Does a lunar occultation of Mars happen twice a year? The Space Exploration SE question When our Moon is hiding Mars, is it possible to create a radio communication between Earth and Mars? links to this short article in the popular press Mars is hiding behind the Moon this morning with the sub-headline

It’s a twice-a-year occultation; like the eclipse, but it’s Mars, and there’s no shadow

As answer(s) to the question mention, the Moon is smaller than Earth so an occultation would not be visible everywhere on Earth, so let's just consider it an occultation if it happens anywhere on Earth.
Question: Does a lunar occultation of Mars happen twice a year? Is the Moon's diameter big enough and it's orbit's inclination with respect to the ecliptic small enough that it can pass between Mars and some place on Earth twice a year?

Does a lunar occultation of Mars happen twice a year?

No, in the sense that the number of lunar occultations of Mars per year is quite variable and not always two.
I do not have a reference handy, but I am sure that there are some years with no occultations.
And we can see that in other years the number can be both less than and much greater than two:

In 2019, there was one lunar occultation of Mars according to the Astronomical Almanac Online 2019 Lunar Occultations.
In 2020, there were five! See 2020 Lunar Occultations.


Why are the wings of many strong lines, including hydrogen lines and most metal lines, stronger in Dwarfs than in Supergiants? What is the reason? Is it because Dwarfs have higher surface gravity (log g), in turn related to pressure, than Supergiants so there is more contribution from van der Waals and Stark broadening?

Collisional broadening - which includes van der Waals and Stark broadening - is more important in the higher gravity, higher pressure/density atmospheres of dwarf stars (a factor of 100-1000 higher for dwarfs vs giants of the same photospheric temperature).
These collisional effects effectively "truncate" radiative emission and absorption processes, leading to broadening of the frequency response of a line absorption cross-section. That means strong lines have appreciable opacity away from the line centres and means the lines sample a broader range of temperatures in the photosphere.
See https://physics.stackexchange.com/a/257471/43351


Where is the sun relative to my lunar rover? I'm a robotics researcher writing orienteering code for a solar-powered lunar rover.
I have two similar problems:

Given the position of a robot on the moon in lunar latitude/longitude, where is the sun? That is, which direction should I orient my solar panel to maximize incidence?
For testing purposes, I need to solve the same problem on Earth. Given Earth lat/lon, where is the sun?

To be honest, I'm not even sure what coordinate system to use to express the answer to these questions. I was considering a coordinate frame centered on the rover with East, North, and Up axes, but I'm open to a better suggestion!

So far, I've found some information about VSOP87 and ELP2000, and I've started reading "Astronomical Algorithms", but I'd appreciate help from the experts.

Edit

Here's what I've got so far. I can get the coordinates of the Earth-Moon barycenter using VSOP2013. If I can get from the barycenter to the center of the Earth, I can probably use EPS2000/82 to find the position of the Moon in heliocentric coordinates.

Sun->VSOP->Earth-Moon Barycenter->Earth->EPS2000->Moon

At very least, I think this would get me the moon coordinates in the sun frame. I'm hoping it would also give me the orientation of the Moon in the sun frame?

If you like Python then see Skyfield; Observing from a Moon location which I believe will do exactly what you need! Also see for example this answer to Ephemeris for lunar body-centered body-fixed coordinates? written by Skyfield's author.

For example, you posted your question at 2020-04-20 15:21:36 UTC. Let's say your rover has just driven up to Surveyor III at 3° 0′ 58.03″ S, 23° 25′ 4.48″ W  (-3.01612°, -23.41791°) (read more about Surveyor III in How did Apollo-12 manage to land next to Surveyor-3? First “Space-Tourists”?)

At that moment the Sun was below the horizon. The output of the script below is

RA:  01h 54m 39.28s
Dec:  +11deg 45' 38.6"
alt:  -38deg 30' 34.6" degrees (below the horizon)
az:  265deg 50' 21.5" degrees around the horizon from north

  I can get the coordinates of the Earth-Moon barycenter using VSOP2013

factoids: VSOP is what PyEphem uses under the hood. It is currently in maintenance mode maintained by the same person who wrote Skyfield.

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# adapted from https://rhodesmill.org/skyfield/planetary.html

from skyfield.api import PlanetaryConstants, Loader

load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
ts = load.timescale(builtin=True)
t = ts.utc(2020, 4, 20, 15, 21, 36)

eph = load('de421.bsp')
earth, moon, sun = [eph[thing] for thing in ('earth', 'moon', 'sun')]

pc = PlanetaryConstants()
pc.read_text(load('moon_080317.tf'))
pc.read_text(load('pck00008.tpc'))
pc.read_binary(load('moon_pa_de421_1900-2050.bpc'))

frame = pc.build_frame_named('MOON_ME_DE421')
Surveyor_III = moon + pc.build_latlon_degrees(frame, latitude_degrees = -3.01612,
                                              longitude_degrees = -23.41791)

apparent = Surveyor_III.at(t).observe(sun).apparent()
ra, dec, distance = apparent.radec()
print('RA: ', ra)
print('Dec: ', dec)

alt, az, distance = apparent.altaz()
if alt.degrees >= 0:
    print('alt: ', alt, 'degrees (above the horizon)')
else:
    print('alt: ', alt, 'degrees (below the horizon)')
print('az: ', az, 'degrees around the horizon from north')


What was the length of year 1 million years back? We know that the universe is gradually expanding and this indirectly means that the gravitational force between sun, earth, planets and other stars (roughly anything in the universe) is gradually decreasing as gravitational force is indirectly proportional to square of distance between the objects.

So I think this also effects to length of the year. If yes then is it possible to know how days does 1 year had 1 million year back?

The Hubble expansion has no bearing whatsoever on the length of the year. This is because the whole Milky Way galaxy (and in fact most galaxies, if not all, and even local groups) has decoupled from the Hubble flow long ago. In fact, it could only form after it decoupled. Note that M31, our sister galaxy, is in fact falling onto the Milky Way rather than receding (as the Hubble flow would imply), demonstrating that the whole of the Local Group (of galaxies) is decoupled from the Hubble flow.

What happens is that any over-density expands at less than the Hubble rate and thereby grows. Galaxies (and larger structures) form from small relative over-densities that eventually grow large enough to withstand the overall expansion and instead collapse under their own gravity to form bound objects, such as galaxy clusters, galaxies, star clusters, and stars. This implies that the Hubble flow has no bearing on the inner dynamics of such systems.

Of course, the number of days in a year was higher in the past than today, but that is only because the Earth is spinning down (due to tidal friction with the Moon), so that the days become longer.

If anything has had an effect on the semi-major axis of the Earth orbit (and hence on its period), then that is gravitational interactions with the other planets. However, weak interactions (secular perturbations) can only alter the orbital eccentricity and leave the semi-major axis unaltered.

