Predicting/simulating magnetospheres of exoplanets? Based on atmospheric parameters, size, or distance from the star, is there any way to simulate or predict the magnetosphere of an exoplanet? Are there programs (similar to climate models for atmospheres) that can simulate magnetospheres?

The magnetosphere is usually not generated by the atmosphere, but by the interior ferromagnetic flows of a planet. I say usually not, because Venus is an exception for the terrestrial planets in our solar system.
But in general rocky planets will exhibit a wide range of interior flow conditions, depending on their individual composition and state of cooling.
Gaseous planets can be a bit easier in that case, if one accepts uncertainties about the properties of metallic hydrogen.

If you then simply go an assume you know all those complicated properties, then sure, you can simulate something. But it's not a scientifically testable prediction only playing around with parameters, as long as it's not tied to a specific object.


Will Saturn's Moons be destroyed? I heard that Saturn has 62 moons with confirmed orbits, 53 of which have names and only 13 of which have diameters larger than 50 kilometers. Some come within 1km of each other.

How do these moons keep in Saturn's orbit without hitting each other, and someday will these moons collide with each other and be destroyed?

The various moons are on orbits that are mostly a lot more than 1km apart from other moons. The orbits are stable, so the moons just keep going round Saturn on the same orbit. The moons are far enough apart that they have negligible gravitational influence on other moons. So the moons never get close to each other and don't collide. There are moons, like Daphnis and Atlas that can come with 1000km of each other, but they are very small. No moons come within 1km of each other.

There are some exceptions to this: Janus and Epimetheus orbit at nearly the same distance from Saturn. If they stayed like this they would collide within each other within about 4 years. However as one moon starts to catch up with the other, it is pulled forwards, causing it to move out, and slow down. Letting the other moon move inwards and faster. The two moons essentially swap orbits. Although there is only about 50km between the orbits, the two moons never approach closer than 10000km, so are not at any risk of collision.

Moon collisions can take place, one theory on the origin of the rings is they they were formed from a moon that broke up, either from a collision or by Saturn's gravity (It's not the most popular or recent theory). Collisions have never been observed, nor are any forecast to occur for as long as we can predict.

Of the 62 moons, remember that some of them are extremely small: a few tens of metres across, and there is no clear division between "small moon" and "large clump of ring material"


Can the moon's gravity influence be observed on the space station's orbit? In browsing through Astronomy Stack Exchange, I came across this question about orbiting the moon (Is it possible to achieve a stable "lunarstationary" orbit around the moon?).  In further looking, I came across (Can I measure the moon's gravity?) which seems to indicate that with sensitive enough equipment, the moon's gravitational pull could be measured on earth.  This prompted me to wonder if the moon's gravitational pull can be observed on the Space Station's or geosynchronous satellite's orbital path?  And if so, if any compensation (either short-term or long-term) needs to be performed due to the moon's gravity.  I would think it to be relatively inconsequential, but am not sure, hence the question.

According to NASA, satellites in Earth orbit don't experience strong enough interference from the Moon to "whisk a spacecraft out of the game". To a satellite in Earth orbit, the gravitational pull of the Sun is actually ~160 times stronger than influence from the Moon. Source: A New Paradigm for Lunar Orbits (see EDIT note below for details).

However, gravitational pull from the Sun, the Moon, planets such as Jupiter (as well as other factors such as solar radiation pressure, atmospheric drag, etc.) do cause enough interference to slightly alter the orbital path of any satellite orbiting Earth. To compensate for the deviations all satellites must periodically fire their thrusters for brief periods of time in order to maintain the correct desired orbit. If a satellite has no fuel to perform these "orbital stationkeeping" maneuvers it essentially becomes space junk. Source: Is the ISS, and all other satellites, affected by the gravitational pull of the moon?

EDIT: Consider a satellite with mass of 3,000 kg rotating around Earth on geostationary (GEO) orbit of 42,164 km radius. Gravitational pull from the Earth is constant throughout the orbit and equals ~673 N (Newtons). As the satellite rotates around the Earth it gets closer or further away from the Moon as well as the Sun and, therefore, their gravitational pulls change. The pull from the Sun is about 17 Newtons but changes only by 0.02 N. The gravitational pull from the Moon varies between 0.12 N (when the satellite is between the Earth and the Moon) and 0.08 Newtons (when the satellite is furthest away from the Moon over the opposite side of the Earth). So even though the pull from the Moon is much smaller than from the Sun, the variation of the Moon's perturbation is about twice as large than from the Sun.

Note: the numbers above are computed under a simplified assumption that all orbits are on the same plane. In reality, due to the inclinations of orbits, the numbers can be somewhat lower, but the overall picture remains about the same.


How can light reach us from 14 billion light years away? One thing that I can't quite wrap my head around is how light is traveling to Earth from 14 billion light years away, aka the beginning of the universe.

The way I see it, the universe itself was very small 14 billion years ago and has been steadily accelerating in size since.  So the photons that left a star that existed 14 billion years ago couldn't travel for 14 billion years because the universe itself was probably only a few million light years total in diameter.  The universe is not expanding faster than the speed of light, so that photon should have hit the end of the universe before now.

What piece of this puzzle am I missing?

If a star 300,000 km away from you explodes, it'll normally take the radiation from that explosion one second to start vaporizing you.
If the space between you and that exploding star inflates to 2.6X10^23km (28 billion ly) one half second after the explosion, the radiation will take 14 billion years, plus a half second, to reach you, and won't vaporize you.
Of course the expansion need not be instantaneous. You can plug in about any function you want there. The important thing is not the distance between you and the star at the time of explosion, but the total distance the photon must travel on its path to you. That distance must take into account the expansion of space between the emitted radiation and you, the observer.


Spectographically determine surface composition of exoplanet We can determine the surface composition of a star by studying its spectrograph. If a exoplanet transits the star, then there is a slight change to the spectrograph, which lets us determine the composition of the planet's atmosphere.
When the planet just become visible from behind it's star, then the star's light will:-

come from the star (known composition)
travel through the planets's atmosphere (known composition)
reflect from the planet's surface (unknown composition)
travel through the planet's atmosphere again (known composition)

Is it possible (either currently, or theoretically) to use this method to determine the bulk surface composition of an exoplanet?
This might be very difficult to achieve as the light reflected from the planet will be swamped by direct light from the star.
How do astronomers determine the texture of an exoplanet? and Can visible wavelength spectroscopy study an exoplanet's chemical composition directly? might be releated

By the title of your question, I don't know if you're asking about the surface composition or the bulk composition. I will answer both.
Although spectrometry can be used to determine the atmosphere composition of a planet, it might also, in theory, helps us to determine the composition of the surface using, as you said it, the reflected light of the planet. This can be done in Solar System planets or moons, because detecting the reflected light is hard but still feasible. In exoplanets tho, it's still very complicated. Maybe with the James Webb Telescope it may be possible in the future. In fact, there are certain papers from which the process of differentiating oceans from land in exoplanets is described, such as Ford et al. (2001), Cowan et al. (2009) or Kawahara and Fujii (2010). I suppose that the actual composition may also be possible. They use the rotation of the planet combined with the difference of spectrometry to make a mapping of the theoretical surface of an Earth-like planet.
Nonetheless, I doubt it about the bulk composition, basically because as you pointed it,  light can't reach into the Interior of a planet, so it's impossible to study its spectrometry in that regard.
If you want to know about the bulk composition of an exoplanet, for example, you can use other techniques such  as the mass radius relationship, which can be often used to determine its bulk composition. Look at Zeng and Sasselov (2013), for example.


Conversion of UBVRIJHK magnitudes to luminosity? So, I know that the standard for conversion from bolometric magnitude to luminosity established in 2015 equates to:

L = 10^(-0.4 * magnitude) * 3.0128e28

Easy enough. But what does one do about luminosities in specific bands? I've seen some information that makes it look like it should be something like:

U: 10^(0.4*(5.61 - U-band magnitude)) * 3.828e26
B: 10^(0.4*(5.48 - B-band magnitude)) * 3.828e26
V: 10^(0.4*(4.83 - V-band magnitude)) * 3.828e26
R: 10^(0.4*(4.42 - R-band magnitude)) * 3.828e26
I: 10^(0.4*(4.08 - I-band magnitude)) * 3.828e26
J: 10^(0.4*(3.64 - J-band magnitude)) * 3.828e26
H: 10^(0.4*(3.32 - H-band magnitude)) * 3.828e26
K: 10^(0.4*(3.28 - K-band magnitude)) * 3.828e26

But I'm not sure about these, because I've seen varying figures from varying sources for the magnitude of the sun in various bands.  I'm trying to extract the luminosity values from PARSEC data.

Speaking of PARSEC data, the documentation is rather lacking, so any clue about the following fields?

McoreTP: Sounds like it should be something about core temperature, but the value is usually either 0, or something small like "0.546"

Mloss: Presumably mass loss, but there's no units!  But I can probably work out the units just by comparing how star masses change over time....

Tau1m: ??? Usually 0.00, but sometimes something small like "0.06"

Cexcess: ??? Generally "-1.000", but sometimes things like "7.356".  Note that there's an entirely separate field for Xc, which is clearly carbon mass fraction.

Luminosities cannot be calculated from magnitudes without knowing the distance to the source (and perhaps something about extinction).

I assume therefore that you are talking about absolute magnitudes in those filter band passes.

Luminosity can also not be arrived at from a single absolute magnitude measurement in the way you suggest in your post, since for a source of a given luminosity, how much of that luminosity emerges in a specific band is entirely dependent on the spectrum of the source.

OK, but now I see what you want - you want to divide up the bolometric luminosity of the Sun between the individual bands using the absolute magnitude of the Sun in those bands. But the luminosity doesn't divide up neatly like that, partly because the bands overlap and partly because the bandpasses are not top-hat functions so it's difficult to see what a "luminosity in the U-band" actually means other than an integration of the intrinsic stellar spectrum (which you don't have) over the (normalised) bandpass sensitivity profile.

$$ L_{\rm band} = \int f_{\nu} S_{\nu} d\nu$$

The formulae you quote contain the absolute magnitudes of the Sun in those bands. That is the first number in the brackets. If you subtract the bolometric magnitude from this, then that gives you (minus) the bolometric correction. The problem is that the UBV etc. magnitude system and the bolometric magnitude system do not share the same zeropoint. The bolometric correction in the V band is about zero, so your formulae would have the entire luminosity of the Sun emerging in the V band. The zeropoints of the UBV etc. system are set by defining an A0 star (like Vega) to have an absolute magnitude of zero in all pass bands.

