orbit - What is meant by matter distribution?

I assume you are referring to the rotation curve of galaxies, i.e. the speed of matter (stars, gas clouds, etc.) as a function of its distance from the center. This speed will depend on the matter distribution, that is, how the galaxies' constituents are distributed throughout the galaxies.

Keplerian rotation

If (almost) all of the mass $M$ were located in the center, then a star at a distance $R$ would have a speed $V = sqrt{GM/R}$. This is called Keplerian rotation, because this is how planets in the Solar system move, which were described by Johannes Kepler. In this case, the mass inside a radius $R$ is (almost) constant, i.e. $M'(<R) = M$.

Flat rotation

Most disk galaxies have a large concentration of visible mass in the center (the "bulge"), so one might think that Keplerian rotation would be a good description of the stars' speeds. But this is not what is observed. In fact, the speed is seen to rise steeply the first few kpc, and then stay constant out to distances where no light is detected. This will happen if the mass $M'(<R)$ inside $R$ increases with $R$, i.e. if the matter is distributed throughout the galaxy. This is one of several pieces of evidence that 1) there exists some matter that we cannot see, in addition to what we can see, 2) that this matter dominates the kinematics of the galaxies, and 3) that it resides in a halo that extends beyond the visible matter. We call it dark matter.

Solid body rotation

If in addition to a given matter distribution the distances between the individual parts are fixed, the object we are discussing would describe a solid body. Of course this can't be the case for a galaxy, but it will be the case for a rocky planet. In this case the velocity of a given point will depend linearly on its perpendicular distance to the axis of rotation, i.e. $V propto R$.

Density profile of the matter distribution

Excactly how the matter is distributed in the galaxy is not easily deducted, and varies from galaxy type to galaxy type. An early description of the matter density as a function of distance from the center was the so-called NFW profile, but other models may be better, e.g. a Sérsic profile or an Einasto profile.

So, to summarize, the term "matter distribution" refers to density as a function of location, i.e. $rho(mathbf{r})$, for some object. This can be for any object — solid, liquid, gaseous, or consisting of several components that are held together by gravity — but in any case it is true that its form determines its rotation, and the three different rotation forms you mention are often discussed in the context of galaxies.

---

The three different cases are illustratred below. Keplerian, solid, and flat rotation are drawn in green, red, and blue, respectively. For comparison, a typical spiral galaxy velocity curve is drawn in magenta.

solar system - Do the planets, asteroid belt, kuiper belt, and scattered disc lie on the same plane?

They are not exactly in the same plane, but roughly are.

For example this graph representing the centaurs, the Kuipier belt and scattered disc objects show that most objects are around the same plane, with few objects with more than 30 degrees of inclination:

This is the inclination of the asteroids:

Again, few have more than 30 degrees of inclination.

The planets are also all roughly in the same plane. Lets take the ecliptic as a reference (which is almost the same thing as Earth orbital's plane):

• The Sun's equator has an inclination of 7.25 degrees.


• Mercury has an inclination of 7.00 degrees.


• Venus has an inclination of 3.39 degrees.


• Mars has an inclination of 1.85 degrees.


• Vesta has an inclination of 7.13 degrees.


• Juno has an inclination of 12.98 degrees.


• Ceres has an inclination of 10.59 degrees.


• Pallas has an inclination of 34.84 degrees.


• Interamnia has an inclination of 17.29 degrees.


• Hygiea has an inclination of 3.84 degrees.


• Jupiter has an inclination of 1.30 degrees.


• Saturn has an inclination of 2.48 degrees.


• Uranus has an inclination of 0.77 degrees.


• Neptune has an inclination of 1.76 degrees.


• Orcus has an inclination of 20.57 degrees.


• Pluto has an inclination of 17.15 degrees.


• Ixion has an inclination of 19.58 degrees.


• Salacia has an inlination of 23.94 degrees.


• Varuna has an inclination of 17.2 degrees.


• Haumea has an inclination of 28.19 degrees.


• Quaoar has an inlination of 7.99 degrees.


