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.)

medical - How important is the size of an astronaut?

In the early days of crewed space exploration, size and weight of crew was an issue. The original batch of Soviet cosmonauts selected were restricted to 1.75m height and 72kg mass, for example, and US astronauts to 1.80m and 82kg.

As launchers got bigger and spacecraft got more complex (and roomier), these restrictions became less important because the crew represented a progressively smaller share of the mass budget.

For example, in the Mercury program, the astronaut's mass was ~5.8% of the spacecraft mass; in Gemini, the crew was ~4.3%; in Apollo earth-orbit configurations, ~1.2%.

At some point in that progression, it makes sense to start relaxing the astronaut height/mass restrictions. China's space program simply hasn't advanced far enough for them to do so yet.

As for inherent advantages of a larger crew member, all other things being equal, larger people in good health are generally stronger, which can be important during certain EVA operations. In most other criteria a smaller crew member would probably be very slightly preferable primarily due to resource consumption rates.

cosmology - What does Stephen Hawking mean by 'an infinite universe'?

A relativistic universe which is expanding faster than light (like ours) is effectively infinite for all practical purposes.

Also due to its relativistic nature and faster than light expansion, even if you assume it's not infinite at some given moment, it still doesn't have any edges or borders - for you. You're in some random place in it, the maximum speed of any possible interaction is speed of light, and the universe is expanding faster than light - then anything that happens at some (real or imaginary) "edge" is outside your realm of existence. It is effectively infinite - for you.

Give it enough time and it would grow as big as you want. Give it asymptotically infinite time and it will grow asymptotically infinite.

As to what the "edge" might be, see Eternal Inflation. This is a model in modern cosmology where local bubbles like ours have stopped inflating (they still expand, but not at the tremendous rate of the initial inflationary phase); however, inflation continues forever outside the bubbles and at the edges. Therefore the bubbles keep growing indefinitely at faster than light speeds. Because speed of light is a limit for any interaction inside the bubbles, for any internal observer each bubble is effectively infinite for any practical purpose.

Be aware that there is no proof that this is actually the case, but this is a model that fits well what we now observe.

---

EDIT: TBH, I'm not even sure this is a question for StackExchange. It's very open-ended, and we don't really have the conclusive answers. All we can say for now is that the universe appears to be infinite for all practical purposes, but we can't know for sure. So often scientists like Hawking just simplify the language and refer to the universe as "infinite" without any of the qualifiers that would be required in a strict context.

I don't think there's any single, final answer here.

orbit - Is there online data on asteroid axial tilts?

From this article, http://hubblesite.org/pubinfo/pdf/2010/33/pdf.pdf, here's the rotation axis for 4 Vesta: RA=307.5°±3.1°, Dec=43.1°±1.2°. Update, per TildalWave's comment: RA=305.8°±3.1°, Dec=41.4°±1.5°

Rotational motion of Ceres.

An old article (1986): Magnusson, P., Distribution of spin axes and senses of rotation for 20 large asteroids, Icarus 68:1 1-39 http://www.sciencedirect.com/science/article/pii/0019103586900722

planet - Spheres in space

I know that gravity will turn a mass into a sphere, which is why
planets and stars are that shape. But then we have asteroids and small
moons which are not spherical. Such as a couple of Pluto's moons. Is
there a size at which they will become spheres? Is it dependent on
their composition?

Google searches provide a wide variety of answers to this and while the other question is answered, it's only answered briefly, so I thought I'd give this a try.

Mike Brown's Planets was the best link I could find on the subject, and I can't swear by his correctness, but his numbers are in the range you find in other articles.

While we can't see most of the objects in the Kuiper belt well enough
to determine whether they are round or not, we can estimate how big an
object has to be before it becomes round and therefore how many
objects in the Kuiper belt are likely round. In the asteroid belt
Ceres, with a diameter of 900 km, is the only object large enough to
be round, so somewhere around 900 km is a good cutoff for rocky bodies
like asteroids. Kuiper belt objects have a lot of ice in their
interiors, though. Ice is not as hard as rock, so it less easily
withstands the force of gravity, and it takes less force to make an
ice ball round.

