Why is Gaia operating around Earth orbit? Why not send it to Neptune's orbit? Gaia is an astrometry spacecraft that's currently operating around the Sun-Earth L2 Lagrangian point. Question: why here? Why not the Sun-Neptune L2 Lagrangian point? By orbiting the Sun at a larger distance, it should be able to get more accurate parallax measurements.

Only reason I can think of is cost. I'm not familiar with estimating how expensive space probes cost, but Wikipedia says Gaia cost ~\$1 billion and this is comparable to the cost of the Voyager program, which also cost about ~\$1 billion. Of course Gaia's instruments should be more sophisticated than Voyager's, but there were also two Voyager probes, not one.

Well, you thought about the spatial aspect of a parallax measurement, but not about the temporal one.

Gaia's intention is to measure 3D positions as well as 3D velocities. For the distance, you need accurate parallactic measurement, which come in with your orbital period.
For a typical Gaia-star with several measurement per year, you'll get 5 values of the parallax after 5 years of time, which you then average. If you'd send Gaia towards Neptune (besides the fact that no one has ever sent an orbiter, to say nothing of a L2 mission that far out) that has a period of 168 years, then after 5 years you'd get... 5/168 th of one paralactic measurement.

It simply couldn't achieve its science goals if put around the L2 behind Neptune. Also no one on this planet has any experience in putting something into a outer system L2 point. This is different than putting it into Earth's L2, because reaching the L2 around one of the giants has vast and very precise $\Delta v$ requirements. This would be a massive technological leap, and things don't work that way in space. Small, incremental technological steps are required in an anyways unfriendly environment, to make sure everything works properly and no millions of dollars have been wasted.
Compare that to Gaia's predecessor, the Hipparcos satellite, which was parked in geostationary orbit.

Now you could still say, why not use Jupiter hypothetically anyways. Well, the orbital period there is still 11 years, and Jupiter's L2 still suffers from the intense radiation environment that is provided by Jupiter's magnetosphere. This would lead to rapid degradation of the CCDs used for scanning across the sky.

Lots of interesting answers. Lots of unnecessary arguing.

1) Neptune sweeps out a parallax of nine billion km every 84 years, Earth sweeps out a parallax of 300 million km every six months so if you want max parallax, then Neptune is a 30 times better place to be. However, Neptune only sweeps out 53 million km in the same six months so you can get better results there but you will have to wait six times longer before you start to get them.

2) Earths L2 point moves with Earth so only requires getting a satellite to a speed of zero (relative to Earth) in the right place 1.5 million km away. remember that when you launch something away from Earth, the Earths gravity will start slowing it down as soon as the engines stop burning so all you have to do is run out of fuel at exactly the right place. Difficult and expensive to say the least but yes, we can manage that. Neptune is 4.5 billion km away, give or take 300 million km depending on the time of year. The good news is that if we could get a satellite out there, we would just require a stable orbit, we would not need to have a zero velocity (relative to Neptune) at Neptune's L2 point. The bad news is we do not know how to put a satellite into any sort of orbit around Neptune at a price we can afford.


What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured? Looking through ALMA on-line documentation and articles, the quantity PWV (precipitable water vapor) is a central theme. I have two linked questions.

What (actually) is precipitable water vapor, and is there a "non-precipitable" component also? Why not just call it "water vapor"?
How is it actually measured? It seems to be monitored by radiometers, but I don't understand how that can be used.

I usually see numbers of the order of 5 millimeter to 10 millimeters for very high altitude locations like Atacama. I assume this is the column-integrated amount, expressed in units of liquid, is that right?

As defined quite clearly, with equations, here

The total atmospheric water vapor contained in a vertical column of
unit cross-sectional area extending between any two specified levels,
commonly expressed in terms of the height to which that water
substance would stand if completely condensed and collected in a
vessel of the same unit cross section.

edit
There's, for example, an ALMA paper which discusses measurements and comparison with meteorological stations.

Source: http://suzaku.eorc.jaxa.jp/GLI2/adeos/Earth_View/eng/adeos02e.pdf

You can measure the error or water vapor by looking at an edge of something over a distance.  This could be possible by looking at the edge of a planet or the edge of something like a satellite.

The International Space Station travels in orbit around Earth at a speed of roughly 17,150 miles per hour (that's about 5 miles per second!). This means that the Space Station orbits Earth (and sees a sunrise) once every 92 minutes.

Its possible to find the ISS using the FREE SkyView on your phone and find the space station and perhaps study the image.

Its interesting to think about humidity measurements that we see from a weather website and compare these to what you actually see.  It maybe that the ideal place for a telescope is just above the tree line?  (for people who don't like numbers) However, there are some really nice urban telescopes and hopefully in the future we will have more telescopes in nice cities even if not all the nights are clear.

Its kinda fun to read an Indian paper on the topic...
https://journals.ametsoc.org/doi/pdf/10.1175/1520-0450%281990%29029%3C0665%3AAPOTWV%3E2.0.CO%3B2

(the Indian and Japanese papers both seem to show that the water vapor disappears at about 10-12 KM)?  However, some of the biggest telescopes in the world are being placed only about ~3000 meters up.  Its sometimes nice to place telescopes in urban areas where we can get to it.  Especially when the vapor doesn't really disappear until 10 KM which is way too high up anyway?!?!

For example: https://www.eso.org/public/teles-instr/elt/

(also remember that temperature is not a linear function with altitude... it goes down then up then down then up then down... i think... just look at the graph...)

Why water vapor is more "linear" is still one of those cosmic mysteries... I liked the x^2 equation the other person posted.


Which came first: Galaxies  Stars  Planets? If a galaxy is defined as a collection of planetary systems (and all matter in between), and a planetary system is defined as a collection of planets circling a star (and all matter in between), and a planet (and other smaller bodies, like asteroids, moonlets, moons, etc) is basically just a very very, very large collection of minerals, stardust and gases (which make up the entire universe), which came first: The planets, planetary systems or galaxies? I've been given to understand the universe came first...but after that?

I thought I understood, at least the layman's basics, of the universe formation. However, today, I’m having trouble conceptualising the order in which they would have been formed, IF they are defined by the smaller unit.

Is this just me getting the semantics of classification and the actual chronology of each feature forming, mixed up?

VERY Basically,

The universe is made up of a complex mix of everything (matter, antimatter, darkmatter, to name just a few), and expanding 'space'.
A very, very, large area of 'space' is covered in very, very, tiny particles: made up of at first gases and later 'stardust'.
A galaxy is formed of gravity-attracted gases, and after the very first stars went supernova and exploded there was also 'stardust', which combines to create 'clumps' of interstellar clouds (cloud nebulae).
The cloud nebulae are the 'nursery' areas for star formation (both initial stars and later generation stars).
The newly created star uses remnants of 'stardust' in the cloud nebulae to form planetary systems and then, a 'short' time later the orbiting planets (and other sized objects).

So then the chronological order would be:

Universe (Largest scale) => gases (very small scale) => galaxy => interstellar clouds => stars => planets (combined to form planetary systems). And the tiny particles of gases and 'stardust', are involved in nearly all scales.

Image taken from spaceexplained

Just writing out this question has helped! I hope you don't mind me posting it anyway, as it might help others and I still need confirmation that I have, hopefully, worked myself out of my conundrum (not deeper).

The structure we see in the Universe has formed from the gravitational collapse of the matter that was once an almost smooth density field of gas ("baryons") and dark matter$^1$. The word "almost" is important here, for if it had been completely — or even non-completely but much more — smooth, then the collapse would not have had the time to happen before the expansion of space has diluted the matter enough to prevent any collapse, and we would never had come into existence.
That is, the density field was slightly clumpy, and these clumps — or overdensities — existed on all scales.
But calculating which clump sizes collapse first — stellar-sized, galaxy-sized, cluster-sized, etc. — is far from trivial. Analytical attempts has to make several approximations, but can still make quite meaningful predictions which have subsequently been backed up and refined by numerical simulations, and by now, although many loose ends still exist, we have a rather good picture of structure formation:
Gravitational collapse
An overdensity is denoted by
$\delta\equiv\rho/\langle\rho\rangle - 1$, where $\rho$ is the density of the overdensity and $\langle\rho\rangle$ is the average density.
The evolution of overdensities under the force of gravity can be calculated exactly for $\delta \ll 1$ using linear perturbation theory, but when $\delta$ becomes of order unity, the non-linear regime is entered, and severe approximations must be made, so one turns instead to numerical simulations. It turns out that if $\delta\gtrsim1.68$ (i.e. if a region in space has a density which is 2.68 times the ambient density), it will collapse. The answer to your question is then given by what size of clumps first reach $\delta\gtrsim1.68$.
First clumps
Primordial quantum fluctuations$^2$ grew in size during the (admittedly still pretty blurry) inflation, a fraction of a second after Big Bang. In the young Universe dark energy was negligible, and the dynamics of the Universe was dominated by matter. Because dark matter comprises $\sim5/6$ of the total amount, we can initially neglect the presence of gas, but when the density becomes very high, gas pressure builds up and counteracts the collapse.
The overdensities amplified as matter started to collapse. It turns out that the density fluctuations are larger for smaller scales, so the smaller the clump, the sooner it will collapse. This results in the so-called "bottom-up" formation of structure, which is in contrast to what was originally thought; namely that galaxies formed "top-down" in a monolithic collapse (Eggen, Lynden-Bell, & Sandage 1962).
However, this approach neglects both gas and the motion of dark matter particles (treating it as so-called cold dark matter). Taking into account the effects of this puts a lower threshold for the masses of the structures of $\sim10^5$-$10^6M_\odot$ (e.g. Naoz et al. 2006; Yoshida 2009). Hence the first structures that formed are believed to be minihalos roughly the mass of globular clusters.
Stars, galaxies, and clusters
Stars consist of collapsed gas and almost no dark matter, and the formation of a star thus needs the theory of hydrodynamics rather than just gravity. In order for matter to collapse to such dense structures as stars, it must get rid of much of its energy. This is not possible for the collisionless dark matter (at least in the normal sense, but this post), but gas, which can collide and cool by radiating, is able to do so$^3$.
Through radiative cooling, the minihalos thus fragmented further into first gas clouds and then the first stars when the Universe was a few 100 million years old. Subsequently conglomerations of stars merged into galaxies, which eventually formed sheets, filaments and clusters of galaxies.
Dust
The first stars were formed out of pure hydrogen and helium (and small amounts of lithium), and the material for making planets did not exist. But when these stars — which were very massive — exploded as supernovae and polluted the interstellar medium with metals$^4$ and stardust$^5$, the formation of less massive stars became possible$^6$.
Planets
Dust particles stick together to form pebbles, rocks, and planets. Planet formation is probably not possible for too massive stars$^7$, but with the formation of small stars this became possible.
Timeline
We can make the following timeline:

