general relativity - A clock travelling "faster than the speed of light"

Let's restrict to special relativity, meaning two inertial frames moving in a Minkowski spacetime.

A clock in the first inertial frame ticks slower, when seen from the second. A clock in the second inertial frames ticks slower, when seen from the first.

Now assume, that you are fixed to one of the two inertial frames. Usually you measure velocities within your own inertial frame, meaning measuring distance and time in your inertial frame to calculate the velocity of the other frame. In this case you'll get a velocity below the speed of light.

If you measure the distance in your inertial frame, and divide this distance by the time you observe on the clock of the moving inertial frame, you'll get higher velocities, which may exceed the speed of light. But this result has only the dimension of a velocity; it is not a velocity with respect of any of the two inertial frames.

The moving observer would observe a slow-down of external events, not a speed-up.
The observer would observe a speed-up only in the sense of a relativistic Doppler blue-shift when moving towards the observed object.

The people on the destination planet age the same as on the Earth, provided they don't move relative to the Earth. It's just you, since you travel that fast, the distance gets shorter, and therefore the time for the travel also gets shorter. By the Doppler shift and space contraction you'll see them age faster.

From Alpha Centauri, it's again the space contraction for the traveller, making the distance shorter. Again we have a combined space contraction/Doppler effect looking the people on Earth aging rapidly for about 4.5 years.

For people on Earth observing the travel takes 4.5 years. For the traveller it may take just a few days. With an acceleration of just 1g you could traverse the Milky Way in about 20 years, seen from the spaceship. The same travel seen from Earth would take about 100,000 years.

More on the twin paradox on Wikipedia.

the moon - Official registry of lunar placenames?

What's the definitive source for names of geographic objects on the moon?

Reason: As a computer graphics project I've rendered a scene based on LROC terrain data. I want to know what I'm looking at.

The most detailed atlas I can find online is this one:

http://planetarynames.wr.usgs.gov/Page/Moon1to1MAtlas

...but it's not brilliant and there are lots of small but prominent craters and ridges which aren't labelled. It may be the some of these simply don't have names, but it's also plausible that there are simply too many to list in that atlas. How can I find out for sure?

Position of Neutron stars in H R diagram

The HR diagram is an observational diagram. Whilst neutron stars could be placed in the HR diagram in the same way as white dwarf stars are, it turns out to be impractical to do so because the photospheric luminosity and photospheric temperature of neutron stars is next to impossible to determine. The reason for this is two-fold: (i) Neutron stars start off very hot (interior temperatures of $sim 10^{10}$K and photospheric temperatures of $sim 10^{7}$K, but they cool very rapidly. Within 10,000 years after the originating supernova they will have cooled below a million degrees, then photon cooling takes over from neutrino losses and they cool to a few thousand degrees within 10 million years (e.g. http://adsabs.harvard.edu/abs/2004ARA%26A..42..169Y ). There are many uncertainties and unknowns in these processes. (ii) the photospheric emission is usually dwarfed by emission from the magnetosphere or luminosity due to accretion from a companion or the interstellar medium.

One can theoretically work out where neutron stars should be by assuming that the emission is like that from a blackbody and that the radius $R sim 10$ km.

In that case neutron stars lie on a locus defined by
$$ frac{L}{L_{odot}} = 1.9times 10^{-9} left(frac{T}{10^{4}K}right)^4 $$
So, contrary to what you you say in your question, most neutron stars could be cool and very, very faint and spend the majority of their (cooling) lives at the bottom-left or even bottom-right of the HR diagram. There is actually a huge uncertainty over where neutron stars would appear on this locus. Their very low heat capacities means that any "reheating processes" could very effectively raise their temperatures. Such processes include Ohmic dissipation of the magnetic field, some kind of thermalisation of their rotational energy or accretion from the interstellar medium. For the latter, luminosities of $10^{20}-10^{21}$ W may be possible, implying effective temperatures of tens of thousands of Kelvin.

A neutron star at the same temperature as Sirius would have an absolute visual magnitude that was about 22 magnitudes fainter, $M_V sim 23$. Another way of visualising this is that the neutron star cooling sequence is roughly parallel with the white dwarf cooling sequence but about 13 magnitudes fainter.

You never see this locus shown on an HR diagram because it is usually way off the bottom of the plot.

solar system - Why can't I see Mars clearly?

Much like Jeremy has answered, Mars will appear just larger than a point, you will be able to discern a disk, however with minimal detail. For an overview of the disk you will see, I two versions of a virtual telescope in Wolfram Mathemtica; one with an orange disk and another using models of the planets. Anyway, using the orange disk variant, I input the CPC1100 details and the resulting view will match my output.