What you can do is convert the magnitudes into some sort of mean flux density over the bandpass and then just multiply this by the "effective width" of the filter. This is very approximate and essentially assumes the spectrum is flat across the filter bandpass.

Page 100 of Zombeck (1992) gives the flux density zeropoints in UBVRIJHK corresponding to magnitude 0.0. So take your magnitude and scale the zeropoint flux accordingly (by $10^{-0.4m}$) and then multiply the flux density by the effective bandpass width. Finally to convert an observed flux to a luminosity, multiply by $4\pi d^2$, where $d=10$ pc, if you are dealing with absolute magnitudes.

By "PARSEC" I assume you are talking about the stellar evolution models produced by the Padova group. In which case a basic output of the model is the stellar luminosity. Any magnitudes that are provided by the models are calculated afterwards by using a particular set of bolometric corrections (the relationship between absolute bolometric magniude and absolute magnitude in a particular passband).

As regards the specific columns in the tables, you will likely have to go to the original papers that describe the models. I cannot comment further without knowing exactly which models they are or where you got them.


What is Charon's ascending node in reference to? NASA's orbital elements given for Charon's orbit are given at https://ssd.jpl.nasa.gov/?sat_elem. The ascending node is defined as being "measured from the node of the reference plane on the ICRF equator". For Charon, the plane being used is given as the "Mean equatorial (IAU equator)". I'm assuming this is Pluto's equatorial plane. What then is Charon's ascending node in reference to? Normally this would be ecliptic vernal equinox, however this obviously cannot be it.

Some sources have the ascending node as undefined, but that then begs the question, what is the argument of perihelion in reference to if the ascending node is undefined?

I'm assuming this is Pluto's equatorial plane.

That is incorrect. Charon's right ascension of ascending node with respect to Pluto's equator is undefined. Seemingly paradoxically, it is well defined with respect to Earth's equatorial plane, and since that has become the universal plane of reference, that is what is used.

Charon and Pluto are tidally locked to one another. Their mean axes of rotation and orbital axes are identical. This means that Charon's orbit has zero mean inclination with respect to Pluto's mean equator, which in turn means that Charon's mean right ascension of ascending node with respect to Pluto's mean equator is undefined.

Charon's orbit about Pluto is however inclined with respect to the Earth's mean equator. In particular, it's inclined with respect to the Earth's mean equator at the J2000 epoch (noon Terrestrial Time on 1 Jan 2000). The Earth's mean equator at the J2000 epoch is the ICRF equator, to within a fraction of a milliarcsecond. The linked data clearly states that Charon's right ascension of ascending node is measured with respect to the ICRF equator.

Reference:

Brozović, Marina, et al. "The orbits and masses of satellites of Pluto." Icarus 246 (2015): 317-329.


accretion discs in h/lmxbs High mass x-ray binaries necessarily accrete via wind, no disc accretion?

What makes the difference between a hmxb and a lmxb?

In what kind of circumstances does an accretion disc form?

How does it form?

When do we think there is a disc?

There is a clear cut between disc and wind accretion?

The physical difference between high- and low-mass X-ray binaries is that the latter has a donor star that fills its Roche lobe. Usually, the compact object is more massive than the donor star. Mass transfer occurs via Roche lobe overflow.

In high-mass X-ray binaries usually the donor star is more massive than the compact object and mass transfer is via a stellar wind.

The reason that a disc may form (in either case actually, but almost always for a LMXB) is that there is too much specific angular momentum (angular momentum per unit mass) in the accreted material to allow it to fall directly onto the compact object (angular momentum must be conserved). Broadly speaking, a condition for the formation of an accretion disc would be that the specific angular momentum of the accreted material exceeds the specific angular momentum of material in a Keplerian orbit at the surface of the compact object. The formation of an accretion disc would be favoured by short orbital periods or slower wind speeds. Material can then only be transferred onto the compact object as angular momentum is transferred outward by (possibly) viscous torques.

This simple picture is immensely complicated by the presence of any strong magnetic fields.


Why would it be very light out at 3 AM? My brother noticed something very odd last night: it was amazingly light out when he stepped outside for a smoke even though it was between 3 AM and 3:30 AM EDST and we are near Toronto Ontario Canada, not in the Far North. He is outside at that time most nights and is no stranger to being up very very early at all times of the year. He has always lived here and is 63 years old, not an inexperienced child. He said the amount of light in the sky was equivalent to the amount of light you'd see in the mid-afternoon of an overcast day. He'd never seen anything like it.
Since I have a lifelong love of science fiction, he asked me if I had any idea why this might have been; he thought there might be some natural phenomenon that might explain it. I asked if he had used some kind of intoxicant but he assured me he was completely sober.
I'm at a loss to think of any plausible source for his experience but I'm not a scientist of either the professional or amateur kind so I thought I'd ask here if there is some rare atmospheric condition that might explain it.

I come from a similar latitude and enjoyed the night skies and two things that can add to "the amount of light in the sky" that I can think of. The first is the near-full moon as mentioned above, and the second is unusually low clouds. Being near Toronto and outlying communities, there is a great deal of light produced, and some goes up. On nights when the clouds or even semi-transparent haze are present and unusually low, they can reflect light pollution more often than normal. Add to that the light from the Moon and it should be enough.
The reason I mention latitude is that when the ground is covered in white, reflective snow, the light pollution is even worse; lights projected preferentially downward are more strongly reflected back upwards. I don't know if there's any snow on the ground still there; I haven't seen snow in way too many years.
Since you didn't mention that your brother noted the full Moon, I'm guessing it was cloudy and so they didn't see it. So I'm guessing that instead they saw clouds or haze lit by it and/or by local light pollution.


Mini disk of a blackhole meaning What is meant by the "mini disk" of a blackhole? Is it the same as accretion disk? If not, whats the difference?

Eg: In the abstract of this paper https://arxiv.org/abs/1806.05697 it is mentioned "When the accretion rate is relatively high, the circumbinary disk, accretion streams, and mini-disks combine to emit light in the UV/EUV bands"

This paper, Quasi-Periodic Behavior of Mini-Disks in Binary Black Holes Approaching Merger, which shares authors with the paper linked in the question, sheds some light on the terminology.

  We present the first magnetohydrodynamic simulation in which a
  circumbinary disk around a relativistic binary black hole feeds mass
  to individual accretion disks ("mini-disks") around each black hole.

They are talking about active binary black hole systems. The system as a whole has an accretion disk, the circumbinary disk, and each black hole has its own accretion disk, which they call the mini-disk.


Our galaxy, the Milky Way, forms part of a galaxy group called The Local Group. What are the parameters that dictate what a galaxy group is? Are there different classifications or just the one? How does our galaxy group compare with others?

A galaxy group is a set of galaxies that are close together and gravitationally bound, i.e. barring outside influences they will stay together indefinitely. That means a galaxy that is in the space occupied by the group but "passing through" it at a high velocity would not be considered part of the group.

The upper limit for calling it a group is roughly 50, bigger accumulations of galaxies are called clusters (which may consist of many groups). The borders between adjacent groups may be uncertain.

These aren't mathematical definitions and actual usage is often based on tradition more than on measurements. It's just like geography: you could debate endlessly whether a given piece of high ground should be called a hill or a mountain, and whether something is a separate mountain or a subpeak, but sensible people recognize it as a waste of time, make up a clear but arbitrary numeric distinction without strongly insisting on its usage.


Buying a telescope for the first time. How important is the focal ratio and is a computerised telescope worth it? I'm wanting to purchase a telescope for the first time. I've narrowed it down to 2 options based on what I've discovered from researching and my price range. I mainly want do planetary observations and some deep space. The 2 options that I've found are the following

Celestron 114 LCM Computerized Telescope
Orion 9851 SpaceProbe 130 EQ Reflector Telescope

From what I've gathered, both do have decent specs for what I want to do but I'm still a bit lost when it comes to the aperture and focal length. The Celestron 114LCM has a 1000mm focal length but the aperture is only 114mm where the Orion 130 has a 900mm focal length but a 130mm aperture. Am I right in thinking that the Orion telescope would be better for me? Also, with the Celestron telescope it is computerised so, although the fun would be taken out of messing around with the scope and feeling satisfied with the results, is it really worth it to have this feature for a first time purchase? Thanks in advance!

We can compare these scopes on a few grounds:

Celestron 114 LCM

Greater focal length
Alt-Az mount (intuitive)
Computerized (easy)

Orion 9851 SpaceProbe 130 EQ

Cheaper
Larger aperture
Greater highest useful magnification
Non-computerized (the experience of learning the sky through one's own methods, alone with the telescope, is one which you should not deny yourself, in my humble opinion)
EQ mount (allows for simple star tracking)

These are just some items which stand out to me as being important. The question you have asked is largely one of opinion—there is no "right" answer here. It depends entirely on what you want. You say you want to focus on planetary observations, a field where extreme magnifications are common and useful. For deep space, larger apertures and shorter focal lengths will aid.

The focal ratio is a quick way to judge the "speed" of the telescope's optics. It is found by taking the ratio between the focal length and aperture. The Orion scope in this instance has an f-ratio of f/6.9, while that of the Celestron is f/8.8. This implies that the view through the Orion scope will be slightly brighter, broader, and less-magnified than that through the Celestron, for any given eyepiece. However, these ratios are quite close and are both within a range commonly used both for deep-sky and planetary observations.

My final piece of advice is that for a beginner, the subtle differences between these scopes in these regards simply does not matter. You are unlikely to experience any significant difference between these two scopes. Because of that, I would personally recommend going for the Orion telescope—it's cheaper, will give you greater magnifications for planetary observations, and its lack of computerization will force you to truly learn the sky. Should you desire easier set-up, less time navigating through the sky, and so on, then the computerized Celestron may be preferable.

Telescopes are beautiful instruments. Whichever you choose will serve you well.

Clear skies!


Galactic extinction as a function of distance I am looking at several galactic sources in the $I$, $R$ and $V$ bands and I want to calculate their absolute magnitudes. I can get the apparent magnitude and I know the distance so I just need to account for extinction to acquire the absolute magnitude. I first used the NED Coordinate Transformation & Galactic Extinction Calculator but the calculated extinction values are too high. I figured that this is probably due to the fact that my sources are relatively close ($\sim 100\,\text{pc}$) and I am thus only looking through a fraction of the dust that these coefficients were calculated for.

