data analysis - Where can I find a catalog of all stars in the Milky Way?

No, such a catalogues does not (yet) exist. There are two reasons.

1 The Milky Way galaxy is about 20kpc (1pc ~= 3 lyr) across and only the very brightest stars are individually identifyable across such large a distance (such bright stars by their nature are very massive and hence young). Astronomers tend to cataloge stars by their apparent brightness, which for stars of identical luminosity declines as $1/d^2$ ($d$=distance). As a consequence, most catalogues contain only stars in the immediate galactic neighbourhood of the Sun. The Hipparcos catalogue (mentioned in another answer), for example, has most stars within a mere 100pc of the Sun.

2 Obtaining distances for individual stars is inherently difficult, in particular the more distant the star in question is. Accurate distances for stare several kpc away can currently only be obtained by indirect methods applicable only to certain types of stars (such as RR Lyrae variables). The classical trigonometric parallax measurement for such distances, however, is subject of ESA's ongoing Gaia mission.

ESA's Gaia satellite launched last year aims at cataloguing about $10^9$ stars across the Milky Way, including their velocity. The first preliminary versions of resulting catalogue, however, will still take some time to appear.

How to calculate full (mechanical) energy on the hyperbolic orbit?

The total energy is the sum of the kinetic energy and the potential energy. It is constant, unless some interaction such as a collision between bodies occurs.

The kinetic energy is easy to understand. For a mass $m$ and velocity $v$ it is given by

$$E_k = ½mv^2$$

The potential energy formula is (technically) given as

$$U = -μm/r$$

where $μ$ is the Standard gravitational parameter. (I say technically because it can be hard to grasp the concept of negative energy. See the linked articles).

In the case of a hyperbolic orbit the velocity and distance from the sun must be measured; one cannot be derived from the other as is the case with a known elliptical orbit.

See also

astrophysics - What happens to the Gas Pressure when working out the Eddington Luminosity?

The Eddington luminosity is defined in this way - but I guess that is not the answer you are looking for!

However, if we ask what isotropic luminosity is required in order to balance the inward gravitational force on a lump of gas then yes we could consider gas pressure as well.

The correct formulation is that the pressure gradient due to gas plus radiation balances the density times local gravity.

The justification for neglecting gas pressure would depend on the application. For example, when using the mass limit to justify an approximate upper mass limit for stars, it is reasonably easy to show that the ratio of radiation pressure to gas pressure $sim 0.1 (M/M_{odot})^{2}$, where $M$ is the stellar mass in solar units. Thus for stars $>3M_{odot}$, radiation pressure dominates, and because $L_{edd} simeq 3times10^{4} (M/M_{odot}) L_{odot}$ and $L simeq (M/M_{odot})^{3.5} L_{odot}$ for a main sequence star, then $L<L_{edd}$ means that
$$ (M/M_{odot})^{3.5} < 3times10^{4} (M/M_{odot})$$
and hence $M < 61 M_{odot}$. At this mass the radiation pressure will be nearly 400 times the gas pressure, justifying the use of a "gas pressure-free" Eddington limit.

I suspect, though I can't give chapter and verse, that for the intense radiation fields normally associated with the Eddington limit this will always be the case.

If we are talking about spherical accretion, then because the luminosity is constant at any radius, then the Eddington limit applies to material at any radius. But at greater radii from the central source, for a constant velocity inflow, the density will fall as $r^{-2}$, and the temperature will be lower at large radii. Thus the ratio of radiation to gas pressure will likely increase with radius and so at some point it will always be appropriate just to consider the radiation pressure.

What this implies is that reaching the Eddington limit could stop the accretion, but that even if it does not (e.g. clumpy accretion), then the increasing gas pressure at smaller radii could limit the accretion if the compressed gas is unable to cool.

Luminosity Schechter function for galaxies

Just a question I am having trouble understanding. I have the Schechter luminosity function for galaxies, given as:

$$Phi(L)dL=Phi_{0}left({frac{L}{L_{star}}}right)^{alpha}e^{-frac{L}{L_{star}}}frac{dL}{L_{star}}$$

I need to consider the case when $alpha=-1$. And then show that the average luminosity of a galaxy is exactly $L_{star}$. Could somebody explain how I could go about doing this and perhaps a hint or some part of a setup would be excellent. I really need to understand this.

Another part, which is related to the above question, is asking me to explain why the total luminosity is a finite number, whereas the total number of galaxies diverges. By this does it mean that the total number is infinite? Any extra comments on this would also be really appreciated.

fundamental astronomy - Which is more rare: Lunar eclipse or Solar eclipse?

The reason that solar and lunar eclipses are about equally common, is that they occur when the moon is "close enough" to the plane of the ecliptic at the point where it is full (lunar eclipse) or new (solar eclipse). Otherwise it completely misses the sun (solar eclipse) or the earth's shadow (lunar eclipse), and you don't get even a partial eclipse.