The field of view using a 40mm eyepiece without a barlow results in a 0.74degree apparent field of view, this equates to 2664 arcseconds. For reference Mars right now has an apparent diameter of 4.5 arcseconds, in your field of view that is .0017% of the apparent field of view (as represented by the orange disk).

eg. 40mm eyepiece @70x magnification w/no barlow

You telescope has a maximum useful magnification of 560x (2x telescope aperture) however it is airy disk limited to 481.2x. This means that a star will only appear 'point like' up to 481.2x magnification, above that the disk becomes increasingly blurrier to a point where you can no longer discern a point it appears as a cloud.

As a point of reference to the size of Mars using different equipment;

40mm eyepiece @210x magnification w/3x barlow

10mm eyepiece @280x magnification w/no barlow

17mm eyepiece @494x magnification w/3x barlow

On another note apart from what you 'should see', is what you can see. Aperture plays the ONLY role in telescope design which defines your clarity. The primary mirror diameter is a light bucket, a larger mirror is a higher image resolution (more discernible features/clearer image). Magnification also plays a role in clarity, each time you increase the magnification by 2x, you reduce the brightness by 4x; increase by 3x, you reduce brightness by 9x. So brightness is directly proportional to magnification. The highest possible brightness you can achieve, is your lowest possible magnification. How aperture works by collecting more light; allows a brighter image for a larger telescope than a smaller one at the same magnification, which is why so many people get 'aperture fever' and think big is best (for clarity yes). Your focal length is the determining factor for image size, meaning a 20inch telescope won't see an object larger than a 5inch at the same focal length.

You are correct in saying that both Mars and Sirius appear brighter with the naked eye than your lowest magnification; referring above, if your pupil diameter is 7mm, then the object appears at a factor of hundreds of times dimmer through your telescope.

I can provide material to calculate your own virtual telescope, it's parameters and viewing expectations if you are an avid user of Mathematica. Hope this has cleared up some information for you.

mars - Could martian meteorites come from Olympus Mons?

Since this has gone unanswered, I'm going to give it a shot.

Probably the answer is no. We know that on some moons, volcanic eruptions do send material into space.

Enceladus for example, feeds Saturn's E-ring. It regularly shoots about 200 KG per second into space.

The other highly volcanic moon in our solar system is Io which shoots material high into it's orbit, and it has some of the largest if not the largest observed volcanic eruptions. Source: http://www.space.com/26732-jupiter-moon-io-volcano-eruption-photos.html

But I couldn't find any references that Io's volcanoes shot material into orbit. short article on Jupiter's rings

The difference between Enceladus and Io is size. Enceladus's escape velocity is 0.239 km/s (860.4 km/h) Source in link above, and it's gravitation is just over 1% of Earth's. Io's escape velocity is 2.558 km/s and it's gravity is similar to the Moon's, about 18% of earth's. If Io has a hard time sending volcanic material into orbit, on Mars, it would be nearly impossible.

Io can shoot material up to 300 miles into the air, Source: https://en.wikipedia.org/wiki/Volcanology_of_Io but that's still only about 1/7th to 1/8th of it's radius, probably well within it's Sphere of Influence. On Mars, it's Sphere of influence is significantly larger and it's gravity twice as much. If a super large Volcano shot material 300 miles into the air above Mars, that's still well below escape velocity. (I could run the Maths if you like, but in a nutshell, escape velocity changes with the square of the distance, so, ignoring air resistance, Io's enormous eruption reaches 14% of it's diameter or 28% of the radius is about 40% of escape velocity - assuming I got my math right. On the Mars, an equivalent enormous eruption would be what, 20% of necessary escape velocity? Less?

Also, while Olympus Mons is enormous, it didn't all erupt at once. Source: http://www.space.com/20133-olympus-mons-giant-mountain-of-mars.html

Olympus Mons is over a hot spot on Mars, subject to regular volcanic eruptions (regular meaning, perhaps every few million years), but because Mars doesn't have plate movement, the eruptions happen in the roughly the same location and that, in combination with virtually no erosion and low gravity, is why Mons is as large as it is. It probably does erupt impressively when it goes, but I'd be pretty skeptical that it erupts violently enough to send magma into space.

Also, most of the Martian meteors on Earth are from a recent impact, about 5 million years ago: http://www.space.com/24962-mars-meteorites-single-impact.html Olympus Mons may have erupted about 5 million years ago, so the timing works, but I still think it's unlikely.

(I could add some links on how far Earth volcanoes can send magma/pumice if interested, I looked that up as well).

