 800,000 year time period (see Figure 1 of the Grand Tack paper)
(Incidentally, the cores of Uranus and Neptune are each $\approx 5M_{⊕}$ at the start of this migration, increasing to values $> 10M_{⊕}$ at the end of it.)

So far, the two models look quite compatible with each other, with the collision occuring after Jupiter settles into its 5.2 AU orbit. But there is one detail I'm not sure about. Here comes the question:

Descriptions of the Grand Tack theory describe Jupiter as scattering early planetesimals as its gravity disturbed their orbits. Some collided with each other, some were propelled into the Sun... Is it more likely that one of these collided with Jupiter itself at the head-on angle needed for the core-warping collision? Which would mean the impact occurred before Jupiter reached its final orbit.

In addition...

The "few million years" timescale is quite vague. Does anyone know any additional detail which might suggest the impact occurred prior to the Grand Tack's timeframe?

References (not paywalled):
Haisch Jr, K. E., Lada, E. A., & Lada, C. J. (2001). Disk frequencies and lifetimes in young clusters. The Astrophysical Journal Letters, 553(2), L153.
Walsh, K. J., Morbidelli, A., Raymond, S. N., O'Brien, D. P., & Mandell, A. M. (2011). A low mass for Mars from Jupiter’s early gas-driven migration. Nature, 475(7355), 206-209.
Liu, S. F., Hori, Y., Müller, S., Zheng, X., Helled, R., Lin, D., & Isella, A. (2019). The formation of Jupiter's diluted core by a giant impact. Nature, 572(7769), 355-357.
Guillot, T. (2019). Signs that Jupiter was mixed by a giant impact.
with accompanying articles:
Wall, M. (2017). More Jupiter Weirdness: Giant Planet May Have Huge, 'Fuzzy' Core. (space.com)
Weitering, H. (2018). 'Totally Wrong' on Jupiter: What Scientists Gleaned from NASA's Juno Mission. (space.com)
(2019). A core-warping impact in Jupiter's past? (Astronomy Now)
A paper cited in earlier versions of this question but which turned out to be incompatible with the Grand Tack theory:
Pirani, S., Johansen, A., Bitsch, B., Mustill, A. J., & Turrini, D. (2019). Consequences of planetary migration on the minor bodies of the early solar system. Astronomy & Astrophysics, 623, A169.
with accompanying article:
Jupiter's unknown journey revealed

I think this question might trigger more an open-ended discussion, than a definitive answer, but let me try my take on it.
Firstly,

Is it more likely that one of these collided with Jupiter itself at the head-on angle needed for the core-warping collision?

In order to collide, a co-orbital configuration is more favourable than a orbit-crossing encounter at high eccentricity (as seen in Liu et al., extended fig. 2), but their 'high angles' don't quantify high eccentricities properly. This is because in their simulations, all planetesimals are initialized at circular orbits, with initial distances of 5-10 mutual Hill-radii (their k parameter).
Why I think they didn't properly compare apples with apples, is because usually an eccentric planet would have more kinetic energy than allowed in their simulations, reducing the collision cross-section.
This would lead to drastically reduced impact rates at high angles, compared to the data they show. Furthermore in the early scattering scenario that you propose, most planetesimals would be of too low mass to trigger the core mixing.
The mass required for the impactors is, I think, the major point why the two scenarios are not compatible. The initial conditions for the impact scenarios are those of 5 densely packed 10 $m_{\rm oplus}$ planets, of which one undergoes runaway gas accretion and becomes Jupiter. Those conditions are explained to arise due to oligarchic growth. For the 5 planets, this means that 50 $m_{\rm \oplus}$ are packed in a region of 5 AU size. This is already 1/3 of a median class 0 disc mass, and it necessitates a 100% high efficiency when translating pebbles to planetesimals to planets.
Seeing those numbers lets me strongly doubt the realism of the initial conditions.
Concerning the other part of your question,

The few million years" timescale is quite vague. Does anyone know any additional detail which might suggest the impact occurred prior to the Grand Tack's timeframe?