• 1992 QB₁ has an inclination of 2.19 degrees.


• Makemake has an inclination of 29.00 degrees.


• Eris has an inclination of 44.04 degrees.


• Sedna has an inclination of 11.92 degrees.

Most of those bodies have a low inclination to the ecliptic, but Pallas is an interesting exception. I don't know why Pallas has such a high inclination but I would bet on a Kozai mechanism due to Jupiter's influence.

Plutinos and Kuipier belt objects are influenced by Neptune, and part by the Kozai mechanism.

Eris is a special and curious case, it is too far from Neptune, although there is still some weak gravitational Neptune's influence there, but I don't know why its inclination is so high. This contrasts with Sedna, which have a more reasonably lower inclination.

Further, the Oort cloud is expected to be full of bodies of essentially unrestricted inclinations and possibly very excentric orbits, as a result of gravitational interactions with passing stars on the past. By the way, a passing star in a distant past might be an explanation for Eris orbit and inclination.

_{Note: All of the data here, as well as both of the images, I found on wikipedia.}

black hole - Linear extent of an accretion disk

I think the outer edge of an accretion disk is not well-defined, and observationally the radius will depend on which wavelength you consider, since the farther you get from the BH, the softer the radiation will be. But if you look in the UV, then Morgan et al. (2010) find the following relation between $R_{2500}$ (the radius when observed at $lambda = 2500$ Å) and the mass $M_mathrm{BH}$ of the black hole:
$$
logleft( frac{R_{2500}}{mathrm{cm}}right) = 15.78 + 0.80
logleft( frac{M_mathrm{BH}}{10^9M_odot}right),
$$
(modulo some uncertainties that you can look up in the paper).

That is, if your BH has a mass of $10^8 M_odot$, its radius will be $R_{2500}sim64,mathrm{AU}$, or roughly 1/3 lightdays.

For comparison, its Schwarzschild radius is $sim2,mathrm{AU}$, so your estimate was actually pretty good.

This result is consistent with Edelson et al. (2015), who find 0.35 lightdays, also in the UV. However, in you look in longer wavelengths, the disk is much, much larger. If you're interested beyond your homework assignment, take a look at the accretion disk theory review by Armijo (2013), who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.

telescope - How much magnification is needed to see planets of solar system?

You're probably asking the wrong question - which I am going to answer anyway, and after that I am going to answer the question you should have asked instead.

---

As a general rule, there isn't much point in pushing the magnification above 2x the diameter of the instrument, measured in mm. 3 inch, that's 75mm, that's 150x max. Beyond that limit, even under ideal skies the image is large but blurry.

After that, seeing (or air turbulence) pushes that limit further down. Your aperture is small enough that it almost never suffers from seeing, but larger instruments are often affected. It varies greatly with time, place and season. There are times when a 12" dobsonian, that in theory could do 600x, is clamped down by seeing to 150 ... 180x. There are times when you could take a 20" dobsonian all the way up to 1000x - but that's very, VERY rare, it's the stuff of legends.

Assuming average seeing conditions and instruments of usual size (refractors of 3...4" aperture, reflectors 6" or larger), here are some rules of thumb:

Jupiter is seen best under mid-high magnification. It's rare that more than 200x is beneficial. This is because it's a very low contrast object, and additional magnification comes at the cost of less contrast, which makes things worse.

Saturn works best at high-ish magnification, bit more than Jupiter but maybe not much more. Around 200 ... 250x usually works. It depends on what you do - if you're trying to see the ring divisions, push it a bit higher.

Mars can use the highest magnification that you could generate, given the instrument and the conditions. It's a very small object, contrast is not bad, so crank it all the way up. Most instruments are limited by seeing when observing Mars.

Moon is the same as Mars.

As you can see, magnification is never an issue for you. More magnification will not make it better. In fact, more magnification always means the image is more blurry, not more crisp - it's always a compromise between size and blurriness that decides the optimal magnification.

Don't worry, everyone begins thinking that more is always better. Soon enough, experience shows them what's really going on.

