gravity - Are black holes orbiting around each other?

A black hole (briefly), is nothing more than a dead star whose mass was more than 3 solar masses. When that star's life ended it collapsed under its own gravity and went through a supernova, losing a portion of its original gas in the process, thus losing part of its mass, and thus its total gravitational pull is diminished compared to the original star (contrary to common believe for a black hole having the "ultimate gravity"). So first of all, black holes vary in size and are not all the same (thing).

If we say the original star was orbiting around another star (A Binary system), (putting other factors aside) the effect on the other star from the black hole's gravitational forces would be less compared to its original star.

However that gravitational pull increases drastically as you go closer to the blackhole, since all the remaining matter is concentrated into a single point (singularity). Gravity becomes too severe, close to a blackhole that even light can not escape, an area called "the event horizon" of the black hole. But that's too close to the black hole. Nothing is special about its gravitational effect on the further objects as I pointed out with the orbiting star.

Regarding your statement: "Is it true also for objects like black holes? Are they even able to be moved by another force?" I don't know what you mean to be moved by another force, but black holes do move as any other object in space, they are nothing more than the remainder of their original star, and they inherited its motion relative to the other objects around. If say the original star was rotating around the center of it's galaxy, it will continue to do so.

Your statement: "bodies with low gravity force orbit around another one with higher gravity force" is not very accurate, since (bodies with higher gravity force - the more massive object), and (bodies with low gravity force - lesser mass) both orbit one another.

For instance, as the Moon orbits the Earth, the Earth also orbits the Moon, they both orbit around a point between the two of them called the "Barycenter". This point is always closer to the more massive object.

In the case of the Earth-Moon system it's inside the Earth's volume. Therefore it's common believe that only the Moon's orbiting the Earth, without the other way around as well.

Topics you must check out besides the link provided in the previous answer:

general relativity - Does one need to take into account finite gravity speed in N-body simulations?

If you're asking whether it's sufficient to use a retarded (time-delayed) positions to calculate gravitational forces, then no, that would be much worse than Newtonian gravity. For example, that would predict that the Earth should spiral into the Sun on the order of about 400 years. See also this question.

Most small-scale N-body simulations (e.g., planetary systems, solar system, stellar clusters, ...) use classical Newtonian gravity.

This is true, but some simulations use relativistic corrections.

The post-Newtonian expansion works in order $epsilonsim v^2sim U$, where $U$ is (the negative of) Newtonian potential, and including the first-order post-Newtonian correction to the N-body equations of motion forms the Einstein–Infeld–Hoffmann equations. Because they contain $3$-body terms, the simulation would now be $mathcal{O}(N^3)$ instead of the Newtonian simulation $mathcal{O}(N^2)$.

The post-Newtonian three-body equations are known to 2PN, while the two-body problem to 3.5PN. Effects beyond the first order can be important in e.g. binary simulations because gravitational radiation losses are seen at 2.5PN order. However, they are not translatable to an N-body problem.

A technique of making the EIH equations more tractable by making a second expansion that depends on the problem at hand can be found in Will (2014), which is applicable when most the relativistic dynamics are dominated by a few contribution (e.g., a supermassive central black hole in galaxy simulations).

Most large-scale N-body simulations (e.g., galaxy clusters, ...) use general relativity.

As Walter says, most of them use Newtonian gravity on a expanding background. Not necessarily directly computer particle-particle Newtonian force, but perhaps modeling Poisson's equation through mesh methods, or perhaps some combination thereof (e.g., direct force calculations for close particles, mesh methods for larger scales).

My question: if I were to make a large-scale N-body simulation using Newtonian gravity, would it be enough to "correct" that to first-order by modifying the equations of motion to take into account the fact that gravitational fields propagate at the speed of light?

If you discount cosmological expansion, it will simply be wrong. However, first-order corrections to the Newtonian gravity part of simulations do exist in the post-Friedmann formalism. I'm not really familiar with PFF, but according to Bruni et al. (2014), the leading-order correction to Newtonian gravity introduces a vector potential in the $g_{0i}$ term in the metric that can produce weak lensing effects but doesn't affect matter dynamics. Which is entirely sensible because that kind of gravitomagnetic effects should be suppressed by a further $v/c$ term for matter, just like magnetism.

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References:

1. Will, C. M. "Incorporating post-Newtonian effects in N-body dynamics". Phys. Rev. D 89, 044043 (2014) [arXiv:1312.1289]


2. Bruni, M., Thomas, D. B., Wands, D. "Computing general-relativistic effects from Newtonian N-body simulations: Frame dragging in the post-Friedmann approach." Phys. Rev. D 89, 044010 (2014) [arXiv:1306.1562]

How can absolute value of azimuth exceed distance from pole?

It turns out I was right when I thought I was wrong.

Consider the azimuth circle for 85 degrees elevation. It's very small. For example, 85 degrees elevation and 0 azimuth is only 10 degrees away from 85 degrees elevation and 180 azimuth.

Therefore, the azimuth circle for a given elevation is smaller than a great circle. It's only cosine(elevation) the size of a great circle.

At 35 degrees, the azimuth circle is cosine(35 degrees) = 0.819152 the size of a great circle, so 4.17 degrees in azimuth = 4.17 degrees * 0.819152 = right around 3.5 degrees on a great circle.

orbit - Could a habitable satellite of a gas giant have a stable subsatellite?

This is a fun question, so I spend some time thinking about it. This is what I came up with.

Is this possible?

Yes, this is certainly possible. Just consider the Solar System itself. We have a massive central body, the Sun, and several tiny subsystems (planets with their moons) orbiting it.