The few-million years are coming from a certain amount of cooling and compactification that is required for the impact to successfully mix Jupiter's envelope. Hence, the impact should not have occurred prior to the Grand Tack.
Taken both those things together, the cooling requirement and the mass/compactness of the initial conditions necessary, I don't think both scenarios are compatible.


Hubble Space Telescope (HST) observing modes I have been reading some literature on using the HST to take spectra of exoplanet atmospheres when exoplanets transit in front of their stars. The literature mention two modes of observations, 'spatial scanning' and 'staring', in order to obtain spectra of the atmospheres of these exoplanets.

Can anyone explain conceptually what the difference is between these two modes?

"Staring mode" is the traditional way of simply pointing the telescope and exposing the detector.

In "spatial scanning mode", you slowly move the telescope, so that the image is slightly smeared out in the spatial direction.

The figure below shows the principle: On the left is how a regular image is taken; in the middle, a regular spectrum is taken in staring mode, and in the right panel, the detector is moved during the exposure, resulting in the spatial scanning mode. The farther you scan spatially, the brighter and object you may observe before over-exposing, but the downside is that more objects will start to overlap, ruining the spectra.

Left panel: "Imaging"; each observed object shows up on the detector at its corresponding position on the sky.
Middle panel: "Staring mode"; when a grism (sort of like a prism) is inserted, the light from each object is spread out according to its wavelength, producing a spectrum.
Right panel: "Scanning mode"; if the detector is moved slightly during exposure, the spectra are spread out perpendicular to the wavelength direction, effective using a larger number of pixel to collect the photons in. However, if the detector is moved too much, the spectra begin to blend.

This picture, from HST's WFC3 handbook shows the difference in a real observation: It's quite small, but particularly for the "$0^\mathrm{th}$ object" (on the left) you can see that in the staring mode it's over-exposed, creating diagonal spikes, which are reduced in spatial scanning mode:

In the picture, the objects are located in the vertical direction, while their spectra are dispersed in the horisontal direction. You can see that in the spatial scanning mode, the objects are slightly broader vertically.

The advantage of the scanning mode is that you can expose for a longer time, without the detector getting saturated (which can be a problem for bright stars). When you expose for longer, you collect more photons and hence obtain a higher signal-to-noise ratio (S/N), meaning that your observation becomes more "accurate". The figure below (from these notes) shows the advantage of a high S/N:

Two spectra of the same object, with the right having been exposed 9 times longer than the left, resulting in a $\sqrt{9}=3$ times higher S/N. Determining for instance the central wavelength of this spectral line becomes much more accurate with higher S/N.

Note that both modes refer to so-caled slitless spectroscopy. A way to avoid the spectra to overlap is to insert a "slit" which allows only the light from a narrow (of the order of an arcsecond) part of the sky to be observed. In that way you will only obtain spectra of objects lying exactly along the slit, but in return you will have a "cleaner" spectrum.

Note also that you can obtain the same effect by dividing your observation into multiple staring mode exposures and then add them together. Scanning mode is a more practical way of not having to do this.


Is there an IRAM satellite that measures thermal radiation at 250 GHz, or was this a ground-based instrument? The Nature Research Letter A Pluto-like radius and a high albedo for the dwarf planet Eris from an occultation (also here and here) says about half-way through:

We now reassess Eris’ surface temperature in the light of our new results. Measurements by the Spitzer22 and IRAM11 satellites imply disk-averaged brightness temperatures of Tb=30±1.2 K and Tb=38±7.5 K at 70 and 1,200 mm, respectively.

11Bertoldi,F.F.Altenhoff,W.Weiss,A.,Menten,K.M.&Thum,C.Thetrans-neptunian object UB313 is larger than Pluto. Nature 439, 563–564 (2006).

22Stansberry, J. et al. in The Solar System beyond Neptune (eds Barucci, M. A., Boehnhardt, H., Cruikshank, D. P. & Morbidelli, A.) 161–179 (Univ. Arizona Press, 2008).