---

That being said, I believe it's not magnification that's giving you trouble, but the general condition of the optical stack that you're using. These are things that are extremely important, and yet are ignored by many, many amateurs - and the results are not optimal. Here are a few things that you should investigate:

Collimation

Is your scope collimated? In other words, are all optical elements aligned on the same axis? The likely answer is no. It makes a huge difference in the scope's performance, especially for planets. Here's a collimated scope, compared to the same scope out of collimation:

Further information on Thierry Legault's site, which is extremely informative.

A series of articles and documents regarding collimation:

http://www.cloudynights.com/documents/primer.pdf

http://www.garyseronik.com/?q=node/169

http://www.garyseronik.com/?q=node/165

http://www.garyseronik.com/?q=node/238

Thermal equilibrium

At 3" aperture, this is probably not a big issue, but there's no reason why you should add another problem to the existing ones. Your scope should be at the same temperature as the air around it, otherwise its performance decreases. Take it outside 1 hour before you start observing, and that should be enough for you.

Larger telescopes (around 6" ... 8" and larger) should use active ventilation for better cooling (a fan on the back of the mirror). More details here:

http://www.garyseronik.com/?q=node/55

http://www.garyseronik.com/?q=node/69

In your case, simple passive cooling for 1 hour should be enough, but it's worth reading those articles.

Focal ratio

A 3" scope, at 300mm focal length, that's an f/4 instrument. That's a pretty steep f/ ratio. Most eyepieces will not do well with such a blunt cone of light, and will start to exhibit aberrations that blur the image. Only very expensive eyepieces work well at such low focal ratios - things like TeleVue Ethos, or Explore Scientific 82 degree eyepieces.

Try and keep the planet in the center - most aberrations are lower there. Even very simple eyepieces do better in the middle of the image.

Look at the stars. Are they tiny and round in the center, and large and fuzzy at the edge? Those are aberrations from various sources (eyepiece, primary mirror, etc).

Coma

Of course, at f/4 even the best eyepieces out there cannot do anything about coma - an aberration coming out of any parabolic mirror, which becomes pretty obvious around f/5, very obvious at f/4, and a major problem at f/3. Again, coma is zero in the center of the image, and increases towards the edge.

A coma corrector is used in some cases, such as the TeleVue Paracorr, but I strongly recommend that you DO NOT use one - I suspect your instrument is aberrating in ways that overwhelm coma anyway. Jupiter would not be too blurry even at full f/4 coma at the edge. This paragraph is for informational purposes only.

Optics quality

An f/4 parabola is not super easy to make at any size. I've made my own optics, and the lower the f/ ratio, the more difficult the process is. Many small, cheap telescopes are made in a hurry, and the difficult focal ratio poses additional problems - as a result, many manufacturers do a poor job. There are even cases where the primary mirror is left spherical, with disastrous results.

This is something you can do nothing about. If the primary mirror is bad, then that's just the way things are. An optician might try to correct it, but it's a difficult process, and quite expensive. I only added this here so you are informed.

---

This is what I would do in your case:

I would take the scope out 1 hour before observing, every time.

I would try and learn how to collimate the scope. I would try to figure out a few simple collimation techniques, and a few simple tests. I would spend a few days / weeks practicing that. I would keep reading about collimation.

When collimation is at least partially under control, I would learn how to properly focus the scope. Seems simple, but it can be tricky. Use a bright star, and try and make it as small as possible. Use the Moon when it's visible, and try and make it crisp and clear. Do not try this with a miscollimated scope, since it's pointless.

After a few months, when I gain confidence that the scope is in better shape, very well collimated, very well focused, I might try to borrow a better eyepiece from a friend. I said borrow, not buy. Something like a 3 ... 4mm eyepiece, good quality, that would give me a comparison for the existing eyepieces. This ONLY makes sense with a scope that is in perfect collimation, perfect temperature, perfect focus. If an improvement is seen, then get a better eyepiece - but do not spend hundreds of dollars for an expensive eyepiece that will then be used in a tiny cheap scope. Second-hand eyepieces often work exactly as well as new ones.