Minihalos
Gas clouds
Stars
Galaxies
Stardust
Planets

However, note that the merging of minihalos to larger "galaxies" (without a significant amount of stars) also may happen at earlier epochs, and in particular that star formation (and dust and planets) is a continuous process which still takes place today (although the bulk of the star formation took place when the Universe approximately 3–6 billion years old (Madau et al. 1998)).

$^1$And radiation, which actually dominated the energy density until the Universe was $\sim50\,000$ years old.
$^2$The term primordial quantum fluctuations might be the coolest term in physics.
$^3$There are hypotheses of so-called dark stars, but that's beyond the scope of this text (primarily because I don't understand them).
$^4$For an astronomer, the term metal means any element heavier than helium. It's easier this way.
$^5$Astronomers just call it "dust", and use only the term stardust when talking to non-astronomers, in order for them not to fall asleep.
$^6$The reason is that with the many possible electronic transitions of metals, the gas has more ways to cool, and hence may collapse at earlier times, before the proto-star clump reaches high masses.
$^7$Because the high radiation pressure of very massive stars blows away and/or destroys the dust (I think, but this is not my expertise).

The short answer is we don't know.

A longer pedantic answer would point out that use of a term without understanding what you mean by that term is never appropriate.

The term I feel is problematical is "stardust". In our standard picture of big bang nucleosynthesis, we are optimistic (because of the good agreement between calculations and observations of cosmological abundances) that we understand that as the Universe cooled hydrogen, helium, and traces of other elements formed (lithium is the one most often mentioned). It is almost NEVER pointed out that if the Universe is composed of 27% dark matter and only 5% regular matter (mostly hydrogen) then these calculations are an abysmal failure: they don't predict the existence of dark matter! All this is ignored and we congratulate ourselves (or rather the popular media lionize science's "accomplishments"). You can't ignore most of the matter in the Universe and expect that your models will be accurate.

Whether the first star preceded or followed the first galaxy is to some extent a matter of definition. To some extent. It isn't at all clear whether or not any black holes existed prior to galaxy/star formation.

If super black holes existed, then anything within their gravitational range would comprise a galaxy (if you chose to use a reasonable definition of galaxy). What would exist then would be gas (and plasma), dark matter, and the beginning traces of what you'd probably say is stardust: the lithium, boron, etc. which would agglomerate as it collided with similar atoms. The density of this dust would be far too low, I think, to ever form planets without other objects gravity intervening. (Although I probably should never say never). OTOH, if the only black holes existing were sufficiently small, they'd probably evaporate with only transient effects on the particle dynamics. So, we can ignore stardust, most of that was formed IN the first (Generation III) stars.

The question is: which came first; galaxies or stars? And the answer is: what exactly defines a galaxy? We believe there were density fluctuations which would lead to gravitational collapse of the gas surrounding them (and composing them). At what point does this motion qualify as being distinguishable from other motions (other volumes of collapse)? Perhaps you can guess that it will depend on both the dark matter present AND the velocity (temperature) of the gas. A careful analysis would probably involve escape velocities (and as you probably know, that depends on the dark matter distribution) and the point at which "most" gas was orbiting rather than just randomly colliding which would in turn rest on what "most" means (51%, 90%, 99%,...).

So define the term galaxy. If you define it in terms of stars, then the answer is easy: (some) stars came first. If you define it in terms of zones of gravitation, then it is possible that the answer is both. That is, in some cases hydrogen had enough time to collapse into a star before "most" gas was bound to a near by "galaxy" (probably proto-galaxy would be a better term here) and in other places the protogalaxy had defined itself before the first star lit up. We don't know.

It is a mistake to assume that the Universe as we see it today is a reasonable approximation of the Universe as it would appear only 10 or 100 million years after the big bang.

Our models aren't good enough yet (and probably won't be until we understand a lot more about what dark matter is and how it behaves) to be confident that protogalaxies existed before the first stars or that the first stars lit up before the zones I'm calling protogalaxies were separate systems.


Does a planet's mass or gravity affect the height of it's mountains? According to this Wikipedia page, the top five highest mountains on Mars (and the highest on Venus) are all taller than Mount Everest (and Mauna Kea as measured from the ocean floor).

Does a planet's mass or gravity affect the height of it's mountains?  Is there an upper limit that a mountain can reach given the mass or gravity of a planet?

Yes, gravity definitely affects the maximum heigh of mountains.

Think on a solid bar of steel. It sticks straight because of the electronic forces. But when you make it larger and larger gravity makes it bend: gravity starts being considerable, but still smaller than electronic forces.

If you make the bar larger, there will be a moment in which the weight of the whole bar will be larger than the short-range electronic force: your bar will break purely due to gravity.

Exactly the same happens to mountains made of solid rock (as opposed to sedimentary ones quoted by Hobbes). There is a point,k depending on the strenght of planetary gravity, where it takes over short-range electronic forces, making the mountain collapse.

This is exactly the force that "rounds up" the planets, as opposed to the non-spherical asteroids.

The mountains actually form as a result of the tectonic plates' movement at the earth mantle. Suggesting that a new mountain would be formed tomorrow, during the tectonic activity rock from the mantle and above moves upwards, only the tip of the newly formed mountain would consist of sediments like sand or soil.

So, I don't think that the gravity or mass of a planet affects the mountains' height.

Plus, Earth is the only known planet to be affected by plate tectonics. So, the 'birth' mechanisms must be a lot different and cannot be compared with the ones' acting on our planet.


Is it certain that dark matter consists of particles? (And not just curved space) The main idea about dark matter today seems to be supersymmetry. That there exists a set of yet undiscovered fundamental particles which (in a mathematically most convenient way) mirror the particles already known.

But has it been established that dark matter consists of particles at all? Couldn't it for example be a continuous property of space itself in some regions? That some parts of space are curved without the influence of any particles or waves. Is anything but particles ruled out?

Also asked at https://physics.stackexchange.com/questions/174080/how-do-we-know-dark-matter-isnt-curved-spacetime

Basically no. Or at least you can't have this idea and General Relativity. GR demands that you have something (matter/energy density) to cause the curvature. Curvature without cause is not part of the model. That's not to say that what you propose couldn't be true, but one would then have to ditch GR, which is a reasonably successful model (though some would argue that the dark matter problem is one of its biggest failures!).

I think a lot of parameter space has been ruled out for dark matter. We know it can't be baryonic from our understanding of primordial element abundances and also from the way it interacts (or doesn't interact) with normal matter in forming the basic structures in our universe on the right timescales. The microlensing surveys that have been conducted rule out large (planetary sized and above) dark matter bodies (and black holes, brown dwarfs, cold white dwarfs etc.), since to account for all the mass, these would produce lots of microlensing events that simply aren't observed.

An excellent primer on these topics has been produced by Garrett & Duda (2011).

In terms of dark matter, there are two notions which are incorrect. One is that dark matter is a clump of stuff traveling with the matter. The other is that dark matter does not interact with matter.

Dark matter fills 'empty' space. Dark matter is displaced by matter.

The Milky Way moves through and displaces the dark matter.

The Milky Way's halo is the state of displacement of the dark matter.

The state of displacement of the dark matter is also known as the deformation of spacetime.

The Milky Way's halo is the deformation of spacetime.

Dark matter is the physical manifestation of spacetime.

'Ether and the Theory of Relativity by Albert Einstein'
http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html

"Think of waves on the surface of water. Here we can describe two entirely different things. Either we may observe how the undulatory surface forming the boundary between water and air alters in the course of time; or else-with the help of small floats, for instance - we can observe how the position of the separate particles of water alters in the course of time. If the existence of such floats for tracking the motion of the particles of a fluid were a fundamental impossibility in physics - if, in fact nothing else whatever were observable than the shape of the space occupied by the water as it varies in time, we should have no ground for the assumption that water consists of movable particles. But all the same we could characterise it as a medium."

if, in fact nothing else whatever were observable than the shape of the space occupied by the dark matter as it varies in time, we should have no ground for the assumption that dark matter consists of movable particles. But all the same we could characterise it as a medium having mass which is displaced by the particles of matter which exist in it and move through it.


Where can I still find solar film? I, like many others, am excited to see the eclipse. I want to view the eclipse through my telescope, however, I don't have any solar safety equipment for solar viewing and I cannot find any online that I can get before the eclipse. Could anyone lead me to a reputable site where I could purchase solar (preferably mylar) film before the eclipse?

Amazon has solar film sheets if that's what you're looking for. It is sold by Amazon for Thousand Oaks Optical. Amazon sometimes has next day or 2 days delivery.