Is it stable?

Again, from the simple observation that the Solar System exists in its current form, we know that such a system can remain more or less stable over a significant portion of the stellar lifetime.

A more difficult problem, I think, is if can it be formed in the first place.

Could it be habitable?

This is a tricky question as it critically depends distances between the several orbiting bodies. Lets consider the orbital mechanism and derive from what we know of the Solar System.

In the following I've assumed the gas planet is Jupiter and the fictional planetary system consists of a Sun', Jupiter', Earth' and Moon' such that the apostrophe denotes the fake object. These fake objects are identical to the real equivalents, except for their orbital configuration and relative distances. If you want a different system you can follow the same line of reasoning, but plug in different numbers for the fake objects.

Now, we know that the Earth-Moon system exists in a stable orbit around the Sun, which means that at 1 AU the gravitational force of the Sun is not strong enough to destabilize the Moon orbit. From this we can find a lower limit on the gravitational force that Jupiter' may exert on the Moon'
begin{aligned}
F_{J'M'} &= F_{SM} \
frac{ M_{J'} M_{M'}}{r_{J'M'}^2} &= frac{ M_{S} M_{M}}{r_{SM}^2}
end{aligned}
Solving this equation we get
begin{aligned}
r_{J'M'} &= r_{SM} left( frac{ M_{J'} M_{M'} }{ M_{S} M_{M} } right)^{1/2} \
&= r_{SM} left( frac{ M_{J'} }{ M_{S} } right)^{1/2}
end{aligned}
which evaluates to about 0.03 AU. So the Earth'-Moon' system should have an orbital radius around Jupiter' of at least 0.03 AU.

Now the question is if the Earth-Moon system can remain bound in its orbit around Jupiter. To answer this we can safely ignore the Moon' and assume it remains tightly bound to the Earth'. The potentially destabilizing force on the Jupiter'-Earth' system is again the gravitational pull of the Sun', but now on the Earth'. We can do a similar trick as before, but now we need to account for a different mass ratio. So lets require that the fraction of forces be equal in both cases:
$$
frac{F_{SM}}{F_{EM}} = frac{F_{S'E'}}{F_{J'E'}}
$$
with some algebra this gives
$$
r_{S'J'} = r_{J'E'} frac{r_{SM}}{r_{EM}} left(frac{M_E}{M_J}right)^{1/2}
$$
which for $r_{J'E'} = $0.03 AU evaluates to a minimum orbital radius of about 0.7 AU.

This greatly surprised me as it means you can comfortably fit the entire Jupiter'-Earth'-Moon' system inside the habitable zone, meaning that in principle life should be possible.

Of course the seasons on such a planet would be rather extreme.

telescope - PowerSeeker 114Q - Astronomy

The PowerSeeker will come with a manual on collimation of the mirrors. As the light entering the telescope hits the primary lens, the light reflects and compresses onto the secondary mirror, in turn reflects and compresses the light onto the eyepiece. When the mirrors are misaligned, the light rays aren't correctly focused onto a point where they are projected through the eyepiece, as such like turning a mirror around your point of view, you will not see directly behind yourself but rather an angle.

This is what is happening even as you look directly down the tube at the primary mirror; if it is not directly facing you, you will not see your face (perhaps part of it) and you will see the sides inside the tube.

There are various method for collimation, personally I use a laser collimater and a barlow to increase the sensitivity. The methods are dependent on what you have available however, you don't need any additional tools to collimate apart from the telescope itself.

The PowerSeeker does not have a central marking on it's primary mirror to indicate the center of the mirror, if you are confident in disassembling the rear of the tube you can remove the mirror and measure the central point and mark it. If not, you will not get a collimation as precise, however generally it will suffice until you gain experience.

When looking through an empty eyepiece barrel, if the primary mirror doesn't appear central in the view, as in you can see the whole mirror and the spacing around the mirror is equal, then the secondary mirror needs adjustment. In the same respect, if the central point of the primary mirror does not line up in the center of view, the primary mirror needs adjusting. You do not need to get the line of sight with the primary mirror 100%, typically users of telescopes may use a star itself for collimation; too lengthy to explain however when viewing a star outside of focus (best to use largest magnification possible/smallest diameter eyepiece) the airy disk of a star will appear distorted to one side, centralizing that airy disk will provide an accurate collimation.

Further information will be available in the manual, or there are a many websites to help finetune collimation. If you need further advice just edit your above post or post a comment.

size - How can the 13.8 billion years old universe have a radius of 46 billion light years?

Far away galaxies are receding from us faster than $c_{0}$ = 299 792 458 m/s, but it does not mean that they are breaking the speed of light limit. These galaxies would still measure the speed of light locally to be $c_{0}$ (assuming the speed of light and the laws of physics are the same throughout the universe), and they would never catch up and move faster than photons moving past them: photons passing by these galaxies in the direction away from us still recede from us faster than these galaxies.

We observe that on average the further away a galaxy is the faster it recedes from us. If we consider that our galaxy is not in a special place in the universe (just like our Sun is not in a special place in our galaxy and like Earth is not a planet around which the rest of the universe revolves), then we think any other galaxy should make the same observation than ours, that is that other galaxies recede away from them and that the further away the faster they are seen receding. We call that the expansion of the universe.

Some people will tell you that this expansion is due to space being "stretched" or "created" between the galaxies, as if space was an entity or substance actually being stretched or created. To me this is a poor explanation because no one has ever detected that entity or substance we call space, for now it is a theoretical abstraction rather than something having a tangible reality.