The caption of Figure 1 of reference 11 (Bertoldi et al. 2006) mentions:

The 117-element MAMBO-2 camera has an effective frequency for thermal radiation of 250 GHz, a half-power bandwidth of 80 GHz (210–290 GHz), and a beam size of 10.7 arcsec (corresponding to 760,000 km at a distance of 96 AU).

and I think it refers to

Welcome to the home of MAMBO; the "Max-Planck-Millimeter-Bolometer"
IRAM/MAMBO2.250GHz

Question: Is the use of the word satellites in "Measurements by the Spitzer22 and IRAM11 satellites..." correct, or does the latter refer to terrestrial measurments?

The Nature paper by Bertoldi et al. (2006) says:

Our millimetre observations were performed with the Max-Planck Millimeter Bolometer (MAMBO-2) array detector at the IRAM 30 m telescope on Pico Veleta, Spain.

This is a ground-based telescope in the Spanish Sierra Nevada mountains at 2850m to try and get above as much of the precipitable water vapor, which has a very large effect on sub-mm observations, as possible. Its website is here
Since it's common to refer to the Spitzer and its predecessor, the IRAS mission, when discussing far-infrared observations, I suspect the authors typed "IRAM", thought "IRAS (satellite)" and made a simple typo that was not caught before publication.


Relationship between metallicity and color? Should Pop. I stars be blue? I have found in numerous places such as this website: http://burro.astr.cwru.edu/Academics/Astr222/Galaxy/Structure/metals.html or in "Introduction to Stellar Astrophysics" by B. Carroll, that state:

  Population I:
  metal rich [Fe/H] > -1
  disk stars
  open clusters
  metal rich makes stars redder

  Population II
  metal poor [Fe/H] < -1
  halo stars
  globular clusters
  metal poor makes stars bluer

And I also found (from: http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html):

  Population I stars include the sun and tend to be luminous, hot and young, concentrated in the disks of spiral galaxies. They are particularly found in the spiral arms. With the model of heavy element formation in supernovae, this suggests that the gas from which they formed had been seeded with the heavy elements formed from previous giant stars. About 2% of the total belong to Population I.

  Population II stars tend to be found in globular clusters and the nucleus of a galaxy. They tend to be older, less luminous and cooler than Population I stars. They have fewer heavy elements, either by being older or being in regions where no heavy-element producing predecessors would be found. Astronomers often describe this condition by saying that they are "metal poor", and the "metallicity" is used as an indication of age.

If Population I stars are metal rich and therefore appear red, how can they be luminous, hot, and young, and concentrated in disk (the region of star formation)? Shouldn't they appear blue? How does color relate to metallicity?

Hot stars should be blue, shouldn't they? As stated here for example:

  A star’s color is critical in identifying the star, because it tells us the star’s surface temperature in the black body radiation scale. The sun has a surface temperature of 5,500 K, typical for a yellow star. Red stars are cooler than the sun, with surface temperatures of 3,500 K for a bright red star and 2,500 K for a dark red star. The hottest stars are blue, with their surface temperatures falling anywhere between 10,000 K and 50,000 K. (http://www.webexhibits.org/causesofcolor/18B.html)

Also, this question is a little similar but not quite: Metal rich or metal poor?

There is a graph: http://science.psu.edu/alert/images/SDSSmetals.jpg/image_view_fullscreen

I'm not sure if the colors there correspond to the colors of the stars, but they used blue to refer to old stars. Why?

I think the misunderstanding (which I've been asked about before) is because Population II stars are redder as a population, even though for a given mass, a main-sequence Pop II (i.e. metal-poor) star would be bluer.  That is, if you compare representative samples of Pop I and Pop II stars, the Pop II stars will on average be redder.

This because Pop II stars are generally old so the blue stars have already died as supernovae or evolved into compact remnants. So looking at the whole population, you'll see more of the longer-lived, lower-mass stars (including red giants), which are redder.

The same argument applies to galaxies that aren't forming stars appearing redder than those galaxies that are still forming stars.  Since new stars are being born, some are blue, but they live fast and die young, so unless the galaxy keeps forming stars, it will become red overall.  Again, this isn't because of the stars being redder for a given property but because the population lacks young, hot, blue stars.


