What is the speed of light relative to (in space)? From what I understand the universe is expanding faster than the speed of light and accelerating so that it's possible that a given galaxy: $A$ will no longer be visible to another given galaxy: $B$.

So, if a light wave is emitted from galaxy $A$ toward galaxy $B$ after it has passed the point where neither galaxy is visible to each other, how can the speed of the light wave be measured if there is no fixed point in space for it to be measure against (as the light wave will never be visible to either galaxy again)?

To any observer between 'A' and 'B' where the light wave passes, it passes with the speed of light. Adding a velocity $v$ below the speed of light $c$ to the speed of light results in the speed of light, simplified $c+v=c$, where the '$+$' means relativistic addition of velocities, not the '$+$' of arithmetics.

Depending on the relative velocity of two observers at the (almost) same point they will observe different Doppler shifts of the light, more precisely of spectral properties of the light. By analysing the spectrum of the light the observers may be able to determine the Doppler shift, meaning relative velocity, to the object which caused the emission or absorption line in the spectrum.


Where might astrometric observations of the Roadster spacecraft be catalogued? above: (see small red lines near center of image) Tesla Roadster. 18 Feb. 2018, 11:28 UT. The image above comes from the average of ten, 300-seconds exposures, remotely taken with the 16”-f/3.75 robotic unit (“Pearl”) of Tenagra Observatories in Arizona. The Tesla Roadster is marked with two red lines. At the imaging time, the car was about 3.5 millions of km from the Earth, that is 9 lunar distances. The object was at mag. 21.0. Image by Gianluca Masi (The Virtual Telescope Project) and Michael Schwartz (Tenagra Observatories, Ltd). From Virtual Telescope Project.

below: Zoomed in and cropped showing Roadster's image.

This excellent answer from @MikeG shows exactly how to find a list of observations for a minor planet. But I'd like to ask about a list of observations of an artificial object.

Currently the NASA JPL's Horizons site is also providing predicted trajectory of an unusual spacecraft which is referred to as Roadster. It is the cherry-red Tesla Roadster attached to a roughly 3.6 x 8 meter white rocket body in heliocentric orbit, and it has been tracked by many different observatories because of its unique circumstances. (some GIFs in this answer) Roadster's 4.6 minute rotational period was also recently reported.

Currently Horizons is reporting solution #7. They have been updating their solutions regularly as new observations are made and reported because as Roadster recedes from Earth it is rapidly dimming and will soon pass magnitude +20, so each new solution helps for the reacquisition of the object as it dims.

  This trajectory is based on JPL solution #7, a fit to 269 ground-based optical astrometric measurements spanning 2018 Feb 8.2 to 13.4.

Would these observations still be catalogued in an astronomical database of observations? Since it's not a natural minor planet, it seems the Minor Planet Center would not be hosting this information, but it seems it should still be made available to astronomers for identification purposes.

But where?

The header of the Horizons output includes this:

*******************************************************************************
Revised: Feb 13, 2018          Tesla Roadster (spacecraft)             -143205
          (solution #7)

 Tesla Roadster (AKA: Starman, 2018-017A)

and this:

TRAJECTORY:
  This trajectory is based on JPL solution #7, a fit to 269 ground-based
  optical astrometric measurements spanning 2018 Feb 8.2 to 13.4.

  Trajectory name                       Start (TDB)          Stop (TDB)
  --------------------------------   -----------------   -----------------
  tesla_s7                           2018-Feb-07 03:00   2048-Jan-01 00:00

The Minor Planet Center inconspicuously puts this under "Distant Artificial Satellites Observation."
DASO Circulars 561-566 have observations of the roadster 2018-017A.
It's also on the short list of artificial objects for which MPC can generate ephemerides.


Asteroids turning out to be Meteorites What causes an asteroid to loose control over it's orbit and enter in to another objects atmosphere?

Asteroids don't have control of their orbit. They are influenced by gravity; they obey the laws of physics.

There aren't any traffic lights for them in space to obey so they know to slow down and stop to avoid a collision with a planet. They'll just carry right on going like someone driving through a red light. If it goes right in front of a planet at the wrong moment then it gets t-boned.

Or it might miss. But here is where the analogy falls apart. Because if someone drives through a red light and misses everyone, they carry on without any subsequent repercussions (except maybe a ticket). An asteroid passing close by a planet will be influenced by gravity between the planet and the asteroid, and it's path will change. That might mean it is flung far away, or it may mean it is captured, putting it on a looping path that eventually draws it in to crash onto the planet.


How did WASP-18b exoplanet form? I am new to this community and not quite sure if this is an appropriate question to be asked here. Anyways its just my curiosity. As far as I could know about WASP-18b is that, it orbits very close to its sun and takes less than 24 hours to complete one orbit. However this was not the case always, as the planet actually was formed in a cold part of the space and later got tug into of its sun's gravitational pull.

Given this is the case, I am curious to know how it was formed in the first place? If I am not wrong planets are formed when the debris orbiting in the gravitational pull of a star stick together overtime gets in to the form of the planetary bodies. So, if WASP-18b was not always so close to its sun and it was formed far away in cold space, then could there be a possibility that, this exoplanet was formed in an orbit of another star and overtime it got away from its original orbit and trapped in the tug of its current sun? (Its my imagination! just bare with it :))

Also could any one point me to good documentations from where I can learn more.

Thanks

@LocalFluff's comment is more-or-less complete. The process of "swapping" planets between stars could happen during the early lives of stars when most of them are born in relatively dense clusters. You can find some more details and references in my answer to this question, but the outcome is that a captured planet would usually be in a very wide orbit.

Approximately 1% of stars have a "hot Jupiter", but even in dense cluster environments it is thought that only a few percent of stars can capture a planet, and because almost all of these end up in wide orbits there are just too many hot Jupiters to be explained in this way.

There are two classes of theory to explain hot Jupiters. One is the dyanmical explanation offered by LocalFluff. This is itself divided into several different mechanisms: "Secular chaos", (e.g. Wu & Lithwick 2011 - this paper also briefly reviews some of the other mechanisms) which is a gradual change in the planetary system that then suddenly drives a transition to a new configuration. Planet-planet scattering where interaction between planets causes one orbit to become highly eccentric and then tidal circularisation results in a hot Jupiter in a close orbit (e.g.Ford & Rasio 2008). The Kozai mechanism is the interaction beteen two planets that results in an interchange between the inclination and eccentricity of the inner planet. This excites an increase in eccentricity that, when combined with tidal circularisation, can  also result in a close-in planet.

The other class of mechanism involve migration through interaction with a disk. This must necessarily happen early in the star's life whilst surrounded by a disk of circumstellar material. Basically viscous drag in the disk itself causes the planet to move inwards (or sometimes outwards). There are three possible subtypes of disk migration and they are briefly defined here.

It is probable that all of these mechanisms play some role in forming hot Jupiters and one of the goals of exoplanetary study is to figure out which may be more important. Typical parameters of interest are the distribution of planetary orbits and eccentricities and whether there is evidence of previous interactions in the form of misalignments between the rotation axis of the star and the orbital axis of the planet.


How would the Tycho Impact have appeared from Earth? The Tycho crater is thought to be the youngest major impact on the Moon, estimated at 108 million years ago. This puts it firmly in the reign of the dinosaurs, I find myself curious as to what our thunder-lizard brethren would have witnessed.

Roughly how bright would the initial flash have been? Would a glowing scar have been left on the Moon for days, weeks or months? Would the Earth have been subject to spectacular global meteor showers from ejected debris? Would any other effects have been felt on Earth? I'd guess that a simulation would be required to accurately answer these questions but I'm hoping that some reasonable estimates can be made by comparison of studies of other impacts.

Finally, was the Tycho impact of greater or smaller magnitude than the Shoemaker–Levy 9 impact in 1994?

I found a partial answer in this article on the smaller and younger Giordano Bruno crater.

  The impact creating the 22-km-wide crater would have kicked up 10 million tons of debris, triggering a week-long, blizzard-like meteor storm on Earth...

So the secondary effects of the Tycho impact would have been even more spectacular. However, I don't know whether the fragments would have been large enough to devastate the Earth's surface.

Experimenting with an online impact simulator it appears that the Tycho impact was somewhere in the order of 1.00x10^23 joules, judging by the size of the crater. This is about four times that of the G fragment impact of Shoemaker-Levy 9. When you consider all the fragments it would be fair to say humans have witnessed an impact of similar magnitude to Tycho.


How to understand Kepler telescope under large aspherical wavefront To my understanding, a Kepler telescope is designed to conjugate one plane to another. For example scale down/up the incoming wavefront onto a wavefront sensor for measurement. It works well for small wavefronts, i.e. the wavefront is relatively flat (e.g. $<10 \lambda$ wavefront RMS).

What would happen if a large aspherical wavefront is impinging onto the same Kepler telescope system? Paper Calibration issues with Shack-Hartmann sensors for metrology applications suggests to place the aperture at the shared focal plane for performance improvement, as below:

How to understand the intuition behind? Why would Kepler telescope fail for large aspherical wavefronts?

Not sure what you're asking here.  Pretty much any telescope will be unable to simultaneously produce infinity (magnified) output and an aberration-free intermediate focus.  Further, an aspherical wavefront can't be produced purely by a radiating point source (all objects are collections of point sources), so no optical system can refocus an ugly wavefront.

The way a Shack-Hartmann WFS works is basically not to care about aberrated images. Each subaperture's image is unresolved to begin with, and all the sensor needs is to determine the approximate centroid location.


If the Wow! signal was not directed at us, then where? We probably all know the Wow!-Signal, happened on August 15, 1977. Assume that it wasn't meant for us. What was it pointing at then?

The signal at that time came from RA= 19h25m31s ± 10s or 19h28m22s ± 10s and the declination becomes −26°57′ ± 20′.

Which stars could it point at and did anybody investigate this?

