spherical geometry - Finding the northernmost latitude in a great circle that passes through two points on the sphere

I think it is a little more straightforward if you work in Cartesian coordinates.

Take an earth of radius one, for simplicity.

Then we can take the longitudes as $pm l$ without loss of generality. Then
the halfway point between the two points, projected to the surface of the earth, will have the highest latitude (equivalently the highest $z$ component).

The two points are $(cos phi cos l, cos phi sin l, sin phi)$ and
$(cos phi cos l, -cos phi sin l, sin phi)$, and the mid point is
$(cos phi cos l, 0, sin phi)$. To project to the surface, we divide by
the norm, to get a $z$ component (on our unit earth) of
$sin delta = {sin phi over sqrt{ (cos phi cos l)^2 + sin^2 phi}} = { tan phi over sqrt{cos^2l+tan^2 phi} }$.

To obtain the $arctan$, we first need $cos delta$, which we get from
$cos delta = sqrt{1 -sin^2 delta} = { cos l over sqrt{ cos^2l+tan^2 phi} } $, and hence
$tan delta = {tan phi over cos l}$, which is the desired result.

universe - Was the cosmic microwave background (CMB) radiation originally gamma rays?

The cosmic microwave background is a result of an almost perfect blackbody emitter. That means the spectrum covers a broad wavelength range with a peak that is given by Wien's law:
$$lambda_{peak} = frac{2.9times 10^{-3}}{T},$$
where the wavelength is in metres and temperature in Kelvin.

The microwave background is formed at the epoch of (re)combination, when protons and electrons combine to form hydrogen atoms and the Uinverse effectively becomes transparent to the radiation field within it. This occurs at approximately 3000 K. So at that time, the cosmic microwave background had a peak wavelength of 970 nm, which in the near infrared part of the spectrum.

The epoch of (re)combination occurred at a redshift of approximately $z=1100$. What this means is that since then, the wavelength of the photons has been stretched by a factor of $1+z = 1101$, so that $lambda_{peak} simeq 1$ mm, and this corresponds to a temperature of $T=2.7$ K. I.e It is near infrared photons emitted at the epoch of recombination that have their wavelengths stretched by the expansion of the universe, and are seen today as microwaves.

However, prior to the epoch of recombination, the Universe would still have been full of radiation, and as we go back in time towards the big bang, that radiation field becomes hotter and hotter, and the peak wavelengths get shorter and shorter.

If we look at models for the thermal history of the Universe, then at about 1 second after the big bang, the Universe would have been at about $10^{10}$ K and the radiation field would have been in equilibrium with the matter at this temperature. The formula above then gives $lambda_{peak}=2.9times10^{-13}$ m, which is definitely in the gamma-ray regime. But let me emphasize that these photons are emitted then absorbed many times and we do not see them now.

The "particles" that you are talking about are photons of light.

supernova - Fusion of elements inside heavy stars

You are correct to say that all the heavier elements in the universe were formed in stars.

Stars like the sun fuse hydrogen into helium. When they get older they can fuse the helium into carbon (it actually takes 3 helium to make one carbon). Larger stars can fuse carbon into oxygen, and neon and elements in the first half of the periodic table.

When the star runs out of fuel, the outer layers are gently ejected in what is called a planetary nebula (though it has directly to do with actual planets). The ejected gas is enriched with the heavier elements that the star has fused. The heavier elements in the planetary nebula mix with the hydrogen and helium gas in outer space and can later go on to be found in new stars. Most of the carbon and oxygen and nitrogen on earth was formed by this process.

Very large stars will fuse all the way up to iron, and then collapse in a supernova. This releases a very large amount of energy, some of which is used to form elements heavier than iron. All the heavier elements (copper gold, uranium for example) are formed in supernovae.

The supernovae explosion distributes these elements back into space, mixed in with the debris from the star, as time goes on, and stars are born and die, the gas in space gets enriched with more and more heavier elements

So the heavier elements are formed in the cores of stars, and are stored there until the star dies, when some of the star's atoms are released into space. The heavier elements are not "radiated out" of the star, until the star dies. Some stars die in a supernova explosian, but most have a more peaceful death in a planetary nebula.

spectroscopy - How to determine atomic number density of an element in a star based on equivalent width measurements

The measurement of a chemical abundance is not a question of using a simple equation.

