Is there a connection between black holes and dark matter/energy? Is there a spatial or gravitational relationship between black holes and dark matter/energy?

No, not really.

As Stan Liou hints at in his comment, massive objects such as black holes (BHs) were previously a candidate to dark matter (DM). But it can be shown that in order to explain the various problems that the existence of DM needs to solve, there would have to be so many, that their gravitational effect ("microlensing") on background sources would be much larger than what is observed. Today, most people think that DM is simply a particles with an extremely small interaction cross section, such that it virtually doesn't interact with other particles, except through gravitational attraction.

You also ask if there's any spatial relationship, and in a way you could say there is. Most, if not all, galaxies sport a supermassive BH, and these tend to lie in the center of the gravitational potential wells created by the galaxies' DM. "Normal" BHs, however, just float around like stars.

BHs also tend to grow by accretion of matter onto them, and of course they can also eat DM, so to some extend BHs also consist of DM. But since the DM doesn't interact, a DM particle would need to fly directly toward the BH, lest it would simply be deflected and escape. Normal matter, on the other hand, is slowed down by friction and spirals toward the BH. Thus, BHs consist mostly of normal matter. It doesn't really matter, though, since after something has fallen into a BH, everything but its mass, charge, and angular momentum will be forgotten.

Dark energy is something completely different, better described as a negative pressure which is a property of space itself. I think. Or we think. Nobody really knows. But it presence doesn't correlate with that of BHs or DM. Ten years ago, there was a hypothesis that a collapsing star should be converted into dark energy rather than a BH, causing a so-called dark-energy star, but I don't think anybody believes this anymore.

No there is no known particular connection between black holes, dark matter and dark energy.


Can matter fly directly into a black hole and avoid the accretion disc? From what my non-scientifically-trained mind understands, matter typically enters a black hole via getting caught up in the accretion disc going faster and closer to the black hole until it passes the event horizon, the point of no return. Using Earth as an analogy for a black hole, I assume the accretion disc would orbit around the equator, similar to Saturn's rings, and the matter might eventually crash into Brazil or somewhere along the equator?

My main question is, can matter simply do a straight shot into the black hole at a higher latitude, say straight into New York City, avoiding the disc altogether?

As you can probably tell by my simplistic analogy, I am a complete layperson. While fancy jargon and artistic looking equations might be expected (which is fine), please also dumb it down to an average intelligent business administration major's level of scientific understanding.

In your analogy: A satellite stays in orbit because it has angular momentum. This always balances the force of gravity, thus when being in the satellite (or in the ISS) it seems as if you're weightless, because force balance. This is very important to understand, because it is the reason that in space all orbital motions can -in principle- go on forever. Because angular momentum is conserved.

Now in order to crash into the Earth/BH you need to have very low angular momentum (which is equivalent to having only linear momentum straight down into the BH) or you have to loose the angular momentum somehow.
Near Earth, this happens to the ISS, because it has a low-earth-orbit between 350-450 km and at this height there are still enough atmospheric particles around to produce drag.

Around black holes it is a very similar thing.
I will define the Schwarzschild-radius $R_s = \frac{GM}{c^2}$ as a relevant length-scale. Then, as long as we are outside $3 R_s$ distance from the BH, we need, as on Earth, friction to dissipate our angular momentum. In the case of accretion discs that form around BH's this can be accomplished through turbulent dissipation as laid out in the famous paper by Shakura and Sunyaev.

Then, only very close to the BH, namely at a distance $< 3R_s$, space-time distortions help us to fall into the BH without the need for any additional source of friction. As you see, this is a region very close to the Black Hole and it is misrepresented in popular accounts of BHs that all matter orbiting it will magically fall into it.

I suppose its certainly possible, however it's highly unlikely. First of all, you'd need your matter to fall in directly, meaning it would have to be traveling directly at the BH. Any motion slightly to the side, and it will start orbiting and spiral into the black hole. That's a pretty small target so that would already make this scenario not very plausible.

However, your main problem is going to be the fact that (1) black holes have magnetic fields usually and (2) black holes also usually spin. What this means is that anything trying to get into the black hole will tend to get swept up and taken along with the rotating magnetic field, causing it to start spiraling in. You could of course say that your infalling material is neutral and thus doesn't interact with the field, but there's no way it will stay neutral all the way in. At some point friction from falling in will make it a plasma and it will then be "bound" to the magnetic field. If you're knowledgeable about your magnetic fields, you may suggest making it so this matter could simply fall into the magnetic poles as it wouldn't then wouldn't be subjected to being dragged by the magnetic field. Unfortunately, the poles tend to be the regions where material escapes from the BH so you're unlikely to get infalling material through that.

In short, it might be possible to contrive some scenario where matter is just "dropped" onto the black hole but the circumstances for such an event to occur are exceedingly rare and will likely be ruined if the black hole is either spinning or has a magnetic field (which almost all do).


Is it possible to know if you are travelling close to the speed of light if you have nothing to compare your speed with? This is a question that has been bugging me for some time. Let's see if I am able to make me understood!

Imagine you are travelling in a spaceship at 99.9999999999% the speed of light and you are somehow within a spaceship without windows.

My question is, is there ANY kind of experiment that would show you any time/space dilation effect as compared to being stationary? Is there ANY lab or thought experiment you could make or think that would tell you there are weird things happening with time and space outside your spaceship? I mean, would physics work differently than they would at home? Would radioactive atoms take longer to decay in a way you could measure and compare with your earth-based decay tables? Anything?

Or the only way to find out would be simply to step out of the spaceship and find out that life on Earth has gone extinct in the last couple of hours?

And another question related to the first. If you were travelling in the spaceship at that speed in the X-axis direction and you handthrew a baseball in that same direction, would that ball's mass or inertia explode asyntotically to practically infinity since you are increasing a fantastically high speed a tiny winy bit more? Or perhaps you could not threw the ball at all because of the huge mass increase being that close already to the speed of light? Or you wouldn't notice nothing special and a regular sized with normal weight baseball would impact the other side of the room at a relative speed to you of 60 km/h?

Velocity is relative. From your perspective, your velocity is zero. Without windows (or sensors) on your spaceship, you cannot know the difference. To measure your velocity, you have to define a reference point, eg a star, the average velocity of surounding stars or the galaxy center. If you have such a huge velocity relative to them, from you perspective, they are this fast and not you. So you can observe the time and space dilation of these objects, while your time appears to be normal. You ship is an inertial reference frame.

As for throwing the baseball: From your perspective, you have the speed zero, and the baseball eg. the speed 0.5c. A "fixed" observer, which observes you at 0.99c will measure the balls velocity at 0.997c. With those relative effects, speeds don't add up, only kinetic energy does. http://curious.astro.cornell.edu/about-us/139-physics/the-theory-of-relativity/special-relativity/1016-why-can-t-relative-velocities-add-up-to-more-than-the-speed-of-light-intermediate The increased mass is only a relativistic observation, resulting from the kinetic energy (relative to the observer).

First of all, common sense should tell you you’re not travelling at that speed; only cosmic rays and accelerated particles can reach such speeds. The incredible amount of energy needed to boost a spaceship to such a speed, far more than is produced by the largest multi-megaton hydrogen bomb, is one of the reasons it couldn’t happen. But let’s assume for the sake of argument that your spaceship has managed to reach this speed, there are ways and means to see what is ahead without having windows. All the photons coming toward you would be dramatically blue-shifted, while those coming from behind would be dramatically redshifted.

Space is not a perfect vacuum, even in the most transparent regions there is a proton every few metres. At speeds close to the speed of light your spaceship would be subject to a headwind of fast protons, and you would feel the walls of the spaceship heating up. If you had the misfortune to run into a cloud of gas or dust, your spaceship would rapidly burn up, and your problems would be over.

Assuming there was no gas or dust in your path and space was apparently perfectly transparent, I still wouldn’t advise stepping out of your spaceship; the proton storm and other radiations would kill you. Only objects which didn’t share your motion would have relativistic mass increase, so the ball you threw would behave as though you had thrown it in the international space station. Inside the spaceship, you would not notice any change in the decay rates of radioactive elements. The spaceship itself, when viewed by an outside observer who saw himself as at rest, would have acquired a colossal amount of relativistic mass increase because both you and the spaceship are now made of relativistic particles. This mass increase couldn’t come from nowhere, it would have to come from the energy provided by the fuel used, and no such fuel exists.


Tidal locking of a moon When a moon is in orbit around it's planet but is so close that it will become tidally locked does it experience a large 'wobble' in between the phase when it still has its own rotation and the final phase when it is always showing the same face to the planet?  I can only imagine that it would 'wobble' very widely, back and forth like a pendulum until finally settling into a locked position.

There's no back and forth motion on the way towards tidal locking.   There's kind of two ways to look this kind of "settling", there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting.    I realize that's just re-writing your question, but it helps to understand the 2nd option.

I'll use the Earth as an example, because the Earth is in the gradual process of becoming tidally locked to the Moon.  It's the same process, but easier to visualize.

Because the Earth rotates faster than the Moon orbits, the tidal bulge moves ahead of the moon.   This is called the tidal bulge offset.

Source.

It's the size of the bulge and the angle of the offset (and the mass and distance of the Moon) that causes the familiar gravitational effects, the Earth's rotation slowing down and the Moon moving away from the Earth.

Source

The current angle of offset for the Earth's tidal bulge to the Moon is about 3 degrees, often exaggerated in diagrams so it's easier to see.  (I should find a source for that - I've not been able to find a good one).

