cell biology - Does electricity cause damage on a cellular level?

Regarding the moss or lichen on the third rail on British train lines: there is no current passing through the moss/lichen as they are not completing a circuit. They are just sitting on one connector. If another connector passed over them, completing the circuit across them, then a current might pass across them for a very short time. But I doubt they grow on the top of the rail (which is the contact part in the UK system) as that is constantly being polished smooth by the contacts from trains sliding across it.

Regarding cell damage: Apart from fibrillation in animals, and burns caused by Joule heating, electricity does cause cellular damage.

At low frequencies (<10kHz), electricity disrupts cell membranes and makes them much more permeable (we actually harness this when we use electroporation to transform bacteria). All organisms rely on electrochemical potential differences across membranes for their metabolism (Berry, 2002). The exact effect is dependent on the previous electrical state of the cells, for example the electrical potential difference across the membrane, and on the surface area to volume ratio of the cell (greater volume relative to surface area leads to more disruption). In any case, extreme electroporation can cause solutes to flow in or out of a cell, and generally disrupt the balance of solute concentrations and cause organelles and other bodies to move out of a cell. When the electrical stimulation stops, the contents are then fixed in a disrupted state. The cell then has to expend enormous amounts of ATP using ion channels and transport proteins in an attempt to reinstate the necessary chemiosmotic potentials, and in doing so exhausts the entire ATP supply and goes into biochemical arrest (no metabolism occurs), which is when a cell is dead. Dead cells break apart because there are is no maintenance occuring.

At higher frequencies (10-100kHz) proteins become permanently denatured. Many proteins carry charges which give them an overall polarity. When placed in an electric field, the proteins reorient themselves and will undergo conformational change to achieve the optimum dipole moment in the direction of the field. Ion channels and pumps are particularly sensitive to these disruptions (since their charges are crucial to their function).

Rather than provide lots of references, there is one excellent review from which I drew all this information, and which you should read for more of the physical detail (Lee et al., 2000).

References:

• Berry, S. (2002) The Chemical Basis of Membrane Bioenergetics. Journal of Molecular Evolution. [Online] 54 (5), 595–613. Available from: doi:10.1007/s00239-001-0056-3 [Accessed: 9 February 2012].


• Lee, R.C., Zhang, D. & Hannig, J. (2000) Biophysical injury mechanisms in electrical shock trauma. Annual Review of Biomedical Engineering. [Online] 2 (1), 477–509. Available from: doi:10.1146/annurev.bioeng.2.1.477 [Accessed: 9 February 2012].

exoplanet - Has the Kepler data been examined for light curves due to phase changes of non-transitting hot Jupiters and other?

I found this article http://www.epj-conferences.org/articles/epjconf/pdf/2011/01/epjconf_ohp2010_03005.pdf from 2011 announcing that two dozen secondary eclipses of "hot Jupters" have been seen in Kepler data. I even think I have seen a light curve showing phase changes of a transiting planet. This makes me think that there should be evidence in the Kepler data for non-transiting planets since the amplitude of the light curve due to planetary phase changes shouldn't be much different from that of a secondary eclipse as long as there is a fairly high inclination (say over 60 degrees).

Is anybody aware of such an examination of the data?

star systems - How many planets does Omicron Persei have?

Currently http://exoplanet.eu/ (which lists more exoplanets than the official NASA archive) lists no known planets of Omicron Persei. So currently we know of none, it is outside of the Kepler field so has no Kepler candidates either.

Though just because none are known that does not mean it has none, it may be that their mass and or orbital plane is such that it is very difficult to identify planets or no one has looked with a suitable instrument.

gravity - Existence of gravitons?

The metric describes the curvature of the space. For space around a massive object this is the Schwarzchild metric

$$
ds^2 = -left(1-frac{r_s}{r}right)dt^2 + left(1-frac{r_s}{r}right)^{-1}dr^2 + r^2(dtheta^2 + sin^2theta dphi^2)
$$

Clearly, if $r>>r_s$ this looks like

$$
ds^2=-dt^2+dr^2+r^2(dtheta^2 + sin^2theta dphi^2)
$$
which is the metric for flat space. So effectively the space gets flatter and flatter at a rate of $1/r^2$, which is the inverse square that you're looking for.