In the opposite direction (RA= 07h 25m 31s ± 10s, +26° 57′ ± 20′, J2000.0, seen from Earth) of the positive horn of the Wow!-signal (RA= 19h 25m 31s ± 10s, −26° 57′ ± 20′, in J2000.0 coordinates), Iota Geminorum, aka Propus, (RA= 07h 25m 43.6s, +27° 47' 53") is within a 3-sigma error ellipse (1.26 sigma in RA, 2.54 sigma in Dec, 2.84 sigma total error).

However, this doesn't reveal, whether the signal pointed to anywhere.


What does the word "compact" mean in astronomy/astrophysics? I have seen countless times the adjective "compact" in astronomy, e.g. compact stars, compact binaries, etc. I would like to know, what exactly is this supposed to mean? From this source, I am led to believe that "compact" is meant to denote objects of the greatest matter densities in the universe, i.e. white dwarfs, neutron stars, and black holes. Is this correct?

Compact stars are stars of great density; stellar remnants in which the fermion constituents have become degenerate. i.e. white dwarfs and neutron stars (though sometimes black holes would be included).

A compact binary is a binary star system containing at least one compact object.


A star a black hole and planets around them Imagine we have a black hole and a star.

There are very close to each other so black hole slowly consumes the star.

It is possible to have planets in this system, and for them not to be consumed by this black hole but just follow their unique orbits. Could there even have life there?

It's possible to have planets orbiting a binary pair of stars, your scenario of a close orbit, sometimes called "short orbit binaries".    See here, also posted above in comments.   In such a binary-system, nothing can orbit an individual star, but at some distance, plants can orbit and some systems like this have even been observed, listed in the link.  The orbital dynamics is the same for a star-black hole short-orbit binary.

Now, there are problems.   Black holes form out of very large stars and the formation is one of the biggest explosions in the universe, a Type II supernovas, and that's not very friendly to any planets in orbit.  A star might survive it, planets would be harder, though it might be possible for new planets to form from nebula material remaining after the nova (I'm just guessing there).

A black hole could also form from a Neutron star accreting matter, but you still have the problem that the formation of a Neutron Star also only happens out of a Type-II supernova, so such a system has a difficult beginning.

Edit:  While some planets have been observed around Neutron stars, these appear to be quite rare.   2 Neutron Stars have been observed with planets, out of over 1,600 Neutron stars observed.    A type II nova is very planet unfriendly.

Theoretically a close gravitational capture is possible, but those are very rare, as stars rarely get that close.   There's many stars that are known to orbit the super-massive black holes at the center of our galaxy (Andromeda galaxy too), but stars and stellar mass black holes are much more rare.   A few have been observed, but they don't appear to be common.   Such a system would be easy to observe, so the fact that there are only a few that have been noticed is evidence to them being rare.  Here's a few mentions of them.  One, Two, Three, Four.

From the 4th article, which is from 2011, so more may be known now, but it says:

  Only about 20 binary stellar systems are known to contain a black
  hole, out of an estimated population of around 5,000 in the Milky Way
  Galaxy.

A 2nd problem is that a star feeding a black hole would create an accretion disk which would be very radioactive and not ideal for life on an orbiting planet.   Maybe the planet could have a very thick atmosphere that might protect it, but that would also likely reduce sunlight reaching the surface.   The star feeding the black hole would also be losing mass, and over time, grow smaller and provide less light and heat to the planet.   Ideally, you'd want it to be a very slow feed.   It's pretty far from an optimal life on planet situation.

  So it is possible to have planets in this system not to be consumed by
  this black hole but just follow their unique orbits

This part is certainly possible.   Things can orbit a black hole at a safe distance without any problem.  As for life, we don't know how common life is in other solar-systems so nobody can say how likely it might be, but it's theoretically possible, but, in my opinion, pretty far from ideal.


Magnification of a telescope I took an image of Jupiter through my 8" Dobsonian Telescope, attaching a DSLR and a 1.25" Barlow Lens where the eyepiece goes, as shown in this video: https://www.youtube.com/watch?v=reFxoF3XoaU

Through numerous online sources, I learnt that to find the angle of view of my image, the magnification of my image needed to be calculated, and that this value along with the given field view for my eyepiece.

http://www.rocketmime.com/astronomy/Telescope/Magnification.html:

The above website answers the following question, the calculations of which I attempted to mimic.

  My first telescope was a Meade 6600 -- they don't make it any more -- it's a 6-inch f/5 Newtonian scope. It came with a 25mm eyepiece. So... what was the magnification I was getting with this scope?

Here are the calculations I did in hopes of getting the above result:

$$\textrm{Diameter} = 8'' = 203.2\ \textrm{mm}$$

$$f_{\textrm{ratio}} = \frac{\textrm{focal length of objective}}{203.2\ \textrm{mm}}$$

$$\therefore \textrm{focal length of objective} = 203.2 \cdot 5.9 = 1200\ \textrm{mm}$$

$$\textrm{Magnification} = \frac{\textrm{focal length of objective}}{\textrm{focal length of eyepiece}} = \frac{1200}{x}$$

As shown in the video (the first link above), I didn't use an eyepiece to take my picture - I used a barlow lens, a couple of adapters, and a DSLR. So at this point, I am not sure what value to use for the "focal length of the eyepiece." How can I proceed to calculate the magnification?

The idea of magnification is not relevant to astro-photography, what is relevant is the image scale.

The image scale depends on the focal length of the objective and the size of the sensor. The Barlow effectively modifies the focal length, so if you have a $B\times$ Barlow and an objective of focal length $f_O$ the effective focal length is $f_e=B\times f_O$.

If your sensor has side of length $l$ then the field of view that corresponds to that side is (in radian) $\theta_{rad}=l/f_e$ or in degrees $\theta_{deg}=(l/f_e)\times(180/\pi)$

So if your sensor has a short side length of $\approx 0.015 \mbox{m}$ and a the telescope has a focal length of $1.2 \mbox{m}$ and you are using a $\times2$ Barlow you have the short side of the image corresponds to a field of view of $\approx 21$ minutes of arc.


Helium rain inside metallic hydrogen within Jupiter, conditions and experiments? In response to today being pi day (also see http://www.piday.org/), NASA has published it's fifth annual pi in the sky day 5 activity set (see also NASA goes the distance and Celebrate Pi Day with NASA).

This year's NASA activity includes the problem set Helium Heist: A 'Pi in the Sky' Math Challenge. Below is a cropped screen shot of the illustration from this PDF version.

The part of the text that I found most striking is:

  It has been hypothesized that the helium was depleted out of the upper atmosphere and transported deeper inside the planet. The extreme pressure inside Jupiter condenses helium into droplets that form inside a liquid metallic hydrogen layer below. Because the helium is denser than the surrounding hydrogen, the helium droplets fall like rain through the liquid metallic hydrogen.

Question: Roughly what are the pressure and temperature thought to be at this depth in Jupiter? Is this regime at all accessible experimentally on Earth to confirm that these two fluids are immiscible, or is this behavior purely theoretical?

below: cropped screen shot from here (NASA). Click for full size.

Different sources give significantly different values for the conditions under which hydrogen shifts from molecular to metallic. See the discussion in the comments associated with this question. Pressures from about 20 to 300 GPa and temperatures very broadly around 10 000 K seem to be the most common.

Experiments can only just reach these pressures, at lower temperatures, for tiny volumes of material, and the question of whether anyone has made metallic hydrogen at all remains controversial. Certainly no one can make it repeatably. (wikipedia has a reasonable summary of the history) or make enough of it and keep it for long enough to check to what extent, if any, helium is soluble in it.

That said, analysing from theory whether it will be soluble shouldn't be that hard, since both hydrogen and helium are relatively simple atoms. An early paper concludes that the hydrogen would be at or close to the limit of how much helium could be dissolved. I'm sure there is more accurate recent work, but I couldn't find it quickly.


Is Planet Nine observable in principle? Is so-called Planet Nine (given it exists) observable in principle? By "observable in principle", I mean "if we knew exactly where to look, would we be able (from a technological standpoint) to get an image of the purported celestial body"?

If they knew exactly where to point Mr. Hubble, then yes, it should be easily visible, though at that distance, it would still be blurry, not a clean image.

The estimated apparent magnitude of Planet Nine varies based on which website you believe, Google provides ranges from 20 to 25 with Wikipedia saying >22.  Higher numbers mean less visible.

Hubble can see things up to apparent magnitudes of about 30, so, yes, it should be easily visible by Hubble or (perhaps) even more visible by an infra-red telescope or radio telescope that knows where to look.  But in any case, with current technology all images would still be blurry.   Hubble's pictures of Pluto are blurry and Pluto's much closer.


Can Earth be spun like a magnet? Could an asteroid or a rogue planet have enough magnetism and/or gravity to spin the Earth near the speed in which they are passing by Earth?

Yes, a little, but not enough to matter.

Magnets can spin other magnets.   Think of magnets as a bar magnet with a north and a south, and you can think of them as linear not spherical if that helps with the imagery, because in a sense, they are linear with a distinct North and South magnetic pole.

Earth's magnetic field, for example has a distinct longitude and latitude that moves over time, but it's fixed at any given time.

For your question to work, it's easier to imagine a very powerful magnetic object flying past, and to think of the Earth's internal magnet as a compass arrow that wants to line up.   Magnets want to line up with magnetic field lines and a force would be created by a passing strong magnetic field that could orient the Earth's magnetic field to align with it.

You might think that this would turn the Earth, but it wouldn't.  Earth's magnetic field isn't "the Earth" it happens in the Earth's core, and it may happen mostly in the region where Earth's inner and outer core meet and the outer core is liquid (dense and viscous but still not solid).  The force would be applied to the magnetism inside the Earth, perhaps giving the Earth's inner core a slight tug and twirl and perhaps re-orienting the flowing charged liquid iron just outside of the core, but this force wouldn't turn the Earth as a whole, it would generate a force on the core and the core doesn't spin precisely with the Earth's surface anyway so any movement would likely not show up on the surface, for some time.

