What do we know about the lifecycle of the Milky Way (or any other spiral galaxy)?

If we look at Newtonian physics, and how galaxies will interact, a central black hole should just be considered as a massive, dense object.

The Milky Way does not fall into its own central black hole, it orbits it - as physics tells us it should.

When we get closer to Andromeda, the gravitational influence of Andromeda will act more strongly on us, and when we get really close, individual masses within each galaxy will have dramatic effects on each other, but as with any such system, the two black holes will not suck everything in.

If the two galaxies end up coalescing - which is not a given - the orbits of the stars and the black holes will be very complex. For billions of years the 2 black holes will orbit each other, getting closer as they shed energy, but during that time the stars around them will suffer many effects, including:

• some will be expelled


• some will hit the black holes


• some will go nova


• and so on

The best way to understand what will happen is to avoid thinking too deeply about black holes being weird, and treat them as dense masses. For most purposes this will help you model galactic collisions.

Have a look at this NASA simulation of the collision between the Milky Way and Andromeda:

Could Venus be a source of Earth's water?

No idea is stupid per se. But in order to place an answer to your question, please consider Venus as a comet. It once had, in that image, a coma (a tail) pointing outwards from the sun, made of evaporated water pushed away by solar wind. How frequently is the Earth exactly inside that coma?

As a comparison, when the Earth strikes a real comet coma we see meteor showers that last no more than one day. So if we consider that Earth may go through Venus imaginary coma once a year, we have an upper limit of 1/365 of Venus' water captured by Earth. This, without taking into effect that Venus had not have a real coma, that evaporation and expelletion of water from Venus was not a so directional process, and surely other factors I can not think about just now.

Besides, do not consider comets the main source of water on Earth. You need to count also with the water vapour generated on volcanoes and that generated on acid-base reactions among Earth's rocks.

Does anyone know why three of Jupiter's largest moons orbit in 1:2:4 resonance?

When the Galilean moons formed, they weren't in resonance with each other. All of them were in slightly smaller orbits than they are now. Over time after their formation, Io's orbit slowly moved outward due to tides from Jupiter. This is the same effect that is causing our moon to slowly move away from the Earth (at about the same rate your fingernails grow). It goes like this. The Moon's gravity causes tides to form in Earth's oceans. This bulge of water gets carried forward with Earth's rotation, because the Earth is spinning faster than the Moon is orbiting. Because the moon is still attracting the bulge, it causes a drag on the Earth, slowing its spin. At the same time, the bulge is attracting the moon, causing it to go faster in its orbit. As the moon speeds up, its orbit gets bigger. So essentially, the Earth's spin energy is getting transferred into the Moon's orbital energy. The same thing happens with Jupiter and Io, with a bulge in Jupiter's atmosphere causing Io's orbit to get bigger.

As Io's orbit expanded, its 'year' got longer, until it approached a 2/1 resonance with Europa. Once they reached resonance, they got 'locked in', their mutual gravity acting on each other reinforced it. Io was still raising tides on Jupiter, though, and its orbit was still trying to expand. As Io's orbit kept expanding, it gave a gravitational kick to Callisto on each pass, expanding both of their orbits until Callisto reached a 2/1 resonance with Ganymede. This is where the inner 3 Galilean moons got their resonance. The orbits are still expanding, but much more slowly because with each one you add on, it gets harder to transfer the energy. Given enough time, all 4 Galilean moons would probably reach resonance, although the sun will die before that happens.

I may have some of the minor details wrong here, but this is the story as I understand it.

launch - Why are spacecraft not air-launched from airplanes

Lots of good stuff on this topic in Wikipedia.:
Air Launch to Orbit

Air Launch

A typical rocket spends the first few seconds going straight up (almost) to get out of the atmosphere. After that it spends almost all of its time accelerating to orbital velocity. Thus getting out of the atmosphere, while hard (rocket is heaviest at this point) is really only a very small portion of the process.

The mass you would need to carry to a higher altitude is so large as to exceed the capacity of the largest aircraft ever built, let alone a balloon.

Consider the case of StratoLaunch. They intend to build a carrier aircraft composed of parts from 2 Boeing 747 (in the top 3 of the biggest aircraft commercially available. A380, B747, and C5 Galaxy are probably the biggest). It will be the largest airplane (in mass, wing length, etc) flying if it succeeds.

