the sun - What constellation does sun occupy by location and date?

You can see for yourself for any date and location using the free Stellarium program (http://stellarium.org/).

The constellation that the Sun is in as seen from Earth does not depend noticeably on the precise location of the observer on Earth. It depends mostly on the time of year. Here are the approximate days of the year at which the Sun enters a new constellation:

• Aries: April 18


• Taurus: May 14


• Gemini: June 21


• Cancer: July 20


• Leo: August 10


• Virgo: September 16


• Libra: October 31


• Scorpius: November 23


• Ophiuchus: November 29


• Sagittarius: December 17


• Capricornus: January 19


• Aquarius: February 16


• Pisces: March 12

The "approximate" part is that the Sun may be up to one day earlier or later than the quoted dates, just like the start of the seasons may be one day earlier or later than the average. (Actually, it is the calendar that is early or late with respect to the phenomenon.) Also, because of the precession of the equinoxes, these dates shift by about 1 day every 70 years.

See http://aa.quae.nl/en/antwoorden/sterrenbeelden.html#6.

Can earth escape sun's gravity with the help of a black hole heading towards our solar system?

I'm thinking, this is the gist of your question

Earth feels a zero net force. Will it help earth to fly away?

First, zero net gravitational force between two large objects is certainly possible, well, not exactly zero, not for more than an instant anyway, but close to zero, absolutely possible, but whether it leads to an object flying away is more complicated. It depends on the relative motion of the 3 objects.

The Moon, for example, orbits the Earth, but from the Moon's point of view, the Sun is about 333,000 times more massive than the Earth and about 388 times further away, (on average) when the Moon is between the Earth and the Sun (390 times when the Moon is on the opposite side of the Earth, again on average. There's some variation in there).

Because gravitation drops with the square of the distance, 388 times more distant means about 151,000 times less G force at that relative distance, but with 333,000 times more mass, the Moon actually experiences over twice the gravitational tug from the Sun than it gets from the Earth, so, even though, from the Moon's surface, the Earth is much larger than the Sun, the sun's mass is sufficient to exert the greater gravitational pull.

So, if, by some magical power, you were to grab a hold of the Earth and stop it from moving and grab a hold of the Moon and stop it too, then let the Moon go, the Moon would fall more towards the Sun than the Earth cause the gravitational pull in that direction is over twice as much. (Ask this great magical being not to let go of the Earth, because if he does, the Earth would fall into the Sun too).

That's not quite the same as your scenario but it points out that zero net gravitation doesn't govern where an object ends up. The Moon orbits both the Earth and the Sun, and it's in a stable orbit around the earth even though it's feeling more gravitation from the Sun. That's because the Moon is inside the stable part of the Earth's Hill Sphere.

In your scenario, however, a passing object the mass of another star could certainly pull the Earth away from the Sun. It wouldn't even need to achieve a net zero gravitation to accomplish that, nor would it need to be nearly so massive.

The picture below covers the Earth orbiting around the Sun. If you bring the net Gravity to zero, in theory the "F" in the diagram shrinks to zero and the Earth continues straight in direction V for that time period, increasing it's distance from the Sun. Source

The Earth's tangential velocity relative to the sun is 30 km/s and it's escape velocity is just the square root of 2 times that, about 42.5 km/s, so an acceleration of the Earth of 12.5 km/s or moving the Earth to an orbit a bit outside Mars' orbit and keeping the velocity the same would both work (or some combination of the two).

The model is a bit more complicated because a gravitational assist, which would also happen in your scenario and a gravity assist can work both ways, increasing or decreasing the orbital velocity. It's possible, depending on direction of the pass, that a passing star could push the Earth closer to the Sun, even passing outside, if it slows the earth's velocity by gravity assist. Drawing it away isn't the only possible outcome.

More on gravity assists here, Short and Longer.

As James Kilfinger points out, stars passing that close is extremely extremely rare so this kind of thing, for all practical purposes, virtually never happens. It's much more rare than a dinosaur killing meteor for example. It's hugely unlikely.

amateur observing - Watching the Mercury transit with improvised devices

Mercury's angular diameter on transit day will be 12 arcseconds.
A camera obscura using a 12 mm aperture could resolve it; one lens from +0.75 diopter reading glasses, if you can get them, will project a bright 12 mm image of the Sun at a distance of 1.33 m.
Note that a larger aperture or a shorter focal length will make the Sun image hotter than direct sunlight unless you add a filter.

Test with sunspots before relying on it for Mercury.
If it counts as improvised, projection with 7x35 binoculars easily showed me the 2012 transit of Venus.

early universe - Is it possible to get a glimpse of the Big Bang through gravitation waves?