If you know somebody who still owns floppy disks, you might consider recycling their inside as filter. Please test before relying on them for protecting your eyes. I did use them during the last solar eclipse as filter for my analog camera with reasonable results.
References

https://www.instructables.com/Floppy-Disk-Sun-Filter/


Can radioactive decay rates be confirmed by looking at supernova? Radiometric dating relies on accurate knowledge of the half-lives of radioactive elements. Skeptics have argued that since many elements' half-lives are too long for anyone to have observed them, we can't know what they were in the distant past.

There are various counterarguments based on terrestrial evidence, like agreement between elements in the same sample, and corroboration from varves, tree rings, and ice layers.

Another counterargument I've seen is by looking at supernova, we get a window "back in time" to when their light began its long journey to us, and that much of this light was generated by radioactive decay, confirming that decay rates were the same then as now.

I don't quite follow this explanation, and would like some help. Distant starlight doesn't give us a rock sample in which to analyze proportions of parent and daughter elements. Does it still somehow give evidence of specific radioactive decay rates? If so, how?

I know next to nothing about astronomy, so please explain this to me like I'm a child.

I think I understand this now. I was trying to imagine how you could take a snapshot of a supernova and infer from it the half-life some some element. In fact, I think the measurement comes from watching the radioactive decay happen over time, in the aftermath of a supernova.
Openstax Astronomy 23.3 "Supernova Observations" shows a graph entitled "Figure 23.13 Change in the Brightness of SN 1987A over Time." It shows brightness increasing during the days right after the supernova, and the rest of the curve not following a consistent fading pattern. It explains:

One of the elements formed in a supernova explosion is radioactive nickel, with an atomic mass of 56 (that is, the total number of protons plus neutrons in its nucleus is 56). Nickel-56 is unstable and changes spontaneously (with a half-life of about 6 days) to cobalt-56. (Recall that a half-life is the time it takes for half the nuclei in a sample to undergo radioactive decay.) Cobalt-56 in turn decays with a half-life of about 77 days to iron-56, which is stable. Energetic gamma rays are emitted when these radioactive nuclei decay. Those gamma rays then serve as a new source of energy for the expanding layers of the supernova. The gamma rays are absorbed in the overlying gas and re-emitted at visible wavelengths, keeping the remains of the star bright.
As you can see in Figure 23.13, astronomers did observe brightening due to radioactive nuclei in the first few months following the supernova’s outburst and then saw the extra light die away as more and more of the radioactive nuclei decayed to stable iron. The gamma-ray heating was responsible for virtually all of the radiation detected from SN 1987A after day 40. Some gamma rays also escaped directly without being absorbed. These were detected by Earth-orbiting telescopes at the wavelengths expected for the decay of radioactive nickel and cobalt, clearly confirming our understanding that new elements were indeed formed in the crucible of the supernova.

So if I understand correctly, the logic would be:

SN 1987A occurred 168,000 light years from earth
Therefore, humans observed it 168,000 years after it happened
As they watched the supernova unfold, astronomers saw radiation that appeared to be generated by the decay of known, short-lived isotopes, and that decay happened at the same rates we observe today
Therefore, the decay rates of these isotopes were the same 168,000 years ago as they are today

When an unstable atom decays, it releases energy. At an earlier time, there are more unstable atoms, so the released energy is larger. At different times, there are different atoms left, so the energy released is different. You can measure the decay rate by following the time-series of this energy.

$$ E(t) = E_0 \exp[-\lambda t]$$

where $E$ is the energy, $t$ is time, $E_0$ is energy at $t=0$, $\lambda$ is the decay constant. This is a very simple picture. In reality, it will be a bit more complicated because of other effects causing the deviation of the observed energy, such as scattering.


How far away is the light that would reveal the Big Bang? I guess theoretically if we could go faster than light, which we clearly cannot, at some point we'd be able to see the big bang itself. Just curious - how far away is that light?

The particle horizon is the point where light arriving at us now would've had to have been emitted at the big bang (though note this is hypothetical as the Universe did not become transparent to light until it was several 100,000 years old).. How far away it is depend on the model you use (not to mention there are several different distance measures in cosmology), but it's proper distance is reckoned to be about 46.9 billion light years.

A note of caution though, when calculating the particle horizon cosmic inflation is usually ignored. What is light arriving (again hypothetically) from usually called the particle horizon therefore would be actually light that was emitted from just the end of the inflationary epoch. If you were to take a model with a big bang singularity and inflation the "true particle horizon" would in fact be much further away.

the light from the big bang is in your living room... because that is where the big bang took place... did you know you were so special? Just kidding, but maybe I misunderstood your question. I think what you are asking is, well if you look into deep space you are looking back in time since it takes light X amount of years to travel to the observer. So look far enough back and you would see the big bang right? We already thought of it, and the problem is you can't see the moment of the big bang, because there were no such thing as photon's at that moment. It took about 400,000 years (an afternoon nap on the scale of the universe) for the intensity of the moment of creation to simmer down enough atoms could form removing the opacity of the universe freeing photons to begin their cosmic journeys.. If you were to look out that far you would just see a big opaque blob like trying to look into the center of the sun.  The light from that moment still exists and you do see it... you just don't SEE it. The spectrum of the sun's light that has the highest intensity is in the visible light wavelength.  One would imagine that all species of the universe see in the same wavelength and thus we would all agree on what visible light is. This is not true. We see in the visible wavelengths because of evolution.  Our eyes adapted to the highest intensity light available. Think about natural selection. Ancient primates who could see better would be better hunters. So those who were utilizing the highest intensity light the best would be ideal mating candidates giving birth to more off spring with eyes like there own (a little theory of mine). If you grew up on a planet with a sun that gave off a lot of infrared light you would see in infrared.  Anyway, what I am getting at is if we somehow managed to evolve to this point on a rock with no companion star (an impossibility for all intents and purposes) in some dark corner of the universe where there was no other really intense light around, theoretically eyes would have evolved to see in the cosmic microwave background because it would be the only light available.  CMB radiation (the afterglow of the big bang) is ubiquitous. It is everywhere at once. It gets cooler as time goes on as space time stretches, stretching the wave length of the light making it harder and harder to detect. I hope that answered your question.


Did atoms in human body indeed come from stars? I think I am not alone who saw videos about that we (humans) are made of same atoms which someday were in stars. In other words, some atoms in our bodies are from stars which exploded billions of years ago.

I wonder if it is indeed true. I mean human's life begins when sperm cell fertilize egg cell. Now does that sperm cell or egg cell indeed contains some of the exact atoms from those stars?

I know little bit strange question, but would be interesting to hear if it is indeed true, that atoms in our body are same which someday were in stars.

In case you are wondering that's the video I am talking about: http://www.youtube.com/watch?v=9D05ej8u-gU

The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.

So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.

After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that's all. No oxygen, iron, carbon etc.

Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies must have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).

But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.

If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the s-process. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with $A<90$) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).

Iron, nickel and many other elements such as sulphur and silicon are also produced during type Ia supernovae. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium.

All these different processes produce distinctive patterns of element abundances.

The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.

Analysis of "presolar grains" found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.

[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my Physics SE answer to this question. ]

I might not be that much conceptual to you, but do you believe that stars are made up of Hydrogen (and Helium too). Hydrogen is a part of all the organic compounds. We humans are made up of billions of organic compounds, which in turn do contain Carbon, Hydrogen and Oxygen as their basic element and other compounds such as Sulphur, Phosphorus and other fats. And I guess a biology student would tell you more clearly what a sperm contains and what and how do these elements ressembles to those who are in Stars.

When a star collapses, it liberates out elements such as Carbon, Iron etc. Which maybe you're talking about. So that sperm consists of the same elements that the Star had at the time of its death (well not death you might know a better word for the star's ending period).

However, this doesn't make sense that these atoms are really the exact same or the actuall ones that were in the Stars many years ago. No such theory has been provided.

What does Star contain and what it liberates: http://answers.yahoo.com/question/index?qid=20080908195830AA5Iheb


How does gravity really work I am only 12 years old and I'm constantly wondering and trying understand how gravity really works. On YouTube everyone always talks about objects warping space time around themselves and uses the analogy of a trampoline. I still don't understand gravity because if space were like a trampoline, then earth would be spiraling in towards the sun along with all the other planets, right? So could someone explain to me how gravity really works without the trampoline analogy?

First of all: "How gravity really works" is a deep question, and any serious scientist would quickly concede that all we have is an incomplete working model. You certainly have heard about  General Relativity; the first image on the page is your trampoline.

Our working model, General Relativity, is working because it explains a lot of observations very nicely. (Careful, here is another deep question lingering: "Explains" means that we can predict some observations from other observations with the model of gravity we have in our  mind. It does not necessarily mean that we understand the "real nature" of the underlying issues.) But we are very confident that the model is working over a wide range of observations. One of the last "first-time" observations which followed the predictions and thus gave us more confidence in the model was the two black holes colliding lately. Lately? Well, billions of years ago. We just learned about it lately.  Here is a link to a New York Times article with an impressive video. (I think one can still read a limited number of Times articles for free, so try it out.)

Our model of gravity is incomplete because it doesn't connect well to the model of nature we have for other things (elementary particles, quantum physics). For a while (like 70 years or so) it didn't connect at all; Einstein himself completely failed to connect the dots, which was probably not encouraging since he had received the Nobel Price for laying one of the foundations of quantum physics and was the obvious authority about gravity. If he couldn't do it, who could?

If I'm not mistaken, the physicists today are making progress, slowly. This connection between quantum physics and gravity is one of the main unresolved problems in modern physics.

Last, let me address your concern about the planets spiraling into the sun. This idea probably comes from actual balls on an actual trampoline spiraling in, I suppose. You probably know that the balls lose speed due to friction, much the same way you slow down on your bike when you stop pedaling. Some of the kinetic energy is transformed into heat.

