cosmology - Why can primordial tensor perturbations of the CMB be ascribed to gravitational waves?

Why can primordial tensor perturbations of the CMB be ascribed to gravitational waves? Is this attribution unique, or are there other michanisms that could lead to the ecitation of tensor modes?

In explanation I have read recently is that gravitational waves turn the E-modes into the now observed B-modes somehow, but I dont understant this. So can somebody give a more detailled explanation of this intereaction of gravitational waves with the CMB?

LaTex and equations are welcome and appreciated :-)

the sun - How many sun-like stars are there in the universe?

This is a question that concerns the initial mass function (IMF) - an empirical (that is, defined by observations rather than theory) function that describes the statistical distribution of stellar masses.

Edwin Salpeter (1955) was the first to describe the IMF, though if you read Chabrier (2003) there are some reasonably comprehensive explanations of the theory and history. However, these lecture notes are a fair bit more accessible.

From the approximations in the UCSC lecture notes I linked above, I get that around 4% of stars are between 0.7 and 1.3 solar masses (92% are between 0.1 and 0.7 solar masses!).

There are perhaps 100 billion stars in a galaxy and 100 billion galaxies in the observable universe, giving something on the order of $4times 10^{20}$ (400 billion billion) stars that are about ($pm 30%$) one solar mass.

Is the white hole a myth or not?

Well, currently, we don't have a shred of evidence that they exist.

White holes are hypothetical objects predicted by certain solutions to Einstein's theory of general relativity. To be specific, they come from the maximally extended Schwarzschild solution. We can deal with them a little mathematically, but nobody will really believe in objects like these unless we have experimental evidence for them.

Do we? I quote NASA here

White holes are VERY hypothetical.

and here

There is no observational evidence for white holes.

This supports this position:

Unlike black holes, there is no real astronomical evidence to suggest that white holes actually exist.

Also, the idea is put nicely here:

A white hole is only a concept for higher levels of thinking. No one has every observed one and no one probably ever will.

Why? This gives an interesting answer:

Once even the tiniest speck of dust enters the part of space-time which includes the black hole, the part which includes the white hole disappears. The universe has been around for a long time and so even if it did start with white holes, they would have all disappeared by now. (Emphasis mine)

Essentially, white holes are unstable. Even if a particle did try to enter one from a black hole via an Einstein-Rosen bridge, the bridge would collapse before the particle could cross, and the particle would hit the singularity.

solar system - How does the Grand Tack Hypothesis explain how Jupiter formed inside the frost line?

A direct cribbing of the wiki's page for the frost line:

In astronomy or planetary science, the frost line, also known as the snow line or ice line, is the particular distance in the solar nebula from the central protostar where it is cold enough for volatile compounds such as water, ammonia, methane, carbon dioxide, carbon monoxide to condense into solid ice grains. This condensation temperature depends on the volatile substance and the partial pressure of vapor in the protostar nebula. The actual temperature and distance for the snow line of water ice depend on the physical model used to calculate it:

• 170 K at 2.7 AU (Hayashi, 1981)


• 143 K at 3.2 AU to 150 K at 3 AU (Podolak and Zucker, 2010)


• 3.1 AU (Martin and Livio, 2012)

Occasionally, the term snow line is also used to represent the present distance at which water ice can be stable (even under direct sunlight). This current snow line distance is different from the formation snow line distance during the formation of Solar System, and approximately equals 5 AU. The reason for the difference is that during the formation of Solar System, the solar nebula was an opaque cloud where temperature were lower close to the Sun,[citation needed] and the Sun itself was less energetic. After formation, the ice got buried by infalling dust and it has remained stable a few meters below the surface. If ice within 5 AU is exposed, e.g. by a crater, then it sublimates on short timescales. However, out of direct sunlight ice can remain stable on the surface of asteroids (and the Moon) if it is located in permanently shadowed craters, where temperature may remain very low over the age of the Solar System (e.g. 30–40 K on the Moon).

Observations of the asteroid belt, located between Mars and Jupiter, suggest that the water snow line during formation of Solar System was located within this region. The outer asteroids are icy C-class objects (e.g. Abe et al. 2000; Morbidelli et al. 2000) whereas the inner asteroid belt is largely devoid of water. This implies that when planetesimal formation occurred the snow line was located at around $R_{snow} = 2.7$ AU from the Sun.

So in all of the three listed models, as well as the snow line inferred from asteroid properties, it seems that the 3.5 AU starting distance for Jupiter in the Grand Tack hypothesis remains comfortably behind the snow line (for water, at least, which is still an advantage over the inner planets).

Since it ultimately fell well within the snow line for a time, one would expect any exposed ices to rapidly sublimate. I would guess that it made up for this by accreting the rockier planetesimals found in the inner parts of the Solar System, which could then shield the remaining ices.