If the moon orbited precisely in the ecliptic plane, then there would be a total lunar eclipse every full moon, and every new moon a solar eclipse that's either total or annular according to the distance of the moon from the earth at that moment. But the moon doesn't orbit in that plane, it intersects it twice a month, at a time that most months isn't when it's close enough to being lined up with the earth and sun.

"Close enough" in the case of a solar eclipse is when the earth, as viewed from somewhere on the sun, is partly obscured by the moon (therefore: part of the sun is obscured by the moon when viewed from part of the earth). Whereas "close enough" for a lunar eclipse is when the moon is partly obscured by the earth from the somewhere on the sun (and therefore part of the earth's shadow falls on part of the moon). This is just about the same size target area.

What the sun "sees" as the moon orbits the earth is similar to what we see when we look at the moons of Jupiter: the moon moves from one side to the other and back. Sometimes it passes "over" the earth, sometimes "under", sometimes through. What's required for a partial eclipse of either kind is that it moves through, such that those two disks intersect as viewed from the sun. The difference from Jupiter, is that Jupiter is really big, so passing through is more likely for the moons of Jupiter than our moon.

If you really want to know which kind is a few percentage points more or less frequent, then you're beyond my ability to predict. Consult a table of eclipses covering a long period of time, or wait for an answer from someone with better qualifications :-)

Why did Venus not lose its atmosphere without magnetic field?

There is an interesting article on the magnetosphere of Venus on the ESA Science and Technology site. You can find the article here and it will probably answer your question.

The article states, like you did, that there are planets, like Earth, Mercury, Jupiter and saturn, have magnetic fields interland induced by there iron core. These magnetic fields shield the atmosphere from particles coming from solar winds. It also confirms your statement that Venus lacks this intrinsic magnetosphere to shield its atmosphere from the solar winds.

The interesting thing, however, is that spacecraft observations, like the ones made by ESA's Venus Express, have shown that the ionosphere of Venus direct interaction with the solar winds causes an externally induced magnetic field, which deflects the particles from the solar winds and protects the atmosphere from being blown away from the planet.

However, the article also explains that the Venus magnetosphere is not as protective as earth's magnetosphere. Measurements of the Venus magnetic field show several similarities, such as deflection of the solar winds and the reconnections in the tail of the magnetosphere, causing plasma circulations in the magnetosphere. The differences might explain the fact that some gasses and water are lost from the Venus atmosphere. The magnetic field of Venus is about 10 times smaller as the earth's magnetic field. The shape of the magnetic field is also different. Earth has a more sharp magnetotail facing away from the sun and Venus has a more comet shaped magnetotail. During the reconnections most of the plasma is lost in the atmosphere.

The article explains therefore that although Venus does not have an intrinsic magnetic field, but the interaction of the thick atmosphere with the solar winds causes an externally induced magnetic field, that deflect the particles of the solar winds. The article suggests, however, that the different magnetic field may cause that lighter gasses are not that much protected and therefore are lost into space.

I hope this sufficiently answers the question.

Kind regards,
MacUserT

Is there a limit to the size of a black hole?

There's no theoretical limit. If you had enough energy to move stars or galaxies, you could in theory keep feeding a black hole until it became enormously large, larger even than the Milky way for example. But there are practical limits past which black holes are unlikely to grow.

The two reasons for this are that 1), black holes aren't efficient at taking in matter. They can spit out as much as 90% of the energy from the matter that falls into them, and 2) once they reach a certain size, black holes are too large to form accretion disks, so matter tends to orbit around them rather than funnel into them.

Source and Source.

As to your 2nd question

Imagine that all the matter in the universe formed a black hole.
Should that be possible or is there a law which forbids creating it?

I've pondered this one myself and I have no idea the answer. Is there a size past which Dark Energy would overcome gravitation? Dark energy operating inside the black hole might overcome the gravitation past a certain size, but that's just my novice speculation and I think the black hole would need to be billions of light years across for that to happen.

I don't know the answer to that one. I'd be curious if anyone does though.

the sun - Could the sun burn a human floating next to it?

From the conditions you give, yes. Absolutely. The photon flux would simply transfer too much energy for the molecules in your body to stay fixed (the radiation would be immense). Another potential way to re-phrase this question would be: "How close could a human, without protection from any form of radiation, get to the Sun before their skin burns?"

gravity - When black holes such as those detected by LIGO merge, in what ways does the Schwarzschild Radius warp?

I was reading the paper about LIGO's detection of gravitational waves, and I wanted to know what the Schwarzschild Radius did. For example, I feel like the simulation here is oversimplified, and that reality wouldn't actually have that simple bump together of the two radii.

human biology - How are the gene sequences of individual sperm and egg cells "randomized"?