The simplest it gets is using a "curve of growth", which relates equivalent width to abundance and assumes you already know the temperature of the star and its surface gravity.

For the Li 6708A line, the relevant tables, that can be interpolated, are found in Soderblom et al. (1993). http://adsabs.harvard.edu/abs/1993AJ....106.1059SS . These are LTE curves of growth based on Atlas atmospheric models.

As an aside, the equivalent width of this line in the Sun is 3 mA. The formula you quote looks like some sort of linear approximation, but this resonance line rapidly enters the saturated (non-linear) portion of the curve of growth for equivalent widths of 50 mA or more. As the equation is couched in terms of ratios, then the units you use are entirely up to you.

What is the distance of earth from 7 planets, moon and sun at a point of time?

I wanted to know the distance of the 7 planets(mercury, mars, venus, jupiter,saturn, uranus, neptune) each (or at least 5 planets) from the earth and the angle that it will make with a line connecting the centers of earth and the respective planet at a POINT OF TIME. Also I need the same for the earth and moon system and sun and earth as they are big influences. At point of time because in the course of revolution of planets the distances with respect to earth will vary.

Being an engineering student I have the wish :) to calculate the mechanical forces acting on the earth at a point of time due to nearby heavenly bodies.

Basically i need the data or in case someone already attempted to calculate the force components I would be delighted to look at it.

What’s the object between the earth and the sun currently showing in Google maps?

That's the Moon alright, and it's definitely real and definitely there. If you go outside and look at the Sun right now, the Moon will be almost but not quite on top of it, though it's impossible to make out due to the Sun's glare. If it were any closer we would have had a solar eclipse around the time of the New Moon that took place 20 minutes ago. Two weeks from now at Full Moon, it'll be in even better alignment, though on the opposite side of the Earth from the Sun this time, which will result in a total lunar eclipse. Mark your calendar!

space time - When you travel fast around a black hole, do you experience high speed yourself?

If you jump out an aeroplane, do you experience high speed? Many people would say you do, since your velocity relative to the ground increases.

Note the importance of the phrase "relative to the ground" since to describe speed you always need to say what you are measuring the velocity relative to.

A simple choice is to measure your velocity relative to a distant observer. In this case then, yes as you pass close to the black hole your velocity relative to the distant observer increases. As you approach the event horizon, you velocity relative to the observer will approach the speed of light.

Depending on how close you pass the the event horizon, you will experience significant time dialation, so that more time will pass in the observer's frame than in yours. This means that if you manage to escape the gravitational pull of the black hole, and return to the observer, you will have found that the observer has aged compared to you.

earth - What are the equations for an orbit with time as parameter?

I have done a lot more research into this and found an answer. It comes down to solving Kepler's Equation, and my main references are:
Kepler's equation and the Equation of Centre which gives an iterative method for solving the equation, and Determination of Position in Orbits which has the formulae for calculating orbital coordinates.

Here is the Processing sketch. If you are not familiar with Processing, it is a language developed for graphic designers, but much more versatile than just that, based on Java and hence C++. It expects two functions, one is setup(), called once at the start up of the sketch, and a second draw() that is repeated over and over until the sketch is stopped. In case you wonder: the whole is encased in a main() before it is compiled. For more details see the Processing home page.

