Could human life thrive on a planet in a pulsar star system? Aleksander Wolszczan of Pennsylvania State University found three planets circling a pulsar called PSR B1257+12.Two of the planets were the size of Earth.The pulsar observed by Spitzer, named 4U 0142+61 spotted a warm glowing disk formed from debris, The disk has a orbit of about 1 million miles and contains 10 Earth mass materials. Deepto Chakrabarty of the Massachusetts Institute of Technology, principal investigator states "Pulsars emit a tremendous amount of high energy radiation".

In a pulsar star system, could human life thrive on a planet with high energy radiation? If answering the question, please provide the latest references to date that are relevant with your statement.

Only five pulsar planets have been confirmed or have garnered enough evidence to make a strong case for their existence. None of them are like the terrestrial planets in the Solar System insofar as the way they formed and their orbital movements.

PSR B1620-26 b. This planet is in a circumbinary orbit around a pulsar and a white dwarf, with a semi-major axis of ~23 AU. It is theorized that it formed around a Sun-like star, which became a binary star system after encountering a neutron star (another partner must have been ejected from the system, as is the case with triple-star encounters). The Sun-like star promptly became a white dwarf, and the planet's orbit would have widened after the encounter with the neutron star, which is a pulsar.

Basically, this planet did not form around the pulsar. The system was captured so to speak, and so neither the Sun-like star/white dwarf or planet would have been affected by the supernova. Additionally, the planet is fairly far away from the two stars in the system, meaning that effects from the pulsar are even less negligible.

The point of this is that life on this planet is possible - especially given the planet's old age - but it would be hard for it to survive. If life formed before the encounter with the original binary system, it would have been disrupted a bit by the orbital changes, reducing the amount of starlight it received. Today, it would receive little light, as the white dwarf has a low luminosity. There wouldn't be enough energy to sustain life - as is the case for most pulsar planets.

Also, as Rob Jeffries pointed out, this planet is a gas giant, not a terrestrial planet, meaning that human-like life could not develop on it and any arriving humans bent on colonization would not be able to live on it normally.
The planets of PSR B1257+12. Three planets orbit this pulsar, at distances of about 0.19, 0.36, and 0.48 AU - fairly close to the pulsar and within range of a decent amount of harmful radiation, relative to PSR B1629-26 b. The major reason that these planets are not a good place for life is that they are thought to have formed after the supernova that led to the evolution of the star to a pulsar, from remnants of debris from the supernova. The star itself is quite young - ~3 billion years - and the supernova would have happened relatively soon thereafter. The planets themselves are less than 1 billion years old, meaning that life has not had time to develop.

My issue with the age of the planets isn't based on the short time period for life to develop so much as on the idea that conditions might be hellish, comparable to those experienced in the Hadean eon. Geologically, these planets would have had different starts than Earth did, and they most likely would not have encountered the problems of the Late Heavy Bombardment, but life might not have a good chance at starting at all for another 500 million years.
PSR J1719-1438 b. This planet is unusual because it is thought to be the remnant of the companion star - severely damaged by the supernova. It is also composed of exotic materials, for a planet - crystallized carbon. The conditions here are not conducive to life.

Looking at this list we see three distinct types of pulsar planets emerge:

Captured planets
Newly formed planets
Former companion stars

The latter two are not good places for life, in part because they may not be composed or compounds life can use. The first group may be the best chance, as their formation could have been relatively normal. The problem lies in the fact that they might have been rogue planets, floating along without good sources of energy from stars.

All of this isn't even taking into account the radiation from the pulsar itself. The only escape is to have an orbit far from the pulsar, as with PSR B1620-26b, but then there's no outside source of energy. The best chance would be to have the planet in a circumbinary orbit around the pulsar and a companion star, as in this planet's case. At least then there would be some helpful light.


How accurate should sun\moon ephemeris be for calculating eclipses? I am writing a program to simulate eclipses for about -2000 to 6000. My question is, how accurate does my ephemeris for the sun and moon have to be to get a decent accuracy for eclipses? I currently have an accuracy of about 1'' for the Sun and an accuracy of about 12'' for the moon. Is this good enough, or should I find a more accurate system?

Edit:
I'm still rather new to calculating eclipses, though I have some experience with calculating ephemerises. I'd like to calculate several things: when they occur, the duration, the path of the shadow on the earth for a solar eclipse, and perhaps the percent covered. I suppose I'd like the accuracy on these to be about $\pm 0.005$ or better. I have access to several resources that should help me do this, but I need to know how accurate to get the ephemeris.

I suspect this is fundamentally a question of error propagation.  My personal favorite reference is "Data Reduction and Error Analysis" by Philip Bevington.  Basically,  once you have the formulas in place,  such as

eclipse_time = f(sun_ephem, moon_ephem)

You take all the partial derivatives and "stuff" them into the error propagator formulas along with the known or estimated standard deviations (error/uncertainty) of your input variables.  The uncertainty in the output  just pops right out.


How to determine the ellipticity of galaxies in SDSS What is the best approach to determine the ellipticity of galaxies in the SDSS DR12. I have read this page. Are those really good methods?

Do flux-weighted second moments (as given in the stokes parameters) really give the correct value for ellipticity?

Also I am not sure what table field they actually refer too (in the Galaxy view) when talking about the method using the Adaptive Moments. For example where can I find the m_rr_cc value for the method described here?

Update

I have calculated the ellipticity of a spiral galaxiy with two different methods using data from SDSS DR7. First I used the method using the stokes parameters and second I used the values for isoA and isoB. Both calculations where done in the r-band.

The data can be obtained with the following CAS Query:

SELECT
    g.objid as objId,
    /*  stoke parameters*/
    q_r,u_r,
     /*  iso_a, iso_b */
    isoA_r,isoB_r
FROM Galaxy AS g
WHERE
    g.objid = 587722982832013381

yielding the result

objId: 587722982832013381
q_r : -0.1206308
u_r: 0.002988584
isoA_r: 169.3949
isoB_r: 84.78848

The ellipticity by stoke parameters is given by

$$
e  = 1 - \frac{b}{a} = 1 - \frac{1 - \sqrt{Q^2 + U^2}}{1 + \sqrt{Q^2 + U^2}}.
$$

The ellipticity from isoA and isoB is very given by

$$
e  = 1 - \frac{isoB}{isoA}.
$$

But I get completely different results when I use isoA and isoB or the stoke parameters. I get:

ellipticity from stoke parameters: 0.2153498356
ellipticity from 1-(isoA/isoB): 0.4994626166

The results can be reproduced with the following python code:

import math

def ellipticityStokes(q, u):
    e = 1 - ((1-math.sqrt((q**2) + (u**2)))/(1+math.sqrt((q**2) + (u**2))))
    return e

def ellipticityNormal(a, b):
    e = 1 - (b/a)
    return e

q = -0.1206308
u = 0.002988584

isoA = 169.3949
isoB = 84.78848

print(str(ellipticityStokes(q,u))) #0.2153498356239003
print(str(ellipticityNormal(isoA,isoB))) #0.4994626166431221

Looking at the image

(and compare with this website from Buta) I would say the value 0.4994626166 (from isoA and isoB) makes more sense. The value calculated from the stokes parameters is just wrong I think.

What is the mistake here? I would expect both calculations to yield similar results.

However for newer data releases isoA and isoB are not available any more because they seem not to trust those value? What should I do with newer data releases as DR12?

Regarding your original question about the "correct value for ellipticity", keep in mind that a galaxy has no objective ellipticity. The inserted image nicely shows that the light distribution of a real galaxy is way more complicated than a smooth profile with isophotes of identical ellipticity. In other words, the ellipticity will in fact be defined through the measurement process.

Flux-weighted second moments (which use the pixel brightness as weighting function) or adaptive moments (which use an elliptical Gaussian weighting function) are two ways of doing this. And there might be nothing wrong with the observed difference in ellipticity measurements, as the galaxy light profile is very different from a Gaussian!

The flux-weighted measurement is more sensitive to the ellipticity at inner radii, while the adaptive moments give more weight to the disk (assuming that the algorithms converge well).


What effect does stellar granulation to have on a chemical analysis of a star's spectrum? The spectrum from stellar granules will obviously be for hotter gas, while the spectrum from the lanes between them will be for cooler gas.

Does this "average out" so that an average temperature distribution can be used for a metallicity analysis?
Will the spectroscopic effects be the same for all lines?