If you know someone in your area who makes mirrors, see if they agree to put your primary mirror on the Foucault tester, and assess its condition. But beware: the results might be very disappointing. Or not. You kind of never know with these little scopes.

EDIT: After the scope is collimated and so on, you could try to increase magnification by using a 2x barlow with your eyepieces, but do not expect miracles - the image will be bigger, but probably rather "mushy". More magnification is not always better, there's always a trade-off.

Good luck, and clear skies to you!

Did atoms in human body indeed come from stars?

The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.

So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.

After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that's all. No oxygen, iron, carbon etc.

Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies must have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).

But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.

If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the s-process. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with $A<90$) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).

Iron, nickel and many other elements such as sulphur and silicon are also produced during type Ia supernovae. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium.

All these different processes produce distinctive patterns of element abundances.

The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.

Analysis of "presolar grains" found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.

[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my Physics SE answer to this question. ]

Orbital eccentricity variation of the other planets?

On Earth, it's fairly well published, mostly in climate change related articles, that the Earth's orbital eccentricity operates on a 413,000 year cycle with roughly 90,000 - 125,000 year variation within that cycle.

Source

The cause of this variation is primarily the other planets, Jupiter and Saturn usually mentioned as the primary causes (same link).

Question is two fold. How stable is the Earth's orbital eccentricity cycle. 413,000 years sounds enormously precise, but logically, I would think small changes in planetary orbits would create some variability. Is the 413,000 years well established and repeating or is it more uncertain?

and, do we have a good estimate of other planets orbital eccentricity variation? I looked, but couldn't anything at all on other planets eccentricity cycles. The closest I came was this gravity simulation chart below.

90,000 year eccentricity chart of the 4 inner planets here

and method used. Source.

stellar evolution - Redshift to calculate age of stars

The redshift that is referred to is not a Doppler redshift, but a cosmological redshift. The difference is that the former is caused by the source moving through space, while the latter is caused by the "stretching" of the wavelength of the light as it travels through space.

The cosmological redshift happens gradually along its journey, and hence it is a measure of the distance to the source. But since traveling through space takes time, it is also a measure of the lookback time to the source — that is, the time that has passed since its emission. This notion of the term can thus be used as a timeline for phenomena in the Universe, and perhaps somewhat confusingly, it is sometimes used this way even when referring to local phenomena.

For instance, Earth was formed 4.54 billion years ("Gyr") ago. If some unrelated galaxy emitted light at the same time, and if that light reaches us today, then that galaxy must be at a particular distance$^dagger$. During its journey it has been redshifted to $z simeq 0.42$. Thus we may say that Earth was formed at redshift $0.42$.

Similarly, stars (in the Milky Way or elsewhere) that are said to have formed at redshift $z=10$ and $z=15$, can also be said to have formed 13.3 Gyr and 13.5 Gyr ago, respectively.

In the figure below, I plotted the relation between cosmological redshift and lookback time (assuming a Planck 2015 cosmology). Now hug each other, Rob and NotSoSN.

---

$^dagger$_{5.4 billion lightyears, actually. Not just 4.54 billion lightyears, because space has expanded in the meantime.}

the sun - why are solar telescopes built on lakes? the site differnece between a solar and an optical telescope

To amplify andy256's comment, the problem that solar telescopes face is that heating of the surrounding ground during the day gives rise to turbulence in the air near the ground, making the observing conditions worse (think of the heat shimmer just above the surface of hot pavement or a hot road -- that's turbulence bad enough for your naked eyes to notice). Since water has a high thermal inertia, it doesn't heat up as much as land does, so you get less turbulence. (Another approach is to put the telescope on top of a tower, above the worst of the turbulence.)