If I lower the extinction coefficients to a certain fraction of NED's value (say 60% in a particular case) then I get the expected answer but I'm looking for an accurate way to choose this fraction. I've looked for a method to calculate the extinction coefficients as a function of distance but haven't been able to find anything. Does there exist a map of extinction as a function of distance (and galactic coordinates) in the galaxy from which I can extrapolate the information required? Is there a way to calculate the extinction coefficients ($A_I$, $A_R$ and $A_V$) for an object in the galaxy a known distance away? I found IPHAS which seems to be the right idea but I'm not sure if I can use it to get what I want (or how).

If your sources are within 100pc then the best thing to do is assume the extinction is zero, unless you are looking for absolute magnitudes that are a lot more precise than say +/- 0.1 mag. I think you would be very unfortunate if the extinction was any more than about ~0.1 mag in the V-band (and less than that at R and I).

Beyond that you are really struggling because the structure of the ISM is very non-uniform, so you would have to go with the kinds of maps that are produced (for example) using IPHAS - but these are only in the galactic plane. But even if you could find some general map covering all positions it would be unlikely to give you an accurate answer out to 100pc, because there isn't enough extinction to give an effect that is big enough to be measured with any precision (i.e. the uncertainty might be bigger than the value).

If you have spectral types for your objects then you could always compare the V-I colour with what you expect for a star of that spectral type. This gives you the reddening $E(V-I)$ and then (roughly speaking) $A_V \simeq 2E(V-I)$.


Is there a website out there where I can find/compare data on stellar parameters for the brightest stars? I'm looking for some kind of catalog/repository which lists stellar parameters (e.g. mass, radius, temperature) in a similar vein to NASA's Exoplanet Catalog (but except for stars), for the brightest stars in the sky, with $m_v<4$. I'd rather not have to refer to Wikipedia as they only list data from one source (or an average of several uncited sources).
I'm aware of SIMBAD, but that site doesn't have a very intuitive UI and doesn't list mass measurements. I've also tried CDS (http://cdsportal.u-strasbg.fr/), but I can't find distance measurements from Gaia (although apparently Gaia doesn't measure distance and I need to put some kind of external study into a Gaia search query which I haven't got a clue how to use).

As you’ve discovered, there are a lot of different catalogs out there, and many ways to search them.  It would help if you could narrow your question a bit - what kind of data are you looking for?   You seem to want stellar masses at least - what else?  As noted above, there are few stars for which mass is measured directly, but there are robust ways of estimating it from other information, just as effective temperature can be determined from spectral type or photometry.
One catalog that might meet your needs is the TESS Input Catalog (TIC) or its brighter subset, the TESS Candidate Target List.  It compiles information from a lot of other sources, and includes magnitudes from optical through infrared, as well as estimates of stellar mass, radius, temperature, metallicity, luminosity, and surface gravity.  It also includes stellar parallax (i.e., distance) and proper motion from Gaia.
In addition to the ways of searching given in the above link, you can search it through Vizier here - I find the Vizier interface a little easier to navigate than MAST, though both have a lot of options.
Now this catalog may be overkill for what you want: it has 471 million stars, though “only” 27 million of those have mass estimates, and “only” 2 million have distances.    :-)  So knowing more about what you want could help narrow things down.
Edit: The above numbers refer to an earlier release of the catalog, prior to Gaia DR2.  The updated version (which is what I linked to from Vizier) has about 1.7 billion stars, of which 1.27 billion have distances and 455 million have mass estimates.
Edit 2:  After clarification in the comments that the goal is finding information only for very bright stars, this isn’t the best catalog to use.  While it does include entries for some bright stars (e.g. Vega, Betelgeuse), the information included is very limited, presumably because they are saturated (overexposed) in most surveys.
In general it is much harder to get homogeneous data for very bright stars because of this.  There is a wealth of data on them individually, but because they are saturated in systematic surveys, there aren’t uniform sources of data on them to which known relations can be applied in bulk.
The Yale Bright Star Catalogue (searchable here) is a good source of basic data (photometry, spectral type, distance, multiplicity), but for derived quantities like mass or radius you would have to apply known relationships yourself (if you need the data in bulk) or go to individual references for particular stars if you don’t need a lot of them. Honestly Wikipedia isn’t a bad starting point in the latter case - it is generally clear about what references the numbers come from.


Do objects that are invisible to the naked eye occlude much of the stars to Earth based observers? I was looking at a star recently and some other object that appeared to be stationary in relation to the star was occluding the light to only one of my eyes. This is why it caught my attention.

I was able to move my head around this shadow and see the star but there was a definite tiny area where the star was clearly hidden. The object hiding the star also appeared to be a similar size to the tiny point of light of the star. Is this common to observe?

For the last few days I have been asking myself how many objects could be hidden like this (though most with shadows that engulf the earth); How thick are asteroid belts etc, that are between our eyes and the rest of the stars.

Is most of it hidden?

From your description, an object that blocks out light from a star in only one location cannot be an astronomical object.

Occultations by asteroids are fairly common. You can see a list at  http://www.asteroidoccultation.com/. When a star is being occluded by an asteroid there will be a fairly narrow strip from which the star will be hidden. But it is still about 50km wide. The star appears to rapidly fade, and it remains hidden for at most a few seconds, then reappears. Your description of "only hidden to one of my eyes" doesn't fit an asteroid occultation.

The object hiding the star must, therefore, have been local, and not moving. It would be speculation to suggest what it could be.

There are may objects that are hidden in visible light, not by asteroids, but by interstellar dust clouds. We can see them by using other wavelengths.


Can Gravitational astronomy look beyond the CMB? In another question dealing with the earliest explored phenomenon in the Universe the observing limit of the CMB was mentioned.  Prior to CMB the Universe was opaque to photons, I believe.  I'm wondering if conditions before "first light" would have allowed sufficient matter clumping to produce currently detectable, or potentially future detectable, gravitational signatures.  The irregularities in the CMB do indicate clumping of some sort, do they not?  I'm wondering if LIGO and it's successors will have a look back time farther than photon-based astronomy.

Yes. Gravitational wave observatories like the proposed eLISA laser interferometer may be able to detect gravitational waves that originate from the early moments of the big bang itself. If some part of the big bang energy goes into gravitational waves then those waves will be redshifted by expansion and waves produced in the first $<10^{-10}$ s should be present in eLISA's frequency sensitivity window (Amaro-Seaone et al. 2012; Bartolo et al. 2016; Ricciardone et al. 2017). This might probe exotic physics like inflation, extra dimensions, phase transitions and cosmic strings.


Is the SOI a spherical region or a oblate-spheroid-shaped region? The Wikipedia article on Sphere of influence states that:

  "A sphere of influence (SOI) in astrodynamics and astronomy is the
  oblate-spheroid-shaped region around a celestial body where the
  primary gravitational influence on an orbiting object is that body."
  (Emphasis added.)

It then gives an expression for the radius of the sphere of influence.

Is the SOI a spherical region or a oblate-spheroid-shaped region?
If it is an oblate-spheroid-shaped region, then why?

Is the SOI a spherical region or a oblate-spheroid-shaped region?

The sphere of influence is neither a sphere nor an oblate spheroid. It is a surface with no name. An approximation of this surface is

$$\left(\frac r R\right)^{10}(1+3\cos^2\theta) = \left( \frac m M \right)^4$$

This is neither a sphere nor an oblate spheroid, and this is but an approximation. Thefull expression is an absolute mess. Dropping the factor of $(1+3\cos^2\theta)^{1/10}$ (which is close to one) yields

$$r = \left( \frac m M \right)^{2/5} R$$

Tada! The equation of a sphere!

The true surface is defined in terms of two ratios. Consider two gravitating bodies, call them body A and body B. From the perspective of an inertial frame, the acceleration of a tiny test mass toward these two bodies is given by Newton's law of gravitation. These two bodies accelerate toward one another as well, so a frame based at the origin of either body is non-inertial.

From the perspective of a frame at the center of body A, the acceleration of the test mass is the inertial frame acceleration of the test mass toward body A plus the inertial frame acceleration of the test mass toward body B less the acceleration of body A toward body B. Denote the acceleration of the test mass toward body A as the primary acceleration and the difference between the inertial frame accelerations of the test mass and body A toward body B as the disturbing acceleration. Finally, define $Q_A$ as the ratio of these two. Now do the same for a frame with origin at the center of body B. The sphere of influence is that surface where $Q_A = Q_B$.


Why don't they make an image of the black hole in the M33 galaxy? Since we see the Triangulum galaxy M33 from a quite vertical position (contrary to our Milky Way and a bit the Andromeda galaxy) it should be easy to image the black hole in the center of it, shouldn't it? Why did they prefer to first image the black hole in a galaxy which is about 20 times farther and thus harder to photograph the black hole in its center?

M33 does not appear to contain a supermassive black hole: in fact there's no evidence that it contains a central black hole at all. The upper limit on the mass of a central black hole based on the dynamics of the core region is a few thousand solar masses.

Merritt et al. (2001) "No Supermassive Black Hole in M33?" derive an upper limit of 3000 solar masses on a central compact object in M33, noting that this is still consistent with the M-σ relationship between the mass of a supermassive black hole and the velocity dispersion in the stellar bulge, using which they obtain a predicted mass of 2600–26300 solar masses.

Gebhardt et al. (2001) "M33: A Galaxy with No Supermassive Black Hole" obtain an even smaller upper limit on the mass of a black hole of only 1500 solar masses (their best fit mass is zero, i.e. no central black hole), which they state is significantly lower than the predicted mass from the M-σ relationship.


Binary pairings that haven't been discovered yet? Question: Are there any kinds of binary pairings that haven't been discovered yet? Any that are particularly significant, or that might shed some light on binary stellar evolution or theories of capture processes, that are sought, but so far no examples have been found or at least suspected?

By "pairing" I mean a binary object where each is a type of star or a black hole.

Stars can be anything from brown dwarfs to neutron stars. Black holes should be very roughly comparable to stars in size. (I'm not asking about a star in orbit around a supermassive black hole in the center of a galaxy)

For example, recent observations of gravitational wave events have suggested a pair of black holes merging, and a pair of neutron stars merging.

As far as I am aware, there are no known examples of black hole+neutron star (BHNS) binaries (e.g. this paper from 2018 notes that "No BHNS binaries (BHNSs) have been discovered so far by radio surveys"). BHNS mergers should be observable with gravitational wave astronomy, so this situation may not last for too much longer. There is a report of a possible BHNS merger detected by LIGO and Virgo (event S190426c). This event has yet to be confirmed, follow-up searches by Hosseinzadeh et al. (2019) and Goldstein et al. (2019) for transients have failed to find a counterpart to the event.