I think we can attribute this to sheer probability. The human genome contains around 3 billion base pairs. When you consider recombination from chromosomal crossover that occurs in germ line cells, there is an astronomically huge number of possible unique combinations that can be made.

Of course, males generate many many sperm, so some of these are bound to be similar. But the number of sperm produced by any individual doesn't come anywhere close to the number of possible unique sperm that he could produce.

planet - Calculating RA/dec from JPL ephemeris data

I'm using JPL planetary ephemerides to calculate the position of planets. Using DE405 ephemeris (using a parser provided by Project Pluto with all tests passing). I am fetching the position of Mars with the observer set as the Earth (geocentric).

My first assumption is that the result is a set of geocentric equatorial coordinates in rectangular form (x, y, z). This may be incorrect.

My goal is to calculate RA/Dec. My second assumption is that I should use a formula converting from Cartesian to Spherical:

$r =sqrt(x^2+y^2+z^2)$
$ra =arctan(y/x)$
$dec = pi/2 - arccos(z/r)$

Using Julian day 2457134 (21st Apr 2015, noon UTC), and cross-checking the result with NASA's Horizons service which apparently uses the same ephemeris (DE405) gets me almost there but my results are a little off:

x 1.721427968227801 y 1.5802812301744067 z 0.6897426775298559

RA 02h50m12.51s Dec +16°26'41.37"

Horizons:

RA 02h50m11.52s Dec +16°26'36.5"

I'm not sure what I'm missing. Do I have to factor distance of the planet due to the time it takes for light to reach us? Am I not using the correct time? Are my assumptions wrong?

Why does the planet Saturn have numerous (62) moons compared to the rest of the planets in the Solar System?

Saturn and Jupiter have many moons for quite a few reasons, one of the main ones being that they have an absolutely immense gravitational pull. During the early stages of the formation of our solar system, there would of been many planet-like objects floating around which our gas giants would have attracted. Furthermore, these planets are so far out in the solar system water would if frozen (which explains Saturn's rings of ice). Infact, we can show that the ice can form moons by looking at some of the moons of Uranus, some of them are half made of ice!

A few of the outer moons of our planets are captured asteroids. Phoebe, which is a moon of Saturn, is believed to have been a captured asteroid.

I haven't heard anything about Saturn having more moons than jupiter.

orbit - Are there accurate equinox and solstice predictions for the distant past?

I wrote
https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13008.c
to find historical solstices and equinoxes. The full results are at:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/solstices-and-equinoxes.txt.bz2

Accuracy:

• I'm using the EARTH_IAU_1976 precession model and the
  EARTH_IAU_1980 nutation and obliquity models, which are
  the only models supported by the CSPICE library I'm
  using:

http://hesperia.gsfc.nasa.gov/ssw/stereo/gen/exe/icy64/doc/frames.req

This is the same library and the same models NASA uses to
do its own calculations, but the models are known to have
limited accuracy. Quoting "http://aa.usno.navy.mil/faq/docs/SpringPhenom.php":

The times given in the tables are accurate to within two or
three hours for 25 to 5 BCE, and one or two hours for 4 BCE
to 38 CE. The uncertainty in these times arises from the
stochastic, that is unpredictable, change in the Earth's
rotation rate.

The accuracy gets worse the further back OR the further
forward you go: not only don't we know precisely about the
Earth's rotation in the past, but we can't even predict the
Earth's rotation precisely for the future.

Sample output:

EQU 511720269.432607 A.D.   2016-03-20 04:30:01
SOL 519734120.174820 A.D.   2016-06-20 22:34:11
EQU 527826138.004142 A.D.   2016-09-22 14:21:09
SOL 535589116.137776 A.D.   2016-12-21 10:44:07

Format:

• The first column indicates whether this is a solstice or an equinox.




• The second column is the ephemeris time of the
  solstice/equinox. If you're doing serious astronomical
  work, this is the column you should use.




• The remaining columns indicate the UTC time of the
  solstice/equinox in a more human-readable format:

  
  
  
  • The CSPICE libraries assume that the Gregorian
    calendar reformation occurred on 4 October 1582
    (meaning the day after 4 October 1582 was 15 October
    1582). Looking at these lines:
  
  

SOL -13191695511.794357 A.D.   1581-12-11 20:07:27
EQU -13183992131.003845 A.D.   1582-03-10 23:57:07
SOL -13175951250.170158 A.D.   1582-06-12 01:31:48
EQU -13167875920.223862 A.D.   1582-09-13 12:40:38
SOL -13160138634.917915 A.D.   1582-12-22 01:55:23
EQU -13152434815.793312 A.D.   1583-03-21 05:52:23
SOL -13144394485.035870 A.D.   1583-06-22 07:17:53
EQU -13136319216.460808 A.D.   1583-09-23 18:25:42

you can see that the dates of the solstices/equinoxes jump
ahead 10 days per the reformation.