Finally, there is a tiny effect from the Sun loosing mass (to the Solar wind). The period of any orbiting bodies is proportional to $M_\odot^{-1/2}$.


Will Gaia detect inactive neutron stars? Will the astrometric precision of the Gaia space telescope be able to detect the gravitational influence of cold old solitary neutron stars on the movements of stars? At least in a statistical sense to estimate how common they are around here, which could say something about how large the population III stars were.

And if an inactive neutron star could be precisely located thanks to its gravitational influence, would it be possible to observe it directly?

I will throw out a suggestion that could hardly be considered an "answer", but it would be interesting if someone could put flesh on the bones of it.

When I first read your question I thought you were asking whether gravitational perturbations from an unseen neutron star would cause a measurable effect on the positions of other stars. I think the answer to this is certainly no unless the neutron star was so close to another star that they were (or could be practically considered as) a binary system.

Binary systems featuring neutron stars could definitely exist (they won't be common though) and Gaia should see them if they are close enough and the companion star is bright enough. A variant of this would be a neutron star that just happens to drift (probably at high speed) close to another star, producing a notable change in the photocentre of the visible star over the 5-year Gaia mission. This is unlikely because close encounters between unrelated stars are very unlikely.

But then I started thinking about lensing. Even the presence of planets in our solar system causes a gravitational lensing signature - the positions of stars are distorted by $\sim 1-10$ microarcseconds close to the positions of planets and this is detectable by Gaia. The distortion can be spotted because of course the planets move with respect to the background stars.

In GR, the lensing angular deflection is
$$\alpha = \frac{4GM}{c^2r},$$
where $r$ is the closest approach of the light to the lensing object of mass $M$. For a lensing object at distance $d$ and a background star at an angular separation of $\theta$ from it, then
$$\alpha = 0.04 \left(\frac{M}{M_{\odot}}\right)\left(\frac{d}{\rm 1\ pc}\right)^{-1} \left(\frac{\theta}{\rm 1\ arcsec}\right)^{-1}\ \ \mu{\rm \ arcsec}$$

Neutron stars are of order 1-2$M_{\odot}$; the closest they might come is a few pc. The astrometric precision of Gaia will be about 20 microarcsec for a star with $V =15$, falling to about a milli-arcsec at $V\sim 20$.
The signature of a neutron star moving with high proper motion might be that it passes very close in the foreground to a bright star, causing its position to wobble backwards and then forwards by an amplitude given by the above formulae.

To get a 20 microarcsecond signature that might be detectable, a neutron star at one pc would have to pass within 0.003 arcseconds of a $V=15$ star.  The surface density of such stars varies according to Galactic latitude but is around 1000 per square degree for stars of $V\leq 15$ (Bahcall & Soneira 1980) at intermediate Galactic latitudes. If the neutron star has a proper motion like Barnard's star of around 10 arcseconds/year, then it traverses 50 arcseconds during the Gaia mission. If background stars are evenly spaced, the chance of coming within 0.003 arcseconds of one is about $10^{-5}$ (if I've done my sums right).

So, almost no chance of seeing this.


Locate asteroid in asteroid belt I was interested in tracking the orbit of 1986 DA. I couldn't find any resource when I looked. Just for fun, so any help appreciated.

It has been observed 1977, and in 1994.
It should be near its perihelion this year (2014) around august 7 (using wikipedia data, perihelion at March 6, 2005 and period 4.71 years), give or take a few days.


Cosmological redshift and comparing past galaxy sizes If you were to divide the present-day
universe into cubes with sides 10 million light-years long, each
cube would contain, on average, about one galaxy similar in size
to the Milky Way. Now suppose you travel back in time, to an era
when the average distance between galaxies is one quarter of its
current value, corresponding to a cosmological redshift of z = 3.
How many galaxies similar in size to the Milky Way would you
expect to find, on average, in cubes of that same size? In order to
simplify the problem, assume that the total number of galaxies of
each type has not changed between then and now. Based on your
answer, would you expect collisions to be much more frequent at
that time or only moderately more frequent?

I am very confused with this question. From my intuition, I take that volume would equal to distance^3. They have given us that the past distance would be 1/4 of current. Therefore, do I cube 1/4distance? Also how does the redshift of 3 play a part in the solution. Thanks for any help! I do not need an answer, just how to start.

Comoving separations go with the scale factor. Comoving volumes scale as $(1+z)^{-3}$.

At $z=3$, that volume was 64 times smaller.


Can a high enough density ignite a gas giant into a brown dwarf? I wonder whether the nuclear fusion of brown dwarfs and more massive stars really depends on their mass only or if it could also start nuclear fusion if it is dense enough but not as massive as brown dwarfs usually are (they need more than 12 Jupiter masses to ignite deuterium). It is like with a nuclear fusion bomb. Since hydrogen bombs explode because of their density and not only their mass, can this also happen to stars? Or did I miss something? Or is a density to ignite deuterium at lower masses simply impossible? If I'm right, what would be the critical density to ignite deuterium at 1 Jupiter mass?

Or did I miss something about hydrogen bombs?

You missed something about hydrogen bombs.
The center of our Sun is the equivalent of a warm compost pile in terms of energy produced per unit volume. A multistage thermonuclear weapon briefly (very briefly) compresses and heats the fusible material in the bomb to conditions far beyond those found in our Sun, let alone Jupiter.


Is it possible for a star to return in the main sequence? Is it possible for a star to return to the main sequence after entered in the post-main sequence phase? I have no clue if there's a mechanism such as acquiring hydrogen from from the environment or from a substellar body might bring again the star into the main sequence. Was it ever explored as a possibility?

The evolution of a star is irreversible process, so strictly speaking, it is impossible to return the star back to its earlier phase (it is impossible to collect all energy it has radiated to space and use it for conversion of helium back to hydrogen).
But it is possible for a star plotted on HR diagram to return to a point corresponding to main sequence phase. These stars are called blue stragglers. They are usually formed when stars merge.


"Early" and "late" nomenclature regarding O, B & K, M stars The Wikipedia article Stellar Classification has a subsection "Early" and "late" nomenclature.
It says:

Stars are often referred to as early or late types. "Early" is a
synonym for hotter, while "late" is a synonym for cooler.

My lecture notes say:

O and B stars are sometimes referred to as ‘early-type’, while K and M
are ‘late-type’.

If I look at a HR diagram, I see that main sequence O, B stars are hotter than main sequence K, M stars. So they are burning their fuel at a greater rate than K, M stars, so they run out of it early. So the stars which we are still able to observe now from the early universe are K, M stars, rather than O, B, which presumably formed rather *late**ly. Yet this is not consistent with the terminology above, so something is obviously wrong here.
What is wrong with my argument that the early and late words should be used in the other way round?