Given that a percentage of earth's magnetism is carried by the liquid outer core, then the effect on rotation would be even less, more of a re-orientation than a direct angular momentum tug.

It's also difficult to generate a magnetic field strong enough to move something as massive as the Earth's core very much anyway.   Something to keep in mind as well is that the magnetic field drops by the cube of the distance, not the square, so you'd need a very powerful magnet, very close, to have any effect at all.   Anything large enough to have that strong a natural magnetic field, the gravity would have a much bigger effect than any interaction between the magnetism.

But to answer your question, a tiny bit of angular momentum would be transferred when two magnetic fields fly past each other in space, so there would be some variation in rotation, but it would be tiny.

Magnetic fields as strong as Earth's are rare, you'd probably need a gas or ice giant planet or large rogue planet, and the interaction between passing magnetic fields is very weak compared to the gravitational interaction.  Compared to it's mass, Earth is a very very weak magnet, so using magnetism to spin it would be ineffective and slow, but but in theory, if we wanted to set up a current and generate a magnetic field around the Earth, such a project could in theory spin the Earth's inner core like a induced magnetic field spins a magnet in a generator, but that would also depend on how permanent the magnetism is inside the Earth and it might not be fixed, but instead, a product of flowing magnetic metals.   Permanent magnets are easier to rotate than charged flowing liquid . . . but I digress.

It would be an interesting calculation if somebody wanted to do the math behind this, but even without doing the math I can tell you that the effect would be teeny-tiny.


Does the size of the atom limit the focal length of telescopes? Does the size of the atom place a theoretical limit on a telescope's focal length (and thus, resolution)?

It seems to me that that the allowed margin of error for parabolic mirrors drops with an increasing focal length. It's already like 25 nm for advanced Earth-based telescopes - which is just 50-250 atoms.

If we try building mirrors with an ever increasing focal length (e. g. as a part of a huge space-based interferometer), will we eventually reach a point where the precision required will be less than the size of the atom and thus no longer achievable? If yes, how far away are we from this limit now?

If we had unlimited resources, would it be possible to construct a space-based telescope for imagining extrasolar planets in great detail, or would this be not even theoretically possible due to the fixed atom size?

Ok, finally I can explain it. Not sure why it took so long.

Bottom line, for visible light atoms don't matter. Let's say the blue side of the visible spectrum of light has a wavelength of 400 nm (round number for simplicity). Let's assume the distance between two atoms in glass is 0.2 nm.

The thing is, light is a wave. Anything smaller than about 1/4 of a wavelength becomes essentially invisible to that wave. It's not a sharp threshold, but as things get smaller and smaller, the wave "sees" them less and less. It just goes around the object as if it's not even there.

1/4 wave of visible light is 100 nm, which is 500 glass atoms.

I make telescope mirrors. I can still see the effects of surface errors down to about 1/20 ... 1/25 of a wave. Beyond that the image is not really distinguishable from a geometrically perfect mirror, either in real-life tests under the stars, or in the lab using the equipment. Maybe an interferometer could push that limit a bit further, but not a whole lot.

Let's say the absolute measurable limit is 1/50 wave, which is 8 nm, which is 40 atoms. That's plenty of atoms. Any defects smaller than that simply do not change the image created by the mirror - when talking about regular telescope mirrors.

The surface level difference from a single layer of atoms does not even register in visible light. It takes far more thickness than that to make a visible difference.

So your ultra-long focal length mirror would be almost flat, slowly adding extra atomic layers towards the edge. As long as it approximates the desired ideal surface (paraboloid, etc), then it will be fine.

I see one exception. If the focal ratio is indeed so big that there isn't enough curvature in the surface to require even a single extra layer of atoms at the edge. Then, indeed, the mirror would not work, because it's flat down to the atomic level. Keep in mind though, that would be a mirror with a humongous focal ratio - and indeed with a cosmic-scale focal length. There are many issues with such a system before you even begin thinking about surface errors. Probably not feasible with current technology.

But then what you would do is make a segmented mirror, increasing the apparent diameter and reducing the focal ratio. You could even have multiple flat segments (flat down to the atomic level) but far from each other to massively increase the total diameter, and slightly tilted according to the shape of the ideal surface. If you can afford a cosmic-scale focal length, then you can surely afford a cosmic scale diameter, and that's how you reduce the focal ratio. Maintaining that precise tilt angle between segments would be a tremendous feat of engineering.


Whats the deal with black holes and "no information from inside the event horizon can leave"? I don't know if this would fall under the other questions but I have always been confused by this... Information has left the black hole; its mass, rotation, location, and a few other properties. My problem has been that a true singularity should collapse the space around it and become separate from the universe as being able to observe the effects of a singularity breaks the fact that it is a singularity.

If I see a black hole I can already tell its mass and be effected by the gravity of an object which I can receive no other information from. This would break things too!

Let's just say I am an all powerful being who can do almost impossible things, but I still have to follow the laws of the universe. I decide to create a gravity observation station in orbit which uses lasers to see very small changes in the black hole. It also has a teleporter of sorts.

I then enter the black hole and construct another station, this one however uses small black holes and impossibly powerful thrusters and materials to do certain things. Its purpose is to manipulate and sustain a small black hole "orbiting" the inside of the bigger black hole (being sustained, barely, by the station) it then changes the direction this smaller black hole orbits in a kind of binary way.

I then step into a "teleporter" converted to binary instructions and encoded into these gravitational and rotational information. (Let's just say the station has no lifespan) and this station over a huge amount of time begins to piece together this code, and reconstructs me in the "teleporter" on board.

Have I just circumvented the event horizon using disclosed information available to me? Why would or wouldn't this work because for what I see being able to describe a black hole violates the properties of the black hole.

You are right in that, if you could send a message of any kind from inside a black hole then there is no magic required rebuild you from the information about you. If you could send even a single bit of information, then the black hole wouldn't be so black. You want to do this using the rotation of the black hole.

This doesn't work, for various reasons.

  Information has left the black hole; its mass, rotation, location, and a few other properties

The mass etc is a property of the black hole, not information travelling from inside the black hole. The reason is that you can't manipulate them in the way you describe, so they can't be used to send a message.

  I then enter the black hole and construct another station

Here is your first problem. From the frame of reference outside the black hole, there is infinite time dilation at the event horizon. Hence you never quite reach the event horizon, nor enter the black hole.

From the frame of reference of the infalling person, the event horizon is crossed. So let's somehow pretend that you are now inside the event horizon.

  Its purpose is to manipulate and sustain a small black hole "orbiting" the inside of the bigger black hole.

You can't orbit, you can only fall. The orbital speed is higher than the speed of light, so you will reach the singularity, as surely as you will reach next Tuesday. Your time has now been infinitely dilated, so events that take happen in finite time for you, take an infinite amount of time in the frame of the outside observer, so that makes sending a message using variation in rotation impossible.

Finally, from inside the black hole you can't change the total mass or rotation. The angular momentum of a closed system is conserved, so if you use your thrusters to change the direction of the small black hole you are carrying, the total angular momentum of the black hole remains constant. You can't change the total angular momentum with thrusters if you include both the angular momentum of the ship and the angular momentum of the gas ejected by the thrusters.

Similarly you can't change the total mass or location of the black hole from inside the black hole. And you can't do anything in a finite time from the perspective of an outside observer.

Thus no information can be encoded in the observed properties of the black hole.


Does the Honey moon have a precession problem? In Steven Colbert's interview with Neil deGrasse Tyson after about 03:16 the following is explained:

But wait, all the moons - all the full moons of the year have names. Some are familliar like the Harvest moon… and the Moon in June never gets too high in the sky, so it takes on sort-of an amber sunset color, the whole trip. And so that's called the Honey moon. So the June full moon is the Honey moon.

Is there a problem here though? For any given month won't the full Moon oscillate between low and high above the horizon at midnight on an 18.6 year cycle?
Video: Is An Asteroid Going To Hit Earth On Election Day? Neil deGrasse Tyson Says Maybe

Is there a problem here though? For any given month won't the full
Moon oscillate between low and high above the horizon at midnight on
an 18.6 year cycle?

No. The full moon is always opposite to the sun (within 5.2 degrees, the angle between the moon's orbit and the ecliptic). So, when in summer the sun is high in the sky, the full moon will be low, similar to the height of the winter sun.
You are correct about the oscillation with the 18.6 year cycle, but that oscillation amounts to only +/-5.2 degrees, while the summer/winter difference results in +/-23.5 degrees. So (in northern latitudes) the typical summer full moon will be 47 degrees lower than the typical winter one, and even the highest possible summer full moon will still be 37 degrees lower than the lowest winter full moon.


What's the farthest point humans have explored? By using powerful telescopes on land and space, what's the farthest point humans have explored or viewed?

This is a really interesting question to many and it deserves an answer post even if folks close it.
In Wikipedia's List of the most distant astronomical objects we can see that the distances that the light travelled to get to us are all around 13 giga light years, and it's no coincidence that the age of the universe is also about 13 billion years.
We can't really call that the actual distance to the object because the size of the universe is changing and the size of the observable universe is changing as well.
So while we like to say we are seeing really really far away, we should really think about it as seeing really really far back in time, because on the scale of the universe the speed of light is actually incredibly slow :-)
No matter how big of a telescope we build with unbelievable light gathering power, we'll never be able to see "farther" than a certain distance. There's a currently unknowably larger universe out there whose light and gravity has not and probably never will reach us
It's almost certainly based on current understanding, but that's subject to change of course ;-)


How moon is visible during sun is still shining, simple explanation please! Sometimes I observed there is moon on sky at noon and sun is also shining.

How is this possible?

I also observed dull moon is still on sky at 7:00 AM to 8:00 AM while sun is rising. My local time is UTC+5.

If possible, please explain it in simple words without using advanced astronomy terminology if possible, thanks!

