Simple Question about Constellation - Astronomy

Taken at face value it's not correct that points on the equator are visible for exactly six months.

For example, Orion is on the celestial equator. It is true that Orion will be above the horizon for half the day (12 hours), but during the summer when it is above the horizon, so is the sun, so it can't be seen.

But Orion is visible from Northern Europe between the middle of August (when it rises shortly before the sun) and the middle of April (when it sets shortly after dusk) That is rather more than six months. Exact dates can't be given as visibility during twilight depends on factors like the clarity of the atmosphere, and stars don't wink off as the sun rises, they fade out.

On the other hand, if you always observe at the same time then points on the celestial equator will be above the horizon for six months of the year, but just being above the horizon doesn't make it visible, when close to the horizon the thicker atmosphere (and more prosaically, trees and buildings) will make a constellation effectively obscured.

It is true that the north celestial pole is always above the horizon, as are many of the constellations around it. From my location that includes Cassiopeia, Ursa minor, most of Ursa major.

More generally, for Northern Hemisphere observers, the further north a point is, the longer it will remain visible.

neuroscience - If a non-mammal is starved of oxygen for some time, how long would it take the animal's neurons to die?

Medically, recovery of brain function after 3 minutes of oxygen deprivation at body temperature is rare.

Further down here in 1 it indicates that at lower temperatures the time can be much longer. At 13C, the record is 80 minutes for people. Animals cooled to 0C have a record of recovery of three hours! Not sure how much like themselves they feel after that though...

What would happen if a rogue planet hit one of the planets in our Solar System?

This really depends on the size, density and velocity the rogue planet was traveling before the collision and which planet it hit. As a side note, the possibility of this happening is quite rare.

If that rogue planet have sufficient velocity and under perfect conditions (such as the angle of collision), yes there is a possibility that it could reach the sun (bringing along with the 'planet' it collided say Pluto). The effect of that rogue planet would be enough to slightly disrupt other planets' orbit if it passes close and could disrupt the entire solar system. Pluto may not just fall to the sun based on the the pull of gravity, but also the velocity from the rouge planet.

Firstly addressing the 'sun' problem, yes if a sizable planet punches in to the sun, it'll create something like a mass coronal ejection which in return would bombard the solar system with harmful rays. Already this has affected the solar system.

If a sizable rouge planet hit one of the rocky planets such as Mars, large chunks of debris would come shooting out of the collision, a very high chance that a few pieces would land on Earth potentially creating another mass extinction; similarly to the other rocky planets such as Venus and Mercury. Similar to one of the solar system's gas giants if say, another gas giant hit it.

To summaries this, even a rogue planet entering our system would have already changed a few things not to mention a collision.

orbit - What would happen if an astronaut orbiting Earth was exposed to radiation?

I guess that depends on how much radiation they're getting. I should also point out that astronauts are not 100% protected from radiation, as discussed in this question. One of the most common affects on actual astronauts is the development of cataracts. They can also receive damage to their nervous system over time, among other effects.

But it appears that you're asking what would happen to a person who was exposed to a huge amount of radiation, such as if they went into space with a suit that did little to protect from radiation. I think to answer that, we just have to look at cases where humans have actually been exposed to huge amounts of radiation.

One interesting example that comes to mind is Anatoli Bugorski who was a Russian particle physicist who accidentally had his head inside a particle accelerator when it was on. His head was subjected to huge amounts of radiation. The results of this are described as

The left half of Bugorski's face swelled up beyond recognition and,
over the next several days, started peeling off, revealing the path
that the proton beam (moving near the speed of light) had burned
through parts of his face, his bone and the brain tissue underneath.
As it was believed that he had received far in excess of a fatal dose
of radiation, Bugorski was taken to a clinic in Moscow where the
doctors could observe his expected demise. However, Bugorski survived
and even completed his Ph.D. There was virtually no damage to his
intellectual capacity, but the fatigue of mental work increased
markedly. Bugorski completely lost hearing in the left ear and only
a constant, unpleasant internal noise remained. The left half of his
face was paralyzed due to the destruction of nerves.1 He was able to
function well, except for the fact that he had occasional complex
partial seizures and rare tonic-clonic seizures.

