planet - Alignment of Semi-Major Axis of Orbits

The current longitudes of perihelion show no alignment. Furthermore, they change because all the planets have orbital precession - that is, the longitude of perihelion changes by between 0.36 arcseconds/year for Neptune (i.e. the longitude of perhelion goes through the full range of angles in 10,000 years) to 19.5 arcseconds/year for Saturn. So not only is there no alignment, but the relative directions of the semi-major axes change comparatively quickly.

Which organisms have the neuroanatomy Roger Penrose supposes play a role in consciousness?

Microtubules are a structure in the cytoskeleton, they are rope like polymers that grow to a length of about 25 micrometers (25000 nm), and have an outer-diameter of around 25 nm. For comparison, the mean spacing between atoms is on the order of 0.1 to 0.2 nm; so the micro tubule really is micro: about 200 atoms across. In terms of quantum effects though, this is pretty big but not unreasonable. Researchers commonly use quantum dots to play with quantum effects, and these are typically spheres on the order of 10 to 50 atoms in diameter. Note, that we don't know how to couple 5000 quantum dots in one coherent chain (how many you would need to get the length of a microtubule).

So microtubules are small, but they are common! Microtubules are found in all dividing eukaryotic cells and in most differentiated cell types. In other words, that mosquito you just smacked and that philosophers always give as an example of something non-conscious is full of microtubules. This should raise some red flags, but we don't need to go into more detail of microtubules to discredit Penrose and Hammeroff. However, if you love cell biology, take a look at Desai & Mitchison (1997).

So microtubules are probably a bad basis, but why did Penrose want quantum effects in the brain? In The Emperor's New Mind, Penrose suggests consciousness is non-algorithmic and suggests that a magical quantum computer could do these non-algorithmic tasks. The reason I use 'magical' is because a real quantum computer is Turing-complete, if a classical computer cannon solve a problem then neither can a quantum one (of course, if a classical computer can solve a problem, then quantum one can as well and might be able to do it qualitatively faster). For a nice computer science debunking of this part of Penrose's argument take a look a Scott Aaronson's lecture notes.

Why did Hammeroff want quantum-ness? To avoid dualism in explaining consciousness. However, he has gone so far down the reductionist rabbit-hole that he popped out on the other side. He arrived at the same 'magic' we feared in dualism except now he called it 'quantum mechanics'.

The biggest irony of this approach is that Penrose was inspired in many ways by Schrodinger's beautiful take on life. Although Schrödinger does bring in quantum mechanics (both as a useful reduction and as an analogy) he uses completely different parts of it (he uses the discritization of energy levels, and specially avoids issues of the uncertainty principle and superposition of states that made him famous). Schrödinger would completely disagree with Penrose and Hameroff::

[I]f we were organisms so sensitive that a single atom, or even a few atoms, could make a perceptible impression on our senses -- Heaves, what would life be like! To stress one point: an organism of that kind would most certainly not be capable of developing the kind of orderly thought which, after passing through a long sequence of earlier stages, ultimately results in forming, among many other ideas, the idea of an atom.

This response can be made precise through quantum decoherence (Tegmark, 2000) and there is little regard for the physical importance of quantum mechanics in the brain (Litt et al., 2006) although Hammeroff (2007) still defends it.

botany - Why are some plants frost tender?

All cells are susceptible to cold, but freezing is particularly difficult. If the water in the cells freeze, it would usually burst the cells (water expands when it turns to ice) and tearing the cell membrane (totally killing it).

Some plants and animals have adapted to colder temperatures by putting antifreeze proteins in their cells. A lot like adding antifreeze to the radiator of a car, this lowers the freezing temperature of the water in the cell.

So the answer is that some plants have adapted to colder climates with the help of additional genes. There are other things that have to change too - the chemical reactions in the cell will go haywire when temperature changes too, so lots of genes must make adjustment for a fully cold adapted plant.

From which exoplanets is our sun the brightest star on the night sky?

I wrote https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13115.m to solve this with the full results at: https://github.com/barrycarter/bcapps/blob/master/ASTRO/brightest-stars-from-other-stars.txt.bz2

As viewed from RigelKentaurusB, the Sun has magnitude 0.48,
and is the 10th brightest star in the sky. This is the only
star system with known exoplanets where the Sun even makes
the top 10.