What's wrong with your argument is that you are using modern theories of stellar evolution. The terminology was introduced in the context of the theory that stars are powered by the Kelvin–Helmholtz mechanism (heat generated via contraction due to gravity). In this model, stars would start out hot and gradually cool down as they evolved, hence hot stars are called "early-type" and cool stars are called "late-type".
This theory was shown to be inadequate because it can only power the Sun for tens of millions of years, while geological evidence indicated that the Earth had been around for far longer than this. The resolution was the discovery of nuclear fusion. However by this point the terminology of "early-type" and "late-type" stars had become established and it survives to this day.
(Incidentally the terms "early-type" and "late-type" are also used for galaxies, where once again they do not correspond to modern theories of how these objects actually evolve.)


What does correcting for LSR mean in galactic velocities? This is a two-part question, but both parts are related:

I'm aware that modern astronomical software can correct velocities for the local standard of rest (LSR), but what does that really mean? As in, if you have measured a radial velocity by hand (e.g., from a plot of a spectral line) and you want to correct for LSR, how is this done?

As an exercise, I want to compare the galactic proper motions (which I presume are another way of saying velocities?) of a group of stars within 2kpc from the Sun between a method described in Eqs.30-32 of Mignard 2000 (Hipparcos data), and Eqs. 1-2 of Bovy 2017 (Gaia data). Apparently I also have to correct these velocities for LSR, but how is that done? Do you add a specific value? I thought these equations already include the components of the Solar motion w.r.t. to LSR, unless the constants included in the equations have nothing to do with that. I'm probably missing a big point here and books haven't helped me at all.

Thank you in advance.

Proper motions are not velocities. They are angle traversed per unit time. They can be converted to tangential velocities (i.e. tangential to a line joining the star and the Sun) if the distance to the star is known, but you need additional information to get the line of sight (radial) velocity and hence a 3D velocity.

Such velocities are with respect to the Sun (or possibly the solar system barycentre - a tiny difference).

The local standard of rest (LSR) is an estimate of the average motions of stars in the solar neighborhood. The Sun has a peculiar velocity with respect to the LSR. To get a velocity with respect to the LSR, you need to subtract the solar motion with respect to the LSR from your heliocentric velocities.

There are disagreements in the literature (at the level of 1 km/s) as to what the solar motion with respect to the LSR is.


Is liquid water on Mars such a big leap compared with ice? I thought we have found bacteria or other kinds of life on Mars.

Now it is liquid water.  Is there such a big difference between liquid water and iced water?

We have not found life, bacterial or otherwise, anywhere other than Earth ... yet. I think you mean the difference between liquid water and water ice. This difference relates to the hospitality to life. Life of the type we are used to needs liquid water, even if this is just trapped moisture, unless the life is temporarily in suspended animation. If there was no liquid water on Mars, there would seem to be no possibility of ongoing life there.

It is important to distinguish hospitable conditions from evidence for life. While having a hospitable environment is necessary for life, it is far from proof of life. The big hurdle is starting life, or bringing it from somewhere else.

There are a variety of hospitable environments in the solar system and on exoplanets, such that if certain life from Earth (such as extremophile bacteria) were somehow placed there they could live. But that does not mean there is in fact life there.

Another important point, Mars was way more hospitable early in it's history, with bodies of liquid water. Which is important because even though starting life on Mars under current conditions is unimaginable, early Mars was rather Earthlike and could possibly start, or receive, life.


Why the Sun-Jupiter barycentre is outside Sun? The Earth-Moon barycentre is said to be inside Earth. But Moon-Earth mass ratio (1.2%) is larger than Jupiter-Sun mass ratio (0.09%). Then how come the Jupiter-Sun barycentre is outside the surface of Sun?

There are three factors that all play into this:

The Jupiter/Sun mass ratio
The Jupiter/Sun distance
The Sun's radius

The barycenter of any pair of orbiting masses lies on the line connecting their centers of mass, and its position depends on the masses of the two objects.

If two objects have the same mass, their barycenter will lie halfway between them.  If one is twice as massive as the other, the center of mass will be 1/3 of the way from the larger mass to the smaller.  The formula is:

DSb=DSP*MP/(MS+MP)

MS and MP are the masses of the Sun and planet, respectively,
DSP is the distance between their centers of mass, and
DSb is the distance from the center of mass of the Sun to the barycenter.

Finally, whether or not the barycenter is inside the Sun depends on the radius of the Sun. How far it is from the center of mass of the Sun is unaffected by the Sun's radius.

For the Earth-Sun pair, MP/(MS+MP) is 3.0×10-6

For the Jupiter-Sun pair, MP/(MS+MP) is 9.5×10-4

So the Earth-Sun barycenter is about 3.0×10-6 of the way from the Center of the Sun to the Earth, or about 279 miles.  The Jupiter-Sun barycenter is about 9.5×10-4 (a number 300 times bigger) times Jupiter's distance (a number 5.2 times bigger) and comes out at 459,000 miles, which is just slightly larger than the Sun's radius of 430,000 miles.

It's outside the Sun because Jupiter is more massive than the Earth and further away from the Sun.  It's just how the numbers work out -- nothing deeper than that.


How does gravity affect the wavelength of light? If, hypothetically, me and my rocket powered flashlight were falling straight toward the center of a black hole. The flashlight is a few kilometers behind me in our travels toward the center of the black hole, but since it is rocket powered, it manages to maintain the exact distance to me for a while.

The point is; The distance between me and my flashlight is constant as long as I am observing it.

The photons coming from the flashlight would obviously not be rocket powered - and they would be affected by the black holes gravitation.

Would the light I see from the flashlight be shifted towards red or blue, even though the distance between me and my dear flashlight is maintained?

If so; switching the positions of me and my flashlight, would it change the color I'd observe?

If we turn off the rocket on the flashlight, I assume it would be redshifted regardless of which were closer to the singularity, and the magnitude of redshift would appear to accelerate?

A partial answer: The lower the gravitational potential the slower the clocks, see Gravitational time dilation. This means, as long as your flashlight is behind you in constant distance, it appears blue-shifted to you. With roles exchanged (light below you in constant distance) it appears red-shifted.
By the heterogeneity of the gravitational field the shift grows with the time, while falling in constant distance.

This is called the Einstein shift, which is not the same as the doppler effect. The doppler effect is caused by an objects velocity relative to the observer.


M87 Black hole. Why can we see the blackness? So, as the title states, why are we able to actually see the 'blackness' of the black hole? I get that what we are actually seeing is the event horizon, or accretion disk. But should this not extend all the way round? Surely the black hole is not a 2D thing so we can "look into it from above" (I use that term loosely as obviously there are no directions in space!), so why are we able to actually see the blackness?