Binaries containing an sdB star and a neutron star have also not been found, though theoretical models predict they should exist in our galaxy (Wu et al., 2019). Sufficiently close sdB+NS binaries should be strong gravitational wave sources, so LISA may discover them.

Depending on how you want to define various "types" of star, you can probably come up with more examples. For example, there is the concept of singleton stars which have never been members of binary systems or had close encounters with other stars, which by definition have not been found in binary pairings.

There are also some weird categories of objects with few known members. For example there are the LP 40-365 stars, which are runaway stars with atmospheres dominated by oxygen and neon, likely the partially-burnt remnants of peculiar type Iax supernovae. None of the four known examples of this class are binary systems, plus the circumstances that lead to their creation makes it seem fairly unlikely that we'll find one in a binary.


How is it that all planets (and moons) in our solar system are in equilibrium orbits? Previously, I was told that in order for any bodies to orbit each other for over a long period of time, the orbital period, distance and masses have to be precisely matched such that the bodies won't eventually collide and forming larger bodies or being swung out of orbit.

How is it then, that all 8 planets (and countless dwarf planets), and their moons, and the Sun are all in such equilibrium for over thousands of years (if not millions)? Could it be possible what we see today are what's left, i.e. countless planets or moons have been swung out of the Solar System and collided, until a stable (relatively speaking) system is reached as we observe today?

I think at our current stage of solar system evolution, because of the fact that we have been able to evolve to our current level of sophistication, it could be considered to be very stable and in a very calm period in its evolutionary history. Unstable objects usually will be flung out very early on in the formation of such a system, hence why we do not see them today. However, just look at the Earth-Moon evolutionary history. It is theorised that our current Moon was captured due to a collision between Earth and perhaps a Mars (or similar size) object. This is anything but stable.

The reason why, is of course, related to gravity. Or more correctly, the gravitational potential of the system. All systems want to thermalise (in the same sense that a room of air molecules want to reach a thermal equilbirium). Although our solar system is anything but thermalised, it is constantly working to achieve this. Hence, why, at our present state of the solar system's evolution, we seem like we are in a fairly calm state. At this state, to go into more depth, we can employ Bertrand's theorem which tells us that for a central potential with an $r^{-1}$ dependance on radial distance the orbits will be stable. The stability of orbits in three spatial dimensions is due to the fact that the gravitational potential decreases with distance $r$ as $r^{-1}$.


Was the progress of astronomy in the 1800s surprisingly slow, and if so why? The 1800's was a century of fantastic scientific discoveries in chemistry, geology, biology, engineering and so on.  Is it correct to say that astronomy did not keep up with this development? And if so, why? If not, what might be some main reasons for me (and others) having this impression?

The late 1700's saw a couple of revolutions in astronomy such as the discovery of Uranus as the first non-antique planet and the measurement of solar system distances thanks to the transits of Venus. As I understand it, the next big leaps in astronomy came in the early 1900's with the HR-diagram and the "discovery" of galaxies and more. The view of the universe did not seem to change much between 1800 and 1900. Nowadays every decade has its revolution such as inflation, dark matter, dark energy, exoplanets.

Spectroscopy, optics, photography, Doppler effect, and electricity all developed greatly during the 1800's but don't seem to have brought any revolution to astronomy until later. Is relativity and quantum physics behind this new flourishing? Had the Newtonian physics reached its limit for potentially revolutionizing astronomy already about year 1800? Or was astronomy just out of fashion, maybe because of more profitable employment for the brightest brains in the age of industrialization?

I feel this question is so broad, it might well be the topic of an interesting book on the history of astronomy, should someone be inclined to write it. :)

Anyway, I think a few points could be made briefly.

1. Collecting data

In astronomy, that means observing the cosmos. That means using an instrument of some kind, typically a telescope, and gathering information through it. Telescope performance is dictated by many factors, but the most important one is size (or aperture).

Telescope size grew rapidly through 1600s and 1700s, from Galileo's 1.5 cm refractor in early 1600s, surpassing the 1 meter aperture in early 1800s - Herschel's 40 foot reflector. There was a steady stream of improvements regularly throughout that 200 year period. One could say that the first golden age of the telescope aperture race both culminated and ended with Herschel and his giant telescopes.

Then there was a lull, briefly interrupted by Lord Rosse's 1.83 meter telescope, the Leviathan of Parsonstown, in mid-1800s. Then nothing again.

The aperture race was resumed only in early 1900s, with the 2.5 meter reflector on Mt. Wilson, the Hooker telescope. Afterwards, throughout the 20th century, and now in early 21st, the race is going strong, with the 10.4 meter Gran Canarias segmented reflector currently in the lead, and the 39 meter E-ELT reflector being under construction at Cerro Armazones.

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

2. Interpreting data

The year 1900 marks the boundary between classic physics and the new physics. After that year, relativity and quantum mechanics took off. This is what enabled the new cosmology to emerge in the 20th century.

In other words, with the 1800s science, even with tons of data, there would have been no way to figure out, basically, everything. Supernovas? The expansion of the universe? Dark matter and the rotation of galaxies? This is all based on 20th century physics. 19th century physics would have been clueless.

Astronomy used classic physics to derive interpretations from data pretty quickly, and that process had achieved great success already well into the 1700s. That's when the structure of the solar system was figured out, way back to Kepler in the 1600s. Herschel found Uranus in late 1700s.

There are some exceptions here. Stellar parallax was detected in early 1800s, which enabled an estimate to nearest stars. Spectroscopy showed that distant stars are made of the same elements like the Earth in the 1850s. Around that same time, Neptune was discovered.

So the 1800s was not quite a completely dry period, in terms of theoretical progress.

In any case, a limit was reached anyway in late 1800s, because what was needed was new paradigms in physics to give new life to the interpretation process. That boost occurred after 1900, with relativity and quantum mechanics.

Cosmology is highly dependent on physics (and vice-versa).


How does glass affect taking solar spectra? In this case I'm using a CCD camera mounted telescope pointed at the clouds to take solar spectra and was wondering how the telescope being pointed at the clouds through a double gazing (two layer) glass window would affect the spectra, if at all?

you'll need to find the spectral transmissivity of each piece of glass.  Most likely they all block UV and large sections of the IR bands.  Further,  unless you've removed the internal IR filter, your camera won't record any IR.  CCDs also cut off in the blue, so no UV will be recorded.

BTW, how are you separating the wavelengths at the camera?  Typically one uses a (expensive) grating spectrometer.


What if we throw two solid objects parallel in space? Do those two objects have any chance to collide with each other? What will happen if just throw two solid objects (for example, solid footballs) parallel in outer space and those two objects are floating and going far away in the universe.

So is there any possibility in which those two football-size solid objects can collide with each other, or will they will float parallel for an infinite amount of time in the universe?

I'm considering gravity also, and I find that other stars's and planets's
gravitational pull can pull them into their gravitational fields and change their direction.

But what if those two objects will never get destroyed by luck? Could those will collide each other?

It all depends on the direction you "throw them" and the space-time they find themselves in.

Case 1 - in a perfectly empty universe (other than your footballs) the two would eventually collide. It doesn't matter what initial velocity you give them. The only thing that matters is there separation and their mass. You can just use Newtonian gravity to compute the two footballs' acceleration toward one another -- and hence the time for them to collide.

Such an 'empty' universe as above is called a Minkowsky space-time.

If you put these footballs on a trajectory into our real universe, well we know that space-time is curved by the matter/energy (just energy density in general) that occupies it. Hence a trajectory that starts out parallel will invariably end up "not parallel". The balls will either collide or diverge depending on the geometry of the space-time they find themselves in. The "geometry of the space-time" is (again) completely dependent on the distribution of "stuff" (matter/energy) in the universe as related to the location of the footballs.

In short, the footballs will follow the geometry of the space-time they find themselves in. That geometry (in our real universe) means they will not remain on parallel paths.


Are the planets in our solar system in free fall? If I’m correct, the International Space Station is in free fall, but is moving fast enough to keep it’s position. Assuming that is true, does that mean that some/all planets orbiting the Sun are in free fall of the Sun?

Yes. Free fall means the only force acting on a body is gravity.

The only force acting on the Earth and every other body in orbit is gravity, so all the planets are in free fall.

(There are very small forces from, for example, solar wind but these are minute in comparison to the gravitational forces)


Finding the radius of an eccentric orbit at any point Knowing the apoapsis, periapsis and therefore period of an orbit, how can I find the radius of an object in an orbit at a given angle or time, whichever is needed.

For example, if I have an object m orbiting around object M with an apoapsis of 100m and a periapsis of 50m, what is the distance between m and M after t seconds, or at $\theta$ degrees. Which ever is needed. I say 'or' because I don't know what you would use to find the radius.

If you know the apoapsis and periapsis, you can find the eccentricity, using $r_a = a (1+e), r_p = a (1-e)$ (https://en.wikipedia.org/wiki/Apsis#Mathematical_formulae).

Using the eccentricity, you can find the radius at any angle $\theta$ from the periapsis (this angle is also called the true anomaly).

$r(\theta) = \frac{a(1-e^2)}{1+e.\cos \theta}$(https://en.wikipedia.org/wiki/Kepler_orbit#Johannes_Kepler)

With time, it is little more difficult. The mean anomaly (M) is defined as

$M = \frac{2\pi t}{T}$ where t is the time since last periapsis, and T is the time period.

This mean anomaly relates to true anomaly through a variable called eccentric anomaly (E). From mean anomaly (which you get from the time), you can get eccentric anomaly using

$M=E-e.\sin E$

Now this equation isn't analytically solvable, but you can solve it graphically or numerically, or using a simple iteration which I would demonstrate here (technically included in the numerical methods, but it's easy enough to show here). It is easy enough to be done on a simple scientific calculator.

$E_{n+1}=M+e.\sin E_{n}$ where $E_0=M$

Using this E, you can find the true anomaly $\theta$ using

$\theta=2 \tan^{-1}[ \sqrt{\frac{1+e}{1-e}} \tan \frac {E}{2}]$

Then you can find the radius as described above.


Black hole collision animation: What are these extrusions? In this video uploaded by LIGO Lab Caltech, two inspiraling black holes are depicted. The video's description explains what is shown and can be summarized by:

  The colored surface is the space of our universe, as viewed from a hypothetical, flat, higher-dimensional universe, in which our own universe is embedded.