But where does the Schwarzchild metric come from? Without getting into the gritty maths, it can be proven that it is the unique metric that possesses spherical symmetry, without which nothing would make much sense. This is called Birkhoff's theorem.

I want to talk about where gravitons come from, but first lets talk about curvature.

If you want to measure the curvature of a space one way of doing it is to move in some closed loop, ending up back where you started. If the space is curved though, you won't be facing the same direction (this idea is called parallel transport)

Let's say we're parallel-transporting a tangent vector, as in the picture. We get a tangent vector from the derivative at a point (a slighly special derivative called a covariant derivative, because the space is curved). Let's take the tangent vector and move forwards then left. And we try again this time moving left then forwards. We end up the same point both ways, but like in the picture, the derivatives will be different in some way. We summarise this up with a commutator (where $D$ is the covariant derivative) like so

$$
[D_mu,D_nu] = D_mu D_nu - D_nu D_muneq 0
$$
It basically means "doing it one way is not the same as doing it the other".

Now let's back up a tiny bit and talk about how electromagnetism and other forces are typically discussed, using quantum field theory.

We describe the theory in terms of a Lagrangian, for a fermion (like an electron) it looks like this

$$
mathcal{L}=barpsi(igamma^mu D_mu-m)psi
$$

If I take the field $psi$ and give it a transformation
$$
psitopsi'=e^{ixi(x)}psi
$$
then the Lagrangian will remain unchanged. This type of transformation belongs to a group called $U(1)$. We say that the Lagrangian possesses $U(1)$ symmetry. Notice that this $D_mu$ is in there again? Its the same thing, a covariant derivative, here in QED as well. We can try taking a commutator again

$$
[D_mu,D_nu] = -iF_{munu}psi
$$
where
$$
F_{munu} = partial_mu A_nu - partial_nu A_mu
$$

From this we form the complete QED (the quantum theory of electrodynamics) Lagrangian
$$
mathcal{L}=barpsi(igamma^mu D_mu-m)psi-frac{1}{4}F_{munu}F^{munu}
$$

Don't get bogged down in the maths. The point is very simple. See the $A_mu$? Its a new field, we had to introduce it to make things work. In QED this field corresponds to a photon (particles are quanta of a field, like a small bump in the field). We had to introduce it becuase we had curvature. How do I know we have curvature? Becuase the covariant derivatives do not commute, just like in GR, above. This time though, the curvature is not of phyical space, its of an abstract object called the $U(1)$ gauge bundle.

So you're totally on the right track when you say that other forces may curve space. Its nice that gravity curves space-time, its very physical and easy to imagine, for the other forces its not so simple to picture, even though its fundamentally the same.

If you want the full picture of Einstein's gravity you do some maths and arrive at something called the Einstein-Hilbert action (an action is just an integral over a Lagrangian), one tidy object that sums up the whole theory

$$
S=int R sqrt{g} d^4x
$$
where $R$ comes (more or less) from the commutator of covariant derivatives we saw at the top. When talking about QED I brushed over the fact that its a quantum theory (it is). This EH action, however, does not describe a quantum theory. So, you might say, lets make it one! Hold on a second though, because it doesn't actually work. The problem is something called renormalisability - QED is renormalisable, GR is not. This is the root of the incompatibility between GR and quantum field theory. If we could carry out the resultant qunatum particle would be a graviton. You're right to doubt their existance as they haven't yet been observed, however...

We saw QED, which desribes particles of light, photons. They are quantised. Then we saw how in many ways GR and QED are very similar. We can't properly quantise GR but if we could we would have gravitons, exactly like photons popped out in QED. The duality between QED (and other gauge theories, QCD, etc) is clear, which leads a lot of people to believe that probably should have gravitons, even if they have not yet been observed, nor consistently formulated.

A note on other theories

There are many theories where gravitons are present from first principles without the problems of renormalisability, string theory or supergravity for instance.

A note on errors in the above

Sorry, I'm tired and rambling. Please point them out if you find them!

How far apart is the dust in the Sombrero Galaxy's dust lane?