Let's look at what the difference is between day and night.

In the day time, the air that's between you and the Moon is in the sunlight, but at night, that air is in the dark.

Day or night though, the air is still transparent (you can see the Moon through it), and the light from the moon is mostly unaffected.

When it's in sunlight, the air scatters some of the sunlight in your direction. For the parts of the Moon that are bright, they are so bright that they look white.

But the dark parts of the moon don't look black to you on Earth because there's still the fainter blue-sky light scattered by the air.

Moon during the day with bright blue sky light

Source

Moon at sunset with faint blue sky light mixed with Earthshine

Source


Moon's path as seen from Earth As observed from the surface of Earth:
Will the moon ever repeat the same path (trajectory)?
If yes, what is the period of repetition?
And also will the moon cover the night sky with its path completely over a period of time?

More explanation:
I was observing the moon from my window and I was wondering will the moon follow the same path (trajectory) as it did before, and if I am taking photos all the different positions of the moon over a long period of time will it have covered the night sky?

There is a beautiful composite picture demonstrating exactly this over at http://www.twanight.org/

This is a composite of 90 photos across the four days, so you can easily see how far the moon moves each day (about 13 degrees)

From https://cseligman.com/

  on average the Moon crosses the sky once every 24 hours and 49 minutes (53 minutes longer than the stellar "day"). As a result, it rises (and sets) later and later every day, until after about 27 days, when it has gone once around the sky relative to the stars, it is back in its original position, rising and setting at its original time(s).


How can I calculate when the new moon will be observed from a particular meridian? I have read that Chinese months start on the day on which the new moon is observed from the 120th meridian east.  From this I am assuming that the new moon is observable from there each time and that it doesn't matter where on the meridian you are (although I suppose it's implied you're in the northern hemisphere).

How can I calculate the time at which the moon is new from a particular meridian?

You need an Orrery. Specifically, one that includes Earth’s rotation.

My noting the meridian on the model Earth, you can see how the Earth may be rotated some amount each time the Moon passes on a line between the Earth and Sun.

The wikipedia page gives a formula.  Take the result (given in UT) and apply the timezone offset for the desired longitude.

The same page discusses Lunar calindars and may have the info you are really interested in.


How can the Earth and the Moon be in synchronous rotation if the Moon won't be in geostationary orbit? It is said that the Moon moves away from Earth and that during the Sun's red giant or white dwarf phase the Moon will be about 40% farther than now and in a synchronous rotation ("hantle rotation") with the Earth, meaning it will always be above the same location on Earth, like Pluto-Charon. But how would this be possible if the Moon wouldn't be in a geostationary orbit which is 35,786 kilometers (22,236 miles) above the Earth's surface?

Although the Earth's rotation is slowing due to tidal interactions with the Moon, the timescale for the Earth to reach synchronous rotation with the Moon's orbit is quite long, even by astronomical standards, and certainly won't happen during the red giant phase.
The current change in Earth's rotation is 1.7 ms/century over the past 100 years, and 2.3 ms/century since the 8th century BCE.  (See Wikipedia for references.)  To be conservative (i.e. to give the Earth the best chance to get to synchronous rotation), let's take the larger rate of spindown.
Converting units, this is
$$ \frac{2.3 \times 10^{-3}\ \mathrm{sec}}{10^2\  \mathrm{yr}} \times \frac{1\ \mathrm{day}}{86400\ \mathrm{sec}}=\frac{2.66\times 10^{-10}\ \mathrm{day}}{\mathrm{yr}}=\frac{1\ \mathrm{day}}{3.76\times 10^{9}\ \mathrm{yr}}$$
So at the current rate of change, the day length increases by one (current) day every 3.8 billion years.  So by the time the Sun reaches the red giant phase in 5–7 billion years, the day will be at most 2 days longer, i.e. about 3 of our current days.  The orbit of the Moon will be longer than its current 27.3 days (it is spiraling outward due to the same interactions), so we won't be anywhere near synchronous rotation at that point.
Assuming the Earth survives being engulfed by the Sun during the red giant phase, how long will it take?  At the rate calculated above, changing from the current 1 day to the Moon's current orbital period of 27.3 days requires a 26.3 day change, which takes about 100 billion years (about 7 times longer than the current age of the universe) at the current rate of change.  We still aren't synchronized at that point, since the Moon will have moved outward at that point, and thus (by Kepler's third law) will have a longer orbital period. (To calculate the right period to compare to, you would have to work out what orbital period gives synchronous rotation given the current amount of angular momentum in the Earth-Moon system.)
There are two major reasons why the above is an underestimate of how long this would take:

As already noted, as the Moon gains orbital angular momentum from the Earth's rotation, the Moon moves into a larger orbit.  This weakens the tidal interactions between Earth and Moon, which makes the rate of change slow down over time.

During the red giant phase (or well before that), Earth's oceans will evaporate.  Without a liquid ocean, the tidal bulge raised on the Earth by the Moon will be much smaller. Thus the exchange of angular momentum, and resulting rate of change of day length, is again slower.

There are various other effects you could speculate about (i.e. will the Earth even have a Moon, assuming it survives the Sun's red giant phase?), but this gives a sense of the long timescales you'd have to consider for this specific orbital evolution.


How did "oddball" Valetudo, Jupiter's new prograde moon, end up in a wider orbit? I just saw this question and that got me to this NASA page which led me to this scientific article.

“Our other discovery is a real oddball and has an orbit like no other known Jovian moon,” Sheppard explained. “It’s also likely Jupiter’s smallest known moon, being less than one kilometer in diameter”.
This new “oddball” moon is more distant and more inclined than the prograde group of moons and takes about one and a half years to orbit Jupiter.  So, unlike the closer-in prograde group of moons, this new oddball prograde moon has an orbit that crosses the outer retrograde moons.
As a result, head-on collisions are much more likely to occur between the “oddball” prograde and the retrograde moons, which are moving in opposite directions.
“This is an unstable situation,” said Sheppard. “Head-on collisions would quickly break apart and grind the objects down to dust.”

My question is, why are most of the moons that are further away from Jupiter retrograde orbits, while the ones extremely close to Jupiter are prograde orbits? What possible situations were required for Valetudo to end up where it is opposed to where the rest of the pro-grade moons are clustered? Pure chance / a series of unlikely events?

My questions are: 1.:  why are most of the moons that are further away from Jupiter retrograde orbits, while the ones extremely close to Jupiter are prograde orbits? 2.: What possible situations were required for Valetudo to end up where it is opposed to where the rest of the pro-grade moons are clustered? Pure chance / a series of unlikely events?

Answer 1. Inner moons usually orbit in the same direction as the rotation of their parent planet, this is favored due to one means of their formation, and the effect of tidal forces if they were close to the planet and in a retrograde orbit.
Source 1: Wikipedia: "Moons of Jupiter - Regular Satellites":

"The inner moons are in a 1:2:4 orbital resonance. Models suggest that they formed by slow accretion in the low-density Jovian subnebula—a disc of the gas and dust that existed around Jupiter after its formation—which lasted up to 10 million years in the case of Callisto.$^{[39]}$
[39]: "Formation of the Galilean Satellites: Conditions of Accretion" (Dec 2002), by Robin M. Canup and William R. Ward.

Source 2: Wikipedia - "Hill Sphere - True Region of Stability":

"The region of stability for retrograde orbits at a large distance from the primary, is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.$^{[3]}$
[3]: "Chaos-assisted capture of irregular moons" (May 2003), by Sergey A. Astakhov, Andrew D. Burbanks, Stephen Wiggins, and David Farrelly.

Answer 2. It is possible that Valetudo was a binary companion and was captured in an exchange reaction.
Source 1: Wikipedia: "Moons of Jupiter":

"These moons were probably captured by Jupiter from solar orbits."

Source 2: "Satellite Capture via Binary Exchange Reactions: Application to Triton" (Dec 2005), by C. Agnor and D.P. Hamilton:

"... exchange reactions during planet-binary encounters (i.e. ~ an encounter in which one member of a binary is expelled and its place taken by the planet) represent a new pathway for the capture of planetary satellites. This new capture mechanism may be relevant to origin and evolution of several small body populations in the solar system (e.g.~the icy and irregular satellites of the giant planets).

Additional Info: Robert Walker explains and offers this image showing how contact and overcontact binaries, ternaries, quarternaries, etc. can form stable and unstable systems. Those can be torn apart by the tidal forces of larger (even smaller) objects.


What do we mean by Sunrise and Sunset? If the Earth rotates on its own axis, then what do we mean by sunrise and sunset?

Yes the earth rotates around its axis,  which gives rise to the motion of the sun as we see it.
To understand sun rise you'd have to start with a reference frame. For us,  the earth is considered to be the inertial frame for any apparent motion in the celestial sphere, it simplifies the computations. Thus any motion of celestial bodies is defined with respect to that frame.
So in that context Sunrise would be the time when for a given latitude, when the sun first appears above the local horizon, and sun set would become the time when the sun disappears below that local horizon.


Can Magnetars destroy planets? Magnetars are the most magnetic things in our universe. Because of how powerful their magnetic fields are, they could split every single atom apart from each other in our body.

Because of this could they destroy planets (rocky or not)? If so, how close would the planet have to be?

Perhaps one way of answering this is to note that the (approximately) dipole magnetic fields of a magnetar fall off radially as something like $r^{-3}$. So at the magnetar surface the fields could be of order $10^{11}$ Teslas (at a maximum), but these are reduced by a factor of $\sim 8$ at 2 magnetar radii, a factor of $\sim 27$ at 3 magnetar radii etc.

Now given that the radius of a magnetar is going to be $\sim 10$ km, then the fields die away pretty quickly.

So let's say the Earth orbited a magnetar. The field strength at $1.5 \times 10^{11}$ m is going to be $\sim 10^{11} \times (10^4/1.5\times10^{11})^3 =  3 \times 10^{-11}$ Teslas - much smaller than the Earth's magnetic field.