Even then, it can only carry a scaled down version of a Falcon (The original payload was to be a SpaceX Falcon 9 but with 5 engines instead of 9 and a concomitant reduction fuel/oxidizer load and thus mass). So the biggest aircraft yet to be built could only carry a smaller version of a medium size booster.

The primary benefit of air launch is not extra mass, but rather launch constraints. If you launch from a fixed site, you have limited launch options to differing orbits, and inclinations. An aircraft can in theory fly to wherever is convenient to hit the right orbital parameters. (Assuming there is a big enough runway for something the size of Stratolaunch vehicle within flight range fully loaded).

Currently, there is the example of Virgin Galactic's SpaceShip Two that uses the White Knight carrier vehicle, whose payload to actual orbit (LauncherOne if they actually build it) would be in the 100 kilogram range.

Pegasus, launched beneath a Lockheed L-1011 (3 engine airplane), maxs out around 500-1000 kilos to orbit.

There was a European proposal to launch an orbital payload (in the 200Kilo range) from the top of an Airbus A300.

The scaling up, that would be required just does not work.

the sun - Did the Babylonians believe in the Heliocentric version?

It's mentioned that the Babylonians decreed that there were exactly 360 days in a year. This statement was based upon the utilization of 360 as a standard (based on the sexagesimal system) i.e., they divided the circle into 360 equal parts.
Although, there is a mention about the Sun making one full circle across the sky in one year (as estimated by them), but, how were they able to measure it? And did they believe in the Heliocentric version of the solar system?

universe - Where is all the antimatter?

To answer your second question first: Yes, antimatter does exist in the same space as matter. In fact, the universe creates antimatter (and an equal amount of matter) every day as a matter of course in events like lightning strikes and supernovae, and even in certain nuclear decays. Humans create it in particle accelerators for research and for commercial/medical applications such as Positron Emission Tomography. The thing is, when we create antimatter, we also create an equal amount of matter.

In the hot flash of energy after the Big Bang, particle-antiparticle pairs were popping into existence and annihilating each other constantly. There were almost exactly equal amounts. For some reason, though, for every 100 trillion (10^11) particles of antimatter, there were 100 trillion and one particles of matter. In the ensuing few minutes, all the antimatter and all but that tiny fraction of matter annihilated each other and turned back into energy. Everything we can see today, all the galaxies, stars, and planets, are made up of that tiny amount of matter that was left over. Particle physicists still aren't sure why there was this tiny imbalance in the amount of matter and antimatter, because all interactions we've seen so far produce equal amounts of both. This is one question particle colliders like the Large Hadron Collider are attempting to answer.

inclination - iteration to cover the whole sky with right ascension, declination, angle

I am sure I get parts of the terminology wrong but if anyone can shed some light in the following:

I understand that for a given right ascension (RA) and declination (DEC), one has defined a ray (half-line) in the sky starting from the center of the earth towards infinity.

Now, if I also provide a angle, say d degrees around that ray, I have essentially defined a cone in the sky. My question is the following:

Which steps (RA_step and DEC_step) should I use in the following loop, as a function of d to ensure that I cover the entire sky and don't leave any patches anywhere?.

for (RA = 0 ; RA <= 360 ; RA += RA_step)
    for (DEC = -90 ; DEC <= 90 ; DEC += DEC_step)
         examine-cone(RA, DEC, d)

Is it safer to be near a star or a black hole?

the candidate star is an "average" 5 solar mass star, and the black
hole is a 5 solar mass black hole

Then their gravity is identical. Black holes don't have magic powers. A 5 M☉ star and a 5 M☉ black hole exert the same attraction from the same distance. The only difference is that the black hole would be much, MUCH smaller (about 30 km diameter in this case), so you could get a lot closer to the center, which is where the extreme gravity happens. But at cosmic distances they are the same.

Hawking radiation from a 5 M☉ black hole is negligible, likely too small to measure - its temperature is about 10^-8 Kelvin.

If it has an accretion disk, that might be a problem, but usually it only generates two relativistic jets at the poles - if you're not hit by the jets, you're fine.

The overall thermal glow from the accretion disk of a 5 M☉ black hole might be intense, but it can't last very long. It's not a continuous burn like a star, unless it's a much bigger black hole with a huge accretion disk and plenty of source material nearby to feed it - which is not the case here.