There is no simple answer to these questions - although I could be brief and say
(i) No it doesn't and (ii) no they won't.
If you make a simple two component atmosphere then the observed spectrum will be the flux-weighted combination of two spectra.
$$ S_{\rm obs} = \frac{A_1 T_1^4 S_1 + A_2 T_2^4 S_2}{A_1 T_1^4 + A_2 T_2^4}\ , $$
where $A_1, T_1, S_1$ are the area, temperature and spectrum of material at $T_1$ and the quantities with subscript 2 are for the regions at temperature $T_2$.
Now, depending on the response of any spectral feature to a change in temperature this could make the equivalent width of that feature become stronger or weaker in the flux-weighted average spectrum. If you then analyse that spectrum assuming that there is a single temperature then what you derive for an abundance based on that feature could get bigger or smaller.
For example most lines of neutral species (e.g. Fe I, Li I) get stronger in cooler atmospheres. If you put cool starspots on a star then the equivalent widths of these lines will become stronger. The average temperature of the star will decline very slightly. The net effect would be to estimate that the abundance of iron or lithium in that star was larger.
On the other hand if the line was very strong such that it was in the saturated part of the curve of growth, then it wouldn't get much stronger in a cooler atmosphere and it could even be that by adopting a lower average temperature, the lack of a significant increase in line equivalent wdth would lead one to infer a reduced abundance.
A more clear-cut case might be lines of ionised Fe II. These would become weaker in a cooler atmosphere, so you would end up inferring a lower iron abundance from the flux-weighted average spectrum.
Then there is the issue of the over-simplification that a simple two-temperature model imposes. Real granulation also involves the movement of plasma with consequent implications for micro- and macro-turbulence and how those affect the line formation and assumption of local thermodynamic equilibrium. These can only be addressed using expensive 3D stellar atmosphere models. Such models do exist and they offer grids of 3D NLTE corrections to abundances determined using 1D LTE atmospheric analyses. The corrections can go in either direction depending on the strengths of the lines and the intrinsic parameters of the star. As an example, Wang et al. (2021) present a grid of corrections for the Li I optical/NIR lines that I have a particular interest in.


How to track asteroids or NEOs? Fast-moving objects near our earth have different paths in the sky. They are different with stars far away.

For telescopes, when we do not know an object's orbit, catching up with its orbit and not losing it is a problem ? How to figure out a new object's orbit parameters quickly before we lose it? For an old object with an irregular orbit, can a telescope track it?

The typical speed in the asteroid belt may be smaller than 1"/min? But there may be faster objects. Then the paths are not so simple and regular.

Specialty hardware will be able to do it, as might clever programming if you have a commercial, computerized mount and feel like hacking away. Either way, an ordinary mount will be able to follow targets in the asteroid belt easily -- their motion across our sky isn't actually very fast. What I'm about to write really only applies to NEOs.

With NEOs, the problem isn't just the apparent speed of the objects across our sky, it's that the objects are usually in irregular orbits while moving at that speed. This is in addition to the apparent motion due to the rotation of the earth.

To track these objects you'd have to project the position of the object onto our sky, using the object's ephemeris data as your starting point, then add the apparent motion due to the rotation of the earth. That position could be calculated by the mount or calculated by a computer tethered to it, which would send updated pointing directions to the mount. The specifics on how to do this depend on your hardware. It's not a small project.


Correct units to use while plotting mollweide projection of sky chart with RA and Dec values, matplotlib I am trying to plot a mollweide projection of a sky chart in Python using the matplotlib package. I am using the RA and Dec values from a database containing all stars in Hipparcos, Yale Bright Star, and Gliese catalogs (almost 120,000 stars).

This is the database: https://drive.google.com/file/d/1sJGUxQpnsw93q48I1v8qo6rbDKCeGYOH/view?usp=sharing

It contains the degree values of RA and Dec (column names are ra and dec respectively) as well as the radian values of RA and Dec (column names are rarad and decrad respectively).

My code assuming I need to use the radian values of Dec and RA is:

figure=plt.figure()
figure.patch.set_facecolor('black') #setting plot background to dark colour
ax = figure.add_subplot(111, projection='mollweide')
plt.scatter(df['decrad'], df['rarad'], s=1, color='red')
plt.show()
figure.savefig("weird.png")

And the output plot I get is:

My question is: Why are all the datapoints plotted within that circle only and not throughout the entire elliptical surface of the celestial sphere? Is it because I have chosen the wrong units (radians instead of degrees) or is the nature of the plot supposed to be this way based on the given star data?

With stars extending to ±90° on the horizontal axis and rather sparse near those limits, I suspect that RA and Dec are swapped.
Try this instead:

plt.scatter(df['rarad'], df['decrad'], s=1, color='red')

Also if you want the plot to resemble the night sky, filter the star list by apparent magnitude, and reverse the horizontal axis so it increases from right to left.


Clear boundary between core and shell of a gas star Does a gas star have a clear boundary between its core and shell? If it does, then what is keeping a helium core distinct from its hydrogen shell, for a main-sequence star that transforming to a giant star?

This depends on the initial mass of the star, but generally the gradient of the core is not sufficiently strong on the main sequence to justify a well defined core-envelope boundary, which is why if a star on the main sequence experiences Roche-lobe overflow the entire star is essentially destroyed, e.g., in a binary system.
The core of a main sequence star develops due to the nuclear fusion of hydrogen into helium. The main sequence ends when the core is exhausted of hydrogen, and then low mass stars enter the giant branch phase and high mass stars enter the Hurtzsprung-gap: in both cases the star begins shell burning of hydrogen around the core. Then, once the temperature in the core is sufficiently high, core helium burning begins which brings the star to the next stage of development - the supergiant phase.
The star chokes like an engine does when it switches from one form of nuclear fuel to another, resulting in contractions of the core and expansions of the envelope, resulting in the star's surface radius to generally increase as it ages.

If it does, then what is keeping a helium core distinct from its hydrogen shell, for a main-sequence star that transforming to a giant star?

So, once a star is on the supergiant phase, its core is sufficiently well defined to support itself under the envelope by the luminosty pressure from the core and the shells surrounding the core, and the core supports its own self gravity by the degenerate matter that develops composed mostly of oxygen and carbon (i.e., the electron degeneracy pressure).
There is also rotational support and its effects, which is an active area of research. The core-envelope boundary is (literally) a turbulent region where chemical mixing occurs which drives convective currents through the envelope and which can result in angular momentum transfer between the core and envelope resulting in spin coupling between core and envelope. This spin coupling can have implications for the late evolution of the star. I asked a question about this that's remained unanswered.
Enhanced rotational mixing of the chemical elements inside a main sequence star can lead to chemically homogeneous evolution, in which there is no longer a core-envelope boundary since the helium has been mixed throughout the star.


Can an amateur astronomer bounce a laser off the moon? In the TV show "Big Bang Theory" episode "The Lunar Excitation", the gang fires a laser from their rooftop, bounces it off mirrors on the moon, and measures the laser coming back on a computer.

Is this really possible?

I know scientists have successfully done this, because it's why NASA put the mirrors on the moon.

Wouldn't this require very precise targeting by the laser to hit the mirrors?

My question is, is it possible for an amateur to successfully perform this experiment?

This experiment is still done frequently, but you need a powerful pulsed laser, precise pointing (error below about 20 arc seconds) and a sensitive detector.

Amateurs will fail most likely, because they won't get access to an appropriate equipment.

Here is a youtube video, which may give an impression of how difficult it is to perform the experiment.


How to calculate the orbital velocity, the high-orbit (quasigeostationary) holding force and the shadow of the disk on Earth? []

The following data are given:

•   The circular orbit of the disk D (quasigeostationary orbit) parallel to the geostationary orbit.
•   Located 4,505 km to the north of the equatorial plane, above the 45 degree parallel. The radius of orbit with respect to the gravitational center of the Earth 41,922 km (calculated by the uhoh we thank)
•   Disc diameter: 10 km.
•   The disc consists of a support frame and solar panels. The specific weight approx. 4.5 kg per square meter.
•   The time when the disc makes a complete, retrograde motion, around the earth: 25 hours.

Required:
1 Orbital speed of the disk so as to achieve a complete circular motion in 25 hours.

2 The force required to  maintain the disk in orbit and to compensate gravitational attraction.
2.1 How can this force be assured?
2,2 with thermochemical propellants (rockets with liquid fuel that could be started periodically?
2.3 electromagnetic, ionic, photonic propellants with cold gases.
2.4 And so on

3   What is the diameter of the disk shadow on the Earth's surface?

Figures 1: 2 and 3 graphically reflect the problem.

Any critical comments or comments are welcome.

enter image description here
enter image description here

Let's just pick up the second point. The disk is 42000 km (more or less) from the centre of the Earth, do the gravitational acceleration is $$9.81\times \left(\frac{6371}{41922}\right)^2 =  0.22 ms^{-2}$$ The component of that along the Earth's axis (Southwards) is $$0.22\times\frac{4505}{41922} = 0.024 ms^{-2}$$

The mass of the disk is $$5000^2\times\pi\times 4.5 = 3.53\times 10^8 kg$$ so the total force which must be counteracted by thrust is $$3.53\times 10^8\times 0.024 = 8.5\times 10^6 N$$

This is roughly the thrust of 4 of SpaceX's new Raptor engines burning continuously, each consuming about 1 ton per second of propellants. The best ion engines I can see quickly on wikipedia produce about $2\times 10^{-5}
 N/W$ (thrust/power consumption), so you'd need over $10^{12} W$ of power to achieve this thrust with  ion engines. The disk collects at most about 200GW of incoming sunlight, so the numbers don't add up that way. You'd also need quite a lot of Xenon and the ion engines would add a lot of mass.