From the Moon's point of view (and if we assume a circular orbit), the tidal offset stays in the same place.  It's a factor of the relative rotation speed.   As Earth slows down, the angle of offset will move towards zero.    At tidal locking, the angle of offset becomes zero and there's no force to continue in the other direction.    In a simple mathematical model, tidal locking wouldn't ever happen completely, it would just gradually get closer and closer, like an object shot from the Earth at exactly escape velocity.

In reality, because objects aren't perfectly symmetrical and because orbits aren't perfectly circular, the "take forever to get to zero", doesn't happen.  But it's still a gradual approach to zero and no overshooting and returning because the force on the rotation approaches zero as the angle of offset approaches zero and the force needed to change a planet's rotation is enormous and there's essentially zero force to make that happen.

With more eccentric orbits, like Mercury and it's 3:2 rotation to orbit resonance, that might be different.   Not sure how eccentricity might play into this.   That's more complicated.

For an animated lesson, see Isaac Author's amazing space engineering anthology, including TIDAL LOCKING EXPLAINED.


How much volume of gases could be added to Mars atmosphere that the planet gravity can retain? This article Mars atmosphere states that

  Today Mars atmosphere is very thin, but it was not always like this. Around 3.5 billion years ago, the red planet had a thick enough atmosphere for liquid water to run smoothly on its surface

Suppose you could increase Mars atmosphere volume by melting frozen CO2 or whatever. How much volume of gases could be added to Mars atmosphere that the planet gravity can retain?

There are hard limits on what types of gas that Mars can retain based on its temperature and mass
(Graph of what gasses an astronomical body can retain).

Volume wise, it's not clear. Currently Mars is still losing its atmosphere, so it can't even retain that amount of atmosphere. But if you continually added gas to Mars, there isn't an end point where the planet could no longer hold the atmosphere being added. It just increases the pressure of the atmosphere at the surface and only slightly extends the height of the atmosphere. And after it stopped being added, Mars would slowly lose the atmosphere again.

Terraforming Mars by melting the frozen  CO2 we know is stored at the poles, mainly the South Pole, is a much more difficult enterprise than some people imagine and I doubt it will ever be done.

If by miraculous means you could restore the Martian atmosphere to what it was 4 billion years ago, it might last a couple of billion years.

You don't stipulate any particular length of time you want it to last, and a billion years is beyond most people's comprehension. They talk earnestly about how humans could avoid destruction when the Sun expands to its red giant phase in about five billion years time, not realising that man will have gone to join the dinosaurs in far less than one billion, so two billon is an almost unimaginably long time.

But the fact of the matter is that Mars is a very hostile place, and we will have to put up with all the difficulty and inconvenience of its thin atmosphere and bitter temperature for as long as we are there.

To look on the bright side, things won't be so bad once we have established a large heated and pressurised base there, but to go outside the base you will always need a pressure suit and space helmet.


Could some bacteria theoretically evolve to survive in space? There is no edge to the atmosphere, and therefor I expect there would be a gradual reduction of concentrations of bacteria as we ascend into space. Logically anaerobic bacteria should be able to obtain its energy from the sun and nutrients from whatever scarce but available organic floating materials. Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them. Could it be that bacteria is evolving/ has already evolved at the edge of the atmosphere to live in space?

Currently there is no life that we know of that evolved in, or can survive for prolonged time in a vacuum.

To say 'Life in space is possible because there is anaerobic life' is a completely wrong point, that ignores the effects of vacuum on life.

Sure, there are extremophiles that can survive for a limited time in space (see also this list) when going into hibernation modes. However all life as we know it needs liquids and/or gases to survive. Those would evaporate once the cell/water bear would become active again. Activity is necessary to sustain and multiply life, so no, nothing that we know of can stay active and alive in space.

  Cosmic rays and high energy particles could damage DNA, but there might be someway life could protect itself against them

Until life comes up with such mechanisms, we just don't know. The problems with liquids and gases evaporating would need some tougher solutions even.

Bacteria are also pretty heavy 'particles' and therefore cannot float anywhere near the 'edge of space'. There are reports of bacteria in the stratosphere, but that requires already strong upwinds to put them there.

Given that we've found bacteria thriving inside "solid" rock a couple km straight down, living on local sulfur and heat, there's no overriding reason that something couldn't evolve on some rock (planet) of sufficient size to hold some heat in, but without any need for an atmosphere.

If you were specifically asking about free-floating life that survives entirely on solar irradiance, then no, but more because of a complete lack of various elements (let alone long organic chains) in most of space.   Veering off into world-building, imagine my intra-rock bacteria slowly growing to the surface, forming a protective layer of dead relatives which absorbs irradiance but seals against vaccuum,  and sometime later a volcanic eruption launches the entire magillah into the vacuum (and the protective layer manages to envelope the entire colony).  Highly improbable.  You'll need The Heart Of Gold here.


What is the temperature inside a Black Hole? Is it insanely hot? Cause of so much gravity?
Is it close to absolute zero? Cause the matter is so closely packed, there is hardly any space for particles to move?
Is it room temperature? Cause Cooper didn't die in Interstellar when he went into a Black Hole?

The temperature of a black hole as seen from the outside is the temperature of the Hawking radiation, which @jyoti has discussed.

The temperature you would experience as you fell freely through the event horizon, is a different question. I think the answer depends on what else is falling into the black hole at the same time. From its own viewpoint, everything that falls into a black hole reaches the singularity in a fixed maximum time (a few days for the very largest supermassive black holes) so you could only exchange energy with things that fell in around the same time as you did. If there was very little such stuff, then I don't think you'd notice anything out of the ordinary -- you'd be exposed to the cold of space, just as you were before (assuming the black hole is big enough that you haven't been heated up the tides of the black hole distorting you. On the other hand if lots of stuff was falling into the black hole with you then the whole lot would get churned up, and you'd be falling in as part of a mass of very hot plasma. Eventually, at least in a non-spinning black hole, you'd approach the singularity and be ripped apart by tides.

What is the temperature inside a Black Hole?

Absolute zero. That's because of the infinite gravitational time dilation. Temperature is a measure of motion of sorts. Gravitational time dilation means things are moving slower. When the gravitational time dilation goes infinite there is no motion, and so no temperature. This is why the black hole was originally known as the frozen star. See the 1939 paper by Robert Oppenheimer and Hartland Snyder on continued gravitational contraction. People often say this is outdated, see for example Brown’s mathpages article the formation and growth of black holes, but I don't think it is. I say that because I've read the Einstein digital papers, and I like to think I know how gravity works.

Regardless of that, in 1972 Stephen Hawking, Brandon Carter, and Jim Bardeen wrote a paper and said "the effective temperature of a black hole is absolute zero”. Robert Wald said much the same in black hole physics: “κ has nothing whatever to do with the physical temperature of a black hole, which is absolute zero by any reasonable criteria”.  Note that Hawking radiation is nothing to do with the temperature inside a black hole.

  Is it insanely hot? Cause of so much gravity?

No. Or should I say I don't think so. I don't actually know for certain.

  Is it close to absolute zero? Cause the matter is so closely packed, there is hardly any space for particles to move?

Yes AFAIK it's close to absolute zero. Or it's actually absolute zero. But not because matter is so closely packed. In fact there might not even be any particles inside a black hole.

  Is it room temperature? Cause Cooper didn't die in Interstellar when he went into a Black Hole?

No, it isn't room temperature. Interstellar is just science fiction I'm afraid, even though it was portrayed as science fact.


What is the difference between asteroids, comets and meteors? Some celestial objects seem to be referred to as asteroids, some as comets, and some as meteors or meteorites.

What is the distinction between all of these different objects? Are any of them the same?

The objects you are refering to are actually two different objects: asteroids and comets. Meteor and meteorite are other names for an asteroid, at a given time of its interaction with our planet. We'll get to that.
So first, what is the difference between an asteroid and a comet?
A comet is a small solar system body that display a "coma" (an atmosphere of a sort) and sometimes a tail passing close to the Sun. They are mostly made of ice and dust, as well as some small rocky particles. We distinguish two kind of comets, with short or long orbital period. The short orbital period ones originated from the Kuiper Belt, a region composed of small bodies beyond the orbit of Neptune. The long orbital period ones originated from the Oort cloud, a scattered disk of icy planetesimals and small bodies laying around our solar system.
An asteroid is a small body, composed mostly of rocks and metals. In our solar system, they can be originated from the asteroid belt, laying between Mars and Jupiter, or from the orbit of Jupiter (the Jupiter Trojans), or from actually almost everywhere in the Solar System.
A small asteroid ('meteoroid') that enters the Earth's atmosphere becomes a meteor, what we also call a "shooting star".
Eventually, a meteor that was massive enough not to be completely distroyed entering the Earth's atmosphere and hitting the ground is a meteorite.

Asteroids and Meteoroids are very different in size. Asteroids are huge usually about few kilometers big.

Whereas Meteoroids can be so small from centimeters to about few meters wide.

Both are very different in size. Sometimes, when two asteroids collide they might throw away few percentage of their rock which falls into planets as meteorites.


How often are supernovas visible by the naked eye from Earth? I heard about the 1054 supernova that was visible with the naked eye.

Of course, these events happen at random. But just like with radioactive decay, it must be possible to determine statistics on these kinds of events.
Wikipedia says

  The total supernova rate in our galaxy is estimated to be between 2 and 12 per century

but this doesn't indicate anything about their visibility from Earth. In fact, quite a few supernovas in the Milky Way probably go unnoticed, so the rate of visible supernovas is likely lower.

In addition, type Ia supernovas can outshine the galaxy they are in, so if one occurred in Andromeda or a Magellanic Cloud, it might be visible. So there possibly could be more visible supernovas than there are supernovas in just our Galaxy.

What is the average time between two supernovas that are visible with the naked eye?