Your conclusion is technically correct. Gravitationally they are the same, but the star has the added problem of generating extra heat. Not that it would matter anyway, because both would completely disrupt orbits in the solar system, and the Earth is going to be flung out into space and would freeze anyway. It's just that the star would bake us first, and THEN we would freeze solid.

Or we would settle into a highly elliptical orbit around either the Sun or the invader body, which will have us alternately baking to death and freezing to death.

There is also the very tiny chance that, due to orbital disruptions, we would collide with something else, either with the invader, or with the Sun, or with another planet. This would mean a quicker death and might be overall "preferable".

Regardless, having a 5 M☉ body entering the system is not a good scenario for life on Earth.

http://xaonon.dyndns.org/hawking/

distances - Diameter of any galaxy

Given the angular diameter $a$ in radians and the distance $d$ in Mpc, you can get the actual diameter $D$ from:
$$D = dtan{a}$$

Using the small angle approximation, you get:
$$D = da$$

$a$ is in radians, so to get the distance in Mpc from the angular diameter in arc seconds you'd need to convert the angle in arc seconds to the angle in radians: $a = frac{2pi A}{360times3600}$, where the factor $3600$ is used to convert arc seconds to degrees, and $frac{2pi}{360}$ to convert from degrees to radians:

To get the diameter in kpc:
$$D = 1000times d frac{2pi A}{360times3600}$$
$$D = d frac{pi A}{648}$$
$$D approx frac{dA}{206} $$

where $d$ is in $textrm{Mpc}$, $D$ is in $textrm{kpc}$, and $A$ is in $textrm{arcsec}$.

Here $D$ is the diameter for round objects (even for disk galaxies seen at an angle). If the object is not round then this would normally be the maximum diameter.

[EDIT (see answer and comments from HDE 226868)] For irregular galaxies you may also need to have more information (viewing angle) to find the real maximum diameter of the galaxy. But that information is (I think) only available for galaxies in the local group.

gravity - A day in earth, a thousand years somewhere else

Assuming you mean a day in terms of an Earth day in both cases, the phenomenon I believe you're referring to is called time dilation, which affects the speed at which time is experienced by one body relative to another. This rate is affected by velocity and gravity, each of which can cause the bending of spacetime that results in the discrepancy between objects. Both gravity and velocity slow one's time as they increase.

The situation you describe is probably not naturally occurring, however is not outside the realm of reality. As referenced by Wiki from Calder's book Magic Universe, Calder claims that an acceleration of a constant 1G (what we feel on Earth all the time) would result in the affects of time dilation allowing you to traverse the entirety of the known universe in a single human lifetime (for the traveler). Conversely, during your trip, the rest of the universe would be aging 'normally'. There are plenty of technical problems with the feasibility of achieving this, but is conceptually possible.

I'm not particularly qualified to explain the equations behind it, so I'll leave that to our more capable members. Coincidentally, time dilation is the excuse I give when asked why I walk so quickly.

which pulsar has the longest spin period so far?

Vela X-1 is an example of an accretion-powered pulsar. These emit pulsed emission because they are accreting material onto their magnetic poles. If the magnetic poles and rotation axis are misaligned this results in pulsed X-ray emission. The power for the pulsar comes from the infalling material.

The classic P-Pdot diagram you show is for radio pulsars, or rather rotationally-powered pulsars. These are objects that are pulsators by virtue of accelerating charged particles from their magnetic poles out along the field lines. These accelerated relativistic particles emit synchrotron and curvature radiation that is beamed and intensified. The energy ultimately comes from the rotational kinetic energy of the pulsar.

In other words, other than the fact that they both involve neutron stars, these are completely different phenomena.

In terms of radio pulsars, your diagram looks reasonably up to date. I think the longest period object on your diagram is PSR J2144-3933, which has a period of 8.51 seconds (Young, Manchester & Johnston 1999).

Your diagram has a line marked as "graveyard". I believe this is a locus defined by Bhattacharya et al. (1992) and has the form
$$P = 2.42times 10^{-6} B^{1/2} s,$$
where the pulsar magnetic field $B$ is in units of Gauss (typical pulsar values would be $B=10^{10}-10^{13}$ Gauss (as also marked on your diagram). The theory behind this "death line" is discussed by Ruderman & Sutherland (1975). Briefly it arises from the requirement of a minimum potential difference to be generated such that accelerated particles produce energetic enough radiation to stimulate the production of further electron/positron pairs. If the magnetic field strength falls or the rotation period gets too long, then this mechanism fails and the pulsar is quenched.