The best estimate for how big an icy body needs to be to become round
comes from looking at icy satellites of the giant planets. The
smallest body that is generally round is Saturn's satellite Mimas,
which has a diameter of about 400 km. Several satellites which have
diameters around 200 km are not round. So somewhere between 200 and
400 km an icy body becomes round. Objects with more ice will become
round at smaller sizes while those with less rock might be bigger. We
will take 400 km as a reasonable lower limit and assume that anything
larger than 400 km in the Kuiper belt is round, and thus a dwarf
planet. We might be a bit off in one direction or another, but 400 km
seems like a good estimate.

Other estimates I've read have 600 KM diameter for rocky bodies and Ceres is ice and rock, not really a rocky body, so I think there's some uncertainty in there.

Does it need a certain mass?

I personally don't like talking about planets by their mass cause it gets rather unwieldy in size. Take Ceres, mentioned above. It's 8.958 × 10^20 kg. A 400 KM diameter ice world would have a mass of, figure a water-ice average density, about 3 x 10^19 and 200 KM diameter, 1/8th of that, about 3.75 x 10^18. The minimum mass is somewhere in those ranges, but I think radius and composition are easier to work with.

Is it only because of the heat of collisions melted the rocky planets
that they became this shape while they were semi liquid?

I can only give an intuitive answer here, but collisions that liquefy a planet are rare and, especially with smaller planets, would be more likely to blow the planet to bits as soon as melt it. Take the giant impact on Mars. Article Here and Here. The 2nd article says that the object that hit Mars was thought to be traveling at about "6 to 10 kilometers per second." and size "roughly 1,600 to 2,700 kilometers across" - so, bigger than Ceres, Smaller than the Moon and this may have melted half of Mars' surface but it didn't melt all of it cause it left a measurably difference in Mars crust one side to the other. If a collision of that enormous size didn't melt Mars all the way around - I'd wager that such melting is pretty rare.

2nd point, as a general principal, the Solar system objects orbit in the same direction and the further out you get from the sun, the slower the orbital speed (though the more likely you get eccentric orbits). The speed of collisions in space are roughly the vector addition of the 2 objects orbital speeds, which are often somewhat in line with each other, plus the escape velocity of the more massive object (or a bit more than that if both objects are massive), so once objects get to be a pretty good size, about the size where they become spheres, collisions that would melt the object grow less and less likely, and if it is large enough to do that, you should also expect a fair bit of debris from impact. My guess, based on just barely enough knowledge to think about this stuff, is that planetoids and large enough to maybe get spherical asteroids are unlikely to form into spheres due to melting from heat from collisions because due to the relatively weak gravity of smaller objects like that (Ceres has an escape velocity of 0.51 km/s, about 1140 MPH), an impact to generate enough heat to melt it would blow half the dwarf planet away, give or take.

Now, if you have a very hot planet that's very very close to the sun and at near liquid temperature anyway or if you have a coalescing and a multitude of successive impacts and lots of heat collecting in the object, then, sure. It's certainly possible but my guess is that a planetoid or asteroid forming into a sphere due to melting it's rare.

That's a layman's answer.

the moon - Day and night temperature on an earthlike planet with longer rotational period

I'm trying to understand the climatic effects of the far future scenario of an Earth-like planet with a reduced rotational speed caused by tidal locking with the moon (day-night period of 28 days, one hemisphere always facing the moon, the other never facing the moon).

I understand that observations of the moon's surface temperature have shown day temperatures of 120°C and freezing cold nights of -230°C.

How would the day and night temperatures of tidal locked Earth vary?

I am also interested in climatic effects caused by the reduced tides, coriolis force, increased evaporation, but this may reasonably be outside the scope of this question and it's answers.

Can a supernova make a new star?

This idea has been around for decades, so I'm not sure who first came up with it. Here's a reasonably sourced article on the involvement of supernova in solar system formation:Exploding Star May Have Sparked Formation of Our Solar System

The shock wave from an exploding star likely helped trigger the formation of our solar system, according to a new 3D computer model, researchers say.
The solar system is thought to have coalesced from a giant rotating cloud of gas and dust known as the solar nebula about 4.6 billion years ago. For decades, scientists have suspected a star explosion called a supernova helped trigger our solar system's formation. In particular, the shock wave from the explosion is thought to have compressed parts of the nebula, causing these regions to collapse.
According to this theory, the shock wave would have injected material from the exploding star into the solar nebula.