Moving on to the shadow. Seen from Earth's surface the disk subtends an angle of about $\frac{10}{41922} = 2.5\times 10^{-4}$ radians. The Sun subtends an angle of about $0.01$ radians (half a degree). So the disk would not come close to blocking the Sun at all. That is there would be no umbra at all. In the penumbra the amount of sunlight reaching the Earth would be reduced by about 0.1%


Disk-integrated vs disk-resolved, what is the difference? I'm trying to find online what is the difference between disk-integrated vs. disk-resolved, but I can't find a website which does explain them, especially in the context of modelling spectral flux density. Also, since the word disk is being used, it is only applicable to spherical planetary bodies? Is spatially-resolved and disk-resolved the same thing?
Source

The way disk-integrated and disk-resolved is used in your cited paper is nothing special, just indicating optical resolving power in relation to the image of the object of interest:
With a disk-integrated signal they mean a single value as measured over the whole disk (of an image a planet makes - or whatever is imaged as a disk).
When they talk about disk-resolved data, they mean that they have a large enough image of the planet they image so that they can start to discover regional inhomogeneity of the signal, thus make out different regions with different properties.
So the distinction is just that of a spatial resolution: either you can resolve the object of interest and see surface (or cloud) structures (disk-resolved). Or your signal is too faint, your optics too small, and you just get spatially unresolved data, thus disk-integrated over the image of the disk, the whole planet (thus you cannot distinguish anything based on location on the planet - you just measure overall albedo, spectral properties etc)


Do we know if there are asteroids leading or following Earth in Earth's orbit around the Sun? A search of the Internet is so clouded with discussions about the asteroid belt or asteroids orbiting the planet Earth that I couldn't see an answer to my question.

Do we know if there are asteroids orbiting the Sun in Earth's orbit (NOT orbiting the Earth).

To be clear: these would be asteroids located somewhere around the Sun at the same distance as Earth, either following it or leading it in the Earth's own orbit around the Sun.

I hope that was clear.

There are several known Earth co-orbital asteroids. The first to be discovered was (3753) Cruithne, which is often mistakenly described as "Earth's second moon". Cruithne has a bean-shaped orbit when viewed in a reference frame rotating with the Earth.
The only known Earth Trojan is 2010 TK7, which orbits the L4 point located 60° ahead of the Earth.


What would it feel like to be orbiting near a merging black hole? The recent 4th discovery of gravitational waves where the black hole masses were 31 and 25 times the mass of the sun, and they released 3 solar masses of energy as gravitational waves, made me wonder what would it feel like to be orbiting this binary black hole system on a planet at 1AU when the merger happened?

It could do serious damage if things were just right.

Three solar masses of energy is about 5x1047 Joules.  1 AU is about 1.5x1013 cm. The area of the sphere is thus about 4x1027 cm2.  That's a flux of 1020 joules cm-2.  That's plenty of energy to do damage...if it can couple to you.

The strain of the GW measured at Earth is around 10-22.  Strain drops off as the inverse distance, and the ratio of a gigaparsec to an AU is about 1014, so the strain at 1AU from the merger would be about 10-8. By itself, this would probably not harm you.

But it's not a simple one-time strain, it is oscillating with the orbital frequency of the merging black holes, and anything which had a physical resonance at that frequency could pick up energy, possibly significant energy, from the GW fields.

(Note that this depends on orientation. The mechanical resonance needs to align with the GW, and GWs are not emitted in all directions, so you'd want (or, perhaps, not want!) to be in the plane of the merger.)

The argument that GWs can't deposit energy is plainly wrong -- if it were true, LIGO couldn't detect them! -- but if you believed that, you were in superb company: It took Einstein decades before he became convinced that GWs actually carried energy!


Can we observe what objects exist at the L3 positiion of planets of our solar system? From what I understand Earth's L3 is obscured by the Sun, so is that true for our viewpoint of the other planets? Or, are there times when other planets are on the other side of the Sun from Earth and we can see their L3 position? For the Hilda's of Jupiter have we observed them in that position or is it that we have calculated their coordinates to be in the L3 position?

You asked

  L3 is obscured by the Sun, so is that true for our viewpoint of the other planets?

No. Each planet orbits with it's own period because each planet is at a different distance from the sun. So most of the time we can see the L3 point of each planet's orbit from Earth. Occasionally it goes behind the sun from our point of view.

  For the Hilda's of Jupiter have we observed them in that position or is it that we have calculated their coordinates to be in the L3 position?

The Hilda family do not occupy the Jupiter L3 point. They generally orbit closer to the sun than Jupiter, and orbit three times for every two orbits of Jupiter. Their orbits are interesting; they have eccentric orbits (up to 0.3) that in some cases cross the orbit of some of the Trojans. You may have noticed that some of the Hildas also approach the Jupiter L3 point. At that point, and when other members of the family approach the Tojans at L4 and l5, the Hildas are moving more slowly. So the Sun's gravity gradually accelerates them, pulling them back to the lower, faster part of their orbits.

I have linked to the Wikipedia Hilda family page. The Lagrangian point page is also helpful.


Why can we still 'see' the CMB I read that the CMB were all the photons left free when the universe was cold enough for electrons got bound with protons (recombination) and the photons free in that moment continued traveling in the directions they were.

So I imagine:

the CMB as a sphere (with incredibly thick walls) increasing its radius continuously away from the 'center' of the universe. When the Milky Way formed, hadn't this sphere of photons already passed the space in which the Milky Way formed? How can we still see them if they are moving away from us.
the rate at which the space grew was bigger than the speed of these photons, so the Milky Way and our solar system had time to be formed before being reached by this photons lagged behind the speed in which the space increased.

On the other hand, how can we continuously measure the CMB if is not being generated continuously as a stream of photons like looking at a distant star. We should be able to see them only once when they passed by earth.

Firstly, there is no centre to the universe and the universe seems to continue indefinitely in all directions. It is best to imagine the universe as infinite in size.

Now we need to explain why the CMB appears to be a sphere surrounding us.

Let's imagine, for a moment that I have filled the entire infinite universe with (magic) mud. This mud glows, and because its magic mud, it doesn't change the orbits of the Moon, Earth or anything else in the universe.  But if I hold up my my hand in front of my face I can't see it, because of the mud.

Now then, at one moment all the mud vanishes, everywhere, throughout the whole universe at the same time. One second later, if I hold my hand up I can see it. But if I look in the sky I can't see the moon. The moon is still hidden in mud.  In fact if I look at the sky, all I see is mud at 1 light-second (300,000km) distant. And because it is glowing mud, the sky appears to glow.  It looks like I am in a middle of sphere with radius 300000, surrounded by glowing mud. But this is an effect of the fact that light travels at 300000km/s

After 1.2 seconds, the moon becomes visible, after about 8 minutes, the sun becomes visible too. But it takes 4½ years for the first stars to be visible. At this point it would look as if I'm in the middle of sphere with radius 4.5 light years surrounded by mud.

Now what really happened is that 13.5 billion years ago the whole infinite universe was filled with hot gas that blocked light. As this cooled this became transparent. The whole universe became transparent at the same time, and while it wasn't instantaneous, it happened pretty quickly. So what we see now is that we seem to be in the middle of a sphere of hot gas 13.5 billion light-years across.

Except that in those 13.5 billion years, space has stretched. This has stretched the sphere from 13.5 billion light-years to 45 billion light years, and it has stretched the light from that hot gas, so it appears to be very cold.

The photons that we see in the CMB now did not come from the hot gas that once filled this part of the universe, but from hot gas that was in part of the universe that is now 45 billion light years away. That being so, The photons that we saw last year came from gas that was in part of the universe that is now 44,999,999,999 light years away. Since we can't detect structures as small as few light years at that distance, the CMB appears as a constant even pattern in the sky, changing very little if at all.


Why aren't planetary bodies static? Why are planets moving in the first place? Gravity causes them to orbit, but why move at all?

note: this is not a question on why planets orbit each other, I know the reason for that is gravity. What I'm asking is why do planets move at all? For example, if you turned off gravity, why would the planets carry on moving apart rather than be stationary?

It's simply because the sun and planets were formed out of a big pile of dust.  Originally the dust was spinning.  So once the dust became planets, it kept spinning around.  That's all there is to it - that's why the solar system is spinning around.

It's was once a pile of dust that was spinning a bit; it's still spinning.

You then ask,

"For example, if you turned off gravity, why would the planets carry on moving?"

THat's just the same as asking "if I spin something around on a string, and break the string, why does it keep moving?"

For that matter, it's the same as asking "What is momentum?"  So, if you push something ...... why does it keep moving?

At this stage in history, we have utterly no clue, at all, what the heck time, space, matter, and momentum are.  "Why does momentum do what it does?" is for now just one of those super-deep questions like "What is time" or "So what caused the big bang" or "What is gravity" or "What's the explanation for this quantum stuff?" or indeed ... "What is momentum"?