One way to answer this is to look at how often naked-eye supernovae were seen in the past. In the last 2000 years, there have been (at least) eight such supernovae recorded: in 185, 393, 1006, 1054, 1181, 1572, 1604, and 1987. Seven of these were in our galaxy, while the last (SN 1987A) was in the Large Magellanic Cloud.
It's possible that supernovae in the far southern sky would not have been visible to Chinese astronomers (the most reliable and comprehensive records we have), so some additional supernovae visible from the far south may have been missed. We could round the total up to ten, which gives us a rate of about one every 200 years. If you want to worry about some being visible but not recorded (e.g., because they happened during a particularly chaotic time in China), then perhaps the rate would be a little higher -- maybe one every 150 years?

This event is very rare and occurs every few 100 years in our galaxy, which makes it unsure if you could even see 1 in your lifespan. you can see how it would look from Earth and read a little about such events here.
Also, there is going to be another supernovae probably visible from naked eye in 2022, which the above link is talking about. Since such a supernovae happens so rarely, that I wasn't even able to find a single image of one, perhaps one happened most recently in the 1600's. This link talks about how you could locate the 2022 supernovae. Hope this helps.


Do interstellar asteroids decelerate and eventually stop? I do wonder, do interstellar asteroids eventually stop at one point in space after they gradually decelerate (or) even do they decelerate?

Though there is no air like on earth and thus asteroids will not be affected by frictional forces, do they have friction with gravitational forces against their trajectories?

If it is so, they might eventually stop. Are there any known objects like that?

An object moving through the interstellar medium will experience a weak drag force. If the drag force is $f(v)$ the velocity will decline as $dv/dt=-f(v)/M$ where $M$ is the mass. In fluids $f(v)\propto v^2$; this appears to work even for low density gas. Note that this differential equation leads to a velocity that declines forever, slowly approaching zero but never reaching it.

We can make a simple ballpark estimate of the timescale of slowdown by estimating how long it takes for it to run into its own mass of interstellar material. If it has radius $r$ it will encounter $\pi r^2 v \rho$ kg of ISM per second. So for mass $M$ the timescale of slowdown is $\tau \sim M/\pi r^2 v \rho$.

If we take a $r=1$ km spherical asteroid of mass $M= 8.8802\cdot 10^{12}$ kg (assuming density 2.12 g/cm$^3$) moving at 26.33 km/s and use $\rho=10^{-20}$ kg/m$^3$ (this varies a lot) then the timescale is about $10^{22}$ seconds, or 340 trillion years. So after maybe a quadrillion years the asteroid would be nearly at rest relative to the local gas if nothing disturbed its trajectory...

However, it is fairly likely for the asteroid to encounter a star during this time. Stellar density is about $\rho_*=0.14$ per cubic parsec, and the timescale for getting within 100 AU from a star is $\sim 1/\pi (100 AU)^2 v \rho_* \approx 10^{11}$years. This is likely to act as a gravity assist giving it some of the star's relative velocity (on the order of kilometers per second). So the true answer is that the asteroid will never settle down as long as there are stars in the galaxy.

The asteroids can be accelerated and de-accelerated depending on which side of a much larger rotating object the asteroid passes.

If asteroid passes on the side in the direction of the rotation of the object, the asteroid experiences a gravity boost.

If the asteroid passes on the side opposite to the direction of rotation, the asteroid experiences gravity breaking.

If the breaking slows down the asteroid sufficiently so it can be captured then it collides with object.

An example of gravity assisted flight, the Mars probes used the gravity assist of the Earth to reach Venus, then the gravity assist of Venus to fling it towards Mar. Landing on Mars used gravity breaking.

But for the Mars lander to land on Mars required fuel and a parachute.

Gravity assisted acceleration and de-acceleration is accomplished by exchanging angular momentum with a larger rotating object and is based on conservation of angular momentum.


Why does gravity increase in star formation? When a star ignites ( ie. fusion starts ), the star maintains its form by balancing gravity's inward pressure, and radiation's outward pressure.

I get that the fusion of hydrogen atoms releases energy... fine...

How does gravity keep it together if the mass is lessening as a result of fusion( mass being converted into energy from fusion) while gravity is weakening( as mass lessens )?

Wouldn't the radiation overpower the force of gravity and tear the star apart?

The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium.

This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way.

Radiation pressure is a feature in stellar evolution calculations. It is almost negligible in the solar interior (at the 1% level compared with thermal pressure). However, it does become more important in more massive stars with hotter interiors and higher luminosities.

As a star runs out of hydrogen fuel, the fusion slows, causing the gravity to overpower the outward force of pressure, thus contracting. Contraction of the star causes high temperature and pressure, to the extent that it is enough to fuse helium into carbon, then the energy released is stronger than the gravity, increasing the size of the star into a red giant. The following paragraph from an article explains this:

  Over its life, the outward pressure of fusion has balanced against the
  inward pressure of gravity. Once the fusion stops, gravity takes the
  lead and compresses the star smaller and tighter. Temperatures
  increase with the contraction, eventually reaching levels where helium
  is able to fuse into carbon. Depending on the mass of the star, the
  helium burning might be gradual or might begin with an explosive
  flash. The energy produced by the helium fusion causes the star to
  expand outward to many times its original size.

The amount of mass lost is more due to stellar wind, rather than to fusion. To answer your second question, the pressure will never overcome completely the force of gravity. When a star reaches the iron nickel stage of fusing, it'll stop, unable to go further. This causes a rapid contraction for either a star to go supernova (which is really tearing most of the star apart except for its core), or cool down to a black dwarf.


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.

There is nothing special about the point in space where the galaxies move out of view of each other. This is just the point where their respective Hubble spheres (the sphere at which the expansion speed is equal to the speed of light for the given observer) overlap. The speed of the light will still be c from the perspective of an observer between the Hubble spheres of A and B


How are galactic rotation curves determined from radial velocity measurements? I have been reading about the deviation of galactic rotation curves from the predicted Keplerian curve to a flat curve at large galactocentric distances.

They were the first indicators of the existence of dark matter but I do not understand how the curve is obtained from Doppler shift data (as mentioned by some papers as the method of determination for the curve.)

Doppler shift data would allow measurement of radial velocity but how can that alone be enough to extrapolate the rotation curve?

Imagine a spiral galaxy where the rotation of gas and stars within the disk is close to circular (generally true for gas clouds, though less true for stars) in a single plane (the disk), and imagine that it is tilted at some intermediate angle with respect to your line of sight. A circular disk galaxy will then project to form an apparent ellipse, with a major axis and a minor axis (see figure).

If you look at points which lie along the minor axis (say, points b1 or b2), you will see stars and clouds which are moving perpendicular to your line of sight, so they have zero radial velocity relative to you, and there will be no Doppler shift.

If, instead, you look at points which lie along the major axis (say, points a1 or a2), your line of sight will intersect gas clouds and stars which are moving (mostly) along your line of sight (either towards you or away from you), so you will observe the maximum possible radial velocity and thus the maximum possible Doppler shift (either positive = velocities away from you, or negative = velocities toward you).

There's one further factor, which is the effect of inclination. If the galaxy is genuinely face-on (inclination $i = 0$), then you won't seem any Doppler shifts, because all the motion is in the plane of the sky, perpendicular to your line of sight. (In practice there will really be some motions up and down with respect to the galaxy plane, but this won't tell you about the rotation curve.) If the galaxy is edge-on, then you get the maximum possible Doppler shift, which gives you the rotation velocity $V$ directly. If the galaxy has an in-between inclination, then the radial velocity is the part of the cloud/star velocity in the direction of your line of sight, which is $V \sin i$. So if you can determine the galaxy's inclination, you can divide the observed velocity along the major axis by $\sin i$ to get $V$.

Up until relatively recently, the practice (in optical astronomy) was to use a long-slit spectrograph and orient it so that the slit lies along the galaxy's major axis. Then at each point along the major axis, from the center outward, you measure the Doppler shift, apply the $\sin i$ inclination correction, and then you have the rotation curve: $V$ as a function of radius $r$ along the major axis.

If you have fully two-dimensional spectroscopy (e.g., from a radio telescope interferometric array observing emission from atomic hydrogen or molecular gas, or an optical or infrared integral field spectrograph), then you can use the data at intermediate locations within the disk (not just along the major axis), where the observed velocity is $= V \sin i \cos \alpha$, where $\alpha$ is the angle between the major axis and the location you're looking at (so for locations on the major axis, the observed velocity is just $V \sin i$). Then you can fit a model to the set of observed velocities (which you get from the Doppler shifts) and work out $V$ as a function of radius. (The standard approach is called a "tilted ring" model, where you assume the velocity is a simple function of radius, but allow for the possibility that the disk might be warped, so that the orientation -- including the inclination -- might change with radius.)

What do you mean by rotation curve? The orbit?

You can derive the theoretical tangential velocity of a body 2 on a circular orbit around body 1. For that you only need the mass of body 1 and the distance. If you know the mass of the galaxy core, you can check if the measured velocities match with it.

The radial velocity of a arm of a galaxy will tell you exactly the tangential velocity to the galaxy core if you look from the side of it, for example.


How well can the Sun's surface be resolved by an observatory? There's no lack of photons, how well could a telescope resolve Sun spots and other surface features of the Sun? How would a high resolution Solar (space) telescope differ from for example Hubble which avoids the Sun by a 50° angle or so?

Solar telescopes typically have filters to reduce the amount of light reaching their detector. Depending on their purpose they may use, for example, neutral density filters to reduce the amount of light to a level where they can resolve details without blocking any of the visible spectrum, hydrogen alpha filters to observe the Sun at a wavelength of 656 nanometres for observing features in the Sun's atmosphere such as solar prominences and the chromosphere, or blocking the solar disc to observe the Sun's corona, flares and other ejecta.

