Has the Hubble ever been pointed at earth? As in taking pix of earth, not just happen to point towards earth during a maneuver or whatever.

xkcd already did the math on this one!
Like Gauti's answer, xkcd also links to this question on Hubble's website. Hubble moves too fast for its minimum exposure time to be able to focus on anything on the surface. He goes one step further, linking to this article on Bad Astronomy that points out that Hubble frequently points at the sunlight side of Earth to help calibrate its WFPC2 camera.

Still, you may be surprised to find out that Hubble routinely points at the Earth! It uses the bright, daylit Earth to help calibrate one of the cameras on board. The Wide Field Planetary Camera 2 (WFPC2, pronounced ``whiff pick 2'') is an electronic device which detects light, and is similar in principle to a normal digital camera. However, it is far more sensitive, and astronomers are very particular about how well they understand such devices. They want to make sure that every part of the detector is very well calibrated.
One way to calibrate the camera is to look at some bright, evenly lit object. If one part of the detector is more sensitive than another, for example, then part of that object will look brighter, even if in reality it isn't. This can be used to correct any later images.
In space there aren't any flat fields you can use, but we do have one right here: the Earth. To calibrate WFPC2, sometimes Hubble is pointed straight down at the Earth. As the terrain (or water or whatever) streaks across the field of view it forms a very crude flat field image. The image is called a ``streak flat'', and looks really weird. Trees, houses, all sorts of objects blur across the image. It takes a lot of sophisticated computer processing to turn this into a real flat field for Hubble to use, but luckily there are a lot of smart people working on the problem.

xkcd also shows an approximation of what it would look like if you tried to take a picture of an average desk from orbit. It is just a smear of colors, devoid of any discernible detail. I tried attaching the pictures here, but SE doesn't like the png files for some reason and I don't have time to do the file conversion. Just check out the first link to see what it looks like, plus a more detailed explanation of why.


At what rate are stars accelerating? How fast stars accelerating away from a central point? Are they increasing in acceleration or decreasing? If possible could you please provide a plot of the acceleration of stars away from the central fixed point.

As has already been mentioned, stars within a galaxy do not generally expand. The stars in a galaxy are gravitationally bound together.

But as you observe more and more distant galaxies, you see a general trend that distant galaxies are moving away from us at a speed that's proportional to their distance.  (There are some small variations due to the random motion of individual galaxies; for example, the Andromeda galaxy is approaching us, and will collide with the Milky Way in about 4 billion years.)

This does not indicate that everything is expanding from some central point.  An observer in a galaxy a billion light-years away would see the same thing, all other galaxies receding at a speed proportional to the distance from the observer.  This can be interpreted as space itself expanding.  Going into more detail would require General Relativity, which I'm not competent to explain.

Your question is about the rate of expansion.  The answer to that is what's called the Hubble Constant, which describes the ratio between a galaxy's speed of recession and its distance from us.  The current best estimate of this ratio is 67 kilometers per second per megaparsec.  A parsec is about 3.26 light-years; a galaxy one million parsecs away will be moving away from us at about 67 kilometers per second.

Source: https://en.wikipedia.org/wiki/Hubble%27s_law#Determining_the_Hubble_constant


Question on Friedman equation involving cosmological constant This is the question:

By considering both the Friedmann and acceleration equations, and assuming a pressureless
Universe, demonstrate that in order to have a static Universe we must have
a closed Universe with a positive vacuum energy. Using either physical arguments
or mathematics, demonstrate that this solution must be unstable.

I started off with the friedmann equation and equated H^2=0 and K>0. How to proceed further and prove that lamda is positive?

When including a cosmological constant, the acceleration equation reads
$$\frac{\ddot{a}}{a}=-\frac{4\pi G}{3c^2}(\epsilon+3P)+\frac{\Lambda}{3}$$
In a pressureless universe, $P=0$, and in a static universe $\frac{\ddot{a}}{a}=0$. The equation becomes:
$$\frac{4\pi G}{3c^2}\epsilon=\frac{\Lambda}{3}$$Energy density is always positive and non-zero, so $\Lambda>0$.

The solution is unstable since after a slight perturbation, if $a$ grows, $\epsilon$ will go down ($\epsilon_m\propto a^{-3}$ for matter for example), which will result in an increase of the acceleration, making $a$ go up even more.


Lorentz transformation for stars If a star that has regular pulses(standard candle) is flying away from us at relativistic velocity(very distant), would that pulses look like very sluggish compared to a non-relativistic velocity(close to earth)...something like a clock ticking in a relativistic spaceship and....... ofcourse red-shifted.Thanks.

They would if we could see them far enough away, but cepheid variables are only visible up to about 20 million light years distant. That may sound like a long way, but in astronomical terms it is nearby.


Where on Earth could dinosaurs feel the impact of their extinction asteroid? Did the possible Chicxulub asteroid impact around 66 million years ago that made the dinosaurs and many other species extinct, cause the Earth to wobble? How large was the area, where they could "feel" the impact's consequences?

The whole Earth "rippled" after the Chicxulub impact. By ripple, I specifically mean surface displacement.  Current models indicate that it rippled enough that most of the dinosaurs and other species with brains certainly would have felt the ripple.
How much did the Earth ripple? Well, that depends on the location. Meschede et al. [2011] in their paper Antipodal focusing of seismic waves due to large meteorite impacts on Earth, visualize a computer model of the surface wave displacement due to the Chicxulub impact in the following figure:

Note that the displacement doesn't get much smaller than 1 centimeter, which is certainly large enough to feel over the order of tens of minutes.  As another point of interest is that their model shows large displacement near the antipodal point, which some geoscientists correlate to the increased volcanic activity at that time in the Deccan Traps. Also note that these are just the seismic waves.  We know from geological evidence at least from the Southeastern United States, that the resulting tsunami was much bigger.
As far as a "wobble", or change in angular momentum, the Chicxulub impact was too small to have a noticeable effect for the dinosaurs.  JamesK has a nice example here: If an Asteroid was to strike the Earth, would it affect the Earth's rotation?.


Why do comets appear to have a continuous supply of dust? Comets orbiting the sun have a tail that points away from the Sun, due to the solar wind.

(Strictly speaking, a comet can have both a type I tail - ion tail, and a type II tail - a dust tail. This question is about the type II tail - the dust tail).

Now I understand that some of the source of the dust is volatile materials in the comet oxidising due to the comet's proximity to the sun. But surely some of these comets have been orbiting for hundreds of thousands of years (if not millions). Why do they still have unspent fuel? Why are they still dusty? Hasn't all the dust blown off by now?

My question is: Why do comets appear to have a continuous supply of dust?

The rate of loss of mass from a comet is perhaps surprisingly low. A paper The calculation of $Af\rho$ and mass loss rate for comets gives a rate for a "typical" comet at 1.29AU as 153kg/s. A typical comet has a mass of about $10^{13}kg$. Even if the comet is constantly active (and they are not) it would have enough material for over 2000 years.

In fact comets are only active when the are near perihelion, and they are only near perihelion for a small fraction of their orbit. Put it together and you have the potential for an active life of hundreds of thousands of years.

Eventually, however, comets do run out of volatile substances, at which point they become inactive, and appear as small dark asteroids. An example of which is the object 2015 TB145, which passed the Earth on Halloween 2015.

Note: not "oxidising", but "vaporising", the ices in the comet are changing into gas, but there is no burning.


What do the lines in this Event Horizon Telescope and Global mm-VLBI Array infographic represent? In the following EHT and GMVA diagram, what do the yellow and blue lines represent? I don't think they're physical communication lines (that would be unfeasible). Do they depict anything specific?

Source: ESO/O. Furtak

I cant see a definitive statement, but I can see some infiormation and make some reasonable suggestions:

The two colours represent the two different collaborations. Blue for the EHT, yellow for the GMVA. We can see this by following the links on the source page of the image. The GMVA page says:

Participating Stations:
in Europe: Effelsberg (100m (Ef)), Onsala (20m (On)), Pico Veleta (30m (Pv)), Metsahovi (14m (Mh)), and Yebes (40m (Ys))
in USA: VLBA (8x25m) - BR, NL, PT, LA, FD, KP, OV, MK (HN, SC are not equipped with 3mm receivers) and the GBT (100m)
ALMA: for VLBI experiments involving phased ALMA the sensitivity depends on the number of antennas which are phased. For N=50 antennas and a typical phasing efficiency of 0.90 a gain of 1.3 K/Jy is obtained, and the effective diameter is ~ 80 m.

This matches the yellow lines.
This page describes the EHT array and more or less matches the blue lines.
So now what are the lines? I'm pretty sure they are "baselines". That is they connect pairs of telescopes whose signals are directly combined in great detail to produce information about the  image. The data from the various baselines is then combined in a later step. This wikipedia article describes some of the physics. This paper refers to the individual baselines used for the EHT and their contribution to the image.


How does moonrise/moonset azimuth vary with time? In my tenacious attempts to observe moonrise (surprisingly difficult in a poor weather heavily urbanized hilly area with frequent bad smog, with busy work schedule and a bicycle) I was frequently trying to find good observation spots that would provide view to open distant horizon at specific azimuth.

While generally similar moon phases correlated with similar moonrise times I found the azimuth to vary in a way I have a difficult time to put my finger on. Sure there's a roughly monthly cycle of "wobble" between deviating about +/- 30 degrees north or south from "pure east" but while the mechanics of correlation between hour when the Moon is visible from behind the horizon and the degree it's lit by the sun are quite clear to me, the way its - or is it Earth's? - orbital plane is "off" relative to the ecliptic plane and Earth rotation plane eludes me. Could someone provide a sort of guide - preferably something that acts more as help in understanding the mechanics, as opposed to providing raw mathematical solutions, what does this deviation depend on, and how it changes with time?

The monthly variations are caused by the change in declination of the Moon as it travels (roughly) along the ecliptic.  This is caused by the tilt of the Earth's rotation axis, which results in an angle of about 23° between the equator (declination = 0°) and the ecliptic.  Around the constellations of Taurus and Gemini the ecliptic lies at much higher declination than near Saggitarius or Ophiuchus.  The same effect causes the change in rise and set times of the Sun during a year.

When you're in the northern hemisphere, a higher declination means that the transit altitude of the Moon or Sun is higher.  In fact, their whole daily (apparent - caused by the Earth's rotation) paths lie closer to the North star.  Because of that, the points where a daily path intersects with your local horizon lie further north, which you see as a change in rise/set azimuth of the Moon.

