Where can I find permissibly licensed database of stars and constellations? I'm working on a game engine of sorts, and one thing I did for a previous game which was under the GPL was I had a night sky with the constellations visibly connected and I'd like to do that again.  I don't know where my data set originally came from but I do have a note that it was GPL, so it likely came from Stellarium, which is GPL licensed.  For this I need the dataset to come from the public domain, us government, permissively licensed, etc.

If possible I'm after the coordinates in the sky, brightness, maybe class and name, but also some mechanism to know which stars would be connected in classic constellations.

I've found some data sets which contain the star data I'm after, though they would take some effort to parse and pull out the data I'm after, but I can't seem to find anywhere online, except stellarium, which provides star constellation connectivity. Any ideas?

I've gotten this list so far from what I can tell:

Stellarium: Has everything. GPL (Really restrictive) But has things like a constellationship file, i.e. https://github.com/Stellarium/stellarium/blob/master/skycultures/western/constellationship.fab
HYG: Great database, has everything I want, but licensed under restrictive CC 2.5 SA https://github.com/astronexus/HYG-Database No Constellations
BSC5: No explicit license, but created from NASA/ADC, data should be public domain.  Good data set, difficult to parse, no star connectivity.  (Best candidate for raw star database)
UCAC5: Navy (public domain), data set not geared for visible light.
HIP (used by Stellarium for stars) - Difficult to parse through, ESA (correct license) http://cdsarc.u-strasbg.fr/ftp/cats/I/239/hip_main.dat.gz

Are there any standards of what stars make up the primary connections in constellations?  Or does it just happen to be whoever puts it together?

P.S. If I ever do succeed, I will release the database (probably 3-4k stars is all I will need with connectivity) under the MIT license to permit inclusion in any commercial or noncommercial projects.

I have an idea how to do this, but it requires a few steps:

Yes, the IAU set down which connections make up the classical constellations, but I didn't see an easy way there to get their data.
Wikipedia has a list of stars by constellation. The stars there per list are many more than you need, BUT: the primary connectors of the constellation will always be made up by the brightest stars in a constellation, which will be the first entries in each list.
To access each list's first entries, I would recommend to modify the simple script from this question's accepted answer, which accesses a website and then datamines it for the interesting stuff.
Now not every constellation has the same amount of stars, but probably mining the 20 brightest ones each should be enough (brightness designators are the greek alphabet in order, $\alpha$ for the brightest, $\beta$ for second and so on.).
Then adding the connections: I don't have a good idea for that one. Possibly you have to bite the bullet there, grab a bottle of wine in the evening, and mark the connections by hand. 88 times 5-10 pair connections. Depending on the efficiency of how you mark the connectors in your database, you could be done in 1-2 hours. Still much faster than without any preselection. Should no other answer cover this step in a more efficient way, at least you could now get back to the IAU website and get the official connectors.


Code/Package for multiple planet simulation I want to do a quick study and some tests on the planet-star dynamics ( including the movement of the star). I'm looking for a code that simulates some planets orbiting a star. I've tried searching online and the ones that I found are either incomplete or do not seem to give realistic results. With that in mind I'm wondering if there is any code/package that does this already (in Python).

There are of course serious Python packages to simulate solar systems out there, one of which might be found in The random walk of cars and their collision probabilities with planets but I think that other answers that will be posted will give a more complete answer.

I looked into a "roll your own" script as described in answers to How to calculate the planets and moons beyond Newtons's gravitational force? If you don't need to propagate for millions of years, it should be pretty easy and extremely instructive to just do the numerical simulation yourself.

update: The Python package Poliastro may also do exactly what you need (poliastro in Space SE), and while Skyfield does not do this, there are rumblings in Github issues about adding an orbit propagator to it in the future. PyEphem has been depreciated (same developer as Skyfield) but I believe it also does orbit propagation of user-specified objects, though I'm not sure if it's n-body or just Keplerian orbits.

I think that Universe Sandbox will work for you. See also Is a General Relativity approximation available in Universe Sandbox (1 or 2)?

Though it is a complete program and not in Python, Scott Manley talks about Universe Sandbox in several videos.

old one from 2014 is Universe Sandbox 2 (Alpha) - Simulate The Universe For Fun!
"The Goblin" and an Exomoon The Size Of Neptune - Universe Sandbox
Universe Sandbox² - Planet Nine, Asteroids Orbiting Earth & more


Distance between stars Epsilon Eridani and Tau Ceti? I am working on a story. It would be helpful if someone could provide the distance in light years between star Epsilon Eridani and star Tau Ceti.

As answered by Wolframalpha, (as linked by chirlu) The distance between Tau Ceti and Epsilon Eridani is 5.45 light years.


What would happen if an astronaut orbiting Earth was exposed to radiation? I know that astronauts are constantly being protected from radiation from the sun and the cosmos. They are normally very safe from the harmful effects of radiation but what if their protection wasn't effective at all? What would be the effects on the human body if it is exposed to radiation?

I guess that depends on how much radiation they're getting. I should also point out that astronauts are not 100% protected from radiation, as discussed in this question. One of the most common affects on actual astronauts is the development of cataracts. They can also receive damage to their nervous system over time, among other effects.

But it appears that you're asking what would happen to a person who was exposed to a huge amount of radiation, such as if they went into space with a suit that did little to protect from radiation. I think to answer that, we just have to look at cases where humans have actually been exposed to huge amounts of radiation.

One interesting example that comes to mind is Anatoli Bugorski who was a Russian particle physicist who accidentally had his head inside a particle accelerator when it was on. His head was subjected to huge amounts of radiation. The results of this are described as

  The left half of Bugorski's face swelled up beyond recognition and,
  over the next several days, started peeling off, revealing the path
  that the proton beam (moving near the speed of light) had burned
  through parts of his face, his bone and the brain tissue underneath.
  As it was believed that he had received far in excess of a fatal dose
  of radiation, Bugorski was taken to a clinic in Moscow where the
  doctors could observe his expected demise. However, Bugorski survived
  and even completed his Ph.D. There was virtually no damage to his
  intellectual capacity, but the fatigue of mental work increased
  markedly. Bugorski completely lost hearing in the left ear and only
  a constant, unpleasant internal noise remained. The left half of his
  face was paralyzed due to the destruction of nerves.1 He was able to
  function well, except for the fact that he had occasional complex
  partial seizures and rare tonic-clonic seizures.

Now imagine what could happen if a person were to be exposed in a similar fashion to highly energetic particles in space, but throughout their body and not just their face. I expect that a person could not survive such an exposure and that their body would be destroyed. Even if they survived, it is highly likely they would live with crippling medical conditions and very likely develop cancer later in life.


What's the difference between apparent brightness and apparent magnitude? Are they the same? I know that the equation for the difference in apparent magnitudes can be represented as:
$\Delta m=-2.5\log\left(\dfrac{I_{1}}{I_{2}}\right)$
and the apparent brightness of a celestial object is shown as:
$\ B=\dfrac{L}{4\pi\ D^2}$

Some people in my class told me that$\ B$ and$\ I$ were the same thing but used in different cases. This explanation didn't help me though. While playing around with the equations, I found this nice relation:
$\ \dfrac{B_{1}}{B_{2}}=\left[\dfrac {\left(\dfrac{L}{4\pi\ D_{2}^2}\right)}{\left(\dfrac{L}{4\pi\ D_{1}^2}\right)} \right]$
$\ \space\ \space\ \space\ ={\dfrac{4\pi\ D_{1}L}{4\pi\ D_{2}^2L}}$
$\ \space\ \space\ \space\ ={\dfrac{D_{1}^2}{D_{2}^2}}$

So if$\ {B_{1}}={I_{1}}$ and$\ B_{2}=I_{2}$holds true, is this following equation also true?

$\ {\dfrac{D_{1}^2}{D_{2}^2}} \space\ = 10^\left(-{\dfrac{\Delta {m}}{2.5}}\right)$
$\ \space\ \space\ \space\ \space\ =\dfrac{1}{10^\left({\dfrac{\Delta {m}}{2.5}}\right)}$

However, if$\ B$ and$\ I$ are not the same, what is the difference between the two measurements? And can either one be used to accurately measure how bright a non-luminous celestial body (ie. a planet) appears from Earth?

Also... does $\ m$ have an SI unit? I've never used any units when I solve for it

Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name "distance modulus" when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for spectral radiance (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the inverse square law, though, so there's no risk that we're confusing them here.


Magnitude, satellite flare and the Heavens Above app On the Heavens above app for Android and maybe iOS, it shows negative magnitude values for Iridium satellites, but for other ones, it shows positive.  Here is a screenshot

Why is this?

The astronomy "magnitude" scale works backwards: smaller numbers indicate brighter objects.  Back in the days before precision measurements of brightness, stars were categorized by eye, with the brightest being "stars of the first magnitude".  When more precise measurement became possible, this scale was retained, and extended into the negative numbers for very bright objects like Venus, the Sun, and a few of the brightest stars.

The Iridium satellites have enormous, mirror-like antenna arrays.  When one of them is angled correctly, it will reflect sunlight straight at you, producing an incredibly bright flare visible even in broad daylight.  Very few other satellites have large reflective surfaces other than their solar panels, and solar panels are kept pointed straight at the Sun, so they never generate flares.

To give some points of comparison, the Iridum flare listed in your screenshot, at magnitude -5.2, is comparable to Venus at its brightest, visible during daylight if you know where to look.  The other satellites listed, with magnitudes in the 2-3.5 range, have brightnesses between that of Polaris, and that of the dimmer stars making up Ursa Minor.


Precise declination of Moon The moon follows 5.14 deg inclined path. the ecliptic and the Moon's path coincide are called 'nodes', and they travel around the ecliptic in a period of 18.66 years.

Once every 18.66 years, the ascending node coincides with the vernal equinox, and the Moon will reach declinations of about +28° and -28° (much more than the Sun, which reaches ±23°); this is called a 'major lunar standstill'. 9.33 years later, the opposite is true, and the Moon will only reach ±18° during a 'minor lunar standstill'.