• The CSPICE libraries use the Julian calendar prior to
  4 October 1582. In reality, the Julian calendar was
  introduced in 46 BCE:
  https://en.wikipedia.org/wiki/Julian_calendar




• Prior to 46 BCE, there were other calendar systems in
  use, but the CSPICE libraries assume the Julian
  calendar goes back indefinitely:

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

My calculations go back to 13201 BCE (the limits of DE431,
the ephemeris I'm using), and it's possible human beings
weren't even using calendars regularly at the time: quoting
"https://en.wikipedia.org/wiki/History_of_calendars#Prehistory":

A mesolithic arrangement of twelve pits and an arc found in
Warren Field, Aberdeenshire, Scotland, dated to roughly
10,000 years ago, has been described as a lunar calendar
and dubbed the "world's oldest known calendar" in 2013.

Notes:

• It takes the Earth 365.256363004 days to revolve around
  the Sun with respect to the fixed stars, but the time
  between vernal equinoxes is slightly less (365.242190402
  days) because the position of the vernal equinox moves
  (precesses) with respect to the stars. Source:
  http://hpiers.obspm.fr/eop-pc/models/constants.html




• The Gregorian calendar's average day length of 365.2425
  is much closer to 365.242190402 days than the Julian
  calendar's 365.25 average day length, but it's still not
  perfect. As noted in
  Do solstices and equinoxes shift over time?
  if we continue using the Gregorian calendar in the far
  future, the equinoxes and solstices will drift
  backwards. By 17090 (the limit of DE431), they will look
  like this:

EQU 476198945887.238159 A.D.  17090-02-22 08:56:59
SOL 476207018540.040894 A.D.  17090-05-26 19:21:11
EQU 476214808218.146362 A.D.  17090-08-24 23:09:09
SOL 476222655067.609985 A.D.  17090-11-23 18:49:59

about a month behind their "regular" times.

MY EARLIER PARTIAL ANSWER FOR REFERENCE:

Since HORIZONS (http://ssd.jpl.nasa.gov/?horizons) and SPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) can compute the ecliptic and solar position back that far, it should be possible to compute equinoxes and solstices with reasonable accuracy. However, I haven't been able to find a site that actually lists these dates (I'm pretty sure USNO did this at one point, but I can't find their list). Other possibly helpful sources/questions:

Simulate an orbit with orbital elements

I want to simulate the orbits of the planets from our solar system. I want to use orbital elements to calculate the current position(xyz) at a time t. The simulation doesn't have to be too exact, but the initial position of the bodies should be somewhat realistic.

Doing some research on the calculation I stumbled upon the following formula:

R,X,Y,Z-Heliocentric Distances
TA - True Anomaly
N - Longitude of the Ascending Node
w - Argument of the Perihelion

R = a * (1 - e ^ 2) / (1 + e * Cos(TA))
X = R * (Cos(N) * Cos(TA + w) - Sin(N) * Sin(TA+w)*Cos(i)
Y = R * (Sin(N) * Cos(TA+w) + Cos(N) * Sin(TA+w)) * Cos(i))
Z = R * Sin(TA+w) * Sin(i)

Source: https://www.physicsforums.com/threads/calculating-elliptic-orbits-in-cartesian-coordinates.712979/

For testing purposes I calculated the true anomaly with the help of the following calculator: http://www.jgiesen.de/kepler/kepler.html
It needs the mean anomaly and eccentricity which I took from Wikipedia. Doing some more research on the mean anomaly, I now believe that the time at which I want the position will be somehow fed into the calculation of the mean anomaly along with the time of the initial position.

Can somebody clarify for me on how to correctly calculate the position at time t with the above or a different formula. With the above formula I also think I need the initial values at a specific time.

I also want to say that I am not an astronomer and I don't have a clue on how to handle the calculation or which formula to use. The result of the calculations should mimic our solar system as closely as possible.

Thanks,
Rene Hollander

neuroscience - Does the Parasympathetic Tract of Colon Sigmoideum Travel with Nervus Vagus and its Nucleus Dorsalis Nervi Vagi?

I found the following from Wikipedia:

Pelvic splanchnic nerves are the primary source for parasympathetic innervation [for Sigmoid colon].

The non-cranial nerve is extending the scope of the cranial nerve by a synapse between the two neurons. The neurons are transmitting the electrical impulse from one neuron to another.
So this way the splanchnic nerve does not need to be with CN 10 in the region of impact.

The parasympathetic tract of colon sigmoideum can travel with nervus vagus by having a synapse between pelvic splanchnic nerve and CN 10. I verified this from my professor.

Why don't Australia/Russia have large optical telescopes?

