What is the minimum meteor size to see a falling star? When a meteor is going through our atmosphere you will see a light beam made by the gas the meteor compressed. But what is the minimum size to see a 'falling star'? Is a 1 mm3 meteor visible?

The International Meteor Organization (IMO) defines a meteoroid as "a solid object moving in interplanetary space, of a size considerably smaller than a asteroid and considerably larger than an atom or molecule." This definition is further refined by Rubin and Grossman, who state "a meteoroid is a $10$-$\mathrm{\mu m}$ to $1$-$\mathrm{m}$-size natural solid object moving in interplanetary space." Additionally, the same authors consider a defintion for a micrometeoroid, which they define as "a meteoroid $10\hspace{2pt}\mathrm{\mu m}$ to $2\hspace{2pt}\mathrm{mm}$ in size."

According to astronomer Karen Masters from Cornell, the typical meteor has a mass of a few milligrams. The American Meteor Society (AMS) also states that the most-commonly observed meteors are roughly the size of a grain of sand. This would seem to imply that many meteors are smaller. In fact, this is also implied by the various statements by many groups, including the AMS, IMO, and IAU, that micrometeoroids commonly cause meteors. Because of this, I would argue that a meteoroid with a volume of $1$ $\mathrm{mm}^3$ would easily cause a relatively bright meteor. Statements surrounding the frequency with which micrometeoroids cause meteors would seem to imply that, on average, any body larger than $10$ $\mathrm{\mu m}$ will be observable as a meteor. These figures will vary largely depending on the entrance velocity of the object.


Active Galactic Nuclei - categorisation on the basis of observer? As we know AGNs are considered to be one of the most luminous objects to be found in the cosmos.Based on the observer, for instance on Earth,  we categorize them on the basis of the angle between the observer and the object. Suppose for instance there are two observers, one on Earth and one on some distant planet (far away from our solar system,  let it be on the opposite side of the Milky Way relative to us). Would both the observers relate an AGN belonging to the same class? In other words, can these categories be relative to their viewing? Like an object can be a QSO to us but to the other observer can the same object be a Seyfert or a Blazar or an RG?

Yes, but most probably not in your mentioned example.
You need larger distances to make a difference, I guess.
However, it is by now largely accepted a unification model for AGNs, where the difference are mostly due to the angle of line of sight.

First introduced by Urry & Padovani, then developed for further observations.

Anyway, this is only half of the truth, since the AGNs evolve, and in this evolution feedback on the host galaxy is exerted. Also, the amount of feeding/obscuring gas is varying with time (redshift) and this has to be taken into account for a correct classification.
For the jets too, it is also possible that they forms only in certain stages of the AGN life...
As usual in astronomy, things are much more complicated than simple catalogs/models !


Is the longitude assigned to sunspots spherical or projected? Sunspot maps are created from a projection of the sun, so the coordinates of sunspots (or active regions in general) are assigned according to it.

Do these coordinates correspond to actual spherical positions, or are they altered as a result of the projection?

Edit: After seeing the answer, I realized that a part of what I am asking about was not clear in the initial question. So here it is: Are the values of the sunspot coordinates as provided officially in spherical positions or in projection?

A commonly used instance of "official" records is X-ray measurements from the NOAA GOES. The text files are found here.

For example, see file for 2000. The output is documented here, under "Output File Specification".

The transformation from projected geometry to heliographic coordinates is straightforward.
As Peter Meadows explains, this can be done either mathematically, or graphically using a latitude/longitude grid template.
Since the solar equator is inclined about 7° to the ecliptic, different Stonyhurst disks are used at different times of year.

W T Thompson 2006 section 2 describes the two main heliographic coordinate systems.
Latitude in both systems is measured from the solar equator.
The central meridian of longitude in Stonyhurst coordinates is fixed facing the Earth.
The Carrington coordinate system rotates rigidly with a sidereal period of 25.38 days, approximating solar rotation at a typical sunspot latitude of ±26°.

The NOAA Space Weather Prediction Center's daily Solar Region Summary provides spherical coordinates in both forms.
Here are excerpts of two reports four days apart:

SRS Number 129 Issued at 0030Z on 09 May 2019
Report compiled from data received at SWO on 08 May
I.  Regions with Sunspots.  Locations Valid at 08/2400Z
Nmbr Location  Lo  Area  Z   LL   NN Mag Type
2740 N08E14   307  0160 Cao  04   03 Beta
2741 N05E52   269  0160 Hsx  03   01 Alpha

SRS Number 133 Issued at 0030Z on 13 May 2019
Report compiled from data received at SWO on 12 May
I.  Regions with Sunspots.  Locations Valid at 12/2400Z
Nmbr Location  Lo  Area  Z   LL   NN Mag Type
2740 N08W40   308  0070 Hax  03   02 Alpha
2741 N06W02   270  0210 Hsx  03   02 Alpha

According to their README,
Location is a latitude and longitude relative to the central meridian, and Lo is a Carrington longitude.
A typical sunspot region moves about 13°/day from east to west in Stonyhurst longitude and less than ±1°/day in Carrington longitude.

GOES X-ray event report columns 29-34 also appear to be Stonyhurst spherical coordinates.
For example, sunspot region 9042 was associated with several X-ray events in June 2000.
If you filter the report by that region number, you can see that its progress from E90 to W90 longitude is roughly linear in time.
In projection, the plot of east-west position vs. time would be sinusoidal, flattening at the limbs.


Elongated Moon crater statistics? Many craters on Moon have circular shape as a result of the explosion upon impact. I got interested in oblong craters created by impacts striking the surface at a very low angle. Is there any statistics on the number of elongated craters on the Moon available?
An article on an Oval Crater captured by the Lunar Reconnaissance Orbiter states:

Non-circular (oval or elliptical) impact craters can form when the impacting bolide trajectory to the surface is <15° from horizontal or when the bolide impacts a sloped region on the (or some combination of both factors).

How well-studied is that threshold impact angle? Is there a phase transition from no explosion if below 15°, to explosion if impact angle is steeper?
References

Are all impact craters circular?
Lunar and Planetary Institute: Shaping the Planets: Impact Cratering

Elliptical craters can be studied three ways:  Laboratory simulations, computer simulations, and observations.
Laboratory simulations have not really been done in detail for this problem in decades (I don't know why, I'm not an experimentalist).  The most recent big modeling study that I know about is Collins et al. (2011) and Eibeshausen et al. (2013).  The former tried to determine what fraction of craters should be elliptical on different solar system bodies, while the latter looked more at that impact angle dependence as a function of other model parameters like pressure.  I suggest going to the Eibenhausen et al. link which is open-access and Fig. 5 shows ellipticity versus impact angle (yes, there is a continuum to answer your question) for a few different pressure regimes.
Observationally, this is rarely studied when building impact crater databases.  Primarily, the reason is that it's of less interest and it requires more effort.  Usually, people will just try to find craters that look elliptical and then, effectively, eyeball the axes and record the major and minor axes lengths.  The problem with this approach is that one must visually recognize the crater is elliptical and then one must subjectively choose those axes.  This then could be compared with a "background" crater population if one has also measured craters that are more circular.  An example of this work would be by Bottke et al. (2000).
A better way to do this (in my opinion) is how I have attempted to study this question.  It begins by tracing the crater rim and then fitting an ellipse to that trace as well as a circle, and doing this for all craters.  I did this with my Mars crater database (Robbins & Hynek, 2012), though the ellipse analysis is in that Collins et al. (2011) paper.  In doing this, there are some problems, though.  One is that any bias in how the rims are traced is going to bias the ellipse results, which are much more sensitive to inaccurate rim traces than circle fits because there are more free parameters.  Another problem is exactly what technique is used to fit an ellipse.  There are lots of different algorithms out there, and many that are the most popular are severely biased in various ways.
I tried to quantify both of those effects in my lunar crater database (Robbins, 2019) by doing a lot of different analyses and playing with a lot of different ellipse fit algorithms.  The supplemental material that discusses most of that is not behind a paywall.  After all those things are factored in, the conclusion is that more craters are "elliptical" than are predicted by simulations, but here again there's a problem: What is "elliptical?"  Is it when the major axis is 10% larger than the minor?  20%?  50%?  100%?  This gets to that problem I mentioned with the Bottke et al. (and similar) studies – when is something visually elliptical enough to count it as "elliptical?"
That said, there is also a problem of studying elliptical craters in the question of: Did the craters form that way?  Differential erosion could cause craters to erode more in one direction and thus become more elliptical as they degrade, especially if they are on slopes.  Another problem is that secondary craters – craters that form from the ejecta of larger, primary craters – form at lower velocities / energy and so are almost always more elongated, so removing those from a primary crater population before studying their ellipticity would also be desired.  This gets to be a problem on the Moon where we might have lots of unrecognized, multi-tens-of-kilometers secondaries from basins that affect what would otherwise be interpreted as primary elliptical craters.
I suppose if I had to summarize all of this, it would be that elliptical craters are, indeed, studied in and of themselves, but there are relatively few of those studies, in part because there is not necessarily too much new science there (is it worth spending a year on a computer to slightly refine percentages? or $millions in a vertical gun lab to refine things slightly?), and in part because the observational study of them is quite difficult due to the time-consuming nature of gathering the data and then understanding the biases in those data before any analysis can be done.


Fate of the Universe Which scenario will be worse for the ultimate fate of the Universe, forever expanding or the "Big Crunch"? Will the Universe be destroyed in both scenarios? At the present time does it appear that it will expand forever or will the expansion stop sometime in the future?

Which scenario will be worse for the ultimate fate of the universe?

Well, that's a wee bit opinionated. It depends on your point of view - that is, how you take a liking to the following two scenarios. Everyone has their own opinion; from the details I give below, I'll let you answer that yourself.

To infinity - and . . . well, more infinity: The perpetually expanding universe

There are actually three types of universes that could fit this model: The accelerating expansion one, the constant expansion one, and the decelerating expansion one. At this point in time, we appear to be in a universe undergoing accelerated expansion, thanks to dark energy. However, all three scenarios give us a similar end result: A dead universe.