Gravitational waves from the big bang may be "heard" but not by LIGO. The waves emitted at or around the inflationary epoch of the big bang are expected to be at much lower frequencies (milli-Hz or lower) than those announced today by LIGO. There are various sources of noise that make LIGO insensitive to GWs at frequencies below about 10 Hz.

It will take space-based interferometers like the proposed LISA, with longer interferometer arms and well away from terrestrial sources of noise to stand a chance of detecting such GWs.

If they are detected - they might "sound" something like this (if upshifted into the audible range) - from the LIGO website. It sounds like white(ish) noise because of the broad continuum of frequencies expected.

size - How thick can planetary rings be?

There is an explanation for why rings flatten out here. The general mechanism is that particles collide, and gets a very uniform momentum. Thus, any set-up giving unusually thick rings is in essence "cheating".

Here are some ways:

Moons can cause spiral waves in the rings, giving them more of a structure in the z direction. The ones known in Saturn's rings has a modes amplitude of just 10-100 m, but larger Moons can easily increase that.

Another way is simply having massive rings. Then they can not get more flattened, as there are no more empty space to remove.

A tilted ring relative to the Planets orbit around the star is going to experience tidal forces, as long as the radius of the rings is some notable fraction of the planet's orbital radius. From the context that sparked the question, that is not a suitable mechanism though, along with the possibility of having a so low density that particle collisions are rare.

However, more promising:

The halo ring of Jupiter is estimated to be around 12500 km thick (about the same as the diameter of the Earth), and are very fine dust kept from condensing into a disc by both the magnetic fields of Jupiter, and by iterations with the Galilean Moons.

We have four planets with rings in the solar system, so the sample size is quite small. Applying some small-sample-size statistical methodology, in this case an unusual application of the German Tank Problem, we can give a rough but realistic maximum thickness of a ring:

$$N approx m+frac{m}{k}-1$$

Where $m$ is the highest observed value, and $k$ the sample size.

Modified slightly to get a non-integer version that makes some sense, we get:

$$max_{thickness} approx 12500km+frac{12500km}{4} approx 16000km$$

By no means a very certain limit, but at least about what can obtain from what we know.

"Supernova" is the explosion or the resulting celestial body? Is it incorrect to call the explosion "supernova"?

Is it incorrect to call the explosion “supernova”?

Yes and no.

Better said, the explosion is the very first part of a supernova. While the explosion lasts for but a few seconds to a few hundreds of seconds, a supernova can last for hundreds of days. What we see visibly as a supernova are the after effects of that explosion. The explosion can create lots and lots of stuff moving at very high velocities, and lots and lots of highly radioactive nuclei.

If the star had outgassed material prior to the explosion, the highly kinetic material produced by the explosion runs into that previously outgassed material and makes it glow. This takes some time to cool down. The radioactive material produced during the short course of the explosion proper takes time to decay to stable elements. This radioactive decay eventually produces gamma rays, some of which is absorbed by the nearby material, heating it, thereby eventually producing thermal radiation.

For example, a type Ia supernova produces a large amount of nickel-56. This decays to cobalt-56 with a half-life of about 6 days, which in decays to iron-56 with a half-life of about 77 days. The heating that results from these decays is what we see, and because it is so predictable, this is makes type Ia supernovae a fantastic standard candle.

exoplanet - A star a black hole and planets around them

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

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

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

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

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

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

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

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

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

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

the sun - What would the Sun be like if nuclear reactions could not proceed via quantum tunneling?

Short answer: Without tunnelling, stars like the Sun would never reach nuclear fusion temperatures; stars less massive than around $5M_{odot}$ would become "hydrogen white dwarfs" supported by electron degeneracy pressure. More massive objects would contract to around a tenth of a solar radius and commence nuclear fusion. They would be hotter than "normal" stars of a similar mass, but my best estimate is that they have similar luminosities. Thus it would not be possible to get a stable nuclear burning star with 1 solar luminosity. Stars of 1 solar luminosity could exist, but they would be on cooling tracks, much like brown dwarfs are in the real universe.

A very interesting hypothetical question. What would happen to a star if you "turned off" tunnelling. I think the answer to this is that the pre-main-sequence stage would become significantly longer. The star would continue to contract, releasing gravitational potential energy in the form of radiation and by heating the core of the star. The virial theorem tells us that the central temperature is roughly proportional to $M/R$ (mass/radius). So for a fixed mass, as the star contracts, its core gets hotter.

There are then (at least) two possibilities.