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This site offers a pretty good detailing of the basic structure of the Grand Tack Hypothesis. The most digestible piece of information is at the very bottom, in the form of an AVI file that models the first few million years of the solar system. The short answer to your question is "Jupiter is assumed to have already formed some fundamental core at the start of the hypothesis, then moves around accreting and scattering things through the inner solar system, until returning to the outer solar system to finish growing". The initial inward migration is caused by interactions with the relatively thick gaseous disc (the majority of it will never become part of any planet), which creates drag on the planets/planetesimals and causes them to lose velocity and fall inward.

But the video's pretty cool, so I'm going to link it and talk about it just for its own sake.

Get the video, as produced by the paper Walsh et. al., 2011

You may find it best to run this at half speed or slower to really see what's going on.

Here's how it starts out.

The solid black orbits are the gas giant orbits, with Jupiter being the innermost one, and Saturn the one after that. The dashed circle represents a 2 AU distance from the sun (modern day Mars is currently at about 1.5 AU, and the modern asteroid belt stretches roughly 2.2-3 AU). This is noteworthy since the model is trying to explain why Mars is so small, and why there is so little material (relatively speaking) in the asteroid belt. The numerous small blue dots are the icy comets/planetesimals that form right around the point Jupiter starts out, and points beyond. The numerous red dots are the rocky asteroids/planetesimals characteristic of the inner solar system.

For the first several seconds we see Jupiter creep in towards the inner solar system. You'll notice a lot of rocky planetesimals get scattered out into the outer solar system, while some of the icy ones are scattered to the inner solar system. Indeed, it seems to be a general feature of this model that the total amount of material within approximately 1 AU undergoes a net increase.

Saturn is also slowly moving in, but not nearly as quicky as Jupiter is (Neptune/Uranus seem pretty stable, but do "pulse" in and out a bit).

In this image we see that Jupiter hits the 2 AU mark at ~3 seconds in, corresponding to a model time of about 68,000 years. At this point you can see a dashed circle at about the 3.5 AU distance, corresponding to Jupiter's starting distance. Saturn is about 4 AU out now, having started at over 5.

The third through fourth second are fairly stable-looking, with Jupiter moving in to about 1.5 AU but the other outer planets staying mostly put.

But in the 5th second (~94,000 years model time)...

We see Saturn make a very rapid move inwards. This is the main reason I suggest playing it slowly, as you can very easily miss it and lose track of where each solid black orbit has come from. By model time 98,000 years (still in the 5th second of the video time), Saturn has already come to within about 2 AU of the Sun; Jupiter is still about 1.5 AU, and there is a thick concentration of mostly rocky planetesimals within 1 AU.

By model time of about 120,000 years, Saturn and Jupiter are rebounding back out into the outer solar system, to much the same distances we see today, scattering most of the planetesimals that remained in the areas that would become Mars and the asteroid belt. This leaves a fairly dense disc of material for Earth, Venus, and Mercury to form and grow from, while leaving significantly less for Mars, and lesser still for the asteroid belt.

The whole 10 second movie covers a model time of 600,000 years.

Walsh, K. J., Morbidelli, A., Raymond, S. N., O’Brien, D. P., Mandell, A. M. (2011). A low mass for Mars from Jupiter’s early gas-driven migration. Nature 475, 206-209.

Walsh, K. J., Morbidelli, A., Raymond, S. N., O’Brien, D. P., Mandell, A. M. (2012). Populating the asteroid belt from two parent source regions due to the migration of giant planets – ”The Grand Tack”. Meteoritics & Planetary Science 47(12), 1941-1947.

What are the Earth-like features of Titan?

A lot - from this space.com article:

Saturn's moon Titan may be worlds away from Earth, but the two bodies have some characteristics in common: Wind, rain, volcanoes, tectonics and other Earth-like processes all sculpt features on Titan, but act in an environment more
frigid than Antarctica.

The article goes on to say that while Titan has these earth like features, it only receives about 1% of the sunlight that Earth receives. Because of this, the average temperature on the moon has an average surface temperature of -180 C (-292 degrees F), and water can only exist there in a super-frozen form as hard as rock.

On Titan, methane takes water's place in the hydrological cycle of evaporation and precipitation (rain or snow) and can appear as a gas, a liquid and a solid. Methane rain cuts channels and forms lakes on the surface and causes erosion, helping to erase the meteorite impact craters that pockmark most other rocky worlds, such as our own moon and the planet Mercury.

As for the planet's core, this Stanford paper has some info:

"The picture of Titan that we get has an icy, rocky core with a radius of a little over 2,000 kilometers, an ocean somewhere in the range of 225 to 300 kilometers thick and an ice layer that is 200 kilometers thick," he said.

I'm not going to quote the whole paper here, but it's a good read on the subject of Titan's core.