To get that kind of velocity, you need significant amounts of concentrated energy. A gun works by creating a controlled explosion with only one place for the energy to go. Rockets work by sustained acceleration over time. While it seems quite different, a meteor strike is similar in some ways to an enormous gun. When a meteor strikes the surface of a planet, it hits with so much force in a localized area that the ground get's crushed and then rebounds with enormous force in a localized area. A volcano simply can't (I think) generate that kind of concentrated energy unless it's very low gravity like Enceladus, 1.1% of the gravity of Earth.

That probably needs tidying up a little. I'll try to clean it up later.

A somewhat related conversations on this subject if interested: http://www.reddit.com/r/askscience/comments/1gkiz3/could_a_volcano_eruption_theoretically_be/

galaxy - Was the Milky Way ever a quasar?

A quasar is simply an active galactic nucleus (AGN) that is viewed from a particular angle; see the picture below, in which quasars are labeled "QSO". This is really a remarkable figure because historically all of the names in the figure were thought to correspond to different types of objects, when really they all refer to the same thing!

Your question really shouldn't be "Was there ever a quasar in the Milky Way?", since the dotted line in the figure would correspond to the Galactic plane and we would not see Sagittarius A* (the Milky Way's super-massive black hole) from the correct angle. A better question might be, "Has Sagittarius (Sgr) A* ever been active?" The answer to that question is yes; according to this page it was probably active (very bright with a jet) about 10,000 years ago. However, at the moment, it isn't really doing anything, since it isn't currently accreting anything (to put it plainly, it isn't eating anything, so it doesn't have enough energy to be active). However, many astronomers (myself included!) are anxiously waiting for a cloud of gas called G2 to fall into Sgr A*. We are hoping that Sgr A* will burp or do something interesting.

cosmology - How likely and severe is the threat of a gamma ray burst to earth?

I'll address WR104 first. The National Geographic article calls it a "potential threat." Yet that potential may be low. There are a slew of articles quoting astronomer Grant Hill on the subject. Hill studied the star and found that it looks like it isn't pointing straight at us. Its axis might be up to 45 degrees in another direction, meaning that we'd be fine if it underwent a burst.

From here:

It would appear the original Keck imagry may not have been as straight-forward as it seemed. Spectroscopic emission lines from the binary pair strongly suggest the system is in fact inclined 30°-40° (possibly as much as 45°) away from us.

So, Earth doesn’t appear to be in the firing line of WR 104 after all…

Here and here Hill is quoted as saying that the star looks like it could be pointed at us - which goes against the evidence he found! Finally, this is another example of Hill's cautious attitude: The evidence says we're good, but he's not making any assumptions.

---

Sabre Tooth, I know you said that you didn't want Wikipedia, but my primary goal here isn't to collect the bounty, so I'm going to reference it in this section. I hope you're okay with that. I may withdraw it, though, if Jonathan indicates that he doesn't want it.

Okay, back to the topic at hand. Wikipedia, of course, has a short tidbit on the frequency of gamma-ray bursts effecting Earth:

Estimating the exact rate at which GRBs occur is difficult, but for a galaxy of approximately the same size as the Milky Way, the expected rate (for long-duration GRBs) is about one burst every 100,000 to 1,000,000 years. Only a small percentage of these would be beamed towards Earth. Estimates of rate of occurrence of short-duration GRBs are even more uncertain because of the unknown degree of collimation, but are probably comparable.

'A small percentage' isn't too specific. The BBC article that is cited in this passage says the following:

Observations of deep space suggest that gamma ray-bursts are rare. They are thought to happen at the most every 10,000 years per galaxy, and at the least every million years per galaxy.

However, the 'small percentage' is never elaborated on.

---

That's all I have for this edit; more to come later. By the way, this Physics.SE question may interest anyone reading this. . .

the sun - What is the angle between the equator of the sun, and the plane of the Earth's orbit

The plane that contains the orbit of the Earth is known as the "ecliptic". The rotation of the sun is tilted by 7.25 degrees to the ecliptic, and this value does not vary over time.

The rotation of the Earth is also tilted, by 23.45 degrees to the ecliptic, it is this angle that causes seasons.

Nasa has a factsheet, that has this and other information about our nearest star.

magnetic field - Why has Venus's atmosphere not been stripped away by solar wind?

Venus has a strong ionosphere that protects it against violent solar winds. So, even though Venus has no intrisic magnetic field, it has an effective, induced magnetic field due to the interaction between the solar winds and the atmosphere, that protects it against solar winds.

Venus atmosphere is thick enough to have a consequent ionosphere, that would be the difference between Mars and Venus (and Venus was able to keep a thicker atmosphere due to its greater mass, contrary to Mars).

Sources:

meteor - What did I see streaking across the sky last night?

I was enjoying a casual naked eye view of the summer sky last night around midnight in a new jersey beach town. The sky was pretty clear for around here.The Milkyway was just visible and there was no could cover or buildings higher than me so I saw from Cassiopeia to Sagittarius.