Now imagine what could happen if a person were to be exposed in a similar fashion to highly energetic particles in space, but throughout their body and not just their face. I expect that a person could not survive such an exposure and that their body would be destroyed. Even if they survived, it is highly likely they would live with crippling medical conditions and very likely develop cancer later in life.

black hole - Where is the flaw in this hypothetical concept of escaping event horizon using another's blackhole flyby?

In short, it doesn't work.

the event horizon is defined as the region of space which you can't escape from, If you are between two black holes, and you can escape then by definition you are not inside the event horizon. In the case when there are two black holes, the black hole is not a Schwazchild black hole (which is a solution of Einstein's equations for a single point, uncharged non-spinning mass) and the event horizon is not spherical. To determine the shape of the event horizon for two masses you need to use numerical solutions of the equations, you can do this yourself with the Einstein Toolkit, but the computing requirements are very high. A supercomputer is required to get very far.

In the exact situation you draw, it is likely that the release of graviational radiation would lead to a merger rather than a pass but you would need to run the simulation to confirm this.

Now you can't just remain stationary inside the event horizon. If you started inside it you can't stay in the same place and wait for a the pass of the other black hole to pull you out. Inside the event horizon spacetime is flowing towards the singularity and even if the shape of the event horizon is changed by the passing of another black hole, nothing passes from inside to out.

stellar evolution - Why do PMS stars on the Hayashi track remain at a constant temperature while they contract?

You are indeed correct. The Hayashi track is one of the surprisingly simple laws in stellar-evolution, where the luminosity is purely dependent on the radius, and not the surface temperature as it remains constant. The reason for this is a bit complicated, but it basically boils down to, as you state, the opacity properties of the hydrogen ion, that becomes especially important for contracting stars with an unusually thin atmosphere, like those on the Hayashi track. Keep in mind that the temperature/luminosity curve is not completely vertical, just very steep.

Special Relativity as Applied to an Interstellar Starship?

In the first example, you certainly could use the calculation of special relativity to measure the difference in the time shown on a clock which was making the high speed journey and one which was at rest. The details may make the calculation trickier, but we can use the flat minowsky space of the clock at rest to do the calculation, so it is relatively straightforward. [see what I did there?]. It would be a factor of about 1.6 so the journey that would take a little over 67 years would, on board ship, seem to take 67*1.6 = 41 years. (I have ignored the time spent accelerating and decelerating, in order to keep the maths simple. Earth would observe the ship leaving 30years after it did (light distance) and arriving after 67 years, however, since the light must travel further, the ship would appear to arrive after 67+50-30 years = 87 years.

For the multiple journey the calculations are just longer. The direction of travel doesn't affect time dialation, only the speed. So it would again be straightforward to calculate the total time dialation.

It would be quite possible for Earth to communicate with the ship. But those the messages would not arrive for up to 50 years. Messages would be certainly out of date.

Could gravity waves be used to confirm the existence of a ninth planet?

The maths on Wikipedia give a way of calculating the the amplitude of graviational waves.

One detail is that for the waves to be detected by a ligo apparatus, one needs to be far enough from the source of the waves. The distance required depends on the frequency. For planet 9, if it has orbit of 10000 years, one would need to be more than $2500pi$ light years distant. That is nearly 10000 light years away. This value is "R"

Then we can use the formula:
$$h_{+} =-frac{1}{R}, frac{G^2}{c^4}, frac{4m_1 m_2}{r}$$

using the above value of R, the known values of the gravitation constant and speed of light, and the masses of the sun and a hypothesised value for the mass and semi-major axis of planet 9.

That gives a strain of $10^{-32}$. That is far far far below what is detectable. It also would have a frequency of 10000 years or so. So you would be trying to detect an oscillation comparable to the plank length, you would need at least 10000 years of observation, and your detector has to be built on the other side of the galaxy.

Not possible.

solar system - What is the effect of the axial precession on the orientation of the planet's orbits?

When talking about the longitude of ascending node you must be very careful to define the reference plane that you are using. As you state, due to the Earth's axial precession, the First Point of Aries moves along the ecliptic over ~26,000 years. This is because the celestial equator is slowly precessing about the ecliptic.

Now, in the case of the orbital elements of other planets, the longitude of ascending node is again with respect to the First Point of Aries. As your intuition suggests, the orbits of the outer planets won't be influenced by the precession of the Earth's equinox. This means that the longitudes of ascending node will not be constant over long time periods. If we suppose that the physical orbits are fixed (i.e., the planets are not perturbing each other), the change in the longitudes of ascending node will be entirely due to the Earth's precession. When you take both these changes into account, you will find that they cancel each other out and the position of these orbits remains fixed with respect to the distant stars.

