Explaining the Drake Equation on a smaller scale So me and my friend were planning a video to explain the Drake Equation (within a time limit of 5 minutes), and we needed some help. This video is aimed at explaining the concept to an age group between 13-18 years, and having gone through loads of articles, we felt that a lot of the common audience would not be able to comprehend the concept.

So our question was, can we take the Drake Equation and try to explain it with maybe a real life example and on a smaller scale? It is an incredibly interesting equation and we felt like doing a good job of explaining it to a common teenager.

Thanks!

No need to make it complicated: what about this...

Just scribble a rectangle on a piece of paper, and say "there are 100 billion stars in our galaxy"....

Then, color off (let's say) 1/3 of the rectangle, and say "only one third of those are the sort of star that could have life, so that's blah billion"

Then, color off (say) 9/10ths of that box, and say "we believe about 90% of those have planets - so that's blah billion"

Then, color off (say) 1/20th of that box, and say "of those with planets, it seems that about 1 in 20 have Earth-like planets. Now we're down to blah billion..."

and so on.

(Note: the Drake equation has a number of fairly silly terms relating to "nuclear war!", which were added as political sops in that era; suggest ignore these unless you want to sound 90 years old!)

So just scribble a box or draw a line on a piece of paper ...  or maybe use "a bag of marbles" as the other answer suggests.

Just BTW there is in fact an entire documentary (I noticed it on "Netflix") called "The Drake Equation" which does exactly what you say...

.. it is not really very good as I remember.  I think the guy simply draws a line in the ground, to do the "fractions" demo, you know?  (ie, they just erase more and more of the line). It doesn't need to be more complicated than that.

It's worth noting that the Drake equation simply points out:

(i) if you multiply those three or four fractions together, you get the number of civilizations in the galaxy. Which is self-evident.

but, the whole point is

(ii) we have utterly no clue - not even vaguely - what most of the fractions are,

You could say it's a written formula, which, helps clarify our thinking on, something we are utterly clueless about. So rather than just vaguely saying "we're utterly clueless," we can speak more clearly about the nature of our cluelessness!

although interestingly,

(iii) very admirably, the issue of "How many stars have planets?" ... one could say that issue has been somewhat settled these very years, as we speak - that's great.


Stability of solar system My question is simple:

  Is the Solar system stable?

You can see this Wikipedia page.
Edit: Sorry, because i think my question is more about mathematics and classical mechanics of planets in billion years scale, than astronomy. But i think this is our fortuity that Earth didn't doomed in past billion years ago and it's possible that we will have less than 8 planet's before Sun will be ruin.

The wikipedia page you linked to tells you that the solar system is gravitationally "chaotic", in part because the mass of the sun is not fixed over time.

But even more simply than that, focusing just on the gravity (ignoring loss of stellar mass, etc.), the solar system is an N-body problem.  We have 8 planets, a sun, and millions of asteroids, comets, and who knows how many individual particles gravitationally bound to our sun (plus ones that aren't and are just passing through the neighborhood, so to speak).  When you have more than 2 bodies, the solutions to the N-body problem are unstable.  What this means is that, say we describe the N-body problem with data $D$ (the "initial conditions", or a perfect description of the state of the system at some specific point in time).  With a given complete data set the (Newtonian) gravitational evolution of the system is completely determined (but so difficult to do we can only approximate it).  What instability means here is that if we have some other data set $D'$ that is only a little bit different from $D$, then the differences between the evolution from $D$ and $D'$ will become exponentially large over long enough time scales.  So what may seem like minor differences now will result in radically different looking solar systems in the long run.

Since all of our observations can never give exact values, but only a range of values, there is necessarily a bit of uncertainty in what the exact gravitational state of our solar system is.  We have very poor data on the exact asteroid and comet content of our solar system, and even planetary data has significant error margins.  All of this means that there are lots of justifiable picks for the data $D$, each differing by a small amount from each.  But due to the instability, eventually these data will produce radically different futures from each other.  Currently we can only predict the solar system's evolution up to a few million years or so (the exact value stated can vary wildly depending on how you opt to define and compute the Lyapunov time).  After that the evolutionary tracks become so disparate we can't really say we're predicting anything other than "it'll definitely do something".

One way or another, it is currently impossible for us to make any clear assertions about what the solar system will look like on a timescale of billions of years.  Maybe all 8 planets will still be there; maybe their orbits will be very similar, but maybe they'll have much different orbits; maybe several planets will have been ejected from the solar system.  At best we can observe a few things that lead certain objects to be most likely to undergo significant alteration.  For example, Jupiter and Mercury appear to have a certain orbital resonance right now which could ultimately lead Mercury to undergo a significant orbit change.  This may ultimately cause it to collide with another planet, or the sun, or be ejected from the solar system entirely.  But maybe it won't.  It's hard to say.


Local Bubble in space Is the Local Bubble alone in space or is there another bubble similar to the Local Bubble in space, i.e., do we live in a special place or is this a more general phenomenon?

The "local bubble" is a region of the galaxy where the density of interstellar gas is lower than average. It has about 50 thousand atoms per cubic metre, compared with 500 thousand averaged over the whole of the milky way.  It was probably formed by one or more supernova explosions (10-20 million years ago) physically pushing gas away.

There are many such bubbles all over the galaxy, carved out by supernova and other energetic objects. See https://en.wikipedia.org/wiki/Superbubble


Why does the path of the moon in an eclipse curve (relative to the sun) Reviewing several simulations of the solar eclipse, it is apparent that the path of the moon relative to the sun curves.

Using https://www.timeanddate.com/eclipse/in/usa/kansas-city?iso=20170821 for example, the moon enters at 135 degrees and leaves at 217 degrees direction - that is almost a 90 degree right turn.

Why is the path curved not straight?

You have misunderstood the website. When the eclipse begins, in the morning, the sun will be in south-east, at a bearing of 135 degrees.

At maximum eclipse (at about midday) the sun will be almost due south, at a bearing of 178 degrees.

When the eclipse ends, in the afternoon, the sun will be in the south-west, at about 217 degrees bearing.

The sun and the moon will move in the sky, due to the rotation of the Earth, this says nothing about the straightness of the path of the moon relative to the sun.


Dark matter emitting EM radiations beyond our currently known spectrum? Could it be possible that dark matter is emitting EM radiations on a scale that is not known or perceivable by humans yet and undetectable by man-made instruments?

A type of radiation that our instruments were just not designed to pick up on? Being insensitive and not advanced enough? Thus we do not yet have all the pieces to solve the puzzle?

Edit: We need to remember that there are infrared and gamma rays on either end of the spectrum and we are aware that dark matter is definitely not emitting visible light, or any of the other type of EM waves. Considering that gamma rays were discovered roughly only a hundred years ago, isn't it possible that the spectrum could be further extended on either side and that we're still not aware of it?

The namesake of dark matter is because it does not emit electromagnetic radiation. Humans have built detectors on all ranges on the electromagnetic spectrum and a single particle will likely not emit wavelengths that are very large.

However, dark matter may produce light in collisions with other dark matter particles or normal matter. That is the premise of large scintillator detectors.


Can a tidally-locked planet maintain an obliquity If a planet is tidally-locked, can it have an obliquity?

e.g. the Earth's rotation is being slowed by tidal forces.  A long way in the future it will be tidally locked.  As it slows, will the obliquity also be eroded or will other forces (such as the orbit of the moon) allow it to maintain an obliquity?

Well, I can't do the math, but if an intuitive answer is OK:

  If a planet is tidally-locked, can it have an obliquity?

I'm going to cheat and give you an answer using moons, because many moons are are tidally locked to their planets.

Io is a funny one, it's slightly eccentric orbit and proximity to Jupiter causes it to resurface pretty regularly.   It's axial tilt isn't listed and might not be known.

Europa.  Axial tilt  0.1°

Ganymede.   Axial tilt 0–0.33°

Callisto.  Axial tilt zero

Titan, to Saturn,  Axial tilt zero

Triton, to Neptune,  Axial tilt zero

Our Moon is a bit of an oddball, it has an Axial Tilt that's 1.5424° to it's orbit.   click here for picture.   Why our Moon varies a bit more than the others, I'm not sure, but it might be the proximity to the sun which keeps the Moon from orbiting more directly over Earth's equator and causes some imperfection in it's tidal lock.  (Kind of a guess, I'm not sure on this part)

But, simple answer, Tidal locking with solid bodies by definition, the same side of the planet or moon faces the object it orbits all of the time.   Obliquity doesn't make much sense if the planet's primary rotation is in sync with it's orbit around an object.   (Tidal Locked planets do in fact rotate).  So, it seems counter intuitive that a planet could have a high obliquity and at the same time, be tidally locked to the object it orbits.  Mercury and Venus, work as examples for this common sense argument too.  Mercury is, in a sense tidally locked but due to it's eccentric orbit it's in a 3:2 ratio.  Mercury has 0 obliquity and Venus appears to be on it's way to being tidally locked to the sun and it has a 177.36 degree obliquity, but that's essentially 2.64 degrees, if Venus wasn't rotating in the opposite direction of the other planets.   As Venus moves into being tidally locked to the sun, it's rotation should approach 0 (or 180) obliquity, the slowing of it's rotation in comparison to it's rotation matching it's orbit suggests that this is the only likely outcome.