How to determine when Sun/Moon is aligned to two location on Earth's surface? I would like to make a photo to Moon/Sun when it is very low on horizon, so if I make the shot far away from an object on Earth surface, it appears as big as the Sun.
But how can I calculate the right moment? Any site or app? Could NASA Horizons help? I made a GUI for it, but don't know if any quantity I can query could give me the information  I need:
http://win98.altervista.org/space/exploration/NHUGUI.html

The Photographers Ephemeris is a very nice tool to give you the direction and time of both Moon and Sunset and -rise, made for exactly that purpose: plan and time shots to be at the right place and time for an awesome scenery.
Other programmes and tools to tell you rise and set times are

Stellarium
heavens above
time and date


Is the ratio between earth's distance from the sun and the speed of light just a coincidence? I was doing some calculations to see how hard it would be to observe the speed of light and discovered an interesting correlation between the speed of light and the average distance from the earth to the sun. In my example, an object would be placed in orbit around the sun so that its  shadow would pass over the earth at the speed of light.  After doing some math, I found that an object roughly 1 km above the sun, orbiting at 2 m/s would achieve this.  I though that these numbers were a bit odd, so I took a closer look at the speed of light (299,792,458 m/s) and AU (149,597,871 km.)  I found that 299,792,458 / 149,597,871 = 2 within a .2% degree of inaccuracy.  To me, this seems too close to be chance.  Is there an established relationship here, or is this just luck?

It is a co-incidence. The units are what you need to look at, in both the speed of light and the distance of the Sun the units of length are the same (m) (your ratio is $\approx 0.002 s^{-1}$ when you use metres as the unit for the distance to the Sun). The definition of the second is in a sense arbitrary and is a consequence of how we have chosen to sub-divide the day into smaller units of time (there is nothing cosmic in dividing the day into 24 units each subdivided into sixtieths which are again subdivided into sixtieths).

If we had chosen to take as our basic unit of time $1/100000$ of a day rather than $1/86400$ the nice round answer vanishes (if my calculations are right the $\approx 0.002$ becomes $\approx0.0023$)


Any plans for TESS after it finishes northern sky survey? TESS has provided the astronomy community with a treasure trove of information.

Once TESS completes its one year survey of the northern sky, are there any plans for an extended mission?
Just seems like 27 days is too short to look for three transits of planets orbiting in the habitable zone of M dwarfs.
Approximately what percentage of planets orbiting M dwarfs in the habitable zone and transiting our line of sight would be missed by 27 day observations?
Are there plans for follow-up observations with telescopes on Earth of 1 or 2 transit events picked up by TESS to confirm planet or not?

Looks like it was a very successful mission with data that will be useful for many years to come.

NASA missions that have gone beyond their original Prime Mission lifetime, such as TESS, go into the NASA Senior Review process every 3 years. The most recent one of these was the 2019 Senior Review which reviewed Hubble, Chandra, Fermi, NuSTAR, NICER, Swift, Newton-XMM along with TESS. This review panel looks at the whole operation of the mission, whether it's doing good science, how the operations are conducted, how much lifetime there is left in the spacecraft and the cost of continuing operations.

The full panel report is available but all 8 missions were recommended for continued funding and TESS was rated as "best of the rest", just behind Hubble and Chandra (which have an enormous science legacy). The panel report is a recommendation to NASA but they normally follow the report as best they can and as the budget permits (Congress can, and occasionally does, intervene in this process by adding or removing funding for specific missions).

As detailed in the NASA response to the panel, the 8 missions were all extended for the Fiscal Year 2020-2022 but as noted, this is contingent on NASA's Astrophysics Division receiving the funding it asked for in the FY2020 Presidential Budget Request and the Congressional budgeting process. Extensions beyond 2022 are possible but depend on the results of the 2022 Senior Review. The TESS spacecraft is in a very stable orbit which allows it to maintain the orbit with minimal use of the propellant consumables on board (which normally sets the end of the mission as was the case for Kepler/K2 for example)

The panel noted:

  The expected planetary return for TESS’s extended mission is high, pushing well beyond the prime mission. Due to the longer time baseline, together with solid Kepler exoplanet demographic statistics, the team makes a strong case that the extended mission will nearly triple the detection of planets smaller than 4 Earth radii in the extended mission, compared to the prime mission, nearly triple the number of detected planets in the habitable zone, and more than triple the number of planets with periods beyond 20 days


What would Aurora look like if no magnetic field was present around the Earth? The greatest part of the Earth is protected from the Solar wind by its magnetic field. This field makes the wind particles (mainly protons and electrons) the beauty of Aurora appear.
Suppose the magnetic field wasn't present. The Solar wind particles hitting the southern and northern parts of our planet would distribute themselves over whole the Earth's surface.
How would Aurora look like?
For certain it would be not slowly swirling (following the slowly changing magnetic field lines). Will we see a sparkling sky? Or what?