From EpsilonEridani, the sun shines at magnitude 2.37,
making it the 79th brightest star in their sky. Things go
rapidly downhill from there:

79 EpsilonEridani Sun 2.36646
207 HIP113020 Sun 3.18865
208 HIP85523 Sun 3.11216
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679

This is based on Mathematica's list of 552 exoplanets
orbiting 464 stars, and a total list of 88,637 stars.

If the naked eye visibility limit is magnitude 5.5, the Sun
would only be visible from 28 of these stars:

10 RigelKentaurusB Sun 0.476137
79 EpsilonEridani Sun 2.36646
208 HIP85523 Sun 3.11216
207 HIP113020 Sun 3.18865
247 HIP106440 Sun 3.29606
396 HIP15510 Sun 3.74083
423 HIP74995 Sun 3.81444
685 Fomalhaut Sun 4.25746
878 HIP64924 Sun 4.48202
949 HIP109388 Sun 4.54456
972 HIP99825 Sun 4.55679
1086 HIP57443 Sun 4.65677
1126 HIP21932 Sun 4.69871
1381 HIP57087 Sun 4.87838
1426 HIP83043 Sun 4.89689
1409 Pollux Sun 4.90049
1617 HIP10138 Sun 5.01833
1649 HIP57050 Sun 5.04143
1676 HIP3093 Sun 5.05658
1716 HIP48331 Sun 5.06528
1831 Gl317 Sun 5.13093
2104 HIP22627 Sun 5.24572
2264 Rho1Cancri Sun 5.31849
2263 HIP40693 Sun 5.32723
2397 HIP27887 Sun 5.36943
2355 HIP80337 Sun 5.37669
2382 GJ1214 Sun 5.3879
2689 UpsilonAndromedae Sun 5.47503

If the limit is magnitude 6.5, 33 more star systems can see the Sun:

2838 Alrai Sun 5.52682
2926 HIP53721 Sun 5.571
3399 HIP79431 Sun 5.69393
3637 MuArae Sun 5.74863
3679 HIP113357 Sun 5.76061
3783 TauBootis Sun 5.7949
3941 HIP98767 Sun 5.83457
4071 HIP85647 Sun 5.86725
4318 HIP71395 Sun 5.92513
4580 HIP6379 Sun 5.95739
4783 IotaHorologii Sun 6.01138
4835 HIP7978 Sun 6.02527
4867 RhoCoronaeBorealis Sun 6.03887
5016 HIP1292 Sun 6.0583
4949 NN3634 Sun 6.08757
4993 HIP55848 Sun 6.10352
5252 HIP49699 Sun 6.11214
5300 HIP83389 Sun 6.11214
5031 HIP65721 Sun 6.11803
5343 HIP79248 Sun 6.12236
5385 PiMensae Sun 6.12986
5453 EpsilonReticulii Sun 6.13303
6158 HIP98505 Sun 6.251
6351 HIP113421 Sun 6.30305
6436 Nu2CanisMajoris Sun 6.31593
6343 HIP99711 Sun 6.32197
6737 Hamal Sun 6.35637
6516 HIP64457 Sun 6.37975
6867 HIP58451 Sun 6.4205
7614 HIP25110 Sun 6.43775
7525 HIP109378 Sun 6.46941
7538 HIP54906 Sun 6.47034
7957 HIP96901 Sun 6.48193

OLD ANSWER FOR REFERENCE (note that Sirius does not have known exoplanets, so the Sirius solution I propose below wouldn't work):

As others have noted, since Sirius is 25 times more luminous than our Sun (though still not luminous as Canopus or Rigel), so it would be the brightest star for most nearby star systems, and all the exoplanets we've found are fairly closeby.

The only possible exception would be an exoplanet orbiting Sirius itself, since Sirius would be considered their sun, and not a star in the night sky.

Unfortunately, Sirius' smaller companion, Sirius B, would be much brighter than our Sun there. I'm also pretty sure Sirius B would be in the night sky (not just in the daytime sky), so it would count.

I'd like to poke around a bit more before declaring this answer complete. I think I can show that any point where the Sun is brighter than Sirius must be less than 4.3 light years (and probably much less) away from our solar system, so, unless we discover an star system MUCH closer than Proxima Centauri, there are no exoplanets where our Sun is the brightest star.

gravity - Is there really no difference between up and down in ISS?

I can think of one place where you would prefer to have your head in a specific direction.