This Wikipedia article has some information about Titan's atmosphere:

Observations from the Voyager space probes have shown that the Titanian atmosphere is denser than Earth's, with a surface pressure about 1.45 times that of Earth's. Titan's atmosphere is about 1.19 times as massive as Earth's overall, or about 7.3 times more massive on a per surface area basis. It supports opaque haze layers that block most visible light from the Sun and other sources and renders Titan's surface features obscure. The atmosphere is so thick and the gravity so low that humans could fly through it by flapping "wings" attached to their arms. Titan's lower gravity means that its atmosphere is far more extended than Earth's; even at a distance of 975 km, the Cassini spacecraft had to make adjustments to maintain a stable orbit against atmospheric drag. The atmosphere of Titan is opaque at many wavelengths and a complete reflectance spectrum of the surface is impossible to acquire from the outside. It was not until the arrival of the Cassini–Huygens mission in 2004 that the first direct images of Titan's surface were obtained. The Huygens probe was unable to detect the direction of the Sun during its descent, and although it was able to take images from the surface, the Huygens team likened the process to "taking pictures of an asphalt parking lot at dusk".

galaxy - How many galaxies have been discovered?

If you want to know the number of spectroscopically confirmed galaxies then the number given by UV-D is roughly correct.

But the actual number of galaxies that have been observed is order of magnitudes larger.

The case I know is the Sloan Digital Sky Survey 3 (SDSS3):
they observed about a third of the sky using five different filters between optical and near infrared. Of all the objects detected 208,478,448 are galaxies. And of those 1.5 to 2 millions will have the spectrum measured by mid next year.
More info here

To these number you have to add the galaxies observed by surveys in different areas of the sky, deeper that SDSS (which means that they can seen fainter and/or more fare away objects).

By the way, if memory does not fail me, estimates gives about 10^10 galaxies in the observed universe

cosmology - Future of CMB observations: How will our knowledge of the early universe change?

The Planck satellite has been presented and awaited for a long time as the ultimate experiments for measuring temperature fluctuations in the cosmic microwave background (CMB) over the full sky.

One of the big questions that still need answer and that Planck might help clarify is about the dynamics and driving mechanisms in the first phases of the universe, in particular in the period called inflation.

Thankfully there is room for improvements at small scales, i.e. small pieces of sky observed with extremely high resolution, and more importantly for experiments to measure the polarisation of CMB.
I know that for the next years a number of polarisation experiments, mostly from ground and balloons, are planned (I'm not sure about satellites).

For sure some of these result will rule out some of the possible inflationary scenarios, but to which level?

Will we ever be able to say: "inflation happened this way"?

gravity - Would time go by infinitely fast when crossing the event horizon of a black hole?

(I will assume a Schwarzschild black hole for simplicity, but much of the following is morally the same for other black holes.)

If you were to fall into a black hole, my understanding is that from your reference point, time would speed up (looking out to the rest of the universe), approaching infinity when approaching the event horizon.

In Schwarzschild coordinates,
$$mathrm{d}tau^2 = left(1-frac{2m}{r}right)mathrm{d}t^2 - left(1-frac{2m}{r}right)^{-1}mathrm{d}r^2 - r^2,mathrm{d}Omega^2text{,}$$
the gravitational redshift $sqrt{1-frac{2m}{r}}$ describes the time dilation of a stationary observer at a given Schwarzschild radial coordinate $r$, compared to a stationary observer at infinity. You can check this easily: plug in $mathrm{d}r = mathrm{d}Omega = 0$, the condition that neither the radial nor the angular coordinates are changing (i.e. stationary observer), and solve for $mathrm{d}tau/mathrm{d}t$.

The conclusion is that if you have the rocket power to hover arbitrarily close to the horizon, you will be able to see arbitrarily far into the history of the universe over your lifetime. However, that doesn't actually cover what happens to an observer that crosses the horizon. In that case, $mathrm{d}rnot=0$, and the coefficient of $mathrm{d}r^2$ above becomes undefined at the horizon: as in the other question, the Schwarzschild coordinate chart simply fails to cover the horizon and so is ill-suited for talking about situations cross the horizon.

But that's a fault of the coordinate chart, not of spacetime. There are other coordinate charts that are better adapted to questions like that. For example, the two Eddington-Finkelstein charts are best suited for incoming and outgoing light rays, respectively, and the Gullstrand-Painlevé chart is adapted to a freely falling observer starting from rest at infinity.

If this is correct, would you see the whole universe's future "life" flash before your eyes as you fall in, assuming you could somehow withstand the tremendous forces, and assuming black holes don't evaporate?