  ... the colors depict the rate at which time flows.

  [Space] is dragged into motion by the orbital movement of the black holes, and by their gravity and by their spins. This motion of space is depicted by silver arrows...

Just before things become quiescent, regions of space around the merging black holes extrude upward. (Actually, the extrusions seem to begin sometime before things get chaotic.) If "space" here is being represented like the rubber sheet analogy, what do these extrusions mean?

If gravitational forces create depressions in the sheet, then it seems to follow that those extrusions are anti-gravitational, which can't be right. Could they be regions where the equations modeling the black holes' interactions spit out nonsense?

(From video, timestamp 0:52)

The shape of the surface shown in the video is a depiction of the spacial curvature of the spacetime. (The relationship with time are depicted seperately by the arrows and the colors.) More particularly, the shape is depicting the curvature of equatorial plane of the binary. The depicted surface has been embedded in a (fictional) 3D space in such away that the curvature of the surface is equal to the intrinsic curvature of the equatorial plane.

Lets try to unpack what this means for the interpretation of the extrusions. First note, that there is no physical meaning to whether something is show as an extrusion or a depression, this does not actually effect the curvature. The video makers could have also chosen to depict the depressions around the black hole as extrusions instead --- without changing the meaning.

What is relevant, however, is that some regions are shown as depressions while others are shown as extrusions. This means that somewhere in between, there must be a saddle point in the depicted surface. Saddle points correspond to regions with negative (spatial) curvature (i.e. an area where if you would draw a triangle its angles would sum up to less than 180 degrees). The extrusions themselves quite clear have positive curvature (a triangle would have more than 180 degrees).

Note that the sign of the spatial curvature has little to do with whether gravity is attractive or repulsive. If you want an indication of direction in which gravity is working, the arrows give a better idea (although that interpretation should also be taken with a pinch of relativistic salt).

EDIT:
Clarifying the last point a bit. The animation depicts three aspects of the spacetime curvature: The rate at which time flows (the lapse) as a color map, the rate at which space is dragged (the shift) as gray/silver arrows, and the spatial curvature as the curvature of the surface. Together these three completely characterize the curvature of spacetime. Consequently, they dictate how a test object would move in the spacetime, i.e. "how gravity acts". Although all three elements are important for the motion of particles, some give a better qualitative indication of the behavior of test particles than others. In this respect, the color map and the arrows are more important than the spatial curvature. The typically a particle will want to move with the arrows, and along the gradient of color towards the redder regions (in both cases this generally means toward the black holes). The spatial curvature plays a somewhat secondary role, and matters mostly for particles moving at high velocities. Hence my comment that the sign of the spatial curvature is not a good indicator for whether gravity is attractive or repulsive at some point.


Flux density from brightness map This is a naive question. I have a temperature map of radio emission at 1420 MHz, in Kelvin. I want to extract the flux density (in Janskys) from an extended region of this map.
Is the flux density ($S_{\nu}$) from this region equal to
$$ S_{\nu} = \frac{2\nu^2 k_B T_{\text{mean}}}{c^2} \times A$$
where $\nu$ is the frequency of my map, $k_B$ is the Boltzmann constant, $T_{\text{mean}}$ is the mean value of the temperature inside my region, $c$ is the speed of light and $A$ is the area of my region in steradians?

This expression is valid for low frequencies, including the case of the 1420 MHz hydrogen line. It arises from treating the source as a black body with temperature $T(\theta,\phi)$$^{\dagger}$, and assuming $h\nu\ll k_BT$:
$$I_{\nu}=B_{\nu}=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1}\approx\frac{2h\nu^3}{c^2}\frac{1}{h\nu/k_BT}=\frac{2k_BT\nu^2}{c^2}$$
where we have made the approximation
$$e^{h\nu/k_BT}\approx1+\frac{h\nu}{k_BT}$$
and I drop the explicit $(\theta,\phi)$ dependence. The flux density is then, by definition (and assuming the source is small, so $\cos\theta\approx1$),
$$S_{\nu}=\int I_{\nu}\cos\theta\;d\Omega\approx\int I_{\nu}\;d\Omega=\frac{2k_B\nu^2}{c^2}\int T(\theta,\phi)\;d\Omega\tag{1}$$
If we calculate the mean of the function $T(\theta,\phi)$ over the source, we get
$$T_{\text{mean}}=\frac{\int T(\theta,\phi)\;d\Omega}{\int d\Omega}$$
and some algebra and noting that $\int d\Omega=\Omega$ should convince you that we can make the desired substitution. We can essentially always take the mean of a well-behaved function when integrating over a patch of sky.

$^{\dagger}$ The brightness temperature $T_b$ is typically defined as the temperature of a black body which would yield the same specific intensity.


Detecting Deep Impact There have been several accounts in media articles of asteroid near misses, many describe that the asteroid was not seen before the near miss.

After reading the question and answer to How can we tell if the asteroid 2013 TV135 will hit Earth in 2032?, where it was discussed about determining the orbits of known asteroids.

The question here is what is the furthest practical distance that a previously undetected asteroid, capable of an extinction level event, can be observed to have a trajectory that will result in a collision course with Earth?

Something to consider might be a large very dark (low albedo) asteroid.  I would say it is quite possible a large dark asteroid (e.g. P-type) "could" be headed for us, and we would not know it until it hit us (e.g. distance 0).  I think you are asking for the closest distance.  It appears the main method of detecting asteriods is with telescopes, but we are starting to scan the skies in infrared too.  Hopefully infrared will reveal these darker asteriods better.  Sources for further research include:

http://en.wikipedia.org/wiki/P-type_asteroid

http://www.space.com/3126-asteroids-data-sheet.html

http://www.nbcnews.com/science/space/how-we-scan-solar-system-find-dangerous-asteroids-f8C11480423

For the furthest distance, that question is trickier.  It would largely depend on what is considered an asteroid, for example, if we consider Pluto to be an asteroid, we can at least detect very large asteroids out to the distance of Pluto, and we could track its path and see if it were on a collision course with earth.

On average, it is said that we have already detected most of the globally threatening asteroids and it does not appear we are in immediate danger.  Still, I'd say we need to stay on the lookout, and especially look out in infrared to detect any dark asteroids we may have missed :-).

Summary:

Closest distance before large threatening asteriod detection: 0 (e.g. upon impact)
Furthest distance before large threatening asteroid detection: Beyond the orbit of Pluto (and most asteroids tracked are in the asteroid belt between Mars and Jupiter)


What is assembly bias in context of cosmology? I see the term "assembly bias" being used quite a bit in cosmology literature? I have not been able to find a good primer on what it means, and why it is important.

Early analytic calculations of the distribution of the masses of dark matter halos (in which galaxies reside) by Press & Schechter (1974), Bond et al. (1991), Mo & White (1996) and several others showed that the clustering of these halos depended only the redshift and the mass of the halos. However, later work showed that clustering depends on several other parameters, such as the assembly history of the halos (Sheth & Tormen 2004; Gao et al. 2005).

This fact is known as the assembly bias. I think the first to use the term was Croton et al. (2007). Later, the term was "expanded" to also refer to the dependence of the clustering on halo spin, concentration, and shape (ellipticity) of the halos.

You can find some more recent good discussions in Lazeyras et al. (2017) and Padilla et al. (2019).


Sun Zenith angle and location on earth Let suppose you are at +3 degrees latitude (3 degrees north of Equator ... for example at Cali in Colombia). At what periods of the year will the sun be at your vertical (at your zenith) at noon?

I know that at Sun is at equator zenith on 21st March and 21st September (spring and fall equinox). So at +3 degree north of equator will it be before or after 21st march? And before or after 21st September?

The Sun will be at your zenith a few days after 21st March, after it crosses the equator heading northwards* going towards summer, and a few days before 21st September, before it crosses the equator heading southwards* after summer has ended.

*  from an Earth-centric viewpoint. We know, of course, that the Earth is rotating about its axis and revolving about the Sun, and that Earth's axial tilt is what causes the seasons and the apparent motion of the Sun across the sky both through the day and through the seasons.


Are we seeing a conjunction these mornings? I have a doubt about planetary conjunction.

These days, we can see Saturn, Mars, Jupiter and the Moon close to each other.

I mean if Saturn, Mars, Jupiter and the Moon were in conjunction with each other, last February, 19th, 2020.

Is this a conjunction or is it an alignment?

Strictly speaking, a
conjunction
is when two objects appear at the same
right ascension or  ecliptic longitude.
These are not exactly the same but are usually close since the major planets remain near the ecliptic.

When Venus or Mercury is in conjunction with the Sun, we distinguish between inferior (near side) and superior (far side) conjunction.
The outer planets can have another type of alignment, opposition, in which they are 180° away from the Sun around the equator or ecliptic.

When two solar system objects appear only a few degrees apart, it's fair to say they are near conjunction.
If they don't quite reach conjunction, their minimum angular separation is called an appulse.

Mars is in conjunction with Jupiter on March 20 and with Saturn on March 31.
Other planet-planet conjunctions are listed here.
On 2020-12-21 Jupiter and Saturn will appear only 0.1° apart at conjunction and should be interesting to see together in a telescope shortly after sunset.


Do we have to take gravitational lensing into consideration when measuring the size of exoplanets? For the method of waiting for a planet to pass in front of a star. rather than just measuring the light that is blocked do we also have to take into consideration the light that may not be coming to us because of gravitational lensing caused by the planet? Is the mass of a planet large enough to cause gravitational lensing significant enough for us to see an effect?

No, there is no effect here.

Why?

Gravitational lensing magnification works by increasing the observed surface area of the lensed object while preserving the surface brightness. An exoplanet angular diameter is much smaller than that of its host star (even if we can't resolve either). Hence, all the magnification happens across a uniformly lit background, for which the measured increased brightness is zero. In effect, you stretch the image of part of the star into equally bright other parts of the star, resulting in zero magnification - the Einstein radius of the lens is much smaller than the radius of the star.

You might get a negligible effect when the star moves across the edge of the stellar disk.

Why does even that not matter?

The geometry is very disadvantageous. Setting aside the small mass of the planet (which is another factor), gravitational lensing works best when the distance between source and lens is about the same as the distance between lens and observer. This is very much not the case in the scenario we look at here.

Summary

Gravitational lensing has no effect on a background of uniform brightness.
The geometry of the lensing system makes any leftover effects tiny.
The mass of the lens is small to begin with.