To put this in context. The field at the surface of a small neodymium magnet are of order 1 Tesla and they don't disrupt your atoms when you pick them them up. To get a field this strong from a magnetar you would have to get to within about 50,000 km of it. The synchrotron radiation field would have fried you before then and any planet would have been ripped apart by tidal forces about an order of magnitude further away. The Roche limit given by
$$ R_{Roche} \sim 1.26 R_1 \left(\frac{\rho_1}{\rho_2}\right)^{1/3},$$
where $\rho_1$ is the density of the neutron star ($\sim 3 \times 10^{17}$ kg/m$^{3}$), $\rho_2$ is the density of the Earth ($\sim 5500$ kg/m$^3$) and $R_1$ is the radius of the neutron star (10 km). This yields $R_{roche} = 480,000$ km.


Does parallax measurement depend on position in the sky Naturally, the parallax is easier to measure for stars that are closer than farther away. But if all stars were the same distance from the Sun, and if there were some other reference to measure the parallax, would all stars show the same parallax?
In other words, does the position of a star (the right ascension and declination) affect the parallax measurement and calculation? For example, are stars near the ecliptic or equator or celestial pole easier to measure than other locations on the sky?
Feel free to include any equations or references if it helps.

It's all about basic geometry.
Base for the parallax measurements is Earth's orbit around the sun, giving you a maximum of 300 Mio km. With a given base dimension, you get the best precision when the base is orthogonal to the star direction. (In the other extreme, you get no parallax at all if the base is in line with the star).
For stars close to the ecliptic you get this optimum base angle only using two specific dates, half a year apart from each other (those when the star appears 90 degrees distant from the sun).
For stars nearly perpendicular to the ecliptic, you can choose any two dates half a year apart, giving you more chances to contribute max-precision measurements.
If doing a continuous observation of the star over e.g. a year, the difference should amount to a factor of sqrt(2), if the other parameters are comparable.


Neutrino producton rate over time Has the rate of neutrino production in the universe varied over time? Are there more or less being created now against the early days?

Likely more Neurinos are produced billions of years ago than today because there were more very large stars in young galaxies.   I'm not going to touch the first few seconds after the big bang.   Someone else can if they want, but post star formation, it's a surprisingly simple problem to work out.

Neutrinos are produced primarily in nuclear reactions in stars.   Large stars produce far more than smaller stars.  A star with 10 solar masses, would, very roughly, produce about 10,000 more neutrinos.   Stars that size are uncommon because they only live about 10-15 million years.   Only about 1 in 800 stars are B-type or greater in terms of neutrino production and most of those are only a few solar masses, producing between maybe 50 and 500 times the number of neutrinos our sun produces.   Very high neutrino producing stars are sufficiently rare that they don't throw off the numbers.  (I'm ignoring supernovas, but I'll get to that).

If we estimate 100 billion stars in the milky way, the larger ones give off far more neutrinos than our sun and the smaller ones, far fewer.   And our sun gives off 1.79 x 10^38 neutrinos per second.   Doing some very bad estimating, all the stars in the Milky way together give off probably in the range of 1 x 10^49 neutrinos per second.  Figuring our sun, being on the large size, gives off more neutrino to mass ratio than average, but not enormously more).   Again, we're ignoring supernovas.  You'll see why soon.

So, bad estimate, 1 x 10^49 per second from all the stars in the galaxy, or 3 x 10^56 per year.   Or 1 x 10^58 in about 30 years.   (Why do I care about 10^58 you ask?)

Because that's the number of neutrinos that the type II supernova observed in 1987 released.  Source.    So one type II nova produces the same number of neutrinos as all the stars in the milky way currently produce in about 30 years, give or take.

The milkyway has an estimated one billion neutron stars and about 100 million stellar mass black holes.   Each of those can only form by type 2 supernovae.   Given the age of the Milkyway, about 13 billion years, if we average it out, that's one type II supernovae every 12 years or so.   The current rate is much slower, so the young milky way had significantly more type 2 supernovae than it currently averages, and, it would be consistant that there were significantly more large stars.   Even with significantly fewer smaller and main sequence stars, the large stars and type II nova are the clear winners in neutrino production.

The current supernova rate (Type 1 and II) in the milky way is one every 50 years, so the rate had to be several times higher when the galaxy was young.

Peak Neutrino production would coincide with peak type II supernova explosions, which would only be 5-15 million years or so after peak very large star formation.   I don't mention Type 1 supernova, because, while more common now, they (I think) produce far fewer neutrinos.  Still a great deal, but well fewer than 10^58, though I had some difficulty finding a specific number.

It's very likely that young spiral galaxies like our milky way produce (very bad guess), maybe 10 times more neutrinos than they produce some 13 billion years after formation, so that would be the most likely answer.   Once large star formation in galaxies was well underway, Neutrino production probably peaked, whether that was when the universe was a few hundred million years old or a couple billion, I don't know, but neutrino production likely peaked early with the formation of many large stars and began a steady gradual decline after the peak.

Corrections welcome.


Change In Direction of Asteroid Approaching Sun A perfectly spherical asteroid on a straight line trajectory with no rotational movement at all, enters the gravitational field of the sun (from elsewhere far far away). When it exits on its hyperbolic path, does it still point in the same direction it entered, or does it rotate to face the direction of travel?

No.

A body in orbit will maintain its angular momentum, it will maintain its rotation rate as it is deflected by the sun. If it is not rotating then the deflection won't change the direction it is facing.

(There may be some minor effects that could cause a little rotation but it won't "steer" to face forward.)


Origin of the magnetic field of neutron stars It seems a little counter intuitive that neutron stars possess such strong magnetic fields. Its electric charge is presumably zero, so however fast it spins, it shouldn't generate any magnetic field. Or is it due to the electric charges of quarks or their intrinsic spins?

The strong magnetic fields in neutron stars are supposed to come from magnetic flux conservation. If we have:

$$\Phi_B = \int B\ \mathrm{d}S = \text{const}$$

where $\Phi_B$ is the magnetic field flux, $B$ is the magnetic field strength, and $\mathrm{d}S$ is the elemental closed surface; then, this integral is constant through the surface.

If we consider the star surface over which take the integral, than

$$S = 4\pi R^2$$

where $R$ is the star radius. This can be translated, altogether with the magnetic flux conservation law,  as:

$$B_f = B_i \left(\frac{R_i}{R_f} \right)^2$$

where $i$ and $f$ are the indices for initial and final stages.
We know that the star implodes from a whatever star size to $\sim10 \; \mathrm{km}$. So the radii ratio is huge. You just need a starting magnetic field of $10-100 \ \mathrm{G}$, to get a final magnetic field of the order of $10^{12}  \ \mathrm{G}$, that is typical in neutron stars.


How do I find stars within milky way using galactic latitude and longitude I have data that contains galactic latitude and longitude of stars. I have to determine all the stars that are within our milky way. How do I find it?

Stars beyond our own galaxy and its immediate neighbors are indistinct without a large telescope, so I assume the question is about the band across the sky, rich with faint stars.

The Milky Way's boundary is fuzzy, so its extent is expressed in terms of isophotes: lines of constant brightness per unit solid angle (e.g. square degree).
It is wider near the central bulge (galactic longitudes around 0° or 360°) than away from it (around 180°).
Most stars within galactic latitudes ±10° are in the Milky Way; most outside ±15° are not.
At galactic latitudes in between, galactic longitude and the choice of isophote determine whether a star is in or out.


Is it possible for a virus to come from a meteorite? So I am not a true scientist and I was just wondering if an extraterrestrial rock from space can have some dangerous unidentified virus or bacteria that can be deadly and wipe out the Earth if it not quarantined. Is this science fiction?.

P.S.: Note I said extraterrestrial rocks not extraterrestrial as in aliens.

There is, believe it or not, a scientific theory regarding this: panspermia.

According to the panspermia hypothesis, microbes "hitch" a lift on bodies leaving a planet. They then travel through space and eventually land somewhere else.
Panspermia involves three difficult phases of travel:

Launch (as well as the impact event)
Travel in the harsh environment of space
Atmospheric entry and landing, with high temperatures

It's difficult to survive these. However, materials related to organic matter have been found that might be evidence of panspermia; see Bell et al. (2015) for one example.

Panspermia is traditionally applied to bacteria known as extremophiles, which are accustomed to living in harsh environments. Space contains temperatures ranging from hundreds of degrees Fahrenheit to hundreds of degrees below zero, and any organism would have to survive those. This may be possible - bacteria have survived in conditions like there - but difficult.

Viruses pose an additional challenge because they cannot survive long without a host organism on which to live. This means they would have to be carried along by an extremophilic bacteria. This also means that they might not be a huge harm to creatures on Earth. Viruses only attack certain bacteria because of proteins on their surfaces; extraterrestrial biology most likely differs from that on Earth, and these viruses might not be equipped to harm terrestrial organisms.


Questions about time and space (from beginner) To be honest my knowledge about astronomy and related physics is close to zero. However, since I studied math a bit, I would like to ask a couple maybe stupid questions, which I am curious about and I don't know who and where to ask.

A) Projection of life

Premises:

As Carl Sagan explained in his "flatlanders" story, we can imagine high-dimension space by making projections into space with lower dimension.
As we learned at elementary school time is the fourth dimension.

Question:

Is it right that what we see/feel in every moment is projection of some 4D reality into our 3D world noticed by our senses.
For example, If my eyes were able to "somehow" see in 4D and I pointed my eyes to a person, would I see that person in all stages of his/her life from birth to death as a single "entity"? In every second we are only a projection of our 4D representation to our 3D world?

B) Parallel realities

Premises:

If we imagine 1D space as a line without ends, 2D space as an unbounded plane and 3D space as an unbounded space, then I could put infinity number of parallel lines into 2D space and infinity number of parallel planes to the 3D space.