You can see in the orbital elements provided by NASA that they are given with respect to the J2000.0 epoch. In other words, the orbital elements are provided for a single point in time, and to get the orbital elements today you must take into account the Earth's precession.

http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt

plant physiology - Why dont "growing degree days" have units of degrees Celsius (or do they)?

From Wikipedia:

Growing degree days (GDD), also called growing degree units (GDUs), are a heuristic tool in phenology. GDD are a measure of heat accumulation used by horticulturists, gardeners, and farmers to predict plant and pest development rates such as the date that a flower will bloom or a crop reach maturity.

GDD is calculated as $frac{T_{min}-T_{max}}{2}- T_{base}$ - so a quick check of the units would indicate that they are in the same units as T, i.e., degrees C. However, they are usually presented without units.

Why is this?

microbiology - Are there animal models for Clostridium difficile that better replicate human infection than hamsters?

So I'm looking for some information on the infectious dose necessary to colonize a human with Clostridium difficile. There's no human challenge studies, and since it's not a foodborne pathogen, little we can get from outbreak data.

Which leaves me looking at animal models, to at least get the shape of the probability of infection based on a given dose. But the papers on hamsters makes it appear that they're infected rather easily, and that this infection is particularly severe.

Is there another animal model that more closely mimics the infection dynamics of C. difficile in humans?

zoology - Will the "frog in boiling water not jumping out" work on warm blooded animals

There is the famous saying about a frog that is put in water that are slowly boiling will not jump out until it's too late. I realize it happens because of the frog's cold blood that adjusts to the temperature change until it's too late.
Will this work as well on an animal with warm blood, or will the animal jump out ones it gets hot, but not hot enough to cause permanent damage?

Edit due to comments:

Cold blooded animals, like the frog, can adjust easier to their surrounding when the surrounding changes temperature gradually as their body temperature adjusts with the surroundings. Will the same happen to a warm bodied creature, and will it find it easier to adjust to the surrounding if the change is gradual.

For example, I don't like saunas, when I get in them I have problem breathing and I can't stand the heat, which means that I can be in a sauna for a very short time until I have had enough and get out. Will putting me in a sauna at room temperature and then turning it on to gradually begin to heat until its destination temperature will allow me to fill more comfortable and stay longer inside once it's at the normal sauna temperature?

biochemistry - What role does a protein's size have on protein-protein interactions?

Protein interactions occur mostly (if not all) through residues that are on the "surface" and exposed to the milieu in which they exist, be it cytoplasmic or extracellular. So, a naive thought would be to guess that a greater surface area means a greater swathe of exposed regions and probable interacting parts. One protein can bind many others, even simultaneously, through different such interaction surfaces. However, other factors also determine whether a given set of two proteins interact, most important being where they localize in a given cell. If one protein is nuclear but the other on the plasma membrane, it is highly unlikely that they ever interact even if thermodynamics highly favors complex formation. Interaction between two proteins also depends on their respective conformations, which can further be regulated by events such as post translational modifications (eg., phosphorylation by a kinase), bound nucleotide state (eg., GTP vs GDP in a G protein). Last, scaffolds can also sequester and organize protein clusters together spatially and increase the probability of them interacting. All put together, I think it is highly unlikely that just looking at 3D surface area of a protein will say much about its "interactability" with other proteins.

milky way - visualisation of galaxy density

For your experiment, you can take it 2D, so let's ingnore the thickness of Milky Way.

So, we have 400 billions stars(the real number is between 100B-400B) in a circle with 100k ly(the real number is between 100k-160k).

The area of the circle is pi*50k*50k ~= 7800M sqly = 7.8 billions sqly.

We have then ~51 randomly stars per square ly.

But you do not sting the Milky Way with your stick, but cut through it.

If you swipe through the middle, you have 51*100k stars in a width of 1 ly.

So, you have a chance to hit a star per 1/5.100.000 ly which means 1.855.000km.

But the diameter of sun(of your stick) is 1.395.000km. Taking diameter of sun as medium diameter of a star(this is not true), you have a chance of 1.4/1.8 = 77% to hit a star. If you go with the "low density" numbers, 100B stars and 160k ly diameter, you'll get ~31%.