So, why do the planets keep moving?: answer "momentum".  If you want to know "what causes momentum?", that is one of the basic total mysteries.

For now, nobody has a clue. You may as well ask ... what is space, what is time, etc.

Regarding momentum, you might like to read up about the so-called "Mach's conjecture". The famous Einstein was, like you, fascinated by the question "WTF is momentum anyway?" One sort of general thinking-point originating with this smart guy called Mach is that momentum could have something to do with "all the other mass in the universe". Nobody has a clue about this, and it's just a vague general idea.

Isaac Newton was a pretty smart guy (if a bit whacky), and in the end he found gravity so mysterious, he just put it down to God. He probably found momentum as mysterious, and he was one of the first guys to think about it clearly.

An interesting point: actually everything astronomers look at (all galaxies, all structures of galaxies) in fact does not (!) behave the way small things (our solar system, as in your question) behaves in terms of gravity and momentum.  This is usually explained by invisible unknown matter ("dark matter") or for a few scientists, that gravity works differently than we think presently.  So the fact is with issues like gravity/momentum you ask about ... not only do we have no clue "why momentum works" but when you look through a telescope, issues like "orbits" work totally differently anyway!


The definition of eclipsing binary star systems In the second minimum (the 3rd step) there is a smaller decrease in light intensity. For this to happen, wouldn't you need to be looking at the plane of orbit from above rather than directly along the plane of orbit? Because if a person was looking directly along the plane, then it would just be a full eclipse of the star behind in step 3?
But then if a person is not directly looking along the plane but just sees the plane from above, doesn't this mean the definition of eclipsing binaries is wrong? The definition is 'when the plane of orbit of stars is in the line of sight from the Earth'. By 'line of sight' I assumed it meant a person from Earth is able to look directly along the plane. So was my assumption wrong?

I assume that the diagram indicates what the observer sees (if they had a big enough telescope!). i.e. The viewpoint is nearly in the orbital plane but not quite. Why then are the eclipses asymmetric, with the secondary eclipse being shallower than the primary? Well probably because the surface brightness of the two stars is different - i.e. they have different surface temperatures. The secondary would be cooler, so when its surface is
obscured by the primary, then that results in less flux being eclipsed. Note that this would be the case whether you were looking along a line of sight exactly in the plane of the orbit or at a small angle to it.
There is no definition of an eclipsing binary that says the plane of the orbit must be in the line of sight from Earth. It has to be close to that in order to get eclipses, but it does not have to be exactly true.


Could someone explain RA/dec in simple terms? The equatorial coordinate system isn't too complicated. However, for people new to this, it seems rather intimidating at first.

Is there an easy way to explain it?

A system projected into the sky

Pretend all the stars are painted on the inside of a large ball, and you are at the center of the ball. The imagined ball is called the Celestial Sphere. If the Earth wasn't blocking the lower half of your view, and the Sun wasn't making the blue sky so bright, you would be able to look at the stars in any direction.

First, the rotation of the Earth causes the imagined ball of stars to appear to move around the sky. If you lay on your back, (in the norther hemisphere,) with your head to the north, the stars slowly slide by from left to right. There are two points however, where the stars don't appear to move: These two points are directly above the Earth's north and south poles. These two points, projected out into the sky are labeled the North, and South, Celestial Poles.

Next, the Earth's equator is projected out into the sky and labeled the Celestial Equator.

Finally, sorting out the east-west positioning, (ie left-right if you lay on your back, head to the north), requires an arbitrary selection: There are an infinite number of circles one can imagine drawing on the sky. (For example, small circles the size of the moon, and also many much bigger circles.) The largest circle you can draw is called a "great circle", from geometry. Point at the North Celestial Pole, (way above and behind your head if you face south in the northern hemisphere,) and draw a line straight south to the horizon, then down below your feet through the South celestial pole and back up to the North Celestial Pole. You have just drawn one Great Circle, through the Celestial Poles, all the way around the Celestial Sphere.

But that Great Circle is just one of an infinite number of similar ones. One could start at the North Celestial Pole and draw a Great Circle a little to the left or the right. So astronomers have chosen one specific Great Circle and labeled one side of that circle as "zero degrees" -- or, "we will start measuring angles from here."

So how do we use this system?

"right ascension" is measured rightward, (from your left to your right if you face south in the northern hemisphere,) from the zero-degrees half of that chosen great circle. So given a right ascension, one can find the corresponding half of a great circle. For a given right ascension, you have half of circle selected from the North Celestial Pole, to the South Celestial Pole. (This is called a line of equal longitude.)

"declination" is measured from the Celestial Equator. The Celestial Equator cuts through the middle of your line of longitude. From the Celestial Equator it is 90° northward(+) to the North Celestial Pole, and 90° southward(-) to the South Celestial Pole. So we can measure from +90° (northward) to -90° (southward) along that line of longitude, from the Celestial Equator. (Don't be confused by "positive declinations"; 45° of declination is northward from the Celestial Equator.)

Notice that the system does not depend on where you are standing on Earth, nor on the time of day. If you have a Right Ascension and a Declination, you have an unambiguous spot specified on the Celestial Sphere. The system is actually very simple, but the rotation (time of day) of the Earth, and where you are geographically located needs to also be figured in.

What's over my head right now?

Face south, (if you're in the northern hemisphere.) What is your latitude -- geographically, how far north are you from the Earth's equator? If you are 40° north, then Celestial declination of 40° north is directly over your head. Declination of 50° is 10° further northward from directly over your head, etc. But for each declination, you have a circle of points in the sky that all have that same declination.  The circle is centered around the celestial poles.  As the earth turns it appears to slowly rotate overhead. Note: If you are at the equator, this circle is just the celestial equator, and it is a great circle.  As you approach either pole, the circle will get smaller.  The poles, this circle collapses to a point.

Right ascension you simply have to look up. That arbitrarily chosen, zero-degree line, appears to spin around the sky every day at the Earth rotates. Also, over the course of a year, the Earth's orbit about the sun makes an extra apparent turn of the sky. (What is overhead mid-winter, is directly under your feet mid-summer!) So there's an offset -- what right ascension is overhead on this date at midnight, and how far (in hours) are we from midnight right now? Combining those two you can figure out what Right Ascension is over head at any time, on any date.

(potential editors: I've intentionally not used acronyms for NCP, RA, etc as that makes it more complicated for people learning this material.)


Will we start seeing galaxies disappear due to Universe expansion? Does the fact that universe is currently expanding at an accelerating rate means that far far away objects (FAO) might start disappearing with time, set aside the "red shift"?

I'm saying this because from what I understand until now:

we receive photons from FAO that were emitted long long ago
now due to space expansion, some of those objects are now moving away at speed greater than the speed of light, but at the time the photons we receive today were emitted, these FAO where not moving away faster than light

So there must be a point in time at which these far away objects disappear from view. How is this concept called?, or is it only a mistake I'm doing in my thinking?

Here's a simple drawing to illustrate:

at t=-11 billion years(By), photons are emitted in a slow expanding universe, and today we can view the object as it was
at t=-2 By, the object moves faster than speed of light, so photon emitted at that time will never reach us, even if we wait the 9 by + "expansion correction" needed...

I couldn't be convinced by the answer I found here, saying nothing disappears out of horizon. I don't see how it complies with that answer were event horizon approaches us.

It is a common misconception that galaxies receding faster than light cannot be observed. There are two versions of this misconception:

Galaxies that are now receding faster than light cannot be seen.
If we observe a galaxy today, it may recede faster than light now, but when it emitted the light we see now, it did not.

Both are incorrect.
Intuitive explanation
This is most easily seen in a spacetime diagram. But before showing that, a somewhat intuitive way of understanding this apparent paradox is to consider the journey of the photon from the reference frame of the emitting galaxy: In this frame, the photon leaves the emitter with a velocity $v=c$ (as it should), but since its velocity is always $v$ locally, then in the frame of the emitter, its velocity increases without bounds. This is not a proof, but may help to show that "light may travel faster the the speed of light".
Spacetime diagram
Now to the spacetime diagram. My favorite version is from the classic paper by Davis & Lineweaver (2004), but I actually like better this version from Pulsar's excellent (!) post:

Real example of a galaxy that has always been, and always will be, receding with superluminal velocity
Such a diagram shows age (or, equivalently, size) of the Universe, as a function of the corresponding comoving distance$^1$ of events. A lot of things are going on in this diagram, and I'll refer to Pulsar's answer for an interesting discussion of all this. For our purposes, consider the only thing the I have added to the figure: the world line — i.e. the "path" taken through spacetime of some object — of the galaxy GN-z11, seen as a vertical dashed line at a comoving distance $d_C \simeq 32.2\,\mathrm{Gly}$. It is vertical, because it doesn't move in comoving coordinates. We observe it at a redshift of $z=11.09$, which means that the light we see (moving at 45º toward our worldline which is at $d_C=0$) was emitted when the Universe was roughly 400 Myr old. At that time, GN-z11 was receding at a velocity $v\simeq4c$!
Hubble sphere and event horizon
The innermost green bubble (labeled "$v_\mathrm{rec}=c$") is the Hubble sphere, defined such that at any point in time, everything outside recedes faster than $c$. Meanwhile, the red line shows the event horizon, which defines the boundary between everything that can eventually be observed (light red region), and everything that cannot. A subset of this is out past lightcone (orange region) which is stuff that we have already observed (or at least have had the chance to).
Everything lying in between the Hubble sphere and the event horizon consists of events that we have been, or will eventually be, able to see, but which recede faster than the speed of light when it emits the light we see.
As you can see the region becomes narrower with time$^2$, meaning that the accelerated expansion of the Universe makes it harder and harder for photons emitted far away to reach us. We will never be able to see galaxies that today are farther away than roughly 17 Gly. But galaxies that are nearer will be seen (if we look), and a subset of these — namely the ones with $14\,\mathrm{Gly} \lesssim d_C \lesssim 17\,\mathrm{Gly}$ — are currently receding faster than the speed of light.
The size of the observable Universe always increases
The observable Universe is defined as the region within which light has had the time to reach us. The "edge" of this is called the particle horizon, and as seen from the blue line, it always increases its distance from us (in comoving coordinates, and thus even more so in physical coordinates). In comoving coodinates, it reach a maximum size of roughly 63 Gly. That means that more and more galaxies will enter our observable Universe, but as time goes they will do so at an ever-decreasing pace, asymptotically reaching a finite value (of 5 trillion galaxies, more or less). Once inside the particle horizon, a galaxy will never leave. It will, however, keep increasing its distance from us in physical coordinates, as well as its speed, and thus redshift without bounds and fade to black.

$^1$Comoving coordinates are defined so that they "expand with the expansion of the Universe". That means that in these coordinates, galaxies lie approximately still wrt. each other. To get the physical coordinates — i.e. what you would measure if you froze the Universe and laid out measuring rods — multiply the comoving coordinates by the scale factor $a$ seen on the right $y$ axis. Since today $a\equiv1$, comoving and physical coordinates coincide today.
$^2$The fact that the diagram is in comoving coordinates makes it a bit hard to see, but this is still the case in physical coordinates.


What is the temperature of the black hole? Are black holes hot or cold? I was thinking because I wanted to know if there any human sustainable temperature so we can travel in it.

See Hawking radiation. The temperature of a black hole depends on its mass, and is given by:

$T=\frac{hc^3}{16\pi^2GMk_B}$

Where $h$ is the Planck constant, $c$ is the speed of light, $G$ is Newton's gravitational constant, $M$ is the mass of the black hole, and $k_B$ is the Boltzmann constant.


Are we so sure global warming is a result of humans burning fossil fuels? The Sun is a fiery mass of colliding particles.  We still know very little about what it is.  Could it possibly be the output of the sun fluctuates, like our weather, from one extreme to another.  From hot durations to cold durations and back again. I think global warming is the result of higher amounts of solar energy being released from the Sun; that's why it's getting hotter here.  I believe the question is important to all people on planet Earth, par excellence; is humanity, namely our consumption of fossil fuels, the cause of our predicament?

We know a lot about the sun. It isn't the main cause of global warming, as we can monitor exactly how much energy it is producing. The sun's output can fluctuate though, and we have yet to measure how badly this affects the climate.

So what is the main cause of global warming?

The main cause of global warming is emissions of C02 (carbon dioxide) into the atmosphere. C02 is odorless, colourless and non-toxic, however, it is the main cause of global warming. C02 is a greenhouse gas and  can stay in our atmosphere for about 20 years.

C02 causes something called the greenhouse effect. The greenhouse effect is the idea that long wave energy which is meant to go back to space, doesn't go back to space, and is instead captured by greenhouse gases such as C02. This causes lots of heat that should go off into space to be trapped inside the earth's atmosphere causing your temperature to rise.

Are there any other greenhouse gases?

Other greenhouse gases do exist, carbon dioxide is just the main one that causes the global warming. Other greenhouse gases include water vapor, methane, nitrous oxide and ozone.

Does global warming happen only on Earth?

Global warming happens all over the  solar system, planets like venus have many greenhouses gases in their atmosphere too, infact, on venus greenhouse gases alone make the planet 465 degrees celsius hot, which is way over the boiling point of water - 100 degrees celsius.

Do I disagree?

Yes, as there is too little evidence to prove that the sun fluctuates enough energy to gradually increase our climate dramatically, and we have also proven the effects of greenhouse gases in a planets atmosphere.


How do we measure the age of the universe? As mentioned in wiki/Age_of_the_universe,

  The current measurement of the age of the universe is around 13.8 billion years (as of 2015) – 13.799±0.021 billion years

When my friend who is not from science background asks me how do we measure the age of the universe? How should I explain to him?

Once we look far enough away to get past "local effects" we see everything moving apart, spreading out in all directions, like a set of dots on a balloon being inflated, or like currants in a loaf of bread rising in the oven.

We can use the laws of physics to follow those movements back in time to work out how far apart all the galaxies were at various dates in the past. If we do this we find that 13.8 billion years ago they were all in the same place (there are a lot of subtle details to get right in this calculation, but a lot of people have worked on it over a long time). That much matter and energy packed into that small a space warps both space and time so that it is not really possible to talk about time "before" that event. Hence that is taken as the origin of the universe.


Black dot visible when looking into the telescope A few weeks a go I bought a telescope.
The problem is I am a total beginner. I was able to setup the telescope correctly and I already saw the moon very clearly. Then I moved on to jupiter. When I looked through the telescope a black dot was visible in the middle of Jupiter. I have already turned the focusing wheel but then Jupiter became smaller and smaller and the point became smaller. But even when I turned the eyepiece all the way out the dot was still visible and jupiter was hardly visible.

What we see in your image is an out of focus object (the white moon shape) with the shadow of the secondary mirror (causing the out of focus object to have a crescent moon shape) and the struts (or "spider vanes") that hold the secondary mirror in place (the vertical and horizontal lines).
Focus on the moon and get that focus sharp.  Then try looking at stars (they should appear as pin-points) then try Jupiter and Saturn.  Don't touch the focus knob, when looking at the planets, since the focus point for the moon, star and planet is the same (they are all at "infinity" as far as focusing is concerned).
Jupiter will be "small" with that telescope.  Perhaps much smaller than you expect. Look a tiny circle of light with four dots (the moon) around it. You may see some banding.  Saturn is often better looking, as the rings should be clearly visible.
As a_donda suggests, the telescope may benefit from collimation (getting the mirrors and eyepiece perfectly lined up) The manual should have details (if not, search the internet) It's hard to give specific advice.  Look through the manual perhaps there's a faceslapper that you're missing, or get a friend to go through it with you (four eyes are better than two). If you got it from a specialist shop, you could try talking to the shop people. You could try the website of the company that made the telescope, they may have a contact number.  You could search for a local astronomy club.


Exoplanets and relativity Ok, so I have read about how exoplanets are found using all sorts of genius methods.  I also understand the further an obect in a telescope is the further into the past I am looking.

So given that we are finding exoplantets with methods that go well beyond any practical visual range are we fnding these exoplanets 'in their past' as well? As far as I understand radio telescopes and other methods can't travel faster than light therfore their results are also from past events, is my reasoning in this right?

Yes, you are indeed correct. We always see what was happening at a given location when the light was emitted there for us to be detected. Let's assume we observe a oscillating star 100 light years away by measuring the shift in frequency of the detected light. That means that the star was actually oscillating a hundred years earlier and we just happen to observe it now; But we (that means nobody) have no way of knowing what is happening now due to the fact that information can't travel faster than light (especially when the information is transmitted via light as in our case).


Can we hear something on Venus, Mars and Titan? Mars, Venus and Titan have an atmosphere and we can theoretically step on them (and die quickly).

So, as an example, let's say we can survive on these objects without a space suit. I say: “Hi” on any of these objects. Would I hear what I said or not?

It depends on the composition of the atmosphere. In a helium (or other light gas) atmosphere, your voice would have a higher pitch. If the gasses are heavier, your voice drops.
The density is also a factor: on Mars, sounds can't be as loud as on Earth. Sound is transmitted as pressure differences, and the largest pressure difference you can get is that between the atmospheric pressure and a vacuum.
The speed of sound also changes with the composition and pressure of the atmosphere. This would change the way you perceive the distance and direction of an object that makes a sound: we normally use the tiny differences in time between a sound reaching our left ear and our right ear. A different speed of sound screws up those time differences.


Can small gas planets exist? Most of the known gas planets (Jupiter, Saturn, etc.) are huge. They are even called "gas giants".

is it possible to exist a gas planet around the size of Earth? If yes, why; if no, why?

According to the article "Minimum planetary size for forming outer Jovian-type planets - Stability of an isothermal atmosphere surrounding a protoplanet" (Sasaki, 1989), not only is this possible, but is suggested to have been an evolutionary stage for the development of Uranus and Neptune.  But there are a few conditions to allow this to happen, primarily due to distance from the sun, hence temperature.