The only thing I can think of is that a black hole, much like our solar system has some sort of ecliptic, that the vast majority of matter is orbiting, and everywhere else just doesn't have enough matter for the light to be visible, kind of like why we are unable to see the Oort cloud.

I hope that made sense, and I could be way off, but it is the only thing I could think of to explain it. If it is the case, then would we be able to get a similar image of Sagittarius A, seeing as we could be in this "ecliptic" of it, so surely we would only be able to see the heated matter around the event horizon and not the blackness?

Rob Jeffries' answer is excellent, I just wanted to add this picture trying to explain the geometry. Here, I assume a non-rotating black hole (BH); for a rotating BH the exact numbers are slightly different.

The photon sphere

Photons move on straight lines, but in the heavily curved space around a BH these straight lines appear curved. Although the event horizon (EH) at a distance of $r = 2GM/c^2 \equiv r_\mathrm{S}$ (the Schwarzschild radius) from the BH marks the region from which no photons may escape if emitted radially, photons on partially tangential orbit will fall back out to a distance of $r = 1.5r_\mathrm{S}$, where photons traveling fully tangentially will stay on the photon sphere (although this is an unstable orbit).

The innermost stable orbit and the accretion disk

Ordinary matter will spiral inward out to twice this distance; hence, inside the innermost stable circular orbit (ISCO) at $r=3r_\mathrm{S}$, matter is pretty much bound to be absorbed. Outside this region matter may orbit, forming the accretion disk, but since friction between the particles will cause them to lose energy, they will slowly approach the ISCO, after which they will rapidly fall into the BH. Note that the M87 BH doesn't have a thin accretion disk like the one depicted in the movie Interstellar; rather a thick "cloud" surrounding most of the BH.

Photons emitted tangentially just outside the photon sphere will spiral around the BH many times, slowly increasing their distance, until eventually they escape at a projected distance of $\sqrt{27/4}r_\mathrm{S} \simeq 2.6r_\mathrm{S}$ from the BH (e.g. Frolov & Novikov 1998).

The shadow

Just as the path of light rays are curved around the BH, so are the sightlines from you toward the BH (you can think of sightlines as reversed photons). That means that all sightlines that are closer than (a projected distance of) $2.6r_\mathrm{S}$ to the BH will, eventually, end up on the EH, even if taking several orbits around the BH. These sightlines comprise the so-called shadow (Falcke et al.(2000); Event Horizon Telescope Collaboration et al.(2019a), ). On the other hand, along sightlines that a farther away, you see the radiation emitted from the matter falling into the BH, both in front of and behind the BH. And since the first sightlines that don't terminate at the EH circle the photon sphere many times, those sightlines are actually very long paths through matter shining its last light before being engulfed, and hence they look exceptionally bright (e.g. Event Horizon Telescope Collaboration et al.(2019b)). This bright ring just outside the shadow is called the photon ring, or the emission ring.

The drawing

The drawing below may help understand. All the red lines are sightlines toward the BH. Only the uppermost one just grazes the photon sphere (and the luminous matter behind). The rest terminate at the EH, and hence look black (except for luminous matter in front). Close to the center, you see the front of the EH; farther out you actually see the back of the EH; even farther out you again see the front of the EH, and so on ad infinitum until you reach the photon ring.

The observation

Despite the observational resolution being an astonishing $\sim25$ micro-arcseconds, the photon ring is smeared out over a larger region, resulting in the doughnut shape you've seen. That is, what you see in that image is not "the EH in front of an accretion disk", but rather "the EH seen from all sides at the same time and enlarged, with light emitted from the photons ring".

Unless the accretion disk is viewed exactly face-on, half of the accretion disk$^\dagger$ has a velocity component toward you, making it brighter than the other half through a special relativistic effect called beaming. This is seen in the southern part of the M87 BH.

The figure below (from Event Horizon Telescope Collaboration et al.(2019b)) shows, from left to right, the actual observation, a model where you see the rather sharp photon ring, and this model blurred to match the resolution of the observation.

$^\dagger$At least the material just before it plunges into the BH, which follows the rotation of the BH. Farther out, the rest of the accretion disk may in principle rotate the other way.


Is Jupiter just a super earth with hydrogen atmosphere? Often I see books which state that Jupiter has a rocky core, if it is true, does it originate from super earth with excessive collection of hydrogen?

Jupiter probably originated with a dense core, and then accreted a gaseous envelope. To call the core "rocky" could be misnomer, since the planet formed much further out from the Sun where it may have been possible to form ices. The exact location of Jupiter's birth and the composition of its core are not known, but the composition was unlikely to have been the same as Earth. It is also probable (again, theory not fact), that Jupiter's core may have reached multiple Earth masses before accreting the gaseous envelope.


How to explain this trajectory of the moon taken from the same spot on Earth over 28 days? Found this picture online, was wondering why the trajectory has taken such a beautiful shape ?

This is a digitally enhanced composite set of images of the Moon, with successive images taken 24 hours and 40 minutes apart. There's also a long gap in time between the left half and right half of the composite. The creator of the photograph took artistic license to enlarge the Moon; she thought that that made the composite image look better. Images of the new Moon / nearly new Moon are also enhanced.
Had consecutive images been taken 24 hours and 51 minutes apart the resulting composite would form a more or less closed curve. Successive images were instead taken 24 hours and 40 minutes apart so as to intentionally create a non-closed curve. The first image is at the top right while the last image is at bottom left.


Are there events in universe which we receive first from their neutrino's instead of their photons? Neutrino's are very small though they have mass. Because of that it is probably that the don't have a speed faster than photons.

But is it in all events, like supernova's or else events, where it is possible that we first receive their neutrino's instead of their light?

Perhaps gravity or other interactions could change something in their way to earth?

Yes.  When a core collapse supernova happens, it takes a few hours for the shock wave to reach the surface and "break out" (when we first "see" it). Meanwhile, the neutrinos can escape nearly directly once they reach the top of the core.

Of course, as you wrote, photons travel slightly faster, so if the SN is far enough away, the photons might overtake these neutrinos.  However, I have not done the calculation to see if the universe is big enough.


Centre of the Universe Why shouldn't the original singularity of the big bang happen to be the centre of the Universe? Assume that the universe is expanding isotropically with a constant speed.

(I can't comment)- See https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point
TLDR, The big bang happened everywhere in the universe at the same time, because it was the universe. The top answer goes more in depth.


Venus transiting behind the Earth’s moon December 7, 2015 Observing from Earth what are the odds that Venus does not line up in conjunction with Earth’s moon and does not transit behind the moon but slightly below or above with a complete visual of Venus during the entire conjunction of the two? What are the chances of a miscalculation in this event?