I was looking through some sites and came across this - https://www.livescience.com/earth-magnetic-field.html
This clearly mentions how, if the magnetosphere surrounding the Earth weakens, the Auroras might penetrate into the Atmosphere and light up the skies closer to the equator. So, I also looked it up but, to my astonishment, I found no sources that answered this question to my satisfaction.
Though, intuition says that If there is no Magnetosphere, there should be no Aurora either.


Why was the Vera C. Rubin Observatory LSST Camera tested with a broccoli and a pinhole? The SLAC National Accelerator Laboratory video Vera C. Rubin Observatory LSST Camera (caution! annoyingly loud music, lower volume before playing) shows a Romanesco broccoli (which I'd assumed was a 3D printed mathematical surface of that name until I checked and found out that this is the name of an actual vegetable i.e. Brassica oleracea) as the object of a pinhole camera projecting its image on to the astoundingly large CCD array.
There are likely several advantages to using the pinhole rather than a well-corrected projection lens, is it possible to say what they are?
Is it also possible to elaborate on the choice of object for the camera?
See also:

slac.stanford.edu Sensors of world’s largest digital camera snap first 3,200-megapixel images at SLAC
BBC: Vera Rubin: Super telescope's giant camera spies broccoli
What is the LSST now? Where does LSST end and Vera C. Rubin Observatory begin?

Caution! annoyingly loud music, lower volume before playing:
https://youtu.be/qc_iscV1uA0

From the first article you linked:

Taking the first 3,200-megapixel images of a variety of objects, including a head of Romanesco – a type of broccoli – that was chosen for its very detailed surface structure, was one of these tests. To do so without a fully assembled camera, the SLAC team used a 150-micron pinhole to project images onto the focal plane.

Which implies:

the advantage of using a pinhole is that it allows the sensors to be tested without requiring the full camera to be assembled yet.
the Romanesco was chosen to test the sensors' ability to image an object with detailed structure.


Books describing first-hand experience of astrophysicists finding planets without skipping actual maths Are there good references of astronomical texts where we get a first-hand experience of astronomers uncovering astronomical gems using maths alone and not using observation. Observation may have been used later to validate the result.

In this book, there are several examples of astronomical jewels that were uncovered by the professionals. However the book shies away from describing the actual maths that aided the astrophysicists to make their conclusions and predictions that were later confirmed by experimenters.

However, how does a professional astronomer come to conclusions that there must be one more planet at so and so location? Are there books that uncover the exact line of thinking and maths that is involved  behind the works of the professionals?

I am not looking for a book which solves a toy problem rigorously using Newton's laws and then concludes that the 21st century astronomers use similar techniques at grander scales in their works. Though this is what is happening, i am looking for a more transparent description of how the pioneer discoverers make astronomical predictions with surety. The ideal place to look would be the research papers of these pioneers but is there a book that does that for many such achievements?

The discovery and recapture of Ceres, and the brilliant discovery and use by Gauss of the Normal distribution, the method of least squares, and a method that would now be thought of as a type of fast fourier transform to predict the position of the planet to within 0.5 degrees, based on just 6 observations is one of the great achievements of "pen and paper astronomy". Gauss, though we now think of him as a pure mathematician, held the post of Professor of Astronomy in Göttingen.

Discovery of the First Asteroid, Ceres: Historical Studies in Asteroid Research by Clifford Cunningham, ISBN 978-3-319-21777-2 tells this story in some detail, but ouch its not cheap! nearly £100 on dead trees.


What does the angle between plane of earth's orbit and ecliptic vary between? What does the angle between the plane of the earth's orbit and the ecliptic (now about 23.5 degrees) vary between, and over what time scale?