In the far future, as long as the universe does not curl back into a Big Crunch, slowly all processes as we know them will begin to end. Star formation will eventually cease, and black holes will slowly start to gobble up a lot of matter. In the even farther future, black holes are evaporate to decay via Hawking radiation, leaving the universe a collection of various subatomic particles. (If protons decay, then perhaps there will no longer be any atoms). The temperature of the universe will drop until everything is about thermodynamically equivalent. Not a great place to be (well, there won't be any life, so nobody will be around to see it, but still. . .).

Beyond this. . . well, there are a few different ideas regarding what could happen. The Big Freeze idea states that the universe will eventually just cool off to this low-temperature state. The Big Rip idea says that matter itself could be torn apart bye extreme expansion of space. Cosmological false vacuum theories suggest that quantum tunneling effects could make some interesting expansion possible. However, I emphasize that we do not know whether any of these long-term theories are correct. They are too far in the future, and there is too much uncertainty among the relevant factors.

In the short term, however, it appears our universe will expand forever.

Back to the beginning: The Big Crunch universe and all its friends

The Big Crunch scenario says that if the density of the universe is greater than the critical density, the universe will fall back onto itself into a singularity - just like the Big Bang. This leads to other theories, such as the epykrotic universe, or a Big Bounce, which postulate that our current universe is just undergoing one of infinitely many stages of expansion and collapse.

This isn't exactly a great outcome, either, but again, I'll leave you to judge.


How do you mount these binoculars and telescope on this tripod? Have this telescope and tripod that is missing the 'tripod head platform' to mount to a binoculars or telescope. Can I buy it somewhere? [Photos attached]

I would like to connect the tripod to the D70 telescope, and also this Celestron binoculars that comes with a tripod mount piece that has two screws in it, you could see in the picture on the right of the binoculars.

EDIT: added additional photos

Photos from the Celestron manual, showing the tripod with all parts:

My tripod, missing the tripod head:

Telescope:

What you want to attach the telescope to a tripod is something like this:

https://www.firstlightoptics.com/dovetail-bars/baader-vixen-style-dovetail-clamp.html

Which is a clamp that holds the standard vixen dovetail used by most small telescopes.  The important thing with the one linked to is that it has a threaded hole to allow you to attach it to a standard 1/4" photo tripod screw. (Not all do - some of them are designed to bolt to a mount, so only have countersunk holes).

Similarly, the tripod adapter that came with the binoculars should have a threaded hole on the bottom part to attach it to the tripod. The captive screw in the adapter attaches to a threaded socket on the front of the
binocular hinge - there's likely to be a circular cover that unscrews or clips off to reveal the socket.

Unfortunately, judging from the picture of the tripod, it looks as though you may be missing the camera platform bit. Normal photo tripod heads will either have a platform with a captive 1/4" whitworth tripod thread thumbscrew, or a platform with a quick release holder for a removable plate which has the screw (the quick release versions make it easier to quickly attach or remove the camera (with the QR plate attached) from the tripod.

From the picture, it looks to me as though that platform part is missing.

If the tripod head was designed to be removable, it will usually be screwed onto a 3/8" bolt attached to the leg section. There may also be some extra
smaller screws on the underside of the platform the head is attached to to
prevent it from unscrewing by accident. If it is removable, you may be able to replace the entire tripod head unit (This is fairly common with more expensive photo tripods, where you can buy the legs and heads separately).

But you might find it cheaper to look for a secondhand tripod.

For supporting a telescope, you usually want a fairly solid and stable tripod. That can often mean heavy, large and expensive. Most of the smalller /  cheaper tripods are designed for lightweight cameras and can be shaky with
a heavier load.

The problem with using a photo tripod for normal binoculars is that when you're observing near the zenith (looking straight up) the tripod gets in the
way. You can also get things called parallelogram mounts that put the binoculars out on a pivoting arm with a counterweight on the other side - since the binoculars aren't over the tripod you can point them straight up without the tripod getting in the way.

[Edit - now that you've added the manual pictures, it looks as though you may just be able to attach the scope directly to a standard tripod, without needing a dovetail clamp, assuming at least one of the two holes in the bottom of the telescope dovetail is threaded. Some tripod heads have a rectractible or spring loaded pin, as well as the main screw, mainly designed for use with movie/video cameras. If the spacing on that matches up with the other hole in the scope dovetail, then the pin helps prevent the scope accidentally unscrewing (since it stops the scope turning round on the platform).]


Where do we define the "surface" of a gas planet? Since gas giant consist of most gas components, where do we establish their "surface"?
My take is basically to take the limit in which all light is opaque. For example, in this photo:

The surface, then, will be the limit of the black blackground with the planet.
Any other way to formally define the "surface" of a gas giant?

There are two common definitions in use for the surface of gas planets:

The 1-bar surface: As pressure increases, the deeper in we go into the gas planet, we will hit a pressure of 1 bar at some altitude. Gas at this altitudes will usually sit deep enough in the gravitational well and be of a near-uniform density and temperature, as to not be influenced by exterior parameters, for example the solar wind. Therefore, the altitude of the 1-bar level will remain essentially constant, for short astronomical times.
The $\tau=2/3$-surface: This is the altitude, from which photons can escape freely into space. This happens at an average optical depth $\tau$ of 2/3. It is essentially what you see in your image as the limit of the black background. For the sun one end of the photosphere is the average $\tau=2/3$-surface, and for transiting exoplanets this is identical to the measured transit radius at that wavelength.

There is no hard relation between those two surfaces, but in general their altitude will not be different by more than a scale height, as at around 0.1-1 bar the gaseous atomic and molecular bands become enormously pressure broadened, which makes the atmosphere quickly opaque at most wavelengths, for the usual gas giant components.


When did we discover that our sun does not orbit its own axis but a barycenter? It's commonly known that the center of gravity of our solar system is not the suns axis, and, that the same goes for other orbital systems like Pluto and Charon, and so on. When was it discovered that our sun has an orbit around a barycenter?

Newton explained the motions of planets in the solar system in Principia in 1687.

This amongst many other things "defines the very slow motion of the Sun relative to the solar-system barycenter;" (https://en.m.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica ).

Since Newton came up with the ideas of gravity and barycentres, surely this is the answer? (Though, I note there is some debate that the inverse square law may have been around and discussed during the previous decade).

Ideas that the Sun moved were of course commonplace before then.


How to improve the precision in a star's distance When I use the Gaia star catalogue, the parallax values often have large uncertainties which, therefore, leads to a large uncertainty for the distances to the stars.
If you were interested in the distance to one particular star and money was no object (i.e. you could buy time on any instrument) - what would be the optimum thing that could be done to improve the precision in the star's distance?

You would have to pay a lot more than buying time on existing instruments.  You would have to pay for the construction of new and very expensive instruments
You could pay to have copies of the Gaia spacecraft built and sent into orbit around the giant planets in our solar system and/or at their L4 and L5 positions.  Thus two or more such observatories could study the same star at the same time with a baseline between them tens of times greater than the baseline resulting from the orbit of the Earth around the Sun.  Thus the parallaxes and distances should be tens of times as precise.
And if you pay for the development of more precise instruments than Gaia has, you can increase the accuracy and precision of the parallaxes and distances even more.
And you could pay to have several such observatories sent out of the solar system in various directions so that they will eventually be separated by even greater distances than the outer planets are, thus making their parallax and distance measurements even more accurate.


Control the behavior of a Black Hole Is it possible to manipulate the size of a black hole or how a black hole behaves if scientists were able to synthesize hawking radiation on demand.  That is, if I could find a way to imitate hawking radiation in a missile would this missile decrease the size of the black hole I fire the missile at?  I am thinking provided I create enough hawking radiation it should have an effect on the black hole size once the missile detonates. That is if it is true that hawking radiation can make the black hole disappear as it does one the mini black holes who die because of hawking radiation.

Hawking radiation isn't some special kind of radiation. It's perfectly ordinary "thermal" radiation, mainly photons, with a few neutrinos and electrons and positrons if the temperature is high enough. Black holes shrink when they emit it, not when the absorb it. So aiming simulated Hawking radiation (or any other kind of radiation) into a black hole just makes it bigger.


Power of the James Webb Space Telescope I have heard that the JWST will be capable of looking at the early stages of the universe. However I have also heard that it won't be capable of detecting the Oort Cloud that supposedly provides comets for the solar system.

How is the JWST able to see millions of light years away but cannot see a giant spherical cloud of rocks around the solar system?

It all comes down to the brightness of objects (not their size). For all intents and purposes we can assume that the most distant galaxies and the small, but much closer, objects in the Oort clouds are unresolved point sources.The Oort cloud objects are too faint to see, with JWST, but it should be able to see bright galaxies and quasars even at 13 billion light years.

JWST will have a magnitude limit of something like 30th AB magnitude at red and near-infrared wavelengths. It turns out that quasars at Z~10 can have brightnesses that enable them to be detected at these magnitudes and brighter.

OK, but now work out how bright an Oort cloud object is seen in reflected light from the Sun. Taking a solar luminosity and calculating the flux at say the nearest hypothesised Oort cloud objects at about 2000 au from the Sun, we get  $1.4\times 10^3/(2000^2) = 3.5\times10^{-4}$ W/m$^2$. Now let's be generous and say that as much as 10 per cent of this is reflected (a very high albedo) and that it is a big Oort object with a radius of 20 km. If this acts as a Lambertian reflector, then it will receive $3.5\times 10^{-4} \times \pi \times 20000^2 = 4.4\times 10^{5}$ W and it will radiate $4.4\times 10^{4}$ W isotropically back into a hemisphere towards the inner solar system.

Now reversing the calculation. After this light has travelled back 2000 au to the JWST, the flux received will be $4.4 \times 10^4 /(2\pi [2000 \times 1.5\times 10^{11}]^2) = 4.9 \times 10^{-25}$ W/m$^2$.

We can compare this with the original flux of the Sun at the Earth. This is $1.4\times 10^{3}$ W/m$^2$. Now assuming that the reflection doesn't change the solar spectrum, we can take the ratio of the solar flux at earth to the reflected light from the Oort cloud object to calculate that the Oort cloud object is 68.6 magnitudes fainter than the Sun as seen from the Earth (or close to the Earth - JWST will not be in orbit around the Earth) in any magnitude band you care to choose.