All this is true even without the 5° inclination of the Moon's orbit w.r.t. the ecliptic.  That inclination only mitigates the effect; the Moons declination does not exactly vary between -23 and +23° every month, but these extreme values themselves can lie between +/-18° or +/-28°, depending on whether the inclination effect increases (23+5°) or decreases (23-5°) the effect of the motion along the ecliptic.

Lastly, the rise/set azimuths do not correlate to lunar phase, as do the rise/set times.  Both phases and rise/set times are defined based on the Sun's position; we roughly define "noon" as the moment when the Sun is transiting in the south, and call the Moon "New" when it's in the same direction as the Sun, so there's no wonder that the New Moon transits around noon.  This is not true for the effects described above, and hence the corresponding periods are different.  The mean lunar period between twice the same equinox point (e.g. Aries) is called the Tropical month and lasts ~27.3d (this is roughly the same as the siderial month, but influenced by precession).  The effect of the inclination of the Moon's orbit has a mean period of one Draconic month, with a period of ~27.2d.  Instead, the Synodic month, the mean period between two cases of New Moon, lasts 29.5d, about 2.2d longer.  Hence, the declination and rise/set azimuths of the Moon do not correlate with the phases.


Could Ganymede's aurora be seen with the naked eye? If one was orbiting in the Jovian system, as depicted in this artist conception,

could the aurora be seen unassisted?

Hubblesite-Illustration Credit: NASA, ESA, and G. Bacon (STScI)

The aurora depicted in the artist's conception is of ultraviolet light.  It would not be seen with the naked eye.

"NASA's Hubble Space Telescope observed a pair of auroral belts encircling the Jovian moon Ganymede. The belts were observed in ultraviolet light by the Space Telescope Imaging Spectrograph and are colored blue in this illustration. They are overlaid on a visible-light image of Ganymede taken by NASA's Galileo orbiter. The locations of the glowing aurorae are determined by the moon's magnetic field, and therefore provide a probe of the moon's interior, where the magnetic field is generated. The amount of rocking of the magnetic field, caused by its interaction with Jupiter's own immense magnetosphere, provides evidence that the moon has a subsurface ocean of saline water."


When was the nearest star discovered? When historically did we realize that the Sun is a star, like all the others?

An answer was posted, then the question was put on hold. I would like to answer:
If indeed someone asserted that the Sun was a star 2500 years ago, that is rather humbling. That it took until about 100 years ago for it to be proven is even more humbling!

People level a lot of claims against science and scientists. But if a field of knowledge takes over 100 generations to prove a true statement, perhaps they have a right to criticize. So, I think that science needs to do a better job of proving its theories. Either that, or people need to have more common sense, not to recognize something that has been staring billions of people in the face for all time. Perhaps, we should all be a bit more humble about the state of knowledge, even in obvious areas?

Maybe if we had taken the people who proposed this fact, and the concept of atoms, seriously at the time, we would be on our way to other stars by now instead of still having only visited that big white thing nearby (forgot the name just now) and that not for the past 40 years.

After a quick google, This page pops up showing that Anaxagoras was the first documented person to make the link between the sun and stars being the same thing.


Why both sunrise and sunset use the position of the sun's upper limb? It is my understanding, that astronomical moments of both rise and set use the relative position of background object's upper limb to the foreground object's horizon.
When were these definitions formulated?
Is this done purely because sunrise and sunset are the main examples of such events and it is easier for the naked eye to tell when sun is completely below horizon?
Wouldn't this offset the symmetry of rotations (since the "day" is always longer than the "night" by the duration it takes the background object to move fully below the foreground horizon)?

When were these definitions formulated?

Multiple religions have rather strict requirements based on sunrise / sunset, for example, when one must start or can stop fasting, or when exactly to sacrifice a lamb (or a human in some religions). The definition has very deep roots.

Is this done purely because sunrise and sunset are the main examples of such events and it is easier for the naked eye to tell when sun is completely below horizon?

Yes.

Wouldn't this offset the symmetry of rotations (since the "day" is always longer than the "night" by the duration it takes the background object to move fully below the foreground horizon)?

Daytime is not longer than nighttime in the middle of winter, in places that do have winter. The definition of sunrise and sunset does make daytime be a bit longer than nighttime on the equinoxes.
Suppose sunrise was instead defined as the time at which the entire Sun first appears above the horizon and sunset was instead defined as the time at which the first bit of the Sun drops below the horizon. Even with these extreme definitions, daytime would still last longer than nighttime on the equinoxes thanks to atmospheric refraction.


What would happen if the earth's orbit were slightly perturbed? Suppose a rogue planet came zipping through the solar system and changed the earth's orbit a little, including the plane of the orbit.

I have the idea that, in a complex system, orbits would tend toward roughly circular and in the same plane. In simulations of star clusters I've seen, they all whip around in a crazy dance, some of them flung from the system entirely, and the remainder in roughly circular orbits. In our own solar system I am vaguely aware that the stability of the planetary orbits has to do with the relations of the orbital periods, resonances, and that the early solar system was a chaotic place, with Jupiter wandering in and then out again, until everything settled down. Just because the primordial cloud had an average angular momentum doesn't mean the planets must have all originally orbited in the same plane.

So I have the idea that in a complex system, where the different bodies can affect each other's orbits, they would tend to become circular and in a plane. Comets and asteroids by contrast, besides having long orbital periods, can't change the planetary orbits to any meaningful degree, they're flying through predetermined gravitational fields that don't adapt to them. But if the earth's orbit were perturbed, I'm supposing that the inner planets would adjust and the earth's orbit would recircularize, even if it doesn't settle down to quite its original orbit.

I have to think that astronomers have simulated solar systems and come to some conclusions about that. Is anyone familiar with that and can say something about it?

You're touching on a lot of related topics which isn't recommended for Stack Exchange.  It's better to keep it to one question and specific.   That said, I'll give this an answer as best I can, as I've been interested in this subject for a while and have read up on it a bit.   The powers that be might close this question for being too broad and/or lacking specifics on the specific details of the rogue planet that affects Earth's orbit.

The subject of the orbital systems around different stars is of interest to astronomers and it has been modeled, many times in fact and observed, though it's very difficult to get a good look at other solar systems.  Most of what has been observed is probably incomplete and a lot of what has been modeled involves some guess-work so it's an ongoing field of study.

  I have to think that astronomers have simulated solar systems and come
  to some conclusions about that. Is anyone familiar with that and can
  say something about it?

Yes, astronomers have studied and modeled this.   It's worth pointing out that models will only give the right answers if you ask the right questions.   The fact that astronomers didn't expect, but found many hot Jupiters is one example of the problem with running models vs learning how real systems are likely to be set up.

This is one of my favorite videos on solar-system formation which touches on your question, though this one focuses on inner-planet formation, it's entirely mathematical modeling and it suggests that 3 gas giants tends towards instability while two can be stable.  It's a little old now, but I think it's still accurate information.   The modeling begins about 20 minutes into the video and the 3 gas giant chaos is shown around 24 minutes.

Scholarpedia also has a nice article on the current solar system's stability.

When you write this:   "I have the idea that, in a complex system, orbits would tend toward roughly circular and in the same plane."

I think it's safe to say that there are both stable and unstable systems and an unstable system, after it kicks enough planets or other material out, will become a stable system.

I think it's also safe to say that the proto-planetary disk is what causes planets to form on the same plane, but planetary migration may change that.   The sum of inclination should remain fixed, but ejected can change that.

Our solar system, the 8 known planets orbit about 6 degrees off the Sun's equator and planet 9 is one possible explanation for that, but an ejected planet could explain it too.

  So I have the idea that in a complex system, where the different
  bodies can affect each other's orbits,

This is probably true and it's still true in our solar system.   Jupiter and Saturn affect Venus, Earth and Mars Milankovich cycles.    Mars, being the closest to Jupiter is especially prone to large axial tilt shifts and eccentricity shifts.     Jupiter might, in time, even pull Mercury entirely out of it's orbit where it could move significantly.  The odds may be low, but if that happens, once it starts, Mercury could pass close enough to Venus or Earth and undergo a gravity assist and basically end up anywhere, possibly crashing into another planet or crashing into the sun.

There is a circularization effect caused by tides, but it only applies in 2 body systems where the rotation is faster than the Orbit and the tidal effect pushes the orbiting object outwards.   There seems to be a circularizing effect for mutually tidally locked systems like Pluto-Charon but planetary perturbations prevent planets from having neat circular orbits in multi-planet systems.   They can still be long term stable, but there is eccentricity variation.

The only planet in our solar system with a nearly perfect circular orbit is Neptune and I suspect it's because it's the outer-most planet.   Outer planets tend to perturb the orbits of inner planets, but less so the other way around.  At least, that seems true.  I can't give a good mathematical reason for that, it's more loosely deducted on my part.

The inner 4 planets have eccentricity that changes over time (one of the Milankovich cycles, driven primarily by outer planets).   Mars' line looks more straight, but it's on a different axis, it actually changes the most.

Source of image

  they would tend to become circular and in a plane

I wouldn't make this assumption.  Though the proto-planetary disk likely begins in a plane over the star's equator, some systems have been found that don't orbit in a plane.   A plane may be most common, but I suspect that's because all systems begin that way more than they tend to become that way.

See here and here

To your main question,

  Suppose a rogue planet came zipping through the solar system and
  changed the earth's orbit a little, including the plane of the orbit.

I would think that if a rogue planet changed Earth's eccentricity, that Jupiter over time would return Earth to about where it was.   Maybe with the inclination too.  I couldn't find any good information on inclination changes over time by orbital perturbation.

If Earth's semi-major axis was changed, that seems to be more constant and that would change the length of Earth's year and might might be a much more permanent change.  Likewise, if the Semi-major axis was moved into resonance, say with Jupiter or Venus or Mars, . . . that might be bad, as it might open the door to repeating perturbations and instability.

That's my understanding anyway.  Corrections welcome.


Is there convection in Mars' mantle? Mars is differentiated, with a core, a mantle and a crust. There are no (longer any) plate tectonics on Mars. Does this mean there is no convection in the mantle? Could there be convection in the mantle without it causing plate tectonics?