I read https://answers.yahoo.com/question/index?qid=20121018091531AArHgNS&guccounter=1 here, the precise declination of moon is at +28.88 deg, over the earth. Where can I find the exact, value from other sources? I don't know how he found this value? "  I get plus or minus 28 degrees 53', over a 18.6 year cycle. Eventually it will cover the whole region, but it will take many thousands of years (depending on how close you consider it has to get to a point to cover it.) " 28 degrees and 53' = 28.88 deg

The 0.88 deg, to time is 53 min, I think it seems to coincide with the fact the moon rises, 53 min sooner, over each day. Is this related to this? 13.2 deg * 4 min/deg = 53 min

The maximum inclination of the moon, for standstill, can it be specified by the eclipse point? On, august 21 2017, an eclipse occurs in 28.88 deg on latitude point. Is this angle the same as the inclination angle specified here: http://www.umass.edu/sunwheel/pages/moonteaching.html - 28.8 deg.

I just want 28.88 deg = 28 deg 53' confirmed.

I just want 28.88 deg = 28 deg 53' confirmed.

Yes, 0.88 degrees is equal to 53 arc minutes, so 28.88 deg = 28 deg 53'.

But the Yahoo source is off in its value. According to this reference, the maximum declination is closer to 28° 44' (28.73°).

Title: Extreme declinations of the moon.
Authors: Können, G. P. & Meeus, J.
Journal: Journal of the British Astronomical Association, Vol. 82, p. 192 - 193
Bibliographic Code: 1972JBAA...82..192K
Available on NASA Astrophysics Data System.

That reference gives a declination of 28° 43' 21" on Sep 15 2006. Using the Lunar Polynomials from the Astronomical Almanac Online, which permit you to calculate the declination to a far greater precision than you can know the time, I calculate a value of 28° 43' 22". Thus, I trust the article.

28° 43' 22" = 28.723 degrees (Sep 15, 2006). The maximum between 1920 and 2050 is 28° 44' 11" = 28.736 degrees (March 15, 1932).


How can I find a date when the moon at zenith over a specific spot on the equator? I would like to know when the moon is directly overhead 0° S, 166° E. Are there any websites that will show me any of those dates?

Unlike the sun, which follows exactly the ecliptic and so is directly on the zenith on the date of the equinoxes, the moon is tilted by 5 degrees relative to the eclipic.  And the direction of this tilt changes over 18 years.
Moreover the motion of the moon is quite fast so even if it is rising due East, by the time it culminates it will have moved and so not be exactly on the zenith.
As you see the moon's orbit is actually one of the more complex orbits in the solar system. It will be quite close to the zenith once a day, and very close once a month, The only way to find if the moon is (within 0.5 degrees) of the zenith is to use planetatium software to model its motion. Possiblities include "stellarium" or if you have programming skills then "astropy".


Why do pulsars turn "off" from rotation? I was looking at a wikipedia article (https://en.wikipedia.org/wiki/Pulsar#Formation,_mechanism,_turn_off)
and I came across an interesting aspect that said pulsars turn off from rotation. What I don't understand is why emission of EM waves would slow a pulsar down?

The first thing you need to recall is that electromagnetic waves do carry momentum as well as energy.  This shows up in effects like light pressure. Specifically a photon of wavelength $\lambda$ carries momentum $h/\lambda$. In and of itself, that doesn't answer your question though, since you are asking about the rotation of the pulsar, and changed to its linear momentum don't affect that. However, there are two ways this can happen. First imagine a rotating sphere which is emitting radiation from its equator. Consider a point on that equator. The radiation it is emitting "forwards" (in the direction towards which it is rotating) will be blue shifted (as seen by a stationary distant observer) while the radiation emitted "backwards" will be correspondingly red-shifted. Since the blue-shifted radiation has shorter wavelength, it has more momentum, the net effect is a force on the source which tends to slow the rotation. (interestingly, the same effect is used at much lower energy levels to trap ultra-cold atoms in what is called "optical molasses").

The second effect is a bit more technical. The radiation which is emitted towards the poles of the pulsar has a property called "circular polarization" and a beam of circularly polarized light carries away angular momentum.


Wondering about a horseshoe orbit Since I'm not a physicist, I want to know how that system works.
For instance, if there was a planet, very similar to the Earth, but with two moons, and the moons where in a horseshoe orbit, how does it work? Is it stable?
I need it explained in an easy way, and also how would the moons look like from the Earth?
Thanks!

A basic orbital rule is Kepler's third law, which states that the closer the moon is to the planet, the quicker is its orbit. A body that is close to the Earth will move in orbit faster than one which is distant.

The fundamental weirdness: If a body is in orbit, and you give it a push in the direction it is moving, it will gain energy, move away from the planet, and slow down. Contrariwise, If you give it a pull back, it will drop to a lower orbit and speed up. This is fundamentally weird because we are used to the idea that if you push something it will speed up, for bodies in orbit the opposite happens. Make sure you understand why this is.

Now suppose we have two moons, a big one and a little one. The big one is in a circular orbit (and is big enough that the orbit won't change) The little one is on the opposite side of the planet, and slightly closer to the planet than the big moon (only slightly closer). The little one is also in a circular orbit, in the same direction as the big moon.

Since the little one is closer to the planet, it orbits faster than the big moon, so will slowly catch up the big moon.

As it catches up the big moon, the pull of the big moon's gravity will pull the little moon forward. Now remember the fundamental weirdness: if the little moon is pulled forward, it will move away from the planet and slow down. The big moon is also pulled, but we'll suppose the big moon is big enough for the gravity of the little moon to be negligible.

Now the little moon has slowed down. It is still going in the same direction as the big moon, but now it is going slower, so it falls behind the big moon. Eventually it falls so far behind that the big moon begins to catch it up.

Now the gravity of the big moon is pulling the little moon back, and because of the fundamental weirdness, this makes the little moon drop towards the planet, and speed up.

The process can now repeat itself, with the little moon repeatedly catching up, then falling back from the big moon. An observer on the Big moon would see the little moon slowly approach then fall back, go behind the planet, approach on the other side, then fall back again. It would appear to move in a horseshoe shape, relative to the big moon. Relative to the planet, both moons move in circles or near circles.

How stable it is depends on all sorts of things: tides, other planets, the effect of the little moon on the big. However such orbit can be stable over long periods of time. But perhaps not over the 5 billion years of the solar system.


How strong are the strongest Ioquakes? Jupiter's moon Io is the most geologically active celestial body in our system. It has the strongest volcanoes and quakes. Of course in practice these quakes wouldn't be as dangerous as on Earth for someone on the surface due to Io's low surface gravity (0.183 g). Let's ignore that however, do we know some of the strongest Ioquakes, what would be their magnitude on the Richter scale?

I did not find any Richter-scale numbers for Io, probably since there are no direct measurements yet, or how atmosphericprisonescape puts it:

Nobody put a seismometer on Io yet, so we don't know. The only other bodies except for Earth, where seismometers have been placed so far are the Moon and Mars.

But I found some interesting comparison at on NASA's SpacePlace: High Tideo on Io!

On planet Jupiter's moon Io ("EYE-oh"), the ground itself moves up and down like an elevator taking people to the top and bottom of a 30-story building!

This would mean amplitudes for the tidal waves (and therefore also for regular Io-quakes) of ${\rm 60 \dots 90 m}$! I am not sure though where that number comes from, but it compares with ${\rm 0.2m}$ tidal waves of the Earth's solid crust which is not really measurable on Earth, except with interferometers like e.g. GEO600.
References

Wikipedia on Tital Heating of Io
Are planet-destroying quakes possible or observable?


Are we made of the stuff of one star or more stars? [T]his coincidence enabled stars in the late stages of their lives to turn helium into carbon, oxygen and most of the other atoms that you and I are made of. [...] [W]e're made of star stuff.

Max Tegmark, Our Mathematical Universe: My Quest for the Ultimate Nature of Reality, p. 64

Assuming that the Sun is not in the late stages of its life, I read that as that we are made mostly of star dust, but not Sun dust. Rather, we apparently are mostly made of atoms produced by earlier (now dead and buried) stars.

My question is whether it is likely that there was one immidiate (o.a. carbon-producing) predecessor star to the Solar System, or whether the Solar system likely has more than one (perhaps many) (o.a. carbon-producing) "parent stars"? (And, there seems to be also also a third "possibility": that the build-up of, say, carbon takes more than one generation of stars.) What's the (hi)story here?

The precise presolar history isn't known. I'll try to tell a likely story backward in time.

Our sun probably has brothers and sisters scattered throughout the Milky Way. Further back in time our sun likely was part of on open star cluster like the Hyades. Open star clusters are not stable and eject stars over hundreds of millions of years. (More about attempts to find Sun's siblings.)

Such star clusters typically form in nebulae like the Orion Nebula. Dense nebulae typically contain short-lived stars with multiple solar masses, which typically end as supernova, adding heavy atoms to the nebula and surrounding interstellar medium. Our sun is rich in heavy atoms, easy to detect in spectral lines of sunlight or in solar wind; those atoms cannot have formed in the sun, because its core is not yet hot enough to form these elements; some will never form in the sun. Earth mainly consists of chemical elements heavier than helium. Stars like our sun, rich in heavy elements, are called population I stars (astronomers call all chemical elements besides hydrogen and helium metals). Population I stars (and Earth) can only have formed, at least in parts, from supernova remnants. Old stars, born in a time, when the immediate predecessor(s) of our sun were born, can still be found in the Milky Way. They are called population II stars. They contain less material made of heavy elements than our sun, but still more than provided by the big bang. Therefore there must have existed even older stars, called population III stars.

Supernovae also can destabilize the interstellar medium due to shock fronts, eventually leading to local collapse especially of nebulae.

Presolar grains are still raining down on Earth; "the elements in these particles were made at different times (and places) in the early Milky Way".

By this, the likely answer to the question is, that we probably consist of remnants of several supernovae, and of several generations of supernovae.


During solar eclipses on Jupiter, can the moon(s)' shadow(s) on the surface be seen from Earth with a telescope? All of the Galilean moons are large and close enough to Jupiter that they can completely eclipse the sun and allow a solar eclipse to happen.

My question is can an observer from Earth, see the moon(s)' shadow(s) on the surface of Jupiter with a telescope or are the shadows too small?

Yes, you can see it (I have seen Io's shadow on Jupiter and we were happy it was visitor night so that we could share the view with guests)

A 50cm mirror and 125x magnification allows you to see it when the air is not too disturbed and when you know where and when to look. Likely a  somewhat smaller telescope will do, too, as light sensitivity is not too crucial - more the resolution and magnification so that you can still see the tiny black dot of the shadow. The shadow has about the same diameter as the width of some major clouds bands on Jupiter.