Well, depending on where you live, the probability is either 0 or 1, modulo cloudy weather.
From the picture below (from this site), you see that you'll need to be in the US to see the occultation. However, only in the regions outlined in cyan (West Alaska, East Siberia, East Canada, and in Caribbean islands east of the Dominican Republic) will it be during darkness (just before sunrise / after sunset).
But with binoculars (or without, if you have very sharp eyes), you should be able to see it over all of the US.

(cyan=occultation at moonrise/moonset; red dotted=daytime occultation; blue=twilight occultation; white=nighttime occultation)

Miscalculation
After your edit, I see that you are not referring to the probability of seeing the occultation, but the probability of a mis-calculation. I'd say that those odds cannot really be calculated, but it can be said that they are extremely small. No physical theory can ever be proved, but after a sufficient amount of verifications, we usually accept a theory as "true for all practical purposes, until disproved". Many, many factors go into a calculation like this (all the way down to mathematical axioms).
But a calculation like this has to do with celestial mechanics, which is very well understood, and which continuously makes accurate predictions and hence is continuously empirically verified. So the odds you request are definitely much, much smaller than, say, the odds of calculating the weather tomorrow, or the mass of a galaxy cluster.
Uncertainties
Of course there are always uncertainties associated with such calculations, which gives an uncertainty in the position of celestial bodies. But if you take a look at this video, you'll see that the predicted path of Venus is more or less through the middle of the Moon. If the actual path should not cross the disk of the Moon, the calculation would be off by $\sim15$ arcmin, the probability of which is virtually zero.


Lunar phases if Earth was tidally locked to the Moon I know that the Earth will never become tidally locked to the Moon, so this is purely hypothetical.

If I understand correctly, in this hypothetical scenario an Earth day is now a lot longer at around 29 current days, and you can only ever see the Moon from half of the Earth.

So standing on the "lucky" side of the Earth you should be able to see all the lunar phases during a single night. The Moon become gradually more illuminated, peak as a full Moon and then darken as the night ends.

Is this correct?

tl;dr: An observer on the "Moon side" would see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. They would also see the Moon during the day (early morning, late afternoon), as we currently can, but they would never see the New Moon or crescent moons at night since these would be only visible during the day.

Background
While the Earth being tidally locked to the Moon would indeed mean that the side of the Earth facing the Moon always sees the Moon, the phases of the Moon result from the angle between the Earth, the Moon and the Sun, and probably wouldn’t change significantly.

The Moon orbits around the Earth-Moon barycentre with an orbital period of 27.3 days and a synodic period (i.e. from Full Moon to Full Moon) of 29.5 days. An observer on the Moon-facing side of the Earth would certainly see the Moon go through its different phases, but it would still take 29.5 days. [For the purposes of this question I'm ignoring the increasing lunar period due to the Moon receding from the Earth.]

As the question notes, one obvious outcome of Earth becoming tidally locked to the Moon is that Earth’s “day” becomes the same as the Moon’s synodic period: 29.5 days. Daylight will last for about 15 days, baking everything other than at the poles. The 360-hour night will be freezing, even in the tropics. Local weather will be utterly different.

Another notable outcome is there will only be really tiny “solar” tides spaced about 15 days apart (due to the Earth’s slow rotation in relation to the Sun). The loss of significant tidal flows would probably have a dramatic negative effect on marine life.

The Earth tidally locked to the Moon is a nightmare scenario!

The core question

  So standing on the "lucky" side of the Earth you should be able to see
  all the lunar phases during a single night. The Moon become gradually
  more illuminated, peak as a full Moon and then darken as the night
  ends.

No: you would only see only half the phases during the fortnight-long night: from waxing half-moon to waning half-moon. You would also see the Moon during the day (early morning, late afternoon), as we currently can, but you would never see the New Moon or crescent moons at night since these would be only visible during the day.

We currently see the New Moon because when it's roughly overhead for some at midday, it's not far above the horizon just before sunrise or just after sunset for others. Since the New Moon phase currently lasts about 3-5 days (depending on how you define "new"), but the Earth rotates in just one 24-hour day, most habitable places on Earth get to see this phase every lunar month. However, in the tidally-locked-Earth scenario, the phase still lasts 3-5 days but the Earth rotates once every 29.5 days. Those viewing the Moon just after sunset will only ever see the Moon in that position above the horizon – it more or less "sits" there, apart from libration and nodal period effects – and they will miss some of the other phases because the Sun will be too bright at, say, the waning half-moon.

EDIT - Two additional things to note:

(1) Earth's phases
While the Earth becoming tidally locked to the Moon is hypothetical, the Moon is already tidally locked to the Earth, which means the above descriptions can be applied in reverse for a real observer on the Moon:

a lunar "day" (sunrise to sunrise) is 29.5 Earth days;
the Earth's position in the lunar sky is relatively constant over the lunar day, within about 15º (due to a slow but noticeable libration);
the Moon's orbit is elliptical, so over much longer periods the Earth would alternately recede and grow larger (a "Super Earth");
unlike the Earth-locked scenario, all of the Earth's phases – from Full Earth at lunar midnight to New Earth at lunar noon - would be visible, because there's no atmosphere and hence no "bright daytime sky" to block out the Earthlight.

Note also that the famous Earthrise experience of the Apollo 8 astronauts would not be available to an observer on the Moon's surface. The Earth "rose" above the limb of the Moon because the astronauts were in orbit. Nonetheless, someone located on the border of the "dark side of the Moon" would see the Earth on the lunar horizon, and libration could result in the Earth alternately setting below the horizon and rising again over a lunar day. For that observer, however, earthrise and earthset occur at roughly the same point on the horizon!

(2) Will the Earth really end up tidally locked to the Moon?
Tidal braking is slowing the Earth's rotation by 23 milliseconds (and increasing the Moon's distance from us by 38 metres) per 1,000 years. At that rate, the Andromeda Galaxy will have collided with the Milky Way long before the Earth's rotation has slowed down enough to approach tidal locking.

In fact, this AstronomySE answer suggests that as the Earth's oceans "are responsible for almost all of the tidal deceleration" (David Hammen's comment), the evaporation of the oceans in a billion years will eliminate the main drag on the Earth's spin and tidal locking may therefore never occur.


How Reliable is Bode's Law? Is Bode's Law just a series of remarkable coincidences,or would it hold good for other solar systems where the star is similar to our sun?

A recent article examines if exoplanetary systems follow the Titus-Bode rule. They find that among the systems with three or more confirmed planets that they studied, the planetary distances follow a logarithmic law roughly 53% of the time. So it is possible that the Titus-Bode rule is more than a simple coincidence.
There is no solid explanation for this, but from Wikipedia:

it is possible that, given a combination of orbital resonance and shortage of degrees of freedom, any stable planetary system has a high probability of satisfying a Titius–Bode-type relationship.