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In case you're wondering, the story as it goes is that the furthest matter we see (the Cosmic Microwave Background or CMB for short) was 42 million light years away from us 380 000 years after the big bang, back when it emitted the light we are detecting now, and this matter is now 46 billion light years away (which we call the radius of our observable universe).

The reason we cannot see light from further back than 380 000 years after the big bang is that before that time, the universe is thought to have been too dense to allow for light to travel freely without being constantly reabsorbed by the surrounding matter.

The reason it took so long for this light to reach us is that 380 000 years after the big bang, the matter that was 42 million light years away (the matter which emitted the light that we call the CMB) was receding much faster than $c_{0}$ from the soon-to-be Milky Way, hence the light it emitted towards us was in fact getting further away.

And the reason this light reached us at all is that during the first 9 billion years following the big bang, the universe is thought to have been decelerating (as opposed to the last 5 billion years when it is thought to have been accelerating). What this means is that galaxies (or soon-to-be galaxies) were moving away from each other at a decelerating velocity during the first 9 billion years. The matter which emitted the CMB didn't slow down enough to recede at less than $c_{0}$ from us, but the light that it emitted 380 000 years after the big bang eventually reached a point where galaxies receded from us slower than $c_{0}$ , i.e. a point where that light started to get closer to us (in other words, the spherical region inside which galaxies recede from us slower than $c_{0}$ enlarged fast enough so that the light of the CMB eventually entered it), and 13.8 billion years later we receive it at last.

The values for the age of the universe and the radius of the observable universe are calculated by solving the equations of Einstein's theory of general relativity, assuming the universe is homogeneous and isotropic, and using cosmological observations to estimate the various parameters in the equations.

If you want to learn more, this is a useful read about galaxies receding from us faster than the speed of light and the expansion of the universe:
Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe

Moving from computerised to normal amateur telescope?

A Celestron Nexstar 130 SLT is not suitable for spectroscopy and the only photography that it will be mush use for would be afocal or DSLR imaging of the Moon (or Sun with a suitable filter).

For more serious work you will need a equatorial mounted scope with a better focuser.

Exactly which one to choose will depend of your budget and what you want to do.

When you say the computer does not work can you be more specific.

orbit - Why don't we have 2 Summers and 2 Winters?

I suspect you're thinking that we'd have a summer when the northern hemisphere, for example, is tilted toward the Sun, and a second summer during the perihelion, when the Earth is closest to the Sun. For one thing, the timing doesn't work; the perihelion takes place in early January, close to the northern midwinter. That probably moderates the effects of axial tilt for the northern hemisphere (and amplifies them for the southern hemisphere), but it's not enough to override them.

The other answers have said that the axial tilt is a more significant
factor than the variation in distance from the Sun, but they haven't
explained why.

The following is a rough back-of-the-envelope guesstimate.

The difference in illumination caused by the varying distance from
the Sun can be computed from the ratio between the perihelion and
aphelion distance, which is about a factor of 0.967. Applying the
inverse square law indicates that amount of sunlight at aphelion is
about 93.5% of what it is at perihelion. Reference:
http://en.wikipedia.org/wiki/Perihelion#Planetary_perihelion_and_aphelion

At my current location (about 33° north latitude), at this time of
year (close to the northern winter solstice), we're getting about 10
hours of sunlight and 14 hours of darkness each day. (Reference: the weather app on my phone.) That's about 83% of what we'd get with 12 hours of daylight during
either equinox, and about 71% of what we'd get with 14 hours of daylight and 10 hours of darkness
per day during the summer solstice. The effect is greater at higher latitudes.

In addition to that, the sun is lower in the sky during the winter
than it is during the summer, meaning that a given amount of sunlight
is spread over a larger area of the Earth's surface, which makes
the ratio even larger.

I don't have the numbers for that, but it's enough to show that the
effect of the axial tilt is substantially greater than the effect of
the varying distance between the Earth and the Sun.

galaxy - Stars in star clusters in SMC and LMC

Is there any catalog or any paper published in any journal that lists the stars discovered under whichever star cluster of the Small Magellanic Cloud (SMC) they belong to? There is one for Large Magellanic Cloud (LMC) in the paper Efremov 2003 "Cepheids in LMC clusters and the period-age relation". But I can't find any for SMC.

It would be better still if the stars happen to be Cepheids.

What would happen if a Black Hole and White Hole Collided?

From Wikipedia:

In quantum mechanics, the black hole emits Hawking radiation and so can come to thermal equilibrium with a gas of radiation. Because a thermal-equilibrium state is time-reversal-invariant, Stephen Hawking argued that the time reverse of a black hole in thermal equilibrium is again a black hole in thermal equilibrium. This implies that black holes and white holes are the same object

The excerpt above was an argument made by Stephen Hawking, explaining that a white hole and a black hole are essentially the same object. With this in mind, we can assume that the collision will be between two black holes. The result will then be the radius of the black hole added to the radius of the white hole.

temperature - Limit of hotness!

Perhaps more fundamentally than temperature, the 'hotness' or 'coldness' of a system can be described by the thermodynamic beta instead, which just like temperature is determined by the rate of change of entropy with respect to energy (at constant volume and particle number):
$$betaequivfrac{1}{k_text{B}}left.frac{partial S}{partial E}right|_{V,N}!!=frac{1}{k_text{B}T}text{.}$$
Intuitively, the thermodynamic beta measures the coldness of a system from $-infty$ to $+infty$. Since temperature per se is just the reciprocal of $beta$, up to a factor of Boltzmann's constant $k_text{B}$, that means the coldest is at $Tto 0^+$, approaching zero from positive temperature side, while the hottest is $Tto 0^-$, from the negative. Treating $+0,mathrm{K}$ as different from $-0,mathrm{K}$ may seem bizarre, but that's what we get for taking a reciprocal.