Stellarium is a great tool to find the right time to look at Jupiter to find these transits.

Find a local amateur observatory and they sure will be happy to share this view with you (possibly not before late northern summer due to the rise time of jupiter in 2020)


What makes protoplanetary disks start rotating? (Initial energy needed to rotate) Planets form from a protoplanetary disk that has been rotating around its star. The initial energy that makes them rotate really matters to me.

Why did the protoplanetary disk start rotating around the star?
Where did the initial energy to rotate come from?
Why hasn't all the disk material been swallowed by the star?

Two rocks placed in space with no relative motion are going to be attracted by gravity, and hit. 3 rocks, placed in space with no carefully rigged symmetry, will likely miss each other, as the gravitational attraction of the additional rock changes their course. Those near misses are the beginning of rotation. Multiply that effect by trillions, and you have a protoplanetary disk. Of course, a lot of those molecules, dust and rock won't even start out with zero relative velocity, so near misses are inevitable.


How does the gravity of one planet affect the velocity of other planets? If there was a massive enough planet outside of earth's orbit to affect it, how would the earth's velocity change? I know it would be small, but would the velocity increase or decrease?

The most famous example of this lead to the discovery of Neptune.   Planetary orbits were neatly worked out by Kepler and Newton, so they had a good idea of how fast planets should be moving.   Uranus wasn't moving as fast as it should have, and that lead to the conclusion that there was an 8th planet.

Assuming the two planets orbit in a similar plane and don't cross, the inner planet moves faster so it's always catching up to and passing the outer planet.   When it's behind the planet and catching up, it moves faster, when it's ahead, it moves slower.   The effect is, as you stated, very very small but in the case of Uranus, with an orbital period of 84 years, so it spends a couple decades being accelerated by Neptune until it passes it, then it spends a similar amount of time being slowed down by it.   The observed speeding up and slowing down over years was enough to lead to a calculation of the approximate location of, and shortly after, the discovery of Neptune.

The 3-body problem is complicated and the math involved is somewhat advanced, but you can do a rough approximation of acceleration and altered distance from the expected orbit using Newton's laws of gravitational acceleration g=GM/r^2 over roughly a fraction of the orbital period.  It's surprisingly small, but over a couple months (for Earth) or a couple decades (for Uranus) it works out to about what you might expect.  A noticeable but tiny fraction of the orbit.

Over time, these variations balance out, so the orbits remain largely unchanged over thousands, millions and even billions of years, unless the planets happen to fall into resonance.  Resonance can lead to the gradual destabilizing of orbits.


What are the equations governing stellar evolution (Luminosity, Mass, Temperature, Radius) I'm looking for ('simplified') equations governing stellar evolution. Especially how mass, luminosity, temperature and radius of a star change during it's lifetime. As well as equations which tell you how long a star stays in a certain stellar stage (PMS, MS, ...).

Extra clarification by example:

...

After that a star with a mass between x and y enters the main sequence for a duration of $\Delta T = formula$. During this phase it's mass changes according to $M = formula(t)$. after that it enters the xxx  phase

...

For those who think this question is to broad, please give some suggestions, 'google terms', ... from which I can move forward.

What you want is hydrodynamic stellar evolution. The following starts with a brief review of hydrostatic case to give the fundamentals. The last paragraph discusses the hydrodynamic case.

For hydrostatic case, see this, or this. Simplified stellar evolution assumes hydrostatic equilibrium (i.e., time-independent fashion). The system of equations compose of i) mass conservation, ii) mass transportation (i.e., hydrostatic equilibrium), iii) energy conservation, and iv) energy transportation. The system also requires i) the equation of state, ii) opacity, and iii) nuclear energy generation for being solved in a close form.
The close-form solution is typically in the form of polytrophs (i.e., Lane-Emden equation). More complicated model, which is still hydrostatic equilibrium, will include other factors such as composition, rotation, and binary interaction. Since it is static and equilibrium case, the time-dependent terms are zero.

For hydrodynamic case (i.e., time dependent) as you wanted, there is no close-form solution, therefore only hydrodynamic simulation is the method for the study. The simulation model uses the set of equations mentioned above without dropping the time-dependent terms. See, e.g., MESA.


As the Moon and the Earth are predicted to get into tidal lock, how slow would the Earth rotate? This answer to Will the Earth ever be tidally locked to the Moon? supports the widely held thinking that during the Sun's red giant phase or later the Earth and the Moon should be tidally locked to each other. The Earth is said to have a slower rotation then.
Do we have any idea how long a sidereal day and the mean solar day on Earth would be in several billion years?

The last part of the answer you linked to actually says (right at the end) that tidal locking will never be achieved, with reasoning similar to what I gave in this answer.
That said, even though the Moon and the Earth will never actually achieve tidal synchronization, we can still do the thought experiment and ask, "If there were enough time for the current Earth-Moon system to achieve tidal synchronization, what would the length of the day be at which the rotation of the Earth and the orbit of the Moon would be synchronized?"
To do this, we can assume that the Moon spirals outward due to an exchange of angular momentum between the Earth's rotation and the Moon's orbit.  The Earth's spin slows down as it loses angular momentum, and the Moon moves into a larger (and thus higher angular momentum) orbit as it gains that same angular momentum.  The rotation of the Moon would presumably stay locked to the Moon's orbital period, so it would slow down as well.
So, using $L$ to represent angular momentum, the key equation is
$$ L_{\rm now} = L_{\rm then} $$
where "then" is some time in the future when lock is achieved.  The total angular momentum in the system is constant.
The angular momentum of any object is $L = I\omega$, where $I$ is the moment of inertia, and $\omega = \frac{2\pi}{P}$ is the orbital frequency, related to the orbital period $P$.   For a constant-density sphere of mass $M$ and radius $R$ rotating on its axis, $I = 0.4 M R^2$.  The Earth and the Moon are somewhat more centrally condensed, so their moments of (rotational) inertia are a little smaller than the 0.4 for a uniform sphere.  The leading coefficient is 0.33 for the Earth, and 0.39 for the Moon.
For the Moon orbiting the Earth, it's a good approximation to just treat it as a point mass (since its size is small compared to its distance from Earth), so it has $I = M_{\rm Moon}R_{\rm Earth-Moon}^2$.
Putting all three of these motions (Earth rotation, Moon orbit, Moon rotation) together, we get
$$L_{\rm now} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_\oplus} +  \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm Moon}} +  \frac{M_{\rm Moon} R_{\rm Earth-Moon}^2}{P_{\rm Moon}} \right) $$
All of the values there represent current, known values, i.e. $P_\oplus = 1$ day, and $P_{\rm Moon} = 1$ month = 27.3 days.  Similarly, at some point in the (hypothetical) future, we would have
$$
L_{\rm then} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_{\rm then}} +  \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm then}} +  \frac{M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2}{P_{\rm then}} \right) $$
or
$$
L_{\rm then}  = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2 \right)
$$
Notice that there's just a single period here, since everything is now assumed to be synchronized.  So we could set this equal to $L_{\rm now}$ and solve for $P_{\rm then}$ – except that we have a second unknown in the equation, $R_{\rm Earth-Moon,\ then}$, the new orbital distance of the Moon from the Earth.   Fortunately, we can use Kepler's third law to relate this distance to the orbital period:
$$ P_{\rm then}^2 (M_\oplus + M_{\rm Moon} ) = \frac{4 \pi^2 }{G} R_{\rm Earth-Moon,\ then}^3  $$
To make life a little easier when we substitute into the equation, we could write this as a proportion with the current values, which makes some of the constants cancel:
$$ \frac{P_{\rm then}^2}{P_{\rm Moon}^2} = \frac{R_{\rm Earth-Moon,\ then}^3}{R_{\rm Earth-Moon}^3} $$
which means that
$$ R_{\rm Earth-Moon,\ then}^2 = R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} $$
Substituting that into our expression for $I_{\rm then}$, we finally end up with
$$L_{\rm then} = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} \right) $$
So in principle, we're done - we set this equal to $L_{\rm now}$ and solve for $P_{\rm then}$.  It's not a simple equation to solve analytically, but not hard to solve numerically.
Symbolically there's a lot going on, but we know most of these values, so if we plug in numbers for everything we know and simplify, this becomes
$$
P_{\rm then} = 0.16809413\ {\mathrm d} + 0.27626727\ {\mathrm d}^{-1/3} P_{\rm then}^{4/3}
$$
where the "d" represents units of days.  Solving this equation gives a period of 46.9 days, so that is how long the day, the month (i.e. the orbital period of the Moon), and the rotation period of the Moon would be if all were to become tidally locked to each other.
If you want to see the calculation done in Python, I've posted the code in a gist here. It's a nice example of the usefulness of Python's quantities and astropy's constants.


How difficult is it for an amateur to discover a new exoplanet or astronomical object? I recently read about precovery, where plates or images of an object predating its discovery are used to precisely determine its orbit. It occurred to me that it might be possible that some objects have yet to be identified in open-access astronomical databases, simply as an oversight or maybe as a misidentification (say, mistaking a planet for a star). Is it possible for an amateur astronomer to go through open-access astronomical image data and perhaps use machine learning algorithms to find new celestial bodies? Has it been done? If not, how likely is it that an amateur astronomer could even do so?

Zooniverse has a citizen science project called Exoplanet Explorers that used volunteers to examine data on exoplanet candidates.  It was so successful that it has now run out of data, but there are other space projects that are looking for contributors


The Luminosity of two stars with same brightness but at different PC away Stars A and B are both equally bright as seen from Earth, but A is 120 pc away while B is 24 pc away. Which star has the greater luminosity? How many times brighter is it?

The absolute luminosity of A is bigger than B. Like two lamps.
$$\frac{120}{24}=5,$$
so A is five times farther B, but as the luminosity decays with the square of the distance, then  A is $5^{2}=25$ times more luminous.


Why do we not find larger quantities of helium in the Sun's atmosphere? The Sun currently shines because H is being burned to He, and it has been doing this for about 4.57 Ga. But when we analyze the Sun's atmosphere, we do not find unusually large amounts of He. Why not?