int wsize = 400;
float rp = 9.0; // radius of planet symbol
float rs = 15.0; // --"-- sun
// Constants of orbital ellipse 
float a = 150, b; // major and minor axes
float e = 0.5; // eccentricity
float halfa = a/2.0;
float c; // focal distance from centre
float E, E0; // eccentric anomaly, temporary values in iteration
float v; // true anomaly = polar coord angle
float M; /* mean anomaly which is effectively the ..
            .. time variable round the orbit*/
float r, x, y; // polar coord radius, rect coords
float k, cosE, cosv; // useful intermediate variables
float twopi = 2*PI;
float PIby2 = PI/2.0;
float Minc = twopi/250; // time proxy increment

void setup()
{ size(wsize,wsize);
  //e = sqrt(((a*a)-(b*b))/(a*a)); kept for reference
  b = a*sqrt(1-e*e);
  c = e*a;
  k = 1-e*e;
  M = 0; // Mean anomaly increases linearly with time
}

void draw()
{ background(255);
  translate(wsize/2,wsize/2); // move origin to centre of display window
  ellipse(0,0,2*a,2*b); // show orbit
  spot(c,0,rs,#FFF303); // draw sun
  translate(c,0); // move origin to focus
  // Calculate Eccentic Anomaly by iteration
  E = M; E0 = 1000; // make sure while starts
  while(abs(E-E0) > 0.000001)
  { E0 = M + e*sin(E);
    E = E0;
  } 
  // calculate polar coordinates of planet
  cosE= cos(E);
  cosv = (cosE-e)/(1-e*cosE);
  v= acos(cosv);
  r = a*(1-e*e)/(1+e*cosv);
  // calculate x,y coordinates of planet
  x = r * cosv;
  y = -r*sin(v);// negative makes it go anti-clockwise
  if (M > PI){y = -y;} // show second half of orbit
  // println(v+" : "+x+" : "+y);
  // line(0,0,x,y);
  spot(x,y, rp, #03ADFF); // draw planet
  M += Minc;
  if(M > twopi){M = 0.0;}
}

void spot(float x, float y, float w, color tint)
{ fill(tint);
  ellipse(x,y,w,w);
  fill(255); // restore white background
}

spectroscopy - Infrared Astronomy for non transiting objects?

Yes it is. As long as light is produced by a source and then passes through a medium, we can perform spectroscopy on it. When considering a planet like Uranus, most of the light that it emits is actually reflected light or re-emitted light from the Sun. Uranus produces very little light by its own merit. For light that is directly reflected (by that I mean the light from the Sun goes into the atmosphere, hits molecules, and gets scattered out of the atmosphere again), this is directly equivalent to the standard transiting planets you mention in your question. For light that is re-emitted, you still see absorption spectra, but the black body background has changed. In this scenario, the light is absorbed lower in the atmosphere and re-emitted at a longer wavelength, but that emitted light must still try to leave the atmosphere and will experience some absorption along the way, creating absorption lines.

orbit - Inclination in Kepler's laws

So, basically you're trying to merge orbital inclination to Kepler's laws. The simplest way is to take the measured orbital inclination of the planet, which is constant, and apply the Pythagorean theorem to any given location and that gives you 3 dimensions of distance from the 2 dimensions defined in Kepler's laws. That's probably what Kepler did. (see picture from the Wikipedia link above)

A much too simple explanation of orbital inclination here.

And a much more complex one, factoring in orbital movement here.

Kepler's laws only work in 2 dimensions because one object orbiting another is a 2 dimensional calculation and bringing a 3rd dimension into his law complicates it unnecessarily. 3 dimensions is essential for accurate planetary tracking, but it's unnecessarily in a 2 body gravitational system that takes into account distance, eccentricity and velocity. Kepler was likely aware of this.

Kepler, in fact, discovered orbital inclination by looking at Copernicus' observations, and that was something he was very proud of and it was a very important discovery. Source here. Kepler deduced that orbital inclination explained away previously not understood and overly complicated orbital oscillations. He didn't just undo Copernicus' circles, he worked things out in 3 dimensions and he might never have figured out his laws without understanding and measuring orbital inclination.

Nice question.

(footnote, my answer is more history of science and mathematics, but I think your question, the way you asked it is more astronomy, so, I think the question should stay here, but the powers that be can decide otherwise if they like).

astrochemistry - What does this information about "upper limit column densities" and "upper limit relative abundances" mean?

A column density is just the integral of a (number) density along the line of sight.

Often, you would get the same observational absorption or emission signature from a number density of $n$ molecules cm$^{-3}$, spread along a line of sight through a cloud of diameter $x$, as you would with a density of $0.1n$ spread over $10x$.