So in the near infrared, the Sun has an apparent magnitude of around -27.5. This means that the Oort cloud object would have a near infrared apparent magnitude of 41 and would thus be 11 magnitudes too faint to be seen by JWST. (And remember we chose a big object that was as close as an Oort cloud object is expected it to be and gave it a high albedo),

Now consider that 30th mag detection threshold. For a solar-type spectrum this equates to a flux threshold of $\sim 10^{-20}$ W/m$^2$. Ignoring the cosmological nuances of proper distance etc., then something at 13 billion light years would have to have a luminosity of a mere $10^7$ solar luminosities to produce this kind of flux - the brightness of a modest galaxy. Of course there is also the extreme redshift to consider, so this calculation is way too optimistic, but even allowing for an order of magnitude or two, one should easily be able to see the brightest galaxies and quasars at 30th magnitude.


How did Jupiter form where it is? Since Jupiter is the biggest planet in our solar system and is made up of mostly hydrogen and helium, (the gases the Sun uses to create energy), how come it didn't form as far out as say Uranus or Neptune since when our star formed it took most of the gases around it to create? Just a little confused on this, thanks for the help!

It's at Jupiter's distance and beyond that ices were able to form out of the disk of material surrounding the early sun.  Go much further in and there is too much energy from the sun for them to stay as solids (and will sublimate into gases); this is why asteroids are principally rocks and metals.  So at this distance more of the materials of the planetary disk can form the base planetesimals.

At this point, your question is answered largely by a geometric consideration (or two) and one of Kepler's laws.

First, the geometric consideration.  A circle of radius $r$ has area $\pi r^2$.  The bigger the radius, the more area.  The material for Jupiter (or any other planet) came from an annulus: stuff outside one circle, but inside a slightly bigger circle.  This annulus had a lot more area for the out planets than the inner planets, and so could contain a lot more mass.

Of course, that could makes us think that Jupiter shouldn't be the largest of the gas giants: it's the closest of them all to the sun, after all.   The density of the disk needn't have been approximately constant throughout these regions, though.  Quite possibly the density was such that Jupiter's region had more mass than the areas for the other planets.  As HDE's answer (posted as I was finishing this) points out, these ices probably also helped stop materials from passing into the inner solar system, maintaining a higher density than you might otherwise expect in the inner solar system, as well as causing materials to sort of "dam up" right around Jupiter's orbit.

Now for the Kepler's law.  The further you get from the sun, the slower your orbital period.  Picking the correct units, we have $P^2=a^3$, where $P$ is the period measured in years, and $a$ is the semi-major axis of the orbit measured in AU.  The further out you go, the slower you go around the sun; indeed, it's not just that it takes you a longer total time, but your actual velocity goes down.  We can also see this as a Newton's law of gravity consequence.  At Jupiter's furthest point from the Sun, the escape velocity is a little more than 18 km/s.  At Saturn's maximum distance, the escape velocity drops to around 13.25 km/s.  So things can go roughly 35% faster within Jupiter's orbit than they can closer to Saturn's, and they have less distance to travel to make a complete orbit.

What this means is that it takes longer for planetesimals to get close enough to each other to accrete together the further out you go, and there is a longer mean time between collisions.

Now, eventually, the Sun "turned on" and started blasting space with it's solar wind (before that, the heat came primarily from thermal radiation from the gravitational contraction of the sun).  This ended up clearing out most of the unaccreted particles out of the solar system, stopping planetary growth (and removed portions of existing atmospheres; a very young Earth probably had a lot of H and He in its atmosphere, until the sun hit it with enough energy to knock any of it not locked up in rocks away).

So Jupiter was probably in a bit of a goldilocks situation.  The average density of the region it formed in was probably higher than where the other giants formed in, it was at the perfect spot for lots of materials to start accreting early, and the accretion process would have been faster.  So Jupiter is growing faster, and this gives it a competitive advantage: the bigger the growing planetesimals get the further their influence extends and the faster they can pull in more materials, and subsequently interfere with the growth of other planets (or planetesimals).  Somewhere around a mass of 10-15 earth masses, the giants can start pulling in large quantities of the hydrogen and helium gasses.  And, again, Jupiter likely hit this mass well before the other giants, and had more material to pull from, so it became much larger than the others could before the solar wind stopped the process.


How is it known that Callisto has no core? My Astronomy book claims that scientists have discovered that Callisto, a moon of Jupiter, has no hot inner core. In fact, it says, Callisto has a core much like the nucleus of a comet.

Is this still accepted theory, and how was this information discovered?

It may have a (silicate) core, but it's very small. There also is the possibility of an internal liquid water "ocean". Currently the most probable composition is that of rock and ice, with the density of rock steadily increasing as one digs deeper (This is in stark contrast with planets like Earth, where there are discontinuities in the density between layers). Planetary differentiation has not occurred appreciably because Callisto is not tidally heated.

The composition can never be completely determined — the composition of a spherical body cannot be fully discerned from just rotation and translation1. However, we can make reasonable guesses. Basically, we can calculate its total mass and moment of inertia by observing its rotation and revolution. Then, assuming that the density ρ varies radially, we get the following two equations:

$$M = \int \rho \pi r^2dr$$

$$I = \int (\rho \pi r^2dr)r^2$$

These do not completely determine ρ, however with other reasonable assumptions on the composition and the amount of planetary differentiation it can have, we can get a good estimate for ρ.

1. It can be determined if we take a larger gravitational field and pivot it, but this ventures into the Aristotlean "Give me a place to stand on, and I will move the Earth." zone of thought experiments.


If the Sun's trajectory were altered, would the solar system follow? I was reading a text state that the sun is not stationary:

"it orbits the Galactic center once approximately every 226 million years as the galaxy spins which is also traveling through space towards an unknown source of gravity"

So does the solar system will also move when the Sun moves? Does it affect the stability of solar system? I am not much known to the cosmology so if my question is wrongly framed, please let me know.

The motion of bodies orbiting anything, is described by orbital elements - their distance from the barycenter (central mass), direction, velocity - and so two bodies of mass significantly lower than the central mass, with identical orbital elements, will move along the same orbit; with very similar elements their orbits will be very similar.
For example, an astronaut on ISS, will circle Earth in nearly perfectly same trajectory as the station itself; both the astronaut and the station are vastly lighter than Earth, they are at the same altitude, the same speed, and so the astronaut is in freefall inside the station, not pushed against the walls or otherwise accelerating relative to the station - they both orbit Earth, their trajectory mostly unaffected by the minuscule differences in location between center of mass of the station and the astronaut - that difference being of order on ~30 meters at most, while the station orbits Earth on an orbit of radius of about 6700km (Earth radius + 400km) - with such small difference in orbital parameters they are both in the same orbit.
Similarly, Sun is about 25,000 light years from the center of the galaxy, and as you can guess the galaxy is much, much heavier than the Sun. Pluto is about 4.6 light hours away from Sun, so it moves 'in' and 'out' relative to the galaxy by maybe 9.2 light hours. So the influence of the galaxy on it practically doesn't differ between these positions - the galaxy influences it just the same as it does the Sun, so they both orbit the center of Galaxy on practically the same orbit.
Now, there is a certain rather significant difference between the orbit described by the Sun, and, say, Mercury. The Sun moves in the orbit around the galaxy at 230km per second. Mercury circles the Sun at 47.36 km/s. Now, when it goes "in the opposite direction" relative to the Sun than the motion around galaxy, that's 230-47=183km/s; when it goes "in the same direction", it's 230+47=277km/s; a body moving at either of these velocities at the same 25,000km from the center of the galaxy would have a much different orbit than the Sun. But as Sun's gravity reverses Mercury's direction relative to the center of the galaxy every 44 days (88 days is its orbital period), it all averages out over the 226 million years to complete the orbit around Milky Way.
So, in short, yes, the entire solar system is affected just the same as the Sun and orbits the center of the galaxy just the same, all the differences either averaging out or too small to be noticeable. And if you tried to change Sun's trajectory by a gravitational influence that is sufficiently strong and sufficiently distant, you'd redirect the entire solar system just fine. Of course if you used a source of gravity that varies more across the width of the solar system - say, another star or a regular (not supermassive) black hole - you'd completely break the precarious balance of orbital motion of our system, planets closer to the new source of gravity captured by it and entering its orbit, leaving the orbit of the Sun, planets farther out possibly escaping into interstellar space.
Gravitational field from a central mass (like a star, or the center of galaxy) varies with square of the distance from the central mass, so you need it acting over relatively short distance in at a very long distance from that mass, to have it "locally" behave like uniform field and act on all the affected objects nearly the same.


What physical features determine if a planet is a major, minor or dwarf planet? Like many, when I was growing up, we always were taught, hence always learned that there were 9 planets.  However, recently, decisions were made and all of a sudden there were 8 major planets and a series of dwarf and minor planets.

What physical features delineate a body being referred to as a major planet as opposed to dwarf planet or minor planet?

Planets

For a body to be classified as a planet it must have a few physical characteristics:

Mass

It must have enough mass to have a strong enough gravity to overcome electrostatic forces to bring it to a state of hydrostatic equilibrium.

Hydrostatic equilibrium is important because early in a planets life it is nearly entirely fluid, crust and all

Internal Differentiation

The life cycle of a planet essentially leads to denser heavier metals being at the centre of a planet, surrounded by a mantle which must at some point have been fluid (it can still be solid and called a planet as long as it used to be fluid.)

Atmosphere

This is usually driven by its mass but a planet should have an atmosphere. This means it should be massive enough to have a strong enough gravity to hold some gasses to its surface.

More massive planets are capable of keeping lighter gasses, such as hydrogen, bound to them too. I.e Jupiter.

Magnetosphere

A magnetosphere suggests that the body is still geologically active. This means they have flows of elements that conduct electricity in their interiors.

As specified by the IAU:

  A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.

Minor Planets

Minor planets  are things such as asteroids, they are usually small mass, rocky bodies. This is different from small mass bodies that might be predominantly ice or water.

Dwarf Planets

Size and Mass

The IAU have not specified an upper or lower limit for mass to be considered a Dwarf Planet, therefore it is predominantly determined by one other feature called:

Orbital Dominance

Orbital dominance is achieved when a body has cleared it's orbit of all other bodies.