Helium is burnt in Sun's core, and it is heavier (specific weight) than hydrogen.
Hence most of the helium stays in Sun's core.
(More details)


If Kepler-444 planets existed for 11.2 billion years, why fear for life on Earth after six billion years? According to recent news, Kepler-444, a 11.2 billion year old star, has a system of five planets. It has a mass of 3/4 of solar mass. So I believe both Kepler 444 and the Sun will share the fate of 'orange main sequence stars'. If so, while scientists fear for life on Earth in the coming billion years, which is only one sixth for our solar system lifetime, how is it that the Kepler-444 planets have existed for a longer period of time?

The rate of evolution of main sequence stars is highly dependent on their mass. Roughly speaking, the time on the main sequence is proportional to $M^{-5/2}$, where $M$ is the mass of the star. Thus if Kepler 444 has $M=0.75\ M_{\odot}$, it can live for $0.75^{-5/2} = 2.05$ times as long as the Sun on the main sequence.

Another way of saying the same thing is that the Sun evolves 2.05 times as fast as Kepler 444. This is important because even whilst on the "stable" main sequence, burning hydrogen in their cores, stars are still changing because the average mass of a particle in their cores is increasing due to the build up of a Helium ash. This slow change means that the temperature of the core must gradually increase, the nuclear burning rate increases and the star becomes more luminous.

In terms of the Sun this means, depending on the atmospheric composition at the time, that the Earth could become much hotter in a billion years or so, and probably too hot for life (as it currently exists on Earth). Thus it is not the existence of Earth that is in jeopardy, it is the surface temperature that may destroy life.

None of the discovered planets in the Kepler 444 system are (or ever were) far enough from their parent star that they are in the "habitable zone" where liquid water could exist. The significance of Kepler 444 is that it shows that rocky planetary systems can have formed as long ago as 11 billion years, giving such systems more time, and therefore perhaps more chance, to have formed life.


How to find a place's latitude when you know a star's minimum and maximum altitude? All we know that a star's altitude at its upper culmination is 38 degrees and 52 arcminutes and that its altitude at its lower culmination is 12 degrees and 24 arcminutes (so it never really sets at that place).
How can we, through this information, find that place's latitude?

As barrycarter mentions in the comments, if you take the average of the two altitudes, you'll get the altitude of the celestial pole—north if you're in the northern hemisphere, south if you're in the southern hemisphere.  In this case, the average is just the average in degrees plus the average in minutes:

$$
\frac{38+12}{2} = 25 \text{ degrees}
$$

and

$$
\frac{52+24}{2} = 38 \text{ minutes}
$$

This is equal to the latitude of the observation point.  One can also obtain the declination of the star in question, by halving the difference between the two altitudes, rather than their sum, and then subtracting the result from $90$ degrees:

$$
\frac{38-12}{2} = 13 \text{ degrees}
$$

and

$$
\frac{52-24}{2} = 14 \text{ minutes}
$$

If you subtract that from $90$ degrees, you get a declination of $76$ degrees, $46$ minutes.  I know of no bright star particularly near that declination, but perhaps this is just an academic exercise.

The foregoing neglects atmospheric refraction.  This turns out to be not entirely negligible at the level of precision provided in the question.  At an altitude of $39$ degrees, atmospheric refraction produces an additional elevation of about $1.2$ minutes (barring local and daily effects), meaning that if a star is observed at $38$ degrees $52$ minutes, it's "really" at $38$ degrees $51$ minutes or so.  At an altitude of $12$ degrees, the additional elevation is about $4.5$ minutes, meaning that the star observed at $12$ degrees $24$ minutes is "really" at $12$ degrees $19–20$ minutes.  If the altitude figures given are as observed, then the corrected figures should be used instead.

ETA: Here's a page with information on the extent of atmospheric refraction.


What (the heck) is a Super Worm Equinox Moon? Google News feed shows me the following.

What does the term "Super Worm Equinox Moon" mean and has it ever been used before this 2019 clickbait instance?

All those adjectives being smooshed together signify an uncommon event. That's why you've never seen them together like that before.

All 3 conditions have to hold true:

It's a supermoon, which means the full moon coincides with the moons perigee or nearest approach. That can make it appear up to 30% brighter than one at apogee (farthest away). These happen about every 13 months and it doesn't have to be exact so usually, we get 2 of them in a row, like we did this year. February had a supermoon and so did March.
It's a worm moon, which means it is occurring in the month of March (see @astrosnapper's answer for a better explanation of that).
It's during an equinox, basically the first day of spring (or autumn).

If any one of those isn't happening then it can't be called a Super Worm Equinox Moon.

Apparently, the term supermoon (all one word, by the way) is a relatively recent thing. I tried to view it on Google N Gram viewer, but...

It is a particularly bright full moon and it does deserve to have its own terminology, IMO.

Update: The rarity of the event is certainly relative. I saw a tweet from National Geographic that we also had a super worm equinox moon 19 years ago.


How much damage would collision with a similar mass primordial black hole do to earth? All this talk of Planet 9 perhaps being a primordial black hole (PBH) made me wonder how dangerous a PBH collision with earth would really be.

Specifically, if earth collided with a PBH of 1 earth mass, traveling fast enough that they didn't end up orbiting each other, what would happen?

Edit:
From a paper asking Can one detect passage of small black hole through the Earth? the analysis was that for an approximately proton-volume and mountain-mass sized PBH

  "It creates a long tube of heavily radiative damaged material, which
  should stay recognizable for geological time."

but is otherwise pretty hard to detect.

What would change if the PBH were the mass of the earth and the volume of a large marble? How slow would it have to be going before it really messed things up?

Consider an initially stationary particle of matter and suppose a 1 Earth mass black hole flies past it at speed $v$ on a trajectory that passes the initial position of the particle at distance $r$. The particle will be mainly affected by the gravity during a time period of roughly $r/v$ (up to some "geometric" constants), during which time it will accelerate at roughly $GM/r^2$. This means that it will move a distance about $GM/r^2 \times (r/v)^2 = GM/v^2$. So if this is small compared to $r$ it will be left "more or less" where it was, rather than being ripped away. So we can expect a hole of diameter a few times $GM/v^2$.  $GM$ is about $4\times 10^{14}$ so to keep the hole diameter to say 1m you could need $v$ about $2\times 10^7 m/s$ which is about $0.07c$.
We can also estimate the typical velocity change achieved by our test particle from this interaction, which is $GM/r^2 \times r/v = GM/rv$, giving a KE per unit mass of about $$G^2M^2/r^2v^2$$.
So, suppose the Earth has density $\rho$  and radius $R$ the cylindrical shell of thickness $dr$ will mass $R\rho rdr$ (still ignoring "small" constants like $2\pi$). That shell will acquire kinetic energy $G^2M^2R\rho dr/rv^2$ from the interaction. Integrating from $GM/v^2$ to $R$ we get a total energy deposited in the parts of the  Earth which are not "ripped away". $$\frac{G^2M^2R\rho}{v^2}\log\frac{Rv^2}{GM}$$
Using $R = 6\times 10^6$, $GM = 4\times 10^{14}$, $\rho = 5000$, and $v = 2\times 10^7$  (all SI units) we get about $10^{25}J$, not enough to actual destroy the Earth (Earth's gravitational binding energy is about $10^{32}J$) but about $10^{10}$ megatons or an earthquake of about 13 on the Richter scale.
A black hole moving 1000 times slower (typical solar system velocity) would essentially destroy the Earth.


absolute speed of bodies as observed by a distant observer What does a distant observer see if two masses with a given velocity are close enough to each other that time is dilated ?

Scenario (using 2 point masses for simplicities sake):
Mass a: 1,00E+030 kg
Mass b: 2,00E+030

Kinetic energy:

mass a: 1,39037910021726E+046

mass b: 2,78075820043453E+046

this would mean a velocity of
Velocity: 149896229 m/s [ = c/2]
for both masses

distance between masses: 3960 m

using the formula for time dilation  the time would pass at a rate of ~0,5 at mass a and at a rate of ~0,8 at mass b

if both masses are flying parralel in the same direction does the observer see mass a flying with a speed of
74948115 m/s [= c/4] and mass b flying with a speed of 119916983 m/s [= c/2 *0.8]
?

In general relativity, then notion of space and time is local. As underlined in the OP, it means that at a radius $r$ of a star of mass $M$, time will run at $$\sqrt{1-\frac{2GM}{rc^2}}$$ the rate at an infinite distance of the object. For example imagine that $1-2GM/rc^2 = 1/4$, so that at a distance $r$, time runs at half the rate compare to an observer at infinity. If you are initially with an observer at infinity, you travel to $r$ (we neglect the time distortion during the travel), stay here an hour and then go back. When you'll reach the observer once again, your local clock will be one hour late, because 1 hour at $r$ will have passed in two hours at $r=\infty$.

So this distortion of space and time is only local. A massive object, like a star is only impacting in its neighborhood space and time. A distant observer is not impacted by the distortion, so that time and distances are unaffected.

To answer your question, an observer far away (for example on Earth) measuring by some meaning the speed of the two stars will actually see $c/2$ for both of them.


Is it possible to filter radio-waves using another radio telescope? When making an observation using a radio telescope, is it possible to filter out noise coming out from a known source by using another telescope that will focus on that noise source?
By filtering, I mean processing the observation on a computer and calculating the filtered image using the second telescope source. Assume that both of the telescopes are connected and can easily sync on time and location and exchange other information.

Coherent interfering signals are usually referred to as RFI (or Radio Frequency Interference).  One could certainly digitally filter out a coherent interfering signal from a telescope by observing from an additional telescope focused on the source of the interfering signal, given the proper geometry.  Here is a research paper on Radio Frequency Interference Mitigation that outlines some of these very techniques in section 3.4: Post correlation studies.
Incoherent interference is typically referred to as noise, and won't be the same at the two telescopes. This means an astronomer can't use one telescope to filter out the noise at the other. Some examples of noise sources are:

Noise that comes from thermal from the telescope's electronics
Ground backscatter into the antenna receiver
Inaccuracies introduced by analog to digital conversion and other signal processing.


Software to convert RA and DEC into ALT and AZ I have implemented the formulas to convert AR and Dec into altitude and Azimuth in C++ following the book Practical Astronomy with your Calculator or Spreadsheet 4th Edition.

In the book there is an example to check if I have do it fine. This example works perfectly but I want to check if my formulas are correct.

I have use this page to check it but I get different minutes values in both altitude and azimuth values.