My conclusion is you often hit a star.

flare - Unknown moving objects

Last night, I used my new Sony WX500 camera to get long exposure pictures of the sky from my backyard. I used the "Heavens Above" app to find out when satellite flares would happen, and I even got a great picture of the ISS passing overhead at roughly -2mag. I was able to identify many flares that I had gotten pictures of when reviewing the pictures today. However, this where it gets odd. Some moving "things" didn't match what was supposed to show, as according to the app. Some were accidentally caught in the pictures I took, and I am sure they aren't any flare that the app showed me. What could they be? Shooting stars? Forgotten/broken satellites?

the sun - Can black holes collide with the Sun? In that case what will happen?

Yes, in theory. But;

No, this is very unlikely to happen during the life of the sun. If it did happen it would be Bad.

Many black holes are much more massive than the sun, even though they are much smaller. If a black hole were to enter the sun, some the matter that forms the sun would form into a accretion disc around the black hole and eventually fall into the black hole. The rest of the sun would be utterly disrupted by the massive release of energy as matter falls towards the black hole.

The solar system would be wrecked. The Earth, even if it survived intact, would not be able to support life.

Fortunately black holes are rare, and space is BIG. The chances of a black hole coming anywhere near the sun in the next billion years or so is very very small. There are more pressing things to worry about.

the sun - Math for calculating the terrestrial longitude directly under the sun with time

I'm trying to calculate the longitude on the earth where it's noon at some time. (That is, the longitude which is coplanar with the plane defined by the sun and the earth's axis.)

Here is my python code:

from math import sin, cos, tan, atan, pi
def sun_longitude(when):
    """Given time in ms since 1/1/1970, return the longitude the sun is over at that moment"""
    # https://en.wikipedia.org/wiki/Position_of_the_Sun
    jdn = 2440587.5 + when / (1000.0 * 3600 * 24)
    n = jdn - 2451545.0 # 1/1/2000
    L = (280.460 + 0.9856474 * n) % 360.0
    g = (357.528 + 0.9856004 * n) % 360.0
    degtorad = 2.0 * pi / 360.0
    lambda_ = L + 1.915 * sin(g * degtorad) + 0.020 * sin(2 * g * degtorad)

    # https://en.wikipedia.org/wiki/Axial_tilt#Short_term
    T = n / (365 * 100.0) # julian centuries from 2000
    # ε = 23° 26′ 21.406″ − 46.836769″ T − 0.0001831″ T2
    epsilon = 23.4392794 - 0.780612817 * T - 5.0861e-8 * T * T;
    alpha = atan(cos(epsilon * degtorad) * tan(lambda_ * degtorad)) / degtorad

    # https://en.wikipedia.org/wiki/Sidereal_time
    # https://en.wikipedia.org/wiki/Hour_angle
    GMST = (18.697374558 + 24.06570982441908 * n) % 24.0
    print "jdn = {jdn}, n = {n}, L = {L}, g = {g}, lambda_ = {lambda_}, T = {T}, epsilon = {epsilon}, alpha = {alpha}, GMST = {GMST}".format(**locals())
    LHA = GMST * 360/24 + alpha;

    return LHA

from datetime import datetime
def sun_long_from_str(whenstr):
    when = datetime.strptime(whenstr, "%Y-%m-%d %H:%M")
    secs = (when - datetime(1970, 1, 1)).total_seconds()
    return sun_longitude(secs * 1000.0)

I'm aware that I'm not correcting for the correct quandrant in figuring alpha, but I have errors elsewhere; for example, for noon GMT Jan 1, 2000, I'm getting 279° rather than close to 0, which is what I'd expect.

This isn't giving me correct answers, and I'm at a bit of a loss for how to debug it. Can anyone find my mistakes or point me to some reasonable sample code or worked-through example for this?

Because, once I get the algorithm right, I'll be reimplementing for an embedded device, I can't just use an implementing package, and I haven't found a library which is straightforward enough for me to understand how I can implement this specific question.

Thanks for a couple of error corrections. Now, when I run it, I get the following for these examples:

2000-01-01 12:00:
  jdn = 2451545.0, n = 0.0, L = 280.46, g = 357.528, lambda_ = 280.375680197, T = 0.0, epsilon = 23.4392794, alpha = -78.7141369122, GMST = 18.697374558
  result: 201.74648145775257

2000-03-20 07:35:
  jdn = 2451623.81597, n = 78.815972222, L = 358.144758099, g = 75.2090537484, lambda_ = 360.006175505, T = 0.00215934170471, epsilon = 23.4375937902, alpha = 0.00566598789588, GMST = 19.4596915825
  291.9010397253072

As you can see, the first query (noon on Jan 1, 2000) now has a correct Julian day number. At noon GMT, I would expect the sun to be above a longitude near 0, so 201 is incorrect.