The article's findings suggest that the critical mass for 'atmospheric collapse' is 0.2 of the current mass of the Earth for distances of 5 AU or greater for our solar system (this value of course would vary, depending on the parent star).  This critical mass increases the closer you are to the parent star - the article explicitly states that Mercury and Mars are too small for this to occur.


What are the dark spots in the "face of Mars" picture? In this picture of Cydonia by Viking 1 (taken from this article), there are a bunch of black spots scattered everywhere. Are they an artifact of the image or are they an actual feature on the surface of Mars?

What are the dark spots in the “face of Mars” picture?

They are called "bit errors" in the NASA JPL image catalog for PIA01141. The explanation seems a bit handy-wavy to me; so I think we'll need to dig deeper to see if they are received noise on Earth or some electrical or imaging problem on the spacecraft.
Read more about the Voyagers' cameras in What is the brown band in which we find the pale blue dot that is Earth?
By the 1980's NASA spacecraft used solid state CCD detectors and end-to-end digital transmission. See Galileo (spacecraft) Solid State Imager
From PIA01141: Geologic 'Face on Mars' Formation

Original Caption Released with Image:
NASA's Viking 1 Orbiter spacecraft photographed this region in the northern latitudes of Mars on July 25, 1976 while searching for a landing site for the Viking 2 Lander. The speckled appearance of the image is due to missing data, called bit errors, caused by problems in transmission of the photographic data from Mars to Earth. Bit errors comprise part of one of the 'eyes' and 'nostrils' on the eroded rock that resembles a human face near the center of the image. Shadows in the rock formation give the illusion of a nose and mouth. Planetary geologists attribute the origin of the formation to purely natural processes. The feature is 1.5 kilometers (one mile) across, with the sun angle at approximately 20 degrees. The picture was taken from a range of 1,873 kilometers (1,162 miles).

Below is a closer-to-raw-data image from Views of the Solar System: Raw Viking Orbiter Image 035a72 of the Face. Notice that the dots are bright, not dark in the raw data. Voyager's cameras are 1970's technology; they are based on vidicon tubes not solid state detectors.

Prior to the design and construction of the Galileo probe to Jupiter in the late 1970s to early 1980s, NASA used vidicon cameras on nearly all the unmanned deep space probes equipped with the remote sensing ability. Vidicon tubes were also used aboard the first three Landsat earth imaging satellites launched in 1972, as part of each spacecraft's Return Beam Vidicon (RBV) imaging system. The Uvicon, a UV-variant Vidicon was also used by NASA for UV duties

If you look closely you can notice both white and black dots. @Planetmaker's answer mentions this looks like salt-and-pepper noise.
From Arizona State University's Space Exploration Resources Control Network:

The majority of Mariner 10 high resolution images were sent back in real time (117.6 Kbits/s) with an attendant increase in bit errors during transmission [Danielson et al., 1975; Soha et al., 1975]. This resulted in severe salt and pepper random noise [Soha et al., 1975] that hampers the automatic identification of reseaux, especially in areas with low signal such as terminator regions and space. Due to the generally noisy nature of the data [Danielson et al., 1975; Soha et al., 1975] tolerances were set high (0.90 correlation) to help avoid spurious reseau identifications. Failure to find an actual reseau at a given position resulted in default to the nominal position for a given camera/encounter combination.

Danielson, G.E., K.P. Klaasen, and J.L. Anderson, Acquisition and description of Mariner 10 television science data at Mercury, J. Geophys. Res., 80, 2357-2393, 1975.
Soha et al., 1975  Soha, J.M., D.J. Lynn, J.J. Lorre, J.A. Mosher, N.N. Thayer, D.A. Elliot, W.D. Benton, and R.E. Dewar, IPL Processing of the Mariner images of Mercury, J. Geophys. Res., 80, 2394-2414, 1975.

From Digital Elevation Model of the Discovery Region on Mercury(Lunar and Planetary Science XXIX, 1849.pdf):

Excessive salt and pepper noise in vidicon images, can cause the mis-identification of reseaus necessary for accurate image rectification.

From book.google.com Yap et al. (2010) Adaptive Image Processing: A Computational Intelligence Perspective, Second Edition (also here):

9.4.2 Models of Impulse Noise
When an image is coded and transmitted over a noisy channel, or degraded by electrical sensor noise, as in a vidicon TV camera, degradation appears as salt-and-pepper noise (i.e. positive and negative impulses).

Emily Lakdawalla discusses the Voyagers' vidicons and problems with the old "treasure trove" of images


How do you carry errors for brightness magnitudes I have a value for the relative flux $F_2/F_1$. With some uncertainty value $a$.
If I use the equation to get relative magnitude: $m_1 - m_2 = 2.5 \log_{10}(F_2/F_1)$
How do you now calculate the uncertainty value $a$ for this equation? Do you just do the same equation but applied to the uncertainty value or is there some other alteration you have to make to uncertainty values?

You need to do what is called “propagation of uncertainties”. You can search to get more information on that, but briefly if you have some function $f(x)$ that depends on variable $x$, then the uncertainty $\sigma_x$ on the quantity $x$ is related to the uncertainty on $f$ by
$$
\sigma_f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 \sigma_x^2}
$$
Here $f$ is your relative magnitude $m_1 - m_2$, and $x$ is your flux ratio.  So taking that partial derivative will give you the factor that relates the two uncertainties.


Can one detect from earth that angular momenta of its axial rotation and its orbital rotation are "almost parallel"? I mean that the angle between these angular momenta is acute as opposed to obtuse (23.5 and not 180-23.5), so that the rotations are "roughly in the same direction".

Related but distinct question -- when was this actually discovered?

Certainly! Ancient astronomers, eg in Babylon ~3000 years ago were sufficiently familiar with the geometry of the celestial equator and the ecliptic to be able to predict eclipses, and discover the eclipse series now known as the Saros series. But of course, in those days astronomers (mostly) didn't conceive of them in terms of the motion of the Earth.

OTOH, there were a few ancient astronomers in ancient Greece (notably Aristarchus of Samos), and even in Babylon (Seleucus of Seleucia), who did believe in a heliocentric system.

The celestial pole is easy to discover, (especially if you're in the northern hemisphere), and it doesn't take a lot of observation to notice that the ecliptic differs from the celestial equator.

Although it's not possible to observe directly that the Sun's path is the ecliptic, it was known that all of the other celestial wanderers stay fairly close to the ecliptic, and that eclipses only occur when the Moon is very close to the ecliptic. And you can get a pretty accurate idea of the Sun's path by observing the rising & setting of stars and planets that are near the Sun's path. So astronomers across the ancient world recognised that the ecliptic is the Sun's path. And by the 2nd century BC, knowledge of these movements was sufficiently detailed that Hipparchus of Nicea was able to estimate the precession of the equinoxes.


Deep space radiation distribution What frequency distribution law applies to a random point in deep space that is assumed to be far away from any particular source of radiation.

Can it be expressed by some form of Plancks/Bose-Einsteins/Fermi-Diracs laws or a mix of all of them?

What would be the mean frequency/wawelength?

Would the mysterious "dark energy" be a major contributor to the distribution?

If  you were in deep space, i,e., a spot somewhere between galaxies, then the night sky would look a lot like the sky seen by astronauts in the ISS, when they look away from both the ecliptic plane and the galactic plane, minus the individual Galactic stars.  That is, it is like our deep sky photos in those directions that are dominated by galaxies, but no atmospheric distortions, no night sky lines, and no zodiacal light.  The spectrum is dominated by different things at different wavelengths, but basically it is the sum of the light of galaxies and perhaps intergalactic stars that escaped their galaxies.  At long radio waves, it is dominated by supernova remnants and AGN as self-absorbed synchrotron radiation, peaking between 1 and 10 Mhz (Simon 1977).  The mm band is dominated by the microwave background radiation at 2.7 C. The far-ir is dominated by cold galaxy dust at 60 microns. The optical is dominated by galaxy thermal starlight at 1 micron.  The UV is dominated by hot stars at 0.2 microns (see summary).  The Xray spectrum from 3 to about 45 keV (Marshall et al 1980, Fabian 1992) is well-fitted by a ~40 keV thermal bremsstrahlung model. Above 45 kev the spectrum falls and is thought to be the sum of many AGN at all redshifts.  The spectral energy density (SED) from radio to X-ray is shown in this diagram from the paper by Hill et al. (2018) The Spectrum of the Universe:

The width of the line in the diagram shows the present uncertainties in the measurements.

Dark energy does not shine, it has an effect on the redshift as a function of time and thereby indirectly affects the shape of the background spectrum.


Why can't we see a solar eclipse every month? In every lunar revolution around Earth, the Moon is at a point between the Earth and the Sun. Then why can't we see a solar eclipse once every lunar month?

Is it because of a tilted orbit of the Moon around the Earth? Is the orbit of the Moon some kind of a sphere?