I have read that there are at least 10 dimensions.

Questions:

Can I put infinite number of 3D spaces into a single 4D space? If yes, then if we for example chose geographic coordinates of Earth-Italy-Caesena as place and 49 B.C. as time coordinate, we would be probably close to event when Caesar screamed his famous: "alea iacta est".

If we considered that 5th dimension exists and that we could put infinite number of 4D spaces into 5D space, then we would end up with infinite number of parallel realities, in which Caesar did not crossed Rubicon, or did not even exist. Is it right or complete nonsense? Can be somehow proved that this is right/nonsense?

A)

  Is it right that what we see/feel in every moment is projection of
  some 4D reality into our 3D world noticed by our senses.

It's more similar to a slice, hyperplane or hypermanifold, less to a projection.

  If my eyes were able to "somehow" see in 4D and I pointed my eyes to a
  person, would I see that person in all stages of his/her life from
  birth to death as a single "entity"?

In a classical world, yes.

B)

  Can I put infinite number of 3D spaces into a single 4D space?

Yes, a possible mathematical description of such a space-time is a 3+1-Minkowski space-time.

  If we considered that 5th dimension exists and that we could put
  infinite number of 4D spaces into 5D space, then we would end up with
  infinite number of parallel realities, in which Caesar did not crossed
  Rubicon, or did not even exist. Is it right or complete nonsense? Can
  be somehow proved that this is right/nonsense?

That's close to the many-world interpretation of quantum theory.
But it's thought of as an infinite-dimensional Hilbert space, as one of the simpler ways of thinking.

Thinking of a 5D space-time as a collection of all possible 4D space-times is less suitable, since it would be a mere set of 4D space-times without meaningful structure. Take Ceasar crossing Rubikon a day earlier or day later or a meter to the left or a meter to the right with or without a leaf falling from a tree, or a huge number of other versions. Which would be the correct order to pile up these versions to a 5D world? The infinite-dimensional Hilbert space model solves this ambiguity, and provides a meaningful metric for the set of all possible worlds.

The 10-dimensional models of some string theories are something else: The 6 extra-dimensions are thought to be curled up, and tiny.

We have no access to parallel worlds, so it cannot be proven by observational evidence. It's a mathematical model. It's possible to prove theorems within the mathematical theory, but that's usually based on axioms, or assumptions. Mathematical theories may, but don't need to match with observation, depending on how the axioms match to physical "reality".


Universe isn't infinite theory Forgive me if any of my terminology is off I am no physicist but I have wondered about this question for a while.

I have read in various books that if the universe was infinite, and uniformly created the whole sky would be white. This is because in every possible location in the sky there would be a star emitting light. Sort of like pixels, filling a screen.

I thought about this and wondered, doesn't this therefore imply that light is never absorbed by another medium?

What if a distant stars light is absorbed, therefore never reaching earth, giving the illusion that there are less stars?

Olber's paradox describes an infinite, static, eternal, and homogenous universe. If there were a cloud of opaque gas situated in some region of the sky, light coming from all conceivable directions via the paradox would bombard the cloud until either it begins emitting light itself via the radiation it absorbs or its particles scatter to allow light to pass through.


Why don't we see evidence of glaciation on Mars, similar to Earth's Ice Ages? With all recent NASA activity surrounding Mars manned missions, it is surprising that given Mars orbital distance from Sun that there is not evidence of massive glaciation similar to Earth's Ice Ages. Also Mars allegedly has erratic Obliquity anything from 11 - 28 degree swings which would obviously affect seasonal changes. Is this a fact (where's the evidence) or "scientific fable".

Interesting replies - due to implications of erratic obliquity, hopefully NASA have a probe recording changes.

Obliquity anything from 11 - 28 degree swings which would obviously
  affect seasonal changes. Is this a fact (where's the evidence) or
  "scientific fable".

I don't like to use the word fact in a situation like this.  What that variation is, is a mathematical model based on well understood but pretty complicated n-body gravitational perturbations between planets, so, it's likely accurate.  Whether it's a fact is perhaps a matter of semantics, but I don't like to use the word fact in this situation.

That said, it's very likely true that Mars undergoes both significant axial tilt changes and eccentricity changes over thousands or tens of thousand year cycles, and if variations like that happened on Earth, they would cause havoc to our climate.
Everything else being equal, that kind of variation should generate considerable glacial movement on Mars' poles, but everything isn't equal between Mars and Earth.

Ice ages on Earth require large land masses on or near one or both poles, where ice can build.  Glaciers can't form in open deep ocean water, they require land, where the summers are cold enough that ice can build up, which, over thousands of years of buildup, becomes a glacier.

Mars lacks the water to expand it's glaciers.   Nearly all Martian water is either underground or already frozen on it's North pole, so there's no water to add.   Literally about 99.99999% of Mars' surface water is already part of one of it's two glaciers on it's North & South poles.

Unlike Earth - where there's always some water in the atmosphere and there's always the possibility of snowfall, so Earth's glaciers can always grow bigger if the seasonal temperature allows it.   Mars has no ice to add to it's glaciers.

That limits Mars' glacial expansion to CO2, and that does happen to a degree.  Mars glaciers do add CO2 during the winter and return it during the summer, but that's seasonal and Mars has only so much CO2, so it's glaciers could only expand as far as it's atmosphere would allow.

Mars' atmosphere is about 2.5 x 10^16 kg.   Mars' surface area is about 1.4 x 10^14th square meters, which means it's entire atmosphere works out to about 180 kg per square meter, which would be about 7 inches high if frozen into dry ice.   7 inches means that Mars can't form glaciers like Earth can, no matter what it's axial tilt or eccentricity variation.  It lacks the material.

If things were very different on Mars, like, lets say it had ammonia-water oceans and a thicker atmosphere and regular snow and rivers, then it might undergo glaciation periods and ice ages comparable or perhaps exceeding Earth, but it can't do that because it lacks sufficient atmospheric material.

Mars can create a thin ice cover and it can do that seasonally:

Mars' seasonal ice, winter to summer, is prone to considerable adjustment.

And, I don't want to say 100% that Mars undergoes no glacial changes.   It still might, as it goes through it's Milankovich cycles.  I'm not sure anyone knows for sure what Mars' ice caps looked like 10,000 or 100,000 years ago, so it's possible there were noticeable changes, but Mars' still lacks the material to form new enormous glaciers the way Earth can.

As a sidebar, and perhaps this could be looked into in more detail in a separate question, but Mars may undergo some kind of below the surface glaciation.  Posting article as a curious sidebar only.


Disc of the milky way For those that haven't seen this link - please have a look as it's a an awe-inspiring 360 interactive panormic image of the milky way (may need a bit of time to load though!). I've found myself idly browsing it several times and as an astronomy layman I wondered if anyone could help me navigate what I'm looking at.

While I can orientate myself and visualise where we are in the milky way, what is the distinctive black band? I'm assuming it is caused by the absence of stars in these areas. However, as I understood it, the disc (as well as the center) are the areas of highest concentration of stars. Are these 'gaps' caused by dark matter?

It's dust!

The black band is not absence of stars, but rather clouds of gas and dust — a significant component of almost all spiral galaxies$^\dagger$ — which block the light of background stars and luminous gas.

In the image, you see both individual stars, scattered all over, and the distinctive bright band of the Milky Way, most of the stars of which are so far away that they blend together. Most of the dust is in the plane of the Milky Way. The individual stars are closer to us than the bright band and the dust clouds, so what you see is the billions of stars in the Milky Way band, some of the light of which is blocked by dust, and then on top of this you see a few 1000s individual stars that are closer.

The components of galaxies

Galaxies consist of roughly 85% dark matter and 15% normal ("baryonic") matter. By far, most of the normal matter is hydrogen and helium, some of which is locked up in stars, and some of it in huge gas clouds, sometimes glowing (the pink clouds you see in the Milky Way image are probably hydrogen clouds being excited by the hard UV radiation from hot and massive stars, subsequently emitting H$\alpha$ light). A small fraction (1–2%) of the normal matter are heavier elements, by astronomers lazily referred to as "metals". Roughly 2/3 of the metals are in the gas phase, but the remaining 1/3 has depleted into dust grains, e.g. silicates and soot. This dust is mixed with the gas clouds and often become dense enough that they block light of stars.

A strong wavelength dependence of the dust's extinction properties

However, dust scatters and absorbs light with short wavelengths (such as blue and UV) much more efficiently than long wavelengths (such as red and infrared). So while it may effectively block visual and UV light, infrared light passes more or less unhindered. For instance, if you look at the molecular cloud Barnard 68 in red and near-infrared, it looks black (left in the image below), but if you look farther out in the infrared, you can actually see the background stars (right below):

Image taken from ESO.

Dark matter

Dark matter cannot be seen. It's… well, dark. It interacts with normal matter and light only through gravity. That means that if you could place a lump of dark matter in front of a star (you can't, really), you wouldn't block its light. It would pass right through. If the lump were big enough, it might gravitationally deflect the light, so the background star would look distorted to you, as would a lump of normal matter (e.g. a black hole).

$^\dagger$In contrast, the interstellar medium of elliptical galaxies tend to be much more gas- and dust-depleted.


What's the reason the Astronomical twilight doesn't have constant time? What's the reason the Astronomical twilight doesn't have constant time although its degrees below the horizon (18 degrees) are constant?

Look at the extremes. At any time of the year, it takes only minutes (several minutes) for the Sun to sink well below the horizon at the equator. On the other hand, it can take days (several days) after the September Equinox for the Sun to sink well below the horizon for locations above the Arctic Circle. As an extreme extreme, at the South Pole, it takes multiple days for the Sun to rise from just the top of the Sun showing to the full Sun showing (or to set, six months later). Astronomical twilight lasts for several weeks at the South Pole.


How can they tell no asteriods will hit earth in the next hundreds years? I have been thinking about this question :

  How can they tell no asteroids (big enough to create damage) will hit earth in the next few hundred years?