Note that Big Bear is at an altitude of 2,000 meters, so it is at a moderately high altitude.

coordinate - Need Simple equation for Rise, Transit, and Set time

I'm not sure it qualifies as "simple", but, using
http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro (and some
additional calculations/simplifications):

$
begin{array}{|c|c|c|c|}
hline
text{Event} & text{Time} & phi & Z \
hline
text{Any} & text{t} & tan ^{-1}(cos (lambda ) sin (delta )-cos
(delta ) cos (alpha -t) sin (lambda ),cos (delta ) sin (alpha
-t)) & tan ^{-1}left(sqrt{(cos (lambda ) sin (delta )-cos (delta
) cos (alpha -t) sin (lambda ))^2+cos ^2(delta ) sin ^2(alpha
-t)},cos (delta ) cos (lambda ) cos (alpha -t)+sin (delta ) sin
(lambda )right) \
hline
text{Rise} & alpha -cos ^{-1}(-tan (delta ) tan (lambda )) & tan
^{-1}left(sec (lambda ) sin (delta ),cos (delta ) sqrt{1-tan
^2(delta ) tan ^2(lambda )}right) & 0 \
hline
text{Transit} & alpha &
begin{cases}
delta >lambda & 0 \
delta =lambda & text{Zenith} \
delta <lambda & pi
end{cases}
& frac{pi }{2}-left| delta -lambda right| \
hline
text{Set} & alpha +cos ^{-1}(-tan (delta ) tan (lambda )) & tan
^{-1}left(sec (lambda ) sin (delta ),-cos (delta ) sqrt{1-tan
^2(delta ) tan ^2(lambda )}right) & 0 \
hline
text{Lowest Point} & alpha +pi &
begin{cases}
delta >-lambda & 0 \
delta =-lambda & text{Nadir} \
delta <-lambda & pi
end{cases}
& left| delta +lambda right|-frac{pi }{2} \
hline
end{array}
$

where:

• $phi$ is the azimuth of the object




• $Z$ is the altitude of the object above the horizon




• $alpha$ is the right ascension of the object




• $delta$ is the declination of the object




• $lambda$ is the latitude of the observer




• $t$ is the current local sidereal time

Note the two-argument form of arctangent is required so that the
results are in the correct quadrant:
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent

Additional caveats:

• If $left| delta -lambda right|>frac{pi }{2}$, the object is
  always below the horizon, and the equations for rising time and
  setting time will not work.




• If $left| delta +lambda right|>frac{pi }{2}$, the object is
  always above the horizon (circumpolar), and the equations for rising
  and setting time will also not work.




• The measurements above are in radians. You can convert $pi to
  180 {}^{circ}$ for degrees.




• Because we use the local sidereal time, the longitude doesn't
  appear in any of the formulas above. However, we do need it to find
  the local sidereal time, as below.




• To find the local sideral time $t$ in radians, we use
  http://aa.usno.navy.mil/faq/docs/GAST.php and make some substitions
  to get:

$t = 4.894961212735792 + 6.30038809898489 d + psi$

where $psi$ is your longitude in radians, and $d$ is the number of
days (including fractional days) since "2000-01-01 12:00:00 UTC". Traditionally, we use $phi$ for longitude, but I'm already using it in the formulas above for azimuth.

If you combine the formula for local sidereal time and
azimuth/altitude and assume excessive precision, you get my answer to
http://astronomy.stackexchange.com/a/8415/21

Additional computations for these results at:
https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m

I was going to add some graphs to show how the altitude is NOT a sine wave and how the azimuth is NOT a straight line (although you might expect them to be), but they turned out not to be terribly instructive/helpful.

You might also be able to get simpler formulas if you set $t$ to be the "hour angle" (which is $alpha-t$ in the current setup).

orbit - Leonid meteor showers and the Tempel-Tuttle comet

When material is shed from a comet, it mostly continues to follow the same orbit as the comet, but drifts out ahead or behind the comet in a complex way due to interactions with solar wind and the gravity of other objects.

If you imagine the elliptical orbit of 55P/Tempel-Tuttle, it will have patches of denser dust in clumps spaced out around the ellipse corresponding to different perihelion approaches. The 1466 clump will be in a different place than the 1500 clump, and they will both be travelling more or less around the comet's orbit.

Since the comet's orbit only crosses Earth's at one place, and we only cross that spot once a year, we will only encounter whatever clumps of dust happen to be at that point at the same time the Earth is.