In other words, all we can infer is the product or, more accurately, the integral of $n dx$, and this is known as the column density.

In the paper you refer to, I doubt they know the thickness of the shock they are investigating, so a column density is all that can be determined.

The relative abundance is more straighforward; it is just the ratio of the number densities of the two labelled species.

Determine the moons of Jupiter through a telescope

Use an orrery that will let you specify a specific epoch and vantage point. There are some quite fancy ones online, for example this Solar System Scope:

What you do in this particular tool is click on the calendar bar below and enter date and time of your observation (if you forgot that, there's a good chance your photograph has a time stamp, either of the file or in its EXIF data). You can adjust epoch later on, too. Then double-click on Jupiter, then its Orbit button, then click on link As seen from Earth below, then click Go back button and use mouse wheel to zoom out so all four Galilean moons display. Then compare with the photograph that you took. Note that it might take a bit of rotation for you image to align perfectly. I use the old neck twisting and turning technique, but photo editing tools or even rotatable screens will do just as well, if not better.

Note that this mentioned tool does offer settings to switch between orrery and realistic model and large and realistic sizes of displayed celestial bodies. But there are many such tools online or downloadable for both computers and smartphones. I think even this one is available as an app for smartphones for free, but last time I tried to install it, it didn't work for me. Maybe they fixed that. Do check because it's a lot easier if you can check orbital alignment of Galilean moons as seen from Earth directly on the field, as you're doing your observations.

Have clear skies and I wish you happy hunting for planets and moons of our Solar system and beyond!

milky way - Sun like star in our milkyway?

Is there any popular star or one which we can see with our naked eyes in our milky way that is atleast 90% sun-like(mass,radius,spectra,luminosity).The important thing is it must not be a binary.
Its okay if its not visible to our naked eyes because our sun's absolute magnitude is +4.5!.

solar system - Is the material that could have formed a "Super-earth" divided among the four terrestrial planets

I think the answer to your question lies with Jupiter, be it directly or indirectly, the gas giant is now believed to have had a large influence on the way the inner planets formed.

Many planetary scientists believe that the reason we only have 4 small inner planets today and not any "super-Earth" sized ones is because of Jupiter, and how it interacted with the circumstellar disk and protoplanets at the time of the solar system formation. There may well have been one or two super-Earths that formed in the early solar system, but models suggest that they would have been sent crashing into the Sun by the inward advances of Jupiter.

I found this blog article which explains it in more detail. And there is also another MNRAS article that goes into more detail about the physics and orbital mechanics behind what might have happened in the early solar system.

How to orient ESO's Milky Way Panorama in a 3D model

I have been trying hard to find the star "Polaris" in the following image from ESO and Serge Brunier:

From peering at other maps and simulation apps, I think it should be in the top-left quarter, relatively close to the middle (vertically), however taking the brightest starts one by one, I cannot piece together the other stars belonging to Ursa Minor (small dipper).

I want to use this image as a background in a 3D model of Earth. The projection is perfect for that purpose, I only need to rotate it in such a way that Polaris lines up with the north pole. Once that is accomplished I can rotate it around the Earth's polar axis to set the right time of day at a certain part of the year (it does not have to be 100% accurate for my purpose, but it should at least look a bit realistic to an astronomer).

One thing that makes this task quite hard with this high resolution image is that stars seem to pop in and out of existence while zooming in and out, probably due to lossy fast re-sampling algorithms used in my image viewers.

If someone with a trained eye could point out the Ursa Minor constellation for me it would save me a ton of time!

cosmology - Is the Drake equation based on a logical fallacy?

I often wonder why the Drake equation gets taken so seriously and why popular scientists in general say "it's probable' life exists elsewhere in the cosmos.
My problem is that the mechanism for abiogenesis is as yet unknown.

We don't know the likelihood that RNA like molecules can assemble from chance or for that matter that this is the source of life on this planet (as opposed to coming via a comet). We do know that we have no record of it recurring and as yet Miller-uri style experiments have not yielded results.