Because, so far, we were unable to get a clear and nice view of a planet of the solar system planet which is not that big, but still bigger than an asteroid. So how can they tell that nothing from the immensity of the Galaxy and Universe, bigger than a few yards, will cross our path and hit us within the next 100 years?

We all remember Chelyabinsk meteor in 2013, or whatever how they called that thing. They never saw that coming but it still caused damage and it was only few meters (± 20 meters according to that page Wiki link), Arround 1,500 injuries and ± 7200 damaged buildings... It's hard to believe that they know how to see something not so much wider than 20 metres wide outside of our solar system.

A couple of points based on some basic orbital mechanics

They don't need to get a "good" view, like the clear, crisp photos of Pluto to see one coming.   They only need to get a picture over time to calculate trajectory.   The unclear snapshots work just fine to calculate if it'll hit us or miss us.

Also, an object as far as Mars at it's closest pass to earth, a bit over 1/2 AU, would still take a few months to reach earth if it's in solar orbit.   Mostly we don't need to track anything as far out as Pluto, they can look much closer to the Earth and still have sufficient warming time.   The hard part, is tracking things that approach from the Sun side, cause those are harder to see.  That's why the Chelyabinsk meteor wasn't spotted.   it was also on the small side, smaller than NASA is currently looking for.

The good news is that, we don't get struck by things that size very often.   The Solar system is pretty enormous and pretty empty and pretty big strikes like that one are rare, like, maybe once a century.

Also, virtually all of the injuries from the Chelyabinsk meteor were from people who didn't know what to do.   If you see a big fireball in the sky, it's human nature to watch it, but use some common sense.  A space rock of that size will make a shock-wave that travels at roughly the speed of sound and the shock-wave can break windows, even knock over trees and buildings if it's big enough.   You don't want to be standing in-front of a window when the shock wave its.  Lay down next to a couch or under a table in case your building gets shaken and cover your ears.   If everyone had done that, there would have been very few injuries.   You only need to wait maybe 2 minutes or so to be on the safe side.

If you're in a car, stop, cause the shock-wave could knock down trees or debris in-front of you and stay in the car, cause that's safer than being outside.     All told, the damage to buildings was tiny compared to natural disasters like Earthquakes, floods or volcanoes which happen to us several times a year.    It's good that NASA is watching for this kind of thing, but it's also a pretty rare event.


Can someone explain a Geostationary Orbit in a simpler way? According to Wikipedia (link),  a geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit (GEO), is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation. An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day), and thus appears motionless, at a fixed position in the sky, to ground observers.

My question is then couldn't any orbit be a GEO? I mean, what about a satellite in an orbit that's say 70,000km, if the angular speed of the satellite was the same as the earth's rotation such that it always stayed in a fixed position when observed from the Earth, then isn't that a GEO orbit as well?

The centrifugal force increases with the radius of the orbit for a fixed angular speed. The gravitational pull decreases with the radius of the orbit around Earth. The geostationary orbit is, where the two forces annihilate for the geosynchronous angular speed. The gravitational pull takes the role of the centripetal force to keep the satellite on the circular orbit.

That's the principle, which keeps satellites, moons, planets on their respective orbit. If there is some additional force, they leave their former orbit.


Is the surface of Venus red hot? I was wondering, is the surface of Venus so hot that it would glow red in the dark (e.g. on the dark side of Venus)?  I am working on making a solar system simulator, and that would make for a great level if it is accurate.

The average surface temperature on Venus is 462 °C. You would need a temperature above 800 °C before you would notice any light.

Added:
Here are some calculated radiation outputs of the surface temperature of Venus, a 1000 °C black body and the sun.


How can the author of this article conclude that earth-shine on the moon is 100 times brighter than moon-shine on earth? I don't understand the reasoning of the author of this article, on page 26. How can he conclude that full earth-shine on the moon is 100 times brighter than full moon-shine on earth?

The author says that the difference between the factor of 100 and the geometric increase due to area and albedo is because of "atmospheric absorption and directional effects". The former is probably only about 20% or so (i.e. about 20% of moonlight is absorbed/scattered in the Earth's atmosphere). The latter effect could be quite big.
The reflectance of solar system objects shows an opposition surge at small phase angles.
It is difficult to see the full Moon at really small phase angles ($<2$ degrees) from Earth because you would get a lunar eclipse. However, it has been measured by spacecraft and according to Buratti et al. 1996, the brightness of the Moon increases by "more than 40%" between phase angles of just 4 degrees and zero.
When the Sun illuminates the Earth as seen from the Moon then it would be much easier to see at very small phase angles because the Moon's shadow is that much smaller.
I suspect this effect plus the ~20% absorption in the Earth's atmosphere is what the author of your reference is talking about.


If objects don't move when the universe expands, how can the expansion result in redshift? I am approaching this topic from a layman's perspective and view this as a logic puzzle, but when someone tries to get me to accept the logic, I'm never convinced by their arguments.

According to the wiki page on the expanding universe, the universe is expanding (which is how we see redshifted light from distant galaxies), however this expansion does not result in the movement of objects across space away from each other. Instead, this expansion is the formation of new space such that all of the objects in space maintain a static relationship with space.

This should mean, that as light produced by a distant light source that is expanding away from us, it is not moving away from the photon. The photon then travels across space, and this trip takes longer as a result of the expansion, but once the photon is in space near to Earth, the earth is not moving away from the light. By the time the light is at the same distance from the telescope as a flashlight held in front of the telescope, the telescope is not moving away from the photon any more than it is moving away from the photon produced by the flashlight.

So, IF the expansion is not the result of motion across space, HOW do we show redshift??????

Another comment, laymans perspective
The whole reason it can make sense that objects far enough away can expand away from us FASTER than the speed of light (which is the reason why telescopes can't see further) is that these objects are not moving across space to achieve this expansion.

Another comment in reference to the Raisin Cake.
understanding the expanding universe as a raisin cake swelling, therefore the actual cake that is in contact with any raisin stays in contact with that raisin,, the raisins do not move through the cake,,, APART. BUT, in space, the raisins actually ARE moving through the cake. Also, the PHOTON that travels across space from the distant light source to the telescope, travels across space. additionally, it's NOT useful to think of this transaction of energy in a raisin cake, because the raisins DO travel across space away from each-other as the cake swells, but this is NOT how we can possibly understand objects in space expanding away from us on Earth faster than the speed of light.

The redshift is indeed not caused by the movement of the objects, but by the expansion itself. This is a theoretical result from considering the FLRW metric for a light ray traveling on a null geodesic.

A derivation can be found here.

The cosmological redshift $z$ is given by $z+1 = 1/a$, where $a$ is the scale factor (the size of the Universe) at the time the light was emitted. Notice that this result doesn't care how the Universe expanded. In principle, you could imagine a universe which was static when the light was emitted, and static when the light is observed, but expanded by the factor $a$ at some point in between, and you'd observe the same redshift as if that universe had expanded gradually in the meantime. This is in sharp contrast to a normal Doppler shift, where in such a hypothetical case, the would be no observed redshift.

Note also that galaxies receding faster than the speed of light $c$ is no hindrance for us to see them; all galaxies with a observed redshift larger than $z\simeq1.4$ recede faster than $c$. The most distant observed galaxy GN-z11 has a redshift of $z=11.09$, and recedes at $v\simeq2.2c$. When it emitted the light we see, it receded even faster, namely at $v \simeq 4.3 c$. See the ant on a rubber rope "paradox" for an intuitive explanation.


Eyepieces and accessories recommendations for a Starblast 4.5-inch scope in a city as well as darker suburbs, and kid friendly? For the Starblast 4.5 Telescope I currently have an Orion 7-21mm Zoom eyepiece attached on top of a 2x Barlow lens. I like them because of the magnification they give stacked together and the zoom eyepiece has a good eye relief which is very kid-friendly, and the zoom function is very convenient to find things quickly by zooming in and out without changing eyepieces or focus.
Are there other eyepieces or accessory (or better ones) I can buy for better viewing of the brighter Messier star nebulae and galaxies specifically under

a) a city environment with lots of light pollution
and/or
b) a darker suburb location?

Please specify if the accessory equipment is good for a) or b), or both. Thanks!
Here are the specs for the Starblast:

Aperture: 4.5 inch aperture
Focal length: 450mm
Focal ratio: f/4 focal ratio
Came with two Explorer II 1.25 inch Kellner telescope eyepieces (17mm and 6mm)

Thank you!

The one thing I would add is an eyepiece which provides the widest possible field for your telescope.
With 1.25" eyepieces this would be an eyepiece with a focal length from 20mm to 26mm and an apparent field of view between 62° and 70°, such as:

Explore Scientific 26mm 62°
APM UFF 24mm 65°
Tele Vue Panoptic 24mm 68°
Pentax XW 20mm 70°

If you want a wider field of view then you have choose a lower focal length - 17.5mm and 76° with a Baader Morpheus, for example, and 16mm and 85° with a Masuyama. These two will give a slightly smaller true field of view.
Your zoom + 2x Barlow should cover all other situations, so I wouldn't go and get any more eyepieces right away.
Later on you might want to buy a fixed focal length eyepiece: maybe you find yourself often using the zoom at 9mm with the 2x Barlow, so you might get a dedicated 4.5mm eyepiece to use instead of the zoom for those situations.

Use the handy FoV calculator at astronomy.tools to compare eyepieces and see the visible area given by each one.
You'll see that all of the eyepieces show roughly the same real area of sky (the 24mm/68° shows the most, and a 40mm Plössl with 43° would show more still).
Where they differ is in the apparent field of view and the size of the image when you look through the eyepiece.
Your Orion zoom has a FoV of about 40° at its 21mm setting. A 24mm eyepiece with a FoV of 68° will give you a low power (about 20x magnification) view which is twice as wide as that given by your zoom.