Optical telescope sites ideally need to be somewhat remote (to avoid light pollution) yet still accessible for construction, engineering, and observing; high in altitude; dry; with exceptional "seeing" (stable atmospheric conditions); and clear weather. Neither Australia nor Russia really have any sites meeting all those criteria: their mountains tend to have some combination of too low altitude, poor seeing, and/or bad weather. (This is also why there aren't any really big, modern telescopes in Europe.)

This site has some discussion of the different criteria that go into choosing the best sites for new, large telescopes.

structural biology - Can protein structure be determined by X-Ray Diffraction in a single image?

Not by analysing a single protein. There is work with x-ray lasers.

You have to take a simultaneous image of millions of proteins and use that to get a structure. It's not quite prime time. People are also doing this with electron beams in electron microscopes.

These methods will reconstruct 3D models of the molecules, sometimes in states which cannot be obtained from crystallography. Examples being the structure of the many megadalton nuclear pore complex, and the f-actin fiber. The classic study is 3d model of bacteriorhodopsin, the first membrane protein structure which was at molecular resolution (this was a crystalline sample though).

While in principle, it sounds much simpler - get a pure sample of your protein, or complex and freeze it down and zap it with an Xray or Electron beam, its a lot more work to reconstruct the image and can take as long or longer than getting an x-ray structure. The resolution is also usually poor as the crystal will reinforce coherence, that is all the proteins are aligned in the same way and have close to the same 3d shape in a crystal.

orbit - The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs

There is quite simple formula that will give you the tidal-locking half-time

$$T = C, frac{a^6 R mu}{ M_s M_p^2}$$

• $a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters


• $R$ - planet radius in meters


• $M_s$ - planet mass in kg


• $M_p$ - parent planet/star mass in kg


• $mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.


• $C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. (See the discussion.)

Solving the problem of interaction between the parent star and between the planets will be difficult. Generally, if tidal-locking half-time between the planets is much shorter than the tidal-locking half-time between the star and the planet, the planets will tidal-lock to each other rather than to the star. So we ask, whether

$$ frac{A^6}{ M_s^3} ll frac{a^6}{ M_s M_p^2};,$$

where $A$ is the semi-major axis of the orbit of the double-planet. For typical planet in the habitable zone of the red dwarf star, $a$ = 9,000,000,000 m (0.06 AU), while the distance of the double-planet could be for example $A$ = 30,000,000 m. $M_s = 6times 10^{24};mathrm{kg}$ and $M_p = 2.9times 10^{29};mathrm{kg}$. We get

$$2cdot 10^{-41}ll 6.3cdot 10^{-36};,$$

which is well-satisfied and the planets will tidal lock to each other rather than to the parent star. Were the distance between the planets $A$ = 300,000,000 m, the inequality would yield

$$2cdot 10^{-35}ll 6.3cdot 10^{-36};,$$

which is not satisfied and the tidal forces of the star would probably disrupt the double-planet. (The planets would not even be in the Hill sphere, which approximately gives the stable orbits.)

So the answer is: a very close double planets can tidal-lock to each other rather than to the parent red-dwarf star, but they would have to be much closer than the Earth to the Moon.

Do the magnetic fields of stars and/or other planets reverse?

Sure. It even happens with the Sun, as part of the solar cycle.

Earth's magnetic field is described by something called the dynamo theory, which states that the movement of fluids in the core generates the magnetic field. A modification of this is used to model the magnetic fields of stars. Given that the cause of geomagnetic reversals is related to the movement of currents deep on Earth's core, it makes sense that the same would be the case in other bodies.

damage from an Asteroid - Astronomy

An asteroid that remains outside Earth's atmosphere, and therefore does not impact Earth, will have no effect on the planet.

Asteroids, even biggish ones like the dinosaur killer, are pretty small compared to the Earth. They do not have enough mass to graviationally affect earth's orbit noticeably. And if they miss earth, the pass quickly.

The concern would be that if an asteroid can pass close to Earth, that it may collide in a future orbit. The asteroid's path will be changed by the earth's gravity. However, a very close encounter will enable astronomers to get a very high quality position on the rock, and so be able to predict its future path with very great accuracy.

Comet tails aren't a worry. We passed through Halley's tail in 1910, the effect on the Earth was precisely nothing. ☄

orbit - How to calculate geocentric conjunction time and moon altitude at a given time

Umm al-Qura calendar of Saudi Arabia determine the new moon if these criteria meet at 29th day:

• The geocentric conjunction occurs before sunset.


• The moon sets after the sun.

I've got the formula to calculate when the sun (seen by the observer) sets, considering observer's latitude, longitude, height (altitude from sea level), and the date (Julian Day). I would like to ask the formula to calculate:

1. When (the time) geocentric conjunction happens.


2. How much (the angle) the altitude of the moon (seen by the observer) at a given time is.

Thank you

gene expression - Determining potential protease sites within a recombinant protein

LC-MS is certainly quantitative and will give you a definitive answer, but it is costly and requires access to such a machine.