If you want to check the axial tilt of other moons, more listed here.

Now, there may be some factors that could, just maybe, lead to exceptions, perhaps A planet with very high inclination might be a wild card.  Or a rotation that's in a 2:1 resonance with the tidal locking.  A binary star system could have some odd orbits too (See Pluto's and Charon's, satellite Nix here.  Now, Nix isn't spherical, so it's prone to much more pronounced oddities, but a planet around a binary star could exhibit some strange orbital resonances, but I suspect using a binary star system to find an exception isn't what you had in mind.

I'm 90%-95% confident in saying that obliquity should tend towards zero as the planet becomes tidally locked to a single star that it orbits.

  A long way in the future it will be tidally locked. As it slows, will
  the obliquity also be eroded or will other forces (such as the orbit
  of the moon) allow it to maintain an obliquity?

This is a tricky one because if the sun wasn't going to go Red Giant, the Earth would eventually become tidally locked to the Moon (in 10-50 billion years), and not the sun.   If the Earth didn't have a moon, then it could get tidally locked to the sun, in, oh, I don't know, 100-300 billion years as a guess.   The Moon is the dominant tidal force on the Earth, so in it's current set-up, the Earth can't get tidally locked to the sun.   If the Moon was smaller, then maybe somebody could run the numbers on this question.   If the Moon was 1/3rd it's current mass then the Lunar and Solar tidal forces would be similar, and . . . then it becomes an interesting question.

Curious sidebar, while the Moon currently stabilizes Earth's obliquity, that's not always going to be the case.  As the Moon moves further from the Earth, the Earth's obliquity is expected to show much more pronounced fluctuations, but that doesn't have anything to do with the Earth being tidally locked to the sun.

Hope that helps.  Corrections welcome.


Mirror telescope: blocked sight Today I try to watch Venus through a mirror telescope. I am no professional. No matter what lens I take, I see this image: What am I doing wrong?

Edit: What happens when I refocus:Sorry for the bad quality, I didn't manage to make better images. :-(

I think you just need to keep winding the focusser knob, you are way past focus in the first picture but you can see the image get smaller as you approach focus. The last image looks like you may be there or thereabouts, I think the colored diffraction patterns are caused by the camera optics.


Why are most planetary orbits nearly circular In our solar system, with the exception of Pluto all planets follow a relatively circular orbit around the Sun, at the same inclination. They also all rotate in the same direction, none are 'retrograde'.

An image of the orbits of our solar system.

How and why have they all formed into these orbits? Is this reflected in most orbital systems? What factors affect the orbits of newly formed systems?

The combination of the conservation of angular momentum and gravity give you the inclination and direction.  Gravity will condense material in the axis direction of the angular momentum, and collisions and gravitational interactions will damp oscillations in that direction, forming a disk of material.  The angular momentum resists gravity in the plane perpendicular to the axis, maintaining the extent of the disk.  Over time, the direction of rotation of the angular momentum is the direction of the vast majority of the material, due to collisions and interactions with that smaller population going the wrong way on a one-way street.

The circularity of the orbits is the result of a more dynamical process.  Hopefully someone else here can explain it better than the following.  My simplistic understanding is that a bunch of orbits crossing each other is not stable.  Planets and planetesimals have their orbits changed continuously until such time as they settle into orbits that have fewer interactions with other bodies.  Sort of a natural selection.  In the long haul, this results in a relatively stable configuration with circular zones that tend to not interfere with the other circular zones.  In fact, this is now part of the definition of the word "planet", in that to be called such, a planet needs to clear the neighborhood of its orbit.  Though its not clear how it could occur, a bunch of elliptical zones that happen to be co-aligned would also not be stable since orbits precess due to various influences.

Where the material was less dense than where the planets are currently must certainly have many objects in eccentric, retrograde, and/or highly inclined orbits.  Pluto is the first hint at that, with a relatively high inclination and eccentricity compared to the planets.  Sedna is far more eccentric.

As for other systems, I'm sure there must be some oddballs out there.  However the Kepler mission has observed mostly circular orbits for planets around other stars.


Radiative Flux, Irradiance and Luminosity confusion First of all I know there is a similar question here, but I feel is wasn't answered at all.
My question is that I have seen many texts which says that Radiative Flux is equal to the Irradiance of a star. In Carroll & Ostlie book "An Introduction to Modern Astrophysics" even say that:

And Wikipedia's Irradiance on Earth's surface says this:

Average annual solar radiation arriving at the top of the Earth's atmosphere is roughly 1361 W/m².

Following this I first I assume that Irradiance and Radiative Flux are the same thing, but when searching for Irradiance on Wikipedia says that:

In radiometry, irradiance is the radiant flux (power) received by a surface per unit area.

Which means that Irradiance is Flux divided by Area, so Radiant Flux has units of J/s and Irradiance has units of W m^-2 which I remember from my Optics class, then again Carroll & Ostlie say that:

That doesn't mean if Irradiance is equal to Radiant Flux divided by Area, so F = I*A that would make Irradiance equal Luminosity / Area^2 ?
In that case the bolometric magnitude should be calculated with the integral of Irradiance, not Luminosity nor Flux right?
Probably I'm making it complicated for myself and it's not that hard to understand. I'm just sharing my confusion with you guys, please help me with that question.
Thanks in advance.

It looks to me like irradiant flux is the same as radiant flux is the same as irradiance, both are energy per second divided by area.  That interpretation would seem to be consistent with everything you quoted.  In more precise terms, these should all be "flux density", not "flux", because flux just means the amount of something per time, but that convention is never followed much any more.


The Matter Anti Matter Asymmetry The matter anti matter asymmetry is still a mystery that still can't be explained. But what if there wasn't an asymmetry? Isn't it possible that there was exactly equal amounts of matter and anti matter which did in fact annihilate each other, and the energy released from it was converted into mass? I know this arises a lot of paradoxes like if so what kind of matter is everything then . I guess what I really want to know more about is, after a Matter and Anti Matter collision, what kind of matter can be created (or rather converted) from the resulting energy outburst?

Energy can indeed transform into matter - for example pair production; but this creates equal amounts of matter and anti-matter.


How complicated is the heating of meteors? When any gas, including our atmosphere, is compressed, it's temperature rises.
When a meteor is falling through the atmosphere, it is compressing the air below it.
Because this makes the air hotter than the meteor, heat moves from the air to the meteor.

Also, some amount of oxidation (burning) might also take place.

What I would like to know, is it really that simple?

Alternatively, what else causes meteors to heat up?

I've heard people say "air friction," which sounds to me like nonsense, but I am not a rocket scientist, so what do I know?  If "air friction" is real I want to know how it works!

Also, I'm fairly certain that heating of air due to turbulence will only affect the air which the meteor has already moved through... not the meteor itself.  If I'm wrong, I want to know!

Your analysis is correct. Compression of the air is the cause of heating.

You call this "simple" but the compression depends on the ability of the air to flow around the meteoroid, and as the modelling the hypersonic flow is far from trivial.

"Air fiction" is the resistance to motion felt by an object moving relative to the air. That includes the reaction to the force that is compressing the air in front of the meteoroid, but also results from work done in moving the air out the way, and is dependent on shape, surface effects, turbulence etc.

The visible meteor is the trail of plasma produced in the air by the meteoroid. The actual piece of rock is too small and dim to be seen. The meteor is glowing air.


Real images (not "artist concept") by NASA Does NASA make publicly available real images of missions in an organized manner? What I mean is: Say I'm looking for the images taken by New Horizons - a simple list of the images in chronological order. Even if it's not a comprehensive list, at least all of those made publicly available. Does that exist?

To be clear - I'm not interested in "artist concept". On the other hand - "false color" pictures of infrared etc. are fine.

NASA mission data is archived at a handful of NASA archives.  If you do not know which archive a data set is in, then you can search the Master Catalog at the NASA Space Science Data Coordinated Archive (NSSDC).
The main astrophysics archives are:

HEASARC (High Energy Astrophysics Science Archive Research Center)

IRSA (Infrared Science Archives)

LAMBDA (Legacy Archive for Microwave Background Data Analysis)

and MAST (Mikulski Archive for Space Telescopes)

Planetary/Solar System Data are at the various Nodes of the PDS.
NSSDC holds data from many of the older missions that predate these archives.

New Horizons data is being archived at the Small Bodies Node of PDS. But, it usually takes several months for data to get into these archives.  So, if you are looking for very current images, you have to rely on what the mission team has put up on its web site.  That is at their discretion.