Also, there will be other effects that will completely dominate any lensing going on.

Edit to emphasize the size argument

In this image you can see a small source in red, and the resulting lensed image in green. Note that the increase in brightness is purely an effect of the size of the lensed image, not more flux per area. If we assume that the blue circle is the size of the star, we can see that any lensing happening while the planet transits its star will simply stretch parts of a bright uniform background into other parts of the same bright uniform background. Of course, the extent of the lensed image here is greatly exaggerated in order to be able to show any lensing features at all. With a planet in front of a star you would not even get multiple images.


Determining effect of small variable force on planetary perihelion precession Is there an analytical technique for determining the effect of a small variable transverse acceleration upon the rate of aspides precession (strictly not a precession but rotation of the line of aspides) of a planet orbitting around the Sun in a 2D plane according to Newtonian gravity law?

I have modelled such effects in a reiterative computer model and would like to verify those measurements.

The transverse acceleration formula is

  $$At = (K/c^2)*Vr*Vt * Ar.$$

Where:-

  c is speed of light,

  K is a constant of magnitude between 0 and +/-3, such that $K/(c^2) << 1$.

  Ar is the acceleration of the planet towards the Sun due to Newtonian gravitational influence of the Sun, ($Ar = GM/r^2$).

  Vr is radial component of planet velocity relative to the Sun (+ = motion away from the Sun)

  Vt is transverse  component of planet velocity relative to the Sun (+ = direction of planet forward motion along its orbital path). Vectorially Vt = V - Vr where V is the total instantaneous velocity vector of the planet relative to the Sun.

  Assume planet mass is small relative to the Sun

  No other bodies are in the system

  All motions and accelerations are confined to the two-dimensional plane of the orbit.

UPDATE

The reason why this is interesting to me is that a value of K = +3 in my computer model produces anomalous (Non-Newtonian) periapse rotation rate values very close within about 1% of those predicted by General Relativity and within a few percent of those observed by astronomers (Le Verrier, updated by Newcomb).

Formula (Einstein, 1915) for GR-derived periapse rotation (radians per orbit)
from http://en.wikipedia.org/wiki/Apsidal_precession
$$
\boldsymbol{\omega}=24.\pi^3.a^2.T^{-2}.c^{-2}.(1-e^2)^{-1}
$$

UPDATE 4

I have accepted Walter's answer.  Not only did he answer the original question (Is there a technique...?) but also his analysis produces a formula which not only confirms the computer-simulated effects of the transverse acceleration formula (for K = 3) but which also (unexpectedly to me) is essentially equivalent to the Einstein 1915 formula.

from Walter's Summary (in Walter's answer below):-

: (from first order peturbation analysis) semi-major axis and eccentricity are unchanged, but the direction of periapse rotates in the plane of the orbit at rate $$ \omega=\Omega \frac{v_c^2}{c^2} \frac{K}{1-e^2}, $$ where $\Omega$ is the orbital frequency and $v_c=\Omega a$ with $a$ the semi-major axis. Note that (for $K=3$) this agrees with the general relativity (GR) precession rate at order $v_c^2/c^2$ (given by Einstein 1915).

You may want to use perturbation theory. This only gives you an approximate answer, but allows for analytic treatment. Your force is considered a small perturbation to the Keplerian elliptic orbit and the resulting equations of motion are expanded in powers of $K$. For linear perturbation theory, only terms linear in $K$ are retained. This simply leads to integrating the perturbation along the unperturbed original orbit. Writing your force as a vector, the perturbing acceleration is
$$
\boldsymbol{a} = K \frac{GM}{r^2c^2}v_r\boldsymbol{v}_t
$$
with $v_r=\boldsymbol{v}{\cdot}\hat{\boldsymbol{r}}$ the radial velocity
($\boldsymbol{v}\equiv\dot{\boldsymbol{r}}$) and
$\boldsymbol{v}_t=(\boldsymbol{v}-\hat{\boldsymbol{r}}(\boldsymbol{v}{\cdot}\hat{\boldsymbol{r}}))$ the rotational component of velocity (the full velocity minus the radial velocity). Here, the dot above denotes a time derivative and a hat the unit vector.

Now, it depends what you mean with 'effect'. Let's work out the changes of the orbital semimajor axis $a$, eccentricity $e$, and direction of periapse.

To summarise the results below: semi-major axis and eccentricity are unchanged, but the direction of periapse rotates in the plane of the orbit at rate
$$
\omega=\Omega \frac{v_c^2}{c^2}
\frac{K}{1-e^2},
$$
where $\Omega$ is the orbital frequency and $v_c=\Omega a$ with $a$ the semi-major axis. Note that (for $K=3$) this agrees with the general relativity (GR) precession rate at order $v_c^2/c^2$ (given by Einstein 1915 but not mentioned in the original question).

change of semimajor axis

From the relation $a=-GM/2E$ (with $E=\frac{1}{2}\boldsymbol{v}^2-GMr^{-1}$ the orbital energy) we have for the change of $a$ due to an external (non-Keplerian) acceleration
$$
\dot{a}=\frac{2a^2}{GM}\boldsymbol{v}{\cdot}\boldsymbol{a}.
$$
Inserting $\boldsymbol{a}$ (note that $\boldsymbol{v}{\cdot}\boldsymbol{v}_t=h^2/r^2$ with angular momentum vector $\boldsymbol{h}\equiv\boldsymbol{r}\wedge\boldsymbol{v}$), we get
$$
\dot{a}=\frac{2a^2Kh^2}{c^2}\frac{v_r}{r^4}.
$$
Since the orbit average $\langle v_r f(r)\rangle=0$ for any function $f$ (see below), $\langle\dot{a}\rangle=0$.

change of eccentricity

From $\boldsymbol{h}^2=(1-e^2)GMa$, we find
$$
e\dot{e}=-\frac{\boldsymbol{h}{\cdot}\dot{\boldsymbol{h}}}{GMa}+\frac{h^2\dot{a}}{2GMa^2}.
$$
We already know that $\langle\dot{a}\rangle=0$, so only need to consider the first term. Thus,
$$
e\dot{e}=-\frac{(\boldsymbol{r}\wedge\boldsymbol{v}){\cdot}(\boldsymbol{r}\wedge\boldsymbol{a})}{GMa}
         =-\frac{r^2\;\boldsymbol{v}{\cdot}\boldsymbol{a}}{GMa}
         =-\frac{Kh^2}{ac^2}\frac{v_r}{r^2},
$$
where I have used the identity
$(\boldsymbol{a}\wedge\boldsymbol{b}){\cdot}(\boldsymbol{c}\wedge\boldsymbol{d})
=\boldsymbol{a}{\cdot}\boldsymbol{c}\;\boldsymbol{b}{\cdot}\boldsymbol{d}-
 \boldsymbol{a}{\cdot}\boldsymbol{d}\;\boldsymbol{b}{\cdot}\boldsymbol{c}$ and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}_p=0$.
Again $\langle v_r/r^2\rangle=0$ and hence $\langle\dot{e}\rangle=0$.

change of the direction of periapse

The eccentricity vector
$
\boldsymbol{e}\equiv\boldsymbol{v}\wedge\boldsymbol{h}/GM - \hat{\boldsymbol{r}}
$
points (from the centre of gravity) in the direction of periapse, has magnitude $e$, and is conserved under the Keplerian motion (validate all that as an exercise!). From this definition we find its instantaneous change due to external acceleration
$$
\dot{\boldsymbol{e}}=
\frac{\boldsymbol{a}\wedge(\boldsymbol{r}\wedge\boldsymbol{v})
     +\boldsymbol{v}\wedge(\boldsymbol{r}\wedge\boldsymbol{a})}{GM}
     =\frac{2(\boldsymbol{v}{\cdot}\boldsymbol{a})\boldsymbol{r}
       -(\boldsymbol{r}{\cdot}\boldsymbol{v})\boldsymbol{a}}{GM}
     =\frac{2K}{c^2}\frac{h^2v_r\boldsymbol{r}}{r^4}
     -\frac{K}{c^2}\frac{v_r^2\boldsymbol{v}_t}{r}
$$
where I have used the identity
$\boldsymbol{a}\wedge(\boldsymbol{b}\wedge\boldsymbol{c})=(\boldsymbol{a}{\cdot}\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{a}{\cdot}\boldsymbol{b})\boldsymbol{c}$
and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}=0$. The orbit averages of these expression are considered in the appendix below. If we finally put everything together, we get
$
\dot{\boldsymbol{e}}=\boldsymbol{\omega}\wedge\boldsymbol{e}
$
with [corrected again]
$$
\boldsymbol{\omega}=\Omega K \frac{v_c^2}{c^2}
(1-e^2)^{-1}\, \hat{\boldsymbol{h}}.
$$
This is a rotation of periapse in the plane of the orbit with angular frequency $\omega=|\boldsymbol{\omega}|$. In particular $\langle e\dot{e}\rangle=\langle\boldsymbol{e}{\cdot}\dot{\boldsymbol{e}}\rangle=0$ in agreement with our previous finding.

Don't forget that due to our usage of first-order perturbation theory these results are only strictly true in the limit $K(v_c/c)^2\to0$. At second-order perturbation theory, however, both $a$ and/or $e$ may change. In your numerical experiments, you should find that the orbit-averaged changes of $a$ and $e$ are either zero or scale stronger than linear with perturbation amplitude $K$.

disclaimer No guarantee that the algebra is correct. Check it!

Appendix: orbit averages

Orbit averages of $v_rf(r)$ with an abitrary (but integrable) function $f(r)$ can be
directly calculated for any type of periodic orbit. Let $F(r)$ be the antiderivative of $f(r)$, i.e. $F'\!=f$, then the orbit average is:
$$
\langle v_r f(r)\rangle = \frac{1}{T}\int_0^T v_r(t)\,f\!\left(r(t)\right) \mathrm{d}t
= \frac{1}{T} \left[F\left(r(t)\right)\right]_0^T = 0
$$
with $T$ the orbital period.