The core becomes hot enough for protons to overcome the Coulomb barrier and begin nuclear fusion. For this to happen, the protons need to get within about a nuclear radius of each other, let's say $10^{-15}$ m. The potential energy is
$e^2/(4pi epsilon_0 r) = 1.44$ MeV or $2.3times 10^{-13}$ J.

The protons in the core will have a mean kinetic energy of $3kT/2$, but some small fraction will have energies much higher than this according to a Maxwell-Boltzmann distribution. Let's say (and this is a weak point in my calculation that I may need to revisit when I have more time) that fusion will take place when protons with energies of $10 kT$ exceed the Coulomb potential energy barrier. There will be a small numerical uncertainty on this, but because the reaction rate would be highly temperature sensitive it will not be an order of magnitude out. This means that fusion would not begin until the core temperature reached about $1.5 times 10^{9}$ K.

In the Sun, fusion happens at around $1.5times 10^7$ K, so the virial theorem result tells us that stars would need to contract by about a factor of 100 for this to happen.

Because the gravity and density of such a star would be much higher than the Sun, hydrostatic equlibrium would demand a very high pressure gradient, but the temperature gradient would be limited by convection, so there would need to be an extremely centrally concentrated core with a fluffy envelope. Working through some simple proportionalities I think that the luminosity would be almost unchanged (see luminosity-mass relation but consider how luminosity depends on radius at a fixed mass), but that means the temperature would have to be hotter by a factor of the square root of the radius contraction factor. However, this could be academic, since we need to consider the second possibility.

(2) As the star shrinks, the electrons become degenerate and contribute degeneracy pressure. This becomes important when the phase space occupied by each electron approaches $h^3$. There is a standard bit of bookwork, which I am not going to repeat here - you can find it something like "The Physics of Stars" by Phillips - which shows that degeneracy sets in when
$$frac{ 4pi mu_e}{3h^3}left(frac{6G Rmu m_e}{5}right)^{3/2} m_u^{5/2} M^{1/2} = 1,$$
where $mu_e$ is the number of mass units per electron, $mu$ is the number of mass units per particle, $m_e$ is the electron mass and $m_u$ is an atomic mass unit. If I've done my sums right this means for a hydrogen gas (let's assume) with $mu_e=1$ and $mu = 0.5$ that degeneracy sets in when
$$ left(frac{R}{R_{odot}}right) simeq 0.18 left(frac{M}{M_{odot}}right)^{-1/3}$$

In other words, when the star shrinks to the size of $sim$ Jupiter, its interior will be governed by electron degeneracy pressure, not by perfect gas pressure. The significance of this is that electron degeneracy pressure is only weakly dependent (or independent for a completely degenerate gas) on temperature. This means that the star can cool whilst only decreasing its radius very slightly. The central temperature would never reach the high temperatures required for nuclear burning and the "star" would become a hydrogen white dwarf with a final radius of a few hundredths of a solar radius (or a bit smaller for more massive stars).

The second possibility must be the fate of something the mass of the Sun. However, there is a cross-over point in mass where the first possibility becomes viable. To see this, we note that the radius at which degeneracy sets in depends on $M^{-1/3}$, but the radius the star needs to shrink to in order to begin nuclear burning is proportional to $M$. The cross-over takes place somewhere in the range 5-10 $M_{odot}$. So stars more massive than this could commence nuclear burning at radii of about a tenth of a solar radius, without their cores being degenerate. An interesting possibility is that at a few solar masses there should be a class of object that contracts sufficiently that nuclear ignition is reached when the core is substantially degenerate. This might lead to a runaway "hydrogen flash", depending on whether the temperature dependence of the reaction rate is extreme enough.

Best question of the year so far. I do hope that someone has run some simulations to test these ideas.

Edit: As a postscript it is of course anomalous to neglect a quantum effect like tunnelling, whilst at the same time relying on degeneracy pressure to support the star! If one were to neglect quantum effects entirely and allow a star like the Sun to collapse, then the end result would surely be a classical black hole.

A further point that would need further consideration is to what extent radiation pressure would offer support in stars that were smaller, but much hotter.

history - How was precision astrometry done before digital imaging?

Prior to digital imaging then photographic plate negatives were analysed with scanning microdensitometers to produce astrometric catalogues. Many of these catalogues are still in use today, they are valuable sources of early epoch positions that enable proper motion measurements.

For more details you could look at the descriptions of the SuperCosmos project http://ssa.roe.ac.uk// (which is based on Schmidt plates) or the UCAC4 catalogue, http://ad.usno.navy.mil/ucac/readme_u4v5, which uses plates to get positions and proper motions for faint stars.