There are several ways to look at this.
One rather simple way is to compute the Rayleigh number of Mars' mantle. The Rayleigh number is a dimensionless number defined as:
$Ra = \dfrac{\rho g \alpha \Delta T d^3}{\eta \kappa}$
where $\rho$ is the mantle density, $g$ is gravitational acceleration, $\alpha$ is the thermal expansion coefficient, $\Delta T$ is the difference in temperature over the convective layer of thickness $d$, $\eta$ is the viscosity, and $\kappa$ the thermal diffusivity. It is generally considered that convection happens when $Ra > 1000$, and higher $Ra$ mean more vigorous convection. $Ra$ for Mars is estimated at $10^6$ by Mulyukova & Bercovici 2020 (compared to $10^7$ for the Earth), meaning that there is convection in Mars' mantle, although less vigorous than on Earth.
Another approach is to look at some geological features of Mars that can provide constraints on the internal dynamics of the planet. For example, recently active point-source volcanoes, such as Olympus Mons, suggest the presence of a mantle plume—thus convection—rising from beneath (Li & Kiefer 2007). Likewise, the presence of mantle plumes/convection can explain variations in lithospheric thickness (Kiefer & Li 2009).
A study by Stern et al. (2017) used volcanisms, faults and impacts to define a "Tectonic Activity Index" (TAI) for 26 planetary bodies, ranging from 0 to 3. They found that Mars has a TAI of 2, less than the Earth and Venus (3), but still inferred to be "tectonically and convectively active" (bodies with TAI < 2 are inferred to be tectonically dead). They think that Mars actually represents the transition between active and dead bodies:

The present tectonic style of Mars represents an almost completely
stabilized stagnant lid [SL]. Here, a single upwelling mantle plume,
pouring lava out at the same sites for hundreds of millions of years,
produced the immense Tharsis rise and volcanoes (Zuber, 2001). When
these last few mantle plumes are extinguished, tectonic style
transitions to the final SL stage – planetoid death. This stage is
typified by Mercury and Earth's moon, bodies with ultra-thick,
ultra-stable lithospheres

Possible evolution of magmatotectonic styles for a large silicate body, like the Earth. Plate tectonics requires certain conditions of lithospheric density and strength to evolve and is likely to be presaged and followed by stagnant lid tectonics. After Stern et al. (2017) (CC BY-NC-ND 4.0)


What is the characteristic time of the evaporation of the galaxies? For a star, to leave its galaxy, it requires probably a 3-body near-collision at the edge of the galaxy. It is unlikely, but possible. And, if a star once somehow got the required velocity to escape the galaxy, then it will escape and never comes back.

Thus, galaxies should have a characteristic time of their evaporation, what could be calculated. I suspect, this time is probably longer, than other related processes (expansion of the Universe, end of the age of the stars, etc), but it exists can it can be probably calculated (most likely, by numeric simulations).

It probably also depends on the size and star density of the galaxy.

Was it calculated? How big is it?

The standard treatment can be found in (Binney & Tremaine 2008), but see also (Adams & Laughlin 1997) for a good treatment.

The overall timescale for galactic evaporation is $$\tau_{evap}= 100\tau_{relax}\sim 10^{19}$$ years.

The relaxation timescale $$\tau_{relax}=\frac{R}{v}\frac{N}{12 \ln(N/2)},$$ where $R$ is the size of the system, $v$ is the typical random velocity, and $N$ is the total number of stars. This corresponds to the time it takes to completely randomize the velocity of a star by interactions with other stars.


How to leave the Universe If the Universe is flat and expanding at the speed of light, that must mean that we could never leave it, because one would have to travel faster than the speed of light which is not possible. However, what about going straight up or down ? The distance is less and if the Universe is not expanding in depth but only sideways then it would be possible to exit. Could we then find another Universe once we have gone out of this one ?

Welcome to Stack Exchange: Astronomy!

When scientists say the universe is "flat," they don't mean that it's shaped like a dinner plate.  It's no easier to travel "up" than "forward."  In fact, off-planet, those terms don't even have any meaning.

What you're referring to has to do with the shape of the universe:

  When physicists describe the universe as being flat or nearly flat, they're talking geometry: how space and time are warped according to general relativity.

Think of it like this: if you draw a triangle on a balloon, is that triangle flat?  In one sense yes, because it never rises above the surface of the balloon.  Relative to the balloon, it is one-dimensional; you wouldn't be wrong to say that the image of the triangle is "flat."  But relative to the rest of the room, it is curved, because it follows the curvature of the balloon's shape.

When most people talk about the universe, they're talking about the triangle.  The discussion you're referring to, of the "flatness" of the universe, is discussing the shape of the balloon.

As for flying "out" of the universe and potentially into another, we're like an ant walking on the surface of that balloon: no matter what direction the ant walks, it'll just keep walking around the balloon.  The surface of his universe may be curved, but there's no direction he can go that takes him off of that curve.  For all he cares, it's essentially a flat surface, because he can only move in two dimensions along it.  Similarly, the geometry of our universe may be one thing or another, flat or curved, but either way that doesn't mean we can point a ship in any direction and "leave."


Does interstellar cloud obstruct or reduce the visibility? We just pictured a medium distant black hole by combining data from various observatories around the world. That's when I wondered if interstellar cloud blocks the visibility at multiple position in the same direction in space such that we never accurately know the exact luminosity and distance between the black hole and Earth? I was not able to find much information about this particular subject so,

Is it possible that the interstellar cloud reduces the visibility of an object behind it?
And if so, is it possible that we are only able to observe interstellar cloud upto a certain distance in space?
If we observe a star that is behind a interstellar cloud, how do we determine the luminosity of the star?
Can we accurately filter out interstellar cloud that are separated by x distance but at the same direction?

Yes it is possible that a cloud reduce the visibility because the matter in it will absorb some of the light which is going through.

Yes, we cannot see all the universe and so we can see interstellar clouds up to a certain distance. Indeed, light takes time to reach us.

If there is a star behind a cloud you will have difficulties to determine its true luminosity. In general, we define the luminosity of stars thanks to a reference star.

I'm not sure to really understand the fourth question ...


How to calculate observer's latitude from declination and culmination of a star? From gcseastronomy.co.uk:
A star culminates at 50°. It has a declination of +20°. What is the latitude from where it is observed?

What is the relationship between a culmination, declination, and latitude? How would one work this out?

This is the answer:
50 - 30 = 20°
Latitude = 20°

Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude.
Neglecting atmospheric refraction, the altitude at culmination is
$$\mathsf{alt_{max}} = 90^\circ - |\phi - \delta| $$

Here are some special cases:

A star whose declination equals the observer's latitude ($\delta = \phi$) culminates at the zenith (altitude 90$^\circ$).
A star on the celestial equator ($\delta$ = 0) culminates at altitude 90$^\circ$ - $|\phi|$.

For the example in the question, if we solve for $\phi$,
$$\phi = \delta \pm (90^\circ - \mathsf{alt_{max}}) $$

Then the observer's latitude is 20$^\circ$ $\pm$ 40$^\circ$, either 20$^\circ$S or 60$^\circ$N.
GCSE Astronomy's answer is incorrect.


Are there any galaxies which fell out of sight horizon due to cosmic expansion? If farthest galaxies run away from us with acceleration making them exceed speed of light, we should expect them to disappear from sky among time with increasing quantity. Did we observe this? Can we indicate next galaxies to eliminate and their time of decline?

My question concerns galaxies moving with all speed ranges, not only ones greater than speed of light.

No. In fact the opposite is the case.

(See the last paragraph for an intuitive explanation.)

It is a common misbelief that galaxies receding faster than the speed of light are not visible to us. This is not the case; we easily see galaxies moving at superluminal velocities. This does not — as I think most people would think — contradict the theory of relativity, which says that nothing can travel through space faster than $c$. Galaxies do not travel through space (except with small velocities of 100-1000 km/s); rather, space itself is expanding, causing distances between the galaxies to increase.

We see "super-luminal" galaxies

The recession velocity $v_\mathrm{rec}$ of a galaxy is given by Hubble's Law:
$$
v_\mathrm{rec} = H_0 \, d,
$$
where $H_0 \simeq 67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ is the Hubble constant (Planck Collaboration et al. 2016). This law implies that galaxies farther away than
$$
r_\mathrm{HS} \equiv \frac{c}{H_0} \simeq 4400\,\mathrm{Mpc} \simeq 14.4 \, \mathrm{Gly}\,\,\mathrm{("\!\!Giga\mbox{-}lightyears\!\!")}
$$
recede faster than $c$. Here, the subscript "HS" is chosen because the ragion within which galaxies recede slower than $c$ is called the "Hubble sphere". Objects at a distance of $r_\mathrm{HS}$ have a redshift of $z\simeq1.6$.

Consider a photon emitted from a distant galaxy (say, GN-z11 at redshift $z=11.1$) in the past, in the direction of the Milky Way (MW). What special relativity tells us is that locally, the photon always travels through space at $v=c$. Initially, the photon thus increases it distance from GN-z11 at velocity $c$. However, even though the photon travels toward us, its distance to MW increases, due to the expansion of the Universe. As the photon increases its distance to GN-z11, the same expansion causes it to recede from GN-z11 at an ever-increasing velocity. Moreover, as it travels toward MW, it will slowly "overcome" the expansion until it reaches the point where $v_\mathrm{rec} = c$. For an infinitesimally small period, it will stand will wrt. MW, after which it will begin to travel faster and faster as measured from MW. Eventually, its velocity — still in MW's reference frame — will reach $c$, at which point it will have reached MW.

Thus, even though GN-z11 and MW recede from each other at $v_\mathrm{rec} = 2.2c$, we are still able to see it. What is perhaps even more counterintuitive is that when GN-z11 emitted the light we see today, it receded even faster, at $v_\mathrm{rec} \sim 4c$.

We see more and more distant galaxies

There is, however, a limit to how fast a galaxy visible to us can recede, given by the distance $r_\mathrm{PH}$ that light has had the time to travel since the Universe was created. Light comes to us from all directions, so we're situated in the center of a sphere of radius $r_\mathrm{PH}$. This sphere is called "the observable Universe", and its surface (which is not a physical thing) is called the particle horizon (hence the subscript "PH"). Galaxies at the particle horizon are receding at $v_\mathrm{rec}\simeq3.3c$.