Do we know what's "behind the Milky Way"? If we do, how? And what are those objects on the other side of the galactic center?
We can't see through the dense band of light and dust, but maybe in other spectrums the band is much less dense? But I doubt it.
Maybe we could detect or have detected those galaxies/satellites through their center's strong X-Ray? Or through their gravitational force on the Milky Way?
If not, then there is a chance, that there is a closer galaxy than Andromeda and even more satellite galaxies than currently observed?
Thanks in advance! :)

The short answer is that we know a little bit about what is beyond the center of the Milky Way, but a lot is still unknown.
The bands of obscuring dust in the galactic plane are only a few degrees of arch wide as seen from Earth.  Thus we can see for millions and billions of light years in directions which are only a few degrees "above" or "below" the plane of the Milky Way, even in the direction of the galactic center.
A number of external galaxies have been discovered which are nearly in the plane of the Milky Way and which are heavily obscured by dust clouds.
In 1967 Paolo Maffei discovered galaxies Maffei I and Maffei II though the galactic plane by their infrared emissions.
Maffei I is a giant elliptical galaxy which is part of the IC 342/Maffei group of galaxies about ten to twenty million light years away.
Maffei 2 is a spiral galaxy thought to be a member of the IC 342/Maffei group of galaxies about 10 million light years away.
I think that Maffei 1 & 2 are about 90 degrees from the direction to the center of the galaxy.
And it is possible that other distant galaxies have been observed through the dust of the Milky Way in directions closer to the galactic center.


Circular velocity in axisymmetrici potential If a matter distribution is spherically symmetric, the circular velocity is straightforwardly related to the mass via:

$v_{circ} = \sqrt{GM/r}$

In an axisymmetric potential, on the other hand hand, we have

$v_{circ}(R) = \sqrt{R\frac{d\Phi}{dR}}\neq \sqrt{GM/r}$

with $\Phi$ the gravitational potential.

What I don't quite understand is the following: the potential of the Milky Way is not spherical. How is the measured velocity then reliably related to a mass?

What you do is predict the orbital velocity for a more realistic hypothesised mass distribution. There is an easy analytic solution for a disc. More complicated distributions require a numerical solution.

If you get far enough outside an arbitrary mass distribution, then that solution will asymptotically look similar to $v_{\rm circ} = \sqrt{GM/r}$.

Astronomers get into trouble when explaining the evidence for dark matter from galaxy rotation curves to the layperson. They often use the analogy of the solar system, where the orbital speed goes as $r^{-1/2}$. The trouble is that that a Galaxy is not like the solar system because the mass is distributed and in general cannot be assumed to be concentrated at a point and Newton's shell theorem does not apply.

In practice what you can do is posit a more realistic distribution of matter based on star counts and the gas that can be observed. You can then predict (using a computer model) what the rotation velocity should be at any point if that is all the mass that exists. This model will converge to the simple $r^{-1/2}$ law once you get well outside the bulk of the mass in the model. The model can be compared to the observed velocities of gas, globular clusters, planetary nebulae and other velocity tracers - some of which do indeed exist well outside the bulk of the visible galaxy matter. What you find is that they move much faster than predicted, indicating that considerably more unseen matter must exist inside their galactic orbits. You can then add that (dark) matter in various ways. Non-spherically symmetric and spherically symmetric dark matter profiles (of the same total mass) would produce different orbital velocities at different radii for precisely the reason you identify. By looking at tracers at different radii you can map out the distribution of unseen mass. Once happy with your model you can integrate the density of seen and unseen matter to estimate the total mass.

The bottom line is that the $\sqrt{GM/r}$ is just a "back of the envelope" demonstration of the effect. Real research uses more accurate models or works in a regime far outside the bulk of the visible matter where the "back of the envelope" approximation works.


My supernovae question I understand the collapse of a star during a supernova, as the outward pressure cannot compete with the gravitational field.

Is it conceivable, that a relaxation of the space-time curvature is what expands the space and looks like an explosion?

I have often wondered, to extrapolate, if a "release" of the curvature of space itself, or in an extreme example a massive black hole releasing all of its potential energy in an instant (In a way not yet understood) could be in fact the (or a) big bang.

No. The precise details of collapse of a supernova are not perfectly understood, in the way we understand (for example) the orbits of the planets. However the energy comes from the release of gravitational potential energy. The "relaxation of spacetime" is not significant. We don't understand supernovae perfectly. But we have a good model of them.

You should re-read Rob's answer to your previous question, not least because he got the order of magnitude of the energy release in a supernova correct. He also points out your error in thinking of the gravitational field changing "in an instant"

Concerning the big bang. No, as has been often pointed out, the big bang was not a explosion. There is no analogy between a supernova and the big bang.


How many galaxies disappear beyond the Hubble Bubble horizon every year now? The accelerating expansion of space means that the space between us and far away galaxies expands faster than light can travel through space. There is a horizon of possible observation beyond which galaxies disappear from us.

At what rate do galaxies, or some other measurement of mass/energy, disappear beyond that horizon at present rate? Will future telescopes actually see galaxies red shift themselves into oblivion? I imagine there being issues about the definitions of "when" and "where" in this context. Answers should try to address how they deal with that.

Since the radius of the observable universe expands at the speed of light, my intuition tells me that there should've been a point in time when the expansion of space balanced the arrival of light of more distant parts of space. if so, when did that happen?

This is a tricky question to answer, because it has some false and imprecise assumptions baked into it. The short version is: I don't know at what rate stuff is currently crossing the cosmic Event Horizon. But future telescopes will not see these galaxies redshift themselves into oblivion, because this will happen in the infinitely far future, literally. Cosmological redshift is equivalent to relativistic time dilation; events that are redshifted from our point of view also play out slower, and infinite redshift means they come to a complete halt.

It is also not correct that the radius of the observable Universe is expanding at the speed of light. There are two different radii that could be candidates for the title "limit of the observable Universe"; they are very different but both expand faster than light.

Your intuition is right, however, that the distance at which galaxies recede from us at exactly light speed - the one we call the Hubble Sphere - has been expanding and since approaching us again. It reached its maximum when the cosmic expansion went from decelerating to accelerating, when the Universe was around half its current age. The Hubble Sphere, however, is not the limit of anything.

One of the two horizons, the one that seems to be the one you have in mind in the beeginning of your question, is the Event Horizon. This is the current distance, beyond which light emitted at this cosmological time will never have the chance to reach us. This horizon receding from us in proper distance, and since it is located outside the Hubble Sphere, these regions of Space are actually receding from us faster than light.
In co-moving coordinates, the event horizon is approaching us. Co-moving coordinates is a coordinate system that stretches along with the expansion of the Universe, such that a galaxy that is not influenced by anything else than cosmic expansion will have constant coordinates in time. Because the cosmological expansion is accelerating, it gets harder and harder for light at a fixed co-moving distance to reach us until it gets impossible and the Galaxy redshifts to infinity. This  is equivalent to saying that the Event Horizon has moved inwards past this galaxy in co-moving coordinates. In proper coordinates, it amounts to saying that the galaxy overtakes the Event Horizon.
Ours, by the way, is currently at a redshift of around 2, around 18 billion light years away. The light we receive from this distance is only slightly redshifted now, but the light that is emitted now will be redshifted to  infinity when it reaches us at infinite time.

There is also another horizon, however, which does expand forever, also in co-moving coordinates. That is the current co-moving distance at which light emitted at the time of Big Bang reaches us just now. In co-moving coordinates, this horizon expands slower and slower, which is why the CMB gets more and more redshifted. Objects currently on the Particle Horizon are a long time out of our Event Horizon - we can only ever see the very beginning of their history, before they redshift into oblivion. And the objects shifting into our particle horizon will show us still less of their history before having left the Event Horizon for good (yes, language gets hairy when talking relativistic time).

So it is hard say what the limit of the observable Universe is, it is simply not well defined. The Event Horizon has always been growing in proper distance but shrinking in co-moving distance. The Particle Horizon has always been growing both in co+moving and proper distance. The Event Horizon has always been expanding faster than light, and always will. The particle horizon started out slower,  but very soon overtook lightspeed in expansion velocity.

Interestingly, all objects with a redshift higher than $z \sim 1.5$ are not just receding faster than light, but they always have been! Yet we observe them with no trouble.


Is it possible for a solar flare in one star to influence the chances of a solar flare in another star? Can a solar flare reach far enough to travel from one ordinary star to another?  If so, how can that solar flare affect the following star?

I'd say it's an open question awaiting some observational evidence that it can happen.  It presumably would depend a lot on what kind of situation you are looking for, and what would you call a flare on one star setting off a flare on the other.  Certainly you'd need two stars that both have active atmospheres and are in a close binary, such that one mass ejection could interfere enough with the magnetic configuration in the other star to set off a flare.  Some stars may be so close together that the magnetic configuration of their atmospheres actually connect both stars, so then when get a flare you might see brightenings on both stars but it would be essentially the same flare.  There are even stars called "contact binaries" where it's not even all that clear if you should think of that as two stars that touch (and share a magnetosphere), or just one star that has two separate cores.  But the bottom line is, it sounds like you'd need something unusual going on, involving two stars that were unusually close together and are interacting in some way.  You might want to read up on "RS CVn stars," for example.


Is there any correlation between the Cosmic Microwave Background (CMB) and the distribution of distant galaxies? The Cosmic Microwave Background (CMB) is remarkably isotropic but does exhibit a distinct dipolar Doppler shift and also much smaller but measurable fluctuations in intensity and polarization.  Presumably some of these fluctuations should carry over into the subsequent formation of early galaxies and other structures.  To what extent is this expected and to what extent has this been observed?
The CMB Doppler shift is attributed to our local velocity with respect to the early universe. Is this Doppler-shift equally observed with respect to distant galaxies and diffuse starlight?
More particularly, is any of the fine structure in the CMB represented in the distribution or properties of the most ancient objects that can be observed optically (infrared)?