Calculating exactly where these clumps of comet dust are is non-trivial. And that is a big reason why predicting how good a meteor shower will be is a rather tricky business.

radiation - Conceptual Doubt regarding the calculation of the Solar Constant

Imagine you have a solar collector measuring one square meter, which can directly measure the power (energy per unit time) striking it. If there are layers of atmosphere above the collector, some of the incoming light is absorbed or reflected by particles in the atmosphere, so you won't measure as much power as if you place the collector in space at the distance of Earth's orbit, with no atmosphere between you and the Sun. But if you place the collector in the upper atmosphere of the Earth, the density of atmospheric particles above you is so small that they absorb/reflect only a negligible amount of power. Orientation is also an issue--you'll get the maximum power if you orient the plane of the collector so that it's perpendicular to the direction of the incoming rays. If it's not perpendicular, then if you approximate the incoming radiation as a set of parallel rays which each carry an equal amount of power (a standard approximation in optics) then fewer rays strike the collector when it's at some angle that's not perpendicular to the rays. All this is illustrated in the following diagram (from this page), with the two possible orientations of the solar collector illustrated in gray, and the light rays illustrated as black lines:

The solar constant is defined to be the power you get on such a surface of unit area if it's not being obstructed by atmospheric particles, and if it's oriented perpendicular to incoming rays so that it receives the maximum possible power at that location.

One aspect of the orientation issue that's important to understand is that if you orient your solar collector so that it's parallel to the ground (so that it would be lying flat if it was actually at ground level), then it will only receive the maximum power if the surface of the Earth is itself perpendicular to incoming rays from the Sun at that point, which is not true in the above diagram. It would only be true at the point on Earth that's the dead "center" of the Earth as seen by an observer looking at it from the Sun's position, or the point where the gray line connecting the centers of Earth and Sun intersects with surface of the Earth on this diagram:

That is also the only point on Earth where an observer on the surface would see the Sun "directly overhead", at the very top of the apparent dome of the sky. So even if you stripped away the Earth's atmosphere, only at this point on the Earth would the power reaching a square meter of the surface be equal to the solar constant, the average power per square meter over the entire illuminated half of the Earth's surface would be less than the solar constant since most of the surface is at an angle relative to incoming rays. Incidentally, the angle of the surface relative to incoming rays is also the main reason why at Northern latitudes, it's warmer in the summer (when the surface at a Northern latitude is closer to being perpendicular to oncoming rays) than in the winter (when the surface is at a more inclined angle relative to incoming rays), as illustrated in a diagram from this page:

It's true that since the Earth is not a point, an observer in the upper atmosphere directly above that point on the surface would not be at exactly the same radius from the Sun as the radius of Earth's orbit (defined in terms of the position of the center of the Earth), but if you actually do the calculation of the solar constant at the radius of Earth's center vs. the radius of the point on the surface that's closest to the Sun, the difference is very small so it can be ignored if you're only stating the solar constant to nearest Watt/meter^2 or even the nearest tenth of a Watt/meter^2. The nominal solar luminosity is defined to be $3.828 times 10^{26}$ and 1 AU is defined as 149597870700 meters, and if you use the formula you mentioned to calculate the solar flux at 1 AU it works out to 1361.1664654 watts per square meter. The radius of the Earth at the equator is about 6378 meters, so at an equinox when the Sun is directly overhead at the equator, an observer seeing the Sun directly overhead should be at a distance of 149597870700 - 6378 = 149597864322 meters, if you use the same formula to calculate the solar flux at this radius you get 1361.1665815, so it only makes a different if you want precision beyond the second decimal place.

light - TIME TRAVEL-Can it be really done?