As time goes by, light from ever-more-distant galaxies$^\dagger$ will reach us; that is $r_\mathrm{PH}$ increases. In other words, the observable Universe always increases in size, and no galaxy visible today will ever leave the observable Universe, no matter its speed.

However, since future observable galaxies will be more and more redshifted, their light will eventually shift out of the visible range and into longer and longer radiowaves. Furthermore, the time between each detected photon will increase, so they will be dimmer and dimmer, and thus in practice, they will disappear.

Intuitive explanation

A good analogy for better understanding why light can reach us from a galaxy that recedes faster than light, is the "worm on a rubber band": Attach an (infinite stretchable) rubber band (of length, say, 10 cm) to a wall and walk away at any constant speed you choose, e.g. 1 m/s. Before you start, put your pet worm at the end near the wall. It wants to get back to you, and starts crawling at 1 cm/s, i.e. 100× slower than you. Will it ever reach you? If you look at it from the perspective of the wall, both you and the worm move away, but whereas you recede at a constant speed, the worm, although slower in the beginning, accelerates because it moves on the rubber band, but the part of the rubber band between the worm and the wall increases in size. The rest of the rubber band of course also increases in size, but that doesn't matter — as long as you have a constant speed, and the worm accelerates, it will reach you (although in this example, it will take the worm $10^{26}$ billion years, at which point it may have lost its patience. But if you walk at only 10 cm/s, it will take just 6 hours).

In this analogy, you're the MW, the wall is GN-z11, and the worm is a photon. Now if you don't walk at a constant speed, but also accelerate (this is an analogy of the effect of dark energy), the worm may or may not reach you, depending on your speeds. Just like there is a limit to how distant galaxies we will ever be able to see.

$^\dagger$Note that since large distances also means looking back in time (since the light has spent a long time traveling), we actually don't see galaxies this far away, as they hadn't formed this early in history. We do however see the gas from which the galaxies were born, as far back as 380,000 years after Big Bang.


When can I see the Milky Way from the Earth? Light pollution is restricting most people from ever seeing the Milky Way from the Earth.

Is there any place in India where I can see the Milky Way?

Is there a database or site with the list of all places where I can see it?

Also, how can we calculate that?

Wherever you have a dark place, you can see the Milky Way. Anywhere in the countryside, far from large cities might work. And it doesn't need to be a very far away - a few km may do the work, specially if there are some hills in between.


What spectral feature was used to confirm that S0-2 is "single and available"? What is Brγ? The Phys.org article Astronomers discover S0-2 star is single and ready for big Einstein test describes the results of a careful analysis of radial velocity measurements of the star S0-2, which orbits the supermassive black hole at the center of our (Milky Way) galaxy with a period of only 16 years. The paper itself (Devin S. Chu et al, 2018) is available for open access at ArXiv, and also ApJ.

The only mention of the word wavelength is in Section 2.2:

  The radial velocity of S0-2 is measured from both data sets by fitting a Gaussian to the Brγ absorption line.

What exactly is a Brγ ("Br" + gamma) absorption line?

below: "The orbit of S0-2 (light blue) located near the Milky Way's supermassive black hole will be used to test Einstein's Theory of General Relativity and generate potentially new gravitational models. Credit: W. M. Keck Observatory (Click for full size)

Brγ is a line in the emission spectrum of Hydrogen, or single-electron (and therefore) hydrogen-like Rydberg atoms in general. In particular, it is the emission due to an electron moving from the 7th to the 4th energy level.

Br is short for Brackett, after Frederik Brackett, and gamma because

$\alpha: \ \ 5 \rightarrow 4,$
$\beta:  \ \ 6 \rightarrow 4,$
$\gamma: \ \ 7 \rightarrow 4,$

...and so on.

Brγ has a wavelength of 2166nm, as you would get from the Rydberg formula, which is well in the infra-red part of the spectrum. It is probably chosen as it has low absorption by the dust towards the centre of the galaxy.

Screenshot from here.


What are the possible star fuels? I always thought the only fuel for a star was hydrogen, which is fused into helium.

But while reading some questions and answers here in ASO, I saw phrases like "This balance stays relatively stable until the star runs out of whatever its current fuel is".

Besides hydrogen, what are the possible fuels in a star’s core?

As a star begins to age and exhaust its hydrogen supply, pressure will build in its interior until temperatures rise enough to begin fusing helium into carbon. If the star is massive enough, this process will continue, using heavier elements each time.

Referring to this link on stellar fusion, depending on their mass, stars use a variety of elements from hydrogen to silicon as 'fuel'. Stellar cores of stars towards the end of their giant stages will have cores of iron, but iron doesn't fuse into anything heavier in a stellar core.


Why is neutron star a star? I am wondering since neutron star is smaller than the average main-sequence stars and it doesn't produce energy via fusion so why do we still call it a star?

It seems Walter Baade and Fritz Zwicky may have coined the term in 1934 (see here) and the name stuck around.

I've known some astronomers to label them 'stellar remnants' for the reason you gave; they are not stars in the sense they do not produce energy via nuclear fusion, and that they are not made entirely of neutrons as the name would suggest.

They seem to be called stars for lack of another layman's term (from our point of view on Earth, not all neutron stars are pulsars), and that they still share some characteristics of stars: they emit their own light, they have masses greater than the Sun and can even have planets orbiting them.


Would bagging an asteroid destroy valuable science about it? NASA has a plan called "Asteroid Redirect Mission option A". It would robotically put an asteroid of a few meters diameter in a bag and push it into reach of astronauts for thorough examination. 44 tonnes of meteorites falls on Earth every day and almost 50,000 meteorites have been found on Earth. So the unique advantage of capturing a pristine asteroid, however tiny, is that it will not have been deformed by entering Earth's atmosphere. It might be a rubble pile and/or its surface might be covered by regolith or maybe even volatiles since it might have formed and always existed in microgravity. Even the Moon is covered by hoovering electrically charged dust near above its surface, ionized by cosmic or Solar radiation. And surface properties are important because it is only the surface which can be observed by telescopes and Earth based radar.

But wouldn't bagging and pushing an asteroid destroy such delicate surface and rubble pile properties?

The simple answer, unfortunately, is yes.

In fact, it is a resounding "yes". Option "A" uses a large "bag" to enclose the asteroid, and then to tow it to lunar orbit (or another feasible location). There's a problem, though, which is that simply surrounding an asteroid with a cylinder won't capture it. You have to "tighten" the "bag". This is implied in a graphic here and is explained in an article in Popular Mechanics ("How to Mine an Asteroid", Aug. 2012), and shown in a video on NASA's website. Basically, a series of "arms" constrict the "bag" to form-fit the bag to the asteroid's surface. The procedure could create many indentations on the asteroid's surface, and, as you said, destroy some of the most valuable information. The footprints of the astronauts who walked on the moon will stay there for many years; any indentations on an asteroid will be impossible to fix.

However, this entire idea is far from finished. NASA will not even decide between options "A" and "B" until "late 2014". So we won't know the precise details for a while.

So why go ahead with the whole thing if we're going to lose some of the best information we could get? Well, the Initiative isn't motivated purely by surface features. Part of it is spurred on by plans to pioneer new technologies needed to get humans to Mars. Other tangential benefits include diverting an asteroid that could be headed for Earth (although the asteroid candidates for this mission are much smaller than any of their more dangerous cousins; also, the Asteroid Initiative is under "Asteroid and Comet Watch" on NASA's website - right near plans to blow up a dangerous asteroid!), or even mining an asteroid for metals or resources for astronauts (such as water).

Also, this mission isn't the only way to study an asteroid (although it might be the best). We could always send a lander to an asteroid - after all, we can land on a comet (erm, well, we're trying - I'm rooting for Rosetta and Philae!), which would make it simple to collect samples before heading home . . . although the distance to travel would be a lot longer than just to lunar orbit.

All of this comes together to mean that NASA won't worry itself over losing data on the surface of the asteroid. Would that information be nice? Oh, yes. But given the other good things coming out of the mission, the pros here outweigh the cons. It's simply one of the trade-offs that might have to be made for this mission.

Sources:

Planetary.org

Space.com

NASA

"How to Mine an Asteroid". Popular Mechanics. August, 2012.


Mars night sky -- Phobos and Deimos? I'm having a debate about how bright Martian nights would be, in terms of both the sky itself and ground illumination. This answer seems to indicate that the stars would be only dimly visible, and wouldn't provide enough light to see by. However, the question remains of how much illumination the moons would provide. I've seen it said that Deimos would appear about as bright as Venus does from Earth, and that Phobos would be about twenty times brighter, but it's not clear to me how that compares to e.g. our own moon's illumination. So how bright would it look? Would it be enough to see by?

Taking Mars' average distance from the sun of 1.52 AU, the sun would be 43% as bright.

Phobos has an albedo of about 0.071 which is pretty dark.  Darker than the Moon with an average albedo of about 0.12, so it reflects about 59% as well as the Moon does.   That gives it a brightness to area of just over 25% the brightness of the moon.

Phobos is an odd shape, about 27x22x18 km.   Averaging that out to 22, then Phobos is 1/158 the diameter of the Moon, but it's closer to Mars' surface.  about 5,986 km, that's about 63.5 times closer than the Moon is to the surface of the Earth.

Working out the numbers, the size of phobos in the Martian sky would be on average 63.5/158 or 40% the diameter, or 16% the size of Earth's Moon from Earth.    Multiply that by 25% the brightness to area and that's just 4% the brightness of a full moon.  That would be peak brightness  (and if you want to fuss about Mars' Perihelion and say 5% the brightness of a full moon, fair enough, but lets not get ahead of ourselves).

Brightness of our Moon varies depending how good a reflector it's surface is, and how much of the surface is in shadow.  The Moon is particularly cratered so a half moon is significantly less than half as bright as a full moon.  Phobos is so close to Mars that from the surface of Phobos, Mars takes up about 40 degrees of arc in the sky.  That casts a huge shadow, so there's no such thing, from Mars, as a full Phobos.  As Phobos approaches full, it enters Mars' shadow and it's brightness drops significantly  (whether it turns red like a blood moon or just vanishes - I'm not sure, perhaps that's best for another question).   But Phobos never becomes full, it gets a bit over 3/4s, then it enters Mars' shadow.