CMB fluctuations
The CMB fluctuations are often analyzed through their power spectrum $P(k)$, which is a measure of the extent to which it is "clumpy" on a given scale $\ell$, with corresponding wavenumber $k = 2\pi/\ell$. The origin of this power spectrum is laid in the very early early Universe, just after the Big Bang, and it is of utmost importance in cosmology, as it specifies how structure subsequently evolves.
Various observations help us constrain $P(k)$. For the early Universe, we use the CMB, while at later epochs, the large scale structure of galaxies and, even later, galaxy clusters are helpful. Moreover, absorption lines in the spectra of background quasars enable us to map out the structure via the so-called Lyman $\alpha$ forest.
The figure below (from Norman 2008) show how a single cosmological model (from Tegmark et al. 2004) fit all epochs.

The power spectrum evolves with time, so to compare difference epochs, the figure shows $P(k)$ for the different observations extrapolated to the current epoch (i.e. today).
CMB dipole
Observing the dipole in the CMB is relatively easy because, to within 1 in ~100,000, it is described by the same Planck temperature in all directions. This is not the case for any astrophysical sources, so measuring the dipole in galaxies is less straightforward. With large samples of galaxies it has nevertheless been attempted. For 1.2 million galaxies in the WISE catalogue, Rameez et al. (2018) find the dipole anisotropy to be consistent with that of the CMB.


Why doesn't Jupiter increase the chances of an asteroid to strike the Earth? I often hear the hypothesis that Jupiter decreases the chances of an asteroid strike on the Earth, but my question is, why does it only decrease the chances ?

While Jupiter can deflect some asteroids that were originally going to hit the Earth, it can also deflect some asteroids towards the Earth. How can we be sure that Jupiter only decreases the chances of an asteroid impact on Earth ?

With regard to collisions with earth, or inner planet collisions in general, you're right that Jupiter has both a positive and negative effect on asteroid and comet trajectories. On balance though, the effect is mainly positive.
From earthsky.org:

Some astronomers believe that one reason Earth is habitable is that the gravity of Jupiter does help protect us from some comets. Long-period comets, in particular, enter the solar system from its outer reaches. Jupiter’s gravity is thought to sling most of these fast-moving ice balls out of the solar system before they can get close to Earth. So long-period comets are thought to strike Earth only on very long timescales of millions or tens of millions of years. Without Jupiter nearby, long-period comets would collide with our planet much more frequently.
...
But Jupiter creates both good and bad conditions for earthly life. Consider that its powerful gravity prevented space rocks orbiting near it from coalescing into a planet, and that’s why our solar system today has an asteroid belt, consisting of hundreds of thousands of small flying chunks of debris... today, Jupiter’s gravity continues to affect the asteroids – only now it nudges some asteroids toward the sun, where they have the possibility of colliding with Earth.


How is the gravitational effect of galaxies outside of the visible universe on galaxies within the visible universe currently modeled? Given currently accepted models of the universe,

How much mass is outside of the visible universe?
What is the gravitational effect of that mass on the visible universe?

The question How strong is the gravitational stretch we experience from the edge of the universe? is possibly related, but it's focused on some stretching effect which I'm not familiar with.

One answer to the related question How much of the Universe is invisible to us, and how does it affect our theories? states:

  there does not seem to be any large mass concentrations outside the
  observable universe, since these would be seen as bias for movement of
  galaxies and galaxy groups, which also supports that the rest of
  universe is at least very close to what we can observe directly

But this talks about the absence of large mass concentrations, while I'm asking more generically how mass outside the visible universe is reflected in current models.

We don't know…

We don't know how much Universe there is outside our observable Universe.

The observable Universe seems to have a "flat" geometry (in the 3D sense, not in 2D). If it really is globally flat, then it just goes on and on forever, i.e. there is infinite mass. However, just as Earth looks flat on small scales, it might very well be that the Universe has another geometry on scales larger than we can observe.

A recent paper (Di Valentino et al. 2019) claimed evidence for a "positively curved" geometry, implying that the Universe is only (very roughly and with high uncertainties) three times bigger than what we see (see this answer and comments for the calculation). However, the authors sort of cherry-picked one out several of the data sets of Planck Collaboration (2018), and the evidence is not convincing.

…and we don't care

However, no matter if the Universe is finite or infinite, we do not take into account the mass outside the observable Universe in any calculations. This mass does not affect us in any way. Gravity travels at the speed of light, and hence hasn't had time to reach us yet. More importantly, due to the inverse square-law, gravity on any object is by far dominated by nearby sources.

Well, we care a little

There is a sense in which you can say that the gravity is taken into account, though: So-called "cosmological simulations" aim to simulate the formation and evolution of the large-scale structure and the galaxies of the entire Universe. Of course, we cannot simulate the entire Universe, so instead we simulate a large chunk of it; large enough that another similarly-sized chunk would not be statistically different from it (preferably several hundred million light-years on each side).

A galaxy close to the edge of the computational box, say its left side, doesn't only feel the gravity of the galaxies on the right. This would make the simulated universe "implode". Rather, we use a computational trick called "periodic boundary conditions" which means that a particle that happens to travel through the left side of the box enters immediately through the right side. And a galaxy close to the left side feels the gravity of matter lying close to the right side.

Snapshot from the Virgo simulations (Jenkins et al. 1998), where I've tiled the box in several copies, emphasizing the periodic boundaries.


How to calculate the altitude of the Moon? With the known latitude coordinate of the observing position, how to find the altitude of the moon when it is high, i.e when it crosses the local meridian?

With a lot of difficulty if you want to do it numerically by yourself. The Moon's orbit is very complex.

Far and away the easiest method would be to use NASA's HORIZONS service for calculating ephemerides, which will get you very comprehensive and accurate information on objects at your specified times.

I don't think that HORIZONS can directly return the time that the Moon is at its maximum altitude. However, if you can write some basic computer code then it shouldn't be difficult to query the ephemerides with a precise (say 1 minute) time interval and find the maximum altitude from there.


Does the sky darken proportionally during a partial solar eclipse? During a solar eclipse, some of the visible area of the sun is obscured by the moon.  The exact percentage varies with time and with the observer's location.  My question is, does the sky darken by this same percentage?

In other words, suppose the sun is 80% covered.  Does sky light diminish by 80%?  Are objects illuminated by 20% of their normal amount?

Assuming the intensity of the sunlight is proportional to the area of the Sun which is showing, we can roughly calculate the intensity of sunlight, $I(d)$ as:
$$I(d) = 1 - \frac{\mathrm{Area}_{\mathrm{overlap}}}{\mathrm{Area}_{\mathrm{Sun}}}$$
Where $\mathrm{Area}_{\mathrm{overlap}}$ is the overlap between the Moon and the Sun (the green area).

(source: uga.edu)
Some calculation finds the overlap to be:
$$\mathrm{Area}_{\mathrm{overlap}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2}$$
Where $d$ is the distance between the centres of the Moon and the Sun. Similarly, the area of the Sun is:
$$\mathrm{Area}_{\mathrm{Sun}} = \pi r^2$$
Putting this together, you get the graph:

Where the Y axis is brightness (ranging from 100% to 0%) and the X axis is $d$ (ranging from $2r$ to $0$).
I hope this answers your question: the brightness does not change proportionally to coverage/time.
Note:
This assumes that $d$ increases linearly with time (the moon doesn't speed up or slow down as it moves across the Sun).


Why does radiation not need a medium to travel? I wanted to know how radiation moves in vacuum of space and why it doesn't need any medium ?
( Like sunlight reaching a planet )

Electric and magnetic fields can and do exist in a vacuum. Electromagnetic waves are just fluctuations in electric and magnetic fields. In a vacuum (meaning no charges or currents, which do require the presence of matter), then the solutions to Maxwell's equations are electromagnetic waves.


How is the maximum density in a circumstellar disk determined? I'm working with some equations to model the evolution of a circumstellar disk. One of the equations is
$$\rho(r)=Ce^{-\frac{(r-r_{peak})^2}{2 \sigma ^2}}$$
where $\rho$ is density, $r$ is the distance from the center, $C$ is a constant, $\sigma$ is one standard deviation, and $r_{peak}$ is the radius at which the density is at a maximum.

If the function was of the form
$$\rho(r)=Ce^{f(r)}$$
where $f(r)$ is a function of $r$, I could find the maximum easily by finding
$$\rho'(r)=Cf'(r)e^{f(r)}=0$$
and solving for $r$. However, this appears to be impossible in the current case because $\rho(r)_{peak}$ is already in the equation, at $r_{peak}$.

How is $r_{peak}$ determined in a given scenario? Is it determined experimentally?

If your $r_{peak}$ is not known a priori then you have two unknowns and only one equation: this cannot be solved. At least not without a second equation or a measured $(\rho,r)$-pair (say the central density $\rho_0 = \rho(r=0)$).