The second query is the time of the spring equinox in 2000, which I expected to result in zeroing out the sidereal part of the calculation, but it does not.

I attempted to consult the NASA HORIZONS web interface for ephemerides for the sun on 1/1/2000 at noon GMT, for both "Geocentric" and Greenwich locations, but I don't know how to evaluate the result and compare it to my work above.

Geocentric:

 Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O
**************************************************************************************************************
 2000-Jan-01 12:00     18 45 09.36 -23 01 59.7 -26.78 -10.59 0.98332762653520  -0.0127281   0.0000 /?   0.0000

Greenwich:

 Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O
**************************************************************************************************************
 2000-Jan-01 12:00 *m  18 45 09.36 -23 02 08.3 -26.78 -10.59 0.98331613178086  -0.0166655   0.0000 /?   0.0000

Thanks again for any help you can provide.

[Edited to correct Julian day calculation and use of degrees vs. radians for alpha and to add examples]

pharmacology - Why is methylcellulose used in pharmaceuticals?

It is a filler/binding agent. Thus MC belongs in the context of a drug to the group of so called excipients. The study of the best suitable excipients (as a tradeoff of factors such as cost, and ease of approving the drug) is called
galenics.

Methyl cellulose is also present in your toothpaste and some of the foods your eat and in the construction industry. It can be considered to be metabolically inert for humans, but can serve as a matrix for enhancing bacterial adhesion and biofilm formation.

The gradual smudging and staining of your bathroom washbowl may be largely attributable to MC and bacteria.

MC is a polymer that is sold as dry powder of various mean chain lengths. It is hydrophilic and can retain large volumes of water.

atmosphere - What is it that distinguishes one atmospheric layer from another?

What is it that defines where one layer stops and another starts?

Temperature. More specifically, it's whether temperature rises or falls with increasing altitude.

In the troposphere, temperature generally decreases with increasing altitude, at an average rate of 6.4 °C/km (the environmental lapse rate). This decrease stops at the tropopause, the boundary between the troposphere and the stratosphere. The ozone layer is in the stratosphere, making temperatures in the stratosphere increase with increasing altitude. While this boundary is a bit fuzzy, it is still very real. It takes an incredibly strong thunderstorm (think hurricanes, storms powerful enough to spawn tornados, and very tall and strong storms in the Inter-Tropical Convergence Zone) to penetrate that boundary.

Temperatures in the stratosphere stop rising at the stratopause, the very fuzzy boundary between the stratosphere and the mesosphere. Like the troposphere, temperatures in the mesosphere fall (and fall very sharply) with increasing altitude. Thermodynamics makes the boundary between the stratosphere and mesosphere very different (and not nearly as clear) as the boundary between the troposphere and the mesosphere.

The boundary between the mesosphere and the thermosphere is similar to the boundary between the troposphere and stratosphere. Temperatures rise with increasing altitude in the thermosphere. This change from falling temperatures in the mesosphere to rising temperatures in the thermosphere makes for a rather stable boundary.

Something else happens near that boundary between the mesosphere and the thermosphere. Long-lived gases are fairly well-mixed in the troposphere, stratosphere, and mesosphere thanks to turbulence. Gases in the thermosphere and exosphere act more like a bunch of individual particles rather than a gas. Unlike the dense layers below, gases in the thermosphere and exosphere are differentiated. The makeup tends toward lighter and lighter particles (e.g., helium and hydrogen) with increasing altitude. Eventually, all one finds are hydrogen and helium. These are the gases that escape from the atmosphere.

assay development - Ideal low-protein binding membranes

I'm looking through the literature on the topic. It seems like hydrophilic PVDF membranes are ideal for low-protein binding but it also sounds like Regenerated Cellulose is an appropriate substitution.