"Enough to see" is a tough question, because sight depends on not just the light source but the reflectivity of your surroundings and there's some variation in how well individuals can see in the dark and, obviously, on how clear the night sky is on Mars on that particular night.  I think 3%-4% of the brightness of a 3/4 moon would be pushing the limit though and probably not bright enough to see for the most people.    BUT reading the comments, and reading the very cool Venus shadow article in the questions comments, my assumption might be wrong.   Limited sight might/(should?) be possible with some recognition of shadows and shapes.


Angles of sunrise and sunset viewed from a surface anyway oriented I need to understand how to calculate the angles of the sunrise and sunset viewed from a surface oriented against the azimuth and  the horizon plane, and for any day of the year.

For example, I have a surface at this position 40° 43' 30.2"N 73° 59' 37.5"W, oriented towards north with an azimuth angle of 174°(O° is exactly the geometric south), and with a tilt angle of 90°,so the surface is vertical, or perpendicular to the horizon.

How could I calculate when it get the sun and when it doesn't during the day?
What are these angles against the azimuth axis?

In my example, sun should hit the surface when it rises, then it disappears, it reappears at the end of the day before of its set, mainly in the summer season. So in this case, I need to know four angles: -ws,w1,w2,ws, where -ws, and ws are the regular angle of rise and set, but how to calculate when sun disappears (w1), and appears again to the surface (w2)?

Do you know the complete formulas to these calculations?

Do you know any websites, or any web app to do exactly these type of calculations?

What about the regular case when the same surface in the same place and position but with its normal towards south?

I'm not very interested in the theory but in the practical calculation. I tried the formulas on this website (http://www.itacanet.org/the-sun-as-a-source-of-energy/part-3-calculating-solar-angles/) but it is not clear what happens if Azs=0°, that is when the surface is towards north. In the calculation of a and b for the formula 3.7 I get at the denominator sin(0)=0, times tan(beta=90°)=inf, if the surface is vertical!! How could I solve this problem?

In other words, at what hour angle $h$ is the Sun's azimuth $\mathtt{Az}_s$ at a right angle to the wall's normal vector (84$^\circ$ or 264$^\circ$)?

Using these formulas, which I've checked, from
Wikipedia: Solar azimuth angle,
$$\sin \phi_s = \frac{-\sin h \cos \delta}{\sin \theta_s} \\
\cos \phi_s = \frac{\sin \delta \cos \Phi - \cos h \cos \delta \sin \Phi}{\sin \theta_s}$$
where $\phi_s$ is the Sun's north-clockwise azimuth,
$\theta_s$ is the Sun's zenith angle (90$^\circ$ - $\mathtt{Alt}_s$),
$\delta$ is the Sun's declination,
and $\Phi$ is your geographic latitude.
Then $\tan \phi_s = \sin \phi_s / \cos \phi_s$, and
$$\phi_s = \mathtt{atan2}(-\sin h \cos \delta, \ \sin \delta \cos \Phi - \cos h \cos \delta \sin \Phi)$$
This formula yields results consistent with the
NOAA Solar Calculator in my tests.
We need to solve for $h$, but how isn't yet obvious to me; maybe someone else can help?

Notes:

The relation between $\phi_s$ and $\mathtt{Az}_s$ depends on whether $\mathtt{Az}_s$ is south-clockwise, south-counterclockwise, or the more conventional north-clockwise.
$h$ here is the same as $\omega$ in the ITACA article.
As you observed, formula 3.7 is undefined for a wall facing due north or south.
I find their azimuth notation confusing (is AAZ an alias for AZS or something else?) and the $\alpha$ on the left side of equation 3.9 is a mistake.
I believe ITACA formula 3.11 is correct. If we let $\cos \theta_i$ = 0, it reduces to something like the above where the surface is vertical, but it has the same difficulty of solving for $\omega$.


What are the last images from the Galileo orbiter before impacting Jupiter? NASA's Galileo orbiter at Jupiter ended its mission in 2003 by intentionally dive into the gas planet. Did it take and transmit any close up images before it ceased to function? If not, why? Are there any images from Galileo, or its separate impactor probe, which show the Jovian clouds from close enough range that their topology is discernible?

On all images I've seen, Jupiter looks like a perfect sphere because of its huge size and far distances from which it has been imaged. But if two spacecrafts have dived into it, there should be close up images of its clouds, right?

The wonderful article linked by Wayfaring Stranger in the comments alludes to the reason for not sending photographs, but does not answer the question directly. As with most spacecraft outside Earth or Lunar orbit, the data rate is extraordinarily low. This page gives data rates for the current Rosetta mission, which are on the order of a dozen or two KiB/s. So a 10 MiB uncompressed image file would take 15 minutes to send. Though it would be an undoubtedly interesting photograph, it would come at the opportunity cost of magnetic readings, temperature readings, radiation readings, and other important scientific data which would help plan our next probe to survive the environment.


Canvas and bowling ball I'm having trouble understanding the basic model for how matter bends space. The classic illustration is the canvas and the ball but that model needs a reference frame where one can define a downward direction to work. For the life of me I can't figure out why matter just doesn't "sit" in space.
I'm sorry this question isn't better formulated, maybe someone can point me towards a video or something?

I'm having trouble understanding the basic model for how matter bends space.

No wonder, because it doesn't. It curves spacetime. But spacetime isn't the same thing as space. See this Baez article: "Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial". Space isn't curved where a gravitational field is. Instead spacetime is.

  The classic illustration is the canvas and the ball but that model needs a reference frame where one can define a downward direction to work.

That's right, it uses gravity to "explain" gravity, so it isn't ideal. However there is a way to understand it, and IMHO it's simpler than you think.

  For the life of me I can't figure out why matter just doesn't "sit" in space.

It does just sit in space. And it causes gravity because it's a concentration of energy that "conditions" the surrounding space. It alters it, and this alteration diminishes with distance. Remember the stress-energy tensor, remember that stress is directional pressure, and think of matter as exerting an outward pressure in all directions in the surrounding space. Something like this:

Earth image courtesy of NASA, lines drawn by me

Then see the Einstein digital papers, where Einstein said a gravitational field is a place where "space is neither homogeneous nor isotropic". This inhomogeneity isn't linear. You can plot it, using say optical clocks dotted throughout an equatorial slice through the Earth and the surrounding space. You plot lower slower clocks as lower down in a 3D image, and higher faster clock rates higher up. When you've finished your plot looks like this:

CCASA image by Johnstone, see Wikipedia

The upper image is a bit like looking at the lower image from underneath. But anyway, your plot is curved, and you made it using clocks in space, so we can say spacetime is curved. Note that whilst we started with a two-dimensional equatorial slice and ended up with a three-dimensional curved plot, there is no extra dimension in any real sense. All we're depicting is the "coordinate" speed of light. This might sound unfamiliar but again see the Einstein digital papers: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". Light doesn't curve because your plot of optical clock rates is curved. It doesn't curve because spacetime is curved. Einstein never actually said that. He said it curves because the speed of light varies. Hence the tilted light-cones on John Norton's course. Light curves wherever your plot isn't flat. Where the speed of light varies. It's actually rather like sonar:

Image courtesy of FAS and the US Navy, see course ES310 chapter 20


Is it possible to achieve a stable "lunarstationary" orbit around the moon? Is there a stable geostationary orbit around the moon?

My feeling is, that the orbit would collide with earth, because of the moon's slow rotation.

First off, such an orbit wouldn't be a geostationary orbit since geo- refers to the Earth. A more appropriate name would be lunarstationary or selenostationary. I'm not sure if there is an officially accepted term since you rarely hear people talk about such an orbit.

You can calculate the orbital distance of a selenostationary orbit using Kepler's law:

$$a = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}$$

In this case, $a$ is your orbital distance of interest, $P$ is the orbital period (which we know to be 27.321 days or 2360534 seconds), $G$ is just the gravitational constant, and hopefully it is obvious that $M_{\text{Moon}}$ is the mass of the Moon. All we have to do is plug in numbers. I find that

$$a = 88,417\:\mathrm{km}=0.23\:\mathrm{Earth\mathit{-}Moon\:Distance}$$

So I at least match your calculation pretty well. I think you were just relying on Wolfram Alpha a bit too much to get the units right. The units do work out right though.

If you want to determine if this orbit can exist however, you need to do a bit more work. As a first step, calculate the Moon's Hill Sphere. This is the radius at which the Moon still maintains control over it's satellite, without the Earth causing problems. The equation for this radius is given by

$$r \approx a_{\text{Moon}}(1-e_{\text{Moon}})\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Earth}}}}$$

In this equation, $a_{\text{Moon}} = 348,399\:\mathrm{km}$ is the Moon's semi-major axis around the Earth and $e_{\text{Moon}} = 0.0549$ is the Moon's orbital eccentricity. I'm sure you can figure out that the $M$'s are the masses of the respective bodies. Just plug and chug and you get

$$r \approx 52,700\:\mathrm{km}$$

A more careful calculation, including the effects of the Sun is slightly more optimistic and provides a Hill radius of $r = 58,050\:\mathrm{km}$. In either case though, hopefully you can see that the radius for a selenostationary orbit is much farther than the Hill radius, meaning that no stable orbit can be achieved as it would be too much perturbed by the Earth and/or the Sun.

One final, semi-related point. It turns out almost no orbits around the Moon are stable, even if they're inside the Hill radius. This is primarily to do with mass concentrations (or mascons) in the Moon's crust and mantle which make the gravitational field non-uniform and act to degrade orbits. There are only a handful of "stable" orbits and these are only achieved by orbiting in such a way as to miss passing over these mascons.


How could we observe a wormhole? A wormhole is a sort of tunnel that could connect two points in spacetime. These are theoretical objects that have not been observed. What observational evidence could we look for to confirm (or disprove) their existence? If they do exist, what should we be looking for to find one?

This answers how wormholes could be created. This is not what I'm asking about. I want to know what observations could be done (potentially with telescopes we don't have yet) to confirm the existence of wormholes.

One plausible way of detecting wormholes (paper) is gravitational microlensing. Light paths are bent by the curved spacetime around the wormhole, similar to what happens near black holes. As a wormhole moves in front of background stars it can cause the brightness to vary in a characteristic way, making it possible to distinguish it from a black hole.

One of the problems is that there are many possible wormhole models, so the light curves can look different. Some can also produce multiple Einstein rings, others none. However, this may in theory be good news for learning what model is right if one sees a really weird light curve. In practice, given noise and lack of observations, multiple explanations will likely be possible for a long time if one sees just one example.


