Formation and capture of Mars' moons, Phobos and Deimos

Mars's moons have the appearance of captured asteroids. How could this capture have taken place. An asteroid in solar orbit would need to loose some momentum to be captured. The gas Giants can capture new moons through 3 body interactions between the asteroid and their existing moons, but Mars doesn't have this issue. Airbreaking would lead to impact, not capture.

An answer here suggests that the moons could have formed recently (millions of years, not billions) after an impact ejected a substantial amount of material in Martian orbit. What is the evidence for this? Is there an impact crater of the right size, and age? Were both moons formed together?

Can a human land on Mercury?

No sir/madam. An astronaut or a cosmonaut cannot walk on Mercury's surface. The reason is plainly simple. During daytime, its temperature surges up to 430 degree C making it 2nd hottest planet after Venus. Whereas, night time temperatures are as low as -180 degree C.

Despite of having ice caps at its polar regions which might seem attractive, its totally not feasible to land there and then mine the frozen ice to make up a base.

For a much more descriptive read, kindly refer this link : What it would be to live on Mercury

big bang theory - Center of the universe

Krotanix : nobody really knows the answer to this. We have good evidence that the universe is expanding, and we extrapolate this back to the big bang. Then some people say the universe was once the size of a grapefruit. See for example this. Since WMAP evidence in 2013 suggested the universe is flat, some people have started saying the observable universe was once the size of a grapefruit. And that the universe was always infinite. But this is a non-sequitur, and a non-answer.

Yes, it looks as if 13.8 billion years space had a much higher density and started expanding. And it may have had a certain volume, so space may have already existed. But nobody knows for sure.

Re your follow-on question, nobody has discovered hollowness or emptiness anywhere in the centre of the universe. The expansion is thought to be like that of a raisin cake rising in an oven. There's no big hole in the middle.

kinematics - highspeed black holes or neutron stars on (almost) head-on collision course and kinetic energy

I think it's a fun question, and I can answer some of it.

If their event horizons don't "touch" I'd assume their courses will
change considerably but it would be a normal flyby - right?

While the relativistic speeds and time dilation might change things a bit, a flyby is still a flyby. They would either fly past each other, the dense objects would change direction as a result, but the combined kinetic energy and momentum would remain constant.

2nd option is they enter into orbit around each other. Pretty much the same with any 2 massive objects passing near each other.

But what happens if their event horizons "touch", if their event
horizons "overlap" just a few meters?

With classic (non Kerr) black holes, if the Event Horizons touch the black holes have no option but to merge. The Event Horizon is directly proportional to the radius, so when the 2 event Horizons touch it, as that happens, no longer 2 black holes but 1 black hole. The mass of the 2 objects combined has an event horizon that circles both of them.

For Kerr black holes this question is more complicated and I'm not sure but I think it's the same. once the event horizons touch it's no longer 2 separate black holes but one bigger one.

On the one hand both objects have a massive kinetic energy with
opposite directional vectors. And on the other hand nothing can escape
below the gravitational radius.

When you're talking about Event Horizons, Gravity always wins. Kinetic energy cannot exceed the speed of light, so it loses.

Where does all the kinetic energy go ? Would the two black holes
extremely deform, like creating a long string before they collapse
into spherical form?

A long string doesn't make any sense. A temporarily non-round event horizon as the holes merge seems possible, but think of a merger of 2 stretched balls, not a string. Each black hole is centered around it's "singularity", those shapes might warp some as the holes collide or spiral into each other. As an FYI, direct collisions would be rare. Spiraling into each other is much more common.

The kinetic energy gets absorbed inside the black hole. It can't, by definition of the black hole, escape. Total momentum is likely preserved.

And what if in a similar scenario one of the two objects is a neutron
star. Would the neutron star be ripped apart when it scrapes the black
hole on high-speed flyby ?

A few points to consider regarding a black hole. Anything that falls within a black hole's photon-sphere is unlikely to ever escape, so one of the magical distances to consider is 1.5 times the radius of the black hole where all orbits are destined to fall into the black hole. If the Black hole and Neutron Star are of similar size and the Neutron Star passes close to the event horizon and relativistic speed, then there's an interesting scenario where some of the Neutron Star (at close to the speed of light) would be outside the photon-sphere and have the momentum to escape and some of it would not. Neutron Stars are bound together very tightly, but the tidal forces would be very strong and it's possible that some of the Neutron Star material could get torn off in this hypothetical scenario. Would the entire neutron star be torn apart? Maybe. It's hard to say. The Math in that hypothetical is way above my pay-grade.

If the Neutron star is much smaller than the black hole, then it's likely inside the photon-sphere and it will spiral into the black hole, perhaps breaking apart in the process into an accretion disk, perhaps not, but it will get absorbed into the black hole if it's entirely within the photon-sphere, even at nearly the speed of light.