For example planets are able to remove small bodies out of their area through impact, capture, or gravitational disturbance.

Any body that is incapable of doing so is therefore classified as a dwarf planet. So if Jupiter was incapable of clearing its neighbourhood of bodies then it too would be a Dwarf planet.


How are microlensing events used to constrain the size of innermost stable circular orbits around spinning black holes? The third paragraph of the introduction to the ArXiv preprint Constraining Quasar Relativistic Reflection Regions and Spins with Microlensing says:

Quasar microlensing has significantly improved our understanding of the accretion disks (e.g., [five refs]) and non-thermal emission regions (e.g., [seven refs]) of quasars, and the demographics of microlenses in the lens galaxy (e.g., [three refs]) . Since the magnification diverges on the caustics produced by the lensing stars, quasar microlensing can constrain arbitrarily small emission regions if they can be isolated from other emission, in position, velocity, or energy. In particular, microlensing can be used to constrain the spin of black holes by measuring the ISCO size. (emphasis added)

ISCO was previously identified as innermost stable circular orbit.
From what I understand from Wikipedia's Gravitational microlensing; How it works, a microlensing observation records a short term brightening of an object due to focusing by an intervening gravitational body, without useful spatial resolution.
See also the Planetary Society's explanation Microlensing; Beyond our Cosmic Neighborhood
Question: How are microlensing events used to constrain the size of innermost stable circular orbits around spinning black holes? What instruments are used and what kind of analysis is done to obtain a ISCO constraint?

Example light curve of Gravitational Microlensing event - OGLE-2005-BLG-006 Source

I think most of the paper is devoted to describing how they do it.  (This may be more of a comment than an answer, because I'm an amateur and not prepared to go through the paper in detatil).

Note that microlensing here is a little different than microlensing as described in the Wikipedia.  In the Wikipedia page and the Planetary Society page you reference, microlensing refers to observing the change in brightness over time of a small, relatively nearby object when a closer but non-luminous object passes in front of it from our line-of-sight.  In this paper, microlensing refers to the extra magnification that can be provided to tiny sub-sections of a "macrolensed" (gravitationally-lensed cosmologically distant object) by the chance placement of a star at the right position in the lensing geometry.

The authors of the paper figure out how this sort of microlensing might apply to the innermost parts of a quasar, where it might highlight small parts of the accretion disk, and so the contribution of those parts to the spectrum (in this case, one particular X-ray emission line characteristic of active galactic nuclei accretion disks) of the "macrolensed" object as seen from Earth.  They apply a statistical model of the likelihood of such microlensing and how it would affect the measured spectrum of a range of possible active-galactic nuclei (resolving a particular part of the accretion disk with microlensing would enhance that particular parts contribution to the whole, and each part of the accretion disk would have a different relativistic shift of the emission line as it whips around the central black hole), and then see how their model matches the actual measured spectrum.  It sounds like they are operating at the limits of statistical significance-- at least for most of the objects in their study-- but got more robust data for one particular object.


two black holes orbiting each other within their event horizon Is there any argument against this situation: two black holes, one within the event horizon of the other, and the system is stable.

It is interesting to me because that if this works, we could have the system's event horizon not in the shape of a perfect sphere. (The event horizon would probably be moving with the orbit)

Inside the event horizon is a weird place to be. Inside a black-hole space-time is flowing towards the singularity faster than the speed of light (as measured from outside the black-hole. The effect is that any thing inside a black hole will, in a finite (and usually short) amount of time end up at the singularity.

Models of merging black holes exist such as video from nasa. It requires super-computers to solve Einstein's equations numerically.

As I noted in a comment, no stable orbit can exist within 1.5 times the radius of the event horizon. The orbital velocity is c at that distance (1.5 Schwarzschild); it's also called the "photon sphere". Within that radius all orbits are unstable. And the radiation of gravitation waves means that as two black holes orbit each other, they will lose energy, and so their orbits will decay. As the black holes approach, the event horizons are distorted, and merge into an droplet shape


Transform pixel coordinates (in FITS file) to equatorial Say I have a FITS file with an observed field of stars, each one with x,y coordinates given in pixels. Assume the image is aligned so that the x axis is the right ascension, and the y axis the declination.

I'm looking for a tutorial to transform the pixel coordinates of each star to equatorial, even if the transformation is approximate.

I'v been looking for a while but found no tutorial/book/site that enumerates the steps needed to produce this transformation.

Assuming that each (square) pixel has the same angular scale (not a given if the field of view is large) of $\theta$ degrees/pixel.

Then the declination (in degrees): $\delta \simeq \delta_0 + \theta y$

The right ascension (in degrees): $\alpha \simeq \alpha_0 - \theta x/\cos \delta$

where $\alpha_0$ and $\delta_0$ are the RA and Dec at $x=0$, $y=0$. [The minus sign is there because right ascension increases towards the left of a sky image.]

This approximatimation becomes poor as the field of view gets larger.

A slightly more complex, but accurate, approach is described in http://gtn.sonoma.edu/data_reduction/astrometry.php


Focusing with Meade Audiostar I recently upgraded from a Meade ETX-RA to and ETX-EC and bought a used Audiostar controller. My EC came with a Meade autofocuser and one of the old (I think it’s 494) controllers that has the focus controls and the four lights on it. Is it possible to just plug the autofocuser AUX cord into the Audiostar and focus from that? I read the manual for the autofocuser and it seems like I should be able to. Just not sure which controls I am supposed to use.

It turns out that if you hold the mode key for about five seconds and have the 1244 focuser plugged into the mount then a screen will pop up that allows you to use the arrow keys on the top of the remote to adjust the focus. Once you are done focusing, you simply press mode to exit.


Why isn't Earth tidally locked to the sun? Given that the moon has long been tidally locked with the Earth, why isn't Earth (or any of our other solar system's planets) tidally locked to the sun?

Really, it's just because the tidal locking timescale is so long for Earth:
$$t\propto\frac{a^6m_{s}}{m_{p}^2R_s^3}$$
where $a$ is semi-major axis, $m_s$ is the mass of the secondary object, $m_p$ is the mass of the primary, and $R_s$ is the radius of the secondary. If we compare the Sun-Earth system to the Earth-Moon system, we see
$$\frac{a_1}{a_2}\approx380,\quad \frac{m_{s_1}}{m_{s_2}}\approx80,\quad \frac{m_{p_1}}{m_{p_2}}\approx333000,\quad \frac{R_{s_1}}{R_{s_2}}\approx3.67$$
where $_1$ denotes the Sun-Earth system and $_2$ denotes the Earth-Moon system. To make up for these differences (and to have Earth be tidally locked by now), assuming similar Love numbers and dissipation functions for Earth and the Moon, we would need Earth's initial spin to be substantially smaller than the Moon's initial spin by many, many orders of magnitude, and this just wasn't the case.


Do all planets have a van Allen radiation belt? Earth has the van Allen radiation belts. Which other planets have one? Can it be determined if distant planets in other solar systems have one?

Mars does not have a magnetic field of any strength and so does not have a van Allen belt (and this is a serious problem for potential human exploration of the planet). Similarly, neither Venus nor Mercury (nor the Moon, which is arguably part of a two-planet system with the Earth) have a van Allen belt.

The gaseous giants do, however, have radiation belts which are analogous to Earth's van Allen belt.

Spotting aurorae on distant exoplanets would be an indication of a strong magnetic field and thus of the likelihood of a radiation belt. I am not aware that such aurorae have been detected but distant aurorae have been seen around a brown dwarf - a substellar object somewhat bigger than a planet (see http://www.theguardian.com/science/2015/jul/29/astronomers-find-aurora-a-million-times-brighter-than-the-northern-lights).

Alternatively, the radio emissions from high energy electrons trapped in planetary radiation belts might be detectable - this was how Jupiter's radiation belt was discovered around 60 years ago. Work seems to be underway to detect such radiation: https://skaoffice.atlassian.net/wiki/download/attachments/22183971/PoS-exopla-AASKA14.pdf?version=1&modificationDate=1403003959989&api=v2


Problem with denoising the BICEP2 data? This question pertains to this article which talks about why the BICEP2 measurements of B-mode polarization in Cosmic microwave background radiation turned out to be noise from galactic stardust. They go on to add that the Plank data of the noise does not lend to getting a good B-mode estimate from the BICEP2 data.

  However, the results of the joint assessment would suggest that
  whatever signal BICEP2 detected, it cannot be separated at any
  significant level from the spoiling effects. In other words, the
  original observations are equally compatible with there being no
  primordial gravitational waves. "This joint work has shown that the
  detection of primordial B-modes is no longer robust once the emission
  from galactic dust is removed," Jean-Loup Puget, principal
  investigator of Panck's HFI instrument, said in the Esa statement.
  "So, unfortunately, we have not been able to confirm that the signal
  is an imprint of cosmic inflation."

Can someone explain why(from a signal processing perspective) exactly the detection of primordial B-modes is not robust even after the emission from galactic dust is removed? And what are the proposed workarounds to tackle this problem?

B-modes in the raw CMB data can be caused by primordial gravitational waves, as well as by cosmic dust. After subtracting the cosmic dust polarization from the BCEP2 data, the residual signal is too weak with respect to noise to be statistically sufficiently significant to claim a discovery.

Or stated in a different way: The observed B-modes can be explained away by cosmic dust.

This doesn't prove the absence of primordial gravitational waves. But it doesn't provide sufficient evidence for their presence either.


V471 Tauri's circumbinary brown dwarf non-observation; Applegate, or over-restrictive assumptions? tl;dr Has the brown dwarf observation been disproven?

I have just started reading about the interesting object V471 Tauri. The first two sentences of the introduction to The V471 Tauri System: A Multi-datatype Probe Vaccaro et al. (2015):

  V471 Tau, a white dwarf-red dwarf eclipsing binary (EB) in the Hyades with orbit period $0^d.52118$, is primarily known for its unique historical role as a stimulus to common envelope evolution theory (Chau et al. 1974; Refsdal 1974; Sparks & Stecher 1974; Ostriker 1976; Paczynski 1976; Alexander et al. 1976; Taam et al. 1978). Other properties include a likely brown dwarf companion to the EB, measured white dwarf spin, mass loss and exchange in a detached binary, differential rotation measured via magnetic spots, spot distributions, accurate white dwarf parameters, and photometric-spectroscopic distance measures that help to pin down the binary’s location within the Hyades.