Hubble, on the other hand, is designed with a very different purpose in mind such as taking long-exposure images to obtain the deep- and ultra-deep field full spectrum images. It avoids the Sun as even a very short exposure to the Sun would damage its instruments.


What range of exit pupils work for observing the full moon? I'm observing the full moon, from a major city, with heavy light pollution and dust.

The objective size is fixed, for this comparison, something between 4 and 6 inches. The question is about the tradeoff between magnification and dimness of the image. I'd like to know what range of exit pupils works best. Again, the goal is to observe the full moon.

To put it differently:

What's the smallest exit pupil at which the moon gets so dim that you can't take advantage of the increased magnification? ("Empty magnification").
What's the largest exit pupil at which the moon is bright enough, so a larger exit pupil doesn't help?

The only reference I could find is this, which claims that 1-2mm works best for the moon. Is that correct?

Light pollution does not matter for the Moon. Even transparency doesn't matter that much. What does matter is seeing, a.k.a. air turbulence.

It is very rare that an exit pupil smaller than 0.5 mm is useful for anything - perhaps for some tight double stars, but that's about it. So take that as a hard lower limit.

In terms of a "soft" limit, it depends. If seeing is excellent, if the optics are very good, if the instrument is in perfect collimation and at perfect thermal equilibrium, then you can push e.p. below 1 mm, especially for a small aperture like the range you mention.

If seeing is less than perfect, if optics are so-so, if the instrument is miscollimated or too hot, then the minimum usable e.p. goes up. Each one of these factors influences the observation, so it's hard to pin a number.

Bottom line is, use what works for you.

  claims that 1-2mm works best for the moon. Is that correct?

2 mm seems too big for a small 4...6" aperture when observing the Moon. It might be okay if you want to capture the whole disk, but it seems large if you're interested in small details.

Again, it's all relative.

I have a 6" reflector with a self-made primary mirror of excellent quality. I always take care to collimate the scope, and to keep it at thermal equilibrium. I live in California, so seeing is often good. It's not uncommon for me to use less than 1 mm exit pupil when observing the Moon. I use 0.84 mm as a starting point, sometimes pushing it down as low as 0.6 mm.

If I use an e.p. of 1.1 mm for the Moon that means seeing is pretty bad - but in those cases I'd rather do something else, instead of being frustrated by bad seeing. But at that e.p. in a wide eyepiece (82 degrees apparent field) you see the whole Moon at once, which is always nice when you do a public demo (sidewalk astronomy).

For much larger apertures, the ideal exit pupils tend to be somewhat larger even when observing the Moon.

To predict seeing, go on the Clear Sky Chart site, find a location near your home, and look at the third row in the chart (labeled Seeing); when it's dark blue, seeing is good.

For the largest practical e.p., don't worry about it. The Moon is so bright, losing light is not something you should pay any attention to at all.

Since the magnification is determined by focal length of the objective/focal length of eyepiece it is difficult to have an eye-piece with the same f ratio as the objective.
Assuming pupil aprature 0.25" then the ideal eye-piece aperature will be 0.25" for most effective light transmission of the eye.A generally acceptable f8 would have a focal length of (0.25x8)=2".
A 2" eye piece x 100 magnification would require an objective focal length of 200",for f8 this means 200/8 which is 25" diameter.This would give comparable brilliance at that magnification.
Any deviation from the ideal means loss of resolution due to lens errors of manufacture,the combination of these errors result higher arc" resolution.
You will find a lower magnification more useful as the lower mag system may effect finer arc",res and additional light to eye/optical system efficiency.
There are no easy answers and sadley cost raises the barriers


If Earth is tilted, why is Polaris always above the same spot? Why is Polaris, the North Star, always above (or near) the North Pole? If Earth is tilted, Polaris' path should be in winter 23 degrees away from its path in summer, or not?

You are correct that the axis of the Earth's rotation is tilted with respect to the plane of its orbit by 23 degrees. But it is incorrect that the direction that the axis points changes by a large amount (it should be 2*23 degrees) over a 6 month time span.

Your assumption: If axis it pointed at Polaris at (1), then it should be pointed at a different star at (2).

The axis remains pointed in the same direction throughout the entire year because the laws of physics are that the axis of a spinning object remains pointed in the same direction unless a torque acts on the body to change its orientation. There are torques acting on the Earth (namely the Sun and Moon), but it takes 13000 years to change the direction from "one way to the opposite way", not 6 months.

Correct: The orientation of the Earth's spin axis remains pointed at the same star throughout the year (ignoring the slow, 26000 year precession cycle).

Kick (or toss) an AFL/rugby football (a prolate spheroid) so that it spins rapidly (around its long axis) at the same as it traverses an arc (falling under gravity). You should be able to observe that the spin axis is parallel transported (at least if the perturbation by air resistance is weak enough to neglect).

The earth is spinning. To change the direction it spins in (with respect to the inertial frame of distant stars) it would be necessary to apply some torque to the earth. A sun's gravity does not apply any torque to a spherical planet (due to the symmetries of the Newtonian forces applied).

Since the earth is not quite a perfect sphere (e.g. is slightly oblate) there is a small torque applied by the sun's gravitational field gradient, so over millennia the planet's north axis gradually precesses to point toward different constellations.


Do comets contain any significant amounts of iridium? I have Googled "iridium content of comets" and can't find anything that suggests comets contain iridium in any detectable quantities. Yet, whenever I read an article about the Chicxulub impactor, the article invariably says something to the effect that it may have been created by an asteroid or a comet. But Alvarez and company first offered the impact hypothesis for the K-T extinction by observing an iridium layer in global sedimentary deposits, which could only have come from an asteroid impact, since iridium is extremely rare in the Earth's crust. Later the Chicxulub structure was identified as the probable site of the asteroid impact. While a comet could have caused the Chicxulub crater, it could not have been responsible for the global iridium layer. So why do science writers offer the comet alternative? If Chicxulub was created by a comet, then another site must have been the source of the iridium layer.

I also googled "iridium content of comets", and the first result was https://news.dartmouth.edu/news/2013/04/dartmouth-researchers-say-comet-killed-dinosaurs

Now there is currently no consensus on the nature of the Chicxulub impactor. The observations of iridium and osmium suggest an asteroid. A minority opinion is that a comet may be responsible. The authors of the article in the linked news report suggest that measurements of iridium concentrations from ocean cores may be erroneously high, due to sorting in marine sediments. If these cores are excluded, the amount of iridium measured is lower, and consistent with some models of the composition of comets.

So, comets do contain some iridium. The amount of iridium is not certain. Some models of comet nuclei contain more dust and less ice, and so have more iridium. The exact amount of iridium in the KT boundary layer is also uncertain. The low end of estimates would be consistent with a cometary impact. The nature of the impactor is not yet settled science, but the balance of opinion currently favours a stony asteroid.

The Chixculub site may not have be the only astrobleme of the collision defining the KT boundary.  Actually, the site of impact is not one but many contemporaneous (or closely) impacts including the Lesser Antilles Islands (now the hotspot under the Galapagos Islands), the eastern end of the Scotia Plate and tracking to under East Antarctica,  on the eastern side of the Klamath Mtns and tracking to under Yellowstone (the Yellowstone hotspot), the Deccan Traps, and others.  This was a very active time and the distribution of the iridium can be accounted for by the wide distribution of impacts from north to south and global circulation.  Additionally, the age of the associated hotspots (at least the Galapagos, Scotia, Yellowstone, and Deccan) can be determined by geomagnetic dating and stratigraphic sequence, as well as, determining their rate of lateral migration.  Another interesting factor is the longevity of the hotspots.  The geomagnetic data suggests they persist for tens and hundreds of millions of years, possibly suggesting a nuclear component to these bodies.   In addition, their density and their energy of impact and intrinsic heat appear to be extreme to the extent of even being capable of moving continents and tectonic plates and fracturing the crust creating spreading centers.  The easterly digression exhibited in the Emperor\Hawaiian chain can be associated with the Caribbean and Yellowstone collisions.  see D.Alt, J.M. Sears, D.W. Hyndman, Terrestrial Maria: the Origins of Large Basalt Plateaus, Hotspot Tracks, and Spreading Ridges, 1988, Jour. of Geology, V.76, no. 6, pp 647-662, Univ. of Chicago (copyright).


Why is magnitude of Beta Lyrae given as 30.0 in Henry Draper catalog? Beta Lyrae (HD 174638) is ascribed visual magnitude 30.0 in the Henry Draper catalog. There are hundreds of other stars in the catalog also given magnitude 30.0. What was meant by these nominal magnitudes? Is there a resource in Vizier or elsewhere that provides corrected magnitudes?
https://vizier.u-strasbg.fr/viz-bin/VizieR-3?-source=III/135A/catalog&-out.max=50&-out.form=HTML%20Table&-out.add=_r&-out.add=_RAJ,_DEJ&-sort=_r&-oc.form=sexa

30.0 mean "variable".
If you click on the "Ptm" column header, you get a pop-up window including "Note (2) : codes used for the magnitudes:"; below that is a list of codes, including "30.0 = variable (var. in published catalog)"

I have a problem to believe, that the visual magnitude of beta lyrae changes from 3.51 to 30 mags. Maybe 3.0 mag. I looked into the new Gaia catalogue and I am wondering about a lot of measurements. In most cases you can read N unbelievable values for one astronomical star and so on. In my opinion the Hipparcos and Tess catalogue conform better to terrestrial measurements until  2010.


Why do we have the cosmological constant? Since the cosmological constant is not required to explain that the universe seem to be expanding, why do we have it?

What other factors cause us to have that constant?