The sizes of dust particles varies enormously, and thus the density of dust particles depend upon which sizes we are talking about. The dust size distribution can be described by a power law with a slope of roughly $-3.5$; that is, the density of 0.1 µm particles of is $10^{-3.5}$ times that of 0.01 µm particles, or roughly 3000 times smaller.

Moreover, the discussion is complicated by the fact that we don't really have a formal definition of what is dust. But one "definition" takes the minumum size as a conglomeration of a few molecules, and the maximum size as "what has time to grow in a formation process" (these dust particles can later "stick together" and grow to form pebbles, rocks, asteroids, and planets. That's why any definition will be arbitrary).

In a dust-dense region of the interstellar medium, this results in a typical mean density of roughly one dust particle per cubic centimeter.

evolution - Why did the process of sleep evolve in many animals? What is its evolutionary advantage?

I found this paper by Benington and Heller that expands on the previously mentioned theory of sleep as a mechanism to renew metabolism. They hypothesise that sleep is necessary to replenish glycogen stores (mainly within astrocytes) in the brain. These stores are normally used to supplement blood glucose due to the high energy demands of the brain.

It is suggested that this may also result in the manifestation of feeling sleepy as a consequence of the exhaustion of glycogen supplies in specific small areas of the brain. Brief and localised depletion of glycogen stores mean that cells are operating with less energy than they normally have to work with. This causes an increase in synthesis of adenosine from the breakdown of AMP. The paper maintains that the increased levels of adenosine are detected by adenosine receptors which then triggers or increases (as seen on EEG scans) the feeling of the need for sleep:

In NREM sleep (which may lend some support to the theory as NREM sleep accounts for 80% of sleep and is most physiologically different to waking), this glycogen is most efficiently replaced. During NREM sleep the release of glycogenolysis inducing neurotransmitters is reduced, allowing glycogen-synthase to predominate and glycogen levels to be restored. However, these same neurotransmitters are key in the processing sensory stimuli (by tonically depolarising neurons in the sensory cortex). Therefore glycogen replenishment will always be associated with a (strongly) reduced response to stimuli.

This leads to finally answering the question as to why sleep has an evolutionary advantage, which I will quote verbatim to maintain the authors momentum:

Glycogen replenishment during waking would be maladaptive because it would impair the organism's ability to to process and respond to sensory stimuli. Sleep has therefore evolved as a state where animals retreat to a safe environment, behaviour is suppressed and glycogen stores are replenished.

How can we tell how many exoplanets a star has?

What you are describing is a basic signal processing problem. The doppler shift that one observes is due to the motion of the star in the system around the system's centre of mass. The star will be influenced by the gravitational pull of each of the planets in that system, each of which exerts a gravitational pull that increases with the mass of the planet and decreases with the orbital radius of the planet.

The overall motion of the star will be the sum of the effects of all the planets. Importantly, the effect of each planet will have its own amplitude and will be periodic with a period equal to the orbital period of that planet.

Let's imagine that each planet is in a circular orbit (elliptical orbits are more complicated, but the principle is the same). Each planet would cause a circular motion in the star about the centre of mass of the planet-star pair, leading to an observable doppler signal which has the form of a sine wave with a period equal to the orbital period of the planet. The amplitude of that signal will increase with the mass of the planet and increases with decreasing orbital radius.

The overall signal is the sum over all the planets in the system. Fortunately the decomposition of this signal back into its individual components is a well-trodden problem in physics, electronics and many other fields and is known as Fourier Analysis. Whether you can successfully recover the original signals from each of the planets depends on how long you observed the system (ideally you want to observe for longer than the longest orbital period) and the amplitude of the signals compared with the noise in your observations.

In general it is easier to recover high-mass planets with short orbital periods and more difficult to recover low-mass planets with long orbital periods.

The image below might be helpful. It shows the track of the solar system centre of mass compared with centre of the Sun over a period of several decades. Notice how the Sun executes a complex trajectory (with respect to the solar system centre of mass) that is mainly caused by the orbit of Jupiter, but then there are smaller, superimposed, signals caused by the smaller planets. In principle, if you observed for longer than the period of Neptune and had a detector which gave perfect measurements, you could reconstruct how many planets there were in the solar system, what their orbital periods were (and then from Kepler's 3rd law, what the planet-star sepration was) and what their masses were (multiplied by the inclination of their orbits with respect to the line of sight of observation, which is generally an unknown in doppler measurements).