The nadir (Earth facing, or "down") Cupola module is a popular place among the cosmonauts, where you can look down at the Earth. Here, you want to stand in such a way that your head is able to watch out of the windows.

Beds do not have a preferred direction, in fact, the sleeping compartments are stacked in a square around the node-2 module in the American segment.

the sun - Does our sun/solar system orbit around any other celestial objects?

The orbit of the Sun around the Galaxy is quite complicated, because unlike the solar system, the mass is not completely concentrated at the centre. So, in addition to the roughly circular 230 million year orbit in the plane of the Galaxy, there are superimposed motions in and out of the plane and towards and away from the Galactic centre. These roughly sinusoidal additional motions, called epicycles, do not have large amplitudes - a few hundred light years - and take about 70 million and 160 million years respectively.

The Sun does not systematically orbit any other Galactic structures or stars (see the flagged potential duplicates for more details - there is no evidence for any binary companion to the Sun larger than a few Jupiter masses) and is unlikely to do so in the forseeable future. The space between the stars in our Galaxy is large enough that they are essentially non-interacting.

The Galaxy itself is in motion with respect to the galaxies around it. The nearest tens of galaxies form the local group and probably have complicated orbits within their summed gravitational potential. These orbits cannot be precisely determined, because although we can measure line of sight velocities using the Doppler effect, the tangential motions require extraordinarily accurate position data over many years. These are now becoming available - for instance we know our Galaxy and M31 will collide in around 4 billion years.

Further afield, the local group is part of a larger galaxy aggregate, called the Virgo supercluster, but here, determining an orbit is impossible.

Even further, large galaxy clusters are arranged into a web of larger superclusters, voids and filaments that are all in motion with respect to each other.

redshift - Why is there no time in the distance equation using Hubble constant, red shift and speed of light?

Are you familiar with Hubble's Law? Because this is just it, in a slightly less familiar form.

Note that the distance in this formula is the proper distance, the distance between us and some source, not the distance travelled.

Other than that, the speed of light in a sense defines our conception of time and space, so it is mentioned in the formula. ;)

atmosphere - Equation for solar radiation at a given latitude on a given exoplanet?

I'm trying to find equations that would help me determine the amount of solar radiation hitting a certain latitude on a certain planet given the following inputs:

• the degrees of latitude of the location in question


• this hemisphere's current season (winter or summer)


• size of the planet


• luminosity of the star(s) the planet orbits

Without taking into consideration wind, air pressure, or any atmospheric effects.

Ideally I would like to determine the average solar radiation of a given location in both the winter and summer.

My end goal is to determine the average surface temperature of a given latitude on a planet using the base solar radiation and the effects of wind, air pressure, and surface ocean currents.

I know it is possible to do this for the entire planet generally, but I would like some way of doing it for a particular latitude and season.

the sun - Will Earth lose the Moon before the Sun goes into supernova?

As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop.

The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer.

---

The current lunar recession rate is 3.82 cm/second, which is outside your one to three centimeters per second window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that
$$frac {da}{dt} = (text{some boring constant})frac k Q frac 1 {a^{11/2}}$$
Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction.

That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further.

If something even funkier happens and the Moon recedes at any average rate of four centimeters per second over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years.

---

What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies.

Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere.

Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill.

What are the mechanism binding histone code with alternative splicing?

According to literature, histone marks seem to be affecting splicing both directly and indirectly. 

The indirect regulation appears to be due to biophysical hindrance, where a certain chromatin structure would cause Pol-II slowdown, allowing the splicing machinery to do its job. 

The direct regulation seems to be essentially dependent on H3K36me3 and H3K4me1, marks that are recognized by MRG15, which in turn recruits the splicing regulator polypyrimidine tract-binding protein to splice sites.

Ref: Luco, R. F. et al. Regulation of alternative splicing by histone modifications. Science 4 Feb 2010

orbit - State vectors of "interesting" multiple stars

I'd like to show a demonstration of Runge Kutta integration of real systems using examples of interesting multiple stars, where "interesting" means you can see through a small telescope that they are at least double. I've selected three:

1. Gamma Andromedae (Almach)


2. Alpha Geminorum (Castor)


3. Zeta Ursae Majoris (Mizar and Alcor)

How can I find a state vector (set of relative positions and velocities) - or something similar - for these multiple star systems? An open access link (for everyone) would be great, but a journal reference is fine as well if that's what it takes.