I have checked on that page that I get the same LST value, but ALT and AZ have different values. This data I have tested:

Altitude: -15° 27' 40"
Azimuth:  273° 12' 34"

This is the data get it from that page:

Do you know if there is another web page or program to check if the result of my formulas are correct?

UPDATE:
I have tried to get Azimuth and Altitude using a date in year 2000 in web page and my program and this is the result (you can find my program's result at the bottom of the image):

Is this the same problem like the previous one (I'm not taking account for precession)?

Based on stellarium, I think you're not accounting for precession (more after the picture):

As you can see, Rigel's J2000 coordinates and Rigel's current coordinates are different by about the same amount as your calculation.


Physical laws in other universes In the theory of multi-verses, we can imagine countless other universes where the constants of nature, including the amount of dark energy in a volume of space and possibly even shapes of hidden dimensions vary from one universe to another. Is there sound reasoning to believe that the mathematics of nature (or laws of nature) would be the same in other universes?

First of all it has to be stated that it is difficult to have a scientific discussion about multiple universes. The problem with multiple universes is that it can be logically deducted from assumptions and hypotheses, but not tested nor empirically verified. This makes discussions around this topic, like string theory, highly "religious".

There are different multiverse theories coming from cosmology, physics and philosophy. One of the theories comes from the cosmological theory of cosmic inflation. Cosmic inflation states that there are always fluctuations on the microscopic level and this results in energy waves that can form into matter and anti-matter pairs. The cosmic inflation now states that at the beginning of our universe there was a cosmological constant or a big vacuum that is by nature unstable and caused the microscopic fluctuations to happen on a cosmic size and thus resulted in a rapid expansion of the early universe. This cosmic inflation is a variation of the (hot) big bang theory, that explains a number of observations in the universe that are not explained by the (hot) big bang theory. What it doesn't explain is that if I have an energy fluctuation at microscopic level, it can happen that this energy wave splits into two matter pairs or an anti-matter/matter pair. Later these pairs annihilate each other and what has happened at microscopic level evens out on macroscopic level. However, when particles appear at microscopic level, what needs to happen to let these particles exist for a longer time to have stars evolve from this matter.

Now back to the multiple universes. The going theory now is that if such a thing can happen to our universe, such a thing can happen in the endless vastness that contains our universe and multiple universes can start and exists next to each other. These multiple universes might also explain that when I have a cosmic fluctuation, I have two universes that come into existence. One with the matter and one with the nati-matter. Bear in mind!!!! These are all philosophical hypotheses and none of it can be verified, falsified or empirically tested.

If I now have made it logically possible to have multiple universes coming from different cosmic fluctuations, I can now have different universes. For instance, I can have for each matter universe a corresponding anti-matter universe to get the energy balance evened out. Now if I have several universes with each different compositions, I can also have universes where, at least, the laws of nature apply differently. In a universe without dark matter, laws of physics apply differently. Whether the "laws of nature" are totally different in other universes, is difficult to scientifically proof. Especially with theories that are build on our laws of nature.

When you think we're already on thin ice here, philosophers can make it even more unreal. If I have cosmic fluctuations that can happen under different  circumstances, which result in universes with different compositions that then have a different application of the laws of nature, I can equally well argue that there will be universes where certain laws of nature don't exists, while there are some particles that don't exist in that universe.

So, to answer your question briefly, there are theories that say there are other universes that have different laws of nature. However, multiverse theories are highly speculative and are difficult to proof. So, whether these theories are right, we'll probably will never know.


Retrieving input effective radius from Sérsic profile I'm playing around with Sérsic profiles, and would like to retrieve the effective (half-light) radius that I put in by integration of the profile. I'm not managing to, so perhaps someone can help me out.

I create the profile using astropy:

from astropy.modeling.functional_models import Sersic1D
from scipy.integrate import simps
Sersic = Sersic1D(amplitude=1,r_eff=1,n=1)

I can then make a light profile:

x = np.arange(1000)
I = Sersic(x)

Then we can perform an integral over this I(r):

integral = [simps(I[0:i],x[0:i]) for i in np.arange(len(I)-1)+1]

From which one would finally find the radius at which the enclosed flux is half that of the total:

from scipy.interpolate import interp1d

f_int = interp1d(integral,np.arange(len(I)-1)+1)
print 'Half radius: ', f_int(integral[-1]/2.)

However, this does not give me the desired result -- the input radius of 1. Why not? Is this a conceptual error or a numerical one - or both?

The conceptual problem is that the half-light radius is the radius of a circle enclosing half the total light, assuming that the object is circular with a radial intensity profile equal to the Sersic function.

So you need to integrate over circular rings, each of which has an area of $2 \pi r \, dr$ and an intensity (or surface brightness) of $S(r)$:
$$L(< R) = \int_{0}^{R} 2 \pi r \, S(r) \, dr$$
where $S(r)$ is the Sersic function evaluated at radius $r$.

Instead, you're just doing the integral of a line of fixed width:
$$\int_{0}^{R} S(r) \, dr$$

In Python terms, the (pre-calculated) function you integrate should be

I = 2 * np.pi * x * Sersic(x)

not

I = Sersic(x)

(Alex is also correct that you have almost no resolution near the value of r_eff that you are trying to find.)


Why did Earth need material from comets to assemble the raw materials for life? It's often speculated that matter from comets played a crucial role in the development of life on Earth. For example, Maps of Time says,

  Many of the ingredients of the early atmosphere (including much of its water), together with many of the organic chemicals that formed the first life forms, may have been brought to Earth by the comets that bombarded its surface for the first billion years of its life (~Kindle location 1922).

Why the presumption that these materials must be brought by comets? Did Earth even have the right atoms to make water and those organic chemicals (hydrogen, oxygen, carbon)? And what are comets doing with heavier elements in the first place? My understanding is that comets are (for most of their lives) very far from the run, and my impression was most of the elements other than hydrogen and helium and are concentrated close to the sun, with T Tauri Wind having push hydrogen and helium further away.

The earth in its early stages was far too hot for water to be in liquid form and the water vapor that would have been around at the time would have been blasted away via solar wind (http://en.wikipedia.org/wiki/Solar_wind). Anyone claiming to know exactly how water came to be on Earth is likely daft, but comets and asteroids seem to be the most common theory for the source. MinuteEarth, a YouTube creator has a great video about it but as a new user on this particular sub-site I cannot link multiple things at once.

https://www.youtube.com/watch?v=_LpgBvEPozk


How can I correct for transit time? I know the exact time a radio telescope detected a transient event. I also know the exact location of the telescope, and the galactic coordinates (galactic longitude, latitude) and right ascension and declination of the beam pointing center.
I conjecture that the event was also observed by another receiver at another location. How can I correct for the time difference between the two locations (i.e. the event reached one observer earlier than the other due to that they are in different locations)? Currently, I’m taking the dot product of a vector between the known observer and the source, and a vector between the known observer and the potential observer, and dividing that by the speed of light. To obtain the vector between the observers, I calculate their ‘n-vectors’ ( https://www.movable-type.co.uk/scripts/latlong-vectors.html  ) and take the difference. However, I’m getting nonsensical delay values.

This may be a partial answer depending on what you do or don't know how to do at the moment, and the level of accuracy you require. Feel free to add some feedback.
If you'd like to use (or learn to use) Python then you can solve this problem trivially using Skyfield!
It depends on the level of accuracy that you need. The time difference will be of the order of 20 milliseconds for a 6000 km difference in light-path distance for example, but in that time the Earth moves (in some direction) about 0.02 * $\sqrt{GM/a}$ where GM is 1.327E+20 m^3/s^2 and $a$ is about 1.5E+11 meters, or about 600 meters.
If you don't want to worry about small corrections right now, just calculate the dot product between the normal vector pointing from the direction of the event and the vector drawn from the known observation site to the proposed site, that's the path length difference. Divide that by the speed of light to get the time difference.


Are the stars in constellation located in a plane or they are in different distances Are the stars in constellation located in a plane or they are in different distances from each other and when we observe them from the Earth we think they are all in a plane?

Constellations and asterisms are generally not proximate in space, but rather happen to be nearby only when viewed from Earth. From Wikipedia's article on asterisms:

  Like constellations, asterisms are in most cases composed of stars which, while they are visible in the same general direction, are not physically related, often being at significantly different distances from Earth.

A recognizable example of this would be the brightest star in Taurus, Aldebaran.  Aldebaran appears in the V-shaped bull's head, mostly formed by stars in the Hyades star cluster. However, while those stars are actually close to one another, Aldebaran is less than half as far away.

  Since the star is located (by chance) in the line of sight between the Earth and the Hyades, it has the appearance of being the brightest member of the more scattered Hyades open star cluster that makes up the bull's-head-shaped asterism; however, the star cluster is actually more than twice as far away, at about 150 light years.

Similarly, the stars of Orion's belt are actually hugely separated in space.

In addition, it's worth noting that the stars have different velocities across the sky, so the constellations will look increasingly different over long periods of time. Here is one prediction for 50,000 years time, and the effect will get proportionately larger as time goes on.


How is the field of view of a radio telescope determined? It’s my understanding that radio telescopes can only receive signals that hit the dish within a certain range of angles. How is that angular range determined i.e. measured and/or calculated?

Single "pixel" dish
If you have a single point feed at the focal plane (one single-mode waveguide or one $\lambda/2$-sized antenna) then the antenna's reception pattern (or radiation pattern) will be similar to it's diffraction pattern. For a circular aperture and ignoring perturbations due to the small secondary or its supports this pattern will be the Airy disk.
The half-angle of the first minimum is given by the familliar equation used to express the resolution of any telescope, optical or radio:
$$\theta = 1.22 \frac{\lambda}{d}$$
where $\lambda$ is the wavelength and $d$ is the diameter of the telescope's aperture.
So if you have a 20 meter dish and you are receiving a signal at 3 GHz, your wavelength is 0.1 meters and the half-width of the angular acceptance is 0.006 radians or 0.3 degrees.
Arrays of telescopes
If you have an array of radio telescopes and you are using interferometry to generate a synthetic aperture (like the VLA or ALMA or the Event Horizons Telescope (EHT)) then your field of view can be much smaller but it is defined in the software running in the correlator computer.
Dish with a focal plane array
Some radio telescope dishes are equipped with an array of receivers, analogous to a CCD array of pixels on an optical telescope, though much fewer pixels.
In this case the half-width of the field of view (in radians) is the half-width of the focal plane array divided by the effective focal length there, or more precisely the arctangent of that ratio.
For more on that see:

What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?
How do ASKAP's focal plane phased array feeds interact with the entire array phasing?