PSR J2144-3933 appears to lie beyond this death line, but there are other ideas and models of how this death line may arise (of which I am not very familiar, but see for example Zhang et al. 2000).

Note that millisecond pulsars are thought to recycled accretion-powered pulsars. That is that they gain sufficient angular momentum from an accretion process that they spin up to become radio pulsars again.

Radio-quiet pulsars are thought to be rotationally powered but where the radio beam is narrower than say the beaming of gamma rays in which they are detected.

Magnetars are something different again. Their power comes from the decay of extraordinarily strong magnetic fields.
The soft gamma ray repeaters and anomalous X-ray pulsars fall in this category.

So you can see that the word "pulsar" might include all sorts of different types of objects and physics.

Viewing a solar eclipse through a leafy tree

Viewing my Facebook feed today, my local news station posted regarding a solar eclipse taking place today:

Note the line about using a leafy tree as a filter:

Scientists say NOT to directly look at the sun (if skies are clearing where you are), but looking at the eclipse through a leafy tree creates a natural filter!

I've heard a lot of people recommend viewing a solar eclipse by looking at the projection of the eclipse caused by light passing between leaves in a tree, in this case looking at the ground or a wall, where ever the light hits after passing through the tree. However, the Facebook posting here seems to recommend looking up at the solar eclipse through the tree.

I suspect that they have misunderstood the projection method (or inadequately explained), and their advice is not only wrong, but potentially dangerous.

Is this method recommendable? Should it be considered dangerous?

Follow-up

Almost immediately upon having seen the posting I had commented on it noting my belief that such advice could be dangerous, including a more clear explanation of how viewing using trees should be done. The staff member who originally posted the status later took note of this and edited the status to less ambiguously explain viewing the projection. Unfortunately, this was done after the solar eclipse had concluded. Hopefully no damage was done.

history - How many constellations in the Zodiac?

There are thirteen modern constellations in the Zodiac. In modern astronomy, a constellation is a specific area of the celestial sphere as defined by the International Astronomical Union. In total, there are 88 constellations.

Astronomy and Astrology are not the same thing. Astronomy is a science while Astrology is not. As such, I'll restrict myself to the historical and modern constellations of the Zodiac.

According to Encylopedia Brittanica

Zodiac (is) a belt around the heavens extending 9° on either side of the ecliptic, the plane of the earth’s orbit and of the sun’s apparent annual path.

In historical astronomy, the zodiac is a circle of twelve 30° divisions of celestial longitude that are centered upon the ecliptic, the apparent path of the Sun across the celestial sphere over the course of the year. Historically, each of these divisions were called signs and named after a constellation: Sagittarius, Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, and Scorpius.

In 1930, the International Astronomical Union defined the boundaries between the various constellations, under Eugène Delporte , who,

... drew his boundaries along vertical lines of right ascension and horizontal parallels of declination. One governing principle was that all variable stars with an established designation would remain in that constellation, as requested by the IAU’s Variable Stars committee.

"Constellations ecliptic equirectangular plot" by Cmglee, Timwi, NASA - Own work, http://svs.gsfc.nasa.gov/vis/a000000/a003500/a003572. Licensed under Public Domain via Commons.

As a result, the path of the ecliptic now officially passes through thirteen constellations: the twelve traditional 'zodiac constellations' and Ophiuchus (which was one of the 48 constellations listed by the 2nd-century astronomer Ptolemy), the bottom part of which interjects between Scorpio and Sagittarius.

Source: journeytothestars.wordpress.com

So, the 13 constellations of the Zodiac are Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius and Ophiuchus.

As seen from the (first) figure, the ecliptic also touches the edge of the constellation Cetus, though it in not usually included in the Zodiacal constellations.

Note about Precession: Because the Earth in inclined (by $23.45^{circ}$), it rotates like a top. This is called precession, which results in a shift in the position of the constellations relative to us on Earth. The result is that the 'Signs of the Zodiac' are off by about one month.

solar system - Could there be another planet between Mercury and the Sun?

No, that is not possible. There are quite a few capable telescopes studying the Sun since some decades. Already Galileo stared at the Sun until he got blind. Don't you think a planet would've been detected if it passed by in images like these? But Sun grazing comets are pretty common visitors in the corona.

It is a bit weird that a couple of percent of exoplanetary systems have hot Jupiters. They probably didn't form there, but migrated. And still stay there somehow.