I was hoping for a meteor and I got one I have never seen before. I will try to describe it.

Appearance: .It appeared out of no where like any other meteor but it had no streak AT ALL. As it traveled it shimmered. It was blue colored and not particularly bright perhaps fuzzy. The shimmering was the best way I can describe it.

Trajectory: It began high in the sky slightly to the east around the "southern" part of the constellation Aquila. It traveled west. It was as fast as any other meteor but as I said had no tail. It lasted very long... say 4 or 5 seconds. It didn't have an arching trajectory like something lower. It traveled straight and as it disappeared, it seemed to travel off into vs down to the horizon.

I was so astonished that I practiced the speed it went with my finger after it finished. It moved from say Vega to Altair in less around a second.

It couldn't have been any kind of craft since it appeared and disappeared and traveled so quickly. I have seen 100's of meteors of many kinds but not any like this. What did i see?

I tried to be as detailed as possible. If I missed anything, I'll add it in the comments.

---

For the record, I am not a believer in visiting aliens or supernatural events. BUT you will laugh at this. Years ago,I had read about an Ozark superstition that if you say "Money" three times when a meteor appears, you will get rich. I liked that one and since I stargaze a lot, I figured I'd get the opportunity. Well I had my chance and then some but I was so astonished, I forgot to say it. Looks like I will have to keep working for a living :)

What measure should I use to help optimise the design of a telescope in a cubesat?

I make telescopes and telescope mirrors. You're asking a hard question, since it depends on so many things. But it's an interesting project, so I'll try to help as much as I can.

The hard limit for performance is aperture, or diameter of the primary optics. Bigger aperture = higher hard limit for performance. On a nanosat you only have room for so much aperture, so performance will necessarily be limited.

Secondly, you'd need some kind of compact design. A reflector for sure, but not newtonian. Some kind of Cassegrain variant, you can make those pretty short and stubby. Unless you make it so that it "unfolds" in space, but that's going to be tricky.

Also, the whole optical stack has to be very rigid and must hold its shape through launch and while in space. The optical system must remain collimated at all times.

Finally, you probably have some weight limits. That will make it harder to create a rigid system.

---

One measure of performance is the resolving power, which indicates the angular size of the smallest details resolved by the telescope. The relationship is linear. An aperture of 100 mm provides a resolving power of 1 arcsec. An aperture of 200 mm provides a resolving power of 0.5 arcsec. And so on.

---

Telescopes operating at ground level are limited by atmosphere in several ways.

One is the so-called "seeing", or air turbulence. This blurs the image and reduces the effective resolving power. Since small telescopes have less resolving power to begin with, they are less affected. A 100 mm aperture is unaffected most of the time. A 200 mm aperture is affected most of the time.

Your telescope will be above the atmosphere, so the effects of seeing will be zero. It will operate at full resolving power all the time.

Another way the air influences telescope is via light pollution and other forms of luminescence and reflected light from the air. This reduces contrast and "erases" the faintest objects from the image; bright objects such as planets are not affected at all.

Your telescope will not be affected by light pollution, so its performance on faint objects (such as galaxies or nebulae) will always be 100% of the maximum for that aperture.

Bottom line: you are going to have a pretty small telescope, but operating completely free of the effects of the atmosphere.

---

Since it's operating in microgravity, the optics will not deform under their own weight, so you can make them quite thin and lightweight.

The instrument will operate under extremes of temperature. You will need some kind of dynamic focusing to compensate for expansion / contraction. Be careful, with some catoptric designs such as Cassegrain you also need to pay attention to the distance from primary mirror to secondary elements, etc. Using carbon fiber for the skeleton would reduce expansion / contraction a great deal, and would also provide excellent stiffness per unit of weight.

Quartz mirrors could deal very well with huge temperature variations, but quartz is expensive to process to optical precision. Borosilicate glass (the material used for most amateur telescopes) should also work well; Supremax 33 for example is cheap and fairly good. It depends on the optician who will make the mirrors. If you find someone who could give a discount on quartz, go for it; otherwise use borosilicate and don't worry about it.

You will also need some kind of inertial elements to change the orientation of the telescope in space. That's not a trivial task.

---

Bottom line: you begin with a performance budget limited by aperture. That's the ideal performance. Then real world issues will push real performance below that level: collimation errors, temperature shifts, optics quality, focusing errors, etc.

Performance of a real instrument is affected by everything.

universe - How can I see a nebula?

Yes, indeed! Many nebulae are visible from Earth in a small and cheap telescope, and even to the naked eye (if you are standing in a sufficiently dark place).