No. I think this is best seen from the Penrose diagram of Schwarzschild spacetime:

Light rays run diagonally. In blue is an example infalling trajectory, not necessarily freely falling. Note the two events where it crosses the horizon and where it reaches the singularity. Shown in red are inward light rays that intersect those events. Thus, the events that the infalling observer can see of the external universe consist of the region between those light rays and the horizon. The events occurring after that won't be seen because the the observer will have already reached the singularity by then.

Now suppose the observer tries a different trajectory after crossing the horizon, accelerating outward as much as possible in order to see more of the future history of the external universe. This will only work up to a point: the best the observer can do is hug the outgoing light ray (diagonally from lower-left to upper-right) as much as possible... but since the observer is not actually allowed to go at the speed of light, seeing all of the future of history will be impossible. The best the observer can do is to meet the singularity a bit more on the right of the diagram.

By the way, since the light ray worldlines have zero proper time, trying to do that will actually shorten the the observer's lifespan. If you're in a Schwarzschild black hole, you would live longer if you don't struggle to get out.

The above is for an eternal, non-evaporating black hole, as that's what you're asking about here. (The 'antihorizon' is there because the full Schwarzschild spacetime is actually an eternal black hole and its mirror image, a white hole in a mirror 'anti-verse', which not shown on this diagram. That's unphysical, but not relevant to the situation we're considering here.)

If it is correct that black holes evaporate due to Hawking radiation, would you be "transported" forward in time to where the black hole fully evaporates?

An evaporating black hole is morally the same as above: only an ideal light ray can reach the point when the black hole fully evaporates; everyone else gets the singularity. (Since this ideal light ray right along the horizon would be infinitely redshifted, arguably not even that.) You can repeat the above reasoning on its Penrose diagram yourself:

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Addendum:

I have thought a bit about this, and does this solution take into account the relativistic time effects near the horizon of the black hole (e.g. is my understanding correct that the observer would observe time in the universe passing approaching infinitely fast when approaching the event horizon)?

How much time dilation happens depends entirely on what coordinates we're talking about (more generally, which frame field). What a given observer will actually see, however, is completely independent of choice of coordinates. In particular, Penrose diagrams illustrate the light cone structure of the given spacetime, and what an observer can in principle see depends entirely on what light rays intersect the observer's wordline. So yes, it's taken into account by default.

If you're actually falling in it, no, your understanding is mistaken, for reasons explained above. For additional motivation, flip the question around: what does the very distant stationary observer see of the infalling object? On the above Penrose diagram, outwardly directed light rays are diagonal, from lower-left to upper-right. Draw some some outward light rays from the blue infalling worldline. You will see that no matter how far into the far future (up on the diagram) you pick an event outside the black hole to be, you can connect that event with an outward light ray originating from the blue infalling worldline before it crosses the horizon. The conclusion would be that an observer that stays outside the black hole would be able to see the infalling object arbitrarily far into the future. No matter how much time passes for someone who stays out of the black hole, the image of the infalling object would still be visible as it was before it crossed the horizon. (In principle at least; in practice it will get too faint to see after a while.)

Thus, the usual result of "infinite gravitational time dilation makes the image of the infalling object hover forever near the horizon" is also straightforwardly deducible from the diagram, and so is completely consistent with the infalling object being able to see a finite part into the future of the external universe. Perhaps it is best to emphasize that the situation is not actually symmetric: what the external observer sees of the infalling object is not some straightforward flip-around of what the infalling object sees of the external universe. The black hole itself breaks that symmetry.

Could there be life beneath the surface of Mars or moons?

And define "life". It's conceivable that lifeforms exist that are for example silicon based, don't need oxygen, and can survive in hard vacuum, even directly use gamma and X-rays the way plants on earth photosynthesise visible light.
Such could well survive there, even thrive. But would we even recognise them as alive?

cosmology - Age of the universe and age of stars

Wikipedia reports the age of HD 140283 as the figure you cited, 14.46±0.8 billion years. Now look at the reference it refers to. This is the paper by Bond et al that presents measurements of some characteristics of the star and the subsequent calculations, as well as a lengthy section on where the error came from (800 million years is, after all, quite a lot - even in Astronomy!).

The paper describes an approach to determining ages of stars without all that tedious mucking around with globular clusters. It suggests determining "ages of extreme Population II subgiants in the solar neighborhood based on direct trigonometric parallaxes, combined with state-of-the-art theoretical isochrones appropriate to the detailed composition of each star." In short, the idea is to test stars to see if there ages are consistent with the calculated age of the universe (13.77±0.06 billion years - note the much smaller uncertainty).

The paper cites some previous estimates of the star's age - ~14 billion years and 13.5±1.5 billion years. It's worth noting this because the authors appear to pride themselves on how they tried to reduce the error of their measurements. The team next measured the star's parallax, also attempting to reduce error. With that in hand, as well as measurements of temperature and visual magnitude, they used stellar evolution models to determine its age. They also compensated for extinction from their measurements. The authors list the error from parallax, photometry, and extinction as 310 million years. The measurements then (page 11) list the age "implied" by the models and measurements to be 14.46 billion years.