Why doesn't sunrise shift as we revolve around the sun (ignoring seasonal variations - so at the equator) If the Earth rotates at a constant speed why doesn't sunrise change as we orbit the Sun?
For example if viewing the Earth's orbit from above at time N let's assume the Earth is at the 3 'o' clock position and so the point on the Earth's surface at the equator perpendicular to the Sun would be in daylight (~12pm local time) let's call this point X (would be at the Earth's 9 'o' clock position).
Six months later the Earth would be at the 9 'o' clock position in its' orbit and assuming the Earth has completed exactly 182 full rotations that point X will again be at the Earth's 9 'o' clock position but would now be facing away from the sun therefore it would be dark but exactly 182 days have passed so it should be (~12pm local time).
I'm probably overlooking something obvious but hoping someone can explain why this obviously isn't what happens in reality.
Many thanks.

This shift is already included in a "day".
24 hours is already the time from sunrise to sunrise, meaning the Earth has performed a full rotation, plus a little bit more since it has moved along a degree in its orbit. That's solar time, the common usage of a "day".
The time the Earth needs to rotate exactly once,so that a star is in the same position in the sky, is a sideral day. It's slightly shorter at 23 hours 56 minutes and 4 seconds.
You can convert between them based on the fact that there's one less solar day than sideral days in a year:
$$d_{sideral} = \frac{year}{\frac{year}{d_{solar}} + 1}$$
Note that the difference between a solar and a sideral day is far from the only subtility arising from the Earth's orbit.


Is it possible for a planet to have multiple moons in a nearly stationary orbit? I have done a lot of searching online and while I have been able to find out that theoretically a moon could exist in a near-synchronous/stationary orbit, I am wondering if it is possible for multiple moons to exist in this state.

I'm not sure if I've explained appropriately or not, so if this needs to be a little clearer, please let me know.

The purpose of this question is that I am writing a novel and I had this idea that the planet has 9 moons and that the area of land that the story takes place in the moons only travel across the sky over the course of many days.  One moon might take 60 days to get from one horizon to the other, while another might take 15 days, and others would be varying times in between.

In my limited knowledge, I have surmised that a moon closer to the planet would take less time to cross the sky than a moon further away.  Obviously, for this to work, the moons would have to be in a supersynchronous orbit, not fully synchronous.  Would the range of distance for supersynchronous orbits allow multiple moons to exist in staggered distances and still be close enough to remain visible from one spot continuously for multiple days at a time?

Disclaimer:
I am not sure that this is the appropriate place to ask this question, since it is mostly theoretical.  I also want to apologize if my English is confusing.  Finally, weird letters and symbols in math make my head hurt, so I am sorry I cannot follow complicated mathematical formulas, but feel free to post them if you believe it may help answer the question.

You may want to try the Worldbuilding site https://worldbuilding.stackexchange.com/

Anyway, what you're describing is more or less possible, with some caveats. A major question is that of stability - how long those configurations remain in place.

There are several notions you need to think about:

Geostationary orbit: it's the orbit where satellites have an orbital period equal to the planet's rotational period, so they appear stationary in the sky.

Roche limit: the lowest orbit where natural satellites remain in one piece. Any lower than that and they break up into pieces due to tidal forces.

Hill sphere: is the region where the planet's attraction dominates and satellite orbits remain stable (as opposed to being stolen by the star, etc).

Usually Roche limit < Geostationary orbit < Hill sphere.

Satellites at the geostationary orbit appear fixed. Those close to the geostationary orbit would indeed appear to move extremely slowly across the sky. Orbits higher than geostationary would appear to move backwards.

With regards to the long-term stability of multiple moons in various orbits, they would be stable if the orbits are far enough from each other, or if they're sitting in each other's Lagrange points. Also see here https://worldbuilding.stackexchange.com/questions/36929/stability-of-multiple-moons


Did Cassini return a photo of Saturn's rings shown from closer to Saturn? A quote from a book, Perelandra by C.S. Lewis: "no eye looked up from beneath on the Ring of Lurga"; now Lurga is Saturn, and no human eye has been to Saturn, let alone at a lower altitude than the rings, and quite probably would not survive the radiation exposure.
But we have an electronic eye; that is the camera, and on board Cassini were two cameras  and these have returned photos of marvelous views of Saturn. I tried a search to find if any photos were taken looking at the rings or otherwise showing the rings while Cassini orbited so close on its final orbits, but I found only artwork.
Were any such photos taken?

Yes, on one of the final orbits it took some pictures of the rings while crossing the ring plane:

More details of that image are here, and this page show some still images.
Here’s another one:

though it’s hard to interpret without reading the description.
All the science done during the “Grand finale” orbits is described here.


Can the Sun be used as a point source of light to achive better focus? There's a limit to how well sunlight can be focused by a parabolic reflector or a lens because the Sun is not a point source. I wonder if the sunlight could be engineered to work as if it was (much more similar to) a point source.

If almost all of the Solar disc was covered by a coronograph, the light which gets through a small hole would be (almost) a point source. Is that correct?

If then an array of such covered reflectors or lenses concentrated their light to one and the same point, would a better focus be achieved than without covers?

No, it won't work. After all, it is recommended to make a pinhole camera to watch a solar eclipse. What you suggested is similar to a pinhole camera. Why it won't work, is that the rays that pass through the pinhole are not parallel, but divergent. Light from a point source would have (nearly) parallel rays.


I'm 8 hours from Greenwich, why is my sidereal time 10 minutes different from theirs? In order to write and run the script in this answer
I typed "Greenwich sidereal time right now?" into google (or at least tried to) and got two numbers.
Greenwich sidereal time was 09:51:07 but my local sidereal time was ten minutes earlier at 09:40:50.
I'm in UTC +8 so the ten minute difference is not because I'm ~8 hours away.
What causes this small difference in sidereal times?

I don't think the Google Search is passing any coordinates to the Heavens Above website to get the results the Search is displaying.
If you click through on the provided link in the search, Heavens Above gives the following default location:

Location:   Bristol
(51.4770°N, 2.5690°W)
Since Bristol is 2.57° West of the Greenwich Meridian, its Local Sidereal Time is about 10 minutes earlier than Greenwich.


what is a Friedmann model? Can you explain what is a Friedmann model to a layman?
And also give some examples of Friedmann models, specially I would like to know if the lambda-CDM model is considered a Friedmann model.