By definition, it is zero; the ecliptic is the plane of earth's orbit (or rather, the inverse: the path the Sun seems to follow along the sky).

But you're probably looking for variations in the axial tilt of the Earth. There's a lot of formulas in the article, and on the (astronomically speaking) short timescale of a few million years, it oscillates between 22 and 24.5 degrees with a period of about 41,000 years. See the graph below, the red dot is the year 2000:

(picture by Tfr000 on Wikipedia, data from Laskar (1986))


Neutrino Modelling in Friedmann Equation I'm trying to model neutrinos in the Friedmann Equation. I've covered the case of the Benchmark Model where we have matter, radiation, curvature, and the cosmological constant, Lambda. I know my coding of the Friedmann equation works because I get the correct plots at different parameters, as you'll see attached below.
Including neutrinos, the Friedmann Equation becomes
$$
\begin{eqnarray}
H(z)^2 & = & H_0^2 \Big[ (\Omega_c + \Omega_b) ( 1 + z)^3 + \Omega_\gamma ( 1 + z) ^4 \\
& + & \Omega_{DE} ( 1 + z)^{3(1+w)} + \Omega_k ( 1 + z) ^2
+ \frac{\rho_{\nu, tot}(z)}{\rho_{crit,0}} \Big].
\end{eqnarray}
$$
To solve for the energy density as a function of the scale factor (or redshift), we can solve for the energy density by the following expression for the energy density of a single neutrino species:
$$
\rho_\nu (T_\nu) = \frac{g}{(2\pi)^3} \int
\frac{\sqrt{p^2 + m^2}}{e^{p/T_\nu} + 1} d^3 p.
$$
The energy density critical is $4870$ Mev/m$^3$. Energy density of the single species can be written as function of scale factor by writing the temperature as a function of scale factor. $T$ is simply the expression shown below divided by a:

In equation 17, we can write $d^3p$ as $4\pi p^2 dp$ and $g=2$ for a neutrino species. Another thing to notice is that (17) is written in natural units where $c = h = k = 1$. I've tried to fix the units and no matter what I do, the density parameter of the neutrino species is always very small (order of $10^{-9}$) where it should be between 0.0013 and 0.007 from Ryden, Intro to Cosmology equation (7.54).
I was really hoping someone can help me with the unit conversion from the natural units to the proper units. Everything else I've figured out, I just can't seem to fix the units for equation (17).
Without neutrinos, I get the following plot consisting of various universe models, and they are correct so coding is not the problem. The problem is the unit conversion to proper SI units of (17).

Once I get the neutrinos figured out, I want to see how they affect the universe models. Any help is greatly appreciated!

The energy density of a Fermi gas is
$$\rho_{\nu}= \int \rho(p)\ dp = \int E(p)F(p)g(p)\ dp $$
$$\rho_{\nu}  = \int \left(\sqrt{p^2 c^2 + m^2 c^4}\right)\left(\exp (E/k_BT) + 1\right)^{-1} \left(g_s 4\pi p^2/h^3\right)\ dp$$
in units of energy per unit volume.

Before neutrino decoupling at $k_B T \sim 1$ MeV, the neutrinos are ultrarelativistic with $pc \gg m_{\nu}c^2$. After decoupling, the shape of the occupation index function $F(p)$ does not change - so $F(p) = \left(pc/k_BT_{\nu} +1\right)^{-1}$ in subsequent evolution.

Thus
$$\rho_{\nu} = \frac{g_s c}{h^3} \int \frac{ \sqrt{p^2 + m_{\nu}^2c^2}}
{\exp(pc/k_BT_{\nu}) +1}\ 4\pi p^2\ dp$$

I don't understand where your $(2\pi)^3$ comes from, other than to suggest that the unit system is actually $\hbar = 1$.


Closest point on Earth to a planet? I'd like to be able to find the latitude and longitude on Earth (assuming a sphere is fine) which is closest to Pluto at the current time (or where Pluto is directly overhead).

Is this possible with PyEphem?