And you know what? You are right. Given enough time, the planets would eventually fall into the sun. Low-flying satellites fall back to earth after a few years, because there are still traces of atmosphere slowing them down out there. The reason is that there is "friction" in the wider sense involved in all large-scale processes in the universe. That is actually one of the fundamental physical principles making up the world we know. It's just that the near-vacuum between the planets doesn't provide that much friction, and the planets are fairly massive bodies with an enormous mass and kinetic energy. It will take a long long time for them to lose enough energy that they'll be so close as to touch the sun. (Perhaps too long to happen at all.) In fact, over human life times the planets, moons and stuff are almost perfect examples for movement without friction. But in the astronomical time scale -- billions of years --, there certainly is friction. For example, the moon is showing us always the same side because friction slowed its rotation so that the rotation is now "locked" with its orbit.

Bottom line: The idea that gravity bends space and time "explains" all large-scale observations so far; the "trampoline" is a good model for a 2-dimensional "space", i.e. a surface, if you ignore friction.

Neutrino String Induction-Refraction is the cause of gravity. Some will say that neutrinos are insignificant, but Dirac, Hawking, and Tyson think otherwise and most discount the effect of a charged particle traveling at the speed of light. Keep in mind that no one can, or has proved that mass is a property of matter, more of an effect.

Go to www.themechanismofreality.com , this site explains exactly how gravity works. Every Physicist who examines this agrees that this is correct. From CERN to the University of Beijing's Physic Department agree that this is a 'Fantastic connection between graviton physics and string theory"! This was also confirmed , indirectly, by LIGO and the gravity wave announcement. Enjoy!


Without using absolute magnitudes or isochrones, how might we tell a star's age and evolutionary status? Usual methods of estimating stellar ages involve isochrone approximations. It can also help to estimate a star's radius by correlating its absolute magnitude with effective temperature and apparent magnitude. In the absence of these measurements or observations of tell-tale variability, how might you guess a star's age and evolutionary status?

Given a single, high-resolution $(R\gtrsim 50000)$ spectrum as your only data point, how easy is it to accurately infer the age and evolutionary status of a star? For example, how would the spectrum differ between a red dwarf and red giant, both with of $T_{\mathrm{eff}} = 4000~\mathrm{K}$? Or between two red dwarfs of ages $2~\mathrm{Gyr}$ and $8~\mathrm{Gyr}$?

A good answer could describe how surface gravity $(\log g)$ affects spectral lines (and how this relates to stellar mass and radius), what elements we might observe more strongly at different stages of evolution, and some observational results of gyrochronology.

The spectra of a red giant and a red dwarf are completely different, so there isn't really too much to say about this and distinguishing giants and dwarfs is simple. For example, alkali lines are almost non-existent in red giants, but strong in red dwarfs. The theory as to why this happens is to do with the surface gravity and pressure broadening; it is the stuff of a standard graduate/undergraduate course on stellar atmospheres, not an SE answer.

The fact is that a R=50,000 spectrum with decent signal to noise ratio will quite easily
give you the temperature (to 100K), surface gravity (to 0.1 dex) and metallicity (to 0.05 dex), plus a host of other elemental abundances (including Li) to precisions of about 0.1 dex.

What can you do with this:

You can plot the star in the log g vs Teff plane and compare it with theoretical isochrones appropriate for the star's metallicity. This is the best way to estimate the age of a solar-type (or more massive) star, even if you don't have a distance and is the most-used method. How well this works and how unambiguously depends on the star's evolutionary stage. For stars like the Sun, you get an age precision of maybe 2 Gyr. For lower mass stars, well they hardly move whilst on the main sequence in 10Gyr, so you can't estimate the age like this unless you know the object is a pre-main sequence star (see below).

You can look at the Li abundance. Li abundance falls with age for solar-mass stars and below. This would work quite well for sun-like stars from ages of 0.3-2Gyr and for K-type stars from 0.1-0.5 Gyr and for M-dwarfs between 0.02-0.1 Gyr - i.e. in the range from where Li  starts to be depleted in the photosphere until the age where it is all gone. Typical precision might be a factor of two. A high Li abundance in K and M dwarfs usually indicates a pre main sequence status.

Gyrochronology is not much help - that requires a rotation period. However you can use the relationship between rotation rate (measured in your spectrum as projected rotation velocity) and age. Again, the applicability varies with mass, but in the opposite way to Li. M-dwarfs maintain fast rotation for longer than G-dwarfs. Of course you have the problem of uncertain inclination angle.

That brings us to activity-age relations. You can measure the levels of chromospheric magnetic activity in the spectrum. Then combine this with empirical relationships between activity and age (e.g. Mamajek & Hillenbrand 2008). This can give you the age to a factor of two for stars older than a few hundred Myr. Its poorly calibrated for stars less massive than the Sun though. But in general a more active M-dwarf is likely to be younger than a less active M dwarf. It should certainly distinguish between a 2Gyr and 8Gyr M dwarf.

If you measure the line of sight velocity from your spectrum, this can give you at least a probabilistic idea of what stellar population the star belongs to. Higher velocities would tend to indicate an older star. This would work better if you had the proper motion (and preferably the distance too, roll on the Gaia results).

Similarly, in a probabilistic sense, low metallicity stars are older than high metallicity stars. If you were talking about stars as old as 8Gyr, these would be quite likely to have low metallicity.

In summary. If you are talking about G-dwarfs you can ages to precisions of about 20% using log g and Teff from the spectrum. For M dwarfs, unless you are fortunate enough to be looking at a young PMS object with Li, then your precision is going to be a few Gyr at best for an individual object, though combining probabilistic estimates from activity, metallicity and kinematics simultaneously might narrow this a bit.

As an add-on I'll also mention radio-isotope dating. If you can measure the abundances of isotopes of U and Th with long half lives and then make some guess at their initial abundances using other r-process elements as a guide then you get an age estimate - "nucleocosmochronology". Currently, these are very inaccurate - factors of 2 differences for the same star depending on what methods you adopt.

Read Soderblom (2013); Jeffries (2014).

EDIT: Since I wrote this answer, there is at least one more promising method that has emerged. It turns out that the abundance of certain s-process elements (e.g. barium, yttrium) are enriched quite slowly during the lifetime of the Galaxy (by the winds of dying asymptotic giant branch stars), and more slowly than enrichment with iron and much more slowly than alpha elements like Mg and Si. Thus a measurement of the relative fractions of these elements, like [Y/Mg] can give the age to precisions of a billion years or so (e.g. Tucci Maia et al. 2016; Jofre et al. 2020). This method is probably the best for solar-type stars older than a Gyr, but remains unexplored/uncalibrated for lower mass stars.

For some very large (and hence relatively cool) red giants you might be able to ascertain something from their spectra, as emission lines are sometimes seen - these are typically brighter central patches seen the middle of the more typical absorption (dark) spectral lines - caused by the large size of (realtively!) hot gas clouds that surround the giants. But that would not be a reliable method of red giant detection.


How close to a brown dwarf would a planet need to be to receive as much light as Earth? How close to a brown dwarf would an orbiting planet need to be in order to receive as much sunlight as Earth receives from the Sun?

Brown dwarfs are born moderately hot and luminous and then they contract and cool. Thanks to electron degeneracy in their cores, they never become hot enough to ignite hydrogen (though there is a brief deuterium burning phase) and as a result their fate is to cool and fade.

The plot below (from Burrows et al. 1997) shows how the luminosity behaves as a function of time since birth (age) for objects of various mass. The brown dwarfs are the green and red tracks (some would call the red tracks giant planets). The curves are labelled in Jupiter masses. The highest mass brown dwarf is 73 Jupiter masses, the next one down is 70$M_J$ and then the green curves count down in steps of 5$M_J$ to 15$M_J$, then the first red curve is at 13$M_J$ and the red curves count down in steps of 1$M_J$.

The axes are logarithmic (to base 10). Thus when the y-axis says -2, it means the brown dwarf is one hundredth the luminosity of the Sun, -3 = one thousandth etc. The x-axis is logarithmic in units of billions of years. Most brown dwarfs observed in the Galaxy will be in the range 1-10 billion years (i.e. between 0 and 1 on this x-axis).

The flux received by a planet in orbit around a star/brown dwarf will be proportional to the its luminosity divided by orbital radius squared. From this we can write that the orbital radius where a similar flux to that of the Earth around the Sun will be received is
$$ R = \left( \frac{L_{BD}}{L_{\odot}} \right)^{1/2}\ {\rm au} $$
$$ \log\left(\frac{R}{1\ {\rm au}}\right) = 0.5 \log \left(\frac{L}{L_{\odot}}\right)$$

This you can get you answer immediately from the graph below. Choose the mass of your brown dwarf and its age. Find the log luminosity on the y-axis and halve this value. That is the the log of the orbital radius in astronomical units where a planet would receive the same amount of flux as the Earth from the Sun.

e.g. Consider a 50$M_J$ brown dwarf that is a billion years old. Find the appropriate green curve and see that the y-axis value for x=0 (1 billion years) is -4.2. The log of the orbital radius (in au) at which a planet would receive the same amount of flux from the brown dwarf as does the Earth from the Sun would be -2.1. In linear units, this is $10^{-2.1} = 0.0079$ au, or 1.19 million km.

A further issue to consider is that the brown dwarf surface will be far cooler than the Sun and also changes (cools) with time. This means that the radiation received from the planet will push further and further into the infrared as the brown dwarfs gets older and hardly any of the received radiation will be in the visible part of the spectrum.

Brown dwarves don't have fusion in their core, thus they don't have their own light. They may have a little heat production from other processes (for example, contraction, or radioactive decay in their core). These aren't enough strong to heat even the dwarf significantly.