Looking at one resource (http://www.zellnet.com/elispotplatescolumn/)

When PVDF filter plates were introduced, some investigators chose to use PVDF plates and some continued to use NC. The reasons for choosing one plate (membrane) over the other are highly varied and won’t be addressed in detail here. The fact that some laboratories and individual researchers feel strongly that one membrane is superior to the other runs contrary to the large body of Western blotting experience: Despite some clear-cut differences in how each of the membranes can be used, there is essentially no reported difference in terms of detection sensitivity or signal to noise on NC versus PVDF in the Western Blotting application. This having been said, it is clear is that the two membranes and their properties are quite different.

The assay that I am looking at in particular is protein/nanoparticle binding to mammalian cells. I am looking for a balance that will enable me to remove free protein without fouling the membrane and/or killing my cells.

human biology - What causes a 'stuffy' or 'runny' nose when you have a cold?

There are different causes and mechanisms behind runny and stuffy nose.
I cover them separately below.

Runny nose

My professor of pathoanatomy and pathophysiology says that the correct answer here is serous inflammation to the runny nose.

Serous inflammation

• Marked exudation by relatively thin fluid


• Cold is caused by Acute respiratory viral infections

Acute viral respiratory infection

• Exudate is very watery and clear - serous inflammation.


• At 3rd day, whitish mucose excretion - catarrhal inflammation.


• At 5th day, greenish colour - a lot of neutrophils and mucose.

The other answers mostly discusses about the mechanisms (secondary), but not the cause itself.
The most upvoted answer describes factors affecting the inflammation but not explicitly mention serous inflammation and its factors.

Stuffy nose

Congestion in nose is a passive and local process resulting from reduced outflow of blood from a nasal tissue.
The nasal epithelium is very thin which makes it vulnerable to flu and cold.
The viral particle passes through the membrane (often because of scarring) and causes the change in the vascular permeability of the surrounding big vein.

Mechanisms of stuffy nose are covered partially by Ram and Aleadam:

• local edema


• because of changed vascular permeability of large veins surrounding the nose


• bradykinin


• histamine (cetiritzin)


• counteracted by epinephrine

which respond against viral particles and cause

• atrophy


• dryness of nose.

Stuffy nose is about the reduced outflow of blood from a nasal tissue i.e. isolated venous obstruction.
Even small obstruction (scarring) can lead to it.
I know no safe systemic medication that can prevent the change of vascular permeability of the nasal big vein.
Local steroids like Nasonex improve the recovery time by reducing the congestion when taken during long term - however, it cannot prevent the vascular permeability change of the large vein.

I have an intuition that the the nasal venous congestion can be relieved by the systemic lymphatic drainage massage by special movements involving

• abdomen (attention to deep breathing with abdomen and exhale too)


• supraclaviculus


• under medial occipitalis


• under auricles

which help to drain the accumulated fluid from the nasal tissues.
I know no substitute or medication to replace the stimulation of lymphatic drainage.

Prophylactics for the symptoms

You can oil your membranes but to oil them all even at the back side is difficult - even more difficult if have to have them oiled continuously.
So there remains some membranes unoiled.
I think it takes rather much time for the venous side to recover.
Therefore, you feel you feel stuffy nose long time after the runny nose has ended.
Local oiling is not enough during cold and allergic times for many patients.

Some medicines for Prophylactics

• Xylometazoline (Olynth) - cause atrophy and dryness, but good for stuffy nose


• Quixx - sea water


• Eucalyptus and vitamin E (Coldises spray for instance)

Are there events in universe which we receive first from their neutrino's instead of their photons?

Yes. When a core collapse supernova happens, it takes a few hours for the shock wave to reach the surface and "break out" (when we first "see" it). Meanwhile, the neutrinos can escape nearly directly once they reach the top of the core.

Of course, as you wrote, photons travel slightly faster, so if the SN is far enough away, the photons might overtake these neutrinos. However, I have not done the calculation to see if the universe is big enough.

amateur observing - Visibility of human activity on the moon

The video is hilariously wrong.

However, the principle of laser ranging is more or less right, and it does require the reflectors left behind by the astronauts on the Moon. It's just that the physics and technology involved are far beyond just pointing a toy laser at the Moon.

Project APOLLO (Apache Point Observatory Lunar Laser-ranging Operation) is actually doing this.

http://physics.ucsd.edu/~tmurphy/apollo/basics.html

You need a fairly large telescope to start with - both for collimating the outgoing light pulse, and for receiving as much of the reflection as possible. APOLLO uses a 3.5 meter telescope at the Apache Point observatory.