Of course the periods are long and there will be uncertainty - I'd just like to use whatever is known and available, for this demonstration.

gravity - gravitational time dilation multiple sources

I wondered how time would be dilated for an object in the middle of 2 black holes, however
https://en.wikipedia.org/wiki/Gravitational_time_dilation only provides a formula for one source of gravity

i tried approximating it by multiplying the time dilation factors, however when i used 1 single source with twice the mass the results differed from 2 objects with half the mass quite drastically

the examples:
mass $1,00E+036$kg distance $3000000000$m
time:$0,7105905766$

mass $2,00E+036$kg distance $3000000000$m
time:$0,099387802$

approximation: $0,7105905766*0,7105905766=0,5049389676
$
which is off by a factor of 5 and would be even more off the closer the distance gets to the schwarzschild radius

so how can the time dilation of multiple objects be approximated ?

amateur observing - Calculate Local Sidereal Time

A highly authoritative source is Explanatory Supplement to the Astronomical Almanac 3rd ed., Urban & Seidelmann, 2013.

Chapter 1 gives the introductory explanations, without all the refinements if you want millisecond accuracy. Page 12 explains that

local sidreal time = Greenwich sidereal time + east longitude

(This book assumes the reader is able to convert between angle notation and time notation whenever necessary.)

If accuracy of 0.1 second is sufficient, the formulas given by the US Naval Observatory should do the job. Note that for subsecond accuracy you will have to determine UT1, which can be up to 0.9 second different from the commonly available UTC. You will have to look up the difference each time you do the calculation because the difference cannot be predicted more than about 6 months into the future.

black hole - Conversion of matter into antimatter

It would be more accurate to say that a black hole eventually turns whatever falls in it into equal quantities of matter and antimatter.

The difference between matter and antimatter is defined by some conserved (or approximately conserved) quantum numbers. For example, matter baryons and quarks have positive baryon number, while antibaryons and antiquarks have negative baryon number. Leptons (e.g., electrons, neutrinos) and their antiparticles have a corresponding lepton number instead.

Semiclassically, a black hole evaporates through Hawking radiation. For astrophysically large black holes, this will be overwhelmingly be in terms of photons, which are their own antiparticles, but once the black hole evaporates to a small enough size, it will also radiate massive particles as well. Because there are no long-range forces that couple to baryon number or lepton number, a black hole is under no obligation to conserve them, and thus should radiate equal amounts of matter and antimatter, regardless of what fell into it.

This is unlike, say, energy-momentum, electric charge, and angular momentum, the presence of which makes a difference in the long-range gravitational and/or electromagnetic fields. Thus the total amount of charge the black hole will radiate should match the total amount that fell into it, etc.

Unless, of course, quantum gravity gives us more surprises, which is definitely possible.

How much is the zoom of Sun-Pluto eclipse image

The "raw" LORRI image is 1024x1024 pixels. The field of view is about 0.29°, see section 3.6 of this paper, or abstract of this paper.
Pluto is a little less than 250 pixels in the raw image, corresponding to 0.29° x 250/1024 = 4'15'', well consistent with your estimate.

zoology - What happens to snakes that swallow rodents as a whole?

Owing to their narrow structure, no chewing mechanism nor limbs, a snake has to swallow its prey as a whole. In fact, their success and diversity is partly due to their ability to swallow prey relatively larger than their own body size. So inevitably, they need to process the parts like teeth, bones, exoskeleton and things of such nature.

Snake saliva not only eases swallowing by lubrication, but also
contains powerful enzymes to break down tissues and even egg shells.
But snakes generally cannot digest keratin (claws, hair) or chitin
(arthropod exoskeleton). Many prey are covered with tough hide and if
a snake relied solely on its digestive juices, it would take a long to
time to get through to the nutrients. So this is speeded up by snake
venom. Venom not only immobilises prey but also starts digesting the
prey from the inside. A study showed that when a fer-de-lance
(Bothrops asper) is deprived of its venom it took 12 days to digest
a rat instead of the usual 2-3 days.

However, there is no consensus among different theories arguing the utility of a snake's venom. The above mentioned opinion is based on digestion hypothesis of venom utility.

Please refer to this excellent review and a summary of a snake's eating habits

Why wasn't there any oxygen in the beginning of the earth?