The Moon orbits in a plane that is inclined with respect to the plane in which the Earth orbits. The intersection between these two planes is a straight line. For an eclipse to occur, this straight line needs to be roughly aligned with the line that goes between the Earth and the Sun.

From this blog post

As the Earth goes around the Sun, the plane in which the Moon's orbit keeps the same inclination compared to the stars (well actually, not quite, because of the precession of the orbial plane, but this gives you the general idea), so it turns with respect to the Earth-Sun system. This is why eclipses occur (roughly) every six months, and not every single month.

The frequency with which you can actually see an eclipse from a given location will be even less than once every six months. This is because to see a lunar eclipse, you need to be on the night side of the Earth while the eclipse is happening. And to see a solar eclipse, you need to be under the track of the Moon's shadow. This is a band a only few hundred kilometers wide. Here are the tracks of a few past eclipses.


Will Saturn's rings become a moon? Planets form from disks of matter orbiting around a star; some moons form from disks of matter orbiting planets. If this were going to happen around Saturn, approximately how much time would it take?

The answer to the headline question is: No. Most of Saturn's rings are below the Roche limit of about 2.5 Saturn radii. Hence tidal forces will prevent that part of the rings to form a (large) moon.

Actually, part of the rings may be caused by loss of material from some of Saturn's moons, as suspected from observations of Enceladus.

Accretion of Earth is still going on. So any number between millions of years and billions of years of accretion time for a planet can be justified.
Half of Earth's mass should have accreted within 10 million years, see this paper.


Why do Jovian moons fade when transiting Jupiter's shadow? When reading an article on using Jovian moons to calculate longitude, I came across this passage:

  Periodically, Jupiter eclipses each of the four large moons as they pass into the planet's mammoth shadow. In a telescope, a moon's brightness takes several minutes to fade to black as it enters the shadow. Reappearances are just as gradual.

How can the moons take this long to fade in and out? Is it simply the shadow creeping over the surface of the moons? What does this look like?

It's pretty simple, really. The shadow of Jupiter, just like the shadow of the Earth, is not razor-sharp. There is an intermediate region, called penumbra, where illumination is partial. This is due mostly to the fact that the Sun is not a mere dot, but has a measurable size. The fact that the planet itself does not have a sharp edge also contributes a small amount.

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

Penumbra can be clearly seen during Moon eclipses on Earth. The next one is on April 15 - a few days from today.

http://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2014Apr15T.pdf

http://www.skyandtelescope.com/astronomy-news/observing-news/aprils-total-eclipse-of-the-moon/


What (actually) does "peculiar periodic spectral modulations" refer to in this preprint? I saw this ArXiv preprint after reading about it in the popular media.
I can not even understand the beginning of the introduction (much less the whole paper):

INTRODUCTION
A Fourier transform analysis of 2.5 million spectra in the Data Release 8 of the Sloan Digital Sky Survey (SDSS) and the SEGUE 2 SDSS was carried out to detect periodic modulations contained in their frequency spectra (Trottier 2012).

where Trottier 2012 is Recherche de signaux périodiques dans des spectres astronomiques, M.Sc. Thesis, Université Laval
The modulation frequency seems to be of order 10${}^{-12}$ seconds, and the SDSS is (as far as I know) not really a particularly time-resolved data set. 1 micron wavelength light (for general reference) has a frequency of about 3x10${}^{-14}$.
I'm not asking (necessarily) if this result is right or wrong, I'd just like to understand how the SDSS can actually be analyzed to detect or even infer THz rate modulation!

The idea of Trottier & Borra is to analyze signals that might be hidden IN the stellar spectrum. I know this sounds weird; it did sound weird to me when I first heard of it too.

You know how you can decompose light in diferent colors, right? Well, we astronomers like to think of colors in terms of wavelength (with $~4500$ Angstroms being blue and $~7000$ angstroms being red) or in frequency ($1.5\times 10^6$ GHz and $2.3\times 10^6$ GHz respectively. To get these numbers, just divide the wavelengths by $c=3\times 10^8$ m/s, the speed of light). In the range of wavelengths in which the Sloan Digital Sky Survey (SDSS) works, as you can see, it is much more clear to work in wavelength, but people, for different reasons (like in this paper), can also work in frequency-space rather than in wavelength-space.

The thing is that the SDSS has spectra for millions of objects, including stars. That is, they have measurements of the flux as a function of wavelength (color) for these objects. Well, the idea in this paper is actually an idea Ermanno Borra published years ago (2012; here is the paper: http://adsabs.harvard.edu/abs/2012AJ....144..181B). In this paper, he proposes that maybe extraterrestials would like to communicate by injecting extra signals in that spectra, i.e., that they would add extra light in different colors to the stellar spectra we observe at Earth. However, he proposes that maybe they would put colors in frequency-space rather than in wavelength-space (because it is easier to modulate that with, e.g., lasers), and that these color modulations will be periodic in this frequency-space spectra. So, basically, this is what they search for in that spectra: periodic modulations in the colors, but in frequency-space rather than in wavelength. This is what they show, for example, in their Figure 4:

The top panel has an added signal with a periodic modulation that is $10^4$ times larger than the one they detected on the real star (lower panel).

As you can see, they don't analyze a time series: they analyze the stellar flux in frequency space. Now, I think I have quite a bunch of experience with stellar spectra, and quite frankly I think this result might just be spurious signals due to instrumental problems and/or data reduction issues. I've seen similar modulations before reducing both high-resolution, mid-resolution and even low-resolution spectra (the process of going from the image to the final spectra easily introduces things like this); I wouldn't be surprised if this is the case too.


In the calculation of GMST, what are these constants? GMST = 6.697374558 + 0.06570982441908 D0 + 1.00273790935 H + 0.000026 T2

From the GMST calculation on the USNO website (and many other places) we encounter these constants:

6.697374558
0.06570982441908
1.00273790935

Can someone enlighten me as to what they represent?

As the comments have said, the second and third numbers are the ratio of sidereal to solar days but it is split into two parts from the original formula which gives GMST at 0h UT1 by including the hours part directly.

The first number comes from the definition of sidereal time as "in general terms, sidereal time is the hour angle of the vernal equinox" (Explanatory Supplement to the Astronomical Almanac, 1992, p48 (referred to on p77 of the 3rd Edition of the Explanatory Supplement)). This is bound up with the original definition of the right ascension of the fictitious mean sun which moves around the celestial equator at a fixed rate which was due to Simon Newcomb in 1895 (Astronomical Papers Prepared For The Use Of The American Ephemeris And Nautical Almanac Volume 6 (1898). This was based on noon and shifting it by 12 hours to the definition used now at converts the 18.64606 to 6.646065. Differences in the measured values of the precession constants and changes in the star catalog reference systems from FK4 to FK5 have resulted in changes in the values from the original.

Aoki et al. 1982 discuss this in more detail and added a $T^3$ term (their equation 13, although note the values there are in seconds not hours as in the original USNO version above) for what became the IAU 1984 system. Capitaine et al 2000 added a $T^4$ term along with another 32 more terms to the so-called equation of the equinoxes which is added onto the formula for GMST to get GAST (Apparent Sidereal Time). The equation of the equinoxes takes account of the accumulated precession and nutation as the fictitious mean sun is not a very good reflection of reality for precise work. Because of these increased complexities and coupling between the Earth's rotation and the background reference frame of the stars needed for precise work, the IAU did away with sidereal time in 2006 and replaced it with Earth Rotation Angle (ERA) which is a simple linear function of UT1 time.


Viewing 2017 Solar Eclipse - How important is it to be in the center(blue line)? I'm viewing the 2017 Solar Eclipse in Graham County, NC.  How important is it to be in the center(blue line)?

It is almost entirely national forest there so it would be fairly easy to get to assuming I can park somewhere.

http://xjubier.free.fr/en/site_pages/solar_eclipses/TSE_2017_GoogleMapFull.html

If you want to see a total solar eclipse, you just need to be between the two red lines on that map.  As long as you're within there, what we call the "path of totality", you will be able to see a total solar eclipse.

If you are outside of the red lines, you will only see a partial solar eclipse.  This map shows what percentage of the sun will be covered the further north or south you are of the line.

Graham County is entirely within the path of totality, so you should see a total solar eclipse.  If you're planning to travel, however, it might be too late.  Most campgrounds and hotels within the path are already full.  If you're already within the path of totality, my recommendation would be to stay put.  Travelling on that day, or even that week, will be difficult.

The center of the path of totality will offer the longest viewing for the eclipse.  The further from the center, the shorter it will last.  The aforementioned map shows timings at different distances from the center.  Notice that it also helps to be close longitudinally as well as latitudinally (that may not be a word).


Is there a way to tell what the surface of a planet is like? Kepler-442b

I'm doing a project in which I need to find a planet within our galaxy that might be habitable. I found this planet that is within its stellar system's habitable zone, and due to research I have found that this planet is one of the closest in similarity to Earth, in terms of size, and temperature. But I don't really know anything about the surface or physical features.

I think there is an equation to find if there is life on another planet, but I cant find anything about water...

If you can help me, thank you so much.