Negative temperatures ($beta<0$) are hotter than any positive temperature ($beta > 0$). They can happen when a system has more high-energy states occupied than low-energy states, such as during population inversion of any laser, or during similar astrophysical phenomena.

gravity - How can Halley's Comet be predictable?

Each time Halley's comet passes us, we can make a pretty good estimate of its current orbit, and determine how close it will get to the massive bodies of the solar system like Jupiter, Saturn, Uranus, or Neptune on its next orbit. We can make good estimates of gravitational perturbation effects, and thereby know where to look for it.

In fact, astronomers have been doing exactly that for centuries with remarkable accuracy:

Halley thus concluded that all three comets were, in fact, the same object returning every 76 years, a period that has since been amended to every 75–76 years. After a rough estimate of the perturbations the comet would sustain from the gravitational attraction of the planets, he predicted its return for 1758.

Halley's prediction of the comet's return proved to be correct, although it was not seen until 25 December 1758, by Johann Georg Palitzsch, a German farmer and amateur astronomer. It did not pass through its perihelion until 13 March 1759, the attraction of Jupiter and Saturn having caused a retardation of 618 days. This effect was computed prior to its return (with a one-month error to 13 April) by a team of three French mathematicians, Alexis Clairaut, Joseph Lalande, and Nicole-Reine Lepaute.

However, Halley's position turns out to be hard to predict for time scales much beyond that of a single orbit; over the long term, the orbit is chaotically unstable and it's possible that it will collide with a planet, or be ejected from the solar system, within some hundreds of thousands of years.

You mention:

Its not a very big object in space so it seems to me that its orbit could be changed very easily.

As it turns out, the mass of an object in space doesn't affect how the gravity of other bodies changes its trajectory. A feather, a baseball, a comet, and a small moon making a close flyby of Jupiter along the same trajectory, one after another, will all exit Jupiter's neighborhood in the same direction and at the same speed. The small moon will have more effect on Jupiter than the feather will, but aside from that there's no difference.

earth - Calculate Day of the Year for a given date

This would really be better off in Stackoverflow, or another page. Presumably, you assumed some of the calculations rely on concepts in Astronomy, which as far as I can tell having gone through them, they do not with the exception of the concept of leap years; which I think is general enough that it doesn't really mandate being in Astronomy. Regardless:

The equation is a little abstract, so probably the easiest way to understand it would be through understanding each bit of data. Let's start with N2, I'll come back to N1:

N2 = floor((month + 9) / 12)

N2 will equal 0 if month is less than 3, and will equal 1 if it is greater. This formula is to determine whether February has passed.

N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))

N3 will be equal to 2 if the year is not a leap year, and will be equal to 1 if it is a leap year. The math here is just to determine whether the current year happens to be a leap year.

N = N1 - (N2 * N3) + day - 30

Now we take N1 (which I haven't covered yet) and we subtract it by the product of N2 and N3, which will equal 0 if we haven't reached March, or if we have it will be 1 on a leap year, and 2 on a non-leap year. We then add the number of the current day of the month, and subtract 30. So what does N1 do?

N1 = floor(275 * month / 9)

Well, lets look at the values returned compared to days in a non-leap year:

Month     N1         Day       Diff
1.        30         31          -1
2.        61         59          +2
3.        91         90          +1
4.        122        120         +2
5.        152        151         +1
6.        183        181         +2
7.        213        212         +1
8.        244        243         +1
9.        275        273         +2
10.       305        304         +1
11.       336        334         +2
12.       366        365         +1

A bit messy, right? Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months prior to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:

Month     N1         Day       Diff
1.        31         31           0
2.        61         59          +2
3.        92         90          +2
4.        122        120         +2
5.        153        151         +2
6.        183        181         +2
7.        214        212         +2
8.        245        243         +2
9.        275        273         +2
10.       306        304         +2
11.       336        334         +2
12.       ---        365         ---

From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later. This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:

N = N1 - (N2 * N3) + day - 30

We take N1 and subtract in by the value of (N2 * N3), which will be 2 on a non-leap year if February has passed, 1 on a leap year, or 0 if we haven't passed February. We add in the days in the current month, and we subtract 30 to get the offset in N1 values.

This should consistently give you the day of the year. As for special significance, they're being used only for mathematical properties. Dividing a number between 1 and 12 by 9, just so happens to equal less than 1 for numbers less than 3, and greater than or equal to 1 and less than 2 all the way up to 12, which is why it is we have started March. The 275 * (month / 9) formula I'm not sure where they came up with. AFAIK, they have no special meaning to time or the Sun-Earth system - rather, they were chosen because they calculate well in this particular usage.

exoplanet - Do any known exoplanetary/solar bodies have "annular" eclipses similar to Earth's?

This will depend the position of the observer and (obviously) the relative sizes of the star and the eclipsing body.

For an intelligent observer standing on the surface of a planet the most obvious and likely candidate for an eclipsing body would be a moon of that planet. The nearer the planet is to the sun the larger that moon has to be, and the further the planet is away from the star the smaller the moon can be.

Finding planets around stars is hard enough, finding moons that will be orbiting quite closely around those planets across interstellar distances is even harder (I'm not going to say impossible, but it's pretty close) with current technology, so at the moment the answer is "None, that we know of".