Where is the extra 490-million year-uncertainty? Take a look at Figure 1, near the end. It is due to a variety of factors, the largest of which is the uncertainty of the "initial oxygen abundance" - although more properly the uncertainty in the ratio of oxygen to hydrogen.

It does come down to what David H said - the error measurements due comply with a universal age of 13.77 billion years. I would note, though, that the uncertainty in the measurement of the age of HD 140283 is much greater (800 million years) than the uncertainty of the age of the universe (37 million years). If the uncertainty due to the oxygen/hydrogen ratio could be eliminated, a more accurate age could be calculated, but until this error is addressed (preferably by another team), I think it's safe to say that it's more likely that the measurements for HD 140283 are off than that the measurements for the universe, and so what could have been a paradox is somewhat resolved.

TL;DR - There's a heck of a lot of error in the age of HD 140283, which fits reasonably with current models of the age of the universe, and I would trust the measurements of the age of the universe much more than the measurements of the age of this star, especially because the method used here is not widely used.

tectonics - Why are most peninsulas oriented north-south?

Weird question, bear with me.
Most large peninsula on Earth are oriented southwards (and most of the rest northwards), much fewer east or west. Some examples:
Florida, Californa peninsula, Yucatan, South America, Antarctic peninsula, India, Kamtjatka, Korea, Malaysia, Scandinavia, Italy, Greece, Greenland. Contrary examples would be east Siberia, Kola, Turkey. (Is it just me, or have this question ever been raised before in history?)

Is this just coincidence, or could conceivably some tectonic, gravitational or rotational forces help cause this phenomena? Maybe because the tectonic ridges in the Atlantic and the Pacific are north-south oriented and stretch almost from pole to pole? And could that in turn be caused by the dynamics of the inner Earth?

Elevation maps of Venus are more fragmented than the continents of Earth. And Mars is half highland half lowland. Neither has plate tectonics like Earth.

cosmology - Why doesn't this paradox disprove (some) multiverse quantum gravitational theories?

An exact answer would require a rather specific and mathematical formulation of the multiverse in consideration.

As a simple approximating example, suppose we have a countably infinite number of (observable) universes of the same mass $M$. Suppose the dimension of the full multiverse is one higher than each individual universe, and suppose the universes are all separated by the same minimum distance $epsilon>0$ from each other. In a 2-D picture, this would just look like a bunch of parallel lines all separated by the same distance.

Pick your home universe and put an observer. Another universe of distance $nepsilon$ away (meaning they're $n$ universes up or down from you in the 2-D picture) exerts a gravitational force on the observer in its direction approximately proportional to $displaystyle{frac{M}{n^2epsilon^2}}.$ With the right units, we can just say "approximately".

The maximum gravitational force occurs when the observer is at the bottom (or top) of the picture, and all over universes are above it (or below): universes on each side will pull in opposing directions and so lead to cancellations. So the net gravitational force from the other universes (in the right units) is at most
$$sum_{n=1}^infty frac{M}{n^2epsilon^2} = frac{M}{epsilon^2}sum_{n=1}^infty frac{1}{n^2} = frac{pi^2 M}{6epsilon^2}<infty.$$

If the observer was "in the middle"—infinitely many universes above and below, with the distribution identical in either direction— the net gravitational force from the other universes is exactly 0.

planet - Why do we not send diggers to Mars if we think there's underground water?

Sending anything to Mars is expensive. Sending a machine capable of drilling to the required depths to find subsurface water would be prohibitively expensive with current rocket payload limits.

There are also far more accessible sources of water on the surface - lakes of ice in craters, water rich minerals, and even an equatorial frozen sea .

There is also water at the surface of Mars on the poles, but the conditions here are much harsher than nearer the equator, and there is less sunlight, reducing the capability of any solar powered rover.

The subsurface water is not necessarily that far below the surface, but it is also not a torrent; more likely trapped as ice particles in between the rock or in secondary minerals.

milky way - What are Flamsteed numbers?

Flamsteed designations are assigned by John Flamsteed. He was an English Astronomer.

Flamsteed Designations (Flamsteed Numbers) are numbers assigned to the stars for the purpose of identification and cataloguing.
Flamsteed designation contained 2554 stars.

Flamsteed designations for stars are similar to Bayer designations, except that they use numbers instead of Greek letters.