In any case, this definition of radial density looks like an ad-hoc assumption to me, and not something that came out of a proper hydrodynamic model. So without knowing why you chose this form or the context of your disk model I can't comment on what equation is best suited to determine $r_{peak}$. A well-informed guess seems like the way to go.


What are "non-Keplerian" orbits? What are some familiar examples in our solar system, and can some still be closed? This excellent answer to Forms of stellar orbits around the galactic center invokes the following concepts:

non-Keplerian orbits
closed orbits

I have a fairly good idea what these mean and so might many of us, but our ideas may not overlap completely.
For the sake of future stack exchangers and orbiteers, how can we best define these?
Question(s):

What exactly are "non-Keplerian" orbits?
What are some familiar examples of distinctly non-Keplerian orbits in our solar system?
What are closed orbits?
What are some familiar examples of distinctly non-closed orbits in our solar system?
Can some non-Keplerian orbits still be closed?

Included should be the understanding that we're talking about soft limits here; no orbit is exactly Keplerian because gravity goes everywhere (there are no "gravitational dielectrics" or opposite charges for field lines to terminate). But for practical purposes we can treat many orbits as Keplerian for many purposes, and some we can't at all.

What exactly are "non-Keplerian" orbits?

Orbits that don't follow Kepler's laws.
Strictly speaking, all orbits are non-Keplerian. In practice, one can model some orbits as basically Keplerian, but with perturbations. Sun synchronous satellites are one example of orbits that are close to Keplerian, but not quite so. The Earth's equatorial bulge makes satellites fail to orbit in a plane. Sun synchronous satellites take advantage of the precession induced by the Earth's equatorial bulge so that their orbits precess by 360° in one year. Connor Garcia's example of satellites in pseudo-orbits about the Sun-Earth linear Lagrange points form another set of examples.

What are some familiar examples of distinctly non-closed orbits in our solar system?

Every planet, every moon, every asteroid, ...

What are closed orbits?

Bound orbits are orbits that stay bound to some central object. Parabolic and hyperbolic trajectories are not bound. Closed orbits are bound orbits that repeat their path. Any attractive central force law can result in circular orbits. These are trivially closed. There are only two attractive central force laws that can result in closed noncircular orbits: An inverse square law (e.g., Newtonian gravitation), and a linear force law (e.g., a Hookean spring). This is Bertrand's Theorem.

What are some familiar examples of distinctly non-Keplerian orbits in our solar system?

Every planet, every moon, every asteroid, ...

Can some non-Keplerian orbits still be closed?

Other than a Hookean spring, no.


Gravity of other objects in black holes? If something falls into a black hole, but has not yet hit the singularity in the center, can we (outside of the black hole) feel its gravity? On the one hand, I've learned that there's no distance limit to gravity and that everything in the universe feels the gravity of everything else. On the other hand, we would be able to measure the gravitational field of the thing and thus figure out where in the black hole it is, which contradicts the idea that no information can escape a black hole.

Gravity, according to the majority of published and non-published scientists, is able to propagate information from one point to another at the speed of light. This means that as soon as any massive body "physically" interacts with a black hole (enters the Schwartzchild radius) there is no longer a way for anything outside of the black hole to know anything else regarding the status of the body. Therefore the new addition to the black hole simply becomes part of the system and can no longer be considered a separate entity. The body's gravity becomes that of the black hole's.

So... No


What fraction of galaxies in the observable universe have we actually observed? There are a finite number of currently observable galaxies due to the finite age of the universe and the speed of light. What fraction of these galaxies have we actually observed (by eye, telescope, etc.)?

There are around 2 trillion galaxies in the currently observable universe according to the latest estimates, obtained by integrating theoretical galaxy stellar mass functions above $10^{6} M_{\odot}$ between $0 \leqslant z \leqslant 8$.

It's difficult to get a precise number for the total observed galaxies as the results from new surveys are being released all the time, and updated analysis of legacy data is revealing more galaxies. For example, the latest SDSS data release identified close to 200 million galaxies, whilst the ongoing Dark Energy Survey seeks to identify around 300 million galaxies. LSST, in Chile, will observe close to 20 billion galaxies, an order of magnitude greater than anything before it. However, all of these surveys operate up to relatively low redshifts. Probing to higher redshifts requires bigger, preferably space based telescopes, and much longer exposure times. As such, the high redshift universe has been poorly documented. Our understanding of galaxy formation at high redshift is not as complete as at low redshift, and so we may be under- or over-estimating the number of galaxies at this epoch.

A vague upper limit: we've observed less than 0.01% of all currently observable galaxies.


Does rusty iron on Mars have tetanus? Since the surface of Mars is covered with lots of iron and became red because of rustiness of iron,

Source from: live.mrf.io

if we do have plan to live there, can we get tetanus on those rusty thing? I don't know much if we can be immunize to this thing but still it might be lethal at some point if they have tetanus, even a slight stumble might be dangerous.

Update: About being off-topic. I thought posting it in Astronomy would be better because they know much and literally know much more about Mars and its atmos, temp, minerals or something, instead of posting it in Medical-related stackexchange. I mean, it's Mars, Astronomy stuffs knows better, isn't it?

No. Tetanus is caused by a soil bacteria. You cannot get tetanus from rusty iron, if that iron has been sterilised.

There are no Earth bacteria on Mars. As far as we can tell, there is no life of any kind on Mars.


What is radio-emitting gas? What is radio-emitting gas? (I think it refers to the Magellanic Stream, but I can't be convinced.)

Referring only to the image you present, the radio part of the image (I think this is a stacked optical and radio image) shows something called the Magellanic stream. Despite what the wikipedia page says, I believe it was discovered by Dieter (1971) using 21 cm observations of neutral hydrogen atoms.

The radio emission in the case of the picture you show arises from the so-called "spin flip" transition associated with changes in the relative spin of the proton and electron in a hydrogen atom. A configuration where the spins are anti-parallel has a lower energy than where they are parallel. (Quantum mechanics only allows these two possibilities). The energy difference is small - only $5.87 \times 10^{-6}$ eV. In hydrogen gas in space it is possible for the atom to make a transition from the higher level state to the lower level, in the course of which it emits radiation with a wavelength of 21 cm (i.e. radio waves). This is not seen in terrestrial laboratory samples because the gas is collisionally de-excited before it has the chance to emit radiation. Hence 21cm radiation is an example of a forbidden line, with a long radiative lifetime and which is only observed from the rarefied neutral hydrogen gas found in the interstellar medium.

The intrinsic width of the emitted spectral line is very narrow. This means that the observed wavelength of this 21 cm line is an excellent probe of the motion of the emitting gas, since the observed radio waves will be doppler shifted from the 21 cm wavelength at which they were emitted according to the line-of-sight velocity of the gas.

The gas in question here has been stripped from the two satellite galaxies of the Milky Way, known as the Magellanic clouds. Although the picture also shows copious 21 cm radiation arising from neutral hydrogen gas in the plane of our Milky Way.

There are many other ways that gas can be a source of radio emission. In addition to atomic and molecular transitions that give radio emission at discrete wavelengths (including the radio equivalent of lasers, called masers), there can be continuum emission caused by the acceleration of electrons in the electric fields of partially and completely ionised ions - so called bremsstrahlung or free-free emission. There is also recombination radiation (free-bound) radiation and continuum radiation emitted by electrons that are accelerated in electromagnetic fields - cyclotron radiation; synchrotron radiation (if the electrons are relativistic); curvature radiation (a variety of synchrotron radiation) or even just the Rayleigh-Jeans tail from cold blackbody emission (e.g. in the cosmic microwave background).

A decent introductory summary can be found here.


In India, what is the best time to view the geminid meteor shower in 2013 As of the time of writing this question, the meteors should have appeared, but so far I haven't seen any. What time would be the best to view them?

They should be visible any time at night (note that longitude has little to do with observing meteors), but there are many things that could thwart your observing.

Bright lights: In a big city you will see far fewer meteors. In a big enough city, you won't see any. On that note, if you keep checking a mobile device with a backlit screen, you probably won't see anything.
The Moon: The Moon will unfortunately be up until something like 4 AM local time (assuming I plugged numbers in correctly to the US Naval Observatory's table). Just as with cities, this will cover up smaller meteors. Just try not to stare directly at the Moon lest you lose night vision entirely.
Limited view of the sky: The key to watching meteors is to see a large amount of the sky at once. Never use binoculars or a telescope. The idea is to get comfortable by laying down (yes, this is important - people who crane their necks get tired after 20 seconds and then stop looking up!) and just look at as much sky as possible. Your eyes naturally catch and follow movement, so you'll instantly focus on a meteor once it comes into view.

As for where to look, the answer is "everywhere." Meteor showers are named after the radiant - the point in the sky from which the trails seem to originate if you trace them backward far enough. This shower originates in the constellation Gemini, which isn't terribly recognizable, but can be found by locating Orion:

Move from his belt to his right arm (the one indicated by the very red star Betelgeuse) and keep going until you come across two reasonably bright stars. That is Gemini. It is important to keep in mind, though, that meteors streak across other parts of the sky. Just because they seem to originate from Gemini doesn't mean they're actually visible there. Knowing the radiant just helps with knowing what direction to expect the streak to be going.


can we see all binary stars as pairs? As it is obvious from the question title
is there any restriction or cases that we can see the binary stars as not in pairs

also is there any binary stars doesn't linked together by gravity ?