Also don't forget that, according to current theories, the moon was formed when a large Mars sized body collided with the nascent Earth, the chances of another Earth/Moon type system is probably remote. This means that most natural satellites are likely to be Phobos sized bodies. Therefore to get an eclipse the planet would have to be further out from the star (so the star looks smaller) and that would tend to put it on the edge of the "Goldilocks" zone, making the likelihood of there being an intelligent observer somewhat rare.

supermassive black hole - What if our galaxy didn't have a SMBH?

The supermassive black hole (SMBH) in the center of the Milky Way (MW) — called Sgr A* [Sagittarius A-star] — has no direct impact on our galaxy. Its mass is only a few million Solar masses, and if you remove it$^dagger$, it will only affect the most central stars, which would suddenly continue in straight paths out through the MW. These stars would almost surely not hit any other stars or something like that (since stars are really, really far apart), but some of them have velocities high enough that they may escape the MW.

If Sgr A* weren't there to begin with, things might look a little different. There seems to be a relation between the mass of a galaxy's SMBH and the velocity dispersion of the stars in its central bulge; the so-called M-sigma relation. so MW without Sgr A* would mean a more ordered center. Our Solar system is located in the disk, far from the center, and their is evidence that SMBHs have little impact on the disk (Gebhardt et al. 2001). However, in their early phase (as an active galactic nucleus), their extreme luminosities cause galactic superwinds which blow out gas and may quench star formation (Tombesi et al. 2015).

$^{^dagger}$^{Removing Milky Way's SMBH is left as an exercise for the reader.}

big bang theory - Is it just the observable universe that is expanding?

There is no evidence to support the idea that some part of the universe (at a cosmological scale) is contracting. Obviously one could construct such a theory but there is currently no observational data to support it.

Until relatively recently there were three broad ideas about the Universe's expansion - that gravitational pull would slow the expansion but by never enough to halt it, that (a boundary case) there was exactly enough matter in the universe for expansion to halt at an infinite point in the future, or that eventually gravitational attraction would first slow and then reverse the expansion, leading to a big crunch.

In fact, though, the observational evidence suggests that the expansion of the Universe is accelerating - that there is some "dark energy" that is speeding up the expansion. This is now the generally accepted view, though arguments continue as to what this "dark energy" is.

What is the shape and size of a gamma ray burst?

When "shape" is used in the context of gamma-ray bursts, it is typically used to refer to the jets emitted on opposite ends of an axis through the source. These jets contain huge amounts of energy, most in the form of photons so high in energy that they are classified as gamma-rays, some of the highest-frequency photons known.

The jets are typically emitted along an axis through the progenitor object (I'm not going to talk about the object itself, as the causes of GRBs are not yet fully understood). A quick search engine search (I prefer duckduckgo, but you might like Google better) can give you hundreds of pictures (mostly artists' impressions) of GRBs. Notice how thin the jets are. They look like long cylinders, but closer inspection shows that they aren't. They are actually cones (truncated ones, but cones nonetheless).

The reason that these jets spread out is relativistic beaming, which also affect the beam's luminosity. The angle at which the beams diverge is so slim - in fact, this pre-print says that one observed GRB had a jet emitted at an angle of one tenth of a radian! That's only 18 degrees. (Note: On page 21, this calls an angle of 3 degrees "typical" for some jets ).

So the answer to your question is that gamma-ray bursts are typically conical, although the jets are often nearly cylindrical. The first paper I mentioned also did an analysis of spherical models, but concluded that they did not fit observations.

space time - What is gravity really?

I can attempt to address the second part of your initial question: "Is it a particle, a wave,...?" Einstein's theory of general relativity states that mass and energy bend space-time. Space-time, in turn, tells matter how to move (John Wheeler put this more elegantly).

This concept is completely different from the theories of the other three fundamental forces (electromagnetism, and the strong and weak nuclear forces). In these quantum theories, forces are mediated by particles called gauge bosons. Electromagnetism is carried by photons, the strong nuclear force by gluons, and the weak nuclear force by W+, W-, and Z bosons. There have been many attempts to find a quantum theory of gravity - that is, to use quantum principles to construct a field theory of gravity. In these theories, gravity would indeed be mediated by a particle, dubbed the graviton. String theory is one example of these theories; the physics community is divided about it.

You probably have heard other terms tossed around that embody interesting concepts. A gravitational wave is essentially a ripple in space-time emitted by an object or system of objects. There are strict restrictions over what sort of objects can emit these waves; binary neutron stars are one consistently cited example. These waves should not be confused with the aforesaid hypothesized gravitons; while gravitational waves carry energy, they do not "mediate" gravity.

Finally, I have heard the term gravity wave used in a completely different context. I can't describe it as well as the other concepts, but I believe it is used to refer to an effect of gravity on other substances. I would advise looking for an answer to that in a textbook. At any rate, it is unrelated to a gravitational wave.

So, basically, the predominant view in the physics community is that general relativity is the best description of gravity; at the moment, gravity is considered to be the bending of space-time, so it is indeed "something else entirely." Many theories of quantum gravity, including string theory, however, attempt to create particles called gravitons as force-carrying bosons. If evidence is found relating to these theories, then we may well learn whether or not gravitons do exist. One more thing about waves: Because of the quantum concept of wave-particle duality, any particle can be described as a wave, which has a wave function. So if gravity is due to a particle, then it is also due to a wave!

observation - Does CIBER Experiment from Caltech suggest that there can be lots of stars which are not in any galaxy?

Does that suggest that about a half of the stars in the observable universe could not belong to any galaxy?

Not really. A key sentence in the article is "The best interpretation is that we are seeing light from stars outside of galaxies but in the same dark matter halo". So the stars are still within the dark matter halo of a galaxy, but are outside the boundary of the galaxy if the dark matter halo is not considered.