Your question is valid and is a common operation. When computing when an object will be visible, many observers want to be able to draw a globe or map that is marked with the position from which an object like Pluto is directly overhead. Such a position on the globe at a given instant (and you do say “at the current time”) is, all other things being equal, the best-situated place from which to observe the object.

Unfortunately, the “libastro” C library that PyEphem wraps seems to only provide one instance of this concept: Earth satellites have .sublat and .sublong attributes, because it is so common to want to draw the path of a satellite on a globe or map.

But libastro does not generalize the concept. From what I can see, there is no way to generate the point on the Earth that is directly below any other Solar System object. And since I have not been in the habit of trying to extend libastro, it is likely that PyEphem will not gain this ability.

However, I have been developing a replacement for PyEphem that is written in pure Python and that I will be free to extend, called Skyfield. I will hopefully have this concept working there soon, and when I do so I plan to make it work for all objects, and not limit it to Earth satellites!

(In the meantime, as pointed out in the comments: you could try asking PyEphem for the geocentric RA and dec and, if you could adjust RA by the Earth's current hour angle, then you could turn those into an approximate latitude and longitude.)


How did Kepler "guess" his third law from data? It is amazing that Kepler determined his three laws by looking at data, without a calculator and using only pen and paper. It is conceivable how he proved his laws described the data after he had already conjectured them, but what I do not understand is how he guessed them in the first place.

I will focus in particular on Kepler's third law, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of the orbit.

I assume that Kepler was working with data about the planets only, plus our own moon, and the sun. I make this assumption because I don't think Kepler had data about other moons, comets, or asteroids, which had not been observed by telescope yet. If this is true, knowing that Neptune, Uranus, and Pluto were not yet discovered when Kepler was alive, this means Kepler had less than 9 data points to work with.

My friend claims that it is totally concievably how Kepler guessed this relationship (although he provides no method of how Kepler might have done it), and also that Kepler's observations are "not that hard".  As a challenge, I gave my friend a data table with one column labeled $x$, the other $y$, and 9 coordinates $(x,y)$ which fit the relationship $x^4=y^3$. I said "please find the relationship between $x$ and $y$", and as you might expect he failed to do so.

Please explain to me how in the world did Kepler guess this relationship working with so few data points. And if my assumption that the number of data points Kepler had at his disposal is small, is wrong, then I still think its quite difficult to guess this relationship without a calculator.

Kepler's account of how the third law came to be is as follows (Caspar p.286; emphasis mine):

  On the 8th of March of this year 1618, if exact information about the time is desired, it appeared in my head. But I was unlucky when I inserted it into the calculation, and rejected it as false. Finally, on May 15, it came again and with a new onset conquered the darkness of my mind, whereat there followed such an excellent agreement between my seventeen years of work at the Tychonic observations and my present deliberation that I at first believed that I had dreamed and assume the sought for in the supporting proofs. But it is entirely certain and exact that the proportion between the periodic times of any two planets is precisely one and a half times the proportion of the mean distances.

Although Kepler does not actually describe inspiration that led him to believe this, the curious phrasing provides a very strong clue when combined with some background biographical information:

John Napier published Mirifici Logarithmorum Canonis Descripto in 1614, which contained the then-new invention of logarithms. Kepler was aware of Napier's work by 1617 (Caspar p. 308), perhaps earlier.
Joost Bürgi published work on logarithms almost at the same time as Napier, and Kepler was similarly aware of Bürgi, even praising his mathematical abilities as surpassing most professors of mathematics.

Thus, Kepler's statement is is equivalent to saying that the data makes a slope of 1.5 on a log-log graph, a which is a very simple linear relationship on this scale.

References:

Caspar, Max, Kepler, (Dover, New York, 1993).


why was the probability of one in 10 billion per planet considered highly pessimistic by a recent study on aliens? A recent article about the existence of extraterrestrial life claimed that "unless the probability for evolving a civilization on a habitable-zone planet is less than one in 10 billion trillion, then we are not the first."
It continues to say that "a probability for civilizations to form of one in 10 billion per planet was considered highly pessimistic."
Here's a link to the actual study. The author of the article is a coauthor of the study.
Could someone go through the study and explain the source of the probabilities and their assessments (optimistic vs pessimistic.)