Cropped from CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background


Why can falling stars only be seen moving down towards horizon not up away from horizon? Are falling stars as we know only just space debris entering and burning up in the atmosphere and not a star at all?

The falling star or shooting star has nothing to do with a star. If shooting stars were actual stars, we'll be burning in their atmosphere, rather than the other way around.

When a meteoroid, comet or (small) asteroid (or space debris in some cases) enters the Earth's atmosphere at high speed (typically in excess of 20 km/s), aerodynamic heating produces a streak of light, both from the glowing object and the trail of glowing particles that it leaves in its wake. This is called a meteor or shooting star (or a falling star).

In general, the meteors travel in a (approximately) straight path. We usually see only the large ones travel down towards the horizon while the small ones are burnt up long before that (giving the impression of coming from the horizon).

Image from Meteors? Even More Remarkable by Bruce Maccabee at brumac.8k.com

Consider the meteor path in the image above. it can be seen that,

The observer will see the meteor only after it passes the initial altitude $H_{i}$. If he sees the meteor at this point, and it burns up in the atmosphere, the impression is that the meteor comes from the horizon.
If the observer notices the meteor after it passes the initial height and tracks it through its travel in the atmosphere, till is exits at the final altitude $H_{f}$, it will appear as if the meteor is moving down towards the horizon.

Another factor is the thickness of the atmosphere itself. The thickness of the atmosphere increases in path of the meteor from $H_{i}$ to $H_{m}$, and decreases from there to $H_{f}$. As the 'glowing' of the shooting star is directly dependent on the density of the atmosphere, its visibility trails of near the observer's horizon. This is another reason for the meteor looking like its falling towards the horizon.

This can be seen clearly in case of meteor showers which appear to radiate from a point.

Image from Lyrids Meteor Shower 2014 by Steve Owens at darkskydiary.wordpress.com

Here, the meteors appear to come away from the (radiant, which is above the) horizon.


Can New Horizons probe turn back and start orbiting Pluto New Horizons probe has found astonishing features on Pluto, while the likelihood of making even greater discoveries is remote in further space. Is it possible for New Horizons probe to turn back and start orbiting Pluto? It may save much time and resources as compared to sending another probe to Pluto. Thanks for your responses.

A Pluto orbiter would have been possible, but then the entire mission would have to be designed around that. New Horizons was designed as a small fast probe that could reach Pluto relatively quickly (almost 10 years), but not be able to stop there. An orbiter would have to take a lot more fuel with it, and approach Pluto slower. Such a mission would therefore take a lot longer to reach Pluto, i.e. it could easily take double the time or more, and would be more expensive.


Spectra of electromagnetic radiation emitted during solar flare During solar flare, electromagnetic energy is released at a wide range of wavelengths but the most dominating are x-rays and extreme ultraviolet rays. Why is that intensity of these wavelengths is greatest?

Flares involve releasing huge amounts of potential energy that are stored in stressed magnetic structures in the solar corona. So-called reconnection events can release this stress but at the same time accelerate charged particles to high energies. These are trapped along the magnetic field lines, spiralling around them, but travelling until they hit the footpoints of the loop that are anchored in the solar chromosphere and photosphere. The much denser plasma here is sufficient to stop the charged particles and this is where they dump all their kinetic energy, some of which is promptly released in the form of non-thermal hard X-rays, which are insufficient to cool the plasma.

The plasma heats up, increases its pressure and it expands to fill the overlying magnetic structures - a.k.a. chromospheric evaporation. This hot plasma radiates and because it is at temperatures between $10^{5}$ and $10^7$ K, most of that radiation is emitted at EUV and X-ray wavelengths in free-free and transitions in highly ionised species.

I guess the root cause as to why the plasma can get up to these temperatures is that the radiative and conductive cooling processes are much slower than the heating processes caused by the accelerated charged particles.


Stellar Information from Hipparcos Is it possible to derive Metallicity/Age/Radial Velocity/Rotational Velocity/Surface Gravity of a star from Hipparcos data files or are do they require individual studying?

Hipparcos measured the parallax (distance) of stars, along with their colors in two wavelength regions (B_Tycho, "blue" and V_Tycho, "green"). Using the color information you can make a reasonable estimate for the stellar temperature, and coupled with the parallactic distance, this would allow you to get an estimate for the stellar radius.

But Hipparcos does not provide us with any good information about stellar metallicities, ages*, radial velocity, rotation, or surface gravity.

* - aside from stars that you can group in position and velocity space to identify as belonging to stellar moving groups with defined ages.


What is supposedly in the center of the Milky Way? It's big. It's bright. It's white. It's so intense. But what is at the center of the Milky Way?

Photos of the galactic center aren't too bright because of all the gas and dust between us and it. For example (in infrared):

I'm guessing, though, that you're talking about other galaxies, because there are no views of the galactic center of the Milky Way face-on. Although the galactic center is pretty luminous, just not in the wavelengths we're used to.

Anyway, the galactic center has lots of stars - many massive, but a handful like our Sun. They're young, though, which is odd, as there isn't much star formation happening.

The main attraction, though, is Sagittarius A*, a radio source that is apparently a supermassive black hole. It's actually a subset of the more complex radio source Sagittarius A. So far, astronomers think that most galaxies have supermassive black holes in their centers.

We know that it has to be really massive because of how it perturbs the orbits of stars nearby:

We know it's massive and we know it's a very strong radio source. A supermassive black hole is just about the only thing it could be.

Courtesy of WayfaringStranger: An awesome picture from NASA:

In this paper by Ghez et al., the orbits of stars were monitored and a mass of the central object was determined to be $4.1 \pm 0.6 \times 10^6 M_{\odot}$. This paper discusses the properties of the star cluster itself.


Can the explanation of Tabby's star KIC 8462852 be an intervening planet in orbit around another star? Can an intervening planet in orbit around a nearby star cause periodic occultation?

Please excuse the crude text graphics:

Tabby ...... PlanetP ......................................... Earth
                |
             StarX

StarX and Tabby's star are close to each other (maybe part of some local cluster, like the Centauri brothers) so the line of sight with our Sun/Earth has maintained over the recent handful of years.
PlanetP orbits StarX with a period of around 750 Earth days.  Its ecliptic is at an angle (somewhat perpendicular) to the line of sight with Earth.
PlanetP's orbit rotates (precession?) a bit around StarX, so the amount of occultation of Tabby's star by PlanetP varies from 15% to 22% to whatever.

The answer is no, because the light curves simply do not match up.

Astronomers have detected planets before using the transit method. Large planets cause dips in a star's light curve when they move between their parents star and Earth; if these dips are regular and strong enough, we can conclude that there is a planet-like body orbiting the star. It's not perfect (best used on conjunction with other methods), but if there is a planet orbiting a star that is detectable like this, we should see a couple key features:

Periodicity, corresponding to the planet's orbital period.
Relative regularity in the depths of the transits.

Transits aren't always easy to see, but sometimes you can rule them out as the cause of light curves. The light curves of KIC 8462852 are one of these cases. Boyaijan et al. (2015) provided some of their light curves from the observations of the star. Here's one from Figure 1(b), showing a number of "dips":

There is neither regularity nor periodicity in the data, meaning that your proposed explanation is not possible. Looking at the dips in more detail (see the rest of the paper) confirms the conclusion.


Why can't we determine the center of the universe I find this baffling. If we can observe objects moving away from us and each other, than it stands to reason that we can track their paths (relative to each other and ourselves) backward to find a point of origin. In this question answers given suggest that there is no center of the universe. I'm looking for clarification as to why we cannot reverse engineer (to some degree) where this point of origin exists, even if a simple general direction relative to us. I understand that paths may be distorted and skewed by forces as time goes on, but I still don't grasp the concept that we have no way to determine a point of origin or general sense thereof. Help wrapping my inferior brain around this is much appreciated.

There is no point of origin! It is a misconception that the universe started as a single point within space. Indeed, measurements since the 1990s seem to show that the universe is infinite in size, which means that the very early universe must have also been infinite.

It has not really been established that the universe originated from a "singularity" at all. We know that in the early universe everything was much closer together, and hence the universe was much denser. The "Big Bang Theory" goes back to the universe being extremely dense, but it is not required that the universe started from a single point.

Really, the Big Bang happened everywhere. Even at these early times it looks like it was infinite in size with basically constant density, and so there's no sensible way for it to have a centre.


How to tell where a meteorite came from I am told in a lecture that some particular meteorite is from Vesta. How can we know that a particular meteorite is from there?

Meteorites have distinct compositions and so meteorites with similar compositions can be grouped into families.  There are, for example the iron meteorites, meteorites with lots of carbon and so on.
One family are called HED meteorites.  They don't contain chondrules and show evidence of igneous processing (the rocks have been melted). The are differentiated.  This suggests that they didn't form from the raw materials in the solar system, but formed on a larger body, and then were ejected into space. The look very like igneous rocks on Earth.
It is suspected that these come from Vesta. They don't come from the Moon or Mars (we know about the composition of these worlds and they don't match). Vesta is big enough to be differentiated, and in the right place to allow for pieces of rock ejected by impacts to be perturbed into Earth-crossing orbits. Observations by the Dawn probe are consistent with Vesta being the source of HED meteorites (By contrast, Ceres has a different composition and so is not a likely source), and there is a large impact crater (the Rheasilvia crater) that could have ejected a huge amount of rock into orbit around the sun.  Some of the debris from this impact would still be falling to Earth, even after a billion years.
So Vesta fits the requirements.  That isn't quite proof that these HED meteorites did come from Vesta, but it is a reasonable belief.


Finding galaxy spectra in SDSS Catalog by object name I am looking to find galaxy spectra of early type elliptical galaxies in the SDSS catalog by the objects name. See: https://dr12.sdss.org/advancedSearch. I dont know how to find out the PLATEID of my objects of interest, e.G. NGC 4125, NGC 4472 (Messier 49) or NGC 4649 (Messier 60), in the advanced search. I have tried to work with the constrains for redshift and sky position, but in the sdss library the objects name never turn up so I dont know if the found spectra are the spectra of the galaxies i mentioned. I am new to working with astrophysical data and the sdss catalogue in particular so please forgive me if this should be approached completely different. Can anyone tell me how to find these galaxy spectra fits files on the sdss website ? I am planning to further work with these fits files in python and extract kinematical parameters.