I know the definition of binary stars that it is  a pair of stars that revolve around their common center of mass but my question is about special cases if it is exist

To answer the main part of your question: Yes, there do exist such systems. They're called visual binaries. We generally need a telescope to tell them apart. Most binary systems look like a single star when viewed with only the naked eye; many cannot be resolved without the aid of a telescope. But visual binaries can.

  also is there any binary stars doesn't linked together by gravity

No, there aren't. You were right when you said that the definition of a binary system is basically two stars orbiting a common center of mass. If the stars don't orbit each other, they aren't a binary system.

By the way, check out Proxima Centauri, the nearest star to the Sun. It's one of the three stars in the triple-star Alpha Centauri system. The reason I mention it is that even though it's seemed for years like it was gravitationally bound to the system, that idea is now under debate.

Let me address something that Mitch pointed out. The classification of star systems as "visual binaries" is based solely on our ability to observe them. If we had better telescope, this class could change. An analogy might be our naming certain wavelengths of light "visible". They only depend on our perceptions, not some objective characteristic that all observers in the universe could agree on.


Rebinning with a non-integer factor I'm in doubt whether I should put this in the astronomy or rather the coding exchange, but since my interest goes out to the former I will try my luck here.

Suppose I have an image with a certain pixel scale $x$. I would like to rebin this to a larger pixel scale, $y$, but this is not an integer multiple - so I can't say, for instance, that every group of 4 pixels in the original grid will now be one big pixel in the new grid, and average their flux.
Rather, I'd like something like i.e. $y=2.3\cdot x$.

Is this routinely done in the astronomical community -- and if so, how?

What you're trying to do is change the pixel scale, but ultimately, you need to decrease the resolution of your image. I'm quite sure this can be done using a variety of image editing tools (even some free ones), but if you insist on doing this yourself, I can show you the general process. Let's say I have the image below, which for convenience is only 6x6 pixels in size.

I'll assume this image has a pixel scale of $p = 1\:\frac{''}{\text{pix}}$ and I want to convert it to an image with a pixel scale of $p'=1.5\:\frac{''}{\text{pix}}=1.5p$. This is effectively saying that I want the resolution of the image to decrease by a factor of $1.5$ and that each pixel in the $p'$ image will be comprised of $1.5^2$ pixels from the $p$ image. This is easily done by defining the new pixel grid over the old one and determining the new pixel "color" as the weighted mean of the old pixels which exist in your new, larger pixel.

For convenience, I'm going to define a coordinate system where the pixels are at position $(x,y)$ where $(0,0)$ is the top left pixel and $x$ and $y$ increase to the right and down respectively. This is a common coordinate system for images in programming.

Let's look at pixel $(0,0)$ in the $p'$ image. This is comprised of pixels $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$ from the $p$ image. Specifically, the fractions of these pixels in our new pixel are $1$, $0.5$, $0.5$, and $0.25$ respectively. These fractions will become the weights.

What you need to do then is look at your four pixels in the $p$ image and determine their color/brightness/property-you're-trying-to-merge. For this simple case, I'll just use a number from zero (white) to one (black) that represents the magnitude of the pixel, but the same process applies no matter what your pixel information contains, be it an RGB value or whatever else. By looking at my four pixels in the $p$ image, I see the magnitudes are $0.05$, $0.05$, $0.25$, and $0.15$ respectively.

The new pixel's magnitude is then the weighted average of these.

$$p'_{00,\text{mag}} = \frac{1\times p_{00,\text{mag}} + 0.5\times p_{10,\text{mag}} + 0.5\times p_{01,\text{mag}} + 0.25\times p_{11,\text{mag}}}{1.5^2} = \frac{1\times 0.05 + 0.5\times 0.05 + 0.5\times 0.25 + 0.25\times 0.15}{1.5^2} = \boxed{0.1056}$$

The process is the same for each pixel in the $p'$ image. Hopefully you can see how this process would generalize for any new pixel scale. The hard part is determining where your new pixels will be and which pixels of your original image will comprise the new pixel. Once you know that, you simply average them together, weighted by the fraction of the original pixels that make up the new pixel. With a little bit of effort, this could all be fairly easily automated by a program and you'd only need to input the new pixel scale and the original image and it could output a new image at the new pixel scale.

One problem you might run into is that your new image can't be subdivided perfectly. E.g., if, in the above example, you wanted a new scale of $1.4$, you'd still wind up with a new $p'$ image of 4x4 pixels, but you'd only be able to cover 5.6x5.6 pixels of the original image and you'd necessarily have to cut off part of your image, thus losing some information. It'd be up to you to decide what you cut off.


Is there a web service providing orbital data of solar bodies (planets, comets, asteroids, dwarf planets)? I would like to be able to connect to a web-service and retrieve JSON data containing information on current (as of today or so) orbits of planets, comets, asteroids, and dwarf planets in our solar system. (Even just planets would be fine but better if I can get info about asteroids, comets, and dwarf planets as well).
I would like to draw planets and their orbits accurately in a 2D animation of our solar system in an Android app in the X-Y coordinate plane as observed from the top of the Z axis.

Where can I find such public web service and how can I access it and get that data?

UPDATE
I tried using HORIZONS Web-Interface but out of all variables I need, I can see only "Semi-Annual Axis" information.  I need following information:

orbital ellipse angle in X-Y coordinate system (i.e 1.5rad or degrees or so)
motion direction (pro-grade or retrograde)
current position of planet/comet/asteroids on that ellipsis in degrees or radians, say 60deg from x (horizontal) axis
time it takes plant to go around the sun (I believe this is called orbital period)
eccentricity of ellipse (orbit)
semi-annual axis

These are the values that give me enough information to draw orbit (ellipsis) and correctly position planet on that ellipsis, and to calculate speed at which it moves around the sun in my 2D model.

So, I am not interested in inclination or any parameters that would be useful for a 3D model.

I can find this data online but it takes lots of searching since I have not been able to find a central place for it.  HORIZONS seems to be such place but out of variables I need (as described above), I can find there only semi-annual axis in HORIZONS.  I am surprised that I cannot fine a public web-service offering this info (REST or SOAP, or whatever, does not matter as long as I can fetch that data and process it for my needs from my application).

I apologize if my explanation is somewhat unclear, I am not an astronomer.

Hope that explains.

Not quite the webservice, but there is a JavaScript library Orb2.js with examples including X-Y(-Z) in ecliptic coordinates. So maybe you can compute it directly, not relying on some external service:

http://www.lizard-tail.com/isana/lab/orbjs/

As far as I do not use it actively, it just a pointer to site to explore, not a definitive solution for you.


How to adapt small Dobsonian to use with a tripod? I am the owner of a small 4", f/6.3 tabletop Dobsonian telescope with a 102mm aperture and 640mm focal length.

It is based on a Dobsonian mount but I would like to use a tripod instead because of the complications of having to cary a table with me everywhere I would like to use the telescope. The alternative is to but the telescope on the ground which makes viewing uncomfortable and is not very stable.

A good tripod could be foldable or collapsible and make my rig much more portable and bring the eyepiece up to a more convenient viewing height, without having to bring along a "telescope table".

Is this possible? If so, are there any specific tripods I should use or other pointers on how best to proceed?

This is not me or my telescope or mount, but it gives an idea of viewing with a similarly-sized tabletop telescope. I'd like to mount the telescope on a tripod instead.

 Source

First, a tripod won't help you point to a particular location in the sky.  For that you would need a correctly-aligned equatorial mount with setting circles, or a mount with digital encoders (you can get these for dobsonians) that when correctly configured can convert the telescopes position to a position on the celestial sphere.

It is possible to convert a dobsonian to a different kind of mounting.  If your telescope doesn't already have a tripod bracket, you would probably need to buy tube rings such as these, and put one on either side of the circles on the sides of the tube that sit in the rocker box, so that the center of gravity of the scope was between the rings.  Then you'd probably need a plate connecting the rings and providing a mount point between them that you could attach to a tripod or a more complex mount (like an equatorial).

My suspicion is that rather than trying to adapt the dobsonian to a different mount, you'd be better served either by buying after-market digital encoders for your dobsonian, or just getting a whole new telescope with the kind of mounting you want.


How to build a powerful home made telescope Simple, I fell in love with the sky when my parents were constructing our new house I was a kid then and one day I ventured out on the terrace, to my amazement it was a full moon, ever since I've felt spiritually connected with the Universe. Its been around 20 years since that day, out of my busy schedule I looked at the stars yesterday, I found love again! Wanted to buy the best telescope out there, but its out of my pocket.

Want to build a home made powerful telescope? Any help fellow sky lovers and star chasers?

You can buy remarkably cheap reflecting telescopes these days on a mount called the Dobsonian mount. They don't (usually) have fancy GoTo automated pointing and tracking, but give fantastic bang for buck as the majority of their cost is invested into the biggest primary mirror possible.