In fact, yesterday I was watching the Orion Nebula with my 4.5" telescope (which is worth $200 or so) from my apartment in the middle of Copenhagen.

The term "nebula" is a bit of a… well, nebulous term, as it covers several distint phenonomena, for instance:

• Galaxies, e.g. the Andromeda Galaxy


• Stellar clusters, e.g. the Eagle Nebula


• Planetary nebulae, e.g. the Eskimo Nebula


• Supernova remnants, e.g. the Crab Nebula

However, you should be aware that even in a large and expensive telescope, the nebulae do not look anything like the beautiful images you find on the internet. With your eye, you will merely see diffuse, whitish clouds. The beautiful colors arise only in images taken through telescopes with long exposure times. If you want to make such pictures, you will need a telescope with a motor (since the telescope must follow the rotation of the sky) and the possibility of attaching a camera to it. People are often disappointed when shown a nebula through a telescope. I think the beauty of it lies in knowing what it is that you are seeing, knowing for instance that it lies 600 lightyears away, that you are looking 600 years back in time, and that stars are right now being formed there.

Googling something like "nebulae visible in a small telescope" should get you started. Good luck!

the sun - How does solar activity (e.g. flares, coronal mass ejections) change over the life cycle of a Sun-like star?

Magnetic activity depends on rotation due to the dynamo mechanism. Younger stars in general rotate faster. They (stars like the sun) spin down because their ionised winds interact with their magnetic fields and gradually remove angular momentum. A star like the Sun at 100 million years old may have had a rotation period of 0.5-5 days - much faster than the present-day Sun.

Roughly speaking, magnetic activity scales with the square of angular velocity, so young stars can be orders of magnitude more active than the Sun, manifested as much stronger coronal X-ray emission, flares and much greater coverage of starspots and strong magnetic fields.

Conversely, stars older than the Sun should be a bit less active, although the rate of spindown slows with rotation rate and so the changes beyond a few billion years are not so great and not so well calibrated.

A useful review article would be Ribas et al. (2005)

http://adsabs.harvard.edu/abs/2005ApJ...622..680R

distances - How can we be sure that we have identified very distant stars correctly?

We determine the spectral type (i.e. temperature) of a star using multicolour photometry, or (ideally) spectroscopy. By guesstimating the temperature, mass and radius of a star, we can say that two stars that have pretty similar observational properties probably are closely related to each other.

Cepheid variables, for example, display periodic pulsations that depend quite strongly on their intrinsic luminosity -- this is why they're reasonably good standard candles. Their characteristic variability makes them clearly identifiable as a Cepheid, and observations of their pulsation are backed up by data that place them in the same region of the HR diagram.

If it looks like a duck, swims like a duck, and quacks like a duck, then chances are it's a duck. If it's as hot as a Cepheid and pulsates like a Cepheid, then it probably is a Cepheid.

So to answer your final paragraph: yep. Though with no spatial resolution and often (in the absence of parallax measurements) only tenuous distance data, classifying stars is a rather messy business.

velocity - Convert orbital elements to positions and velocities

I wrote a similar N-body orbital program a few years ago.
To obtain the initial velocities and positions for Solar System bodies I used a free program called PLANEPH.
Download available at:- ftp://cdsarc.u-strasbg.fr/pub/cats/VI/87/

It can give you 3D xyz position and velocity of solar system bodies at a particular point in time with resolution of a few seconds of time as I remember.

The user interface is a bit old-fashioned (DOS-style) but it seems to work OK. I ran it on Windows 2000.

I used PLANEPH to extract Heliocentric Coordinates then I had to convert to Solar System Barycentre frame using data on planetary masses obtainable elsewhere (e.g. NASA).

If an Asteroid was to strike the Earth, would it affect the Earth's rotation?

To have a noticeable effect the impactor needs to be BIG.

Most questions about "what would happen if ... hits" can be answered by the "Earth impact effects program" (http://impact.ese.ic.ac.uk/)

Here are calculations for a 100km stony asteroid...

A brute like this would have a good chance of wiping out most complex life on the planet. There has been nothing like this in the last 4 billion years (or so)... It could cause the length of the day to change by "up to 2.42 seconds"

As gerrit said, it would be the last of our worries.

planet - Elemental abundances in the crust of exoplanets and accessibility to them

No, it doesn't work like that. A rocky planet, to first order, will probably contain a similar mix of elements to the Earth.

The most likely reason that the star has a low metallicity is because it is old, and formed when the interstellar medium was less metal rich. But because rocky planets do not incorporate (much) hydrogen and helium, what you are really asking is how do the abundances of other elements scale to the iron abundance (which is often what is meant by metallicity)? The r-process events, such as type II supernovae that produce the heaviest radioactive elements (beyond lead), enrich the interstellar medium "promptly", because the progenitor lifetimes are short compared to the age of the Galaxy.