You mention viewing under light polluted skies as well as darker skies, and the wish to see nebulae and galaxies.
Light pollution and galaxies are incompatible, unless you use image enhancement technology of some kind (night vision, live stacking, or full-on astrophotography).
Except for the biggest and brightest galaxy - Andromeda, M31 (the core of which should still be visible beneath all but the worst inner-city skies) - your search for galaxies will be a frustrating one.
Even under dark skies, seeing galaxies with the naked eye is not easy, requiring the use of averted vision and memory in order to see any detail.

Some nebulae are visible from the city. The Orion nebula, M42, is very bright and always worth a look, no matter where you are or how many times you've seen it before. In the summer the Triffid nebula can be seen in the southern skies, and the Swan (aka Omega) nebula comes out with the help of a filter.
There are also some bright planetary nebulae which you can see from the city. The Saturn nebula comes to mind as one I've seen under light pollution.
Generally, though, the best sights for the city astronomer are:

The inner planets and Jupiter and Saturn
The moon
Bright star clusters
Bright double stars and carbon stars

(Disclaimer: my urban skies could be a lot worse, and my telescope could be a lot smaller, so what I can see won't necessarily be the same as what you can see)

Also, you could get yourself a couple of books. A paper star atlas is a nice supplement to an electronic one, especially for those trips to darker skies.
Turn Left At Orion is a classic book which is frequently recommended. Nightwatch is another.
Get a red light as well, to preserve your night vision in the dark.


The Moon drifting away from the Earth wiki says that average distance from earth increases 3.8 cm per year, can you tell how much of that is due to the expansion of the universe?

wiki also says:

  Through the action of tidal forces, angular momentum is slowly being
  transferred from Earth's rotation to the Moon's orbit. The result is
  that Earth's rate of spin is imperceptibly decreasing (at a rate of
  2.3 milliseconds/century),[20] and the lunar orbit is gradually expanding.

Can you give some details about this process, is the orbit changing shape, eccentricity? is the average orbital speed decreasing, by what value? is the spin angular velocity increasing, by what value?

If I did not misunderstand it, the current view on tides says that tidal bulges exert a torque on the moon, and L is tranferred to it, if that is correct, that L being transferred to its spin is increasing or decreasing it? Can you clarify also how a torque on the spin can produce a change of the orbit, which only implies a change (a decrease, in fact) of its orbital speed and not of the spinning speed? Can the spin (L) influence the orbital L, and how?

edit

I asked specific questions, which are not asked or answered there. In particular I am asking: if as they say the tidal/equatorial bulge increases spin angular momentum, how is this transferresin its turn to the orbital angulat momentum? Or, alternatively
can gravity directly increase the orbital momentum of a satellite?
Actually the answer, besides being incomplete, may be wrong because momentum/KE is not added , but subtracted to the moon, since its orbit gets larger. There can be no transfer, since both earth and moon are slowing down
torque is not correctly explained

Can you give some details about this process, is the orbit changing
  shape, eccentricity?

This isn't a simple answer.   Also, do you want to know this for a simple 2 body system or the more complex, essentially 3 body system of Sun-earth-Moon.

If the Earth and Moon were alone in the cosmos, far away from everything else, then over time, the tidal forces would circularize the Moon's orbit as it pushes the Moon away.  The reason for this is that the same force that slowly pushes the Moon away from the Earth pushes hardest when the Moon is closest, and least hard when the Moon is furthest away, so over a long period of time that circularizes it's orbit.  All 2 body systems where the moon is being pushed away tends to circularize.  This can be observed in our own solar-system.   Two body systems that are very far away from orbital disturbances, like Pluto-Charon or Neptune-Triton or even, Sun-Neptune, are highly circular orbits.

But when you bring the Sun into the picture, everything changes.  The Moon orbits the Earth and the Earth Moon system orbits the sun, and in a very real sense, the Moon also orbits the sun, but it's within the Earth's stable region of it's hill sphere, so the Moon's orbit around the Earth will remain stable for billions of years.   But while it's stable, it's measurably inconstant.  The moon's orbit wobbles a great deal, being stretched and contracted based on it's position in relation to the Earth and Sun.  3 body mechanics is quite complex, even with stable orbits.  Isaac Newton was aware of this but he wasn't able to work it out.  Mathematicians who came later were able to figure out orbital interactions with more than 2 bodies, called perturbation theory.

So, your question depends on whether you mean a 2 body system or a 3 body system.

The Earth's eccentricity, for example, changes on roughly a 100,000 year cycle, mostly due to Jupiter, but that effect is pretty slow, on a roughly 100,000 year cycle.

The Moon's apogee and perigee happen every 27.3 days (Sidereal orbit) and it's eccentricity changes measurably every orbit.  In a sense, it's not very accurate to say the Moon even orbits around the Earth in an ellipse, it's too squashed and stretched for the elliptical model to be accurate.

While the Earth takes about 100,000 orbits to go through one elliptical period, the Moon needs only 6 or 7 orbits to do that.   Note how measurably it's perigee and apogee change every orbit.   See table.

There's a fair bit of information available on the Moon's irregular orbit, but I suspect that has almost nothing to do with your primary question.   In general, in the 3 body system, as the Moon moves away from the Earth it's orbit grows more erratic, so in a sense, because of the sun, the Moon's orbit grows more eccentric, over time, where, if the sun wasn't there, it would grow more circular.

  is the average orbital speed decreasing, by what value? is the spin
  angular velocity increasing, by what value?

On average, as the Moon moves away it's orbital speed decreases.  More distant orbits have less kinetic energy, but again, with a 3 body system it's more complex.

OK, I think I see what you're asking on Spin.

  from Wiki "angular momentum is slowly being transferred from Earth's
  rotation to the Moon's orbit."

and

  There can be no transfer, since both earth and moon are slowing down

You should look at the Earth-Moon as a closed system.  For a closed system, angular momentum remains constant, but angular momentum in this sense, is around the barycenter.   The Moon's angular momentum, in the sense that Wikipedia means isn't the Moon's spin but the Moon's orbit.  As the Moon moves away from the Earth, it's velocity decreases but it's angular momentum around the barycenter increases because angular momentum is a product of both velocity and distance from the center.

Jupiter, for example, has about 71% of the total angular momentum of the solar system.   See article, which I think is related to your question.

Finally,

  wiki says that average distance from earth increases 3.8 cm per year,
  can you tell how much of that is due to the expansion of the universe?

Expansion is best understood if objects aren't orbiting around each other.  If the Moon was stationary and the Earth was stationary, and the only way to do that in our thought experiment is to give one of them a mass of zero so there's no orbit and no gravitational attraction.   We're breaking almost every rule in the book with this thought experiment, but bear with me.

If we take the measured expansion of space to be 74.3 km/s over 3 million light years.   Source.

Average lunar distance 384,400 km.  And, 9.461 trillion km in 1 light year, so about 2.8 x 10^19 km in 3 million lightyears.  Roughly 74 trillion lunar distances. which works out kind of neatly, 74.3 km/s divided by 74 trillion gives us one trillionth of a KM/s, 31.5 million seconds in a year gives us about .0000315 km/year expansion between a stationary, no gravitational attraction moon and Earth, which equals about 3.15 cm per year of expansion of space over 1 lunar distance, which is curiously not far off from 3.8 cm per year increase in distance due to tidal forces, but the similarity is purely coincidental as well as irrelevant.

(too long?)


Galaxy rotation curve and dark matter I am reading "The Essential Cosmic Perspective" by Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit.
In Chapter 14, it is stated that an evidence of the presence of dark matter in our galaxy is that the rotational curve does not match that obtained from calculation.
In the calculation, the following formula is used
$$M_r=\frac{rv^2}{G}$$
where $M_r$ is the mass enclosed inside a radius $r$ from the galactic center.

My understanding is that the above equation is a result of the shell theorem. But we know that shell theorem applies to spherically symmetric mass distribution only, and most galaxies are disks. So why can we still do this?

We can't. That is an over simplification only used in elementary treatments simply to provide the the flavour of the argument. If you see it done somewhere in the refereed literature, it is probably incorrect. Of course it may be true that the mass is almost spherically symmetric, especially if it is dominated by a spherically symmetric dark matter component, but that the visible light is not. Or, it may be true that at large distances from even an asymmetric mass distribution, a Keplerian potential is a good approximation for the orbits of distant objects (e.g. satellite galaxies to the Milky Way). A measured galaxy rotation curve makes no such assumption, it is merely a measurement of rotation speed as a function of radius. It is only the interpretation and modelling that needs to deal with the mass distribution.

The real situation is much more complex. See for example http://ned.ipac.caltech.edu/level5/March01/Battaner/revision.html

However, even if one assumed all the mass was concentrated into a disk-like shape, the only way you can get flat rotation curves is to assume that the mass in the disk does not "follow the light" - that the mass-to-luminosity ratio increases vastly with radius - which is essentially still saying that you have "dark matter", just in a disk.


When will we be able to detect Earth-sized exoplanets? I have heard that the only planets able to be seen outside the Solar system are Jovian-sized planets with the occasional detection of planets three times the Earth's size. But, as far as I know, we haven't seen any Earth-sized (due to the range of our telescopes?) extrasolar planets. When will we be able to detect them?

In 2013, the smallest detected exoplanet was Kepler-37-b, which is actually smaller in mass and size than Earth, so we already had a limited ability to detect these size exoplanets.

It is worth noting that there wasn't any new technology that allowed this advance, the paper linked to by NASA indicates that the same methods were used as they would usually use. Making it remarkable that they were even able to discover something this small.

Now, even more Earth-sized planets have been detected. One method used to aid in their detection is described as follows:

  We account for Kepler's imperfect detectability of such planets by injecting synthetic planet-caused dimmings into the Kepler brightness measurements and recording the fraction detected.

Sources:

Kepler-37 - NASA
Prevalence of Earth-size planets orbiting Sun-like stars - Petigura, Howard, and Marcy