Background:
Without the cosmological constant, distant stars should be affected by a great redshift. The amount of redshift is a function of their distance from us. This is due to gravitational time dilation. We are looking 13 billion years into the past, where the universe was very dense. Those stars should be experiencing extreme gravity, causing Einstein shift.

Since we DO have the cosmological constant, we are now looking for other explanations for the redshift.

Reason 1:

Let's look at the Friedmann equations without the cosmological constant.

$$  \frac{\dot{a}^2 }{a^2} = \frac{8 \pi G \rho}{3}-\frac{kc^2}{a^2}$$

The term on the LHS is just the Hubble constant squared $H^2$ which can be measured the direct measurement of recession velocity of galaxies

The density term can be said to be a combination of $\rho_{matter}+\rho_{dark- matter}$ both of which can be measured directly;$p_{matter}$ by observation of matter in our galaxy and other galaxies while $\rho_{dark- matter}$ by rotation curves of galaxies.

The curvature constant $k$ can be estimated today by the anisotropy measurements in the CMBR.

As it turns out the parameters don't fit and we need more mass-energy in the universe(almost 2-3 times of that we had estimated).

So comes along Dark energy or basically the cosmological constant. Cosmological constant or the dark energy are just two ways of looking at the equation,either as just a constant or a form of mass-energy(though we have solid reasons to believe the latter).

And this is our picture of the universe today:

Reason 2:

Now historically the cosmological constant was necessary for an altogether different reason.

The second Friedmann equation without the cosmological constant looks:

$$   \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) $$

Now this predicts for normal type of matter,the universe must decelerate.($\ddot{a}<0$)

Now,people measured the redshift of the type-1a supernovae and found out the quite paradoxical result that the universe was being accelerated in its expansion.

Since normal matter can't explain this type or behaviour,we again have to look at Dark Energy(or the cosmological constant).And so with the cosmological constant the equation becomes:

$$ \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$

Thus $\ddot{a}>0$ is possible.

Therefore the cosmological constant is necessary to both explain the current rate of expansion and the accelerated expansion.

So finally the accelerated expansion can be explained and today we have the $ΛCDM$ model of the universe.

References:

1:http://en.wikipedia.org/wiki/Friedmann_equations

2:http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html

3:http://en.wikipedia.org/wiki/Lambda-CDM_model

I propose the following model:

The universe consists already of a giant central black hole attracting our milky way and all other galaxies. The observed red shift can be explained by the 1/r² law of gravitation: galaxies that are nearer than our one to the central black hole have higher speeds towards it than we have. So we see them moving away from us. Galaxies that are further away from the central black hole than our one don't move so fast towards it as we do. So looking at them we see them escaping from us as well. The black hole and our distance to it are so big, that the field gradient is quite low, so we experience no tidal forces. Adjusting Newton's 1/r² law by Einstein's general relativity field equations does not make a big difference to this model. The general pattern of galaxy motion relative to each other remains the same.

How can these claims be verified? Here is the principle algorithm. It can be implemented and executed even on a PC:

Take all recorded quasar spectra from the known quasar catalogs. Normalize their spectra to zero red shift. Compare each spectral signature which each other. Try reasonable compensation for differences according to dust or other noise adding effects. If you find two identical or similar ones, check the respective quasar's positions. If their angular positions differ by several degrees, you have some evidence. Because you found the same quasar twice: Once seen in direct (curved) line, and once its light wrapped around by the central black hole's gravitational field.
Twin quasars are already known but turned out to be the result of "small" (means: a massive galaxy in our line of sight to them) gravitational lens effects. In these cases their seeming positions differed by some arc seconds or minutes. A number of twin quasars with significant (several degrees) angular position differences or even at opposite sites of the universe would be a proof for the above model


How to calculate the stellar visibility? Please could you let me know a (basic) formula for calculating at what time a star would become visible of an evening in clear weather conditions.  I presume the formula would involve the stars apparent magnitude, the location of the observer, and how far the sun has dipped below the horizon (so, the sky's limiting magnitude).  Working with these three parameters, I guess there must be a simple enough formula, however I cannot find it!  Could you help with this?  For example, assuming clear weather and good eyesight, how could I calculate at what time the star Deneb will become visible on the 1st May
2019 for a rural location?
I am asking this as I want to create an art piece where something happens each time a new star (or stars) becomes visible around twilight time....

Maybe that piece shouldn't consider necessarily individual stars rising from the East but entire constellations/parts of constellations. Trying to express the sky movement as a whole sounds really nice to me. There could be planets setting (like Mercury or Venus right after dusk in the West) or others rising (quite bright) Jupiter, Saturn Mars. Even large deep sky objects portrayed there could be amazing - like the Orion Nebula, Andromeda Galaxy, The Double Cluster, Pleiads, M44 Cluster, meteors, etc. I found a good expression of items of the night sky here.


What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere? Overview:

Taking a planet Ganymede for example (moon, technically I suppose). Ganymede is huge (well, in contrast to other bodies) and I've noticed a pattern between larger bodies and atmospheres. Usually it seems that the larger a body is, the more likely it will have an atmosphere (I know this isn't true, but it definitely seems true, with obvious exceptions). Even when playing Kerbal space program I came up on Vall expecting to be slowed due to its huge size. Big mistake, slammed into the surface. So, if size isn't the end-all-be-all of why a planet would have an atmosphere, what really causes it? Why wouldn't something like Ganymede, doubly protected by it's own magnetic field and Jupiter's field not have an atmosphere? What causes the planet to lose it once an atmosphere is created?

You can ignore my overview rambling, I suppose my one-liner focused question is:

What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere?

Technically speaking, every planet has an atmosphere (though most drop Mercury from the list because it's so thin). Let's start with the big guys, which are pretty straightforward. Gas giants just keep their atmosphere from when they gathered the remnants of gas from the nebula that their parent star formed from. If you're like the gas giants in our solar system, you usually keep this atmosphere for a very long time because you have a strong magnetosphere and you also are fairly far from your star. Your star also isn't particularly powerful. Some exoplanets are categorized as hot Jupiters because they are similar to jupiter in composition but are orbiting very closely to their star. If this is the case, despite the strong magnetosphere, their star may quite literally boil their atmosphere off, either adding it to the star's mass, or blowing it away, sort of like a comet. Eventually they may become what we call chthonian planets, which are hypothetical planets that are the rocky cores left over from boiled gas giants.

How about rocky planets like Earth? They get their atmosphere from a process called outgassing, which basically means that volcanoes burp out gases. Earth is unique in that we believe our atmosphere was at least partially created from comets and wet asteroids. Venus burped out too much greenhouse gases and got a hellhole for an atmosphere. At least it rains there...rains sulfuric acid...and even that boils before it reaches the surface! Mars probably had a thicker atmosphere than today, as evidenced by big volcanoes like Olympus Mons. What happened to it? Well first, Mars doesn't have a very strong magnetosphere because it is simply smaller than the Earth. Because it's smaller its core cooled down faster, and its mantle no longer supported a good magnetic field. More charged particles means a higher chance of atmospheric stripping, which Mars suffered from at least partially. However, we think that a large impact probably knocked a lot of Mars's atmosphere off. Finally, that leaves us with Mercury. Mercury is pretty small and very close to the Sun. It just can't retain an atmosphere because its gravity is weak. The same can be said for most moons in our solar system, as they are typically smaller than Mercury. They are simply too small to retain an atmosphere.

Wait, then how come some moons have atmospheres and others don't? Well the ones that don't, as I said before, are too small to retain one, and they also have to deal with interactions with the solar wind, atmospheric stripping, and interactions with their planet. Some moons have volcanic activity, so their thin atmosphere can be replenished with burped up gases. In fact, some may have weather! I believe Io and Triton both have weather systems, despite having very thin atmospheres. Titan has the thickest atmosphere of the bunch, with one that is about 50% denser than Earth's. Thus, it also has complex weather systems. But why? Titan is similar to Ganymede in structure. Why does Titan get to have an atmosphere? Short answer: we don't fully know why. We think it might've been comets, but we really aren't 100% sure on anything at this point. Go check out the wikipedia article for more.

So to answer your question, atmospheres are usually formed from volcanic activity or, for gas giants, retained gas from nebula remnants. Atmospheres are lost due to low gravity, high temperatures, solar wind, atmospheric stripping, large impacts, and planet-moon interactions.

Its a balance of stellar wind, gravity, ionosphere, magnetosphere and volatility of the planetary surface material: stellar/cosmic wind eliminates the atmosphere, gravity, ionosphere keeps atmosphere on, and volatile surface material generates new atmosphere.
The volatile material at the planet's surface, which is partly dependent on the above two variables because it can get complete obliterated by solar wind like the moon and mercury.
Pluto has a billion times higher atmospheric pressure than the moon, in 900 times less solar radiation, while the gravity at it's surface is 35% that of the moon.
Venus has 90 times the atmospheric pressure of earth, and it's protected by an ionospheric magnetotail.
Rank    Name    Surface Pressure (bar)

Saturn  >>1000
Uranus  >>1000
Neptune >>1000
Venus   92 (huge magnetic field generated by ionosphere, boiled oceans)
Earth   1.014
Jupiter 0.2 - 2 (same range of atmosphere pressure as our planet)
Mars    4 - 8.7 x 10-3
Sun 8.68 x 10-4 (a lot of solar wind, 1000 times less pressure than terra)
Pluto   3 x 10-6 (30% gravity of moon, billion times more atmosphere)
Mercury 10 x 10-15 (slow rotation, no magnetism, near sun)
Moon    3 x 10-15

The moon has 25 tons of exosphere www.space.com/amp/18067-moon-atmosphere.html
Near a fierce star 10000 times stronger than the sun, our planet and perhaps even Saturn could be mostly stripped of atmosphere over billions of years, even though the have a strong magnetosphere/ionosphere.