It's the "likely brown dwarf companion" I'd like to ask about. The paper with the catchy title "The First Science Results from SPHERE: Disproving the Predicted Brown Dwarf around V471 Tau" Hardy et al. (2015) SPHERE is a new, advanced adaptive optics system at the Very Large Telescope (VLT). The image below (figure 3) is part of the argument, and the suggestion is that if there is no brown dwarf seen in the band between the two white circles, then the predicted brown dwarf does not exist.

This would be interesting, because an alternative explanation for the slow, periodic drift in the eclipse timing would have to be found. One possibility is the Applegate Mechanism which I don't understand, but may ask a separate question after this.

Back to Vaccaro 2015, section 9, titled "On the Reality of the Third Star" is over six pages of discussion of the underlying assumptions, and if I understand it correctly provides several possible ways in which a suitable brown dwarf could exist but not show up in the SPHERE image. In essence disproving the existence disproof.

I would like to know: Is my understanding of the current situation correct? Have there been more recent developments?

above: Left panel of Fig. 3 of Hardy et al. 2015: "Figure 3. H-band image of V471 Tau obtained on the SPHERE IRDIS instrument at the VLT. Left panel: Resulting image after angular differential imaging (ADI). The area in-between the white circles denotes the 5 sigma predicted position of the brown dwarf..."

above: The monster currently known as SPHERE from here.

For all of these direct imaging results, the critical parameter is the contrast as a function of separation. This lets you know how much fainter an object you can see around the much brighter primary object whose light has been suppressed by the coronograph (the black circle in the center of the star).

From the change in eclipse timings (Figure 1 in their paper) you can predict the expected mass and separation of the third object (proposed brown dwarf) using their equation 1 since you know the mass of the binary. From evolutionary models of brown dwarfs, you can make predictions of how bright the proposed brown dwarf should be for its predicted mass.

You can then look for an object at the right separation in the SPHERE images. This is shown in their Figure 3 (right hand figure of the 2) where they show how bright an object they could see as a function of separation from the primary star; everything above the solid curve is something they should see. The vertical dotted lines show the boundary on the separation of the brown dwarf, predicted from the eclipse timings. The diamond symbol is the predicted brightness of the brown dwarf from the evolutionary models given its mass.

Since this is well above the contrast curve (by a factor of about 15x) and nothing is seen in the images at that separation, this strongly suggests that the proposed brown dwarf does not exist. The only "get-outs" are if the SPHERE team have measured their contrast curve incorrectly (unlikely at the level needed) or our models of how bright brown dwarfs should be are also wrong by about a factor of 15x and the brown dwarf is much, much fainter than predicted.

The Applegate mechanism is a bit weird. The idea is that as the star goes through magnetic activity cycles (as the Sun and a lot of other stars do) and the magnetic field strength grows and shrinks, this causes the shape of the star to change, bulging more or less at the star's equator as the cycle progresses. This changes the amount of angular momentum in the star, which has to couple into the binary orbit as the angular momentum has to be conserved, causing the binary orbit to shrink or expand. This would then explain the change in timings of the eclipses without the need for the third body brown dwarf tugging on binary. Unfortunately the Applegate mechanism, although it may work in V471 Tau, can't explain the variations in a bunch of other binary systems (magnetic field strength not strong enough) so it can't be the only solution for eclipse time variations.

The latest paper I could find that references the Vaccaero et al. 2015 paper, is Vanderbosch et al. 2017. They argue that because of the timing of the eclipses change but the spin period of the white dwarf secondary doesn't change, a third body such as a brown dwarf can't explain the changes as it should change the timing of both "clocks".


Why does light from objects take time to reach us? Why does the light of objects such as the Sun, stars etc. take time to reach us?

E.g. Why does sunlight take 8 minutes to reach us, instead of reaching our eyes as soon as it rises.

For the very simple reason that light travels very fast, but not infinitely fast.

It takes light about 1 nanosecond to travel 30cm. That is so fast that you don't normally notice it. However it does take some time for light to reach your eyes, even from just across the room.

Objects in space are a lot further away. It takes light about 1.2 seconds to reach the moon, and as you note it takes light 8 minutes to come from the Sun. The reason is that the sun is 150 million km away, and it takes about 8 minutes for light to go that far.

If light did have infinite speed, it would be a weird universe. Effects would happen at the same time as causes, so the basic nature of cause preceding effect would be broken. The "speed=frequency*wavelength" formula for light would be broken. Refraction and diffraction wouldn't exist. A photon would be emitted and absorbed at the same time. Could something that is absorbed at the same time as it is emitted actually exist?


If Jupiter is a gas-giant then why don't its features change? A naive question. When we look at Jupiter, we see that its features didn't change largely over many years, for instance, the red-spot. If it is composed of gases and liquids, then why aren't the effects of mixing of these fluids visible?

My intuition is that due to very low temperatures ($-145\, ^{\circ}$C), diffusion of fluids doesn't occur and therefore the superficial appearence of Jupiter remains the same.

Believe it or not, Jupiter isn't too consistent. Take a look at these pictures, the first taken in 2009 and the second taken in 2010:

and

Quite the difference, eh? Why?
Jupiter's atmosphere is made of zones and belts. Zones are colder and are composed of rising gases; they are dark-colored. Belts are warmer and are composed of falling gases; they are light-colored. The reason the two don't intermix is because of constant flows of wind, similar to the Jetstream. These winds make it hard for bands to mix.
There are two types of explanations for the jets. Shallow models say that the jets are caused by local disturbances. Deep models say that they are the byproduct of rotating cylinders comprising the mantle. At the moment, we don't know which explanation is correct.


What kind of things I could "see" with an amateur radio telescope? There are apparently not many reasonably priced radio telescopes available for the amateur users. I only could find a SPIDER 230C 2.3 meter diameter compact radio telescope, which costs ~10 k€. Reading the page, I don't get a good overview what kind of things I could "see" with a radio telescope that has a 2.3 meter antenna.
Is the instrument already a (semi-)professional one?

Rather than looking for ready-made systems, take a look at projects. Right now, plenty of amateurs are using software defined radio coupled to various antennas for astronomy. Start here:
http://www.rtl-sdr.com/rtl-sdr-for-budget-radio-astronomy/
And while it has nothing to do with imaging, there's plenty of radio astronomy that amateurs can do using simple (albeit sometimes large) antennas:

http://radiojove.gsfc.nasa.gov/
http://radio-astronomy.org/pdf/qex/radio-jove-proof.pdf
http://www.radiosky.com/project.html


Helium discovered on an exo planet Why is there so much excitement over helium having been discovered on an exoplanet? Isn't it oxygen that we should be looking for, since this is what is needed for life? What do we need helium for?

You are right that helium around a gas giant is no kind of surprise, and oxygen on an Earthlike planet would be spectacularly more important.  However, the excitement is simply that if you can detect one element, it means you might be able to detect another.  It just gives you more ways to learn about the planet, and offers promise for more new ways to learn.  One might find analogy with the first detection of exoplanets, which were "hot Jupiters."  No one was surprised stars can have planets, and no one thinks hot Jupiters are conducive to life, but if you can discover one kind of planet, it bodes well for the later discovery of others.  Anything we can learn about composition is an important step in the right direction.


Spacing of G and K class stars I was looking at the Wikipedia article "List of potentially habitable exoplanets" and I noticed that many of the closest planets listed(tens or hundreds instead of thousands of light-years away) orbited M-class stars. I have also read that scientists are more certain about the possibility of habitable planets around G and K class stars, because the habitability of red dwarf systems is still debated. I was wondering whether there is something that makes the larger stars be farther away from each other, or whether it is just a coincidence. So is it possible to have many G and K class stars tens of light years near each other?

The solution to your mystery about why habitable planets are found around nearby low-mass stars, but distant G/K stars is all to do with observational selection effects and biases. There are many G/K stars within 100 light years of the Sun, but almost none have been examined in the detail required to reveal small, habitable planets around them.

Almost all the small "Earthlike" planets in habitable zones have been found by the transit technique. It is much easier to find such planets around small stars for two reasons:

The transit technique yields a signal that depends on $(R_p/R_{*})^2$. Thus small planet transits are easier to see around small stars.

The probability of a transit depends on the ratio of the star size to orbital radius. This sounds like it works against finding planets around small stars; and it does at a fixed orbital radius. But, because the habitable zone is much closer-in to a smaller, less luminous star, then the habitable planets around such stars are more likely to transit. In addition, their orbital periods will be tens of days rather than the months to a year around G/K stars and this makes it much easier to observe repeated transits.

In fact, the combination of these two selection effects makes it possible to find habitable zone planets around small stars all over the sky from ground-based observatories.

The downside is that small stars are low luminosity, so the planets that can be found are found around nearby examples at tens of light years.

In contrast, habitable zone planets around G/K stars are mostly found in months-year orbits by Kepler, which stared at a single small field for 4 years. The majority of the stars targeted by Kepler for exoplanet searches were 10th-15th magnitude G/K stars (there are few brighter examples in such a limited patch of sky), which are at distances of hundreds to thousands of light years.


How to send information from the far side of the Moon to the Earth? Chang-E has just sent a shot after landing on the far side of the moon. It looks tricky, how  can a device send information from a place that is oriented opposed to Earth?

You need a satellite that has a view of the landing site and the Earth. There are several possibilities for doing this. According to the NASA Space Science Data Coordinated Archive, the communications are relayed through the satellite Queqiao which is in a halo orbit around the Earth-Moon L2 point.
You could also have a satellite in lunar orbit, transmit data from the lander while it's over the far side, transmit data to Earth while it's over the near side.
Or relay via a satellite in interplanetary space, etc.


Which are stars and which are noise in this comet photo? The Forbes news article NASA's Hubble Telescope Snaps Best Images Yet Of Our First Interstellar Comet shows the image below (which I have since annotated) of comet 2I/Borisov (C/2019 Q4) and links to this series of Hubble images: https://archive.stsci.edu/proposal_search.php?mission=hst&id=16009

Looking through the previews of those images, they all seem to have streaks of similar length and direction, so I am assuming that the telescope is tracking the comet's motion in each exposure and the streaks are stars, suggesting that all the small dots are noise and not stars.