The problem with this hypothetical is that it's pretty much impossible. Objects the size of stars don't accelerate at nearly the speed of light relative to other near by stellar mass objects. What actually happens is described in the first answer to your question, the two dense objects would either have sufficient relative velocity to fly past each other or they would get caught into a mutual orbit that would likely lead to a merger. In the actual universe, a close to the event horizon fly-by between a Neutron Star and a black hole would result in a merger of the two. In theory, with enough kinetic energy the Neutron Star could fly close to and still fly past the black hole, but that much relative velocity probably never happens.

People smarter than me should feel free to correct anything I got wrong, but I thought this one was a fun one to answer.

naming - Why was this asteroid (4864 Nimoy) chosen to be named after Leonard Nimoy?

I'm not entirely sure, but I would guess that it's simply because the discoverer suggested it. After a few steps in the process of characterizing an asteroid's orbit, the discoverer gets to suggest a name (within reason) to a committee. The full guidelines are available at the relevant IAU webpage. (It's longer than is worth reproducing here.)

Note that there are plenty of "celebrity" asteroids, drawn from a wide range of human endeavours: Adamsmith, Balzac, Beethoven, Megryan, etc. If anything, I'm surprised that an asteroid wasn't named Nimoy sooner. There's already a Mr. Spock!

Size of Sun as seen on Pluto

New Horizons encountered Pluto when it was about 33.5 au from the Sun.

The solar diameter is $1.39times 10^{9}$ m.

An astronomical unit is $1.496times 10^{11}$ m.

The angle subtended by the Sun at Pluto (in radians) is the ratio of the diameter of the Sun to its distance from Pluto. Converting to degrees, minutes, seconds, gives 0.01589 degrees, or 0.95 arcminutes.
The atmosphere of Pluto is way too thin (much thinner than the Earth's) to make any significant difference to this at all.

The diameter of Pluto is $2.37times 10^6$ m. In order for Pluto and the Sun to subtend a similar angle in an image, you just need to move far enough away from Pluto to make the same angle.

Thus
$$ D_P/R_P simeq (D_P + D_{PS})/R_S,$$
where $D_P$ would be the distance from New Horizons to Pluto, $D_{PS}$ the distance from Pluto to the Sun, $R_P$ the radius of Pluto and $R_S$ the radius of the Sun. Because I will show that $D_P ll D_{PS}$ we can say that
$$ D_P simeq frac{R_P}{R_S} D_{PS} simeq 0.057 au$$.

New Horizons is travelling at about 14.5 km/s. So a simple bit of Maths tells us it would takes about 6.5 days to get into this position.

The picture I think you are talking about was taken on July 15th 2015 only 7 hours after the flyby. The Sun and Pluto do not share the same angular size in this photo. You are not seeing the Sun at all in this picture. It is eclipsed by Pluto completely. All that is being seen is sunlight refracted (or more likely scattered) through Pluto's atmosphere. The Sun acts more-or-less like a point source of light at more-or-less infinite distance.

The angle that the light needs to deviate in order to be seen through the atmosphere is about $0.19$ degrees (the half-angle subtended by Pluto at the Sun [negligible] plus the angle subtended by Pluto at New Horizons after 7 hours). If we were to model the atmosphere as a glass prism. The deviation angle is $delta simeq (n-1) alpha$, where $n$ is the refractive index and $alpha$ is the opening angle of the prism. If you let $alpha$ be quite large (let's say 1 radian) to simulate rays coming through the atmosphere, then $n = 1 + delta$, where $delta$ is in radians. So $nsimeq 1.003$. NB: This assumes there is no scattering in Pluto's atmosphere, but I suspect that is not true and that my estimate of the refractive index is a large overestimate. In fact I'm sure it is an overestimate because this is larger than the refractive index of air on Earth. i.e. I am sure that what we are seeing is scattered light, not refracted light.

spectra - Why does spectral class and U-V colour correlate slightly differently for main sequence, giant and supergiant stars?

In this diagram the $x$-axis is spectral type for stars and the $y$-axis is B-V colour.

Taking Series 1 as the Main Sequence, Series 2 as Giants and Series 3 as Super-giants. Why the spectral class for all these stars differ from the other ones?

If we were to say differences, for instance, how would one express that information?

distances - How can I find out which known stars lie within or close to M8

I'm looking for a way to find stars within a certain region of space, the Lagoona Nebula in this case, based on what we currently know. Are there some databases that allow you to make such queries?

Please note that I'm not looking for stars that can be seen as being near M8 from earth, but for a list of known stars that are thought to actually be near or part of M8 itself.