Will Pluto ever clear its path? I am not quite sure if it would, considering i'v looked through countless websites and found nothing and would like to know if it could clear its path and join the other planets?

In addition to the answer above, Pluto is a Neptune crossing trans neptunium object or TNO as it's often abreviated.  Being a TNO doesn't exclude the possibility of an object being a planet but crossing the orbit of Neptune does.   So, even if Pluto cleared out it's entire orbital neighborhood, it would still cross Neptune and naming it a planet would be problematic.

If it's orbit changed to being entire outside of Neptune, it would still face the problem of being one of several minor planets in that general region of similar size.

If you really wanted to push the theoretical to the limit, if Pluto's orbit moved so it stopped crossing Neptune's and if it merged with all the other similarly sized TNOs, then you could make an argument that it stands out as unique for it's region and a strong argument could be made that it should under those conditions, be called a planet, but that's a stack of improbabilities so high that winning next weeks mega millions top prize seems likely by comparison.

Also, if all the large known TNOs were somehow combined, they would still have a mass less than Mercury, though perhaps a slightly larger diameter, but that would be large enough where it just might get planet status.   It won't happen though.  Space is too big and object's orbits are too far from each other.   Planet formation happened within the proto-planetary disk at distances much closer to the Sun.   Uranus and Neptune likely formed much closer to the Sun than they are now.   There's not enough material and too much space in between for a planet to coalesce at Pluto distance from our Sun.

Space is super empty, and Pluto has a very small mass. There's a extremely small chance that it could attract any objects in its "neighborhood", so if it ever does happen, it would be a very, very long time from now. Perhaps, it would take so long that by the time that happens, Pluto would have been ejected from the Solar System after the Sun dies.


Clearing the Neighbourhood: Definition of 'Orbital Region'? I am trying to determine how wide a 'neighbourhood' a planet is expected to clear.

However, the Wikipedia article on the matter provides no clarity, simply stating that a planet is expected to clear it's 'orbital region' with no definition provided.

I suspect that the correct answer is that a planet is expected to clear a region around it's orbit equal to the radius of its Hill Sphere, but I am not certain of it.

Furthermore, for a planet of Earth mass, the Hill Sphere radius is a measly 0.01AU, which doesn't seem like much of a neighbourhood!

Is there a formal way to define the 'neighbourhood'?

The IAU gives no definition, it leaves the term somewhat vague.

Various calculations of the "planetariness" of solar system bodies use slightly different notions of the "neighbourhood"

Soter's $\mu$ considers bodies to be in the neighbourhood if they share a common radial distance (ie the orbits cross each other) and the orbital periods differ by less than an order of magnitude

Margot's $\Pi$ uses $2\sqrt 3$ times the Hill radius.

Stern and Levinson's $\Lambda$ in the original paper consider the neighbourhood to be bodies that cross the orbit of the planet, while not in an orbital resonance, to be in its neighbourhood, in a somewhat complex way that requires knowledge of the relative inclinations and and orbital velocities of bodies that cross orbits.

Changing the details of these values doesn't change the conclusion of the calculations: There is a clear and measurable difference between bodies like Mars and Neptune that are (to use Stern and Levinson's term) uberplanets and Ceres and Pluto that are unterplanets (or dwarf planets as termed by the IAU).

No, it is a vague term that isn't defined anyhow by the IAU. If you consider the object's hill sphere as what is meant, then Pluto and Eris "cleared their neighbourhood" too. If you consider it the way that Pluto's orbit overlaps with Neptune for instance, then no planet cleared their orbits. There are Mercury-crossing asteroids, Near-Earth asteroids, Trojans crossing Jupiter's orbit et cetera. If we consider it as being meant that the bodies must be the most massive ones in their area (which again is vague) then Eris and Ceres should be planets and perhaps Pluto too (if you look three-dimensionally on the ecliptic). If you set an arbitrary border, either Ceres would be a planet or there wouldn't be planets at all.

So the 2006 planet definition is pure nonsense and shouldn't be considered valid and it isn't by many.


Will there be any more planets come into our solar system and make it their "home"? Could we have more planets come into our system and get trapped by our suns gravitational field and rotate?

Short answer: Yes, but it is very unlikely.
The probability that a random object drifts into the Sun's sphere of influence (SOI) is low (excluding tiny stuff like space dust and small rocks). But the chance that a full-blown rogue planet enters our Solar System is many orders of magnitude lower. Let's set up a Drake equation:
Let $R_n$ be the number of rogue planets in the galaxy, $R_{ns}$ the fraction of rogue planets that may enter the solar system, $R_e$ be the probability that one enters the Sun's SOI, $R_i$ be the probability that the planet will survive an encounter with a Solar System planet, $R_s$ be the percentage of all possible orbits for the rogue planet after the interaction that are long-term stable, and $R_o$ be the probability that the rogue planet gets captured into an orbit. Therefore, $$R_o=R_nR_cR_iR_s$$


Why could a gas planet not get much larger than Jupiter? What is the maximum size for a gas giant?

Usually they can get is Almost twice.  Here's the evidence:
 It is difficult for them to become even bigger than that as once they reach 15 Jupiter masses or so there is enough pressure in the core to ignite deuterium fusion, so those are considered “brown dwarfs” rather than planets.
An example is TrES-4, which is located 1,400 light years away in the constellation Hercules. The planet has been measured to be 1.4 times the size of Jupiter, but it only has 0.84 times Jupiter’s mass. With such a low density, it is dubbed the puffy planet.
It was discovered in 2006 orbiting a brightish star 1500 light years from Earth. Its diameter of about 1.8 times Jupiter’s makes it the largest accurately measured planet. Strangely, though, TrES-4 is very light for its size. It has only 88 per cent of Jupiter’s mass, giving it a density of roughly 0.2 grams per cubic centimetre, less than that of cork. Just how a planet can be as fluffy as TrES-4 remains a mystery.


Why doesn't the Sun explode? I know stars explode because of the fuel causing a fusion compound pushing it apart and the fuel runs out and it 'bounces' for lack of a better term.

Given the fact that more massive stars are supposed to explode more quickly: why doesn't the Sun explode considering its massive size?

There are two things to discuss here: (a) why the Sun does not explode; and (b) why the Sun will not explode.

(a) An explosion occurs when the timescale for the energy release by some process is much shorter than the timescale on which a system can adjust to damp the energy release process. In the present day Sun, nuclear fusion is a very slow process: on average it takes many billion years for a proton to fuse with another. This timescale is quite temperature dependent, so you might have thought the centre of the Sun might heat up quickly, leading to a runaway "explosion". However, an increase in temperature leads to an increase in pressure that would expand the Sun, reducing the core density and temperature and decreasing the rate of nuclear fusion again. The timescale for the Sun to react in this way is just millions of years, so this acts like a thermostat that keeps the reactions under control.

(b) Stars more massive than the Sun burn through their hydrogen and other heavier fuels, and end up with an inert iron core from which no further energy can be extracted. The core subsequently collapses and the gravitational potential energy released by the core collapse powers a supernova explosion. The reason that the core does not quietly collapse into a black hole (at least in the supernovae we see!) is that the core "bounces" when neutrons in the core are squeezed very close together. This provides a quantum mechanical effect called degeneracy pressure that resists a complete collapse.

Ironically, it is degeneracy pressure in the solar core that prevents our Sun reaching the supernova stage. After burning through its hydrogen and helium the core of the Sun would consist of carbon and oxygen; but it will never become hot enough to commence fusion of these elements, because electron degeneracy pressure will support it from contracting and getting any hotter. Only more massive stars will ever attain the core temperatures required to ensure progress towards a terminal iron core.

Someone else asked a similar question that was linked here as a duplicate. It seems that question isn't quite the same, but similar enough that it makes sense to put it here.

That person asked:

I know that the Sun is made up of hydrogen and helium.
Also, hydrogen explodes when it touches a flame.
But why does the Sun not explode?

The kind of explosion that happens when hydrogen touches flame is an exothermic chemical reaction in which hydrogen is rapidly combined with oxygen (oxidation).  When heat is applied (flame) to hydrogen (H2) in the presence of oxygen (O2), it causes a molecule of O2 to separate and each oxygen atom to bind with a molecule of H2.  When this happens, it releases a large amount of energy ( believe this is caused by the breaking of the O2 bonds), and you end up with water (probably steam) and heat (fire).

The kind of "buring" that happens in stars is thermonuclear fission. When two atoms of hydrogen are brought together by immense heat and pressure, they fuse into an atom of helium.  In this process, enormous amounts of energy are released.  This is the same nuclear reaction that takes place in a thermonuclear fusion bomb.  It does not involve a chemical change, but a nuclear one.

Let me take a half-step back on that.  A chemical change or reaction is when the atoms rearrange themselves in molecules.  In most such reactions, energy is either absorbed (endothermic) or released (exothermic).   Sometimes fairly large amounts are released.  Explosives such as gunpowder, TNT, C4, and the like are all chemical explosives and mostly their reactions involve very rapid oxidation of the substance being burned. The burning of a match, candle or campfire is the same type of chemical oxidation reaction, but at a much slower pace.

Nuclear reactions come in two main types: fission and fusion.  Fission is the splitting of larger atoms into smaller ones.  The isotope of uranium with an atomic mass of 235 (92 protons and 143 neutrons) has a half-life of about 700 million years (which is far shorter than the 4.5 billion year half-life of the more common U238).  Atoms naturally decay over time, and in a span of 700 million years, a mass of U235 will naturally split on its own and the number of atoms will reduce by half (hence the term half-life).  When an atom of U235 splits, it releases a bunch of energy, produces an atom of krypton and one of barium, and releases a few extra neutrons. When one of these neutrons meets another atom of U235, it joins that atom, which destabilizes it, and causes the reaction to happen again.  Note: because we are talking about reactions in the nucleus of the atom and resulting changes to the atoms themselves, we are talking about nuclear reactions, not chemical ones.