Many brown dwarve were found orbiting a star closely, they may have surface temperature even in the order of some thousands K. But they get their temperature from their stars.

For example, the Jupiter's temperature is only 40K more warm as it would be reasoned by the Sun.

In the case of red dwarves, the situation is better. The distance depends on the star. You can very easily calculate it: check its absolute luminosity, so you can see its total power output, compared to the Sun.

The radiation what a planet receives (per surface area) depends quadratically from its distance.


Why don't we have in-between planets? The planets in our system are most often grouped into two categories:

Terrestrial:

Mercury
Venus
Earth
Mars

Gas Giants:

Jupiter
Saturn
Uranus
Neptune

Why is it that we don't see "in-between" planets in our system - large, rocky worlds larger than Earth but with thick atmospheres somewhat similar to those of gas giants? Are they possible? Did they not occur by random chance?

Super-Earths and Mini-Neptunes are the "in-between" types of exoplanets you're looking for. A sweeping generalization would put most in the range of $\sim1$-$10M_{\oplus}$ (Earth masses), with some outliers a bit above that. They may have significant quantities of hydrogen and helium in their atmospheres, as well as water, in liquid or vapor form. The latter should also have what are referred to as volatiles (colloquially called "ices") - compounds such as ammonia and water that are present in the atmospheres of proper ice giants like Uranus and Neptune.

One reason we don't have any of these in the Solar System is simply that no worlds with the proper mass formed. I explained here that there is a transition region of sorts, but most super-Earths require minimum masses of $1.5$-$2M_{\oplus}$ to keep their large hydrogen/helium envelopes (which terrestrial planets may accrete early in their lives but later lose). There would have to be a planet just in that mass range and at the right distance from the Sun (see D'Angelo & Bodenheimer (2016) for general constraints on semi-major axes). The environment in the protoplanetary disk may simply not have been suitable to produce the right body at the right time.

Since 2016 a ninth planet is hypothesized beyond the Kuiper belt which, if it exists and if it's a planet, is either a super-Earth or a mini-Neptune.

But you might want to consider Venus a planet in-between. While it isn't bigger than Earth, on the surface of Venus the air pressure is almost a hundred times that on Earth which makes the atmosphere a supercritical fluid there like those of the gas giants. For comparison, the air pressure within Uranus and Neptune is more than a thousand times that of Earth and within Jupiter and Saturn more than a million. So Venus is kind of inbetween. From space, Venus looks similar to the gas planets since you can't see its surface at all, but its cloud layers.


A good telescope for the viewing of Nebulae, Stars and Planets So, I am a 15 year old interested in astronomy. My father used to have a reflector telescope, but was never so serious about it that he could recommend me a good telescope. I am looking to buy a telescope up to £400 (possibly £500) and would like one that is mainly great for looking at nebulae. Stars and Planets would be great also, but are not necessary. Light Pollution isn't too bad. I have been recommended a Newtonian Reflector, but am not too sure if this would be suitable for my conditions. Also, does anyone recommend a good, British website where I could buy such a scope from? Finally, is there much point in me buying a computerized scope, will this allow me to align the scope better, or is it just going to be a waste of money.

This is a very common question, yet very hard to answer if you prefer a clear, concise, uncontroversial answer that applies to all situations. So I'm not going to do that.

Instead, I'm going to describe your main options, and let you choose. Be aware that you'll make the choice while still not knowing much about optics. So, in a sense, it will be just the beginning of a learning journey, and you'll have to keep learning and trying things out. Hope you like that sort of thing, because it's kind of mandatory in this hobby.

Also, there are two kinds of people in this field: those who enjoy using the instruments, and those who enjoy making the instruments. I'm 25% of the former, 75% of the latter kind (I make telescopes and mirrors, and I do some optics design). So that's the perspective you'll get.

First off, a few words about the objects you'll observe. There are two kinds of objects:

HIGH RESOLUTION, HIGH BRIGHTNESS

These are the planets, the Moon, many double stars, and the like. These are typically pretty bright, so the amount of light you capture is not essential. But these objects show high-resolution detail, so the resolving power of your scope is crucial.

You get more resolving power if the aperture (the diameter) of the scope is bigger. But there's a problem - seeing (the inverse of turbulence).

http://en.wikipedia.org/wiki/Astronomical_seeing

When air is turbulent, it will reduce the resolving power of your instrument. If your scope is 200 mm (8") or bigger, this will happen A LOT. If your instrument is 100 mm (4") or smaller, it's not so noticeable because you don't have that much resolving power to begin with.

There are ways to forecast seeing, because it depends on weather:

http://cleardarksky.com/csk/

So, for high resolution targets, the point of having a large instrument (like a huge dobsonian) is because once in a while seeing is so good (turbulence is so low) that you can actually use all its huge power - but most of the time you'll be using it at much less than its theoretical power, as if it's a 4" instrument or whatever the seeing is at the moment.

These objects are not affected by light pollution. You could observe the Moon from a brightly illumined parking lot in the middle of the city. You don't need to be in deep darkness while observing them - in fact, it's beneficial if you're not adapted to very deep darkness.

LOW BRIGHTNESS

These are most of the nebulae, star clusters, and galaxies. These are distant, faint objects. Their light is very faint, so it's essential you capture as much of it as possible. Again, bigger aperture is better.

These are the kind of objects that huge dobsonians (half-meter aperture, or more) are made for. In a very large scope, there's galaxies everywhere.

But again, there's a catch. These objects are affected by light pollution.

http://www.jshine.net/astronomy/dark_sky/

You need to take your scope far and away from any city, town and industrial complex (at least a 1 hour trip at highway speed). This is pretty hard to do in Europe; in the US, population density is low and this is somewhat easier. Anyway, just stay away from city lights and the view will improve.

A few of these objects will be visible from the city: the Andromeda galaxy, the M13 cluster, the Orion nebula, etc. But many of them look much, much better away from light polution. In any case, you should be able to observe the whole Messier catalogue (over 100 objects) from the city - but out in the boondocks these things look better.

http://en.wikipedia.org/wiki/List_of_Messier_objects

It's important that your eyes are adapted to very deep darkness when observing these things, so stay away from any source of light. Even the screen of your phone is far too bright. Street lights 500 meters away are too bright. And turn off that flashlight and keep it off. One second of exposure to a bright light destroys a dark adaptation that took 20 minutes to build.

Since these objects are so faint, the eye doesn't see a whole lot of high-resolution detail, so they are not affected by seeing (as far as the human eye is concerned).

Okay, now here are the main types of instruments that you're most likely to encounter:

NEWTONIAN REFLECTORS IN GENERAL (DOBSONIANS IN PARTICULAR)

Many people will say "just get an 8 inch dob as your first scope", and that's usually correct. This design gives the biggest aperture per price. Biggest bang for the buck. It scales up easily to very, very large sizes. It's the ideal instrument for observing the "faint fuzzies" (nebulae, galaxies). It's great for learning, because you can easily observe its guts. When seeing is good, a large dob will open up the sky for you - you can crank up the magnification a lot (depending on aperture).

On the flip side, due to the low price, you will see many dobs out there that are shoddily made.

You need to remember two things with a dob: collimation, and thermal management.

Collimation means keeping all the optics aligned. The mirrors will move a little, due to the way a dob is built, and they'll lose collimation. So don't forget to re-adjust it once in a while. In time, you'll get so good at it, you'll give it a two-minute quick adjustment every time you observe, and it'll be good to go; that's what I do.

This is why collimation is important:

http://legault.perso.sfr.fr/collim.html

This is a good introduction to newtonian collimation (which is a vast topic, you could write a whole book on it):

http://www.cloudynights.com/documents/primer.pdf

Thermal management means keeping the scope at the same temperature as the ambient air. If you don't do this, internal convection will destroy the image. This is why:

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/

Small dobs (up to 6...8"), just take it outside at least 1 hour before you observe - or better yet, 2 hours. Let it "breathe". Dobs 10" or larger, you MUST use a fan on the primary mirror to force cooling.

REFRACTORS

These instruments are very convenient and very low maintenance. Take it out of the box, and it's good to go. It's collimated from factory, and it doesn't care about thermal issues. A refractor makes a great "travelscope", and a great grab-and-go scope.

On the flip side, they are the most expensive, in terms of how much aperture you get for the money you spend. And they are typically made in small apertures; most commercial refractors are not bigger than 6" or so; those that are bigger are extremely expensive. The small aperture limits the maximum performance of these instruments.

There is this myth out there that refractors are somehow magically better for planets (and other high-resolution targets), whereas dobs, even very large ones, are not so great. This is false; load up the designs in Zemax, do a basic performance analysis, and you'll see that aperture is the dominant factor, with everything else being secondary. But there are several reasons why this myth is around, and it's instructive to see what these reasons are:

refractors perform at 100% of their potential no matter what, right out of the box. Dobs require maintenance and fiddling to perform at 100%.
many dob owners simply don't do the required maintenance - they don't collimate their scope, and don't give it time to reach thermal equilibrium. Then of course it will perform miserably on planets. There are 12" dobs out there that perform worse than 4" refractors because of how out of collimation they are and how hot the mirror is.
many refractors out there are quite well made (they ought to be, given the higher price you're paying), whereas many dobs are cheaply made in a sweatshop. So you're not comparing apples to apples.

Bottom line: don't buy a refractor because you "want to watch planets". But do buy a refractor if simplicity of use and the low maintenance are very important to you (and you don't care much about price).

A decent refractor on an excellent mount can do pretty good astrophotography. More on that below.