The "Friedmann model" is a model of the Universe governed by the Friedmann equations, which describes how the Universe expands or contracts. These equations are a solution to Einstein's field equations, and with two very important assumptions they form the basis for our understanding of the evolution and structure of our Universe. These assumptions, together called "the cosmological principle", are that the Universe is homogeneous, and that it's isotropic. This is a layman's explanation, so I won't write the equations here, but farther down I've added the equation and go a bit more in detail what it means.

The cosmological principle

Homogeneity

That the Universe is homogeneous means that its "the same" everywhere. Obviously, it's not, really. For instance, under your feet is a dense, rocky planet, whereas above your head there's thin air. We live in a galaxy full of stars and molecular clouds and what not, while 100,000 lightyears from the Milky Way, there is virtually nothing. But on very large scales, say above half a billion lightyears, the Universe actually looks the same all over.

Isotropy

That it's isotropic, means that it looks the same in all directions. Again, obviously it doesn't on small scales, but on large scales, it does. If it didn't, it would mean that we occupied a special place in the Universe, and we don't think we do.

So, neither of these assumptions have to be true, but observations tell us that apparently, they are. See this image, where every dot is a galaxy (Maddox et al. 1990):

You might think that a homogeneous Universe would also be isotropic and/or vice versa, but that's not the case.

Three possible solutions

It turns out that, for these assumptions, there are three possible solutions to the Friedmann equation. We call the three possible universes flat, positively curved (or "closed"), and negatively curved (or "open"). Which of these possible universes we live in, turns out to depend upon the average density in the Universe, so by measuring this, we can determine the "geometry" of our own Universe. And it seems that it's "flat".

A flat universe

The reason it's called flat is that the geometry is like that of a flat 2D table, only in 3D. That is, a triangle has 180º, parallel lines never meet, etc. And it's infinitely large. Intuitively, we'd think that this is the way the Universe is, and definitely on small scales (say within our own Galaxy), it's an adequate approximation. In the 2D analogy, the 2D surface of Earth seems flat locally, and for all practical purposes the parking lot outside is flat. But if you draw a triangle from Congo→Indonesia→the North Pole→Congo, you'd measure it's sum of angles to be roughly 270º. That's because the geometry of the surface of Earth is not flat, but is "closed".

A closed universe

If the Universe were "closed", it's geometry would, in the 2D analogy, correspond to that of the surface of a ball, i.e. a triangle has more than 180º, lines that start out being parallel will at some point meet, etc. But like the surface area of a ball is finite (but doesn't have a border), so is the Universe. So if you take you spaceship and fly straight away from Earth, you would end up back here (assuming the Universe doesn't collapse before you get back, or expand too fast for you).

An open universe

If it were "open", it's geometry would, in the 2D analogy, correspond to that of the surface of a saddle, i.e. a triangle has less than 180º, lines that start out being parallel will diverge, etc. And it's infinitely large.

This picture from here visualizes the 2D analogies.

In 3D, only a flat geometry can be pictured, and this doesn't look "flat" by any means; it is simply your good old 3D ("Euclidian") space that you know from your everyday senses.

Expansion of the Universe

The Friedmann equation, together with the densities of the constituents of the Universe (radiation, normal matter, dark matter, and dark energy) tell us how the Universe expands. So, again by measuring these densities, we can prediect the evolution of the Universe. And it seems that the Universe not only expands, but actually expands faster and faster.

Beyond the layman's explanation

Here, I'll expand a bit on how to understand the equation:

The first Friedmann equation is intuitively most understandable, I think, when written like this:
$$\frac{H^2}{H_0^2} =
\frac{\Omega_\mathrm{r,0}}{a^4} +
\frac{\Omega_\mathrm{M,0}}{a^3} +
\frac{\Omega_k}{a^2} +
      \Omega_\mathrm{\Lambda}.$$
This equation tells us the connection between the expansion rate of the Universe (the left hand side), and the density of its components and its size (the right hand side). Below, I'll go through the components of the equation.

The Hubble parameter
In the equation, $H$ is the Hubble parameter describing how fast a galaxy at a given distance recedes (or approaches, for a collapsing universe), at a given time in the history of the Universe. $H_0$ is its value today, and is measured to be roughly $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$. This means that a galaxy at a distance of, say, 10 Mpc ($\simeq33$ lightyears), moves away from us at a current speed of 700 km/s. A galaxy 20 Mpc away recedes at 1400 km/s, and so on (and galaxies farther away than roughly 4.3 Gpc recede faster than the speed of light, but that's no problem and we can still see them).

Size
The size of the Universe is unknown, and quite possibly infinite (lest it wouldn't be homogeneous, but to be fair, we only know that the observable Universe is homogeneous). Thus, we cannot talk about its absolute size. But we can talk about how much a certain volume of space expands in a given time. We us the parameter $a$, called the expansion factor. Defining $a$ to be 1 today, that means that at the time when the Universe was so small that all distances between the galaxies was, say, half of today's values, $a$ was equal to 0.5 (this happens to be 8 billion years ago).

Density parameters
Whether the geometry of the Universe as described above is flat, closed, or open, depends on whether the total density $\rho_\mathrm{tot}$ is exactly equal to, above, or below a certain critical threshold $\rho_\mathrm{cr} \sim 10^{-29}\,\mathrm{g}\,\mathrm{cm}^{-3}$. It is customary to parametrize the density of the $i$'th component as $\Omega\equiv \rho_i / \rho_\mathrm{cr}$.

Matter
The term "matter" includes "normal" matter (gas, stars, planets, bicycles, etc.), and the mysterious dark matter. As the universe expands, the volume grows as $a^3$. That means that the density falls as $\Omega_\mathrm{M}=\Omega_\mathrm{M,0}/a^3$.

Radiation
Photons redshift as space expands, and this redshift goes as $1/a$. This is in addition to having their number density decrease, so the total energy density of radiation decreases faster than matter, namely as $1/a^4$. Today, the energy density of radiation is dominated by the CMB, and can be neglected, but in the early times, they would dominate.

Curvature
If space is not flat, its curvature contributes to $\Omega_\mathrm{tot}$. The reason is that the curvature affects the volume in which we measure densities (thanks to John Davis for this explanation). This scales as $1/a^2$.

Dark energy
Finally, there's the magical dark energy, of which even less is known than the dark matter. If it exists, it's thought to be a property of space itself, i.e. its energy density grows proportionally to the volume of the Universe, and thus there's no $a$-dependency.

Interpretation

From the equation, it is readily seen that if we're able to measure all the Omegas, then we know how fast the Universe has been expanding at all times. That means that we can integrate backwards in time and calculate when $a$ was 0, i.e. we can calculate the age of the Universe. Also, from the $a$-dependencies we can see when the Universe went from being radiation-dominated to being  matter-dominated. We can also see that not only is it now dominated by dark energy (since $\Omega_\mathrm{\Lambda}\simeq0.7$, but $\Omega_\mathrm{M}\simeq0.3$), but due to the $a$ factor, it will only get "worse". That is, all other densities will keep decreasing, but $\rho_\Lambda$ stays the same, and since dark energy has a repulsive effect rather than attracting, the expansion of the Universe accelerates.