How to see other celestial things using stars for direction? I have a simple question. I can see orion constellation clearly in the night sky. I read it somewhere on this website that gemini is on the left-hand side of orion constellation. I was able to spot gemini later that night. I was wondering on how to spot other constellations using orion and gemini constellation.

Use a constellation map! That will provide much of the information you need.

Here are however some things you can try to spot:

On the lower left hand side of Orion, you can spot a very bright star, Sirius, in fact the brightest one in the sky. It is part of the constellation Canis Major (great dog).
A bit further left and up you probably see a star nearly as bright as Sirius. That is Procyon in Canis minor (lesser dog).
Up, and a little bit to the right from Orion you can see a very red star. That is Aldebaran in the constellation of Taurus (the bull). In the same constellation you can see the Pleiades, a bunch of stars in a cluster.

Taurus with Aldebaran and the Pleiades:


Is axial tilt direction correlated with semi-major axis direction? Earth comes closest to the Sun every year around January 3. Winter solstice is at 21-22 December, merely some 12-13 days away. That means projection of Earth rotation axis onto the plane of its orbital movement almost overlaps with its orbit's semi-major axis; the axial tilt direction coincides with the direction Earth orbit is elongated.

I couldn't find corresponding data regarding other bodies in the Solar System.

Is this merely a coincidence, or is there a regularity, with direction of the axial tilt coinciding with eccentricity of the orbit? How does that look like in case of other bodies than Earth?

It's mere coincidence. In about 10,000 years, perihelion will coincide with the northern hemisphere summer solstice.

There are two precessions of interest here. One is the 26,000 year period axial precession of the Earth's rotation axis, caused mostly by the Moon and the Sun; the other is the 112,000 year period apsidal precession of the Earth's orbit, caused mostly by Jupiter and Venus (but also with a pinch of help from general relativity). This result in three different concepts of a year:

The sidereal year is the mean time it takes the Earth to make a complete orbit with respect to inertial space ("the fixed stars").
The tropical year (aka solar year) is the mean time it takes for the Sun's ecliptic longitude to increase by 360 degrees, and also the time between one vernal equinox to the next. This is the year used as the basis of our calendar. This year is shorter than the sidereal year by about 20 minutes and 25 seconds.
The anomalistic year is the mean time between one perihelion passage of the Earth to the next. This year is longer than the sidereal year by about 4 minutes and 43 seconds, and hence longer than the tropical year by about 25 minutes and 8 seconds.

That 25+ minute difference between the anomalistic and tropical years means that time of perihelion passage advances by about a day every 57 years.


Successor to the Hubble Telescope Are there any plans for the 'Hubble2'? Maybe with an even bigger mirror capable of even finer resolution, to look even deeper into the origins of the universe?

If so, when can we expect to see it in action?

The James Webb Space Telescope (JWST) will be Hubble 2.0, and, my personal favorite, it could get a better look at planets in other solar-systems than we've ever had.   Scheduled for launch in 2018.  (Footnote, as Rob Jeffries points out, it's infra-red through long-wave visible light only).   Hence, no Blue/Indigo/Violet or Green, but some yellow.   So, it's not quite like Hubble, but it's still often called Hubble's replacement.

Chandra observes X-rays.

Spitzer observes infra-red.    They both followed Hubble, but they're not the same type of telescope.


How to calculate how much of the Earth's (or another planet's) surface is visible? My latter three questions remain unanswered but I won't wait with this one, I'd like to learn the formula and hope this question gets an answer.
As you're going higher respectively farther away from the Earth, more and more of the Earth's surface is visible. At infinite distance the percentage of the surface that is visible would be 50%. Now I'd like to know how you calculate the percentage that is visible from altitude/distance, and how you calculate the altitude/distance from how much of the surface is visible.
For example, how much of the Earth's surface is visible at 3,000 mi (4,800 km) altitude or at what altitude do you see 38% of the Earth's surface?

The Earth's surface which is visible when you look at the planet from a certain distance is a spherical cap in terms of geometry. Here it is, in blue:

$A$ is the position of the observer, $H$ is the distance from the observer to the surface of the sphere, $O$ is the center of the sphere,$r$ is the radius of the sphere, $AB$ is the distance to the true horizon, $\angle ABO = 90°$,  $\angle \theta$ is the angle between the rays from the center of the sphere to the apex of the cap (the pole) and the edge of the disk forming the base of the cap.
The area of the cap $A_c$ can be found according to this formula:
$$A_c=2 \pi r^2(1-\cos \theta)$$
$\cos \theta$ in the right triangle $ABO$ is the ratio of the adjacent cathetus $OB = r$ to the hypotenuse $OA = r+H$, that is, $\cos \theta = \frac{r}{r+H}$, so
$$A_c=2 \pi r^2(1-\frac{r}{r+H})$$
The percentage ratio $R_\%$ of the visible area $A_c$ to the full area of the sphere $A_s$ is $$R_\%=\frac{A_c}{A_s}\times 100\%$$
Since the full area of the sphere is $A_s = 4\pi r^2$, we have:
$$R_\%=\frac{2 \pi r^2(1-\frac{r}{r+H})}{4\pi r^2}\times 100\%$$
$2 \pi r^2$ above and below the line are canceled, so the final formula is like this:
$$\bbox[7px,border:2px solid red]{R_\%=\frac{1}{2}  \times (1-\frac{r}{r+H}) \times 100\%}$$

Answering your question, from the altitude of 3,000 mi (4,800 km) one can see $21.4842\% \approx 21.5\%$ of the Earth's surface.
I have typed it in the Google calculator here, so you can use it, just substitute the Earth radius of 6371 km and the altitude 4800 km for the numbers you would like.
Also, here is a graph of how the visible area of the Earth ($r=6371 km$) changes with the altitude, scale and drag the graph with the mouse.
As for finding the altitude from which the given percentage of the Earth's surface is visible, here it is, it is just a transformation my final formula so that $H$ is on one side of the $=$ and all the rest is on the other, only $R$ here is just a ratio, no percent, so you will put 0.38, not 38% into the formula:
$$H=2r/(\frac{1}{R} - 2)$$
The calculator says that to see 38% of the Earth's surface you need to be 20,174.83 km (12,536 mi) above the surface.


Can any satellite train still be seen? I "missed the train" on the Starlink satellites last year, and now it looks like they've separated, and the conservative voices convinced Space X to darken new ones with visors or paint.
Is it still possible to see 2 or more satellites moving in the same direction with the naked eye? (Actually, any celestial bodies visibly moving in the same direction will do.)
I'm located in California and I tried Find Starlink and the Heavens Above app but wasn't able to see anything.

I can personally attest that the Starlink satellites are visible with the naked eye even after they have spread out, however they are quite dim - with apparent magnitudes near 5.5. Here is a brightness comparison drawn from a widefield capture where two Starlink satellites are can be seen (albeit barely) juxtaposed with another satellite with apparent magnitude near 3.

The region in which this was taken is Class 1 on the Bortle Scale and the entire 60-satellite train was visible, though only 4-5 were visible at any given time. In brighter regions the satellites would not be as readily visible. Furthermore, since they have such low orbits, they were only visible for ~105 degrees in the sky before they passed into Earth's shadow.


Strange speck of light in the picture sequence of the approach to 2014 Mu69 (Ultima Thule) In the picture sequence which was taken by New Horizons during the approach to Ultima Thule on the 3rd of Dec there can be seen a speck of light in the lower left corner, not as bright as 2014 MU69 but clearly visible.

Source: Here (German) or here (English), NASA / Johns Hopkins Applied Physics Laboratory / Southwest Research Laboratory / Henry Throop

Is there any theory what may have caused this signal? A particle from the probe itself, coincidentally passing in front of the lens? A variable star? A small object in the right distance and angle to be glimpsed upon by New Horizons? I think the latter would be an astronomically improbable event...

Here are some single frames of that image sequence, the speck is clearly visible on the 3rd of December.

It's probably a cosmic ray strike, a common artifact on New Horizons images.  The image, however, is of very poor quality which makes it difficult to analyse the exact nature of the spot.  I'll go into this more  at the end of the answer.
Possibilities:

Cosmic ray hit
A common occurance (see below); the most likely cause without evidence to the contrary.

Background star (e.g. a nova)
Novæ (and variable stars) brighten and dim over several days.  Were this a nova it would have shown in images from previous and later days.

Kuiper belt object
To not show in images from adjacent days it would need large relative motion and thus a sizable trail would be expected (none is evident).

Aliens
Possible?  Yes.  Likely?  Far, far, from it.  Not worth serious consideration.

Here's the New Horizons team's standard disclaimer:

This image contains one or more objects whose brightness exceeds the detector's saturation level. This sometimes produces a "tail" of bright and/or dark pixels to the right of the object. You may also notice a faint vertical white stripe passing through the saturated object; this is an artifact called "frame transfer smear" and is associated with the incomplete removal of signal produced when the image is transferred from the optically active region of the detector to the storage region of the detector. If the target is badly saturated, you may also notice a faint, X-shaped feature nearly centered on the object; these are optical diffraction spikes.
This image contains one or more streaks associated with cosmic rays passing through the detector. Nearly every LORRI image has at least one cosmic ray strike, but most are "single pixel" events (i.e., they only appear to be in single pixel and can easily be mistaken for stars). But sometimes a cosmic ray is energetic enough that it leaves a "trail" as it passes through the LORRI detector.

Example of cosmic ray hit and saturation trails:

Source: here (via "Learn more..." link), NASA/JHUAPL/SwRI
Note that despite being a PNG, JPEG compression artifacts can be seen in this image
The frame in question:

Source: Frame K-29 of this image taken from here, NASA/JHUAPL/SwRI/Henry Throop
Enlargement below is from frame K-29 of this image
The image in question is almost certainly a stacked composite of ~10 source images and appears to have been 2×2 binned (compare here).  To make faint objects more visible, the image was also likely contrast-enhanced.  The image also shows clear signs of lossy compression and quantization (from 12 bits per pixel to ~5), see enlargement below.  While these are (mostly) acceptable choices for the image's intended use it tends to obscure detail from the original images.
Crop showing artifacts:

Looking at the spot in question (enlargement below) a couple of things are noticable:
It is compact and relatively dim.  Its overall size in pixels is about half that of the stars and of MU69.  In addition, the drop-off in brightness moving away from its center is much more rapid than for the other objects.  This is consistent with a bright but compact object appearing in only one image in the K-29 stack.
This object also exhibits a less-blurry structure with two distinctly brighter spots or lines.  This is consistent with spray from a cosmic ray hit.
Crop showing mystery spot:

All in all, I think a cosmic ray hit is the most likely explanation.  Given the very poor quality of the image in question though it's difficult to give a definitive answer.  To make a proper determination one would need access to the yet-to-be-released raw imagery.