In theory it is possible to do time travel. There are several possibilities from theoretical point of view. Two aspects are there, travelling to the future and travel back to past. Although many scientists including Stephen Hawking said that travelling to the past is not possible (as there are several paradoxes including "the grand father paradox") that results in some weird phenomena (like in grandfather paradox, if you are allowed to travel back in time and you went to the time of your grandfather's youth and kill him then you are not going to born in the first place in future, thus a contradiction). But travelling to the future is seems possible. There are several ways to do that, thanks to The Theory of Relativity. As Einstein said, time flows at different speeds in different part of the universe, faster in Earth as compared to the near vicinity of a black hole, due to the stronger gravitational pull of the black hole then the Earth. So if you travel near to a black hole and keeping a safe distance and starts revolving around it for few hours and return back to Earth, then you will find the many decades has been passed here, due to the different speed rate of time in the two places, so essentially you are in future (but alas you cannot go back!). Another way is to travel at a very high speed (almost near the speed of light) in a spaceship for few years also take you to the future back on Earth, this is because time slows down in case you speed up. And many more concepts of time travel exists involving Wormhole and others.

Grandfather paradox
Time travel
See this

Would all neutron stars inevitably collapse into black holes due to quantum tunneling?

The materials inside a neutron star are densely packed (extremely close to each other) and as time goes by say about an eon later, most of the neutrons would eventually tunnel into each other

Based on what ?

The material's collapse was halted from an astonishingly high speed collapse by the neutron degeneracy pressure that exists precisely because the universe doesn't want to let two neutrons fuse. The enormous forces involved in the collapse and the enormous forces keeping the mass compact are not high enough to overcome this. That's why it exists in the first place.

Over time "tunnelling" works in many, many ways. If two neutrons were to fuse in this manner the list of possible by products is enormous, and a black hole is most unlikely, IMO. We've a lot of experience of whacking particles into other particles in extreme conditions ( think CERN ) and we haven't seen any sign of a black hole yet ( and believe me, they're looking for them ). So I think the odds of a fuse into a black hole are extremely low. Over time it's more likely the neutron star would "evaporate" into other particles. But over those time scales you're looking at the cold death of the universe and protons decaying and so on.

And if, for argument sake, a mini black hole formed from two neutron's "fusing" it's life expectancy would be minuscule as tunnelling effects ( predicted by Hawking ) suggest it would "evaporate" in the blink of an eye. So isolated fusing events, even f they happened, would be unlikely to produce a black hole that grew, but instead one that just blew up as fast as it was created.

and voila! the neutron star becomes black hole? sound suspicious to me but can it happen?

No evidence I am aware of suggests neutrons will fuse in the manner you're describing.

"Neutron Star" is probably something of a misnomer. Apart from neutrons, current models ( "educated guesses" ) suggest that the core of such a body might be a lump of something called quark-gluon plasma, which is essentially mush made from neutrons. This is what you get when you fuse neutrons together - a form of mush made up of the stuff that neutrons and protons are made from - quarks.

So theory and evidence suggests that you don't get a black hole from just waiting for a neutron star to change into one.

solar system - Is there a ninth planet?

After Pluto's demotion as a planet, we have currently eight planets in our solar system. But Sun's gravitational pull can be felt well beyond Pluto, so is it possible to have a ninth planet beyond Pluto? I asked this question because we have Kuiper belt beyond Pluto, so there might be some Kuiper belt object which may qualified to be a ninth planet, isn't it possible?

black hole - What star system/Galaxy did LIGO observe the gravity wave?

The location of GW150914 was rather poorly constrained by the 6.9 ms time difference between the event when it was observed by the two separated LIGO detectors at Livingston and Hanford.

Further details are here
https://www.ligo.caltech.edu/image/ligo20160211b and this site provides a visualisation of where the waves came from.

The location OF GW150914 can only be constrained to be somewhere within 600 square degrees with 90% confidence. There are millions of possible host galaxies within this region.

space - Why can falling stars only be seen moving down towards horizon not up away from horizon?

The falling star or shooting star has nothing to do with a star. If shooting stars were actual stars, we'll be burning in their atmosphere, rather than the other way around.

When a meteoroid, comet or (small) asteroid (or space debris in some cases) enters the Earth's atmosphere at high speed (typically in excess of 20 km/s), aerodynamic heating produces a streak of light, both from the glowing object and the trail of glowing particles that it leaves in its wake. This is called a meteor or shooting star (or a falling star).