Building a telescope would be a great hobby and learning experience, and there are many books which will show you how. Just don't expect to save a significant sum of money over a commercially built scope - it could even be more expensive to build your own.

Fancy GoTo tracking mounts are not usually worth the expense and hassle (in setting them up) for a first-timer, though the Orion Intelliscope series is an intriguing compromise. For an extra $100 or so, the telescope has no guiding motors, but tracks where you push the telescope and a hand controller will tell you how much further you have to go till you reach your target.

My recommendation? Get the biggest Dobsonian mount scope you can afford. With a mirror greater than 10" things begin to get large and unwieldy, so 10 inches (mirror diameter, not tube length) is my recommended maximum size for a first telescope. Just go for the biggest aperture you can buy and easily carry. Don't invest much in GoTo systems, as although they can be quite useful they are also a pain to set up if you don't have the scope permanently mounted.


How can I safely observe a Solar Flare? Solar Flares obviously release extreme amounts of energy and extend thousands of miles out into space.

Because they are so big I would like to be able to observe some of these events through a telescope but I am wary of damaging both my eyes or the sensitive telescopic equipment.

How might I safely observe a solar flare through an amateur telescope? Might this be solved by using light filters to dim the amount of light entering the telescope or is a more specialist piece of equipment required?

One truly safe way of solar flare observations is to not directly observe them (with your eyes or imaging technology) at all, such is the case with radio-astronomy.  There are a number of examples of how this could be done, for example, the Sudden Ionosphere Disturbance Monitor, this device measures disturbances in the ionosphere that includes the effects of solar flares.  This operates in the VLF (3-30kHz) range.

From the Society of Amateur Radio Astronomers, the advice is

  For VLF solar flare observations you will need a strip chart recorder and a radio receiver capable of operating in the noisy 20 to 100 kilohertz radio band. These receivers are quite simple and may be home constructed.

(Plans for these can be obtained from that group)

A repository of further links (far too many to list here) can be found on the website Amateur Radio Astronomy on the Internet.


Is the universe macroscopically transparent to CMB? Is the fraction intercepted by stars and dust so tiny that it doesn't have a correction factor? Background
The question Why do some electromagnetic waves continue travelling while others disappear? is interesting, and in addition to the answer there I started to write:

This is a supplementary answer to @ConnorGarcia's answer using the example in the question. It's not very accurate but it does address part of the question:

Cosmic background radiation emitted when the Universe was very young still exists. But my wifi signal seems to disappear a short distance from my apartment. Why?

Imagine walking around in the middle of a city with a receiver that can pull in 900 MHz to 5 GHz. It will tell you that you are in a "soup" of radio photons; no matter where you go inside the city you'll...

and then realized I didn't know what I was talking about.
Question
So instead I'd like to ask about the following:
I think that CMB photons are believed to be "pristine photons" from soon after the big bang rather than absorbed and re-emitted and thermalized, and I seem to remember recently reading an answer here or in Physics SE saying that a measured CMB spectra (in any given direction) is fit with a Planck distribution with unity emissivity; there was no "fudge" or scaling factor in front for the absolute spectral radiance measured.
But now I can't find that answer anywhere!
Together those suggest to me that macroscopically the universe is essentially transparent to CMB, and the fraction intercepted by stars and dust between us and it is so tiny that it doesn't require a correction factor.
Question:  Have I got this right?

For the most part, the CMB photons travel directly to our telescopes from the surface of last scattering.  Some corrections need to be made to determine the blackbody nature of the spectrum, but they are not corrections for absorption of the photons.
The two main corrections are shown clearly by this COBE image sequence:

(from here).
First, the motion of us as observers, relative to the rest frame of the CMB, distorts its single-temperature blackbody shape, making it appear hotter in one direction and cooler in the opposite direction.
Once this is removed (by a sky-position-dependent frequency shift), the foreground emission of the Milky Way becomes apparent (middle panel), and that must be modeled and subtracted. The dust in our Galaxy is on average hotter than the CMB, so its contribution can be measured well by observing over a range of wavelengths, and of course it is most heavily concentrated toward the Galactic plane.
After that is subtracted, the remaining emission (bottom panel) is that of a blackbody to very high precision, with of course the remaining structure that gives us so much interesting information about the early universe.
Here is part of the key concluding paragraph from the Mather et al. 1992 paper on measuring the spectrum:

The FIRAS spectrum of the cosmic microwave background radiation  agrees with a blackbody spectrum to high accuracy.  The CMBR spectrum is the result of fitting a model including a dipole and a dust map derived from 240 micron data, excluding a region around the Galactic center. The final temperature is 2.726 +/- 0.010 K (95% CL systematic), where the error is dominated by our estimate of the thermometry errors.

So they are shifting the emission and subtracting the foreground emission, but not applying any overall correction for absorption of the emission.


How to install the Barlow lens on this telescope? I've brought a 4.5inch astronomical telescope recently which has an aperture of 114mm and a focal length of 1000mm, and two eyepieces. The two eyepieces were of 25mm and 10mm. I enjoyed stargazing using the 25mm eyepiece. But the stuff that I could see with my 25mm eyepiece was not visible through the powerful eyepiece. I'll post some pics. Can someone tell me how to install the Barlow lens?

Based on the length of the optical tube (which doesn't appear to be roughly 1 meter even though you mention it is a 1000mm scope) this is probably a "Bird Jones" telescope and not a typical Newtonian reflector.

  A Jones-Bird reflector telescope (sometimes called a Bird-Jones) is a mirror-lens (catadioptric) variation on the traditional Newtonian design sold in the amateur telescope market. The design uses a spherical primary mirror in place of a parabolic one, with spherical aberrations corrected by sub-aperture corrector lens usually mounted inside the focusser tube or in front of the secondary mirror. This design reduces the size and cost of the telescope with a shorter overall telescope tube length (with the corrector extending the focal length in a "telephoto" type layout) combined with a less costly spherical mirror. Commercially produced versions of this design have been noted to be optically compromised due to the difficulty of producing a correctly shaped sub-aperture corrector in a telescope targeted at the inexpensive end of the telescope market.

Also see Astrowiki

The "Bird Jones" designs use a spherical primary mirror and then use (or attempt to use) a corrector installed in the focuser assembly.  It also acts as a 2x Barlow.  A 1000mm "Bird Jones" design scope would really be roughly 1/2 meter in length.

The optics are usually not great (not as good as a Newtonian) and not a good candidate for using a Barlow (because now you're basically putting a Barlow on a Barlow).  The image will get faint and very fuzzy.

Have you tried using the scope with just a 10mm and no additional Barlow?

Remove the eyepiece (and Barlow if you've inserted one) from the focuser.  When you look through at the secondary mirror... do you see any "glass" in the path (other than the secondary mirror?)  If yes, then you have a "Bird Jones" design telescope and not a typical Newtonian reflector.

While you are certainly able to try using a Barlow, I suspect you wont be happy with the results no matter what you do.  I suspect a Barlow with a 10mm eyepiece would be pushing the magnification well beyond the practical limits of the optical system of that telescope.


Could someone see anything while being inside black hole? If we managed to survive in a black hole and move inside the event horizon then could we see the surroundings of the black hole inside the event horizon by source of light? Can the light not come up to the event horizon or does it have to travel through different spheres of the black hole and enter a sphere which won't allow the light to return to its previous sphere? If the we reach up to the singularity then can we ever reach up to the event horizon or would we get stuck in the point of the singularity?

The answer is most definitely yes, or at least yes, as far as our current understanding of how gravity works goes. It is observationally untestable (let's be more specific - nobody could report the results of an observational test!) since no signal can emerge from inside the event horizon.

The scenario is treated in some detail by Taylor & Wheeler ("Exploring Black Holes", Addison, Wesley, Longman - highly recommended) in terms of what an observer would see on a direct radial trajectory into a non-rotating "Schwarzschild" black hole. I won't bore you with the maths - it is fairly complex.

A star situated at exactly 180 degrees from the observer's radial path will always appear in that position as the observer looks back - right down to the singularity. The light will be gravitationally blueshifted by an ever-increasing amount - essentially tending towards an infinite blueshift at the singularity.

For stars at an angle to the radial path, their positions will be distorted such that they appear to move away from the point at which the observer has come from (and are also blue-shifted). In the final moments (it takes less than $1.5\times 10^{-4}$ seconds of proper time to fall from the event horizon to the singularity of a 10 solar mass black hole, but a huge $\sim 60$ seconds for the black hole at the centre of our Galaxy) the light from the external universe will flatten into an intense ring at 90 degrees to the radial direction of motion. So you would end up seeing blackness in front of you, blackness behind and the sky split in two by a dazzling ring of light (almost seems worth it!).


Alignments of planets I am going to ask an odd question, It is: In most of the images of our solar system we see that the planets are arranged in  a straight line. As the paths of the planets are elliptical, could it be possible for planets be arranged other than in straight line (a bit up and down) ?

The orbits of the planets are coplanar (in the same plane) because supposedly during the Solar System's formation, the planets formed out of a disk of dust (ha, ha)  which surrounded the Sun. Because it was a disk, all in one plane, all of the planets formed in that one lousy plane as well.