For the orbit averages required in $\langle\dot{\boldsymbol{e}}\rangle$, we must dig a bit deeper. For a Keplerian elliptic orbit
$$
\boldsymbol{r}=a\left((\cos\eta-e)\hat{\boldsymbol{e}}+\sqrt{1-e^2}\sin\eta\,\hat{\boldsymbol{k}}\right)\qquad\text{and}\qquad
r=a(1-e\cos\eta)
$$
with eccentricity vector $\boldsymbol{e}$ and $\hat{\boldsymbol{k}}\equiv\hat{\boldsymbol{h}}\wedge\hat{\boldsymbol{e}}$ a vector perpendicular to $\boldsymbol{e}$ and $\boldsymbol{h}$. Here, $\eta$ is the eccentric anomaly, which is related to the mean anomaly $\ell$ via
$
\ell=\eta-e\sin\eta,
$
such that $\mathrm{d}\ell=(1-e\cos\eta)\mathrm{d}\eta$ and an orbit average becomes
$$
\langle\cdot\rangle = (2\pi)^{-1}\int_0^{2\pi}\cdot\;\mathrm{d}\ell = (2\pi)^{-1}\int_0^{2\pi}\cdot\;(1-e\cos\eta)\mathrm{d}\eta.
$$
Taking the time derivative (note that $\dot{\ell}=\Omega=\sqrt{GM/a^3}$ the orbital frequency) of $\boldsymbol{r}$, we find for the instantaneous (unperturbed) orbital velocity
$$
\boldsymbol{v}=v_c\frac{\sqrt{1-e^2}\cos\eta\,\hat{\boldsymbol{k}}-\sin\eta\,\hat{\boldsymbol{e}}}{1-e\cos\eta}
$$
where I have introduced $v_c\equiv\Omega a=\sqrt{GM/a}$, the speed of the circular orbit with semimajor axis $a$. From this, we find the radial velocity $v_r=\hat{\boldsymbol{r}}{\cdot}\boldsymbol{v}=v_c e\sin\eta(1-e\cos\eta)^{-1}$
and the rotational velocity
$$
\boldsymbol{v}_t = v_c\frac{\sqrt{1-e^2}(\cos\eta-e)\,\hat{\boldsymbol{k}}-(1-e^2)\sin\eta\,\hat{\boldsymbol{e}}}{(1-e\cos\eta)^2}.
$$

With these, we have [corrected again]
$$
\left\langle \frac{h^2v_r\boldsymbol{r}}{r^4}\right\rangle =
\Omega v_c^2\,\hat{\boldsymbol{k}}\,
\frac{e(1-e^2)^{3/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta}{(1-e\cos\eta)^4}\mathrm{d}\eta
=\frac{\Omega v_c^2e}{2(1-e^2)}\hat{\boldsymbol{k}}
\\
\left\langle \frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle = \Omega v_c^2\,
\hat{\boldsymbol{k}}\,
\frac{e^2(1-e^2)^{1/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta(\cos\eta-e)}{(1-e\cos\eta)^4}\mathrm{d}\eta=0,
$$
in particular, the components in direction $\hat{\boldsymbol{e}}$ average to zero. Thus [corrected again]
$$\left\langle 2\frac{h^2v_r\boldsymbol{r}}{r^4}-\frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle
=\frac{\Omega v_c^2e\,\hat{\boldsymbol{k}}}{(1-e^2)}
$$


What are the differences between matter, dark matter and antimatter? I thought dark and anti matter were kinda the same, but after saw a video, they mention that dark matter is not antimatter but their explanation is a little fast so I got doubts.

What are the differences between matter, dark matter and antimatter? Are they related? How they interact each other? Where can I see an example of this interaction? and also, in terms of percentage, how much of matter, dark matter and antimatter exist in the universe?

Thanks!

Matter is the stuff you are made of.

Antimatter is the same as matter in every way, looks the same, behaves the same, except its particles have electrical charges opposite to matter. E.g., our electrons are negatively charged, whereas a positron (an antimatter "electron") is positively charged. The positron is the "anti-particle" of the electron.

When a particle meets its anti-particle, they "annihilate": the two particles disappear, and gamma photons are released carrying off their energy. For this reason, should a lump of matter touch a lump of antimatter, they would annihilate, and a giant explosion would result because of the huge energy released (E=mc^2).

Matter and antimatter are definitely related: same thing, but with opposite signs. Twins, but opposites.

It is not clear why, but it seems like there isn't that much antimatter out there, more like trace amounts. Definitely not as much as regular matter as far as we can tell. This is puzzling to physicists and cosmologists, because you'd expect the Big Bang to make roughly equal amounts of matter and antimatter. Scientists agree that the paradox of "excess matter" will advance physics even further once it's solved.

Dark matter - we don't really know what it is. It's not even sure it's "matter" in a conventional sense, or related to it in any way. We just know that galaxies are rotating in such a way that indicates there's a lot more mass out there, but it is mass that we cannot see and cannot be accounted for in the usual ways. Hence the name "dark" (as in invisible) matter.

Dark matter doesn't seem to interact much with regular matter, except gravitationally. Right now dark matter could be passing through you and you wouldn't notice. Dark matter also does not interact with light, so you can't see it. It doesn't seem to interact much with itself either, so for this reason dark matter cannot form "clumps" such as planets or stars. Instead, it probably exists in a diffuse form. Bottom line, dark matter interacts pretty much only via gravity.

The shape of galaxies is a proof of the existence of dark matter, and is a result of the interaction between matter and dark matter. Without dark matter, galaxies would be much less massive, and the outer parts would rotate much more slowly compared to the center. Due to dark matter, galaxies are quite massive, and they rotate almost as solid objects - the outer parts rotate approximately as fast as the central parts.

Estimates vary, but it seems like there's something like 5x to 6x more dark matter out there compared to regular matter.


What is the last day a waning moon is visible in daytime (the afternoon)? From the fourth quarter onward, can we see the moon in the afternoon sky?

At third quarter the moon rises at about midnight and sets at about midday The exact rising times depends on the position of the moon relative to the ecliptic and the season.

So in spring, at third quarter, the moon is low and sets quite early (around 10:00 in my location)  In Autumn the moon is high at third quarter, and sets at about 15:00 (though daylight saving time affects this)

As the month progresses the moon rises later and sets later. The moon is above the horizon in the afternoon, but visibility depends on many factors, and a very thin cresent, close to the sun, may not be visible with the naked eye. There is some analysis of this with respect to the appearance of the new moon, (which has calendrical significance to some religions) There are too many uncertainties in determining visibility to give a simple answer, much depends on the observer.


When was it first determined that the Sun is a star? Just looking at the sky, it is not at all obvious that the Sun is a star: stars are fixed on the celestial sphere, they are point-like and not very bright, whereas the sun is a big (compared to a star) sphere that moves through the sky, making plenty of light. To know that they are actually the same kind of object, I guess you would first have to figure out that stars are much further out.
When was it first hypothesized that the Sun and the stars might have the same nature? What was the first element of observational evidence used to backup that idea?

I'm going to be vague on the timing because it depends on what you mean by "determine". Which in turn depends on what is meant by "know" in science. There is a very good model for stars which predicts that balls of plasma that are prevented from collapse by fusion in their cores will look like the sun up close and will look like stars from a distance. So people believe that both the sun and the stars have the same nature. However, while there is direct evidence of nuclear fusion in the sun (neutrinos) there isn't the same evidence of fusion in stars.  So it can be argued that this hasn't been "determined", and given the nature of scientific discovery, might never be.
It also depends on what you mean by "the same nature". What if we suppose that two things have the same nature, but are completely wrong about what the nature of the object actually is.  We didn't have any real understanding of the core of stars until the 1930s, and we are still filling in the details.
The early claims that the sun is a star were based on general evidence and some incorrect, or at least odd beliefs.  Later discoveries showed how starlike the sun actually is but there was no Eureka momement; there is no single event that you can point to in which someone discovered this.
Anaxagoras in the 400s BC suggested this or something like this.  He suggested that the sun was a ball of molten metal, and the stars were fiery stones, so something of "the same nature"  He proposed that the sun and the stars were torn from the Earth, and then ignited.  He probably supposed that meteors were "falling stars" and might have linked meteorites to meteors.  He almost certainly believed that the sun was bigger than the stars. So "of the same nature" but the sun was the greatest of the stars. He is significant because he is probably the first to give a physical theory for the nature of an astronomical body.  He made the first steps to bringing the study of the heavens within natural philosophy, or "physics".
Anaxagoras was the first to hypothesise that the Sun and the stars might have the same nature.
The notion that the sun is not special is suggested by Copernicus, and there were mutterings about it in the period after Copernicus' publication; it is implicit in some of Galileo's work. However, it was Giordano Bruno who gave full weight to this idea. His evidence seems merely to have been that, if the sun was very distant, it would appear as a star. The observation was naked eye viewing of the sky. He does not seem to have estimated how distant the stars would be, as he had no way of measuring the actual brightness of stars of the sun, but he most clearly declared that:

“The composition of our own star and world is the same as that of as many other stars and worlds as we can see.”

(And by "our own star" he means "the Sun")
It wasn't until the middle of the 18th century that any actual evidence for stars being of the same nature as the sun was found. Angelo Secchi used spectroscopy to analyse starlight, and found that it was very similar to sunlight. By this time though Secchi was confirming the orthodox position rather than proposing a new discovery.
With the measurement of stellar parallax, another piece of the puzzle settled into place, now that the distance to stars could be measured, the were found to comparable to the sun in brightness as well as in spectrum. And developments in nuclear physics in the '30s-'50s showed how stars and the sun could be powered by similar nuclear engines in their cores.
So this question doesn't afford a simple single answer.  This wasn't a sudden paradigm shift. Rather it was a gradual process.  However, if you need just one name to hang on this idea then Bruno would seem to be the first to clearly assert that stars have exactly the same nature as the sun.
(Much taken from http://www.astronomytrek.com/who-discovered-the-sun-is-a-star/)


Pulsars: How do astronomers measure minute changes in period (~picoseconds per year)? I've been to some talks that mention how stable the period of a millisecond pulsar is over long periods of time. Recently, it was mentioned that astronomers have calculated the change in period over time to be less than 10^-12 seconds per year for several pulsars. No one I've talked to seems to know any details of this calculation.
How do we calculate such small differences in period? How much data must be collected and what are the exposure times for imaging such rapid phenomena?
A source/paper would be excellent. I apologize that I don't have a citation for the 10^-12s figure, but the lack of citation is mostly my reason for posting this question.

Let us suppose that the pulsar is spinning down at a uniform rate. So it has a period $P$ and a rate of change of period $dP/dt$ that is positive and constant (in practice there are also second, third, fourth etc. derivatives to worry about, but this doesn't change the principle of my answer).