In general, the meteors travel in a (approximately) straight path. We usually see only the large ones travel down towards the horizon while the small ones are burnt up long before that (giving the impression of coming from the horizon).

Image from Meteors? Even More Remarkable by Bruce Maccabee at brumac.8k.com

Consider the meteor path in the image above. it can be seen that,

• The observer will see the meteor only after it passes the initial altitude $H_{i}$. If he sees the meteor at this point, and it burns up in the atmosphere, the impression is that the meteor comes from the horizon.




• If the observer notices the meteor after it passes the initial height and tracks it through its travel in the atmosphere, till is exits at the final altitude $H_{f}$, it will appear as if the meteor is moving down towards the horizon.

Another factor is the thickness of the atmosphere itself. The thickness of the atmosphere increases in path of the meteor from $H_{i}$ to $H_{m}$, and decreases from there to $H_{f}$. As the 'glowing' of the shooting star is directly dependent on the density of the atmosphere, its visibility trails of near the observer's horizon. This is another reason for the meteor looking like its falling towards the horizon.

This can be seen clearly in case of meteor showers which appear to radiate from a point.

Image from Lyrids Meteor Shower 2014 by Steve Owens at darkskydiary.wordpress.com

Here, the meteors appear to come away from the (radiant, which is above the) horizon.

saturn - Am I using my Celestron 8 Schmidt-Cassegrain telescope correctly?

I became interested in astronomy a few years ago and started with a Celestron AstroMaster 70AZ refractor telescope. That seemed nice, and I wanted to upgrade to a go-to telescope that was also more much more powerful, so I bought the 8 inch Celestron Schmidt-Cassegrain.

It's great, but honestly I'm a bit disappointed because the magnification doesn't seem all that much stronger than my 70AZ. I haven't done a side by side comparison. But for example when looking at Saturn, I was expecting to see more than just a white ball with a ring around it - I was able to see that with my 70AZ. I thought I'd see colors, and maybe be able to differentiate between some of the bigger ring bands.

I haven't tried any deep sky objects. I'm using various lenses starting from 30mm down to 6mm.

Am I doing something wrong?

The effect of gravitational wave is like tidal forces?

It depends what you mean by "very strong".

The announced first detection (14th September 2015) was considered a very strong signal and that still only stretched the 4 km detector arms by less than the width of a proton (a tiny particle inside the nucleus of an atom). This wave would have no effect on a person.

You could hypothesis a stronger wave that would stretch and squish a person, I'm not sure what effect this would have but there are no objects in nature capable of producing waves this strong, even something as close as our sun becoming a supernova wouldn't be that strong, so it will never happen in real life.

observational astronomy - What study profiles could land me the job of astronomer?

My career path:3-year Bachelor's degree in "Physics with Astrophysics"; PhD in X-ray astronomy; 5-years as a postdoctoral research assistant (two separate posts); got a lectureship at a UK university doing teaching and research in Physics and Astrophysics.

This is reasonably typical. These days, the content of the first degree is not so important - Physics, Astrophysics, Applied Maths all would be ok. "Astronomy" would put you at a disadvantage, since the implication is a non-physical, observational approach; but you would have to look at the course content.

A masters degree or 4-year first degree is usually necessary to get onto PhD programmes in the best places (this has changed since my day). Doing your PhD quickly and writing several publications is usually necessary to proceed any further.

The normal next step is to get a postdoctoral position; preferably somewhere other than your PhD institute. Then after 2-3 years of producing more research papers (2-3 per year), you could try for personal research fellowships. If you can get one of these, or perhaps a second/third postdoc position, and your research is going well and is productive, then there is a few year window in which to get into a tenured or tenure-track position. Getting some teaching experience at this stage is probably important.

For someone on a "normal" career path, it would be unusual to get a University position before the age of 30 (i.e. 8-9 years after your first degree). The large majority of people with a PhD in Astrophysics do not end up doing that for a living.