[ Bayer designation is a stellar designation in which a specific star is identified by a Greek letter, followed by the genitive form of its parent constellation's Latin name. The original list contained 1,564 stars.
For the most part, Bayer assigned Greek and Latin letters to stars in rough order of apparent brightness, from brightest to dimmest, within a particular constellation. Since in a majority of constellations the brightest star is designated Alpha (α), followed by beta for the 2nd brightest and so on.
But this was not the rule always followed by Bayer.
Somtimes by position of a particular star in that constellation, sometimes the order of rising up in the sky (the star rises up early will get earlier number), sometimes he follows mythological stories or data or sometimes he even uses his own way of assigning the number.]

In Flamsteed designations, Each star is assigned a number and the Latin genitive of the constellation it lies in.
e.g. 52 Cancri is star in the Constellation Cancer and assigned a number 52 according to star's right ascension. i.e. there are 51 stars in the Cancer constellation to the West direction of 52 Cancri star or there are 51 stars that rise before/earlier of 52 Cancri star.

The numbers were originally assigned in order of increasing right ascension within each constellation.
[Right ascension is the angular distance measured eastward along the celestial equator from the vernal equinox (it is one of the point of intersection in the space of two imaginary circles: Celestial Equator and Earth's orbit around the Sun. Our Earth is at this point on 21/22 March each year while orbiting around sun.) to the the position of the star. It is measured in degrees but for simplicity and ease of use, mentioned in the form of hours. That the 360 circle is divided into 24 hours (1 hour = 15 degrees). ]

But due to the effects of precession (Top like rotation of Earth's axis of rotatation in the space mataining a constant angle of 23.5 degrees with verticle) they are now slightly out of order in some places.

Flamsteed catalogue don't cover all the stars visible from Earth with naked eye. Flamsteed's catalogue covered only the stars visible from Great Britain, and therefore stars of the far southern constellations have no Flamsteed numbers.

Reference/Information Source/Quoted From:

1) List of constellations for a list of constellations and the genitive forms of their names: https://en.wikipedia.org/wiki/88_modern_constellations

2)Bayer Designation: https://en.wikipedia.org/wiki/Bayer_designation

3)Flamsteed Designation: https://en.wikipedia.org/wiki/Flamsteed_designation

How do we find the exact temperature of a star?

This question is very broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do of course have line of sight velocities which give their spectrum a redshift (or blueshift). It is a reasonably simple procedure to correct for the line of sight velocity of a star, because the redshift (or blueshift) applies to all wavelengths equally and we can simply shift the wavelength axis to account for this. i.e. We put the star back into the rest-frame before analysing its spectrum.

Gerald has talked about the blackbody spectrum - indeed the wavelength of the peak of a blackbody spectrum is inversely dependent of temperature through Wien's law. This method could be used to estimate the temperatures of objects that have spectra which closely approximate blackbodies and for which flux-calibrated spectra are available that properly sample the peak. Both of these conditions are hard to satisfy in practice: stars are in general not blackbodies, though their effective temperatures - which is usually what is quoted, are defined as the temperature of a blackbody with the same radius and luminosity of the star.

The effective temperature of a star is most accurately measured by (i) estimating the total flux of light from the star; (ii) getting an accurate distance from a parallax; (iii) combining these to give the luminosity; (iv) measuring the radius of the star using interferometry; (v) this gives the effective temperature from Stefan's law:
$$ L = 4pi R^2 sigma T_{eff}^4,$$
where $sigma$ is the Stefan-Boltzmann constant. Unfortunately the limiting factor here is that it is difficult to measure the radii of all but the largest or nearest stars. So measurements exist for a few giants and a few dozen nearby main sequence stars; but these are the fundamental calibrators against which other techniques are compared and calibrated.

A second major secondary technique is a detailed analysis of the spectrum of a star. To understand how this works we need to realise that (i) atoms/ions have different energy levels; (ii) the way that these levels are populated depends on temperature (higher levels are occupied at higher temperatures); (iii) transitions between levels can result in the emission or absorption of light at a particular wavelength that depends on the energy difference between the levels.

To use these properties we construct a model of the atmosphere of a star. In general a star is hotter on the inside and cooler on the outside. The radiation coming out from the centre of the star is absorbed by the cooler, overlying layers, but this happens preferentially at the wavelengths corresponding to energy level differences in the atoms that are absorbing the radiation. This produces absorption lines in the spectrum. A spectrum analysis consists of measuring the strengths of these absorption lines for many different chemical elements and different wavelengths. The strength of an absorption line depends primarily on (i) the temperature of the star and (ii) the amount of a particular chemical element, but also on several other parameters (gravity, turbulence, atmospheric structure). By measuring lots of lines you isolate these dependencies and emerge with a solution for the temperature of the star - often with a precision as good as +/-50 Kelvin.

Where you don't have a good spectrum, the next best solution is to use the colour of the star to estimate its temperature. This works because hot stars are blue and cool stars are red. The colour-temperature relationship is calibrated using the measured colours of the fundamental calibrator stars. Typical accuracies of this method are +/- 100-200 K (poorer for cooler stars).