In terms of what we could currently see if we observed the Sun as a star: basically we would (assuming we observed for 20 years) detect Jupiter quite easily with a doppler amplitude of about 13 m/s. We would also see that there was a drift in Jupiter's signal due to the influence of Saturn, but we would have to observe for >Saturn's orbital period in order to confirm the presence of Saturn, its orbital period and mass. The inner planets produce an amplitude that is too small to be visible using the technology currently available. e.g. The Earth would produce a doppler wobble of amplitude $<8$ cm/s, but the current precision of doppler measurements is limited to about 50 cm/s.

The doppler amplitudes in m/s due to each of the planets (assuming we view them edge on) are:

Mercury <0.01
Venus 0.08
Earth 0.08
Mars <0.01
Jupiter 12.5
Saturn 2.6
Uranus 0.28
Neptune 0.26

Thus with current technology, only Jupiter and Saturn are detectable.

Below I simulate what the doppler signal due to these two planets would be. I hope you can see that the overall signal consists of the superposition of two sinusoidal signals with different periods and amplitudes. Myriad computational tools are available to do the fourier decomposition to establish these.

cosmology - Light that travels eternally because of inflation?

What happens with electromagentic waves that are emitted into
intergalactic space, but reach not any object because of inflation? Lets assume that photon A and photon B are emitted. Photon B hits some object, photon A enters a gigantic void or simply has nothing in its path, and heads for the horizon, so to say.

Relativity tells that A and B do not "feel" time passing. From our point of view B travels a finite time, while in the frame of B the object is reached instantly. However, since A will be eternally underway, what is going to happen? We have kind of a zero*infinity travelling time in the frame of A.

Do our models make statements about this? Or is there some explanation

astrophotography - Are there are any photographs online that approximate what the Milky Way looks like to the unaided eye?

It may be difficult to find proper photos that emulate this, both because given the tools and the subject, most people would opt for longer exposures, and also because of the general difficulty in emulating how our brain processes visual information into an image, especially in low-light or high contrast situations like this. I think the photo at the bottom of this article may be a decent recreation of it, though it's a bit fuzzy and not so wide. Note: There are star trails forming in the image, which is indicative of it being a longer exposure. The image (unless cropped) doesn't appear to be from a wide-angle lens - I would guess somewhere in the 35-50mm range. Perhaps a 10 secondish exposure, but my judgement is more based on result than technique.

What you can see will depend a lot upon the conditions of the night. You're probably not expecting a lot of light pollution, but you also have to keep in mind that your eyes will adjust to light as well. A cell phone or a full moon can be enough to prevent you from being able to see well. If conditions are perfect - little to no light pollution, no light from cars, flashlights, etc, a new moon, low humidity/no clouds, etc, and the milky way is in a good position, you will be able to pick out the galaxy quite well. You should be able to notice both the high concentration of stars as well as the 'milky' lighter shade caused by the many stars too dim or distant to appear bright.

orbit - Determining distance from semi-major axis and eccentricity

Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:

$frac{T^2}{a^3} = frac{4pi^2}{G(m_E + m_b)} approx frac{4pi^2}{Gm_E},$

where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):

$M(t) = frac{2pi}{T}t$.

Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)

$M = E - esin E$

and then use the following equation to derive the true anonaly $nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:

$ cos nu = frac{cos E - e}{1 - ecos E}$
and
$ sin nu = frac{sqrt{1-e^2}sin E}{1 - ecos E}$.

The distance from the Earth is just given by the orbit equation

$ r = frac{a(1-e^2)}{1 + ecos nu}$.

If I'm not wrong with calculation, it should be:

$ T = 9364 s = 2.6 h.$

$ M(90min) = 207.60°$

$ E(90min) = 203.11°$

$ nu(90min) = 198.95°$

$ r = 11'366 km$

To get the covered distance you should calculate the line integral of the orbit equation.

molecular biology - How do you knockout an E. coli gene without disrupting the rest of the gene cluster?