I've seen a paper with a more thorough discussion of New Horizons image artifiacts but don't seem to have saved a copy.  I'll add references from it when I track it down.


Solar and lunar eclipses -- how rare I was fortunate enough to see the solar eclipse in southern Illinois last year. The clouds were not friendly, but we did have one spectacular moment of perfect eclipse. It is an eerie sensation, and I cannot imagine how utterly terrifying it must have been for our ancient ancestors to see the moon eating the Sun.

However, it got me wondering. A solar eclipse can occur in the spectacular fashion it does because the angular dimension on the moon and the Sun are almost exactly the same -- this despite the fact that they are dramatically different in actual dimension. This seems a remarkable coincidence to me -- the ratio of the distances being almost exactly the same as the ratio of their diameters.

I was wondering if there are any other known examples of such a phenomenon occurring anywhere else in the universe? Can we make any estimate of how often something like that might occur in a galaxy?

TOTAL solar eclipses and TOTAL lunar eclipses are actually not that common in places outside of the solar system. For these TOTAL eclipses to occur, and I'm emphasizing total bc partial eclipses happen all the time, there needs to be the perfect distance so that an object perfectly covers the Sun.

In other words, you are correct - it is such a remarkable coincidence that this occurs in our Solar System. I'm not entirely sure about how many exoplanetary systems also experience this phenomenon, but I assume that it won't be a lot. I do know that it DOES HAPPEN in other places, but it just isn't common.

Just for a fun fact - the Moon is actually moving away from us, meaning that in a few million years, total solar eclipses won't be possible anymore - and we'll only see partial eclipses.

Hope that helps! :)


At what distance wouldn't the Sun cast a visible shadow anymore? How far from the Sun would a (dwarf) planet have to be so that the Sun is dim enough that it doesn't cast a shadow when you're standing on the planet? What magnitude is the border in brightness beyond which a shadow isn't visible anymore?

This depends on eye biology, so there's no purely astronomical answer.  I'll note that Venus is just about capable of casting a shadow (at greatest elongation, with ideal conditions) so let's say that a magnitude -5 object is at about the biological limit of casting a shadow.
The sun has a magnitude -27, so that is 22 magnitudes brighter than Venus.  5 magnitudes is a factor of 100 in brightness, so 20 magnitudes is a factor of 100000000 (100 million)  and 22 magnitudes is about 600 million.
But light follows an inverse square law, The square root of 600 million is about 25000.  So at about 25000 times the distance of Earth, or 0.4 light-years, the sun would be as dim as Venus at its brightest and perhaps marginally able to cast a shadow for someone with acute vision, on a good matt white surface. This puts our astronaut in the Oort cloud.
There's a good deal of uncertainty in this estimate, based on eye acuity. So let's say "something between 0.1 and 1 light-years"


Very Old Stars in the Universe How could a Star possibly be older than the Universe ? Is this possible or are the calculations off when determining the age of such Stars ? Can a Star exist that formed on its own independent of a Universe ?

It can't. If we find a star that's "older" than the universe, then either the measurements of the star's age are incorrect, or the measurements of the universe's age are incorrect.


Metalicity and age of bulge stars vs halo From what I understand of current models, the bulge of the galaxy formed first, and thus, would contain older population II like stars. Currently, however, the halo has a higher population of stars with lower metalicity than the bulge. Is this because the density  of stars in the bulge is much higher? My thinking is that this is a densely populated region of massive hot stars that die quickly and create a newer region of next generation population I stars that are metal rich. The inverse would be true for the halo where stars are less densely packed. Is this correct?

Your intuition is largely correct: the key is that the proto-bulge region had a deep enough potential well so that the supernovas couldn't expel the remaining gas, and so new stars could form out of the gas (enriched by the supernova ejecta) in a continuing cycle. In the low-mass, isolated protogalactic clouds which probably contributed to the halo, the initial round of supernovas ejected most of the gas (including the original gas that hadn't yet formed stars) -- thus, little opportunity to form more stars (out of higher-metallicity gas) on a continuing basis.

It's really the total mass in a given region that matters. For example, the density of stars in the central regions of a globular cluster is pretty high, but the mass of the cluster as a whole isn't enough to keep most of the gas when its massive stars go supernova.

(Side note: the traditional terms "Population I" and "Population II" aren't used all that much any more, since age and metallicity can vary continuously and aren't always strongly associated.)


What is the ratio of cosmic microwave background radiation to normal radiation? I would like to know the ratio of cosmic microwave background radiation to normal radiation in the universe.  I am considering cosmic microwave background radiation to include the microwave, and any other radiation that is being emanated from near the "edge of the universe", while normal radiation is radiation emmitted by stars, nebulae, and other sources within the universe (excluding the cosmic microwave background).  Since we are in a galaxy, I know that the normal radiation here far exceeds the cosmic microwave background, but I am interested in the "average" value over the universe.  For example: total cosmic microwave background radation in the universe / total normal radiation in the universe.

Photons from the CMB outnumber all other photons by more than 200:1.  This is contained in Pela's excellent answer to: Does the number density of photons  nγ≈108m−3  refer to CMB photons only?.  Pela uses this graph from Hill et al. 2018 to derive this ratio.  Note that the CMB forms the largest contribution to the electromagnetic spectrum.


Which side of earth will face toward sun when earth and sun will be get closer and be bonded by Gravitational Force? We all know that one day the distance between the Earth and the sun will be zero. They will be bonded with gravity. My Question is which side of the earth will face toward the sun? I can't calculate that. Someone please calculate and don't say it's impossible.

We all know that one day the distance between the Earth and the sun will be zero.

This may be true, as the sun will get much bigger and engulf the Earth.  But this is irrelevant to the rest of the question, which seems to be about tidal locking.

  They will be bonded with gravity.

The Earth is already gravitationally bound to the sun. But it isn't tidally locked. The Earth+moon has too much angular momentum to ever become tidally locked to the sun. It will be engulfed by the sun before it becomes tidally locked.

  which side of the earth will face toward the sun

As noted above, the Earth will continue to rotate and will not become tidally locked.

Even if it would become tidally locked it would be impossible to calculate. It is like putting a battery into a clock and asking exactly what time the battery will run out.  You can't measure the present nor predict the future with enough accuracy to forecast such things

In other words. This won't happen, and if it does we can't calculate it.  It's impossible.


What is this thin, square-ish object on asteroid Bennu? How big is it? What could this square thing be? Roughly how big is it? From BGR's Our first up-close look at the Bennu asteroid is a messy one:

rotated, 2x zoomed, and sharpened. unscientifically:

In this composite of the rotating asteroid you can see that it is not thin, but actually protruding:

This series of images taken by the OSIRIS-REx spacecraft shows Bennu in one full rotation from
a distance of around 50 miles (80 km). The spacecraft’s PolyCam camera obtainedthe thirty-six 2.2-millisecond frames over a period of four hours and 18 minutes.
Credit: NASA/Goddard/University of Arizona

It looks like an 'ordinary' boulder.
(You can see several more in that sequence)

Simply comparing the 'height' from the second picture with Bennu's diameter of 492m shows this 'boulder' to be 28 meters high.


Are there any alternative explanations for the gravitational force we attribute to dark matter? It always seemed counter-intuitive to me that we observed an measured a gravitational force and since the universe has not enough mass to account for it, the conclusion was to say there's a different kind of matter wich doesn't interact with ordinary matter nor can be seen o measured.

What if the force was applied by other adjacent or overlapsed universes or by something completely different? Are there any other theories propossed? or would we just call whatever we find out to be the source of this force "Dark matter" even if it´s nothing alike matter?

There are certainly people who study alternative (non-General Relativistic) theories of gravity. The most popular theories have so far been:

Modified Newtonian Dynamics (MOND) - which essentially postulates that Newtonian Mechanics break down on some scale, leading to the rotation curves we see in galaxies.
Tensor–vector–scalar gravity (TeVeS) - this is a relativistic generalization of MOND.
Falling into the MOND category is f(R) gravity, which alters the general relativistic expression relating the curvature of space-time to the stress-energy tensor (the 'source' of gravity - like mass and pressure). These theories relate different functions of the Ricci scalar to this stress-energy tensor.

TeVes seems to be the most promising alternative to General Relativity (it may be able to explain things like rotation curves, predicts gravitational lensing, and some other things), but there are still many problems it doesn't do well with. So far it seems like modified theories of gravity are creating more problems than they are fixing - though I must admit, the existence of things like dark matter and dark energy are non-trivial issues cosmologists must come to terms with.

When it comes to things like alternative universes and the like as the cause of our current mysteries, I'm sure these things have been proposed. The problem here is coming up with an experiment to test them. Dark matter has been named as such because we know that it doesn't emit or absorb any radiation. As far as detecting such a particle, many real experiments have been conducted and will be conducted, and is an exciting field of research.

Below shows a few of these experiments and the range of cross-section/mass parameter space it has/will probe:


What does the active area on a typical CCD based detector look like? Pixel based detectors, particularly optical CCDs like what was used in the SDSS camera, are ubiquitous in astronomy. Is there any dead area on the detectors? Not the obvious gaps between each individual sensor, but on the sensor itself. That is, does a typical detector have any gaps between the pixels from microscopic wires laid on the front, alternating doping regions in the silicon, or just areas where a photon can hit and the photo-electron is unlikely to be collected? What fraction of the area is dead?

An astronomical CCD is in principle sensitive across its entire active surface; there are no physical gaps between pixels. The pixels themselves are controlled and maintained by the underlying electronic circuitry, which sets up potential barriers between individual pixels. In practice, there are variations in the actual sensitivity within individual pixels: sensitivity is highest in the center and falls off towards the edges, although it never goes to zero anywhere within the pixel.