Why did the moon abruptly change positions in the sky?

Compared to the planets the moon changes it's rise and set times very quickly. Using the calculator on this page it can be seen that if one were looking at the sky on May 1st, 2014 from Irvine, California the moon set at 10:08PM (which at 8PM would have had the moon most of the way across the sky as noted in the question.)

The moon rises and sets approximately one hour later each day. So, by May 14th the moon did not even rise until nearly 8PM - which also correlates with the observations noted above.

Contrasting that with the other context point made in the OP, using this page we can see that on May 1st Saturn rose at 8:07PM and on May 14th rose at 7:11PM.

So while the moon changes its rise a set times by almost an hour each day, the planets change much more slowly by comparison, almost an hour over two weeks.

If one weren't continually watching the moon's progression it would appear to abruptly change positions when comparing it to the planets' movements.

orbit - Calculate apsides without knowing eccentricity

You mention this is a system in which a vessel is orbiting a planet. In that case it's fair to assume the mass of the vessel is negligible compared with the planet mass and
the planet will sit in one of the focal points of the ellipse.

You can then use Kepler's second law to find the semi-minor axis as
$$ frac{1}{2} r v = frac{pi a b}{P} $$
where $r$ the current distance and $v$ the current velocity, $P$ is the period and $a$ and $b$ give the respective semi-major and semi-minor axes.

From $b$ you find the eccentricity as
$$ b=a sqrt{1-e^2} $$

And the apsides follow as
$$ mathrm{periapsis} = a ( 1 - e ) $$
$$ mathrm{apoapsis} = a ( 1 + e ) $$

venus - Why do rocks on other solar system bodies that have an atmosphere seem to be flat?

Images taken by landers on Titan and Venus and Mars show landscapes where rocks, to me at least, are surprisingly flat. Being used to walking around in forests with roundish meter sized boulders, I'd be very surprised to find myself in a landscape of flat rocks.

Their atmospheric density range between about 100 times, half and 1/100th of Earth's atmosphere. So that can't be the main explanation. Actually, I think that the landscape of the Moon is more Earth like than that of the atmospheric worlds out there.

Is this my impression of "flatness" a real phenomenon, and if so, why and how?

What record do we have of the length of supernovas?

recently I heard an astronomer on the radio claim that the supernova we have observed in our own galaxy lasted about 6 months, while in distant galaxies they last about 7-8 months, due to relativity and the expanding universe. He didn't give any references. Just to be clear, I don't mean to imply that he implied the actual duration was different, but that it only appeared different because of expansion.

I've found records of supernova, but only 1 mentioned how long it lasted, the first recorded one was said to last 8 months according to wikipedia.
Can someone point me to a list that contains both the location and duration of supernova's .

mars - Why should space probes have to orbit the Earth before being launched at other planets?

There are several reasons why Satellites need to orbit Earth before they go interplanetary...

The first reason:
The launch site is very rarely in the right position to start an interplanetary flight. Earth rotates on a tilt, so a launch has to be timed when Kennedy Space Center crosses the ecliptic plane (the general plane that most planets orbit on). Also, it has to be in the right season, so that when the probe goes out to orbit it ends up headed the right direction when it leaves Earth's SOI. All of this is possible with a direct ascent profile, it just takes really good timing - but these perfect windows come very rarely.

Any probe that makes ecliptic orbit around Earth, first, has a launch window pretty much every 45 minutes, as apposed to a couple times per year.

The second reason:
The amount of Delta-V required to escape Earth's SOI is pretty big. While it is possible to build rockets large enough to do so - the limiting factor is really the efficiency of typical rocket fuel and rocket engines.

To lift a probe out of Earth's orbit with a rocket takes a pretty heavy rocket. That heavy rocket must be lifted into Low Earth orbit, which takes a massive rocket.

One way to improve this fact is to make your rocket much more efficient - but we are already close to the theoretical efficiency limit of a chemical rocket. So NASA started using ION propulsion, which is way more efficient than a chemical rocket - it's also very weak - which is the thirdreason...

The third reason:
Now that most probes use Ion propulsion, they don't have the thrust to just eject themselves from earth in a direct ascent - they spend weeks with the Ion thruster thrusting for a little while (a few minutes) at a key point in the orbit. Each time the Ion engine does this, their orbit gets closer and closer to Earth escape velocity.

Once the probe is out of Earth's SOI, it can basically turn the Ion engine On and leave it there for as long as it wants to complete interplanetary maneuvers. Usually, most maneuvers between planets are minor course corrections to take advantage of a planetary fly-by to sling-shot to much higher velocities.