The main reason for this is the new data we aquired with the Kepler telescope and our findings about the goldilocks zone (where temperature is right for water to be liquid on the planet).

In the last decade a lot of discoveries, like the highly likely liquid water on the moon europe (which is way outside of the goldilock zone), the findings of amino acids in stelar clouds, etc, lead to the understanding that the chemical ingrediants and the chemical processes for life are possible in way more places than we thought.

Adding to this, the data from the Kepler telescope suggests that around 25% of all solar systems have a potentially habitable planet in some way (which makes that around 60-100 billion potential habitable planets in the milky way alone)


Redshift, velocity, distance How are redshift, (recessional) velocity and distance of a galaxy related?
I understand what v=HD states. So I assume that Hubble measured a direct relation between the distance and recessional velocity of a galaxy.

I get confused about the relation between the redshift and the other two. Is there a direct relation between the redshift and the distance and between the redshift and the velocity of a galaxy?

Welcome to StackExchange. Good question. Hubble's Law says that an object's velocity away from an observer is directly proportional to its distance from the observer. In other words, the farther away something is the faster it is moving away from us. The redshift tells how fast a star is receding from us and we can therefore get the distance. Hubble's equation states that $v\ =\ H_0\cdot D$ where $H_0$ is Hubble's constant.  It makes sense that the further away a star is the faster it has been moving.

The redshift is measured for a star and for small velocities relative to c it can be written that $z \approx \frac{v}{c}$. For larger speeds the equation is $z \approx \frac{H\cdot D}{v}-1$

A good general description is given here.


Where can I find a list of retrograde Milky Way stars? I read in papers that the Milky Way contains some retrograde stars (retrograde to the Milky Way rotation). Does anybody know where I can find a list of them, possibly including data on their distance from the Milky way center?
many thanks

The Milky Way's outer halo has many globular clusters with a retrograde orbit (about 40% of all clusters in Milky Way). One of the more prominent example include Kapteyn's star which is highly retrograde due to it being ripped from a dwarf galaxy and merging with the Milky Way.
However, the structure of the halo is a topic of an ongoing debate. Several studies have claimed that the halo consists of two distinct components. The "inner" halo consists of more metal-rich, prograde stars and "outer" halo consisting of metal-poor, retrograde stars. These findings have been challenged many times due to the argument on the topic of "duality of motion".
More readings here:

Carollo et al. 2007
Kravtsov 2001
Kordopatis et al. 2020


What would our night sky look like if the solar system was 100 light years from the centre of the Milky Way When we look at photo's of the centre of the Milky Way, it looks extremely bright due to the concentraion of star systems...

If our solar system was only 100 light years from the centre of the Galaxy, would we still have a dark night sky?

I have tried to collate some info from various other answers.
Density of stars where we are:
About 0.15 per cubic parsec
Density of stars "at the middle of the galaxy".  Unfortunately I don't know how "100 ly" affects this. It's probably about this figure.
About 50-100 stars per cubic parsec
So in this QA it's "about 500x as dense".
It would seem that this is in fact:
Not very bright. About the same amount of light overall as when there's a full moon.
However the overwhelming factor is
If one particular star was "very close" to us?
How likely would it be, in the region under discussion, that another star is "very" close to us - close enough that that one particular star is extremely bright?
The answer seems to be
No, the stars there are not so close that you'd typically be close enough to a really bright star that it would be bright.
That seems to be the bottom line.
(There's a completely pointless, if pretty, image here showing what it's like "inside a star cluster" {where? At what depth?} here  Unfortunately that image does not simply state the star density they are simulating - so it is of no help at all to this excellent QA!)
the bottom line then:
"If our solar system was only 100 light years from the centre of the Galaxy, would we still have a dark night sky?"  Answer, yes, there would just be "more stars". It would be hardly any brighter.
To improve on this answer, the next thing we need to quantify is one piece of information:
What actually is the star density at "100 ly from the center"?
Is it 50, 100, 25, 300, 72.275 stars per cubic parsec?
I don't know.

A fun point, would it be different (slightly? a lot?) if inside the galactic bar?  IDK.