Furthermore the "intrahalo light" explanation is just one of two possible explanations according to Updated analysis of near-infrared background fluctuations which explains:

Two scenarios have been proposed to interpret the clustering excess. The first advocates the contribution from intrahalo light (IHL), i.e. relatively old stars stripped from
their parent galaxies following merging events. These stars
therefore reside in between dark matter halos and constitute a low-surface brightness haze around galaxies. The IHL
is expected to come mostly from low redshifts (1 + z <
∼ 1.5)
systems (Cooray et al. 2012b; Zemcov et al. 2014).

The second scenario is instead based on the presence
of a class of early, highly obscured accreting black holes
of intermediate mass (∼ 10^4−6M⊙) at z >
∼ 13 (Yue et al.
2013b, 2014). As a suitable mechanism to produce such
objects does exist – the so called Direct Collapse Black
Holes (DCBH, for a concise overview of the problem see
Ferrara et al. 2014), and the interpretation of the super-
massive black holes observed at z = 6 seemingly requires
massive seeds (Volonteri & Bellovary 2011), such hypothesis seems particularly worth exploring.

Both scenarios successfully explain the observed clustering excess, albeit with apparently demanding requirements. In fact, if the excess is to be explained by intra-
halo light, then a large fraction of the stars at low-z must
reside outside systems that we would normally classify as
“galaxies” (Zemcov et al. 2014). On the other hand, in the
DCBH scenario the abundance of seed black holes produced
until z ∼ 13 must represent a sizeable fraction of the estimated present-day black hole abundance, as deduced from
local scaling relations (Kormendy & Ho 2013) and recently
revised by Comastri et al. (2015). However, it is important
to outline that both scenarios are not in conflict with any
known observational evidence

photography - How tall are the "fractured" linear structures on Europa's surface?

What are the altitude differences in the high resolution part of this image of Europa?

How high do those ridges rise over their neighboring valleys? And how large is the area shown in this image?

And are the craters believed to have been formed by impacts or some kind of "eruptions" from below? It looks to me as if the areas which have the most (tiny) craters is on top of and thus newer than the less cratered areas. Well, here and there anyway.

orbit - Does the Sun turn around a big star?

Does the Sun turn around a big star?

No. Such a star, if it existed, would easily be the brightest star in the sky. You would have been taught about it early on in school if it existed. But it doesn't.

For a while it was conjectured that the Sun had a small companion star to explain a perceived periodicity in mass extinction events. This too has been ruled out by the Wide-field Infrared Survey Explorer.

What are all the intermediate subsystems up to motion around the center of the Milky Way?

Our Sun, being a single star, is a bit of an oddity. Most stars are members of multiple star systems, typically pairs.

Some stars occur in clusters. The Pleiades is a relativity nearby (440 light years) cluster of stars. Someone with extremely keen eyesight and exceptionally good viewing conditions, might be able to see 14 stars of the over 3,000 stars that form this cluster. Open clusters such as the Pleiades don't last long. The stars in an open cluster are only weakly bound to the cluster and are eventually dispersed.

A key feature of the Milky Way is its spiral arms. Our Sun is currently in a lesser arm of the Milky Way, the Orion Arm. Stars however are not gravitationally bound to spiral arms. One widely used explanation of the spiral arms is that they are gravitational traffic jams in space.

We could also ask the same beyond...

Our galaxy is a member of the Local Group, which in turn is a member of the Virgo Supercluster, which in turn is a part of the Laniakea Supercluster. Even larger scale objects include galaxy filaments. And that's where the hierarchy ends. The expansion of space overtakes gravity at such immense distances.

Why are there no Oe or Ae stars?

Well, there might be.

I strongly disagree with your statement that circumstellar discs are what differentiate B(e) stars from other related stars. In fact, there are four primary criteria for stars that satisfy "the B(e) phenomenon":

1. Strong emission lines in the Balmer series


2. Emission of lines of certain (low ionization) metals such as Fe II


3. "Forbidden emission lines" in the visible part of the electromagnetic spectrum such as O I


4. Hot circumstellar dust

The dust is not the key to the B(e) phenomenon, though it can change the spectrum slightly.

HAeB(e) stars are a class of stars (and a subclass of B(e) stars) related to Herbig Ae/Be stars, which have not yet left the main sequence. There are additional criteria that a star must satisfy to be considered an HAeB(e) star, too. These stars can be of spectral type A, meaning that you can have B(e) stars that are not of spectral type B, if you use a certain definition.

This answer is based largely on Lamers et al. (1998), a fascinating paper on the classification of B(e) stars.

expansion - On what scale does the universe expand?

The short answer is: gravitationally bound system will not be ripped by the accelerated expansion.

A longer answer: The current standard model ($Lambda$CDM model) says that everything started with the big bang. It released a lot of energy that pushed the Universe and is since then expanding. While it's expanding, gravitational attraction has been working to create the structures that we see today (planets, stars, galaxies, cluster of galaxies).

As the Universe expands, the density of matter and radiation within the Universe decreases. At about 5 billions years after the big bang, these densities became smaller than the density of Dark Energy (Dark-energy-dominated era). The strange things about Dark Energy are that, 1) its density remains a constant regardless of the expansion of the universe, and 2) it has anti-gravitational properties - it pushes rather than pulls. Once Dark Energy came to dominate the density of the universe it began to dominate its overall dynamics, and has caused the expansion of the universe to accelerate.

But in structures which have already formed, gravity is much stronger than dark energy and so they don't feel it.

Why does Pluto's Orbit underlap Neptune's Orbit

There are a few peculiarities of Pluto's orbit. These are:

• Its high orbital eccentricity (e = 0.25) causes Pluto's perihelion to be ever so slightly smaller than Neptune's perihelion.