So yes, material from supernova can end up triggering the formation of new suns, and material from supernova do end up mixed in with the nebular material that makes the new stars.

the sun - Is it possible to move a planet out of its orbit? At least a lighter planet?

yes it is possible by very few different ways.

https://www.uwgb.edu/dutchs/pseudosc/flipaxis.htm

Nothing acting solely from on or within the Earth could change its orbit or seriously alter its rotation.
One way to move an object is to throw mass in the opposite direction, the way jets or rockets do.
If we think really big and imagine blasting a chunk out of the Earth as big as North America and 100 miles thick so that its final speed, after escape, with respect to the Earth is 25,000 miles an hour, we will have expelled only 1/500 of the total mass of the Earth. The Earth would move in the opposite direction 1/500 as fast or 50 miles an hour. The speed of the Earth in its orbit is about 67,000 miles an hour. We will not change the orbit of the Earth very much--if we apply the impulse to speed up the earth in its orbit we would put the Earth into a new orbit with its most distant point about 70,000 miles further from the Sun than now--and the Earth's distance from the Sun varies now by three million miles over the course of a year! Exactly the same arguments apply to changing the orbit of the Earth through the impact of a large asteroid. The largest asteroid, Ceres, about 600 miles in diameter, is only about as massive as our hypothetical chunk of Earth above. Changing the orbit of a planet is a tall order. An impact big enough to have even a tiny effect on the Earth's orbit or rotation would almost certainly destroy all life on Earth as well.

http://usatoday30.usatoday.com/news/science/astro/2001-02-15-orbit.htm

hope this two links provides enough satisfactory explanation to you. if not do say so i will make further search to clear you the answer.

Can you see man-made lights on the dark side of the Earth from the surface of the moon with the naked eye?

It's highly doubtful you could see any normal light source on the surface of the earth.
Using $$text{brightness} = frac{text{luminosity}}{4 pi times text{distance}^2}$$
(with brightness in watts, and luminosity in watts per square meter.
and distance to moon of $3.84 times 10^8$ meters.)

---

Try a hypothetical light source 100 megawatts output, all visible light, no heat.

$$text{brightness} = frac{100 times 10^6}{4 pi times 1.474 times 10^{17}}$$

$$text{brightness} = 5.4 times 10^{-11} text{ watts per square meter}$$ at the lunar surface.

---

That's pretty dim. By, contrast sunlight at earth's surface runs about 1300 watts per square meter.
In reality it'd take about a gigawatt to produce 100 megawatts of light in the visible range.
That's about what it takes to power a city of a million homes.
Cities also bounce most of the light they do produce off the ground, which'll have an albedo of somewhere around 0.3. So with ordinary city lights it'll take over 3 gigawatts to reach $5.4 times 10^{-11}$ watts per square meter on the lunar surface.

You might fare better with a big laser. The Apache Point Observatory Lunar Laser-ranging Operation picks up multi-photon signals from the Apollo retroreflectors using only a 1 gigawatt laser and a 3.5 meter telescope. As the article states, the laser beam only expands to 9.3 miles in diameter on the way to the moon, so you might see it wink at you.

At 10 parsecs, the sun has a magnitude of 4.83. It'd be visible on an average night. That magnitude corresponds to a brightness of 3X10e-10 watts per square meter, about 5.6 fold brighter than our hypothetical earth based light source. That puts our light at magnitude 6.5 to 7. Naked eye visibility runs to about 6.0

fundamental astronomy - US observations relevant in UK?

If they mention times of night then you have to figure out to what extent the time zone (BST/GMT vs the US time zones) makes any difference, but it shouldn't make more than an hour difference either way for an object that is moving sidereally with the rest of the night sky.

The main difference would be that the USA is at lower latitudes than the UK. Therefore there could be instances where an object with a small or southerly declination is better seen from the USA than the UK. Conversely, more of the sky is circumpolar from the UK.

Finally, if you are trying to see a specific event at a given time (UT), then of course it may well make a difference. The Sun could be up in the UK and down in the USA, or vice-versa.

space time - traveling past the speed of light

According to the theory of relativity (If I understand correctly), nothing is supposed to be able to travel faster then the speed of light. I believe, to my limited knowledge and grasp of the concept, that quantum theorie speaks against, i.e. extend, the theory of relativity, hence my question:

Suppose we find (or imagine) a place in the universe where there is no obstruction for this thought experiment to take place. We take a spherical object and attach a spire at both sides so you get an object shaped like this: -O-

We start turning the object so the ends of the spire travel in an orbital mather according to the spherical object. What if the sphere starts to orbit near the spead of light, wouldn't the ends of the spires travel faster than the speed of light?