I presume you're analyzing your protein based on western blotting.

The first thing you should always do is verify your DNA sequence is coding for the protein product you want. Once you're sure of this, the western blot will give you an indication as to where your protein is (approximately) being cleaved. Say you protein has a predicted size of 40 kDa and you see a band at 20 kDa, then your cleavage is somewhere in the middle of your sequence. There are a TON of proteases that could be potentially cleaving your peptide and you need to have a hypothesis as to what that could be. Your peptide could be cleaved by an endopeptidase (cutting within), or a carboxypeptidase (C-terminal cleavage) or an aminopeptidase (N-termainal cleavage). To go back to the 20/40 kDa example, it could be your peptide was cleaved N-terminally o C-terminally down to a size of 20 kDa, or that it was literally cleaved in the middle by an endopeptidase.

Something you may consider is the use of a general protease inhibitor cocktail (Roche makes a really good tablet product called Complete and Complete mini tab). These mixes have a bunch of general protease inhibitors which will stop most cleavage events. If you still see cleavage after using an inhibitor cocktail, you can reasonably expect that you are not inhibiting the protease with the tablet (thus narrowing down your search) and then you can better comb through Expasy's PepCutter data. You should also do a literature search to see what has been reported about your peptide or motif.

gravity - Are black holes really singularities?

There's many good answers to this on a physical level, but I don't see this one addressed, so I'll give an answer briefly.

I'm just having a hard time accepting that anything can exist with an
infinitely large property as it would lead to infinitely large mass
leading to infinitely large forces...that would just destroy the
universe infinitely fast, wouldn't it?

If black holes have infinities, which is only possible if they have a singularity, then those infinities are, by definition, infinity small and not infinite from any distance away, even a fraction of an atom's distance away. So "destroy the universe by infinite forces", sure, if the universe was smaller than an atom, or maybe, inside the event horizon where it couldn't do anything but fall towards the singularity, but at any safe and reasonable distance, black holes aren't dangerous and not a threat to the universe.

So I think a supermassive object can be a badass just as well without
having to be a singularity

and

Gravastars to the rescue! Just what I needed. Finally, some theorists
who are not comfortable with infinitely nasty objects populating our
universe!

OK for starters, the laws of physics are what they are, and they don't care what we think. As to things being "nasty", that's in the eye of the beholder.

2000 years ago, "hellfire" was magma seen coming out of the occasional volcano and hell was inside earth. Today, the inside of the Earth generates a magnetic field that protects us and gives us plate tectonics, which is really useful for life bearing planets. The inside of the Earth hasn't changed, but our perception of it has changed enormously. Now we love the inside of the Earth, but 2,000 years ago people feared it.

100 years ago, everyone thought the Universe was the Milky way and 80 years ago, they thought the Universe was eternal, until that lousy Hubble suggested the Universe had a beginning and, well, so much for things being eternal. If Hubble had said that 300 years earlier he'd have been burned at the stake, no question about it, with a plaque that stated "blasphemer"

That's the problem with point of view. It's not a complete picture. Black holes make a great "boogie man", as something that eats everything and cannot be escaped, but that's only part of the big picture.

Einstein himself found black holes a distasteful idea (He wasn't too keen on quantum mechanics either), so Einstein imagined the Universe had some physical law that prevented black holes from ever forming and that might even be the case, but honestly, how different is it to be slowly squashed around the event horizon over an eternity vs quickly squashed in the singularity or whatever goes on in the center. From a certain point of view, it's pretty much the same thing.

Some theories have gone so far as to suggest black holes are magical places with entire baby universes inside of them. I find that more creative thinking than science, but the truth is, what happens inside the event horizon stays inside the event horizon and nobody knows.

On black holes in general:

They're very useful. The formation of black holes, and the gravitational collapse and rebound off the collapse, creates and distributes heavy elements across the galaxy and in the collapse that creates a black hole or Neutron star, something like 90% of the matter of the star gets blown off, recycled, if you will, back into the galaxy and only 10% or so, forms the collapsed core.

You can think of the black hole as a nasty little remnant of a star's death, but I find the fact that large stars recycle and distribute so much of their matter across the galaxy to be highly cool. Supermassive black holes also help in the formation of galaxies, so the simple truth is, black holes are very useful, even if you wouldn't want to cross paths with one.

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Now on what happens inside a black hole, there's some good answers to that already and I don't want to make this too long, especially since I'm a layman, but I find the speculation of that exotic region inside the event horizon to be great fun to think about.

I personally don't believe in singularities. I (THINK), that the wave and field nature of quantum mechanics and the fact that empty space has properties which, for example, particle anti particle pairs can form essentially out of nothing, (which make hawking radiation possible), I think there's probably some kind of exotic space of never a full singularity.