One thing you might try doing is to use the SkyServer Navigate interface to see if the object was imaged by SDSS. Enter the name in the "name" box in the upper left and then click on the "Resolve" button. If an image with the galaxy shows up, click on the "Object with spectra" checkbox in the "Drawing options" panel on the left, and red squares will appear for objects with spectra. If there's a red square associated with the center of your galaxy, then there's a spectrum. (You can click on the links on the right-hand side to find out more; there may even be a little graphic of the spectrum you can click on.)
Here's an example for the elliptical galaxy NGC 6066, which does have an SDSS spectrum (as indicated by the spectrum-plot graphic to the lower right):

Something you should be aware of: in many cases, very nearby and luminous galaxies like your three examples simply don't have SDSS spectra. (This is true for your three galaxies, in fact.) Since such galaxies were known to already have plenty of spectra (including redshift measurements), there was little need to get SDSS spectra of them, so they were ignored. So you may end up needing to focus on more distant elliptical galaxies (NGC 6066 has a redshift of 0.034, while your example galaxies all have redshifts $< 0.005$.)


Spacing (2422 days) between solar eclipses - unrelated saros numbers I just made an observation that is very puzzling.  The observation concerns total solar eclipses and their spacing in time.

I know that the saros cycle dictates recurrence of same saros eclipses roughly every 18 yrs 11 days apart.  However, the two American eclipses (Aug 21, 2017 and Apr 8, 2024) are spaced by 2422 days.  These are saros 145 and 139 respectively - seemingly unrelated eclipses.

There is another pair of eclipses that are upcoming (Aug 2, 2027 and Mar 20, 2034), and they are also spaced exactly 2422 days apart.  These are saros 136 and 130 respectively.  Again, they are unrelated eclipses from each other - OK expected.  The only thing that these two pairs of eclipses have in common is that their paths CROSS each other 2422 days apart. So, here is my question:

Why are these two pairs of totally unrelated eclipses "dancing" with each other with exactly the same spacing of 2422 days?  What mathematical phenomena could account for this?

Thank you for your time.

Since a solar eclipse is a coincidence of a new moon and
an ascending or descending lunar node,
the interval between solar eclipses must be a common multiple of
a whole synodic month
(or lunation, 29.53 days between same lunar phases)
and half a draconic month
(27.21 days between same lunar nodes).
These line up around eclipse seasons,
173.3 days apart on average.

The saros cycle of 18.03 years is 223 synodic months, 242 draconic months,
or 239 anomalistic months
(27.55 days between lunar perigees or apogees).
This multiple coincidence of phase, node, anomaly, and time of year makes the saros especially useful for prediction.
However, there are other relationships between eclipses.

Between consecutive saros series there is another interval called
the inex:
358 synodic months, 388.5 draconic months, or about 28.94 years.
George van den Bergh showed in 1955 that the interval T between any two eclipses can be expressed as T = aI + bS, where I and S are the inex and saros periods, and a and b are integers.
Some such combinations have names:

                   Synodic  Draconic  Anomalistic  Eclipse
Name      Formula  Months    Months     Months     Seasons

Semester  5I - 8S      6      6.511       6.43       1.02
Hepton    5S - 3I     41     44.493      43.94       6.99
Octon     2I - 3S     47     51.004      50.37       8.01
Tritos     I - S     135    146.501     144.68      23.00
Saros      S         223    241.999     238.99      38.00
Inex       I         358    388.500     383.67      61.00

Source: F. Verleben

Between the eclipse pairs in question, 2422 days = 82 lunations = 10S - 6I or 2 heptons.
The two pairs also happen to be 3 heptons apart.
The same hepton family includes southern hemisphere eclipses in Dec 2020, Nov 2030, and Jul 2037.
Tabulated in a
saros-inex panorama,
all seven eclipses would fit on a single diagonal line of slope 5/3.

            Greatest eclipse  Sun      Moon
   Date      UT     Lon Lat   Decl  EcLat Node  Saros  Inex

2017-08-21  18:26   88W 37N   +12   +0.43 asc    145    39
2020-12-14  16:13   68W 40S   -23   -0.29 desc   142    44
2024-04-08  18:17  104W 25N    +7   +0.35 asc    139    49
2027-08-02  10:06   33E 25N   +18   +0.14 desc   136    54
2030-11-25  06:50   71E 44S   -21   -0.39 asc    133    59
2034-03-20  10:17   22E 16N     0   +0.29 desc   130    64
2037-07-13  02:39  139E 25S   +22   -0.71 asc    127    69

Sources: F. Espenak,
JPL HORIZONS

From a Moon-centric point of view, the umbra of a solar eclipse moves straight along the ecliptic, but several factors shape its path on the Earth's surface.
The terrestrial longitude of a path is directly related to the time of day (UT).
A path's terrestrial latitude is very sensitive to the Moon's
ecliptic latitude
and is also affected by the Sun's declination.
Due to the tilt of the Earth's axis, an eclipse path
slants southwest-northeast near the March equinox,
bends southward near the June solstice,
slants northwest-southeast near the September equinox,
or bends northward near the December solstice.
The Moon's motion in ecliptic latitude also contributes a slight slant,
northward at ascending node or southward at descending node,
as the umbra passes west to east.

Both the 2017-2024 and 2027-2034 eclipse pairs occur near opposite equinoxes and at similar times of day, so their paths naturally have opposite slants and similar ranges of longitude.
The 2030 and 2037 eclipse paths have similar ranges of latitude but barely cross because they occur 4 hours apart in the day.
Eclipses in the same saros series, if not widely separated in longitude, occur at the same time of year and have nearly parallel paths with few opportunities to cross.


How does the geometry for constructing a declining vertical sundial work? I understand how to construct South-facing vertical an horizontal sundials and the geometry behind their construction by projecting an equatorial sundial into a plane, but I can't quite get how to do that for a declined plane.

You might find the following useful; scroll down to "Vertical Declining Sundial".

http://www.walkingshadow.info/Publications/Scratch.pdf


International Space Station Recently, I have been going out to observe the ISS when it is within sight of my location.  I have noticed that the ISS, based on the time that I have to observe the space station during any given transit, moves at a fairly constant speed across the sky, depending on the degree of elevation and maybe some other factors related to the seeing conditions.

My question relates to the different course the space station takes when traversing over my location. There may be as long as several weeks when the ISS is not within viewing parameters for my location. Why does the course across the sky vary from time to time?

I recommend you to take a look at Wikipedia about the ISS's orbit: Wikiepdia - ISS: Orbit

There is also a very interesting video that I think will answer to your question: ISS Orbit animation


What is the reason for high lithium concentrations in Sun-like stars? In reading the European Space Organisation paper Lithium depletion in solar-like stars: no planet connection (Baumann et al. 2010), several conclusions based on their observations include:

Lithium depletion in Sun-like stars is independent of having planets (an earlier theory), but rather is a function of the star's age.  From the article:

  For solar-like stars, the lithium vs age trends for planet-hosts
  and stars where no planets have been found are statistically
  identical. Thus, the presence of a planet does not influence
  the observed surface lithium abundance.

However, they observed that

  A number of solar-like stars with unusually high lithium
  abundance for their age are present in the field.

and that these stars also have an anomalously low surface gravity.

What is the current theory as to why some Sun-like stars have a relatively high lithium abundance and low surface gravity?

I could offer a couple of speculations - but there is no "theory". Incidentally, I was looking at the refereed journal paper by Baumann et al. (2010).

The atmospheric parameters of these stars are not correct. Given that the ages are derived from the gravities and temperatures of the stars and that the lithium abundances are also dependent on the gravities and particularly the temperatures, there will be correlated errors. I'm actually struggling to understand how you get stars which are "solar analogues", all of which have similar gravities, yet seem to have a spread of ages (estimated from the Teff vs log g plot) from 1-10Gyr? It is notable that there appears to be just 1/117 in Baumann's sample, compared with 10/82 in Israelian's work...
Maybe they are actually PMS stars, so the estimated ages are completely spurious since they assume that the stars are on the main sequence. I would however have expected the abundances to be 3 or higher.

Neither of these explanations seems completely satisfactory.

I mention in passing that the question of planets and lithium is far from settled. The latest salvo in the war was fired by Figueira et al. (2014) who claim there is statistical evidence for enhanced Li depletion in solar analogues with exoplanets.


Is the nose of the solar system and the solar apex the same thing? Results from the IBEX satellite indicate the nose of the solar system is pointing to the constellation Scorpius.  Various sources say the solar apex is pointing to the constellation Hercules.

Are the two different, or is simply the information about the location in conflict?

Is the nose of the solar system and the solar apex the same thing?

No, they're roughly 60 degrees apart.
The Solar Apex.
The solar apex does indeed indicate a component of direction of travel within our corner of the galaxy:

The coordinates as obtained by visual observation of the
apparent motion is right ascension 18h 28m 0s and declination
of 30° North in galactic coordinates: 56.24° longitude, 22.54°
latitude. The radioastronomical position is 18h 03m 50.2s and dec
30° 00′ 16.8″ galactic coordinates: 58.87° longitude, 17.72°
latitude.

There are two different precise measurements here which conflict, I won't speculate as to why Wikipedia is unreferenced on the subject. The galaxy is a place containing a great deal of complex movement and the Local Standard of Rest is moving relative to the galactic plane, the centre of the galaxy and the other local stars:

In astronomy, the local standard of rest or LSR follows the mean
motion of material in the Milky Way in the neighborhood of the Sun.
The path of this material is not precisely circular. The Sun follows
the solar circle (eccentricity e < 0.1 ) at a speed of about 255 km/s
in a clockwise direction when viewed from the galactic north pole at a
radius of ≈ 8.34 kpc about the center of the galaxy, and
has only a slight motion, towards the solar apex, relative to the LSR.

We're traveling relative to an ever shifting cloud of stars towards Hercules (also shifting), southwest of the star Vega.

The sun's motion in the Milky Way is not confined to the galactic
plane; it also shifts ("bobs") up and down with respect to the plane
over millions of years.

So you see, there are many components of direction of travel to consider.
There's an unquotable abstract from a Harvard article here, which may give you an idea of how much needs to be taken into account in determining the relative vectors.
The Nose.
The ""Nose" of the solar system is sometimes used to refer to the direction of travel of the local galactic wind. Just as a comet's tail points away from the sun, the tail of the Heliopause points away from the wind, the nose points in the direction it's coming from.