Sorry if this is the wrong place for this question.

the sun - What is the air pressure in the heliosphere (Sun's atmosphere)?

I am going to assume that what you mean here is "what is the pressure" in the solar wind? There is no "air"!

The solar wind is pretty complex, consisting of a "fast wind" observed predominantly at high solar latitudes and a slower more variable wind a low latitudes. Both components essentially consist of an expanding stream of protons, electrons plus a small fraction of helium nuclei.

I think all the information you want is contained in this paper by Ebert et al. (2009), but I will attempt a broad-brush summary.

The fast wind has a typical proton number density of $n_p=3$ cm$^{-3}$, a temperature of $T=2times10^{5}$ K and a speed of $V=700$ km/s. Note that "temperature" is a slippery concept here since the velocity distribution of the particles is non-Maxwellian. However, ignoring this and assuming the protons are matched by numbers of electrons, then the "thermal pressure" $P_{th} = (n_p + n_e)kT simeq 10^{-11}$ Pa. However, this is negligible compared with the dynamic pressure $P_d = rho V^2$, where $rho$ is the mass density given roughly by $rho = n_p m_p$. Thus $P_d simeq 2times 10^{-9}$ Pa (i.e. about $2 times 10^{-14}$ bar).

The "slow wind" is quite variable, but mean values could be $n_p = 6$ cm$^{-3}$, $T=8times 10^{4}$ K and $V=400$ km/s. Here again, the thermal pressure $P_{th} simeq 10^{-11}$ Pa is dwarfed by the dynamic pressure $P_d = 1.6 times 10^{-9}$ Pa.

The thermal pressure component falls as distance from the Sun cubed, ie. as $P_{th} propto R^{-3}$. This is because for mass conservation, then a sherically expanding wind must have a density that falls as $R^{-2}$, but it is experimentally found that the proton temperature also falls as $simeq R^{-1}$.

The dynamic pressure does fall almost as $R^{-2}$. This is because there is very little radius dependence for the proton velocity (i.e. the protons do not slow down, right out to the heliopause - as measured for instance by Voyager 2).

exoplanet - What are the analysis steps in taking raw data from Kepler to a planetary system determination

I wish to get a concise list of the analysis steps required to take raw light data from a Kepler data set of a star through the steps needed to get to an analytical determination of the existence of a planetary system. I would also like any software analysis tools that would be suggested for each data analysis step. A good reference specifically outlining these steps would also be helpful.

I would first like to replicate the complete set of steps needed to corroborate the existence of a well excepted planetary system. After that I would feel confident that I could move on to analysis of some Kepler KOI data sets that NASA has published or even some data from a non-KOI entity. I am an amateur and I realize that astronomical knowledge needs to be applied at each data step and that this might not be an automated process but the steps of this analysis would open up new areas of study for me.

I have tried to assemble the basic data analysis tools from various astronomical python sites and other light transit tool sites and get a little familiar with them but I am not clear as to the actual steps as to how these programs feed to each other. I have looked at the NASA Kepler pipeline to see the steps but it is not clear to me what specific software they apply. I have read several papers confirming exoplanetary discoveries but from them I can't get a complete picture of the data analysis steps and software employed.

It is disappointing that the actual code listings are not presented in the papers or in a separate reference doc. In the past (50 years ago) , it was required to publish code listings in reports and references to the code listings in any papers. But, this discussion is probably another subject.

Tom Kosvic

orbit - Earth's gravitational pull on ISS

Your question presumes that the ISS is beyond Earth's gravity, that it has escaped earth's gravitational pull. This is not correct. All objects with mass in the universe affect all other things with mass in the universe, the effect just gets weaker with distance. So the ISS is feeling the effect of gravity from Earth significantly more than the moon is.

The reason the ISS doesn't just fall to the earth, either directly or gradually spiralling towards the earth, is that it is travelling fast enough around the earth that it is continually "missing" earth. It is sometimes described as 'falling' constantly around earth.

If I am to be properly correct though, the reality is that ISS is in fact falling towards the earth, getting closer and closer to Earth all the time. It needs occasional boosts to push it further back out in it's orbit.