I don't think there's what might be called a physical material inside a black hole. I think things would behave differently than that, more like the exotic nature of a proton or electron than the physical nature of a surface, but that's just my thoughts on the subject. Without a quantum theory of gravity, it's a bit like the blind looking at a map though. Nobody knows. (too long?)

What is the temperature inside a Black Hole?

Temperature of a black hole is determined by 'black body radiation temperature' of radiation which comes from it (if something is hot enough to give off bright blue light,it is hotter than something that is merely a dim red hot.) For black holes the mass of Sun, radiation emitted from it is so weak and so cool that temperature is only one-millionth of a degree above absolute zero. some black holes are thought to weigh a billion times as much as the Sun and they would be a billion times colder, far colder than what scientists have achieved on Earth.
Even though these things are very cold, they can be surrounded by very hot material. As they pull gas and stars down into their gravity wells, material rubs against itself at a good fraction of speed of light. this heats it up to hundreds of millions of degrees. Radiation from this hot, in-falling material is what high-energy astronomers study.
A black hole behaves as though its horizon has a temperature and that temperature is inversely proportional to the hole's mass: T = (6 x 10^-8)M. Here M is in units of solar mass (2 x 1033 grams). Temperature is in degrees Kelvin.
This means that a hole recently formed by gravitational collapse of a star (which has to have a mass larger than about 2 suns) has a temperature less than 3 x 10^-8 centigrade above absolute zero which is very cold.
Mass with a finite temperature radiated energy. Anything which radiates energy is also losing mass.As black hole loses mass, emission of energy from black hole increases and its temperature increases and thus rate of mass loss increases.As mass of black hole gets small,one has unstable 'runaway effect'. Black hole gets hotter and hotter which causes M to decrease rapidly.When the hole is reduced to a fraction of the size of an atomic nucleus, it will be trillions of degrees. Hole will burn up and disappear. Lifetime of a black hole is greater than age of the universe.

vision - How do we know the brain flips images projected on the retina back around?

The basis of this question is a common misconception, and unfortunately the accepted answer by @CHM is also based on this common misconception. The misconception is based on the homunculus falacy and the tendency for people to think that the image that lands on the retina is somehow 'assembled' and presented for something (the 'consciousness') to view. This is not the case.

As the comment by @mgkrebbs expains, there is no orientation (up or down) in the brain, there is only neural firing. The infromation of the visual scene is distributed over the brain, and information does not have physical properties like orientation. Although as @nico pointed out the neurons that process the information do have a spatial structure that mimics that of the retina, this is a topological property (i.e. simuli that are close on the retina are processes by neurons that are close in V1) and such a topological property does not induce an orientation.

The root of the problem is really that the question "How do we know the brain flips images projected on the retina back around?" is a pseudo-question. Although it is grammatically well-formed, it makes no semantic sense. When the image is 'in the' (i.e. being processed by the) brain it no longer has physical properties like orientation. Thus you cannot ask if it has been flipped or not.

radial velocity fitting of a binary

The radial velocity curve of a star in a binary system (with another star or a planet) is defined through 6 free parameters
$$V_r(t) = Kleft(cos(omega + nu) +e cos omega right) + gamma,$$
where $K$ is the semi-amplitude, $gamma$ is the centre of mass radial velocity, $omega$ is the usual angle defining the argument of the pericentre measured from the ascending node and $nu$ is the true anomlay, which is a function of time, the fiducial time of pericentre passage $tau$, the orbital period $p$ and the eccentricity $e$.

To proceed you estimate what all these parameters are - i.e. an initial guess.

Then, for each time $t_i$ of a data point in your RV curve you:

1. Calculate the mean anomaly
   $$M(t) = frac{2pi}{p}(t - tau),$$




2. Solve "Kepler's equation"
   $$M(t) = E(t) - e sin E(t)$$
   numerically (its a transcendental equation, you could use Newton-Raphson or similar) to give $M(t_i)$, the eccentric anomaly.




3. Use
   $$tan frac{E(t)}{2} = left(frac{1+e}{1-e}right)^{-1/2} tan frac{nu(t)}{2}$$
   to calculate the true anomaly $nu(t_i)$.




4. Calculate $V_r(t_i)$

You then calculate some figure of merit (e.g. chi-squared) for how closely the model and data agree and go through an iterative process to adjust the parameters and optimise the fit of model to data.

A more sophisticated discussion can be found in this paper by Beauge et al.

If you have the RV curves of both stars, then you can fit them both simultaneously. Obviously, they have $p$, $e$, $gamma$ and $omega$ in common, but their RV amplitudes $K_1$ and $K_2$ will be different. The ratio of $K_1/K_2$ gives you the ratio of the two stellar masses.