Is that right?
Is the fuzzy one an extended object?
What causes so many isolated pixels to be so much brighter than the background? Is this just the tail of a statistical distribution of shot noise, or are there other mechanisms that can produce single pixel noise many standard deviations above  the statistical shot noise distribution?

Is that right?

Yes.

  Is the fuzzy one an extended object?

That would certainly be my guess (probably a distant galaxy).

  What causes so many isolated pixels to be so much brighter than the background? Is this just the tail of a statistical distribution of
  shot noise, or are there other mechanisms that can produce single
  pixel noise many standard deviations above the statistical shot noise
  distribution?

Some of those might be "hot pixels", which are manufacturing defects that cause individual pixels to overproduce electrons. Most of them are due to so-called "cosmic rays", a generic term used in astronomical imaging to refer to the effects of energetic charged particles hitting the detector. In ground-based images, these are often not actual cosmic rays, but instead are particles from the decay of radioactive isotopes in or near the detector. In the case of space-based images like these, they more often are proper cosmic rays, though many of them are really charged particles trapped in the Earth's magnetosphere. (There is a part of Hubble's orbit which takes it within the South Atlantic Anomaly, where the Van Allen belt dips down toward the Earth's surface; observations are not scheduled during those periods, because of all the extra cosmic-ray hits the detectors receive.)

For example, here is a preview image from one of the other exposures in the sequence of observations. If you look closely, you can see that many of the cosmic-ray hits are slightly elongated, because the particle was traveling at an intermediate angle and passed through two or more adjacent pixels. The bright linear streak right next to the comet is an extreme example of this, where the particle was traveling almost parallel to the detector and so passed through an extended sequence of pixels.

The usual approach to dealing with this is to take several short exposures, one right after another, and combine them afterwards by rejecting extreme outlying pixel values from a single exposure (stars and other genuine objects will have similar pixel values from one exposure to the next, while cosmic rays are randomly located and do not repeat). This is why the HST images you see usually don't have cosmic-ray artifacts: they are the result of combining multiple exposures. (See here for a discussion and an example.) There are also statistical analysis tricks one can use to try to clean up individual exposures, though this necessarily involves interpolating data.


Sun like star in our milkyway? Is there any popular star or one which we can see with our naked eyes in our milky way that is atleast 90% sun-like(mass,radius,spectra,luminosity).The important thing is it must not be a binary.
Its okay if its not visible to our naked eyes because our sun's absolute magnitude is +4.5!.

There's a whole Wikipedia page about it.

https://en.wikipedia.org/wiki/Solar_analog

If you don't want Alpha Cen A, then 18 Sco might be the one.


Are there bodies in the Universe that are not spinning? Do we know of a celestial body which doesn't spin at all on its own axis? If not, why is it so?

Many red and brown dwarfs must have planets that are in full tidal lock with their host star, that means that the same side of the planet is always facing the star so there is always day while on the other side there is always night. Our Moon and many other moons are also in such a tidal lock with their planet but not with the Sun so that the often-used term "Dark side of the Moon" is verbally incorrect since on the far side there is a day too (when on Earth it is a New Moon).

If you mean non-spinning in the sense that on the same side of the planet it is day for half a year and night the other half I'm quite sure such planets exist too. On SpaceEngine I've seen similar planets. Nothing speaks against the existence of such planets I believe. There also might exist moons whose same side faces their planet for half a year and the other half faces away from it.


How would a flat mirror on the Moon reflect sunlight I was solving a task that said (paraphrasing):

  There is a flat mirror on the Moon, whose reflection coefficient is 100%, and observers on the Earth observe it as a star whose apparent magnitude is $3^m$. Knowing the magnitude of the Sun is $-27^m$, and its angular diameter is $0.5 {^\circ}$, calculate the diameter of the mirror.

The solution used Pogson's formula to state that the Sun is $1e12$ times brighter than the mirror ... and then the author said it is "obvious" that implies the surface of the mirror's disk in the sky, in $({^\circ})^2$, is $1e-12$ times that of the Sun.

How is it obvious, how do we derive this relation?

The (weird for historical reasons) defintion of magnitude is that a difference of 5 magnitudes corresponds to a factor of 100 in the brightness of the source.

So a difference of 3-(-27) = 30 magnitudes is a difference of $10^{12}$ in brightness.

The surface brightness of the mirror and the sun is effectively equal, since the mirror reflects 100% of the sun's light, so the only reason for the sun to be brighter is that it is larger. The brightness is in direct proportion to the apparent area of the source.

This is because brightness is a simple linear scale. Two suns of the same size and brightness would be twice as bright as one. So if the sun is $10^{12}$ times brighter it must cover an area $10^{12}$ times larger.


Why do stars born in a cluster finally disperse? In other words, if a group of stars born within the same nebula, e.g. the Pleiades, are initially gravitationally bound - what causes them to finally get separated and proceed individually? Is it the influence of  "external" mass? Or, are they now and we're they always, even at birth, having the escape velocity? (If the latter is true then I think they should not be considered truly bound from the beginning...)

The boundedness or otherwise of clusters remains to be established in most cases. The vast majority of clusters become unbound and disperse at a much younger age than the Pleiades. Or they may be born unbound as you suggest.

The stars in a cluster have a distribution of velocities and there will always be a tail of high speed stars that will be able to escape the gravitational potential in an "evaporation process", even in a cluster which is globally bound (i.e. the sum of kinetic and gravitational potential energies is negative).

It turns out that this isn't usually sufficient to explain the disappearance of clusters. A more effective disruptive mechanism are the Galactic tides. Stars that stray beyond a tidal radius can be stripped from the cluster. The reduction in the cluster's mass then reduces the tidal radius allowing more stars to escape in what eventually would become a runaway process. The tidal radius for a cluster near the Sun is about $5 M^{1/3}$ light years, where $M$ is the cluster mass expressed in units of the solar mass. For the Pleiades, $M \sim 1000 M_{\odot}$, so stars that stray beyond 50 light years can be lost.

In some cases, the evaporation process causes a cluster core collapse. The escaping stars take away kinetic energy and the remaining stars become more tightly bound. This is what has happened in most of the massive and long-lived globular clusters. Less massive open clusters like the Pleiades are much more susceptible to tidal dissipation.


The newly discovered comet NEOWISE has been given the name C/2020 F3. What is the meaning of the F3 and the C in this name? The new comet C/2020 F3 was first spotted by NASA's NEOWISE satellite in March 2020.
It is now visible in the NE sky just before dawn. The C may refer to it's non-periodic nature (so far). What does the F3 mean in its name ? The orbit may be parabolic.

C means "comet": it has a coma which means that volatiles are being released due to solar heating.  Other possible letters are "A" for asteroid,  "P" for (short) periodic comet, "D" for disappeared comet, "I" for interstellar object and "SN" for supernova.
2020 is the year of discovery
F represent the half-month of discovery, so "A" is the first half of January, and so on. "F" means "the second half of March".
3 means the third comet identified in that half month, and "NEOWISE" is the name of the discoverer, in this case it was discovered by a space telescope that scans the sky looking for Near Earth Objects.
Such designations are considered provisional and can be replaced with a permanent number if the object is found to be in a permanent orbit. So a full designation of Halley's comet is "1P/1682 Q1 (Halley)" as it has permanent number 1, it is periodic, it's orbit was the first published in the second half of August 1682 by Edmund Halley.
See https://minorplanetcenter.net//iau/lists/CometResolution.html


If Mars orbited the Earth how distant would it have to be to cause the same tides? If it were possible to replace the Moon with Mars, how distant would it have to be to essentially create the same oceanic tides as the Moon currently does? Mars seems to be roughly 3 times the mass of the Moon, so does that mean it should be 300% the distance?

Furthermore, would the increased distance cause any issues with Earth's orbit around the Sun or potentially cause some sort of situation like Charon and Pluto where the centre of mass is between them?

The Tidal acceleration between 2 bodies is calculated with this formula:

$$
a_T = \frac{2GM_MR_1}{Dm^3}
$$

Where $M_M$ is Mar's mass, $R_1$ Earth Radius and $D_m$ the distance to the Moon. If you equal this to Moon's Tidal acceleration you will get $D_M$ as distance to Mars to get the same Tidal acceleration having $\frac{M_M}{M_m}=8.73328184501$:

$$
D_M = \sqrt[3]{8.73}D_m\simeq2.06Dm
$$


What is it about the layout of the three gratings in a Bahtinov mask that makes its indicate focus in this way? The question Lucky imaging with Celestron 14 - is this result reasonable? includes a beautiful example of a focus test using a Bahtinov mask and the image below from Deep Sky Watch's Home Made Bahtinov Focusing Mask & Templates shows how the middle spike moves left-right with respect to the crossing point of the pair of spikes on either side.
Question: But optically speaking how and why does this happen? What is it about the layout of the three gratings in a Bahtinov mask that makes its indicate focus in this way? Why exactly does changing focus produce a lateral shift of one pattern with respect to the other two? What is it about the mask's peculiar design that makes this happen the way it does?
update: This answer to How does a Bahtinov mask work? explains more what happens, but it does not even begin to address Why exactly...
 Source

Star image produced by Bahtinov focusing mask at different focus positions

The movement of the diffraction spikes is similar to the phase detection autofocus that's been used in SLR cameras for decades. Before the days of autofocus, many SLR cameras had a split-image viewfinder that used the same principle as well1.
To keep things simple, let's consider a mask with just two slits out at the very edge of the aperture.