In the good old days, before even these catalogues existed, astronomy groups would have copies of the whole Schmidt and Palomar sky surveys. You would put the relevant plate on a massive, concrete-based, hydraulic X,Y measuring machine, with a binocular microscope. You would measure sets of Standard stars, get a 6-coefficient fit to convert x and y into RA and Dec,then find your objects on the plate, measure x,y, calculate RA and Dec. Then snap a polaroid to use as a finder chart at the telescope.

I was doing this as late as 1995 before the advent of the Digitised Sky Survey.

the sun - If there are neutron stars, would most stars be considered "proton stars"?

Protium is a proton + an electron.

Under enormously high pressure, it's energetically favorable for electrons to merge with protons and become neutrons - see here.

are stars mostly protons

By mass, yes, at least before they get too old.

The mass of the universe is more complicated, but anything solid that we think of as matter is made of atoms, which are by mass, mostly protons and neutrons (you can break it up further than that if you like, but that's best for another question).

Most hydrogen has no neutrons so any hydrogen rich object (the sun, most young stars, gas giant planets) are by mass, mostly protons. That's no longer true when a star gets close to the end of it's life and has burned much of it's hydrogen.

Jupiter, by mass is roughly about 80% protons. The sun, because it's been turning hydrogen into helium for about 4.5 billion years, is roughly about 67% protons by mass. The Earth, mostly other elements, Oxygen, Silicon, Iron, etc, is a about 50% proton by mass.

Would stars like our Sun be considered "proton stars"?

I suppose you could use that term, but I don't see any benefit to it. It's not too different than calling the sun a "hydrogen" star. All stars start out as hydrogen stars.

galaxy - Was the Milky Way ever a quasar?

A quasar is simply an active galactic nucleus (AGN) that is viewed from a particular angle; see the picture below, in which quasars are labeled "QSO". This is really a remarkable figure because historically all of the names in the figure were thought to correspond to different types of objects, when really they all refer to the same thing!

Your question really shouldn't be "Was there ever a quasar in the Milky Way?", since the dotted line in the figure would correspond to the Galactic plane and we would not see Sagittarius A* (the Milky Way's super-massive black hole) from the correct angle. A better question might be, "Has Sagittarius (Sgr) A* ever been active?" The answer to that question is yes; according to this page it was probably active (very bright with a jet) about 10,000 years ago. However, at the moment, it isn't really doing anything, since it isn't currently accreting anything (to put it plainly, it isn't eating anything, so it doesn't have enough energy to be active). However, many astronomers (myself included!) are anxiously waiting for a cloud of gas called G2 to fall into Sgr A*. We are hoping that Sgr A* will burp or do something interesting.

astrophysics - In what units to quote the thermal Blackbody temperature

Im not entirely sure what you mean, but the (planck's) formula for blackbody radiation is given by

where h is in [J*s], c in [m/s], lambda in [m], k in [J/K] and T in [K].
So, the temperature is just in Kelvins, not in energy.
This formula, with these units, gives S in [W/m2/m], which describes the amount of energy per temperature and wavelength. There is no need to convert a given temperature to energy.

source

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.

Does gravity bend light, and how much time does it take for light to cross gravity of a Black Hole?

You are quite right: Einstein's theory says the curvature of space is locally deformed. The essence of this is captured in the spacetime 'metric', a mathematical tool that tells us what space looks like and, derived from this, what is meant by 'a straight path', which photons take. If there is no source of gravitation present, the path of a photon will be what you know intuitively as a straight line. However, for some mass concentration (e.g. a black hole, as you say), this path will be bended such that the mass concentration acts as a lens. This is immediately clear from this image from the CFHTLenS survey:

Crucially to your question I think, you must remember that photons do not experience time and their speed is equal to $c$, the speed of light. Photons are not unaffected by their movement through a gravitational field, mind you: but this shows up as a gravitational redshift (for a time-varying potential), rather than a time delay (other than the slightly elongated path, perhaps).

Besides that, the age of the Universe is typically not quite measured as you say, but rather through parameter estimation in e.g. the cosmic microwave background. The effect of gravitational lensing needs to be taken into account for that, but not in the way you presume.

Also, remember that an extreme gravitational field such as that of a black hole is relatively rare, and even if it did delay the photon for a 1000 years that is still a tiny fraction of the age we would, according to your way of thinking, infer; the photon would have to encounter a LOT of such black holes for this 'effect' (which does not occur) to have a huge impact.

As a general point, you seem to have some (Interstellar-induced?) misconceptions about the universe, but I think that's not for me to address here.