I need to know the physical and chemical features on a planet's surface. I know the planet is within the habitable zone and is thought to contain water, but I need more detail, if there is any way to get the specific or at least highly probable features of the surface. Sorry I didn't make this clear enough.

The answer to your question for that specific planet is No.

There is no way at this point in time to tell if Kepler-442b is a big barren rock or a water world or like earth for that matter. They only know it is there because it passes in front of the star it is orbiting and that dims the light. That is the only reason they know it is there.

This planet is not visible.


Is there any way for a planet orbiting a red dwarf in the habitable zone to not be tidally locked? Is there any way to avoid the tidal locking of a planet orbiting a red dwarf in the habitable zone?

For example, could a planet with a 90° obliquity and large moon avoid such a situation?

Leconte et al. (2015) suggested that the presence of an atmosphere could prevent or at least slow tidal locking. The star should exert two separate torques: one on the atmosphere and one on the solid body of the planet:
$$T_a=-\frac{3}{2}K_ab_a(2\omega-2n),\quad T_g=-\frac{3}{2}K_gb_g(2\omega-2n)$$
where
$$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\equiv\frac{GM_*R_p^5}{a^6}$$
for stellar mass $M_*$, planetary radius $R_p$, mean density $\rho$, semi-major axis $a$, mean motion $n$, rotation rate $\omega$, and response to torques $b_a$ and $b_g$. The two torques could be equal, and assuming that the atmosphere transfers some angular momentum to the surface of the planet, this could prevent tidal locking. There are several equilibria at which this could occur:


Telescope buying guide for a beginner in India I am located in the southern part of India and am looking forward to buy a telescope to gaze up to look ay farther planets, moons in our solar system and take pictures. How should I go about getting a telescope as I know nothing about telescope and things I need to know?

If you are a beginning astronomer, there is not much point worrying about all the bells and whistles you can get with a good telescope.

Instead, the key points to look for are:

light-gathering diameter - this gives an indication of how dim an object you will be able to see
supports/stand/mount - a basic tripod will be fine if you want to see no further than Jupiter, but it will wobble too much for longer distance viewing, so a solid mount will be essential
motor/computer drive - at high magnification, objects will pass the eyepiece very quickly, so you will want to look at tracking drives that can follow objects smoothly

But to be honest, you could start with a $100 telescope and really enjoy yourself learning the sky.


How strong are the Lagrange Points of Earth? How strongly are objects bound to the Lagrange points of Earth that they inhabit in them in the same way Jupiter's Trojan meteors are entrapped in its Lagrange points?

In the co-moving frame with the two main objects, how deep is the potential energy well of L4/L5?

Because The triangular points (L4 and L5) are stable equilibria, provided that the ratio of M1/M2, moon 0.166 g and earth 9.8 g  so 0.0169 g in L4 and L5


Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole? I've heard that the actual density of a black hole (within the volume of the event horizon) is pretty low - does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

Forked from "Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this?"

... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

I'm not completion sure what distinction you're drawing between concentration and density, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.

For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is Birkhoff's theorem: outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's shell theorem. Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.

So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be careful about how you define 'volume' when comparing overall densities.

What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what will fall into the black hole in the future.

This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.

Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the no-hair theorem. In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the singularity theorems guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.


What has GAIA found out about the dwarf galaxy heritage of stars in the Milky Way, if anything? I ran across the image below in the August 2000 ESA Bulletin # 103 GAIA – Unravelling the Origin and
Evolution of Our Galaxy

  Figure 3. This simulation shows how a galaxy halo, like our own, may have been
  built up by the accretion of 50 dwarf galaxies colliding with the galaxy at various
  times during the last 10 billion years. The unit of distance is the kpc, or kiloparsec, where 1 kpc is a little more than 3000 lightyears. GAIA would be able
  to detect the fossil streams of these ancient merging events (Courtesy of Paul Harding)

Twenty years later, have results from GAIA been used to generate any data of this nature, identifying groups of stars that are likely to have come from nearby dwarf galaxies merging with our own?

Malhan et al. (2018) summarise progress so far (the Gaia astrometry is improving all the time as more data are collected) using Gaia DR2 data. In particular they produce "panoramic maps" of streams in the Galaxy halo, which I believe is exactly what you are looking for (see below).

Coherent tangential velocity structures, a.k.a. "streams", seen in the southern Galactic hemisphere by Gaia DR2 - from Malhan et al. (2018).


Orbital Elements Transformations I'm a third year undergrad, and for the project I'm doing this summer, I need to figure out the Earth's position and velocity from the moon's point of view. From the reading I've done, it seems like there are formulae to convert from the orbital elements of the earth-moon system to a state vector describing the moon's position and velocity in the earth's reference frame. So, I'm wondering if because of the symmetry of the problem (i.e. from the moon's POV, the earth orbits it in an ellipse), there is a way to convert the easy-to-find orbital elements of the earth-moon system to the orbital elements of the earth as seen from the moon? I could then just do the conversion to the state vectors that I'm looking for.

Any sources/help is much appreciated!

You can get moon centered position and velocity vectors at the Horizons Ephemeris Generator.

Here's a screen capture:

Code for moon is 301, Code for earth is 399


What is the Galactic Latitude and Galactic Longitude? How is different from the latitude and longitude we use on Earth? It has to something with the positioning of sun and galactic center of the milky way but I wasn't able to understand how for a given star, its co-ordinates are calculated.

Some explanation would be appreciated.

The Galactic Coordinate system is a longitude-latitude coordinate system that is used to define the positions of objects in space, most commonly objects within our own galaxy. It uses the center of our galaxy as the focal point (i.e., where we consider $(0^{\circ},\:0^{\circ})$), much like we use the position just off Africa as the focal point of the Earth's lon-lat system. You can see a visual representation of this system below.

The reference point is of course from our solar system. Longitude is measured in the plane of the galaxy, going positive as you sweep to the left (thus following a right-hand rule). The symbol $\ell$ is generally used for galactic longitude. Latitude is measured above or below the plane, going positive "above" the plane and negative "below". We define above and below based on the right hand rule as applied to our galaxy's rotation. The symbol $b$ is generally used for galactic latitude.

If someone says an object is at galactic coordinate position $(\ell,b) = (180^{\circ},\:+45^{\circ})$, then the object will be located directly opposite the galactic center and above the plane of the galaxy.

Aside from being a lon-lat system, just as the Earth's is, it does not share any features with Earth's lon-lat system. That is, there's generally no relation between the two and the definition of one is not affected by the other.


Can Jupiter's nightside be classified as spectral type Y? As far as I am aware, the latest spectral types that have been assigned are around Y2, for objects like WISE 0855-0714 that have temperatures around 250 K or so. I've also seen several directly-imaged exoplanets have had their spectra classified as L- or T-type (e.g. Bonnefoy et al. 2016 who find that the giant planets HR 8799 d and e are good matches for L6–L8 dwarfs).
Jupiter is colder and less massive than the observed Y dwarfs and differs from isolated brown dwarfs because it is being illuminated by the Sun, but there is thermal emission which could be observed from the nightside of the planet. Has Jupiter's nightside spectrum (i.e. the spectrum of Jupiter excluding contributions from reflected sunlight) been observed in sufficient detail to compare it to the spectra of Y dwarfs, and if so does it fit the trends observed in spectral type Y or is it so different that it cannot be spectrally classified as a Y dwarf?
For clarity: I am asking purely in terms of spectral classification, i.e. the classification of the spectrum. I am not asking about whether Jupiter is a brown dwarf making the Sun+Jupiter a binary system. That has been asked elsewhere.

This Wikipedia page lists Jupiter as a disputed sub-brown dwarf because its mass reportedly is sufficient to qualify for one. Distance to another star like the Sun doesn't matter, nor does the visibility in the infrared spectrum either since every planet is emitting some infrared light. The fact that Jupiter emanates more energy than it gets from the Sun can't be accounted for either, since that's also a matter of distance from the Sun, and Saturn and Neptune do so too. But, as the article states, Jupiter obviously has enough mass to possibly count as a sub-brown dwarf. Sub-brown dwarfs are still planets because they don't undergo nuclear fusion unlike brown dwarfs and hydrogen-fusing stars and never did.
Edit: There is no reason to only classify the night side just because the day side is illuminated by a(nother) star. Spectral classifications are given to the entire object. Since sub-brown dwarfs reportedly have a mass of at least one Jupiter mass, you may want to classify Jupiter a Y9V sub-brown dwarf.


Sky from Equator at 8,000 years Before Present? The Stellarium software allows future views of the sky. I need an online resource (planetarium projector?) that shows pre-historical views, and from the Equator at 8,000 years before the present.

You can equally well just type in a negative year and Stellarium will navigate you into the corresponding year B.C. Don't forget that the year 0 didn't exist. 1 A.D. follows directly after 1 B.C.

A while ago I've confirmed that Stellarium is very accurate down to 1500 B.C. using a babylonian list of lunar occultations of stars. Thus, it should be reasonably accurate for the year 6000 B.C. as well.