Now let's assume you can measure the period very accurately - say you look at the pulsar today and measure its radio signals for a few hours, do a Fourier transform of the signal and get a nice big peak with a period of 0.1 seconds (for example).

With that period, you can "fold" the data to create an average pulse profile. This pulse profile can then be cross-correlated with subsequent measurements of the pulse to determine an offset between the predicted time of "phase zero" in the profile, calculated using the 0.1 s period, and the actual time of phase zero. This is often called an "O-C" curve or a residuals curve.

If you have the correct period and $dP/dt=0$, then the residuals will scatter randomly around zero with no trend as you perform later and later observations (see plot (a) from Lorimer & Kramer 2005, The Handbook of Pulsar Astronomy). If the initial period was in error, then the residuals would immediately begin to depart from zero on a linear trend.

If however, you have the period correct, but $dP/dt$ is positive, then the residuals curve will be in the form of a parabola (see plot (b)).

If you have second, third etc. derivatives in the period, then this will affect the shape of the residuals curve correspondingly.

The residuals curve is modelled to estimate the size of the derivatives of $P$. The reason that $dP/dt$ can be measured so precisely is that pulsars spin fast and have repeatable pulse shapes, so changes in the phase of the pulse quickly become apparent and can be tracked over many years.

Mathematically it works something like this. The phase $\phi(t)$ is given by
$$\phi(t) \simeq \phi_0 + 2\pi \frac{\Delta t}{nP} - \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$
where $\phi_0$ is an arbitrary phase zero, $\Delta t$ is the time between the first and last observation and $n$ is the integer number of full turns the pulsar has made during that time. If the period is approximately correct, then $n = int(\Delta t/P)$.

The "residual curve" would be given by
$$\phi_0 - 2\pi\frac{\Delta t}{nP} -\phi(t) \simeq  \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$

For example, if the period of a $P \sim 0.01$ second pulsar changed by a picosecond in a year, then there would be an accumulated residual of almost $10^{-4}$ seconds after 1 year of observation. Depending on how "sharp" the pulse is, then this shift of about 1% in the phase of the pulse might be detectable.

Perhaps needless to say, but there are a host of small effects and corrections to make in order to get this very high precision timing. You need to know exactly how the Earth is moving in its orbit. The proper motion of the pulsar on the sky also has an effect. These and more can be found in Lorimer and Kramer's book, but there is also a summary here.


If a person were to die on the Moon or Mars, would the body decompose? There wouldn't be enough oxygen for any bacteria to decompose the body, right? Not to mention, the radiation of space might kill off most organisms on it. So would it decompose, given millions of years?

Space, as Randall notes, is really dry. Mars, (recent discoveries notwithstanding) is not much moister.

In these conditions, bodies mummify.

The microbes that live in you, wouldn't survive the freezing, dessication and radiation. There is no real upper limit on how long a mummified body could exist in space. There would be nothing to cause the body to change, and so it would remain. There would be a slow breaking down of surface proteins, due to UV light, and eventually micrometeorites would erode the body, but these processes would take many millions of years.


What is spacetime 'made' of? General relativity is often explained as saying spacetime is curved by gravity, what does this mean? How could we perceive a curve in spacetime when there is no external "straight" reference frame for instance?

Space-time is not "made" of anything, it is merely a medium or coordinate system. Think about the grid lines on a map, they aren't "made" of anything, they're just a representation of the geometry of the Earth. Space-time is a concept envisaged by Einstein when he wrote his theory of Special Relativity that the properties of space and time become intrinsically linked at relativistic speeds. One of the consequences of this is that you cannot go forward in space without going forward in time, this can be visualized in a light cone:

This is by no means a representation of what space-time looks like, only how it behaves in 2 dimensions of space and one of time. Gravitational waves are caused when two massive objects (binary stars) orbit each other, and they cause "ripples" in space-time. Again another 2D visualization is needed:

This shows how the gravitational potentials of two stars interact in the fabric of space-time as they orbit, causing the radiation of gravitational waves.

Both of these are only visual guides, we will never be able to "see" gravitational waves or space-time itself with our own eyes, that is why we need detectors such as LIGO or VIRGO to infer their existence.


Where can I learn these Astrophysical techniques? [Read below] Whenever I read any Astro papers in http://arxiv.org/ they usually talk about Data fitting or modelling or sentences like 'we fitted the spectral region by analytic functions  i.e. by multi-component Gaussian model plus continuum'.

Is there any websites, online books or maybe YouTube videos which teaches these techniques in detail?

Any info will be much appreciated!

If you are going to do data analysis, you need to understand how the fitting procedure works. This means a lot of statistics, and new terminology and so on, which is hard and takes much time.
If you want to start to fit, e.g., a spectrum, you should definitely read the XSPEC guide. You can find a pdf online as well.
In short, fitting is to take a model and measure how well this model fits to your data. To quantify this "how well", usually the chi-squared distribution is considered:
$\chi^2 =  \sum\limits_{i=1}^n(\frac{X_i - \mu_i}{\sigma_i})^2$
where $X$ is your data (observed value), $\mu$ is your expected value (which, in this case, corresponds with the prediction of the model), and $\sigma$ is the variance on your data point (the error).
This is the most general information that you need.
You can read more here, and especially on the Numerical Recipes (don't be scared by the huge format of the last document, the pages you need are only few around Chapter 15).
For fun, you can just try to play with XSPEC, to take confidence, and see at least how things change when you change your model or your data (especially if you know which model is the best-fit for your data). To understand everything will take years, but if you never start, you'll never arrive ;)


What I can observe with a 80/900 refractor telescope? What can I study? Recently I bought a telescope after a very good conversation with an astronomer friend of mine. The main goal is to see planets.

However, Mars is so difficult and is far to find it in opposition. I really like the views of the Moon and there is where I spend most of my time observing. But I really want to go view the planets, but I seem a bit without an objective. I don't want only to see them at random; I want to do some kind of study using my observations. So, can you suggest something to do with my telescope to start some serious study?

I have a good base of Physics (I'm a computer engineer) and I really like Math, but I'm a complete amateur in astronomy.

There is a wonderful intersection of Physics, Mathematics, Applied Math (numerical analysis) and Observational Astronomy in a systematic analysis of the movement of four bright Galilean moons of Jupiter!

They are easy to see in a small telescope, they have well defined orbits around Jupiter, and they have some trick up their sleeves that you can analyze.

Here's an exercise you can do right now. When Jupiter is anywhere near quadrature, it's shadow extends behind it by an angle of almost 11 degrees, so the moons will often be eclipsed. That means instead of appearing or disappearing into Jupiter's overwhelming brightness, they can "pop" in or out of visibility as they pass out of or into Jupiter's shadow.

                a (km)
Io             421,800
Europa         671,100
Ganymede     1,070,400
Castillo     1,882,700

Jupiter radius  71,492

You can predict when these happen, and then watch to see them and confirm your calculations. You can start timing them to get more accurate information.

From a mathematical perspective you can check the periods of their orbits (from your measurements, or from a table) to see if there are any surprising numerical relationships!

You can check the inclination of their orbital plane to see if the eclipses always happen or sometimes don't.

You can read the answers to When will the next series of mutual eclipses of Jupiter's moons begin? and start planning for 2021 when these begin again!

You can learn to do numerical integration of solar system bodies to simulate their orbits, or use the Python package Skyfield or the JPL Horizons website for high accuracy positional information.

These can also help you calculate the eclipse times by Jupiter, and the mutual eclipses of Jupiter's moons.

Finally, you can try to measure the speed of light by seeing that the eclipses come sooner or later than average when Jupiter is closer or farther than average from the Earth. This is one of the early ways that the finite speed of light was verified.


How can a supernova cause a gravitational wave? With the recent results of LIGO, I read a lot how limited the detection capabilities where and how one could only expect to measure the most extreme events like black hole mergers and supernovae.

I get the black hole mergers, since here we see an ever higher frequency gravitational interference of two large bodies. But how can a supernova cause a gravitational wave? All the mass is still there, it is just thrown out  isn't it? So from far away shouldn't the mass still be concentrated on a relatively small place and thus the gravity from the recently exploded star should remain unchanged, right?

The key here is that to produce gravitational waves, the supernova explosion must not have perfect spherical symmetry. Technically, what you need is an "accelerating" gravitational quadrupole moment.

It is quite unlikely that supernovae explosions will be symmetric. Asymmetries can be produced by binary companions, magnetic fields, rotation...

The characteristics of a "burst" source of gravitational waves will be quite unlike the signature of coalescing binary black holes. The magnitude of the signal is expected to follow the collapsing core of the star and reach a peak at something called the "core bounce", where the equation of state of the gas "stiffens" and a shock wave propagates outwards. The signal is thus likely to gradually increase in amplitude for some tens of milli-seconds, followed by an abrupt spike at the bounce, followed by a "ring-down" as the core (a proto-neutron star) settles down towards a spherically symmetric configuration over the course of a further 10 milli-seconds or so.

The frequencies of the waves will have a broad range governed by the characteristic frequency implied by $(G\rho)^{1/2}$, which is the inverse of the freefall timescale, where $\rho$ is the density at the time when the GWs are emitted. For a typical core collapse, the densities could range from $10^{14}$ to a few $10^{17}$ kg/m$^3$ over the course of the event, producing GW frequencies of tens to thousands of Hz - exactly in the range at which aLIGO is sensitive.

Because GWs pass through the envelope of the star unimpeded then the GW signature of a supernova probes the very heart of a supernova explosion and should be seen some hours before the visible signature is apparent. On the other hand any neutrino emission from a core collapse supernova should be detected at a very similar time to the GW waves. Any delay here could fix the currently poorly known neutrino mass.


Other than liquid water what factors could make way for life to be on Europa? I know that Europa has liquid water, but I just don't know if there is anything else on it that could give way for life.

Well, Europa is one of the closest moons to Jupiter. Jupiter is an extremely massive planet (317.8 the mass of Earth) which makes strong tides on Europa and other close moons, so Europa is torn a bit which results in so-called tidal heating underwater. Through this heating, the temperature underwater makes the subsurface ocean habitable despite the Jovian system not being in the habitable zone. There are so-called hot spots underwater for which plumes erupting from under the surface are responsible. Life could thrive around those plumes.

So in other words, there is ideal temperature for life under Europa's surface due to tidal heating by Jupiter. On Ganymede and Callisto there may be a subsurface ocean too (though that's not confirmed) but they are farther away from Jupiter so there maybe isn't enough tidal heating, contrary to Europa.