Just to blow your mind a little bit: The ISS is pulling on the Earth with the same force that the Earth is pulling on ISS.

neutron star - Pulsars with accreting disk in binary system

Consider a binary system with a neutron star and a companion star. It is fair to assume that the two stars have had tidal interaction, such that their angular momentum is aligned. This is just saying that both stars rotate in the same direction as their orbital motion.

Now say that the companion is filling its Roche Lobe and steadily transferring mass into the gravitational potential of the neutron star.
At this point there are two things we need to realize

1. The transferred matter has velocity and angular momentum which is set by the motion of the companion star.


2. The matter is being transferred in a rotating system, so from the neutron star's reference frame it is subject to the Coriolis effect.

Then, as the matter stream moves down the potential well, it will fall behind the neutron star (Coriolis effect), only to loop around and settle into an orbit (angular momentum). The disk then rotates in the same direction as the orbital motion, making it prograde with with neutron star rotation.

Note that this explanation is rather simple and by no means comprehensive. There may very well be peculiar situations in which a retrograde disk forms. I know that at least for black hole binaries there are some systems suggested to be retrograde.

For more details on mass transfer you can look at this file.

galaxy - Are there heavenly bodies between galaxies?

Certainly "empty" space between galaxies has some things in it. Space has properties in and of itself, such as dark energy and virtual particles appearing in pairs and then disappearing, but ignoring the virtual and the not very well understood, empty space is full of photons, more specifically, Cosmic Background Radiation and all kinds of wavelengths of light from galaxies as well as cosmic rays and a boatload of neutrinos.

Most of the matter that came out of the big bang was gravitationally drawn into galaxies, but likely not all of it, so there likely is some primordial matter between galaxies too, just not very much of it. (precisely how much is a bit over my pay grade).

Also, as Conrad Turner points out, 3 or more body gravitational interactions can from time to time, kick a planet or star out of the galaxy but gaining sufficient velocity to escape the dark matter halo completely is probably quite rare, outside of galaxy on galaxy collisions, but rogue stars/planets outside of galaxies probably happens from time to time.

the moon - Is a spotting scope or binoculars a better choice for astronomy?

I'm interested in observing the moon and planets, and maybe some nebulae and star clusters.

I have a pair of binoculars:

• Nikon OceanPro 7x50 binoculars (around $300)


• Celestron 20x80 binoculars (around $100)

I'm looking at a step up, to 100mm. Specifically, I'm trying to choose between 25x100 binoculars (either a Celestron or an Orion) and a 100mm spotting scope.

The only reason I'm considering a spotting scope is that it has a magnification of 22-66x, compared to the binoculars' 25x. But I want to check if such high magnifications really work for astronomy.

Let's say I'm observing the full moon. What's the highest usable magnification, without resulting in too dim an image?

An Amazon review says that:

I have found that while the readability improves from 22x to about 38x
the resolution actually decreases after that and I never use it from
40-66x, there is no point. Maybe it would have been better to have
optimised it for 22-40x.

If the scope is limited to 40x in daytime use, I expect it to be worse at night, even observing the full moon.

I'm considering buying the scope only for the increased magnification over the binoculars, but if I end up using the scope only at 22x, it defeats the point of buying the scope in the first place. Should I stick with the binoculars?

the moon - strange moonset around 10 pm (3.2 2014, NZ) - explanation?

during a visit to New Zealand we observed (on 3.2 2014) a strange moonset in Te Anau.

Around 10 pm, shortly after sunset, the moon appeared, but then quickly began to go down, and about 1 hour or less it "touched" the mountains on horizon and shortly after it completely disappeared and then we did not see it anymore.

I did not bring a camera or camcorder, so I only have this bad photo from a mobile phone:

The whole process was pretty fast, so I only have one picture. I've never seen anything like it - usually (in Europe) the moon comes out after sunset rather than going down :-)

There was also one guy living there more than 10 years, but he was apparently surprised too.

Can anybody explain this? Is it something normal in this location?

thanks

cosmology - The largest discrepancy in the history of science

There are roughly two possibilities: either there isn't a large vacuum energy, this would imply that there is something missing from quantum field theory. We don't know what it could be.

Or the zero point energy really is as large as QFT predicts, but there is something else that prevents it from having a large cosmological effect. We have no idea what that could be.