TL;DR?
Many reasons:
Timing - launch position and orbital orientation mean few good launch windows for direct ascent, reaching orbit first allows for many more options.
Too much fuel needed - Getting a probe way from the earth takes a lot of chemical propellant, so we now use Ion engines instead.
Ion engines are weak - it takes a long time (weeks!) for these highly efficient engines to do their job.

observation - Point Spread Function size: Full Width Half Maximum (FWHM) vs Sigma

FWHM is the indicator of the width of a Gaussian that is easiest to measure, and is least error prone in terms of actual physical measurement (due to the slope of a Gaussian being the highest near half maximum - not exactly sure about this, but it does seem like that's the case - so error in y contributes least in the measurement of width, relatively. If this is not clear, look at a noisy Gaussian - something like this: http://www.astronomie-amateur.fr/Documents_Supernovae/Mesure_Vexp_gauss.PNG - and you'll know what I mean. It seems most sensible to estimate width using FWHM). And it has a simple correlation to the sigma with a factor of 2.35 if the distribution is actually Gaussian (which is often the assumption for PSFs - correct me if I'm wrong). Measuring actual standard deviation of data with errors is much more complicated than just simply measuring FWHM, and often it is enough to know the FWHM estimate. Hope this is the answer you were looking for.

big bang theory - Matter and Antimatter Interaction in the early Universe

Well it is said that during the Big Bang, things were created in pairs: one matter and one anti-matter. I think its a part of the Big-Bang Theory.

It is also said that matter + anti-matter - Energy

Then shouldn't it be so that there would have been no matter or antimatter in existence now since all of them should have reacted to form pure Energy. I mean how were we created then?

And also where are all the antimatter gone now? Are they still reacting with us?

Thanks

orbit - Supermoon Lunar Eclipse?

Lunar Eclipses occur when a full moon occurs when the the moon crosses the ecliptic. The ecliptic is the plane in which the Earth orbits the sun, and so also the track which the sun appears to trace in the sky relative to the stars.

It appears from the diagram that there are two nodes each month (ascending and descending), but this is only approximately correct. The influence of the sun causes the nodes to move.It takes 29.5 days between two full moons, but only 27.2 days between two ascending and descending nodes). This means that a node can occur on any day of the year, and any day of the month.

A "supermoon" occurs when the moon when it is closest to the earth, the perigee. This point of perigee also moves, so the perigee can also occur at any time during the month, and the time of perigee is not related to the time of the node. Unlike terms like "node" which astro(nom|log)ers have used for many years, Supermoon is a neologism.

Thus is is possible, but rare for a full moon, a lunar node, and a perigee to occur at about the same time, and if it does, a supermoon eclipse will occur.

supermassive black hole - What is the galaxy M87 doing these days?

The answer to your question is relatively straightforward: Not much. I'll delve into it in two parts: Star formation and the activity of the supermassive black hole at the center.

Star Formation

M87 is, according to Wikipedia, home to many Population II stars. They have little hydrogen and helium in them, and the clouds of gas and dust in the galaxy don't have much, either. In fact, many dust clouds can be destroyed by the intense radiation emitted by the accretion disk of the central black hole. This combines to make it seem unlikely that there will be much star formation in the galaxy.

That said, just because a galaxy doesn't have a lot of materials for star formation doesn't mean it can't form lots of new stars - at least, no under certain special circumstances. Galaxy collisions can induce high rates of star formation - starburst galaxies are good examples of this. While M87 might not be at a high risk for a collision, there is still the possibility of one. M84 is nearby, and may have had an encounter with M87 in the past. While it is a long shot (and M84 has a low star-forming rate), an interaction with the two could spawn new stars.

The Black Hole

You seemed particularly interested in the black hole at M87's center. It is a supermassive black hole, with a mass billions of times that of the Sun. Quite a monster, by all standards. Yet it is not doomed to swallow the galaxy whole anytime soon. It is sucking in gas and dust at a rate of 0.1 solar masses each year - and there is over 70,000 solar masses of dust in the galaxy. You would think that, therefore, the dust should be gone in 700,000 years, right? Well, the dust isn't all concentrated at the center, near the black hole. It is spread throughout the galaxy, and it would take a long time for all of it to reach the black hole - let alone the billions upon billions of stars in the galaxy.

In summary: Star formation in M87 is at a low rate, and while a galactic interaction could create a new wave of star formation, it is unlikely. The central black hole is sucking in lots of gas and dust, but it will take billions upon billions of years before it eats up a substantial amount of matter.

I hope this helps.

Sources:

https://en.wikipedia.org/wiki/Messier_87

https://en.wikipedia.org/wiki/Messier_84

http://arxiv.org/pdf/astro-ph/0202238v3.pdf (Beware: Wikipedia cites the paper by saying that one solar mass is absorbed every ten years, while the paper says that 0.1 solar masses are absorbed each year.)