Now let's consider what that does as we're focusing. I'm going to draw things in terms of a refractor, but using a reflector doesn't really change things. So, here's a simplified (and drastically not to scale) diagram of light coming from a star, going through the two slits above, and coming into focus on a plane behind the lens:

So, we have a star on the left, two rays of light leaving the star and traveling to the two slits, going through the slits, being refracted by the lens, and coming into focus on the plane. The image is in focus when the two rays of light come back together precisely at the surface of that plane. So what we see when we look at this should be basically a single line. We really have two diffraction spikes, one from each slit, but since we've brought it into perfect focus, they'll line up with each other perfectly. So, in simplified (and hand drawn) form, what we see looking at this, is a picture like this:

Two diffraction spikes, but they align perfectly with each other. I've drawn a boundary between the two, but in real viewing, there won't be an obvious boundary at the edge of each diffraction spike--we'll just see basically a single line.
So let's consider what happens when we move that plane, so the image is out of focus. The rays of light continue to be refracted to exactly the same degree, so we end up with something like this:

In this case, we can see that the two aren't hitting the focus plane at the same point, so what we'll see will be two mis-aligned diffraction spikes:

That explains how (some of) the spikes appear to move as we focus. That leaves another obvious question: how do we get the other spikes that don't move as we focus? That's actually pretty simple. If you look back at the first picture above, there's a crucial fact to note. The slits are (at least approximately) tangent to the aperture.
To form the X-shaped diffraction spikes, the Bahtinov mask uses slits that are radial instead. Technically, these do still move (a little) but it involves the diffraction spike moving along its length instead of sideways, so it's much less visible even at best. There's also little or nothing at the end of each diffraction spike, so you don't have much to compare against to see the movement.
Beyond that, it's all "user interface" enhancements to make it easier to use: More slits, carefully aligned with each other, project diffraction spikes on top of each other, so you get what looks like a single diffraction spike that's a lot brighter. The X-shape guides you eye toward the place you should be looking to see whether the spikes are properly aligned, and so on.
Now don't get me wrong: those are important, and definitely part of the genius of the design--but the basic principle of how/why the spikes appear to move relative to each other remains the same (and why some move but others don't appear to).

1. Note that many rangefinder cameras also used split-image focusing, but it worked rather differently--what I'm talking about here is strictly the version used with SLRs. An SLR with a split-image viewfinder used the same principle as shown here. The difference is that instead of using an aperture with slits, it used a pair of prisms at the center of the viewfinder to collect light from near the edges of the lens. But it still used the basic idea of collecting light from two opposite edges of the aperture, and showing them next to each other, and your focusing by aligning the pictures showing from each. It didn't use diffraction spikes, because in normal photography you typically have objects in the picture with obvious lines in them. Diffraction spikes just let us take subject matter that's mostly just dots, and creating lines from them.


Names of specific galaxies in Hubble's XDF I'm working on a little side project that traces and identifies galaxies in Hubble's XDF using OpenCV.

This is my current progress with tracing the more prominent galaxies:

I was wondering if there was some kind of database for the galaxies that appear in the XDF. Like their scientific names, coordinates, distances, and etc.

Here's a link to the image I am using for the project: https://www.nasa.gov/mission_pages/hubble/science/xdf.html

The XDF data release
has images in FITS format, which includes a world coordinate system to map image coordinates to celestial coordinates.
Then you can query SIMBAD
to see what galaxies are near those coordinates.

For example, I loaded a WFC 60mas/pixel image into a FITS viewer,
which indicates that the face-on spiral at center left of the PR image
is near J2000 3h32m40.8s -27d46m16s.
(To align the PR image with the equatorial coordinate grid, rotate it about 50 degrees clockwise.)
SIMBAD found this galaxy at that location, B-band magnitude 22.

To do this in Python code, you could try using
astropy.wcs
to get celestial coordinates from the FITS image and
astroquery.simbad
to look up objects of interest.


Safety while using a telescope to view a total solar eclipse I have an 8-inch reflector telescope and a solar filter that goes on the front aperture to view the Sun (most recently to view the transit of Mercury). If I were to use this telescope during a total solar eclipse, would it be safe to remove the filter during totality? Let's assume I also get the filter back on before totality ends.

Yes, you can remove the solar filter during totality - otherwise you will have no way to see the corona (or anything at all) through the telescope during totality.
The dangerous time is (as you also say yourself) towards the end of totality - and it's easy to get lost in time and the joy of the moment. Set a timer to wake you seconds before end, have others observing with you do the same. You absolutely must put back the filter in time prior to end of eclipse or risk your eye sight looking through. If you (only) use it to take images, it's not as critical as it's not your eye being fried but your camera, but don't go beyond diamond ring without filter and camera attached either.
Thus either case: Put back the filter preferentially before the end of eclipse.


Why does the Parker Solar Probe slow down as the distance from the Sun increases? Why does the Parker Solar Probe slow down as the distance from the Sun increases?

Image credit: Wikipedia user Phoenix777, CC BY-SA 4.0

Why does the “parker solar probe” lose speed as the distance from the sun increases?

Because energy and angular momentum are individually conserved quantities in the two body problem. Except for where the Parker Space Probe has close fly-bys with Venus, the gravitationally interactions between the Parker Space Probe and the solar system are very closely modeled as a two body problem (the Sun and the probe), plus very small perturbations from the planets.

One way to express conservation of energy in the two body problem is the vis-viva equation, $$v^2 = \mu\left(\frac2r - \frac1a\right)$$

where

$\mu = G(M+m)$ is the sum of the central body's standard gravitational parameter and that of the orbiting body,
$r$ is the distance between the two bodies,
$a$ is the semi-major axis length (a constant), and
$v$ is the magnitude of the velocity vector.

Note that the mass of the Parker Space Probe is so much less than that of the Sun that one can drop the Parker Space Probe's mass from the expression $\mu = G(M+m)$, resulting in $\mu = GM_{\text{sun}}$.

Note that the only variable on the right hand side of the vis-viva equation is the radial distance. As radial distance increases the square magnitude of the velocity vector (and thus the magnitude of the velocity vector) decreases.

Without mathematics, conservation of energy dictates that the sum of an orbiting body's kinetic energy and gravitational potential energy must remain constant. As the orbiting body moves further from the central body, the orbiting body's potential energy increases, which means its kinetic energy must correspondingly decrease. This in turn means the orbiting body's velocity must decrease.


What's the difference between the Julian and the Julian Ephemeris Day? I'm making a code where I calculate the sun position in a given time. But I also want to learn astronomy with it. I'm following a .pdf to make the program, and in this .pdf it says

  The Julian Day (JD) is calculated using UT and the Julian Ephemeris
  Day (JDE) is calculated using TT.

Is it the only difference?

Is it the only difference?

By definition, yes. It's important to keep in mind that the difference between TT and UT (UT1) is not constant.


What scientific evidence is there to support or refute the Iron Sun Hypothesis? Professor Oliver Manuel,  a professor of nuclear chemistry at the University of Missouri-Rolla, claims in the article Sun Is Made Of Iron, Not Hydrogen, Professor Says that

  iron, not hydrogen, is the sun's most abundant element.

Specifically, he and his colleagues postulate that

  "We think that the solar system came from a single star, and the sun formed on a collapsed supernova core," Manuel says. "The inner planets are made mostly of matter produced in the inner part of that star, and the outer planets of material form the outer layers of that star."

Is there evidence to support or refute Professor Manuel's theories?

If the sun was principally composed of Iron, then this would be apparent in the spectrum of light from the sun (which is how we know what stars are made of) (but see comment by @Keith Thompson below, the spectra only tells you about the surface, rather than the core).  The fact that all solar physicists bar one seem to think it is made up of hydrogen and helium strongly suggests this is not the case.

If the sun were made of iron, then I suspect that would require a revision of our understanding of nuclear fusion, as iron is the product of the final fusion reaction that stars can support, unless of course the theory is that the sun is no longer undergoing fusion reactions and all we have been seeing for the last 4 billion years+ is the residual heat of the core.

Apparently heliosiesmology suggests there isn't that much iron in the suns core (although the physics is beyond me, so caveat lector).

As a further point, if the sun is primarily made of iron, it becomes difficult to explain why the planets (especially the gas giants) have so little of it.  It is reasonably straightforward to explain why the inner planets have lost much of their original hydrogen and helium, but rather more difficult to explain how the outer planets could have become hydrogen enriched.

Update.  Dr Manuel's website claims that "About 99% of the sun has the same composition as ordinary meteorites.".  The density of carbonaceous chondrite meteorites is about 2100-34700 kg/m3 (with porosity being an issue).  The mean density of the sun is 1410 kg/m3.  Given the temperature and pressure within the suns' core, if it had a similar composition to a meteorite its' porosity would be zero, and hence even more dense than an ordinary meteorite.  There is a problem with accounting here, if 99% of the sun had a composition similar to that of a meteorite, it would have to be full of voids to have the suns observed density.  The figures just don't add up as far as I can see (although how materials may behave at ultra high temperatures and pressures means intuition may be invalid here).

For Manuel to be right, pretty much all we know about stars must be wrong.


Which stellar properties can we describe as "first principles" in which we can derive the rest? Mass, size, temperature, luminosity, chemical composition, the initial abundance of the molecular cloud, distance, brightness, age, and evolutionary cycle can all be used to characterize a star. A star's brightness is dependent upon its distance and luminosity, so brightness is not a "first principle"; however, there are some properties that seem less obvious in determining their dependencies. For example, chemical composition and temperature seem to be at a constant interplay. Does one always inform the other? Chemical composition and mass influence a star's evolutionary cycle, so it's evolution through the HR diagram is also not a first principle (from my understanding).
For fun, if we wanted to describe a star in the most simplistic way possible, which properties would we need to know? My assumption would be the initial chemical composition when the star was made and its initial mass. If we wanted to know the star's brightness for observational purposes we would also need to know the distance, but this isn't a fundamental property of the star. In other words, what are the basic properties we need to know to extrapolate the rest? I'm, betting on mass and initial chemical composition, but am unsure about temperature. We can throw in distance for observing properties but I was mainly using this as an example.

Starting from a protostar, one would hope to be able to predict everything about its future development if we knew its initial mass, chemical composition and angular momentum.
Mass is fundamental because it determines how much fuel the star will have and the pressure at its core. Composition is key because among other things it determines the opacity of the stellar material. Angular momentum is important because rotation can alter the structure of the star, if fast enough, and can also enhance internal chemical mixing.
Of course extrinsic factors like interactions are also possible, so the properties of any companions are also fundamental, as might be the birth environment that can do things like hasten the dispersal of circumstellar material from a protostar.
If you are talking about figuring out the past and future of a star observed midway through its life, then in addition to the above, we may need to know how the chemical composition varies with depth inside the star, since it will not be uniform.