We might get more of an idea if we find out what Dark Matter consists of, or a good quantum theory of gravity we might know more.

the sun - Solar Noon: meridian crossing time versus time of maximum elevation

There are a couple of things you have to think about in order to understand the problem at hand. The first is that there are three coordinate systems at play. The first is the celestial coordinate system, which is a coordinate system based upon the latitude and longitude of the Earth (the celestial equator is the Earth's equator projected out into space). See this post.

The second coordinate system is that of the solar system: The plane of the solar system (which contains the sun and planets, and traces out a line in the sky called the ecliptic). Now because of the tilt of the Earth's axis with respect to the normal vector of the plane of the solar system, the celestial equator is not the same as the ecliptic. What's more, is that the relative position of the ecliptic and the celestial equator change throughout the course of the year.

Lastly, there is the altitude/azimuth coordinate system which locked to the observer. The zenith (the point at which all horizons are equidistant from the point you are looking at) is $+90^{circ}$ in altitude and is a degenerate point in azimuth, and all horizons are $0^{circ}$ altitude but lay on a unique azimuthal arc stemming from the zenith and intersecting the horizon.

Below is a diagram showing the relationship between the first two coordinate systems, the celestial coordinate system and the coordinate system of the solar system.

Why does all of this matter for the question you've asked? It's because as you look due East and due West, you're looking at $0^{circ}$ declination (the celestial equator intersects the horizon at exactly due East and due West) whereas the ecliptic intersects the horizon at different points. These points change throughout the year as the relative position of the ecliptic changes with respect to the celestial equator. Now, the amount the sun travels on the ecliptic over a 24 hour period is $0.9856^{circ}$ ($360^{circ}/365.25 text{ days}$; actually I may want to use a sidereal day here but nonetheless it's about a degree).

Due to this, there is potentially a little bit extra in altitude that the sun may travel throughout the course of a day (since the path of the ecliptic is neither aligned with the celestial coordinate system, nor aligned with the local alt/az coordinate system of the observer; unless you're at the north or south poles in which case your local alt/az coordinate system is the same as the celestial coordinate system, however both are still mis-aligned with the coordinate system of the solar system). I think the word "approximate" needs to be there.

All told, I don't personally know how to quantify the extra bit of altitude the sun may gain from the misalignment of the ecliptic with lines of constant altitude and from the sun's motion along the ecliptic, but I would welcome references and calculations from other people. This little bit extra may not be a really small number. In other words, you might want to see how the rate of change of altitude of the sun at that point changes simply due to the Earth's rotation and how it may compare to the rate of change of altitude due to the motion of the sun along the ecliptic.

EDIT:
I also just looked to see what the ecliptic looks like on the sky (as seen from Philadelphia), and how it compares to the local alt/az coordinate system. Below is a picture with all coordinate systems present to try to illustrate the point I was trying to make.

The green line is the meridian, the orange(ish) lines are alt/az coordiantes, the blue line is the celestial equator, and the red line is the path of the ecliptic. Hopefully you can make this out sufficiently well.

fundamental astronomy - Speed of light through the ISM and Wavelength

You can find a neat description and some examples of the effect here. This is known as the pulsar dispersion measure. As you correctly say, waves with longer wavelength (lower energy photons) are delayed with respect to shorter wavelength radiation from the same phenomenon.

When electromagnetic waves travel through a plasma, they excite currents in the free charged particles. In such cases it can be shown (using Maxwell's equations) that the waves propagate with a relationship between their frequency $omega$ and "wavenumber" $k = 2pi/lambda$ given by
$$omega^2 = omega_p^{2} + c^2k^2,$$
where $omega_p$ is known as the "plasma" frequency and equals $(4pi n_e e^2/m_e)^{1/2}$ for the electrons in the plasma (i.e. it depends on the electron number density $n_e$.).

Now, if you have a bunch of photons emitted as a pulse, the relevant velocity is the group velocity given by $v_g=domega/dk$. So
$$v_g = frac{c^2 k}{(omega_p^{2} +c^2 k^2)^{1/2}} = cleft(1 - frac{omega_p^2}{omega^2}right)^{1/2}$$

This converges to $c$ when the frequencies are high (wavelengths are short), but is slower when frequencies are low (wavelengths are long).

In terms of an intuitive physical picture, yes you could think that the refractive index is frequency dependent, but the difference here is that the reason for this dispersion is that waves travelling through a conductive medium are "lossy" - that is the currents induced also encounter resistivity and therefore the waves heat the plasma.

jupiter - Can magnetism escape a black hole?