• Pluto has an orbital resonance with Neptune. Its orbital period is exactly 3/2 of Neptune's. This orbital resonance is the cause of its orbital eccentricity that you ask about.




• Pluto has an orbital inclination of 17 degrees to the ecliptic. No one knows why, we can only guess.

The origin of Pluto's eccentricity explained:

During the later stages of planet formation, in this early Solar system, the planets were still forming. Neptune exchanged angular momentum with the remaining planetesimals, and its orbit expanded outwards. If Pluto were in a near-circular orbit larger than Neptune's (at about 33 AU), there is a high chance that Neptune could capture Pluto and lock it an orbital resonance (this probably happened when Neptune was at about 25 AU). As Neptune's orbit continued to expand outwards, this expansion drew out the eccentricity of Pluto's orbit.

Source and further reading: The Origin of Pluto's Orbit: Implications for the Solar System Beyond Neptune

exoplanet - Motion of rogue planets

1) this question has no real answer as it depends on the reference frame being used. It is very unlikely that they will be stationary except in their own reference frame.

rephrased 1) In theory their paths can be calculated precisely if you know their speed and the positions and motions of all other bodies (also gas and dust clouds) that might influence their motion. Practically, you will not have that knowledge, as most non-stellar bodies that the rogue planet may encounter will be very dim and you will not be able to detect them. On the other hand space is for the most part empty, so you might reasonably expect their paths to be quite predictable (within a limited time frame). They'll probably be in orbit (either an elliptical, parabolic, or hyperbolic) around the centre of the milky way.

2) yes, they will probably have some angular momentum left from where they formed.

Space is very large so it is not very likely that one will appear in our solar system in our lifetime. And of course we do not know if none ever moved through the solar system.

Calculate angle of inclination for a certain angle in a planet's orbit?

In short, no, you don't have to solve Kepler's equation. If you define a viewing longitude and you know the full set of Keplerian orbital elements, you can calculate if a planet will show a transit.

Let me first clear up some of the jargon, though. The term inclination is traditionally reserved for the angle between the orbital plane and the reference plane. So unless the entire orbit is wobbling (for instance due to precession) the inclination will be constant for all positions of the planet.

The angle that varies as the planet moves through the reference plane is the latitude. However, in this particular problem calculating the latitude isn't particularly useful. The more practical coordinate to look at is the height of the planet away from the plane, I'll call this $r_z$. If along your observers line of sight you find $r_z$ to be smaller than the solar radius, the planet will show a transit.

Working out a fully general elliptical orbit in 3 dimensions is somewhat tricky, so I'll start out with a simple Cartesian coordinate system aligned with the orbital plane. In subsequent steps I work my way back using a number of coordinate transformations.

First a list of the Kepler parameters and their meaning:

• $nu$ - true anomaly - The anglular position with respect to perigee in the orbital plane.


• $omega$ - argument of periapsis - The angular position of perigee with respect to the ascending node in the orbital plane.


• $lambda_{asc}$ - the longitude of ascending node in the invariable plane ($Omega$ in the picture).


• $i$ - inclination - the angle between the orbital plane and the invariable plane, such that $lambda_{asc}$ gives the rotation axis.

Now, in the orbital plane, the radius of the planet orbit is given as
$$
r(nu) = frac{a(1-e^2)}{1+ecos(nu)}
$$
If I define the x-axis to align with the ascending node, I can write the position vector as
$$
vec{r}_{orb} = r(nu)
begin{bmatrix}
cos(omega+nu) \
sin(omega+nu) \
0
end{bmatrix}
$$

To move to a Cartesian coordinate system aligned with the invariable plane I have to apply a rotation of angle $i$ about the x-axis, so
$$
vec{r} = R_x(i)vec{r}_{orb}= r(nu)
begin{bmatrix}
cos(omega+nu) \
sin(omega+nu)cos(i) \
sin(omega+nu)sin(i)
end{bmatrix}
$$
So now I have an expression for $r_z$, I just need to know how $nu$ relates to the planet longitude in the invariable plane, $lambda_{planet}$.

I can find the longitude of the planet relative to $lambda_{asc}$ as
$$
lambda'_{planet} = arctan( frac{r_y}{r_x} )
$$
such that the actual longitude is
$$
lambda_{planet} = lambda_{asc} + lambda'_{planet}
$$
Inverting this last expression gives
$$
nu(lambda) = arctan( frac{tan(lambda - lambda_{asc})}{cos(i)} ) - omega
$$

Now I have all the equations to answer the question:
If the planet is along the observer's line of sight, then is has a true anomaly
of $nu(lambda_{obs})$, which gives a height above the plane
$$
r_z(nu) = r(nu)sin(omega+nu)sin(i)
$$
If $|r_{z}| < R_{sun}$ the planet can be seen in transit.

iss - Could a harpoon-like gun be used by an astronaut to stop drifting away from a ship?

Problem with what you propose, a kind of harpoon with probably a soft-tipped and magnetic head tethered projectile so it doesn't damage / penetrate the station's hull yet still holds onto it once it would reach it, is that the shooting of a kinetic projectile in one direction would propel you with equal force in the opposite direction since there's nothing to hold on to in a freefall orbit around the Earth and the law of conservation of momentum as implied by Newton's third law of motion states that: "when one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction to that of the first body".