If we count Avogadro's Number of stars that are closest to Earth, how big that space would be? I think that the answer to this question will be useful as a scale to appreciate the size of the universe.

So, If we count Avogadro's Number of stars that are closest to Earth, how big that space would be? And how many galaxies/clusters will be engulfed in that space? And for a start, let's just consider only three dimensions.

It's going to be all (or perhaps nearly all) of the observable Universe. Roughly speaking, there are several hundred billion stars in the Milky Way. And extrapolating the number of galaxies in deep Hubble images suggests something like a hundred billion galaxies in the observable Universe.  Put these together and you get about $10^{22}$ to $10^{24}$ stars in the observable Universe.  Avogadro's number is $6.022\times 10^{23}$, which is in the upper end of that range.  So, given current estimates, you'd need at least most, if not all, of the observable Universe to get $N_A$ stars.


When did the Moon become tidally locked to Earth? According to various theories the Moon was created around 4.5 billion years ago. About all of these theories suggest that it was rotating around its axis at that time though. Currently, Moon is at tidal lock with Earth, despite some monthly "wiggling" a flat zero on the long-term rotation speed relative to it.

I wanted to ask when did that stop occur - relative to Moon's age, how long was the period of rotating Moon?

The answer would shed some light on my other question - Why are most lunar maria on the visible side? as Earth tends to catch or deflect many bodies heading for Moon surface from "our" direction - still, there is no erosion on the Moon, so craters once formed are extremely slow to vanish - if that period was relatively long, Earth's "protection" wouldn't explain the maria, as rotating Moon would get 'cratered' uniformly all over its surface.

"Protection" isn't the only effect of Earth. Here is a different POV: Earth may have accelerated impactors by gravity assist.

A different approch is the thinner crust, as suggested for the near side, which may have allowed asteroids to penetrate Moon's crust, such that lava could flow into the basins, or which may have favoured volcanism on the near side (see "Lunar interior" on this site).

A third approach is the protective property of Earth preventing the near side to be covered with many new craters, hence leave the maria visible.

According to Wikipedia the time to lock tidally is about
$$t_{\mbox{lock}}=\frac{wa^6IQ}{3G{m_p}^2k_2R^5},$$
with $$I=0.4m_sR^2.$$
For Moon $k_2/Q = 0.0011$, hence
$$t_{\mbox{lock,Moon}}=121\frac{wa^6m_s}{G{m_p}^2R^3}.$$
With Earth's mass $m_p=5.97219\cdot 10^{24}\mbox{ kg}$, Moon's mass $m_s=7.3477\cdot 10^{22}\mbox{ kg}$, Moon's mean radius of $R=1737.10\mbox{ km}$, $G=6.672\cdot 10^{-11}\frac{\mbox{Nm}^2}{\mbox{kg}^2}$we get
$$t_{\mbox{lock,Moon}}=121\frac{wa^67.3477\cdot 10^{22}\mbox{ kg}}{6.672\cdot 10^{-11}\frac{\mbox{Nm}^2}{\mbox{kg}^2}\cdot{(5.97219\cdot 10^{24}\mbox{ kg})}^2(1737.10\mbox{ km})^3},$$
or
$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$
Parameters are $w$ the spin rate in radians per second, and $a$ the semi-major axis of the moon orbit.

If we take the the current simi-major axis of the moon orbit of 384399 km
and a maximum possible spin rate of
$$w=v/(2\pi R)=\frac{2.38 \mbox{ km}/\mbox{s}}{2\pi\cdot 1737.10\mbox{ km}}=\frac{1}{4586 \mbox{ s}},$$
with $v=2.38 \mbox{ km}/\mbox{s}$, Moon's escape velocity, 1737.1 km Moon's radius,
we get
$$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}\cdot \frac{1}{4586 \mbox{ s}}\cdot (384399\mbox{ km})^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}\\
=501416\mbox{ s}^{-1}\cdot \mbox{ km}^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}=
5.01416\cdot 10^{14} \mbox{ s}.$$
That's about 16 million years, as an upper bound.

If we assume a higher Love number for the early moon, or slower initial rotation, the time may have been shorter.

The time for getting locked is very sensitive to the distance Earth-Moon (6th power). Hence if tidal locking occurred closer to Earth, the time will have been shorter, too. That's likely, because Moon is spiraling away from Earth.


Is this a real photo or an art work? I've never seen this Hubble picture, is it an artist impression? It's fascinating because there are voids apparently cause by supernovae.

From the video 1 Hour of Epic Space Music: COSMOS - Volume 1

The deviant art mentioned in the information below the video is by Natasha Sharma, she has a different style than that. There are also real photos from hubble in the video, I can't tell if it is a genius space artist or a real photo.

It is 'From Nothing', made by cosmicspark, a.k.a Tim Barton.

  This picture was painted from nothing, in roughly 12 hours while I listened to a lot of interesting lectures on physics, including Dr. Lawrence Krauss' famous lecture "A Universe From Nothing".


Will I be able to see the Leonid Meteor Shower during mid night? Most of the articles online say that the peak will be in a couple of days and around 2:00 AM in the night. I am currently living in Dubai, UAE.

When I use Google Sky Maps, it shows both the Leonids Meteor Shower and Leo Constellation below the horizon, when using the search feature. It only comes above the horizon at around 6:00 AM, when it is 45 minutes since sun rise.

So, I am not sure whether it'd be dark enough to see them. What is my best course of action.

EDIT:
This link: https://www.timeanddate.com/astronomy/meteor-shower/leonids.html shows Leonids will appear at 65° Azimuth and 1.7° altitude. How'd I interpret that?

Leo starts to rise at about midnight, local time, so you should be fine, weather permitting.  Note also that you may be able to see meteors even before then, as Leo is merely that part of the sky from which the meteors appear to originate, but you will likely be able to see them some distance away from Leo.

I don't know where the Google rise time comes from.

EDIT; the link you added shows Leo rising at about midnight, confirming that the google source is incorrect, or not being correctly interpreted.


Flux density to flux The definition of flux and flux density is fairly straightforward. However, I'm musing over the practical usage of it.

For one thing, if you observe a source through a filter, how would one go from the total observed energy/number of photons to the flux per unit wavelength? It's not a matter of simply dividing by the width of the filter -- a filter can have any exotic form of course.

So, how is this derived?
Related, given a $f_{\nu}$ in some filter, what would be the way to go to total flux within that filter?

If the normalised filter response function is $R_{\nu}$ then the measured  flux is
$$ F = \int f_{\nu} R_{\nu}\ d\nu  $$
The integration is done over the frequency range of the filter.

If you measure a flux through a filter then the process cannot be inverted exactly. However the average flux density can be found by dividing the total flux by the effective frequency range
$$\langle f_{\nu} \rangle = F/ \int R_{\nu} d\nu $$


Can the Universe be spatially closed and expand forever? In the lambda-CDM model, the Universe is spatially flat, and will expand forever. Is possible that a spatially closed Universe expands forever?

Yes, if the Universe is matter-dominated and spatially closed, then a cosmological constant:

$\Lambda > \frac{4\pi G \rho}{c^2}$

Where $\rho$ is the critical energy density (of matter) of the Universe, will cause it to expand forever.

NB Where $\Lambda$ is equal to this value you get the (unstable) Einstein static Universe.


How to calculate the temperature of a star I need a way to calculate the effective temperature (surface temperature) of a star for a stellar model. I need something in the form Te=....

I have:

Radius in m
mass in kg
the composition of particles (eg H 90%, He 8% etc)
the combined stored thermal energy of the body in J

Constants (any really but I'm using these for now):

G=gravity constant=6.67408E-011
k=kbolzmann=1.3806485279E-023
s=sbolzmann=5,67036713E-008
PI=pi ~3.14...

Example of the sun:

mp=average mass of a particle=1,7E-027
M=total mass of the body=2E30
r=radius of the body=700000000

I'm using this equation to estimate the core temperature :

(G*mp*M)/(r*(3/2)*k)

which nets 15653011 for the sun which is close enough given that that is the only star core temperature known (afaik).

I'm using this to estimate the luminosity L:

4*PI*(r^2)*s*(Te^4)

which results in an error of ~1-5%  with 90% of my sample stars which is close enough. For the sun this results in 3,95120075975041E+026 W which is only 2,7% off.

The problem is I need Te for the 2nd formula which I don't have in my scenario.

Due to the formula for L being dependent on the surface temperature to the power of 4 this value has to be relatively precise.

Assumptions of my model:

uniform distribution of particles: so every slice of the body has the same composition as the entire body.
perfect sphere: every body is a perfect sphere, no handling for elliptic bodies needed.

My sample values (first line is the sun with a core temp of 15000000):

     emitted energy Surface temp    radius       mass
     (in Lsun)      (in K)          (in m)       (in Msun)
     1              5800            700000000    1
     8700000        53000           25200000000  265
     6300000        50100           23100000000  110
     2900000        42000           23660000000  132
     2000000        44000           16800000000  80
     1260000        13500           140000000000 45
     57500          3600            618100000000 12.4
     78             5700            6440000000   2.56
     78.5           4940            8540000000   2.69
     15100          7350            51100000000  9.7
     1.519          5790            858900000    1.1
     0.5            5260            605500000    0.907
     370000         3690            994000000000 19.2
     123000         33000           7560000000   56
     2200000        52500           12600000000  130
     200000         10000           151900000000 22
     446000         19000           43330000000  42.3
     25.4           9940            1197700000   2.02

Errors in luminosity to actual value (the maximum error is about 100% which I can live with since it might just be inaccurate measurements for the sample stars)

2.74%
6.71%
-1.13%
11.29%
-2.00%
-4.27%
106.76%
3.99%
2.51%
-6.50%
1.12%
4.00%
-8.27%
2.10%
1.57%
113.75%
1.64%
2.15%

Empirically (I fit a regression on log(mass) vs log(surface temp)), using the table of values in the article on Main Sequence stars, I get a fairly well-fitting formula:  $\mathrm{estTemp} = 5740*\mathrm{mass}^{0.54}$, where estTemp is in C and mass is in multiples of the sun's mass.  Seems to work very well for all but the largest and smallest main sequence stars (and not TOO bad for those).