This paper (PDF) is a recent analysis of a (frontside-illuminated) CCD; Figure 13 shows the measured pixel response functions (how sensitive an individual pixel is across its surface), and Figures 14 and 15 show the photometric sensitivity map (what the sensitivity is for uniform illumination). In the latter case, you can see that the sensitivity never varies more than about 20% or so, depending on the wavelength of the light. Some of the variation corresponds directly to wiring within the CCD absorbing some of the light, but this has a very minor effect on the sensitivity.

The comparison made in Florin Andrei's answer to a consumer-camera-style CMOS sensor, as shown in the image, is misleading, both because CMOS pixels are physically somewhat more discrete in ways that CCD pixels are not, and because astronomical CCDs do not have per-pixel filters glued on top of them.

(The argument that "there has to be a narrow slice of silicon in between pixels to prevent short-circuits" is not correct for CCDs; the electrical separation between pixels is maintained by the underlying electronics, not by any material barrier. In fact, you have to be able to create temporary "short-circuits" between adjacent pixels in a column in order to transfer the accumulated electrons during readout. That's what the "charger-transfer" in "charger-transfer device" means.)


What is that light of the middle of the milky way galaxy? I'm Being confused what is that light in the middle of milky way galaxy?

Image is from Charting the Milky Way From the Inside Out

The text with the image explains a lot about the spiral arms and the location of the Earth, but the central bright area isn't explained in great depth.

The galactic bulge is formed by tightly packed stars and interstellar dust
Also most stars are in the direction towards the galactic center
Thus it would appear brightest if it were viewed from outside the plane of the galaxy as this image is intended to show.
However, we don't see this from Earth, and the reason why is explained in  Phys.org's Why can't we see the center of the Milky Way? explains why we don't see this at night.
A brief outline of what is described there is as follows:

When it is dark enough, and conditions are clear, the dusty ring of the Milky Way can certainly be discerned in the night sky. However, we can still only see about 6,000 light years into the disk with the naked eye, and relying on the visible spectrum. Here's a rundown on why that is.

Size and Structure
Low Surface Brightness
Dust and Gas
Limited Instrumentation

However we could see a bright spot if we were outside the Earth's atmosphere and could see in certain wavelengths of infrared. The article shows the following image from COBE

Milky Way in infrared. Credit: COBE Source

False-color image of the near-infrared sky as seen by the DIRBE. Data at 1.25, 2.2, and 3.5 µm wavelengths are represented respectively as blue, green and red colors. The image is presented in Galactic coordinates, with the plane of the Milky Way Galaxy horizontal across the middle and the Galactic center at the center. The dominant sources of light at these wavelengths are stars within our Galaxy. The image shows both the thin disk and central bulge populations of stars in our spiral galaxy. Our Sun, much closer to us than any other star, lies in the disk (which is why the disk appears edge-on to us) at a distance of about 28,000 light years from the center. The image is redder in directions where there is more dust between the stars absorbing starlight from distant stars. This absorption is so strong at visible wavelengths that the central part of the Milky Way cannot be seen. DIRBE data will facilitate studies of the content, energetics and large scale structure of the Galaxy, as well as the nature and distribution of dust within the Solar System. The data also will be studied for evidence of a faint, uniform infrared background, the residual radiation from the first stars and galaxies formed following the Big Bang.


Speed comparison of both voyagers I came across this page which provides (mostly extrapolated) speeds of both voyagers. Assuming they are fairly accurate, any reason why the speed of Voyager 2 (5-6 km/sec) is less than that of Voyager 1 (13 km/sec).

If you notice carefully, you'll see that their speeds away from the sun are fairly similar (7-9 km/s), and the difference is only in their speeds away from the earth. This is because of a small component of earth's motion around the sun (which is at about 30 km/s) adding to Voyager 1's motion and subtracting from Voyager 2's motion.


Do gravitationally bound (e.g. orbiting) objects really resist the expansion of space? Thinking about the expansion of space, it is commonly explained that gravitationally bound objects (like galaxies and galaxy clusters) will not be affected by the expansion of space (dark energy).  I personally don't get why this would be, I would think galaxies, and things orbiting one another (e.g. solar systems) would be impacted.  If space doubled, the orbit distance should double.  I do get that things bound together gravitationally (perhaps a second definition / originally meant definition of "gravitationally bound") (e.g. planets, stars, asteroids, etc...) would not get ripped apart as space gradually doubles.  They would just pull themselves back together, and resist the very weak force of the expansion of space.  Is my understanding correct that things held together by gravity (like stars and planets) will hold together as space continues to expand, but things in orbit (e.g. stars orbiting within the galaxy, and planets orbiting stars) will be moved further out with the expansion of space?  It seems this could be fairly easily verified or refuted by seeing if ancient galaxies (those really far away) are denser than closer galaxies.  If they are denser, then this is strong support for the idea, and if not, then it largely refutes it.

I have a feeling this might be why quasars stopped happening at some point, and are now just observable in great distances (where we are seeing them in the past).  My understanding is that a quasar is the supermassive black hole at the center of a galaxy feeding on large amounts of material.  Back to my main question though, does the expansion of space thin out galaxies / solar systems?

Expansion of space does affect orbits, but the effect is really small.

Rate of space expansion is currently about 2cm/s per light year. Sun orbits Milky Way at about 230 km/s, and space expansion contributes about 0,5km/s. It is going to be difficult to measure other variables, such as stellar masses, with such detail.


Can you see the starting and the ending of a light beam passing in the distance? This question arose to me when I saw a SciFi movie where they shot with laser guns and you clearly could see the dashes of light beams travelling from the shooter to the target. Nonsense of course, reality is much more boring.

But I wondered if you would just need more distance:

Given a limited light beam traveling from A to B, where "limited" means that there is a start and / or end of the emission source causing that beam.
Given enough distance from us to A, B and the line between them
And add some kind of nebular smoke in between just enough to disperse as much light as needed so we can see it here but not as much to prevent it from reaching B.

Couldn't we observe the starting and / or the ending of that light beam, progressing on it's way from A to B, like a jet trail in the sky?

If it is possible, where can this be observed? Or do the laws of physics forbid such a thing?

Nothing forbids this, and it is actually observed astronomically. You need a very bright source of light: such as a supernova (which isn't a beam, but radiates in all directions) and very large distances. The flash of light can be seen spreading out from supernova in a circle, as it illuminates the dust and gas ejected by the supernova progenitor star in the years before

The effect is known as a light echo. Over time the "beams" of light move further out, and so the echo widens, as shown in this series of images (from an Australian telescope so you have to read from the bottom up, 507 days means 507 days after the supernova in 1987.) For scale, 5 arcmin is quite small: The moon is about 30 arcmin. source


Do all galaxies have a black hole in the center? Would I be correct if I stated that all galaxies have a black hole in the center? Since other galaxies orbit around a center point in a galaxy - the barycenter, right? - I would assume at this point there must be some insane gravitational pull. Which only black holes can create, right?

(If this is a bit vague, please tell me so I can clarify!)
I am a complete newbie to astronomy in general, so sorry if this question makes no sense whatsoever.

The answer here is certainly no. Many dwarf galaxies (e.g. the Magellanic Clouds) don't contain supermassive black holes, although some may contain less massive intermediate-mass black holes (Mezcua et al. 2018). As they are the most numerous type of galaxy in the universe, it's quite possible that most galaxies, in fact, do not contain supermassive black holes. This shouldn't be too surprising; if the bottom-up theories of supermassive  black hole formation are correct, many intermediate-mass black holes must coalesce to form a massive one. Dwarf galaxies simply wouldn't have (or be able to form) the requisite population, in most cases.

That said, a number of dwarf galaxies have been observed to host supermassive black holes (notably including M60-UCD1 and Hen 2-10, among others). These have provided, and continue to provide, insights into the formation and evolution of low-mass galaxies.

Some users here have also identified some massive galaxies without supermassive black holes, including M33 and A2261-BCG. The former has a bulge, but the central stellar velocity dispersion is low, implying there is no black hole at its core (Merritt et al. 2001), assuming the required parameters are known well enough.  In A2261-BCG, it's possible that one or more central supermassive black holes were ejected, leaving a "puffy" core without a black hole. Keep in mind, of course, that this is only one possible explanation for the observed distribution of stars in A2261-BCG.

In general, a central supermassive black hole does not influence the large-scale kinematics of stars and gas within a galaxy. Most supermassive black holes are much less massive than their host galaxy; I believe Sag A*, at the center of the Milky Way, is on the order of about 0.001% of the total mass of the galaxy. Even massive ellipticals like NGC 4889 have black holes about 0.1% of their total mass. As far as I know, ultra-compact dwarfs have the highest black-hole-to-galaxy mass ratios, with an extreme example being the previously-mentioned M60-UCD1.


Diamonds in Neptune? In this publication from 1981, was shown that Neptune is raining diamonds:

  Many of the current models of Uranus and Neptune postulate a three-layer structure, consisting of an inner rocky core, a middle ‘ice’ layer of fluid, $\rm H_2\rm O$, $\rm{CH}_4$, $\rm{NH}_3$ and an outer hydrogen–helium layer of solar composition. The estimated pressures and temperatures of the ice layer range from about 6Mbar and 7,000K at the inner core-ice boundary to $\approx$ 0.2 Mbar and 2,200K at the outer ice/hydrogen–helium boundary. I point out here, that shockwave experiments on these liquids, as well as theoretical studies, imply that the $\rm H_2\rm O$ and $\rm{NH}_3$ in the ice layer are almost totally ionized, and the $\rm{CH}_4$ has been pyrolysed to carbon, possibly in the metallic or diamond form.

Question: In the decades since this work was published in 1981 have these predictions been addressed further, or even proven or disproven?

If this turns out to be true, would these diamonds have different physical properties than those on Earth because of the different formation mechanism?

People have demonstrated high-pressure diamond formation using laser-driven shocks, much in line with the original idea. The diamond rain idea seems to be doing well (popular science article about it, with references).

However, it is quite likely that the diamonds could have different properties because of formation mechanisms. There are a lot of unusual allotropes of carbon, and local conditions of formation (not to mention impurities) may cause the carbon to turn into these forms rather than the standard diamond lattice.

Simulations show that beyond 1000 GPa diamond has a BC8 structure. However, this is a pressure unlikely to occur inside Neptune's atmosphere: it occurs deep in the mantle, so sinking diamonds may shift crystal form.