CATADIOPTRICS (CASSEGRAIN, MAKSUTOV, DALL-KIRKHAM, RITCHIE-CHRETIEN)

Broadly speaking, these are for astrophotography (not always, but usually). But can you do visual astronomy with them? Sure. It's just that it's easier to optimize them for photo, with this design.

Usually, they come installed on a motorized mount. All but the very cheapest ones have a computer on-board; you just punch the name of the object into the remote control, and the scope will turn around automatically to face the object. Sounds cool, right?

Well, I am going to say something a little controversial here: this is not a good place to start.

First off, astrophotography has an extremely steep learning curve. It's very easy to take a little blurry photo of Jupiter that will impress nobody. It's very hard to take a great, high resolution photo like these:

http://www.acquerra.com.au/astro/

The skills and knowledge required to take that kind of photos are more easily acquired by doing purely visual astronomy, on a dob or whatever, for a couple years or more. Also, the equipment required for that sort of stunt costs many thousands of dollars - your cheap little Cassegrain cannot do it.

So why buy a small Cass? Well, if you have an overwhelming interest in astrophoto, and are prepared to deal with the difficult learning, then go ahead. But realize that the first images are not going to be very impressive, and the motorized mount will not teach you some valuable skills that you would acquire more easily pushing a dob around and keeping your eye plugged into the ocular for a while. Also, money spent on the motorized mount is money NOT spent on aperture, and the aperture is so important for overall performance.

For astrophotography, the mount is of huge importance; the scope per se is relatively less important (I'm generalizing). A so-so scope on a great mount is much better than the other way around. But a great GEM mount can be very expensive.

Well, that's it. Do some more research on your own, then make a decision. Don't forget that learning only begins with the moment you acquire the scope.

Above all, have fun, and keep looking up.

Clear skies to you!

I have had all three "styles" of scopes. The "Best" scope is the one you get out and go outside with:

The 4.5" (114mm) reflector on a very cheap equatorial mount was my first scope. Not very good at all. It worked and I did see brightest planets, stars, Andromeda, Pleiades, and of course the Moon. No it wasn't a great scope, but I used and abused it.
The next was a cheap 90mm refractor on a cheap equatorial mount. Same viewing but I think just a tad better as far as clarity, but it had the exact same mount, and just as unsteady.
Next was the 8" Dob, which I love, and still have, and the last scope I'll probably ever buy is my Stellarvue Access SV102. The mount I purchased separate, which is the Vixen Optics Advanced Polaris-M Motorized EQ Mount, with Tripod. I sold the 4.5" (114) Reflector.

My experience has been that I stayed around long enough, went to enough star parties and evens, checked out all the astronomers in the area, went to McDonalds Observatory on Star Party Nights, and eventually got around to putting together a scope package that I use every night....well almost. There are pro's and con's to all scopes, but if I was a beginner again, I'd take the Dobsonian hands down for a starter scope. I still use it frequently. Just my opinion...


If fusion only happens near the center of the sun, what's with all the "fire" on the surface? I understand that it isn't "fire" in the combustion of oxygen sense, but I can't find an answer online about what we're witnessing happening at the visible surface of the sun if all actual fusion explosive stuff is occurring near the center.

Would love a primer or a link to an explanation.

99% of nuclear fusion of hydrogen nuclei into heavy nuclei happens at the core of the Sun. This is because only at the core is the temperature high enough to overcome the coulomb barrier caused by the electrostatic repulsion between the nuclei. The solar core produces a lot of energy $(3.828×10^{26} W)$ by fusion and since all that energy has to go somewhere, it eventually reaches the surface making the surface of the Sun hot (6000K).

Most objects like a red hot iron or tungsten filament of a bulb glow when heated. An idealized version of this is called black body radiation. The colour at which the body glows or the frequency of the electromagnetic radiation depends on the temperature. Since the surface of the Sun is hot, it glows. This is not combustion but natural heat glow. Moreover glowing hot plasma rises up in columns while cooler denser gases settle at the surface of the Sun at speeds upto 7km/s in a process called solar granulation causing the Sun to look like a fierce fire ball.

The fusion reactions only happen in the central regions, because immense heat and pressure are essential for these reactions to take place. The mutual repulsion of
like charges has to be overcome before the various nuclei can fuse. Then the heat and energy generated has to find a way out, otherwise the star would explode, so this heat and energy gradually makes its way to the surface, taking some time to do it. Fortunately some energy is instantly carried away by neutrinos, which reach the surface at the speed of light. However, scientists are baffled by the surprisingly high temperature of the corona, which mysteriously is much higher than at the sun's surface. So far, no one has come up with a mechanism which explains this.


Does "spectral type" and "stellar classification" refer to the same thing? Should the tags stellar-classification and spectral-type be merged? has been asked in meta two weeks ago. It requires some careful consideration but so far no response has been forthcoming, so I'm asking this question here to get a definitive answer that can then be applied there.
Wikipedia's Stellar classification begins

In astronomy, stellar classification is the classification of stars based on their spectral characteristics.

and while there is one subsection labeled simply Spectral types to my untrained eye it looks like there may be other classification schemes besides O, B, A, F, G, K, M, L, T.
Question: Do the astronomical terms "spectral type" and "stellar classification" refer to the same thing? Or at least always when applied to stars? Might "spectral types" also refer to a way to classify asteroids?

Spectral type is only based on the star's color. You need additional the mass or luminosity to make a stellar classification form it.
Example: Giants may belong to the same spectral class as main sequence stars.
Of course, if you don't simply look at the color but measure the spectra you see the differences. Spectroscopy is one of the most important tools for astronomers.


Is the duration of a sidereal year stable? According to this essay by Phil Plait, the duration of Earth's sidereal year is 31,558,149 seconds.

Is this figure stable? If I were to meaure Earth's orbit several million (or billions) of years in the past or future, should I expect to see roughly the same duration?

If not, what would be the cause of any change observed?

It changes some.   The solar energy pressing against the Earth from the sun, the kinetic energy from coronal mass ejections hitting the Earth, tidal forces from the Earth-Sun relation and even gravitational tugs form other planets and meteors and space dust that hit the earth all change the Earth's orbit a teeny tiny bit, also, changes in the Sun's mass either by the sun accumulating mass from space dust or comets or losing mass, which most suns do over time.   Each tiny orbital change changes the sidereal year (and all the other years) a tiny little bit.    Perhaps this only amounts to millionths of a second per year but some of these forces are consistent and can lead to predictable changes over time.  Others variable, leading to uncertainty.   The long term changes to the Earth's orbital distance are currently unpredictable.

It's generally believed that the Earth is slowly moving away from the sun, but I don't believe there's much agreement on how fast this is currently happening.   Articles here and here.   The 2nd one estimates 1.5 CM per year or 15 KM per million years which even over a billion years (15,000 KM) is quite a small distance compared to 150 million KM.   But estimates are difficult for a variety of reasons.

I asked a related question here, with a bounty and didn't get any definite answers.   It's possible to mathematically model the movement of some planets (Jupiter to an extent and Neptune/Uranus and perhaps Saturn based on the Jupiter model).  It's thought Jupiter moved both towards then away from the sun over the first few billion years of the solar system.  The Moon's distance from the Earth is also mostly predictible and can be modeled back into the past based on the Oceanic tidal bulge, but modeling the Earth;s distance from the sun over the past is much harder.  It's a tiny object compared to the Sun and Jupiter and N body systems have a lot of chaos and unpredictability to them.   4 billion years ago it's possible that the Earth was quite a bit closer to the sun, but I don't think anyone has a good model for how much closer.   The faint young sun paradox and the theory that young suns eject much more matter and have more active coronal mass ejections suggests to me that the Earth was likely closer to the Sun billions of years ago and not further away.

The earth's orbital eccentricity also changes with time, on a cycle of roughly every 100,000 years but according to Wikipedia, that doesn't change the length of the year, only the shape of the orbit.

Finally, and just for fun, it makes logical sense to me that the Moon orbiting the Earth creates small apparent fluctuations at least from the point of view of the Earth, because the Earth in effect orbits the Earth-Moon barycenter in an ellipse roughly 1/81st the size of the distance to the moon, or about 4,480 KM in radius every 29 days.     Depending on whether the Moon is ahead of the earth or behind it, the Earth year can change by as much as 5 minutes.    I suspect that this is largely ignored when measuring Earth years though and the Earth-Moon barycenter is used.   If we measured the Earth's center we should get much larger variations year to year.  (see picture).

Source.

(if anyone can improve on this, feel free, I have the feeling it's only a partial answer, cause logically it makes sense that the Earth-Jupiter positions or Earth-Venus should change, if only slightly, the Sidereal year).*

It fluctuates a little now and then.

The Moon's gravity is responsible for tides. This means that the oceans are not constantly in synch with the planet's rotation. The Moon pulls the oceans back a little and then releases them. This creates drag - a form of friction. However minimal, it is still telling over the millennia. However, certain types of volcano eruptions, along with the tsunami it creates, speeds Earth's rotation up a little bit.

If the mantle and core are in fact liquid, they also create a degree of drag.

This changes (a little bit) the duration of each second. Thus, ancient Roman seconds were a trillionth of a fraction faster than ours, give or take.

That said, the orbit itself fluctuates a little bit. It is believed today that our planet is moving slowly away from the Sun. This would make the orbit longer without changing Earth's velocity. Thus, yes, each year must be getting longer, although the difference is ridiculously small.


When did people discover that the Earth is not an ideal sphere, but a little flattened? When did people discover that the Earth is not an ideal sphere, but a little flattened? And if you know, who was the first person to discover it?

It wasn't one person. René Descartes had theorized that planets must be in the shape of prolate spheroids. Christiaan Huygens and then Isaac Newton theorized that a rotating, self-gravitating fluid body would relax itself into an oblate spheroid. So which view was correct? Giovanni Cassini argued strongly that Descartes' view was the correct one, and he argued with measurements.