You need a laser that can generate a high-energy light pulse that is very short. The pulse is injected into the telescope's optical train and sent to the Moon. This is not a laser pointer; it's a high power Q-switched laser for research, a device the size of a refrigerator.

On the receiving end, you need a very good detector also plugged into the telescope. Of the many, many photons sent to the Moon in the pulse, only between 1 and 5 photons make it back down to the detector. You need a detector that can tell the extra 1 to 5 photons from the background noise of light coming from the Moon.

Using this technique, the distance from Earth to Moon can be measured with very high precision.

This is the APOLLO system in action:

---

Regarding observing human artifacts on the Moon with terrestrial telescopes, I wish it was doable, but it's not. Again the YouTube video is wrong.

The biggest telescope accessible to amateurs has an aperture (diameter) in the range of 1 meter or a little bit larger (the aperture of the largest amateur-owned telescopes currently). The resolving power of a telescope (the size of finest details discernible) depends on aperture - if aperture is measured in mm and the resolving power in measured in arcseconds, then the formula is:

resolving power = 100 / aperture

So a 1 meter telescope has a resolving power of 0.1 arcseconds.

The distance from Earth to Moon is 384000 km (3.84 * 10^8 meters). With a resolving power of 0.1 arcseconds, the size of the smallest detail discernible on the Moon is:

detail size = distance_from_Earth_to_Moon * arctan(resolving power)

or

detail size = 3.84 * 10^8 * arctan(0.1 arcsec) = 186 meters

Anything smaller than 186 meters would be blurred into a single dot. There's nothing we've done on the Moon that's as big as that. It's not possible to see traces of human activity with amateur-level telescopes, even with extremely large meter-class dobsonian telescopes. Even with professional telescopes, we just don't have the aperture yet to resolve that kind of detail.

However, satellites in orbit around the Moon, such as the LRO, were able to image traces of the Apollo missions. That's because they are a lot closer to the site.

http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html

dust - What is the distribution of organic compounds in the Milky Way Galaxy?

The answer depends on which you are talking about (you mentioned both organic compounds and organic matter, which are two completely different things).

Organic compounds we do not have a clear definition of, but is most agreed upon to contain carbon atoms, either on their own (C), bonded with at least one hydrogen atom (C-H), or bonded with at least one other carbon atom (C-C). Compounds such as Methane (C-H4), Ethanol (C2-H6-O), and Glucose (C6-H12-O6) are textbook examples of organic compounds. Methane has been discovered on several objects within our solar system, including Mars, Venus, and Saturn. Methane has also been discovered on the exoplanet HD 189733b (though it is the only example we have of methane on an exoplanet), and often exists in interstellar clouds, showing that it is quite common throughout our galaxy.

Organic matter refers to matter composed of organic compounds, as part of an organism. While we are searching for organic matter as proof of extraterrestrial life, we have yet to discover any organic matter or clear evidence for such matter outside of our own planet.

comets - How did water get on Earth

The key finding why we think Earth's water came from Asteroids (big rocks) and not Comets (small rocks) is the Deuterium/Hydrogen ratio that we can measure in several sources.
When a star forms, it has an initial value of D/H that came from the nucleosynthesis in its progenitor nebular / star.

In a protoplanetary disc, as dust grows to rocks grow to planets your nebular gas will be trapped with initial D/H ratios in gas giant atmospheres. But the way to get Water onto Asteroids and Comets (those don't have substantial mass to retain atmospheres themselves!) is sublimation and maybe adsorption.
The latter two processes are strongly sensitive to the gas-mass and therefore different D/H ratios from the protosolar ones are expected.
And in fact we find them to be different:

This was in the news lately, as ESA managed to touch down and measure 67P's D/H ratio which gave another hint on the asteroidic origin of water on earth.

This finding, however does not resolve the question:

• There maybe other isotopical tracers like D/H that give a hint on the history of water.


• There could have been a decrease in D/H and subsequential increase again, or the other way round.


• We know that the protosolar nebula must have had massive amounts of water (Oxygen is the third-most abundand element in normal stellar fusion!), so again the question is, why did Earth retain so much water, while the lighter planets Venus and Mars retained the heavier element $CO_2$
  ...

I could continue this for quite some time, but the bottom line is: We only have hints, not definitive answers.

To refer to the rest of your question: A hot plume in a colder atmosphere will not necesarily leave the whole gravitational potential well.