However, because a lot of the iron and nickel in the interstellar medium is not produced/disseminated by type II supernovae, but by type Ia supernovae with longer-lived, lower mass progenitors, then if anything I would expect a larger ratio of elements produced/disseminated by type II supernovae to iron/nickel. This will also be true of the "alpha" elements like Silicon, Oxygen and Magnesium.

The planet might be a bit short in those elements predominantly produced by the s-process in lower mass giant stars - so elements like carbon (!), nitrogen, barium, rubidium, strontium, zinc and many others may be scarce.

For a star of this age/metallicity I would expect these variations to be plus or minus a factor of two or so compared to the Earth.

Note, I am discussing bulk abundances. The accessibility of the material, where it is etc., is more of a speculative geology question, and I don't know the answer. A further complication is where exactly the planet forms compared with the condensation temperature of various elements and materials. The above discussion implicitly assumes that the planet forms in a protoplanetary disc material at a similar temperature and density to the conditions in which the Earth formed. However, a change in these conditions will alter the detailed compositional mix.

telescope - What is meant by Electronic Beam Steering and how is it achieved?

This can actually mean a couple of different things, but in the case of steering an MWA tile, Jeremy is quite correct. Each tile has a 'beamformer' box that all of the individual antenna cables plug into. This box physically delays each signal appropriately (by having a longer path to go through for more delay) based on the signals sent to it for each pointing. Because of the quantised path lengths there are what are called 'sweet spots' that are used, in other words arbitrary pointings don't normally happen you just dial in the closest 'sweet spot' to where you'd like to point.

There are also version that do this in software (and in fact in a special phased mode the MWA does this between tiles sometimes), where 'turns of phase' are used instead but it is essentially the same idea.

Accelerating universe expansion and standard candle

Most type Ia supernovae are thought to arise from the thermonuclear detonation of white dwarfs that are composed almost entirely out of carbon and oxygen.

These white dwarfs are the cores of relatively low-mass stars that have lived their lives, gone through stages of core hydrogen and helium burning, leaving behind degenerate carbon/oxygen cores that become cooling white dwarfs after the outer envelope has been shed during the asymptotic giant branch and planetary nebula phases. As such, their composition, at least to first order, is almost independent of the initial composition of the star from which they were formed. That is, even if the progenitor star had a very low initial metal content, the white dwarf produced would still be almost exclusively a carbon/oxygen mixture, which had a similar Chandrasekhar mass and a similar explosive potential.

It is well known however that not all type Ia supernovae are the same. It has long been known that their light curves are subtley different and there is a so-called stretch factor that can be applied to get a "corrected" peak magnitude. a.k.a The width-luminosity relation.

More recently there has been a realisation that type Ia supernovae could arise from both accretion or mergers and there is clear evidence that the amount of radioactive Ni varies from explosion to explosion. A very recent paper by Milne et al. (2015) has however challenged the view of metallicity independence. They claim there are two populations of type Ia SNe, connected with progenitor metallicity, and that these populations become more apparent at high redshift when looking at rest-frame ultraviolet emission. The gist of their conclusions is indeed, as your question supposes, that this may go some way to ameliorating (but not eliminating) the need for dark energy.

impact - How gently could a comet/asteroid/meteorite "hit" Earth?

Yes, spacecraft do it all the time by using air resistance (and sometimes rockets) to slow down.

Meteorites enter the atmosphere at high speed, typically 10-70 km/sec, but the smaller ones are slowed by air resistance, so they typically hit the surface at just a few hundred kilometers per hour. Reference: http://csep10.phys.utk.edu/astr161/lect/meteors/impacts.html

Larger bodies are not affected nearly as much by the atmosphere. For any such body, the velocity at impact is going to be at least several kilometers per second. A body moving obliquely in the same direction as the Earth's rotation would have a slightly lower speed at impact, but the impact still could not be described as "gentle"; the Earth's rotational speed is still a small fraction of orbital or escape velocity.

Here's one way to think of it. Freefall trajectories are reversible. If you watch a movie of an incoming body in reverse, it still makes physical sense (ignoring air resistance). Any meteorite hitting the surface has to have been in deep space at some point before the impact. If there were a trajectory that allowed such a body to have a "gentle" impact speed, then it would be possible to start with the same body near the surface at the same "gentle" speed, but in the opposite direction, and have it reach deep space. Unless the meteorite has its own propulsion system, that's just not going to happen.

You can get to space with a lower starting speed by moving along with the Earth's rotation - which is why most rockets are launched to the east, to take advantage of that. Reversing such a trajectory can result in a slightly slower impact, but only slightly.

jupiter - Are a black hole's jets caused by the black hole's magnetic field?