The French Academy of Sciences was split on this important question in the 1730s. The question was of interest in the world of commerce because knowing the shape of the Earth became of increasing import with regard to accurate ship navigation. To this end, the king of France financed two expeditions by the Academy, one to Peru and the other to Lapland.

The group that went to Lapland included Pierre Louis Maupertuis (the head of the Academy), Alexis Clairaut, Charles Camus, Anders Celsius, and others. This group left after but returned before the group that went to Peru. Their measurements agreed with the Newton's concept of an oblate spheroidal shape and disagreed with Descartes' concept of a prolate spheroidal shape. While their observations tilted the discussion in the favor of the Newtonian view, some of those who argued for a prolate spheroidal shape remained unconvinced. Jacques Cassini (Giovani's son) strongly disagreed, arguing that the measurements the group had made in Lapland were flawed. Cassini de Thury (Jacques' son) continued the family argument.

The group that went to Peru took ten years to return. (The voyage to Peru took over a year, all by itself.) The group was plagued by bad luck, lack of supplies, and inept and corrupt leadership. The leader of this expedition wanted no part in the diplomacy and management tasks required of an expedition leader. Even worse, he spent a good part of the expedition's funds on a hottie, and then bailed on the team entirely.

The reluctant second in charge, Pierre Bouguer, took over and made the expedition work. He and his team of over a dozen other scientists made astonishingly precise measurements of the distance between two Andean mountaintops. On returning, it was those measurements that convinced the remaining oblate spheroid skeptics that the Earth was indeed more or less an oblate spheroid.

  And if you know, who was the first person to discover it?

From the above, it should be obvious that it wasn't one person. It most certainly was not Descartes or the Cassini family. (But all four are remembered for their positive contributions to mathematics and science.) But it's very hard to say who was the first to discover that the Earth is an oblate spheroid (more or less) is given that so many people contributed to this discovery.

This, by the way, is how modern science works. Most modern scientific discoveries are team efforts. In the case of physics at the Large Hadron Collider, the list of authors on key scientific articles takes an entire page. It's been argued that Newton was the first modern scientist. It's also arguable that he wasn't. He worked alone. The discovery of the figure of the Earth was very much a collaborative effort that spanned over half of a century.

References:

W.W. Payne, "Attraction and figure of the Earth." Popular Astronomy 9 (1901): 117-123.
A rather old but very nice article on this old subject.

P. Lynch, "That’s Maths: Earth’s shape and spin won’t make you thin."
A lay article on the expedition to Lapland, which was led by Maupertuis.

D. Main, "The Remarkable Story of the First Accurate Measure of the Earth"
A lay article on the expedition to Peru, which was ultimately led by Bouguer.

The Wikipedia article on Clairaut's theorem
Clairaut was one of the key members of the expedition to Lapland. His equation that describes gravitational acceleration as a function of latitude was supplanted by a slightly better expression, the Somigliana equation, almost two hundred years later.

Wikipedia articles on Descartes, Huygens, Newton, the three Cassinis (three links), Maupertuis, Clairaut, and Bouguer.
Take these with the grain of salt one should always take when reading Wikipedia articles.

It was probably in the mid-17th century, and may have ben Isaac Newton, as PM 2Ring suggests, or one of several other people. Galileo devoted a lot of time to studying Jupiter and Saturn, and if he had good eyesight would probably have noticed that both planets were oblate spheres. Christian Huygens a little later in the century also studied both planets and observed that they were oblate spheres. Both men were astute enough to realise what was the cause of this oblateness, and knew that because the Earth rotated fairly rapidly the same principles would apply to Earth. We will never know with certainty who first made this analogy between Earth and the two gas giants, but it must have been in the 17th century.

We will also never know who first mentioned in print that the Earth too was an oblate sphere; that information has also been lost in the mists of time. By the time it was first mentioned in print, a lot of people must have been aware that theory dictated that this must be the case and all that needed to be done was to measure what was already a foregone conclusion.


Why don't Neutron Stars form event horizon? Trying to compare density of Black Holes and Neutron Stars I came up with the following:

A typical neutron star has a mass between about 1.4 and 3.2 solar masses1[3] (see Chandrasekhar Limit), with a corresponding radius of about 12 km. (...) Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m^3 [1]

and

You can use the Schwarzschild radius to calculate the "density" of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x10^16 g/cm^3) x (Msun / M)^2 (...)
The value of the Schwarzschild radius works out to be about (3x10^5 cm) x (M / Msun) [2]

Let's take a neutron star from the top of the spectrum (3.2 Msun) and same mass black hole.
Converting units:

Neutron star: 5.9×10^17 kg/m^3 = 5.9 × 10^14 g/cm^3
Black hole: 1.8x10^16 g/cm^3 x (1/5.9)^2 = 5.2 x10^14 g/cm^3

The radius of the black hole would be (3x10^5 cm) x ( 5.2 ) = 15.6km
The 3.2Msun Neutron Star of this density would have volume of 1.08 x 10^13 m^3 which gives radius of 13.7 kilometers
According to Shell Theorem, spherical objects' gravity field strength at given distance is the same for spheres as for point masses so at the same distance from center of same mass (point - black hole, sphere - neutron star) the gravity will be the same.
That would put the surface of the neutron star below the surface of event horizon of equivalent black hole. Yet I never heard about even horizon of neutron stars.
Either I made a mistake in my calculations (and if I did, could you point it out?) or... well, why?

As Francesco Montesano points out, using the wrong mass leads to the wrong answer.  Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.

Since the density scales as ρ ~ M/R^3 and the Schwarzschild radius as Rs~M, the density of BHs scales as ρ~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.

It is said that neutron stars bend space/time so strongly that parts of the back are visible from the front! Of course a neutron star is essentially one very very large ball of neutrons with all the light elements on the surface. Some scientists now believe that simple neutron star collisions do not generate all heavy elements but the existence of elements heavier than iron is due to black hole-neutron star collisions. If so then they don't have an event horizon despite their enormous gravity because the matter is too spread out, whereas for a true black hole its all concentrated in one place.
In fact it is believed that the escape velocity for a typical neutron star is around 1/3 to 1/2 the speed of light, still a large number and incidentally life may be possible on a planet orbiting a neutron star with radiation tolerance sufficient even in a bacterium like Deinococcus radiodurans as long as the planet's orbit kept it well away from the jets.
A variant of this concept is when a neutron star hits a red supergiant briefly igniting helium fusion if the whole thing doesen't blow up first.
https://arstechnica.com/science/2014/06/red-supergiant-replaced-its-core-with-a-neutron-star/


Is there any practical use for astronomy? Although astronomy is very cool and the things we are learning are awesome, is there really any practical use to knowing the things we know about the universe?

Do other fields of science draw from the current tome of astronomical knowledge?

This question begs the question, does everything need a practical use? The answer is a resounding no. What's the practical use of the Louvre, or of your local neighborhood public park where you enjoy weekend barbecues?

There are some things that are very worthwhile that have little or no economical gain. Your local neighborhood public park in fact has negative economic gain. Admission is free, but maintenance is not. Think of how much money your city would make if they sold it to a condominium developer, and how much money it would save by not having to pay to have the park maintained.

Despite having no obvious economic gain, some things are nonetheless worth quite a bit. Many of the sciences fall in this category. For example, what is the practical use of archeology? (There are some, but that's not the point.)

Astronomy, like archeology, the Louvre, and your local public park, doesn't need a practical economical purpose. The purpose of the science is good enough.

That said, there are practical applications of astronomy. The key application has been and still is navigation. Knowing the location of a ship at sea or the orientation of a vehicle in space requires astronomy.

A less direct but still very important application of astronomy is in how it informs physics. Kepler was an astronomer, not a physicist. (Those two disciplines were very, very distinct in Kepler's day). Yet Kepler's work informed Newton on how to describe gravitation. More recently, astronomy has informed physics that its standard model was not quite correct. The observed neutrino flux from the Sun (see http://en.wikipedia.org/wiki/Solar_neutrino_problem) was a third of what physics at the time said it should be. This resulted in a change to the standard model. Neutrinos have a small but non-zero mass, and they oscillate from one form to another.

Astronomy continues to inform physics to this day. Physicists (and astronomers) remain clueless with regard to what constitutes dark matter and dark energy. But whatever they are, they certainly do exist.

No, there isn't a practical value. How much was wasted on the Hubble Telescope, then how much more was wasted correcting the lenses to actually get a clear picture? How much was wasted on a Mars Lander that crashed upon arrival?

If people want to pay for astronomy out of their pocket, fine. Don't dig into mine just so I can hear about how some galaxy was doing 10 billion years ago. If anything spend the money on oceanography- an area that is of much greater concern to those of us living here and not in the clouds.


How much magnetic attraction is there between the Sun and Earth? How much magnetism is there between the Earth and Sun? Would the distance from the Sun differ for not the Earth iron core and how much?

Warning: I am only rather mediocre on astronomy. It was a hobby of mine when I was in second grade to now, so I pretty much know a decent amount of info. But my speculation for this question may be some distance off.

First of all, let us look at the Sun and Earth. The Earth has a [25~65 microtesla] magnetic field due to the motion of convection currents its molten [mostly iron and nickel] outer core. These convection currents are caused by heat escaping from the solid inner core. Heat from below heats up the liquid, which rises, cools, sinks, heats up, rises... you get the point. These convection cells creates charge/electricity in the outer core. Now, when you hear electricity, and then, iron/metals, you will obviously know that iron would conduct electricity. Sort of like an electric magnet nowadays. That is exactly so in Earth's outer core. The electricity generated by the convection cells is [what I believe] conducted by the outer core and amplified/extended to form the magnetic field. This is the Dynamo theory, which contains three parts:

  There are three requisites for a dynamo to operate:

  1) An electrically conductive fluid medium

  2) Kinetic energy provided by planetary rotation

  3) An internal energy source to drive convective motions within the fluid.