Nothing "escapes" a BH - in the sense that a signal originating inside the event horizon remains forever inside. If something is observed moving away from the BH, then it was generated outside the event horizon. If it was generated inside, it would never be observed at all, forever and ever.

Gravity itself does not "escape" a BH - and neither does "not escape". Gravity is simply a characteristic of the metric of spacetime. If spacetime is warped in a certain way, gravity can be measured to exist. A BH is simply a very powerful distortion of spacetime, nothing more, nothing less. It is generated by a concentration of mass/energy, which warps spacetime, and then that concentration becomes trapped by this distortion that it has produced.

In that sense, gravity is simply part of the BH, because it's spacetime being warped, and because a BH is essentially just that. Its gravitational field is part of itself, extending to infinity (but getting weaker with distance). It doesn't "escape" because there's nothing escaping.

It's like having a plastic bag tied into a knot to keep water inside, and someone asks "so how does the plastic escape the knot?" The plastic does not "escape" the knot, the knot is part of the plastic.

This all becomes easier to understand when you realize that gravity is not a thing, it's just an effect of spacetime being distorted.

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EDIT: I think what you were really asking was - can a BH have its own magnetic field? The answer is yes.

A BH can have 3 characteristics: mass, spin (rotation), and electric charge (a.k.a. the no-hair theorem). All other characteristics of the matter falling in it are lost, except these three. If you drop a proton into a neutral BH, then the BH acquires a charge equal to one proton, and that's a measurable electric field.

Now consider a spinning BH with an electric charge, the Kerr-Newman metric. You have a charge, and you have spin. That means you have magnetism. So, yes, a BH can have a magnetic dipole. However, the rotation axis and the magnetic dipole axis must be aligned - a BH cannot be seen as "pulsing". Again, no signal from inside the event horizon can be observed outside.

However, you should not imagine the electric (or magnetic, same thing) field as "escaping" the BH. It does not escape. What happens is, when the charges were swallowed by the BH, the lines of electric field remain "glued" to the BH, which then acquires a charge. Those lines of electric field have existed forever, they don't "escape" anything, and continue to exist after the charge is trapped by the BH.

Note: electric fields and magnetic fields are one and the same. One could appear to be the other, depending on the motion of the observer.

Which kind of properties can we get for cosmic ray particles hitting on an optical ccd?

Cosmic ray "hits" are artifacts caused when high energy particles (cosmic rays, often muons) slam into atoms in the CCD itself and liberate large numbers of electrons, which then show up as bright spots and streaks when the CCD data is read out. They can also occur as a result of radioactive decay processes in the material in the detector and instrument itself and also the immediate surroundings.

They are a tremendous nuisance that can be removed to a certain extent by median stacking images - though this also fails if you get into a situation where cosmic ray hits are found in the same pixels of $>1$ images in your stack. This limits the lengths of exposures that can safely be made with astronomical CCDs. Depending on exact usage, this limit is usually between 30 minutes and an hour.

The flux of muons and the distribution of angles at which they reach the detector is actually very well known already. It is largely unaffected by local shielding (unless your telescope is buried far underground!)

It is possible to distinguish different types of events (muons, locally produced gamma rays etc.) from looking at the pattern made on the detector (e.g. Groom 2004), but I don't think you can tell much about their energies, unless it could be by looking at the flux as a function of how much shielding (lead?) or how far underground the detector was, since often only a small fraction of the cosmic ray energy ends up deposited in the CCD.

star - is pan-starrs' data available to public users?

Looks like you've to be member of the PS1 Science Consortium to be able to log in.
And you'll need some database query knowledge, at least by using the PSPS Query Builder, or more advanced programming capabilities (XML, SOAP, Python, Perl). A short documentation of the database tables can be found here.
If you like to dig deeper into the database, this FAQ page may be a starting point.

The image processing pipeline (IPP) is described in this paper.

planet - The role of gravity during planetesimal accretion

The very first stage of planet formation involves purely inelastic (i.e., sticky) collisions between pairs of small particles of dust to make slightly larger particles of dust, pairwise collisions between these slightly larger particles to make even larger particles, and so on. So far, the only gravitation that comes into play is that of the central protostar and the disk as a whole. The gravitation of these little objects is very small.