The nose is as you have stated is pointing towards Scorpius.


Selecting a Telescope for Viewing Planets I hope to observe planets like Jupiter and Saturn, and moons. I am a novice when it comes to telescopes and I know images in magazines are not taken with at least 12-inch telescope in a middle of nowhere. Is it possible to view Saturn in little yellowish and Mars in little reddish using following telescopes?  I am going to buy one of them. Which one is worth more for the money with the price difference?

I am living in Sri Lanka (a country close to the equator) in a suburban city.

Option 1
EQ-Reflecting telescope

Aperture: 135mm
Focal length: 800mm Focal ratio: f/6
Max power: 265x when the air is steady
Price: 300USD approx
Link: http://www.ksonoptics.com:8081/Product/PrdDtls/121/

Option 2
EQ-Reflecting telescope

Aperture: 114mm
Focal length: 900mm Focal ratio: f/8
Max Power: 228x when the air is steady.
Price: 220USD approx
Link: http://www.ksonoptics.com:8081/Product/PrdDtls/119/

Go with Option 1 (135mm) as it has bigger light collecting area, i.e. diameter of primary mirror. It will enable you to see faint objects and it will also help you to see the dim objects under a greater zoom eyepiece, e.g. 4mm or 10mm.

Also, don't fool yourself with the bigger numbers advertized by telescope vendors like 238X zoom or 300X zoom.

First, these are zooming power using Barlow lenses also the more you zoom in on any object, the less light you gather on the mirror and hence objects will look much fainter in appearance. I will suggest the 10mm eyepiece with a 135mm telescope (Option 1) will be sufficient for you to observe planets like Jupiter, its Galilean satellites, Saturn's rings and also many Messier objects like Andromeda Galaxy M31, the Honey Bee Cluster, and open star clusters in the Auriga constellation.

Also bear in mind that you live in a country area closer to sea like here in India and the weather will be hot and humid.
So ask the vendor about special coating which protects the mirror from rusting (spots on mirror due to moisture) during the humid monsoon season as at least for 4 months it will be humid.


Question about stellar remnants I know that after a star undergoes the process of mass loss, depending on the mass of the core the stellar remnants gets converted into a white dwarf star, neutron star or a black hole. Hence, if the mass of the stellar remnant exceeds the Tolman–Oppenheimer–Volkoff limit, then it gets converted to a black hole. So is there currently any equation which helps us to calculate the amount of time taken by the leftover core to get converted to a black hole or the rate at which the mass is gravitationally collapsing?

It depends entirely on the circumstances of black hole formation.

The remnant may be a stable neutron star, but if it accreted enough matter later on (by fallback in a supernova, or from a binary companion), then it may collapse. This could happen at any time after formation.

If the core of the collapsing star forms a proto-neutron star above the TOV limit of stability, then no force can prevent black hole formation. It will happen on the cooling timescale of the proto-neutron star at the centre. Once it is "cold" then the collapse would occur on a free fall timescale - which is about a millisecond in these circumstances. The cooling timescale of a proto neutron is governed by how long the copiously generated neutrinos can be trapped in the core, which could be as long as(!) a few seconds


How often are there five planet apparently aligned within 25°? I chanced upon this in Stellarium.

Is it common for five planet appear to align within 25°-30°, viewed from any of the inner planets?

It is not super-unlikely. Let's look at alignments of the three interior planets and two superior ones.

From Mars, the three interior planets (Earth Venus and Mercury) all appear to remain quite close to the sun, and so quite close to each other.  Moreover they will seem to move slowly as they reach greatest elongation. The chance that two interior planets will be on the same side of the sun as each other is 50%, and they will be "close" quite often. I've not done careful calculations, but let's say there is a 1/3 chance that two interior planets are close.

So to get the third planet close is another 1/3 chance, it would have to be close to both other two planets but the probability of all three interior planets aligning somewhat is around 1/3^2 or roughly 10%.  (This is a Fermi estimate so I'm taking the liberty of rounding)

Now the other planets can appear anywhere on the ecliptic. If you want them to be within 25 degrees of the others, that is a 25/360 chance (ie roughly 7%) and to get two planets both aligned is 7%×7% or about 0.5%. But there are four such planets, so six possible choices of two planets. That means that there is roughly a 3% chance that two superior planets will align with the inferior ones.

In conclusion, there is roughly a 0.3% chance of an alignment in the morning (or evening) sky on Mars. It is notable, but not super-unlikely.  Your particular alignment includes Uranus (which is almost to faint to see) and Mercury (which is all but lost against the sun.)  Close alignments of Jupiter Saturn Earth and Venus would look more spectacular.  A similar four-planet alignment was visible on Earth a few years back. How to watch at best this week's planet alignement?


How do you draw the ecliptic on a celestial sphere sketch? I don't really understand how it's drawn on sketches of the celestial sphere. At first I thought it goes through east and west, but it does not. How do you know where to draw it?

Example from Roy and Clarke, Astronomy: Principles and Practice:

The ecliptic is the plane of the Earth's orbit around the Sun.
From the Earth's point of view, the Sun appears to migrate along the ecliptic, eastward one cycle per year.
The ecliptic and equator intersect at the equinoxes.
The Sun appears to cross the equator northward at the vernal equinox (♈) around March 20 and southward at the autumnal equinox (♎) around September 22.

Equatorial coordinates are linked to the equinoxes; ♈ is at right ascension 0h, and ♎ is at RA 12h.
Where these points appear in the sky depends on the local sidereal time:

♈ rises in the east at 18h, crosses the meridian at 0h, and sets in the west at 6h.
♎ rises in the east at 6h, crosses the meridian at 12h, and sets in the west at 18h.
At sidereal times between the above, apply a polar rotation of 15° per hour to interpolate.

Once you've determined where on the celestial equator to mark ♈ and ♎, then you can draw the ecliptic through them as another great circle 23.4° oblique to the equator.
The ecliptic runs north of the equator east of ♈, and south of the equator east of ♎, in the sense that eastward is counterclockwise as seen from the north celestial pole.


Why has the orbit of Venus such a low eccentricity? The eccentricity of Venus is almost 14 times smaller than that of Mars, and 2.5 times smaller than that of Earth, and 30 times smaller than that of Mercury. Is it just a pure fluke? Or is there some “physical reasons”? I realize those are the ratios as of now, and that they change over time: will the evolution of those eccentricities dramatically change those ratios? If not, then the question of a “physical reason” emerges again.

It has a low eccentricity, but there may not be a particular reason.

Image by Kheider on wikipeda using Gravity Simulator by Trevor Dunn

In a simulation of the solar system, both Earth and Venus had orbital eccentricities that were much below those of Mars and Mercury (note the two axes. Mars and Mercury are on a scale that is 10 times bigger). But in 25000 years, Earth will actually have a smaller eccentricity, and 20000 years ago Venus's eccentricity was much larger.

So there may not need to be a special explanation for this. Planetary eccentricities do change over 1000s of years due mostly to perturbation of other planets.


Mass of a potential black hole in a binary system So I've been given the velocity curve, parallax and apparent magnitude of a star in a binary system with what is potentially a black hole. I've calculated from the apparent magnitude and parallax that the star is a type F5V, which puts the mass at about 1.4 Solar Masses. The velocity curve has an inclination of 90, and oscillates back and forth between +/- 75km/s. There is no data on the companion of this star, just the fact that it could be a black hole. I'm supposed to estimate the mass by numerically approximating a polynomial.
So far I've used this equation

$\frac{M^{3}}{(m+M)^{2}} = \frac{Pv^{2}}{2\pi G}$

where M is the mass of the thing I don't know, m is the mass of the known companion (1.4 solar masses) P is the period (5.59 days) and v is of course the velocity (75km/s)

I got lazy and wrote $\frac{Pv^{2}}{2\pi G}$ as $k$ and arrived at

$M^{3} - kM^{2} - km^{2} = 0$

Using python's optimize library I found the mass of this unkown partner to be about 0.018 solar masses. My question here is where did I go wrong, and if I didn't go wrong anywhere is that a realistic mass for a super small black hole / other very small, dense and invisible object?

Your first equation is incorrect. The left side has dimensions of mass, the right side has dimensions of mass × time × length-1. The velocity semi-amplitude (usually denoted $K$ rather than $v$) should be raised to the third power.

As noted by @PM-2Ring in the comments, your second equation is also incorrect as you didn't expand the $(m+M)^2$ term correctly.


Where did TRAPPIST-1a go? According to the NASA announcement and the Wikipedia article on TRAPPIST-1, the planets around the TRAPPIST-1 star are named TRAPPIST-1 b through h. What happened to "TRAPPIST-1 a"?

To add to Rob's answer, I wanted to expand on where this naming convention comes from.

The International Astronomical Union (IAU) is the organization which generally sets conventions and definitions. They're the ones who demoted Pluto to being a dwarf planet in 2006. Anyway, before any exoplanets were found, there existed a convention for naming multiple-star systems. The rule was that you gave the system a name, for example Alpha Centauri, and then the brightest star in that system is designated by "A". Any other objects are then given the letters "B", "C", etc. The Alpha Centauri system has three stars in it and so they're named Alpha Centauri A (The brightest of the three), Alpha Centauri B, and Alpha Centauri C (a.k.a. Proxima Centauri).

This convention then continued on to naming exoplanets. The star, according to this convention is "A", whereas all the exoplanets are given the letters "b" and onwards. Note that generally the exoplanet letters are lowercase, whereas they'd be uppercase for stars. Also note that if you have a star (say TRAPPIST-1) around which planets are found, the star's name doesn't then become TRAPPIST-1 A. The "A" is implicit.

Just to point out another point of interest with this naming convention, the letters of the planets don't always signify their distance from the star. That is, the "b" planet isn't necessarily the closest, followed by "c", etc. The planets are labeled at their time of discovery. Someone could find a planet, label it as "b" and then later find a second, closer planet and label it "c". It just so happens that the lettering of TRAPPIST 1 is in order because the first three were found at the same time and labeled according to distance, then the last 4 were found at the same time and labelled according to distance.

You can read more about the full set of rules concerning this on Wikipedia's exoplanet naming convention page. There are a lot of caveats such as what to do when your planet is orbiting a binary system.