Let's say that we have gene A, followed by gene B so that they overlap with 100 bases. If you want to knock out only gene A, you can do the following:

• PCR gene A and clone it into a vector


• PCR gene B and clone it into another vector


• Insert a antibiotic resistance marker at the overlapping region of gene A and B
  
  • now you don't have functional gene A and gene B
  
  


• Transform your cells only with the vector that contains gene B

Now you have disrupted only gene A but you have a functional gene B.

the sun - How can the sun burn without oxygen?

As you are suspecting, the sun burns in a different sense, not by chemical reaction with oxygen.

Atoms consist of a tiny, heavy nucleus, surrounded by an almost empty space, populated by electrons. Burning by chemical reaction with oxygen doesn't change the nucleus of atoms, but takes place in the hull of atoms: Atoms may assemble to form molecules; electrons change their orbitals (the way they surround the nucleus), and release some energy as heat.

Atomic nuclei are (positively) electrically charged, and repell each other.
But if small nuclei, like those of hydrogen atoms, come close together, they can fuse and form a larger nucleus. This nuclear fusion of hydrogen to helium (in this case) releases much energy, more even than fission of uranium in a nuclear power plant. The notion "burning" is used sometimes for reactions of atomic nuclei, too, if they release energy as heat.

To overcome the electrostatic repulsion of hydrogen nuclei, high pressure and temperature are needed. These conditions occur in the core of the Sun.

gravitational waves - Binary stars system ultimate fate

Well, yes apparently that is how it works and a Nobel prize was won in 1993 by Hulse and Taylor for demonstrating this phenomena in the object PSR B1913+16.

PSR B1916+16 is a pulsar (a rapidly rotating neutron star with a strong magnetic field that emits beamed radiation along its magnetic axis). The pulsar provides an extremely precise ticking clock with which to study the motion of stars in a binary system. The pulsar is in orbit with a companion that is also a neutron star (though not an observable pulsar). The orbit is elliptical with a period of 7.75 hours.

The period of the orbits is decreasing. That is, the two stars are gradually getting closer together. This effect is a consequence of general relativity, that predicts that such massive objects that are so close to each other will act as a source of gravitational waves. The gravitational waves have not (yet) been directly observed, but GR makes a prediction for how much energy they carry away and hence at what rate the orbital period changes.

The rate of decrease in the distance between two bodies is given by
$$frac{dr}{dt} = -frac{64G^3}{5c^5}frac{(m_1 m_2)(m_1+m_2)}{r^3}$$
and leads to a merger between the two objects in a time of
$$t = frac{5c^5}{256G^3}frac{r^4}{(m_1 m_2)(m_1+m_2)}, $$
where $m_1$ and $m_2$ are the masses of the two orbiting bodies, $G$ is the gravitational constant and $c$ is the speed of light.

In slightly more friendly units
$$ frac{dr}{dt} = 7.8times10^{-19} frac{(M_1 M_2)(M_1+M_2)}{r^3} au/yr,$$
where $r$ is in astronomical units and the masses are in solar masses, and
$$ t = 3.2times10^{17} frac{R^4}{(M_1 M_2)(M_1+M_2)} yr. $$

Thus you can see that the effect gets much stronger (and the merger times shorter) if you have two high mass objects (it depends on the product of the masses) with a small orbital separations.

For "normal" stars you can never get them closer than their stellar radii and generally means that the effect is too small to be important. However, compact objects (white dwarfs and neutron stars) can be brought much closer together and this is why the effect has been seen there. The Hulse-Taylor binary components
have masses of $sim 1.4 M_{odot}$ and are separated by an average of about 2 million km (0.013 au). Inserting this into the formula above give a merger time of 1.7 billion years. A more accurate calculation that takes account of the elliptical orbit gives 300 million years.

However, if you made the neutron stars orbit with a separation of 0.1 au (more typical for a normal stellar binary system), the merger timescale would increases to well beyond the current age of the universe.

genetics - Predicting progeny of recessive mutations using recombination

First, the recombination equation is:

cM = recombinants/(recombinants AND parentals) * 100

Let's assign the following genotypes for clarity's sake:

RR = black bristles
Rr = black bristles
rr = red bristles
SS = pebbly eyes
Ss = pebbly eyes
ss = shiny eyes

F1 cross: rrSS x RRss = RrSs (all progeny)
F2 cross: RrSs (F1) x rrss = ?