The reaction I mentioned above takes place in U235 naturally. If you don't have much of it and it's not in a dense enough state, it just sits there like a lump of rock and seems to do nothing.  Most of the neutrons it releases don't find another atom to play with, and end up just flying off into space.  But if you squeeze it down enough or get enough of it together, more neutrons are meeting up with other atoms than are flying off into nowhere.  This creates a chain reaction, which is the basis for nuclear power and nuclear weapons.

A reactor works by controlling the rate of this reaction, keeping it from going too fast.  Lumps of uranium are held just close enough together to react,but not rapidly, and rods of neutron-absorbing material (usually graphite) are slid between the lumps to help absorb the neutrons and prevent them from causing more reactions. A nuclear meltdown occurs when the reaction gets out of control and heats up so much that it literally melts the framework holding it all together.

This isn't enough, however, to create a nuclear explosion.  That occurs when you apply pressure and squeeze the material down, increasing its density.   U235 isn't actually suited for this, but element 94, plutonium, is. When you squeeze plutonium down to a high enough density, this reaction happens extremely rapidly.  First one atom of Pu239 splits, releasing its energy in the form of light and heat along with additional neutrons.  Then those extra neutrons find other atoms of Pu239 and cause them to split.  Every split creates at least twice as many extra neutrons as splitting atoms.  The first atom splits to cause 2 to split, which split and cause 2 each for four, then eight, sixteen, 32, 64, 128, 256, 512, 1024, 2048, 4096... each "generation" of these splits is double the previous. This all happens at a time scale measured in tiny fractions of a second, the time it takes light to go only a few millimeters.  This all builds up tremendous amounts of heat and pressure, which, unconstrained, erupt out of the core of the bomb.

As it turns out, only a small percentage of the original mass is actually lost to fission. The heat and pressure build up so great so fast that it forces the atoms away from each other in the explosion, thus ending the reaction.  In fact, in a fission bomb, before anything other than high-energy x- and gamma rays leaves the bomb case, the reaction is complete, all that's left is for it to expand outward due to the immense pressure.

The other kind of nuclear reaction, fusion, is also used in bombs.  In fusion, as I mentioned above, huge amounts of pressure are used to force atoms of lighter elements, typically hydrogen, together.  Every time two atoms of hydrogen fuse into one, a huge amount of heat and energy are released (to be honest, I've never understood why energy is released here and not absorbed, but I'm not a nuclear physicist, just a guy who likes to read a lot).  In bombs, the 'trigger' for the fusion is actually a fission bomb.  The fission bomb creates the heat and pressure necessary, and when hydrogen atoms are added to the mix (usually in the form of tritium or deuterium), the fusion reaction begins and starts adding its heat and pressure to the pot, which helps sustain the reaction long enough to put out an amount of energy several times that which the fusion bomb started.

If you remember back to your high school chemistry lessons, Boyle's Law states $P_1V_1 = P_2V_2$. P is pressure, V is volume.  This formula shows that pressure and volume are inextricably interconnected.  Charles' law states that $V_1T_2 = V_2T_1$, showing the interrelation between volume and temperature.  Gay-Lussac's law, $P_1T_2 = P_2T_1$ shows the relation between pressure and temperature. These all come together into the combined gas law and Avogadro's law.  They show that if you increase the pressure, volume, or temperature of a mass, the others are affected as well. An increase in pressure results in an increase of temperature.  A decrease of volume creates an increase in pressure which create an increase in temperature.  If you increase the volume, the temperature and pressure decrease.  Etc...

Ok, so, now here's where it all applies to stars.

In the result of the Big Bang, most of the matter created was hydrogen. There was some helium and a little lithium, but the overwhelming majority was hydrogen.

When the hydrogen molecules floating through space started to clump together through gravity, they started to condense down into smaller and smaller volumes.  Separately, the gravity of a single atom of hydrogen is minuscule.  But as more and more of them interacted, the combined gravity began to build up.  The clouds became more and more compact - denser, and their gravity brought in more and more mass, which added their gravity to the total.  Things began to heat up, due to the pressure caused by the volume.  The gravity started to get to be enough to continue to pull things toward the center of mass, into the "gravity well" caused by the cloud.  As the heat increased due the the pressure caused by gravity, it caused outward pressure.  The forces continued to balance each other.  As more mass fell in, gravity increased, causing more heat, which caused more pressure, which pushed outward... and so on.

At a certain point, the gravity gets so great, that it exerts enough pressure on the center of the mass to cause nuclear fusion.  This, in turn, causes more pressure outward, but the force of gravity of the entire mass is great enough to keep it all together.  This is when a star is born.  The pressure of gravity has caused a gravitational collapse to ignite into a star.  The balance of the forces of gravity and heat and pressure are what keep it going.  If there wasn't enough pressure outward, it would collapse further into a singularity (a black hole), if there's too much pressure outward, it would either explode (type II supernova) or at least expand outward until balance, what we call hydrostatic equilibrium, is achieved again.

As a star ages and burns through its supply of hydrogen, the equilibrium changes and at some point it begins to burn helium. After that, carbon. Depending on the original mass, it may fuse other materials up to iron.

In some stars, the mass is enough that when the fusion can no longer be supported, there is no longer enough outward pressure, and the core collapses in on itself.  The gravity increases as it shrinks, and there's not enough heat or pressure to overcome it.  It collapses to a point of infinite density and becomes a gravitational singularity and a black hole is created.  In other stars, the mass isn't quite enough to keep collapsing, and the pressure overcomes it in a massive explosion (type II supernova).  In other stars, the star belches out wave after wave of gas as it expands and contracts, finally belching enough away to no longer have enough reactive mass.  It shrinks down into a white dwarf - a hot ball of atomic particles that slowly cools over billions and billions of years until they are no longer luminous, becoming a black dwarf (the time for this to occur is believed to be so long that there are, as yet, no black dwarfs in the universe... but there someday should be, long, long in the future).

So, to circle back to the original questions... why doesn't the sun explode, and why doesn't the hydrogen in the sun explode... the main reason is gravity. It reaches a balance.  The heat and pressure pushing  are balanced against the gravity pulling in.  There is no chemical oxidation going on, because the atoms of hydrogen are being fused into helium and there's little oxygen to be consumed in the oxidation process to begin with.  Someday we expect the sun will run out of hydrogen to burn.  As it starts burning helium, it will expand out past the orbit of Mars, then at some point it will shed its cooler, outer layers and contract into a white dwarf which will slowly cool down to a dying ember in the ever darkening universe.


Will the SKA's total bandwidth still exceed the Earth's internet's bandwidth? In the early 2010's there were many popular articles that said that the total bandwidth of the the Square Kilometer Array's optical fiber network for raw data from each dish would be larger than the Earth's internet. Here is just one example:

Wikipedia's Square Kilometre Array
:

  It would require very high performance central computing engines and long-haul links with a capacity greater than the global Internet traffic as of 2013.3

As you can see that link is broken.

The SKA project has several Precursors, pathfinders and design studies but the SKA proper has not been built yet, and perhaps is not even finalized.

Still, with available data, by the time it might be built (2025?) is it likely to still exceed the Earth's internet's bandwidth?

Will the SKA's total bandwidth still exceed the Earth's Internet's bandwidth?

No.

First some conversions:

1024 Gigabytes = 1 Terabyte
1024 Terabytes = 1 Petabyte
1024 Petabytes = 1 Exabyte
1024 Exabytes = 1 Zettabyte

From the SKA Telescope website: "The Operational Model" (.PDF), on the webpage: "Info Sheets", this explanation and graphic is offered:

"The data journey

Basic data products will be produced and stored in Cape Town for SKA1-mid and Perth for SKA1-low. From there, they will be delivered to a global alliance of SKA Regional Centres for further processing and archiving and access by the user community.

New data models will need to be developed as current software packages for radio astronomy data reduction don’t have the capabilities to handle the SKA’s large bandwidths and Field of View dataset.".

"Major dates

SKA1 OPERATIONAL CONCEPT DOCUMENT - Operational Principles

late 2027 - Commencement of Key Science Projects (KSPs)

late 2026 - Commencement of PI-led programmes

2024–2027 - Key Science Project (KSP) planning & proposals

2022 - Start of Observatory & Science commissioning

2020 - Start of construction activities

2017–2018 - Prototypes deployed at the telescope sites"

See also: "SKA Community Briefing – 18 January 2017", page 12:

Nothing will flow to /dev/null.
Source: Cisco - Global - IP Traffic: https://www.cisco.com/c/m/en_us/solutions/service-provider/vni-forecast-highlights.html#

Globally, IP traffic will grow 3-fold from 2017 to 2022, a compound annual growth rate of 26%.

Globally, IP traffic will reach 396.0 EB per month by 2022, up from 122.4 EB per month in 2017.

Global IP networks will carry 13.0 EB per day by 2022, up from 4.0 EB per day in 2017.

...
Internet Traffic

Globally, Internet traffic will grow 3.7-fold from 2017 to 2022, a compound annual growth rate of 30%.

Globally, busy hour Internet traffic will grow 4.8-fold from 2017 to 2022, a compound annual growth rate of 37%.

Globally, Internet traffic will reach 350.8 EB per month by 2022, up from 96.0 EB per month in 2017.