How can the observable universe be so small if there are so many stars in it?

If you know the total number $N$ of galaxies, and you know the size $V$ of the (observable) Universe, then you can calculate the number density $n = N/V$ and mean distance (of the order $1/n^{1/3}$) between galaxies, and convince yourself that it actually works out fine.

However, the reason we know how many galaxies are in the observable Universe in the first place is the other way round:

1. Observe an average number density of galaxies (which of course varies quite a lot from dense clusters to voids).




2. Calculate the size of the Universe by integrating the Friedmann equation (using as input observed values of the densities of the constituents of the Universe; dark energy, dark matter, gas, stars, radiation, etc.).




3. Finally, multiply the two numbers to get the total number.

Note that when observing the number density, we must not look too far away from the local Universe, since this means looking back in time to a period where the number density was different from present-day's value.

big bang theory - Is nucleosynthesis responsible for the expansion of the universe?

Is it just a coincidence that the two major expansionary periods occur close to periods of nucleosynthesis?

1. Big Bang nucleosynthesis occurred very close to the inflationary period.




2. Supernova and stellar nucleosynthesis created heavier elements as the stars died, and the universe underwent increased acceleration.

Further thoughts:

The metaphor for the expansion of the universe is often described as an inflating balloon, but could it be equally valid to think of expansion as matter falling into its own gravity well, away from infinity minus other matter within its gravity field. The effect being that matter radiates space itself as gravity pulls itself away from infinity. Just as heavier elements create a stronger gravity field, heavier elements would accelerate away from infinity minus other matter within its gravity field, at a faster rate.

Eventually, on a cosmic scale, atomic decay would release the confined energy and the energy would return to infinity, creating the appearance of a contracting universe that restarts the Big Bang. If this idea is reality it would create a geodesically complete cosmology.

can we use connect a long pipe with space station and use vacuum for fuel transfer?

A pump can lift water about thirty feet. That's the height where the column of water in the pipe weighs the same as an equivalent column of air (and that's why deep wells use submersible pumps: they push the water up from below, instead of pulling from above and relying on atmospheric pressure to provide the lift). While fuel would be less dense than water, it's still much more dense than air, and the pressure of the atmosphere simply doesn't provide enough force to raise that column of fuel very far.

What visible differences do Geminids have to other meteors

Well done on your observations, I've been completely clouded out.

You are quite right, the Geminids are noted for a variety of colours, and a good number of fireballs, though most observers note a predominiance of yellow meteors This is partly due to their composition and density (coming from a peculiar "rock comet") and partly to their velocity (35km/s) which is fairly fast.

There is a good chance of seeing some good trails with this shower. Brighter meteors and most fireballs will leave a trail, and the Geminids have more of these than most other showers.

senescence - How do caspase proteins kill a cell?

Caspase do not directly kill the cell, but rather activate a process known as apoptosis, or programmed cell death. The programmed part is there to distinguish it from other types of cell death, such as necrosis, which are more aspecific death processes.

Coming back to caspases, they are a series of proteasis, that can activate in cascade in response to a pro-apoptotic signal.

To simplify matters a lot, the pro-apoptotic signal first activates initiator caspases, such as caspase 8, 9, 10 (plus a series of other proteins).

These, in turn, cleave other effector caspases, which are the ones who effectively do the dirty job: caspase 3, 6, and 7.

Now there are many things happening at the same time (see this PDF from AbCam to get an idea of the complexity of the system).

At the nuclear level you have, for instance:

• activation of DNA cleaving enzymes, such as CAD (caspase-activable DNAse), which is normally inactive, due to binding to its inhibitor I-CAD. The latter is a target of effector caspases and its degradation causes CAD to go in the nucleus and starting fragmenting DNA.




• cleavage of PARP (Poly(ADP-ribose) polymerase), a nuclear enzyme involved in finding and repairing single-strand breaks on DNA.




• cleavage of lamins (by effector caspases + other proteases), proteins that are important for the formation of nuclear lamina, and stability of the cell nucleus




• cleavage of U1, a nuclear protein necessary for processing of mRNA.

Very complex events also happen at the mitochondria level: here you have a series of protein, which belong to the Bcl-2 family that control apoptosis mostly by regulating the levels of Ca2+ in the cell. There are 25 members of this family, that can be divided into two "sides": pro-apoptotic proteins such as Bax, Bak, and BAD and anti-apoptotic proteins such as Bcl-2, Bcl-xL, and Bcl-w.

The functioning of the Blc-2 proteins is very complex and I am not up-to-date with the last literature but to simplify, essentially what happens is that when the balance between anti-apoptotic and pro-apoptotic proteins is shifted towards the pro-, there is:

• liberation of cytochrome c from the mytochondria into the cytoplasm, which can activate effector caspases.