You would have the following gametes:

Parent 1, RrSs = RS, Rs, rS, rs
Parent 2, rrss = rs, rs, rs, rs

Draw your punnet square and you get the following genotypes of the offspring.

So your GENOTYPES of your F2 cross will be RrSs, Rrss, rrSs, rrss

Let's translate those to phenotypes.

RrSs: black pebbly (parental)
Rrss: black shiny (recombinant)
rrSs: red pebbly (recombinant)
rrss: red shiny (parental)

Now let's look at recombination frequency.

15cM = (recombinants/(recombinants+parentals)*100
recombinants/(parentals+recombinants) = 15/100

This tells you for every 15 recombinants, you have 85 parentals. If you scale this up to 1000, that's 150 recombinants for every 850 parentals.

Your 150 recombinants comes from 75 black/shiny and 75 red/pebbly.

Your 850 parentals ceoms from 425 black/pebbly and 425 red/shiny.

Anyways that was my attempt which appears to be directly opposite to what your answer key says. Did you copy it down correctly?

Are there real uses of abstract algebra in astrophysics?

It is known that a physicist must have, at least in general, a good level in mathematics in order to research and understand other people's works. Linear algebra is a good example, as ODEs and calculus.

I suppose that some astrophysicists have deep knowledge of abstract mathematics and there are some applications of it (for example, I've been told that there is a connection between topology and cosmology). I wondered if there are some field in Astrophysics in which abstract algebra (i.e. fields, rings, category theory...) has been applied.

human anatomy - Mechanism of Tear Gas (chloropicrin/Trichloro methane)

The general belief is that these agents are irritants which do not have a very specific effect. That is to say that there is no specific cell or biological function that they set off, but that they irritate the cells, so the cells respond by putting up general defenses to insulate themselves from the noxious chemicals.

Mucous and tears wash away the compounds. I'm sure there are all sorts of things going on, but its hard to be more specific. For instance, onions have many sulfhydryl (-SH) compounds, many many of them. In particular propanethial s-oxide causes the tears we get when we cut an onion. This compound is very reactive and reacts with all sorts of proteins and disrupts the membrane, etc. Its perhaps an example of a specific irritant that is so reactive that it creates tears.

An example of a specific reaction would perhaps be capsaicin. There is a specific protein that is a heat receptor and the chemical capsaicin activates it, causing the feeling of heat/warmth.

Tear gas and mustard gas appear to fall into the former category, which are chemicals so reactive that they just tear into the cell, damaging all sorts of things in there.

the sun - Do we still use the term "astronomical unit" nowadays?

Certainly. Astronomical Unit is probably one of the most used distance units used in astronomy. It is of course only used when discussing the distances within a stellar system, such as the distances between the Sun and its planets or other bodies in the solar system. It is also used to discuss distances in other stellar systems, e.g. the distances between stars and their (exo)planets.

It is easier to get a sense of the distances between bodies in the solar system if you use astronomical units. It doesn't say much when we say that the distance between Neptune and the Sun is $5times10^{-4}$ light year. You would have to know the distance between Earth and the Sun ($1.5times10^{-5}$ light year), and then calculate the difference (i.e. 30 times = 30AU) to get a sense of scale.

As a side note, in professional astronomy, the light year is hardly ever used. Professional astronomers use the parsec (=3.26 light year), kilo parsec (kpc), megaparsec (Mpc), and Gigaparsec (Gpc) to specify distances between stars, galaxies, and galaxy clusters.

orbit - What is the largest object on which the Yarkovsky effect has been observed?

The Yarkovsky effect is responsible for changes in the rotation and orbit of some celestial bodies, most notably asteroids. It has been measured on asteroids, such as 6489 Golevka and 1999 RQ36.

What is the largest object on which the Yarkovsky effect has been observed? Scholarpedia has an excellent article and list of asteroids, but the list isn't necessarily complete, and I don't know if the effect has been detected on other (non-asteroid) bodies.

singularity - Do black hole singularities actually merge?