You appear to have a thing about linking the magnetic field generated by Jupiter to that around a black hole (note that a field does not "belong" to an object) - the two things are entirely different, the only connection being that magnetic fields are generated in both cases by currents caused by the motion of charged particles.

In Jupiter's case, the magnetic field is (likely) generated via a dynamo process, involving currents that flow in its metallic hydrogen interior.

In the case of a black hole, as material travels inwards, it forms an accretion disk around the black hole (because angular momentum must be conserved), in which magnetic field can be generated and amplified. The twisting of the magnetic field lines around the rotation axis of the black hole could create a kind of magnetic funnel lined up with the rotation axis, along which charged particles can be accelerated by a strong magnetic field gradient. The field lines would not look like you have shown for Jupiter, they would be twisted into a helical topology along the rotation axis. This is part of the Blandford-Znajek effect, which I do not claim to properly understand.

So the answer to your question is maybe, but its very much a topic of contemporary research.

debris - What happens if an astronaut is hit by a tiny micrometeorite?

A micrometeorite (including space debris in LEO) hitting the metal wall of the ISS would likely cause a shower of secondary high speed fragments. But there are other "softer" materials proposed which would just let the object through without much interaction. For example by Bigalow Aerospace.

What would happen to a human who is hit by a micrometeorite directly? Take a 1 mm diameter metal object at 30 km/s impact speed as an example. Much smaller but much faster than a bullet from a gun. Would it just make a thin hole through the body, or would it like most military bullets tumble and tear up a much larger cavity in the tissue? Does it matter much how fast the projectile is spinning?

To my knowledge, micrometeorites have never caused any actual injury or even any important damage to equipment, although it is a much talked about potential threat.

the moon - Gravity Assist - Orbiting Satellites & Speed Increases

So, a gravity assist in very simple terms is when an object uses a planet's gravity and rotating momentum to increase the objects speed.

If this is true, why doesn't the ISS get further and further away from us as it gains angular momentum from Earth? As I understand every once in a while, the ISS has to do a burn to get back into stable orbit around Earth because of the very thin atmospheric drag on the ISS. That would mean the ISS isn't increasing it's speed, it's slowing down.

On a separate but related question, if gravity assists do indeed work that way, isn't our moon stealing angular momentum from Earth every second it's in orbit around us? As I understand, the moon is gradually being tugged away from Earth, not by stealing Earth's angular momentum, but because the tidal effects of the moon on Earth bump the moon into a higher orbit.

Please, no math, keep it to worded answers. I just want to understand the concept behind the explanations.

Thanks!

the moon - Do other planets in our solar system experience eclipses or is this unique to Earth?

Pretty much every planet with a moon can have eclipses - you'll have seen photographs of Saturn with shadows cast from its moons, like this one from wodumedia.com:

If you were in a balloon where that shadow is, you would see a solar eclipse. It wouldn't be as exciting as from Earth, as the Sun would appear so much smaller from this distance.

If you want the planet to eclipse the moons, this happens far more often than on Earth, as the moons are so much smaller than Saturn. They often are completely shadowed.

light pollution - Do Noctilucent clouds cause problems with telescopes?

Noctilucent clouds are not a problem for space telescopes because their orbits are always more than 85 km. The Hubble Telescope orbits at about 570 km. Noctilucent clouds are a problem for ground-based telescopes (although only at high latitude sites,> 50$^{circ}$), especially if you are trying to get accurate photometric brightnesses. However, typically you can see them reflecting city lights or moonlight, or you can tell from the large variance in your photometry counts that they are there and you should try again another night.

How do I calculate the inclination of an object with an amateur telescope?

You just need a telescope with a "wedge" mount, (ie one for polar coordinates,) and not one which is simply a pan-and-tilt mount like commonly used for cameras. (BUT, it might be more fun to do it with a sextant -- see below.)

With a telescope that has a polar mount, you just need to set it up correctly. That means orienting the base of the mount in the correct north/south direction, and then setting the "wedge" angle to account for your geographic latitude. (Any amateur level scope with a polar mount will have instructions.) When you pan the scope it will sweep celestial lines of equal latitude, and when you tilt the scope it will sweep celestial lines of equal longitude. Point the scope at your target, and read the declination from the scope's mount.