• induction of MPTPs (Mitochondrial Permeability Transition Pores), proteins which increases mitochondrial permeability which causes all sorts of trouble, eventually leading to swelling of mitochondria and loss of their function

Different events also bring to an increase in cytoplasmic calcium concentration, which can, for instance, induce the activity of calcium-activated endonuclease or other pro-apoptotic calcium-dependent proteins.

This is far from being an exaustive list of what happens, but it gives you the idea of the complexity of the system.

observation - When do Mercury/Venus reach greatest elevation at sunset/twilight for a given location?

On what day does Mercury reach its greatest elevation (in degrees from
the horizon) at sunset a given location?

The obvious answer is the day of Mercury's greatest elongation from
the Sun, but, since the ecliptic is slanted with respect to the
horizon, I'm not convinced this is correct.

In other words, on the day after greatest elongation, Mercury's total
angular distance from the Sun will be smaller, but it's vertical
distance in elevation (for a given location) at sunset might be
higher.

Same question for Venus, and for when the sun is 6 degrees below the
horizon (ie, civil twilight), and for sunrise/dawn.

I'm guessing the date might vary based on position (mostly latitude) since the ecliptic's slant varies at different locations.

I googled and found nothing. My (preliminary and possibly wrong)
expierments with stellarium show that Mercury's elevation at sunset IS
higher 1-3 days after its greatest elongation, but by less than 1/2
degree.

So, it's possible that the date of greatest elongation is a close
enough approximation.

What would happen if those gravitational waves were much stronger?

Based on a post by an Astrophysicist my understanding is as follows:

1. Gravitational waves would still be faint to detect irrespective of the size of the Black Holes. BTW aren't black holes meant to be more concentrated rather than bloated?


2. The distance from ground zero is what matters rather than the size of the colliding black holes and even if they were a billion times closer they would still not measure up to 1 mm.
   
   
   3. There wouldn't be significant impact from the waves themselves rather the Black Holes are the ones that would do the damage.
   
   
   4. For the shift to be 1 mm we would require a force enormously stronger than that of a gravitational wave.
   
   

Here's the link to the article: http://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#1605e7fa4aac

Disclaimer: All my points are a result of my learning from the above link rather than any native knowledge or original research that I profess to have done.

volcanism - Volcano activity on moon

The grey areas are known as "maria" (singular:"mare") and they were formed from flood basalts. These are a type of volcano and they have formed on earth too, examples include the lake eruption on Iceland, and the Deccan traps, that played a role in the mass extinction 65 million years ago.

On the moon, the flood basalt is old. About 3-4 billion years old, with some regions probably younger, but nothing less than 1 billion. Volcanism is driven by the heat of radioactive decay. On the moon there are no longer enough radioactive elements to melt rock, and so the moon is no longer a volcanic active body.

There are no active volcanoes on the moon, and there is no prospect of any in the future.

Why do post main sequence stars enter the red giants branch?

I am an early graduate student in astronomy and have hard time understanding why do post-MS stars move up the RGB.

Here is what I understand about post main sequence evolution of stars. As their hydrogen core is exhausted, the core shrinks under its own gravity. This lets a region of previously too cold hydrogen enter hotter regions, thus starting a hydrogen shell burning process. While this process can keep the luminosity constant, it expands the envelope, hence lowering the temperature and moving the star to the right in the HR diagram.

Then suddenly, the star starts moving up on the RGB: slight or no temperature change, but sudden increase in luminosity. What makes the star move up the HR diagram all of the sudden? This is before the helium flash when the core starts burning helium.

Thank you!

How to calculate the expected surface temperature of a planet

The formula

$$
4 pi R ^ 2 ơ T ^ 4 = frac{pi R ^ 2 L_{sun}(1 - a)}{4 pi d ^ 2}
$$

is correct, if you want to calculate the radiative equilibrium temperature. You only need to use the right units. We can further simplify the formula to

$$
T ^ 4 = frac{ L_{sun}(1 - a)}{16 pi d ^ 2 ơ};.
$$

You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as
$$
σ = 5.670373 × 10^{−8} ;mathrm{W}; mathrm{m}^{−2}; mathrm{K}^{−4}.
$$

The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:

$d = 149,000,000,000 ;mathrm{m}$

$L = 3.846×10^{26} ;mathrm{W}$

Albedo of Earth is 0.29. (The Bond albedo should be used.) You will get

$$
T ^ 4 = frac{ 3.846×10^{26}(1 - 0.29)}{16 pi times (149,000,000,000) ^ 2 times (5.670373 × 10^{−8})}=4,315,325,985 ;mathrm{K}^4;.
$$

After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.