The argument I keep hearing from Dawkins, Sagan and many others is that the universe is vast and life occurred relatively quickly in our planet's history which suggests it is common.
I detect a logical fallacy based on the inductive reasoning of using one occurrence in order to establish a pattern.

Think of it this way. If I've never played cards and get dealt my first hand which is a royal flush I might think its quite likely based on my past hand that I will get another in the next deal. This doesn't change if the deck contains a hundred suites and the odds of a royal flush is astronomical - if my first hand is a royal flush.

Similarly the vastness of the cosmos tells you nothing about the likelihood of other life if you don't the probability of its occurrence.It could be very likely or so rare as to only occur once.
Does anybody have an opinion on this?

the sun - Where is the materials of comet after it crashed into the sun?

It depends on the size of the comet. If it's only $0.1$–$1,mathrm{km}$, it will evaporate before even reaching the Sun (Li. et al. 2013; Knight & Walsh 2013), and the released gasses will be ejected by the Sun's radiation pressure. Larger comets, so-called Sungrazing comets, can survive close passages, although they probably will be torn apart by tidal forces.

If the comet comes so close that it crashes into the Sun (I'm not sure this has ever been observed, despite various blogs discussing this, but there's no reason it couldn't happen), it would reach the escape velocity of the Sun, i.e. $sqrt{2 G M_odot / R_odot}sim600,mathrm{km},mathrm{s}^{-1}$. This New Scientist article claims that the impact would cause an explosion and possibly a coronal mass ejection, but the underlying scientific article by Brown et al. (2015) seems to indicate that the large ablation of the comet will result in less dramatic velocities.

Anyway, some of the released gas would stay at the surface of the Sun, just like some of the Sun's "own" gas stays. And since the convection cells, or granules, are quite shortlived (of order 10 minutes; Bahng & Schwarzschild 1961), the gas will soon be engulfed by the Sun, being carried deeply into the body of the Sun.

The convective zone of the Sun is in the outer ~30% of the radius, and there isn't much mixing of material between that and the inner, radiative zone. In the last phases of the Sun's life, however, when core hydrogen burning stops and the Sun becomes a red giant, convection reaches so deeply into the Sun that some mixing takes place. This is called dredge-up. Eventually, roughly half of the Sun will be ejected due to the large mass-loss during the red giant phase, while the other half becomes a white dwarf. If the dredge-up didn't happen, the comet material would all be a part of the ejected material, but due to the dredge-up, some of it will end up in the white dwarf.

amateur observing - What were the 2 satellites I saw this morning?

This morning at 6:03am Israel time (3:03 UTC) I saw, simultaneously, what appeared to be 2 satellites streaking across the sky directly overhead. If I remember correctly, they were both coming from NW (but it was early).

Does anybody know what they were, or how I can find out?

I live in Mitzpe Netofa, lon/lat about 35.39/32.80.

jupiter - How big would the asteroid belt planet be?

The largest main belt asteroid is 1 Ceres, which alone contains almost a third of the total mass of the whole main asteroid belt.

Ceres is large enough to be in hydrostatic equilibrium, i.e. its own gravity is strong enough to pull it into a roughly spherical shape. Since the mass of a spherical planet scales as the cube of the diameter (assuming constant density), piling all the other main belt asteroids together onto Ceres would only increase its diameter by a bit under 50%. It would still be a roughly similar type of body — a small sphere of partially differentiated rock and ice, with no atmosphere to speak of (as it'd be way too small to hold onto one).

Thus, to answer your question, a hypothetical planet containing all the matter currently making up the main asteroid belt would look pretty much like Ceres already does, just a bit bigger.

general relativity - What you would see in a wormhole?

Around the worm hole would be a region with an intense gravitational field, and there would be gravitational lensing occurring, similar to that around a black hole. The difference would be that light that has passed through the worm hole would would be visible. So, a distorted view of the "other side" could be seen.

This physicsbuzz article has some images and discussion. Note that the wormhole would be a spherical region in space, and not a 2d "stargate"