The hardest part is going to be that the satellite will be moving pretty fast, and its orbit is going to have a continuously varying declination. (Unless you target a satellite in a circular, equatorial orbit. :)

Fun with a sextant

It might be easier to "shoot" the satellite with a simple (search for "Davis Mark 15" on ebay) sextant and something called an "artificial horizon." You need any liquid reflecting surface... a swimming pool or kiddie pool might work for shooting a moving satellite. (You can buy a small "artificial horizon" that's a few inches square, but you'll never catch the satellite's reflection in that.) Using the sextant you measure the angle between the satellite and its reflection, also noting the compass direction. Then you crunch a bunch of trigonometry. But this is exactly how you practice astronomical navigation star sighting on land.

star - Sun from SuperNova

Actually yes, gravity is too weak to do the job by itself.
But as you mention, another force is acting on the system too, in a very strong way: it is the pressure that pushes outwards. In fact, the energy released by the explosion of the supernova, helps to compress the layers of released material (gas and clouds). You have shocks between the expelled gas (fast) and the ISM (slooow). Also, you need cold gas to produce a collapse event, and the ejected material from an explosion is not cold at all!

So, you need an efficient mechanism to induce collapse of the gas cloud, and this mechanism is furnished by compression instabilities.

You start from the Virial theorem: $2K + U = 0$

with K kinetic energy and U potential energy. What you want here, for your collapse is that the gravitational potential energy is larger than the kinetic energy (otherwise particles energy will overcome the gravity and avoid collapse - not exactly true, but you can think as if high energy particles pushes outward).

Then you can rewrite the Virial theorem as:

$$3 N K T = frac{3}{5}frac{G M^2}{R}$$

where K boltzmann constant, N atoms number, M mass of the cloud, R its size.

In our case (collapse case) the left term must be less than the right one.

Then you can transform $N = M/m$, with m mass of the single particles, and $R = frac{3 M}{4pirho}$, with $rho$ cloud density (assumed constant).

Than you can calculate the Jean mass as:

$$M_J = (frac{5KT}{Gm})^{3/2}(frac{3}{4pirho})^{1/2}$$

This is the limit after which your cloud can collapse.
Look at the dependence on the temperature, and the inverse dependence on the density.

Just to give an idea of the typical values encountered, we can still rewrite the critical mass as:

$$M_J = 2M_{sun}(frac{n}{10^5 cm^{-3}})^{-1/2}(frac{T}{10 K})^{3/2}$$

Source: this lesson

PS: a similar mechanism is triggered by the rotating arms of spiral galaxies.

photography - Sky Glow calculations

Well, light, and the resulting sky glow comes in a lot of different wavelengths, so the first question you need to ask yourself is what range are you looking for? If this project is intended to give someone a good idea as to whether it is worth their while to drag out the telescope on a given night, chances are you only care about human visible wavelengths (but don't underestimate the amateur radio astronomers in your area!). My phone's camera will pick up light in the near-infrared, so you want to be careful.

Secondly, you need a camera with very low noise. If you go into a windowless room with the lights off, and take a picture, you want to have as close to 0x000000 rgb measurement from every resulting pixel as possible. A magnitude 0 star will only produce 2.08 microlux, so electrical noise can alter your results significantly.

Third, and this may be obvious, make yourself a cardboard "horizon shield" (a name I made up) to block any direct sources of light.

And if it isn't of vital importance to use an actual camera, since you are using a raspberry pi, you might just want to go with a lux meter instead. This one has a separate meter for the human visible wavelength, and the noise should be very low, depending upon how you wire it and what voltage you supply.

the sun - Amount of EM radiation (and particles) Earth "receives" from other planets in our solar system?

Of course all the planets have been observed by many telescopes at different wavelengths. Yes, they do emit energy at wavelengths other than in the visible part of the spectrum.

Broadly speaking, the peak of their output depends on their temperatures. Most of the light from the outer giant planets emerges in the infrared for example.

However, some planets also are strong radio emitters. For example Jupiter has a strong magnetic field that accelerates charged particles which then emit radio waves.

Neutrinos are hard to pinpoint - telescopes currently struggle to say exactly where they come from in the sky. The planets are not expected to be strong neutrino emitters.

observation - Calculating area of visible sky

Approximating the Earth as a sphere with radius $R$, then when viewing from a height $h$ above the surface, the Earth blocks out a cone of some opening angle $2vartheta$, where $cscvartheta = 1+frac{h}{R}$. Thus, the visible portion has a solid angle of
$$Omega = 2pileft(1+cosvarthetaright) = 2pileft(1+frac{sqrt{h^2+2Rh}}{R+h}right)$$
steradians. Divide this by $4pi$ to obtain the fractional area of the the visible sky, compared to what you could have if the Earth wasn't blocking your view, since a full sphere subtends a solid angle of $4pi$ steradians. That is probably a more natural measure of the visible sky than a literal area.

For an actual area, you need some sort of reference distance $r$ to measure from, with the visible sky a distance $r$ away having area $A = Omega r^2$.