  In the case of the Earth, the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth.

The sun's [100 microtesla] magnetic field compared to Earth's is a really large and complicated. Since the sun is THAT LARGE :P, its magnetic field extends beyond the planets forming the heliosphere, a protection against cosmic rays from interstellar space. Unlike but similar to the Earth, the magnetic field of the sun is derived from the electricity conductive plasma inside the sun, but also from convection. Since the sun is not a stable ball, but rather a giant ball of fluid in a medium, it's rotation speeds on the equator and the poles are different, and this bends/stretches the magnetic field, creating sunspots, CMEs, Solar Flares, and many other "space weather" events.

To answer what you are asking, the magnetic attraction between Earth and the Sun should be rather strong, but not to an extreme extent. If you were to take 2 magnets of the same magnetic strength ratio, the stronger magnet the readily attract the weaker magnet. This same affect should be that of what the Sun is sort of doing. But even along with gravity, the Sun's pull on Earth cannot force it into a collision course due to Earth's sideways momentum. As commented on by uhoh, the interplanetary magnetic field is the magnetic fields of all objects in the Solar system. The sun, being the strongest one, takes dominance. But what is rather confusing is that there is a thing called Solar Wind. The Solar Wind is a rather "pushing" stream of charged that will "push" away particles if the objects doesn't have a magnetic field to repel that force. The Solar Wind carries the Sun's own magnetic fields away from it, but since magnetic lines cannot cross celestrial objects, they "drape" themselves onto the magnetic field-less object and form the "magnetosphere" of that object. I would imagine this would have little effect however on the attraction of the Sun and that object(earth, asteroid...) itself.

If you are looking for numbers, a hefty equation to calculate the electrical
magnetism oh an object is the Boit-Savart Law.

  If you have a
  current $I$ flowing through an infinitesimal length
  $\mathrm{d}\vec{l}$ then the field at a point $\vec{r}$ is given by
  \begin{equation} \mathrm{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{I
> \;\mathrm{d}\vec{l}\times\vec{r}}{|\vec{r}|^3} \end{equation} The
  field does still drop of with the inverse square of the distance, but
  the vector nature of the current makes the formula rather more
  complicated. Generally this expression has to be integrated over the
  length of a wire carrying the current. This can get quite complicated
  and so this method is generally only used for simple situations such
  as the field around a magnetic dipole.  Having found the field, the
  force on a infinitesimal length of current can be found from.
  \begin{equation} \vec{F} =
> I\;\mathrm{d}\vec{l}\times\vec{B}\end{equation} This is again
  analogous to the electrostatic case but with added complication due to
  current being a vector.

    https://physics.stackexchange.com/questions/122258/what-is-equation-to-find-force-of-magnetic-attraction

This can be used to calculate the force of Earth's magnetic field, and the Sun's magnetic field, since they both are electrically generated. The resulting numbers can be put into the magnetic attraction equation:

$$F=\frac{\mu q_{m1} q_{m2}}{4\pi r^4}$$
where

$F$ is force attraction in newtons.
$q_{m1}$ and $q_{m2}$ is the magnitudes of magnetic poles in ampere-meters.
$\mu$ is the permeability of the intervening medium is tesla meter per ampere, henry per meter or newton per ampere squared.
$r$ is the separation in meters.

Thus, the attraction between the Earth and the Sun can be calculated in the following means:
$$\implies \operatorname{B}(\text{Earth})=q_{m1}=7.84027\times 10^{22}$$
$$\implies \operatorname{B}(\text{Sun})=q_{m2}=3.48526\times 10^{29}$$
$$\implies \mu\approx 10^{-6}$$
$$\implies r=\operatorname{AU}=149597870700$$
$$\implies F=\frac{10^{-6}\cdot (7.84427\times 10^{22})\cdot (3.48526\times 10^{29})}{4\pi\cdot 149597870700^4}$$
$$\implies F=\boxed{4.34165\text{ newtons}}$$
If the Earth doesn't have a liquid, conductive outer core, it wouldn't have a magnetic field. This would be similar to Venus, which has a super duper weak magnetic tail and magnetosphere from the Solar Wind(suprising is that it still retains an atmosphere. I will ask why...). However, we can use a real world example of such. Let's take Mars as the example. Mars used to have a magnetic field from its liquid outer core 4 bya, but this field was destroyed, either from repeated bombardment from large astroids which would have disrupted its internal structure, or from the cooling of its liquid core. Thru recent evidence from tests on Mars' crustal fields, it has shown that Mars' earlier magnetic field is more powerful that of Earth's. Now all that's left is some umbrella shaped fields concentrated on it's southern hemisphere. This offers little protection and would sufficiently answer your second question. You asked if not for the conductive, LIQUID, outer core, what would happen to the distance as possibly the attraction between the Sun diminishes and Earth wanders more outwards. I would suggest not such. Mars would fit your speculation, but its 4 bya orbit and current orbit has minute difference. I would postulate that the magnetic field degration would have some effect, but such effect would be rather unnoticeable, because even though electromagnetism is a very strong force, realize that gravity is the strongest force in the universe(black holes). The attraction between the celestrial bodies orbiting the Sun and the Sun itself is almost completely gravity.

I may be off, or way wrong, but to what I know, I believe this is more or less the right answer.

In SI units the force between magnets is
$$\approx 10^{-7}\frac{m_Sm_E}{r^4}$$
$$=~10^{-7}\frac{(3.5 \times10^{29}) (7.84\times 10^{22})}{(1.5 \times 10^{11})^4} = 5.4 $$ Newtons.

Magnetic dipole moments $m_E$ from here and $m_S$ from this impeccable "scientific" source. From Figure 4 here, $m_S$ is about $10^7 \times m_E$ so the value for $m_S$ looks about right.


How much iron does a star fuse before going supernova? I understand stars fuse atoms into heavier and heavier elements. When the star starts to fuse iron, the amount of energy released is less than the amount of energy it to takes to fuse the atoms. This creates an energy deficiency and the core is unable to maintain its outward pressure against gravity's inward collapse. The star collapses, rebounds against electron degeneracy pressure (I believe?) and goes BOOM. But how much iron is actually fused? And, for how long is the star fusing iron before the collapse?

Also, as a little bonus, I read somewhere recently that the presence of nickel is possible prior to going supernova. I can't find where though. Was that person wrong?

Wikipedia says the iron-nickel core is inert, and doesn't mention any endothermic iron fusion reactions, just the core collapse, which converts the core to neutrons (and a not insignificant proportion of protons & electrons), and releases a lot of neutrinos. The nickel is radioactive, decaying to cobalt & iron.

The silicon burning phase leading to nickel only lasts about 5 days. The inert core collapses when it exceeds the Chandrasekhar limit of about 1.2 M$\odot$.

The collapse of the stellar core has actually 5 steps, and only the first is well known.

1. exotherm fusion

The first is the fusion of the nuclei until iron/nickel. Also the normal life of the star belongs here.

2. endotherm fusion

While the iron-nickel core is inert on the Wikipedia, it is true only in "normal" circumstances. In fact, also iron nuclei can fuse, but it is already an endotherm process: they get away energy, thus they cool the core, accelerating the contraction.

Until iron (and nickel) is created due to nuclear fusion, the nuclear reactions are working against the collapse, because they generate energy (photons) and their pressure tries to avoid the collapse.

After that, this obstacle stops and the core can contract. From a point, also the iron (and nickel) nuclei start to fuse, to create all known elements of the periodical system, and probably even much higher, which can't exist in our low-pressure environment.

However, it is already an endotherm process, thus it takes away energy from the core. Thus, instead of decelerating the contraction, it accelerates.

3. exotherm fission

But it can't balance the temperature elevation due to the gravitational contraction, and over some billions kelvins, the nuclei start to decay.

It is the reverse process what the star did in the previous millions of years. It is exotherm for nuclei higher than iron/nickel and endotherm for nuclei below it.

4. endotherm fission

Thus, first the fission of the nuclei decelerate the contraction, then (after iron nuclei decayed), they accelerate again.

The produced energy radiates away (more exactly, it hits the upper layers of the star, causing the supernova explosion what we can see), and the contraction goes, this time with elevating temperature.

Over 60billion degrees, even the helium nuclei decay, which is still a highly endotherm process. From that point, there is no more way to get away energy from the core by nuclear processes: there is too hot to fuse the nuclei and there is no more way to decay.

The result is a mess of protons, neutrons and electrons. Essentially what the star did millions (or hundred millions) years long, is reversed in some seconds. The difference is that this mess is compressed into a size somewhere between a neutron star and a white dwarf, and it is 60billion degrees hot (normal stellar core temperature is in the order of some ten millions).

This is the point where there is no more obstacle before the further contraction.

5. proton-electron fusion

More exactly, the next obstacle will be that the beta decay of protons becomes energetically favorable. It can happen either by the fusion of a proton and an electron to a neutron and an anti-neutrino, or by the fission of a proton to a neutron, a positron and an anti-neutrino. The positron finds an electron quickly and annihilates.

Although neutrinos can permeate normal matter very quickly, their mean free path in such dense matter is only some hundred meters. They explode away the upper layers of the forming neutron star, dispersing all the elements of the periodical system in the Universe.

This all happens quicker than a minute, the steps 2-5 happen in seconds.

Thus, the answer is, that in a supernova explosion, practically all matter in the stellar core will be iron. Not even once, but twice. Although both is happening only for a very short time, they might be iron nuclei maybe some tenth of seconds long.