Infinities are hard to bend the mind around, but in this case, the merge is not impossible. Yes, the distance between them must reach zero in order to make the black holes merge, but the rate of which energy is lost to gravitational waves also increases when they get close to each other.

We are therefore dealing with a $frac{mathrm{infinity}}{mathrm{infinity}}$ problem, where a finite limit may, and in this case does exist.

Do not forget your $lim$'s!

Calculate right ascension in Zenith at this moment in my location

I've been searching on Internet about how to calculate which RA is in zenith at a given location in a given time but I haven't find anything (or maybe I don't know how to search).

Using my star finder I have not draw any conclusion.

How can I calculate which RA is on zenith on a given location in a given time?

amateur observing - What are some night sky objects I could see with my Celestron UpClose 20x50 Porro Binocular?

I bought this binocular because I've read countless times that one should start with a pair of binoculars before diving into telescopes. I've seen details of the moon and I can notice Venus round shape on some nights. I also saw Jupiter and it's two larger moons, as faint as the faintest stars, but this was on a beach trip so the sky was clearer than what I'm used to.

I live on a place with a 7 or 6 on the Bortle Scale, is there something else I should try to see?

milky way - What parameters determine whether galaxies colliding will result in a merger or a hit and run?

The main factor is the velocity of the encounter. The higher the relative velocity between the two galaxies, the easier it is for them to pass through each other without being slowed down enough for a proper merger to take place, and without being strongly distorted by the encounter. (In a very high speed encounter, the two galaxies will spend almost no time close enough for tidal forces to be effective.)

In clusters of galaxies, where the typical velocities are around 1000 km/s, mergers are rare; in small groups, where the velocities are around 100 km/s, mergers are more common. So the Milky Way and Andromeda are pretty much destined to merge.

During a merger it's possible for stars in the outer parts of the galaxies to be ejected by tidal forces, but the central regions of both galaxies should definitely merge. Thus, the supermassive black holes (one in each central region of the original galaxies) will end up in the center, too. (Unless you have something exotic like a merger where one of the galaxies already has an unmarked binary SMBH; then you might get one of the three SMBHs ejected via a 3-body interaction. But I suspect that's pretty unlikely.)

galaxy - Why can't we see distant galaxies with the naked eye?

Not at all a dumb question, but actually you can see distant galaxies with the naked eye. From the northern hemisphere, the Andromeda Galaxy, our biggest neighboring galaxy, is visible if you know where to look, and is at a reasonably dark place. From the southern hemisphere, the two smaller, but nearer, irregular galaxies called the Small and Large Magellanic Clouds are visible.

The reason that more distant galaxies are not visible, is due to the inverse-square law: As the light particles (photons) recede from the galaxy (or any other light source), they are distributed over an ever-increasing surface. That means that a detector (e.g. your eye) of a given area will catch less photons, the farther it is placed from the galaxy. The law says that if in a time interval Δt on average it detects, say, 8 photons at a distance D, then in the same time interval, at a distance 2D it will detect 8/2² = 2 photons. At a distance of 4D, it will detect 8/4² = 0.5 photons. Or, equivalently, it will need twice the time to detect a single photon.

The bottom line is that in principle you can see the very distant galaxies, but the photons are so few and arrive so rarely, that your eye is not a good enough detector. The benefit of a telescope is that 1) it has a larger area than your eye, and 2) you can put a camera at its focal point instead of your eye and take a picture with a large exposure time, i.e. increasing the Δt.

astrophysics - Distance of a planet to the star?

3 parsecs is the distance from us (Earth, Sun etc) to the star.

At a distance of 1 parsec, the apparent movement of a star wrt the background, as the Earth orbits on its 1 AU orbit is 1 arc second.
So, from 1 parsec distant, the earth would move by 1 arcsec each 6 months. It woukd have an apparevt angular radiusof 0.5 arc sec.

Now this star is 3 parsecs distant, so an orbit with radius 1au would appear to have a angular radius 3 times smaller (here I am approximating, it is valid because the angle is small) or 1/6 arc sec. The planet in the question has a smaller orbit, 6/10 of the Earth, or 0.6 au

The convenient units, and tge the small angles mean that we can avoid trionometry in the solution.