Global Internet traffic will be 11.5 EB per day by 2022, up from 3.2 EB per day in 2017.

Global Internet traffic by 2022 will be equivalent to 1 trillion DVDs per year, 88 billion DVDs per month, or 120 million DVDs per hour.

By 2022, the gigabyte equivalent of all movies ever made will cross the Internet every 1 minutes.

Global Internet traffic by 2022 will be equivalent to 193x the volume of the entire Global Internet in 2005.

...

No, it's not even close.


If the hammer and feather move at the same speed why do comet and the tail particles move at different speeds? From this exhibition by an astronaut on the moon dropping a hammer and a feather and showing they move at the same speed, why do the particles of the comets tail move away from the comet, slow down and fall away as it disintegrates? Shouldn't they move at the same speed of the comet?
For example if the feather was attached to the hammer and it was falling and broke off wouldn't it fall at the same speed, not start to trail behind and eventually fall away?

The tail of a comet is not actually "slowing down and falling away" from the comet, like you might expect to see when smoke streams out from behind a moving object on earth. The tail of a comet is actually being pushed away from the sun by the solar winds and radiation. That's why the tail of a comment always points away from the sun, and doesn't stream out behind it.

If there were no other forces at work, the coma would actually travel at about the same speed at the comet. As the material surrounding the core of the comet vaporized, it would form a coma ("atmosphere") made up of sublimated gas mixed with dust, moving along  at the same speed (more or less(1)) as the comet… like you observed in the video.

(1) In reality, without the solar winds, the coma of the comet would continue to move outward, as its expulsion velocity would easily overcome the comet's weak gravitational pull.

There are a few more elements which you might consider. Like, when dropping a hammer along with a feather in earth, the feather descends slower than the hammer. Reason: although the force of gravity is same, their mass, and the resistance from external forces faced by them is an entirely different story. The air between the object and the ground must also be taken into consideration. Whereas, moon lacks atmosphere and hence they fall at relatively same speed (with only low gravity acting upon them).

As in http://en.wikipedia.org/wiki/Comet_tail

  As a comet approaches the inner Solar System, solar radiation causes
  the volatile materials within the comet to vaporize and stream out of
  the nucleus, carrying dust away with them. The streams of dust and gas
  thus released form a huge, extremely tenuous atmosphere around the
  comet called the coma, and the force exerted on the coma by the Sun's
  radiation pressure and solar wind cause an enormous tail to form,
  which points away from the Sun.

Similar to the air acting on the detached feather (as said above), the solar radiation pressure and solar winds act on the tail debris of the comet. For instance, you drop an article from a moving vehicle, it falls to the ground rather than flying in the air at the same speed. Deceleration and opposite forces like solar winds cause it to slow down.

UPDATE: The reason why mass is taken into consideration is due to the fact that mass is directly proportional to gravity. So, increasing the mass means increased gravitational pull. That is why heavier objects like hammer fall faster than lighter objects like feather. Explaining this would be a matter of physics and not astronomy which is irrelevant to this website and should be discussed in the physics stack exchange website.


Do all objects get larger in an "expanding universe"? I'm envisioning an analogy of the universe as a plane with marbles in that plane to represent different planets.

Now, if this plane expands how does this work? I assume that it is not like 'nothing' becomes 'something' then but rather expanding/stretching the universe; but doesn't this imply that everything around us is actually stretching/becoming bigger (like every atom/quarks) becomes a little bit bigger? Or is it just the space between those subatomic particles and combinations of these become bigger?

An example could be the structure of carbon dioxide becoming larger, but only because there is more space between the carbon and the oxygen atoms.

It would, if there were no interactions.

However, everyday matter is held together by other kinds of forces (electromagnetic, mainly). These are quite strong, and any "stretching" of space will be counteracted by these forces "pulling" the atom/molecule/object back to its original size.

So the expansion of the universe only matters on the larger, cosmic scale, between objects like galaxies which are separated by a large enough distance that they don't affect each other too much, gravitationally speaking.

Ummm this question made me think about a lecture from my teacher. When he was teaching us about the Big Bang theory!

From Wikipedia: The big bang states: The Big Bang theory is the prevailing cosmological model that describes the early development of the Universe.1 According to the theory, the Big Bang occurred approximately 13.798 ± 0.037 billion years ago, which is thus considered the age of the universe. At this time, the Universe was in an extremely hot and dense state and began expanding rapidly. After the initial expansion, the Universe cooled sufficiently to allow energy to be converted into various subatomic particles, including protons, neutrons, and electrons. As stated here.

Example of balloon:

So what you can imagine, there is a baloon with some dots on it. When you are blowing it, the balloon gets bigger, its size expands. But the dots stay where they were. There do comes a change in their size too. Galaxies expand too, but that doesn't mean that there is amount being added, but that means there is some space some extra space being added to it.

Objects expand:

Objects around you aren't expanding. Because this change would occur in years, so its almost ~0 this size as the changes occur inside the atom, atom contains magnetic field by the movement of electrons that would lesser down expansion and bla bla which is Chemistry so skip this. However, galaxies, stars, and on whole universe is expanding! And its been years it was expanding, and will expand! :)

Your plane example:

And your plane? Imagine that the plane is expanding. Hey wait! try this one (if possible, as this one would explain it fairly), place some magnets over a magnetic field, make sure north faces the north so they keep floating. Now when the surface expands, the magnets would stay there! They won't move backward or forwards just as the universe (plane) is moving. Similarly, as universe expands, planets don't expand! As they are not stitched to the universe or clamped to the universe. They are free moving.


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.

Of course; how would we find it otherwise? This is how we look for it: image one part of the sky with a telescope, about where the planet is presumed. At another time we image the same part again and see whether a star moved or is no longer there. This star is the wandering star = planet we're looking for.
The hypothetical object has a very elliptical orbit, and perhaps can even be seen with the naked eye when at perihelion since it may be a primordial black hole (which itself isn't visible but its accretion disc would be). About 4,000 years ago, when the black hole (if it is one) was closer to perihelion, the Sumerians  may have seen it and called it Nibiru. But we dunno whether that's the currently looked-for Planet IX.


Why do astronauts move so slowly in zero gravity? When astronauts are floating about in a spaceship or space station, they nearly always move very slowly. After doing a bit of research I can't see why being in zero gravity would restrict movement to such a degree. It's almost as if there's resistance to their movement, like they're moving through water.

It's more for safety than anything else. Space is a very dangerous place for so many reasons. And making mistakes can very easily cause death.

Being weightless does not mean you lose mass, so momentum is just as difficult as ever. But whereas on the ground you can easily use friction to stop, in space if you try to stop against the floor you will just move off it. You can only stop by holding something, or pressing against something close to perpendicular to your movement.

As an example, imagine you jumped with all your force from one wall in the ISS. You will notice as you approach the other end that you are travelling at speed, head first, with no safe way to stop. Even reaching out to a handhold on a side wall will whip you round and into that wall, possibly injuring yourself or damaging instruments on the wall.

Look at any video from the ISS to see how carefully they move.

Similarly, outside the ISS, you want to do everything slowly so you don't damage your suit, miss a handhold or otherwise cause death.

There is no air resistance in space. So if you increment your own speed you'll continue with that speed until you crash (considering the laws of thermodynamics).

For safety, you should move slowly.


Is it coincidence that Einstein predicted exactly twice the outcome of Newton for the gravitational deflection of light? During a solar eclipse Sir Arthur Eddington measured the deflection of star light by the Sun. Einstein predicted a deflection of 1.74 arc seconds, but according to Newton the deflection should have been 0.86 arc seconds. So is it a coincidence that the prediction of Einstein is twice as large or is there a kind of law involved which makes Einstein's outcome of his GR twice as big as that according to Newton?

As far as I know Newtonian prediction is based on Soldner's work in 1801 and Einstein 1911 also predicted around $0.87$arcsec (he was also accused for plagiarizing Soldner's work the article:Historical Note on the Problem of Light Deflection in the Sun's Gravitational Field explains why the accusations were wrong).

The Newtonian approach accounted for the space curvature, as did special relativity which Einstein first used, but it missed the part that not only space is curved but space time is curved. So the factor of two was because of the spatial curvature, that's because by Minkowski metric we consider that light travels flat so that $\rm{d}s^2=0$.The correct answer would be to use the Schwarzschild metric and get the correct answer.

This paper gives more insight on the formulas and the correct approach. Gravitational Bending of Light

Einstein predicted a declination of 1.74 arc seconds, but according to Newton the declination should have been 0.86 arc seconds. So is it a coincidence that the prediction of Einstein is twice as large... ?

No. Einstein's figure is twice Newton's because of the wave nature of matter. Check out Ned Wright's deflection and delay article:

He says this: "Newtonian mechanics predicts that a particle traveling at the speed of light which just grazes the edge of the Sun will be deflected by 0.875 seconds of arc". But that's a particle such as an electron, which we might have created from light in pair production. Pair production "often refers specifically to a photon creating an electron-positron pair".

Obviously an electron isn't the same thing as a photon. One essential difference is mass, another is charge, and another is the way it's deflected by gravity. If you threw an electron fast past the Sun, nearly as fast as light, its deflection would be half the deflection of a photon.

You can understand why by thinking what would happen if you trapped the photon in a gedanken mirror-box such that it was going round and round a square path. Then if you put it in a gravitational field the light would be deflected like this:

The horizontals curve down, and because of this the position changes. The photon in its gedanken mirror-box falls down. Then if you were to throw it fast past the Sun you'd say it was deflected - it's moving fast from left to right, but it falls down too. But note that the verticals are not affected. Only half the light path is curved by the Sun. The electron is affected in a similar fashion because of the wave nature of matter. This is why the deflection of matter is half the deflection of light.