Single rings and disks are common in astronomy. Jupiter's moons are coplanar too.

The common explanation is that orbits are unstable and they all eventually tend to get into the same plane and stay there. Wouldn't it be fun to have a planet orbiting the Sun at a right angle to the other planet's orbits? You'd never know when the damn thing's going to hit what.


Were parts of the Universe deflating during Inflation? I was reading about the history of the early universe in Carroll & Ostlie and specifically looking at Inflation, which, as I understand, occurred because our [observable] part of the universe entered a false vacuum state containing negative pressures, thus fueling expansion. (Carroll & Ostlie state the opposite, but as per the discussion below, that appears to be incorrect).

Does that mean that there are other parts of the Universe (outside of our horizon) that experienced deflation, as inflationary parts pressed against them? Would there be boundaries between separate parts of the larger universe where inflationary parts encounter one another? Any thoughts on what those parts of the Universe are like, such as containing their own CMB that's NOT close to uniform, or where there is a much higher density of matter?

That sounds... confused, or else just wrong. Negative pressure is what causes inflation: something (inflaton field, cosmological constant, or "dark energy" in general) with an energy density that stays approximately constant as space expands has negative pressure and will drive the exponential expansion of space.

The transition to "true" vacuum is what shuts off inflation, after which you get  the standard Big Bang, during which the expansion continues but slows down.

(The cosmological constant -- or "dark energy" -- also has negative pressure, and so the universe is gradually moving towards a state where the space between galaxy groups and galaxy clusters will again be expanding exponentially.)

The "eternal inflation" model has our visible universe -- almost certainly a tiny part of a much larger region of true vacuum -- as one of many such pockets of true vacuum embedded within a vastly larger false vacuum that's still expanding exponentially.


Where is the intervening light in the M87 black hole? The M87 black hole is at the centre, more or less, of an elliptical galaxy of 1 trillion or more stars. So certainly there are stars and other light emitting objects (emission nebula e.g.) in the intervening 53 million light years between earth and the black hole.

How is it that the centre of the black hole image is perfectly dark, as if it is being observed directly, with not one point of light between the black hole and earth? I appreciate that space is mostly empty, but do we really have an unobscured line of sight to a black hole in the centre of an enormous elliptical galaxy?

Does it have something to do with the type of light being observed?

Based on the comment from @SteveLinton

The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years.  Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.

Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km).  That's 0.004 light years.

If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years.  The volume of that cylinder is just over 3 cubic light years.  Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.

Whether or not any star would even be bright enough to be visible is another matter.

1 Technically a conic frustum but I don't think the difference in volume is significant.


Luminosity L(t) for a homologously contracting star I'm dealing with a homologously contracting star with Mass M, Radius R and a gravitational binding energy of

$E = -a GM^2 / R $

(a is a constant).

I was looking for a way to find an expression for L(t), the luminosity as a function of time. Unfortunately, my attempts using time derivatives have not taken me far.

Thank you for your help!

Fairly sure this is standard bookwork if you are talking about a PMS star on the Hayashi track.

Differentiate your GPE wrt time, assuming that the mass is constant. Take half of this as the luminosity (via the virial theorem). This gives $L$ in terms of $dR/dt$.

Then if you are on the Hayashi track you can assume surface temperature $T$ is constant and Stefan's law gives you $dR/dt$ in terms of $L$, $T$ and $dL/dt$.

Substitute that in and integrate the resulting differential equation to get $L(t)$.

Using this approach, I managed to get
$$ L = \left(\frac{\alpha GM^2}{6}\right)^{2/3} \left(4\pi \sigma T^4\right)^{1/3} t^{-2/3},$$
where $T$ is the temperature of the Hayashi track in question.

If you don't want to make the Hayashi track assumption, you can say that $L = AM^B T^C$ defines tracks in the HR diagram and this can be used in addition to Stefan's law to eliminate $dT/dt$ and give your result in terms of constants $A$, $B$, $C$. These can come from polytropic theory or by fitting the $L,M,T$ function to numerical calculations. (E.g. see Jackson & Jeffries 2014 https://arxiv.org/abs/1404.0683 ).


Are binary neutron star mergers needed to explain the abundance of gold? The NPR News item Astronomers Strike Gravitational Gold In Colliding Neutron Stars mentions and quotes "Daniel Kasen, a theoretical astrophysicist at the University of California, Berkeley:"

He spent late nights watching the data come in and says the colliding stars spewed out a big cloud of debris.
"That debris is strange stuff. It's gold and platinum, but it's mixed in with what you'd call just regular radioactive waste, and there's this big radioactive waste cloud that just starts mushrooming out from the merger site," Kasen says. "It starts out small, about the size of a small city, but it's moving so fast — a few tenths of the speed of light — that after a day it's a cloud the size of the solar system."
According to his estimates, this neutron star collision produced around 200 Earth masses of pure gold, and maybe 500 Earth masses of platinum. "It's a ridiculously huge amount on human scales," Kasen says. He personally has a platinum wedding ring and notes that "it's crazy to think that these things that seem very far out and kind of exotic actually impact the world and us in kind of intimate ways."

Has the merger of neutron star binaries been necessary to explain abundances of the heavy elements such as gold and platinum, or is this just an anecdotal item? How important are binary neutron stars for the abundances of heavy elements such as gold? Is there a particular or notable paper I can read on this?
I have already read this answer but I'm looking for a better explanation of the need for this kind of merger to explain abundances. I'm pretty sure there is nothing in any observed gamma ray events that shows spectral lines of gold or any identifiable heavy element (due to the incredible doppler broadening), so the connection must actually come from simulations.

The creation of some very heavy neutron-rich elements, like gold and platinum, requires the rapid capture of neutrons. This will only occur in dense, explosive conditions where the density of free neutrons is large. For a long time, the competing theories and sites for the r-process have been inside core-collapse supernovae and during the merger of neutron stars.

My understanding is that it has become increasingly difficult for supernovae to produce (in theoretical models) sufficient r-process elements to match both the quantity and detailed abundance ratios of r-process elements in the solar system (see for example Wanajo et al. 2011; Arcones & Thielmann 2012). The conditions required, particularly a very neutron-rich environment in the neutrino-driven winds, are just not present without the fine tuning of parameters (see below).

Instead, the models invoking neutron star mergers are much more robust to theoretical uncertainties and successfully produce r-process elements. The question mark appears to be only over their frequency at various times in the evolution of a galaxy and exactly how much enriched material is ejected.

The announcement of GW170817 makes this all more plausible. A neutron star merger has been seen. The behaviour of the optical and infrared emission after the event matches the expectations of merging neutron star models (e.g. Pian et al. 2017; Tanvir et al. 2017). Of particular note is the developing  opacity and fading in the blue and  visible, with the spectrum
becoming dominated by the  infrared with broad spectral features. This is the expectation for an expanding cloud of material that is heavily polluted by the presence of lanthanides and other r-process elements (Chornock et al. 2017). The reasonable agreement between the observations and models suggests that indeed a large quantity of r-process elements were produced in this explosion.

To go from there to the claim that the origin of gold is solved (as claimed in the press conference) is a step too far. The amount of r-process material produced has large uncertainties and is model-dependent.
The rate of mergers is only constrained to about an order of magnitude in the local universe and is not measured/known in the early universe. What could be said is that this channel for r-process production has been directly observed and so must be taken into account.

On the other hand, r-process production by the supernova channel is not yet ruled out. Some simulations at least, that involve rotation and magnetic fields appear to be still "in the game" (e.g. Nishimura et al. 2016). It could be that the presence of significant r-process material in very old metal-poor stars requires a supernova channel, since the merger of neutron stars takes some considerable time to occur (e.g. Cescutti et al. 2015; Cote et al. 2017).

The overall picture is still uncertain. A review by Siegel (2019) concludes that the best fit to the available evidence is that some rare type(s) of core collapse supernovae (known as "collapsars") are still the best bet to explain the Milky Way r-process elements. The primary evidence for this is the presence of Europium (an r-process element) enhancements in some very old halo stars and the general trend of decreasing Eu/Fe with increasing Fe, suggesting a more alpha-element-like production site for the r-process - i.e. supernovae.


Weight of a celestial body How do astronomers find out the correct weight of a planet even though there isn't any direct means to weigh them? What technique do they use? Just curious!

I don't know if this is what most astronomers use, but it is certainly a method that could be used. It also makes for some interesting photos.

Gravitational lensing was one of the great predictions of Einstein's theory of general relativity, and was actually one of the first pieces of evidence for it - see the solar eclipse of 1919. Gravitational lensing is the bending of light by a massive object. Because mass and energy warp space-time, we know that if you can calculate how much the light is bent, you can calculate the mass of the body bending it. Wikipedia gives the following formula:

$$\theta=\frac{4GM}{rc^2}$$

where $M$ is mass, $r$ is the distance the object is from the light, and $G$ and $c$ are, of course, constants. So if you can measure how much light is bent by an object, you can calculate its mass. It's hard, and you would need good conditions for it to work, but it could work, nonetheless.