                                  

                      ^{Newton’s 3rd Law of Motion: For every action there is an equal and opposite reaction}

Now, the projectile would probably be a lot lighter than your own and the extravehicular suit's total rest-mass, so the dart would still propel relative to your initial point faster towards the station than you would be pushed the other way (with magnitude of linear momentum being $p = mv$, where $m$ is the mass of the object and $v$ its velocity), but since you're already moving with some velocity away from the station, otherwise you wouldn't have these problems to start with, the difference might not be enough and you possibly wouldn't have a tether long enough to reach the station. And you should really try not to tug on the tether while still flying towards the space station and unwittingly reduce its velocity and make it twist and bend.

Targeting wouldn't be easy either, since you'd have to fire it from the center of your mass perpendicular to the remaining two axes than the one you're firing in, not to merely end up in an uncontrollable spin. At the exactly the right time too, since you'd most likely already be spinning relative to the station as a result of your unexpected separation from it. To put it differently, for example, if you were standing on ice and fired a gun from your right hand at your waist, even if you managed to keep your balance after the gun recoils (something you wouldn't be able to in microgravity of space, remember, there's nothing to really hold on to), you would end up in a clockwise spin on the ice with only a small drag of the ice on your shoes eventually stopping you. There is no meaningful drag in outer space (at least not meaningful for our case), so you would spin indefinitely or until some other body with mass absorbs your angular momentum. The physics behind this are explained a bit more in the LiveScience article on What Would Happen If You Shot a Gun In Space?

Anyway, there are better ways though and a lot of research went into developing various astronaut propulsion units that are constantly improved on throughout the manned space exploration history. What is common to all of them though is that they try to control the relative movement on all three axes, and your pitch, roll and yaw.

   

   ^{Edward H. White II during Gemini-Titan 4 egress from the spacecraft carrying a Hand-Held Maneuvering Unit (Source: Wikimedia)}

Early versions of astronaut propulsion units were handheld and that didn't prove too intuitive to control and had too limiting amount of propellant to be really of much use except for providing short entertainment for some and mild annoyance for others of its early astronaut testers (they were only ever tested while the astronauts were tethered to the space vehicle), but later versions (e.g. the Manned Maneuvering Unit) evolved to be easier, more intuitive to maneuver and carry useful amounts of onboard propellant to allow for shorter excursions in the space vehicle's vicinity.

The latest development in astronaut propulsion units (Simplified Aid for EVA Rescue or SAFER) is even more advanced propulsive backpack with ability to automatically stabilize uncontrolled rotation of its carrier.

   

   ^{Astronaut Rick Mastracchio working with a SAFER system attached. (Source: Wikipedia on SAFER: Simplified Aid for EVA Rescue)}

It is now a standard equipment carried by every spacewalking astronaut in an Extravehicular Mobility Unit (EMU) during their Extravehicular Activity (EVA) at the International Space Station (ISS). It is however a self-rescue device and would only be used in an emergency. It is not meant as a mobility device to assist astronauts during normal EVA conditions, and ISS spacewalkers are tethered to the space station and use safety grips on its outer hull to push against.

What are Trojan Asteroids and where are they located?

A Trojan Asteroid is defined by the Swinburne University Astro page "Trojan Asteroids" as:

Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points.

These are often divided into 'leading' and 'trailing', as per the diagram below (from the source linked above):

Even though the term Trojan asteroid is often associated with the 4800 or so that lead and follow Jupiter, it actually applies to any asteroid that displays this behaviour, associated with any planet.

There are numerous Trojans co-orbiting with Mars and Neptune. Recently, single Trojans have been found co-orbiting Uranus and even Earth.

Do Pluto and Charon have unusual Langrange points?

The usual examples of Lagrange point mechanics one most commonly encounters, Sun-Earth-L# and Earth-Moon-L#, are examples of 3-body problems where 1 >> 2 >> 3 in terms of their mass. The Pluto-Charon system, however, are much closer in their relative masses, so much so that their barycenter is outside Pluto's surface. From the wiki:

Pluto and Charon are sometimes described as a binary system because the barycenter of their orbits does not lie within either body.[23] The IAU has yet to formalise a definition for binary dwarf planets, and Charon is officially classified as a moon of Pluto.[24]

How does this affect the orbital stability of the 5 PCL points?

telescope - When is Gamma Draconis closest to the zenith in London on April 4th

I am reading Roobert Hooke's paper An attempt to prove the motion of the earth from observations (1674, faksimile). Hooke writes

The principal dayes of doing which will be about the 4 of April, when our Zenith passeth by the said Star at midnight

The said star is Gamma Draconis, and the observations are made in London. As far as I see it, the star is closest to the Zenit at 6 in the morning in London at April 4th.

So, why does Hooke say at midnight? Is this an expression of the 17th century?

At what density does helium burning start in a star?

From this webpage, I have a few statistics regarding required density and temperature to burn a specific element.

Burning phase; Required temperature; Required mean density; Duration

---

Hydrogen burning: $4 times 10^7 text{ degrees K; } 5 text{ gm per cubic cm; } 7,000,000 text{ years}$

Helium burning: $2 times 10^8 text{ degrees K; } 700 text{ gm per cubic cm; } 700,000 text{ years} $

Carbon burning: $1 times 10^8 text{ degrees K; } 200,000 text{ gm per cubic cm; } 600 text{ years} $

Neon burning: $1.2 times 10^9 text{ degrees K; } 4,000,000 text{ gm per cubic cm; } 1 text{ year} $

Oxygen burning: $5 times 10^9 text{ degrees K; } 10,000,000 text{ gm per cubic cm; } 6 text{ months} $

Silicon burning: $2.7 times 10^9 text{ degrees K; } 30,000,000 text{ gm per cubic cm; } 1 text{ day} $

This is data for a star of 25 solar masses.