Then some magic happens that makes the little rocks that result from this very first stage grow into planetesimals that do gravitate. How do little rocks (a few centimeters in diameter) grow into little planetesimals (a few hundred meters in diameter)? This is the key open problem in the theory of planet formation.

The core of the problem is that objects the size of a spec of dust to an object the size of a small pebble move with the gas that forms the bulk of the disk. The outward pressure from the gas in the disk makes the gas and small objects move at a slightly less than orbital speed. Once an object reaches the centimeter scale or so, the pressure from the gas no longer provides the support needed to keep those objects moving at that slightly suborbital speed. The objects start to become separated from the gas. By the time an object reaches a meter or so in diameter, the gas is equivalent to a hurricane force headwind.

One consequence of this is that collisions become ever more violent as objects grow beyond a centimeter in size, and hence are more likely to break colliding objects apart than to make them coalesce. An even bigger problem is that that headwind makes centimeter- to meter-sized objects spiral into the protostar, and very quickly (a few hundred years from 1 AU). A number of proposals have been put forth to overcome this meter barrier problem. So far, none appears to have stuck. (Sorry for the pun.)

Once objects reach some critical size, the drag from the gas reduces in importance thanks to the square-cube law and the gravitational attraction increases. That critical size varies from hundreds of meters to tens of kilometers, depending on which paper one reads (experts vary). These objects are "planetesimals," and there are lots of them. These once again combine readily to make planetary embryos. From then on, its just a mopping up process as gravitation either makes these planetesimals / planetary embryos combine or ejects them from the star system.

Does Pluto have impact craters?

It was predicted by some, before the images arrived, that Pluto's craters would be much shallower than perhaps on other rocky objects/satellites. The reason is the icy composition and the relatively low impact velocities in the outer solar system (Bray & Schenk 2015).

There are indeed reported quotes in recent days that there are no impact craters on Pluto (e.g. here or here), based on the highest resolution images.

However, I'm with you - the things you have marked look like very large impact craters, or the remains of them, and so what we are seeing is that there are no small ones.

The gist of what I have heard/read is that this indicates geological activity on Pluto, which is effectively removing the evidence of relatively small impacts and that the surface we see is mostly younger than 100 million years old.
Watch this space, a lot of people are just flapping their lips at the moment (including me).

Another possibility - raised by Wayfaring Stranger - could be that the expected number of impacts was (hugely) overestimated. This is discussed by Singer & Stern (2015). There are indeed uncertainties, and it is hoped that cratering on Pluto & Charon would help constrain the number densities and size distribution of small impactors. My impression is that the uncertainties could be at levels of up to factors of 10 - and this seems to be discussed quite carefully in Greenstreet et al. (2015) (behind a paywall, sorry), who also comment in the abstract that surface ages judged from cratering are " entirely dependent on the extrapolation to small sizes [of the impactor distribution]".

Further Edit: LDC3 suggests all the craters could be too small to be seen. The high resolution images that have been released could resolve craters of a diameter of a couple of km or more and cover a few percent of the surface. According to Singer & Stern (2015), the number of craters at this size and above on Pluto they expected in such an area is $sim 1000$. So I think that either the planet has been resurfaced recently, the craters have been eroded in some other way or their estimates for the number of impacts/size of the impactors is wrong.

the sun - Name and language of symbol in astronomy book

I've been working on a programming project involving Peter Duffett-Smith's Practical Astronomy With Your Calculator (3rd Ed.), which has a large number of mathematical/astronomical formulas in it. There's a part at the end of the book for symbols and abbreviations, which contains this:

After doing some Google-fu, I was unable to find the name of the symbol. Would anyone smarter than me (which should be quite a few of you... heck, quite a lot) know what this is? Hopefully there's Unicode for it...

solar system - Pluto's orbit overlaps Neptune's, does this mean Pluto will hit Neptune sometime?

No. From 1979 to 1999, Pluto was the eighth planet from the sun. In 1999, it slipped beyond Neptune to become the ninth. But Pluto's 248-year orbit around the sun takes it 17 degrees above and below the plane in which Neptune and the other planets travel.

So their paths don't actually cross as they swap positions. Imagine you are the sun in the middle of your back yard. The fence is Neptune's orbit. You toss a boomerang way out over the neighbor's houses and it comes back, being on both sides of your fence during its travels without hitting the fence. Of course, activity like that can be frowned upon, and in Pluto's case helped lead to its demotion.

Reference: Will Pluto Neptune Hit