If you only have one RV curve you are limited to estimating the mass function of the binary system.
$$ frac{M_2^{3} sin^{3} i}{(M_1 + M_2)^2} = frac{p K_1^{3}}{2pi G},$$
where $i$ is the inclination of the orbit with respect to the line of sight.
This can only give you a lower limit to $M_2$ unless $i$ is known.

Taking your specific case study. If you know $M_1$ and $i$ (this could be the case for a transiting exoplanet, or maybe a binary featuring an eclipsed black hole candidate), then the primary radial velocity curve gives you $K_1$ and hence $M_2$. If the masses and $p$ are known then Kepler's laws give the orbital separation.

There are a number of options if you want an off-the-shelf solution to fitting RV curves. Perhaps the best free one is Systemic Console.

There is no fundamental difference between analysing the RV curves of stars with exoplanets and stars with unseen (stellar) companions.

Plausible? Brown Dwarfs are rogue celestial bodies because they absorb all light due to their thick layer of matte black soot

They are so dark they absorb almost every bit of light from any star and receive no photon velocity momentum or electromagnetic radiation.

The momentum that a planet receives or does not receive from photons is negligible for its orbit.

These dark masses are extremely high in mass and do not interact with light like other planets and moons.

A dark object very much interacts with light: it absorbs light. An object that does not interact with light would be transparent, not dark.

I believe different levels of light interaction on a mass causes planets to move away from stars or stay at bay.

Planetary orbits are governed by gravitational effects. The effect of light interaction on orbits is negligible.

Finally, none of this has anything to do with planets in orbit or not. Planetary orbits are due to gravitation.

Only another collision from a celestial body can interfer with the direction they are heading they outweigh most of the competition so distraction is rare.

Indeed, the only force massive enough to alter the orbit of a planet would be another planet, either hitting it full force, or passing closely thus altering its angular momentum. You'd need an awful lot of this to cause a planet to reach a solar system escape trajectory, however!

And it has nothing to do with colour or photon interaction. It's all about mass.

observation - How can apparent magnitude be negative?

Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us. The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6). Each successively lower magnitude was twice as bright as the one before, meaning the scale was logarithmic. We still use magnitudes for historical reasons, though the scale was later standardized to use the formula

$m_x - m_{x,0} = -2.5log_{10}(frac{F_x}{F_{x,0}})$

where $m_x$ and $F_x$ are the magnitude and flux of the object of interest and $m_{x,0}$ and $F_{x,0}$ are the magnitude of flux of a reference object (where usually Vega is used to define the 0 point in magnitude). This means any object that appears brighter than Vega has a negative magnitude. There is no limit to how bright an object can appear, so there is no lower limit to magnitudes. The sun, for example, being the brightest object in our sky, has a magnitude of roughly -27.

botany - How deep in the soil can a seed be placed and still develop into a plant/tree?

To answer your specific questions...

Does nutrient availability limit emergence depth?
Yes, the size of the nutrient store in the seed does impose a theoretical limit on the maximum depth at which seed germination and emergence can take place.

What is the best depth for a given species?
The specific depth which gives best germination for any given species is very unlikely to be the maximum depth at which seeds of that species can germinate. This is because other factors also impose limits on germination depth.

Probably the most important other factor is light - most seeds require the perception of red light in order to trigger germination. For most species then, you are unlikely to find seeds germinating and successfully emerging from depths greater than the penetrance of light.

Another important factor in some latitudes is temperature - many seeds (from 26 families) use dormancy as a way of preventing germination in the wrong seasonal conditions. In species where temperature serves as a cue to alleviate dormancy, there will be a maximum depth at which a given seed can detect the temperature fluctuations at the surface. Beyond a certain depth, surface temperature changes will not perceptibly affect local temperature.

If you ever need to find out the best depth of germination for a given species, many can be found on google, or failing that it is trivial to perform a simple experiment.

Is there a mathematical relationship?
There is a general mathematical relationship, an allometric relationship (i.e. relating size to another trait), between seed size and maximum depth of emergence, described by Bond et al. (1999):

                                     
where:
dmax = maximum depth of emergence
w = seed weight
c = a constant which varies with phylogeny and environmental conditions.

Many species however will not submit to this equation. For example some large-seeded species (incl. coconut, coco-de-mer, avocado), have large seeds because they may need to enable long-distance growth along the ground in order to locate optimum growth conditions. Seeds of these species will therefore not conform to the depth-size relationship above.

Reference

Visibility of earth from moon during day-time of moon

Inspired by this question. I am curious whether earth, besides being nearly fixed on one place on the moon's sky, is it visible during the day-time on moon too?

My understanding is that earth should be visible as moon has no atmosphere. Also, if the NASA didn't edit the following photograph, it suggests that the day-time sky of moon is all black and earth should be visible in it.

I think, Stellarium don't take into account the atmosphere once you are on other planet.