A complete answer gets long and above my pay-grade, but when the earth and solar system were young, it's unlikely that the Earth had much atmosphere at all. The young sun was hot enough to burn off most gases and ices in the inner solar-system and when the earth coalesced it was largely metals and silicates. The oceans and atmosphere came later by comet and asteroid belt impacts, and, as Jeff Y points out, after that it's just heat and chemistry. Oxygen or O2 binds so easily with other molecules, like Methane or Hydrogen or Iron that it goes away pretty quickly until those are used up.

Like O2, N2 wasn't nearly as present in the young atmosphere as it is today either. More on that here: http://sciexplorer.blogspot.com/2012/01/earths-atmosphere-part-4-evolution-of.html The four most common early ices/gases on earth were CH4, CO2, H2O and NH3. Over time, the NH3 and CH4 get used up by bacteria or chemical processes, lightning, volcanoes, UV light from the sun, etc. The H20 and CO2 are much more stable, though photosynthesis did slowly pull most of the CO2 from the atmosphere, returning O2.

Would a plant need light if the chemicals gained by photosynthesis were given through the roots or as a foliar spray?

I doubt this would work for the vast majority of plants. I think it would cause root rot as the microorganisms in the soil would out-compete the plant. Also the transport systems of the plant might not be efficient in this direction.

I mean might there be a plant somewhere where this might work? Sure. Fungi that grow in the dark would be a lot like such plants, so its biologically possible. There might be a primitive plant that doesn't need its chloroplast to be active to live. I've never heard of one and wikipedia is not helping here...

gravity - Could we verify the structure of a black hole by observing an orbiting object?

The comments from @userLTK, and @Lacklub are correct.

Lets assume there is an object of radius $R$ and mass $M$, from a Newtonian point of view, if you are at another radius $r$, such that $r > R$, then there is no difference in the gravitational field experience by an object at $r$ if the mass spread across a shell of radius $R$ or if its concentrated anywhere between $r=0$ and $r=R$. GR doesn't do much to change this, and in fact if $r >>R$, then the result is of course exactly the same.

Now to black holes, all taken from from Wiki https://en.wikipedia.org/wiki/Black_hole

Physical Properties
"The simplest static black holes have mass but neither electric charge nor angular momentum...This means that there is no observable difference between the gravitational field of such a black hole and that of any other spherical object of the same mass."

So basically if you're outside the event horizon (not that'd you know where it was) you're experience with the black hole is the same as with a planet or star of that mass.

Again from same wiki article:

Singularity
"At the center of a black hole, as described by general relativity, lies a gravitational singularity, a region where the spacetime curvature becomes infinite... It can also be shown that the singular region contains all the mass of the black hole solution".

The "it can be shown" references page 204 of Carroll, Sean M. (2004). Spacetime and Geometry. I don't have my copy with me right now so I can't look it up, but I would say I remember reading at in Carroll back in the day.

Finally, let me just add that, once something passes into the event horizon of a black hole, there's no getting back. So we really have to ask ourselves how any information about the "stuff" inside the event horizon would come to us? I like your idea of getting some indirect evidence and perhaps gravitational waves will shed some light on this topic but I don't suppose there is any known way to getting direct access to anything beyond the event horizon.

neuroscience - Density of neurons/cells in the mouse brain

The newest and most accurate method (far more accurate than older extrapolating/manual counting methods (Stereology) and yielding some surprising results) to estimate number/density of neurons/cells in brains is Isotropic Fractionator to my knowledge. Using this keyword you find some recent papers, comparing different brain areas (cerebral cortex, cerebellum,...) among rodents:

Combining our estimates of total cell number and percentage of NeuN-containing nuclei in each brain region, we find that adult rat cortex contains ∼80 million cells, 40% of which (∼30 million) are neurons. In comparison, rat cerebellum contains more than twice as many cells (∼170 million), >80% of which are neurons (Table 1, top). Therefore, the adult rat brain contains almost five times as many neurons in the cerebellum (∼140 million) than in the cerebral cortex. When all of the brain regions are taken into consideration, the cerebellum thus accounts for more than one-half of the cells and ∼70% of all of the neurons of the entire rat brain (Table 1, bottom). Overall, we estimate that of all of the cells in the adult rat brain, 60%, or 200.13 ± 21.17 million, are neurons. Glial cells, therefore, contrary to common belief, are not the most numerous cell type in the rat brain.