Observationally, it is found (in several independent ways) that all the $\Omega$s add up to one, i.e. that the total energy density of the Universe happens to be exactly equal to the critical density. This is pretty amazing. This figure, taken from here, shows the contribution from the different components now (top), and at the time of the CMB emission (380,000 yr after the Big Bang; bottom):


Is it possible to mount a telescope on a plane? Is it beneficial? I was wondering if any planes had mounted telescopes with the intent to observe the stars. I understand that the atmosphere itself can warp and hinder incoming light and even completely obscure views on cloudy days. Would it be possible to have a telescope which could account for the speed of the airplane and remove motion blur? Would it even be worth it, seeing as the size of the lens would also be limited? Could we see anything interesting that a terrestrial scope of the same size would miss?

Hobbes has mentioned SOFIA as a great example of this being done. Does anyone have more information on the benefits of a plane mounted telescope over one on the ground for noninfrared observation?

This has been done.

SOFIA is an infrared observatory built into a Boeing 747 SP:

SOFIA takes advantage of the fact that some infrared bands are visible at atltitude, these are attenuated by water in the atmosphere so they're less visible on the ground.
There have been infrared observatories before SOFIA:

The first use of an aircraft for performing infrared observations was in 1965 when Gerard P. Kuiper used the NASA Convair 990 to study Venus. Three years later, Frank Low used the Ames Learjet for observations of Jupiter and nebulae.[20] In 1969, planning began for mounting a 910 mm (36 in) telescope on an airborne platform. The goal was to perform astronomy from the stratosphere, where there was a much lower optical depth from water-vapor-absorbed infrared radiation. This project, named the Kuiper Airborne Observatory, was dedicated on May 21, 1975. The telescope was instrumental in numerous scientific studies, including the discovery of the ring system around the planet Uranus.[21]
The proposal for a larger aircraft-mounted telescope was officially presented in 1984 and called for a Boeing 747 to carry a three-meter telescope. The preliminary system concept was published in 1987 in a Red Book. It was agreed that Germany would contribute 20% of the total cost and provide the telescope.

Other airborne astronomy is more incidental. Around solar eclipses, you'll often see some astronomy flights. Some of these are tourism, others perform science. These are usually passenger aircraft temporarily modified (scopes installed that look through the existing windows). These take advantage of the fact you can lengthen the eclipse by flying along its path, and you can reach eclipses in places otherwise inaccessible.


How does Hawking Radiation work exactly? I know that a particle and anti particle, virtual particles, spawn at the edge of an event horizon, and one particle falls into the black hole, and the other goes out, but how does the other know how to take energy from the black hole? I mean how does it? Does it just know to take energy from the black hole or does the antiparticle fall in, and the blackhole loses mass?

... but how does the other know how to take energy from the black hole?

In order to understand this, you need to be familiar with the essence of this picture$^1$ including negative energy states inside the horizon, creation of virtual particle-antiparticle pairs, and conservation of energy.$^2$ I try to intuitively answer your question with simple words.
Assume that a virtual particle-antiparticle pair is created near the black hole's horizon. According to the principle of conservation of energy, the total energy of the pair particles must be zero. So, one of the particle has positive energy and the other one has negative energy. On the other hand, it can be shown that there exist negative-energy states inside the horizon of a static black hole and the negative-energy particles can occupy these states. In order to have a pair of real particles with zero total energy, the only physical possibility is that the positive-energy particle can escape to infinity, while the negative-energy particle falls into the black hole. In this way, the strong gravitational field of the black hole can convert a virtual particle–antiparticle pair into a pair of real particles with zero total energy. This is the reason why the virtual particle that's sucked in always get negative energy.

Does it just know to take energy from the black hole or does the antiparticle fall in, and the blackhole loses mass?

Note that, in this picture, when a virtual particle-antiparticle pair is created near the black hole's horizon, each of them may fall into the horizon or may escape to infinity (Always, one of them falls in and the other escapes.) So, it is not correct that only the antiparticle falls into the black hole. This means that a static observer outside the event horizon will observe both the particle and antiparticle spectrums.

In my opinion, you should not take the above argument about virtual particle-anti particle pair etc. too seriously since this picture helps to naively have an intuitive understanding. A rigorous treatment of Hawking radiation using quantization of quantum fields in curved black hole background does not need such a naïve picture.

$^1$There exist more rigorous treatments for understanding the Hawking radiation but I restrict myself to this framework (picture) that you are interested in and ask about.

$^2$ Here, for simplicity, I restrict this discussion to the case of static black holes. The general conclusion is still valid for the rotating black holes.


Are these real paths of planets as traced from Earth? Few days a go, I came across these two pictures of paths of planets as observed from Earth. They have a perfect geometric shapes.  Actually, to me they look so beautiful and perfect, that they appear fake or made by a pseudo-scientist or some propaganda society (e.g. flat Earth like community people) who talks about divine geometry etc.

Auxiliary questions: Is this the reason, why Earth was consider to be in the center of the solar system?

From this answer here are views of the inner and outer planets. I used the python package Skyfield which uses the NASA JPL ephemerides to generate positions of all the planets, then I subtracted the Earth's position to generate the geocentric positions.
I agree with your guess that the first image is too perfect.
The paths will not be closed, because there is not a rational number relationship (e.g. 5:9, 2:3, 1:1) between the periods of the planets.
The ones in the image are reminiscent of the Spirograph drawings I used to make a very long time ago.

...plotting using the Python package Skyfield.
The "inner" plotted bodies (from outer to inner) are Mars, (Earth), Moon, Venus, Mercury, The Sun. The data is for 2000 days.
The "outer" plotted bodies are Uranus, Saturn, Jupiter, Earth, and The Sun. The data is for 2000 months.
Distances are in AU.


In a planetary system close to the galactic core, would it be possible to see the supermassive black hole? I understand that radiation on planets near the galaxy’s center make life on these planets nearly impossible, and that one cannot truly “see” a black hole. However, if you could stand on the surface of a planet that orbits a star close to or within the galactic core, could you theoretically look up at the sky and see an absence of light/stars indicating the location of the central supermassive black hole?

Would it be too far away to see, obstructed by debris, or too small to notice?

If the black hole is active, meaning it is still capturing matter from its surrounding, it will have a large accretion disk all around, which is the only way to dissipate angular momentum for the matter falling into it.

As a result of this dissipation all the matter will warm up and emit radiation. This disc will be fairly large and thus clearly visible as a bright object in the sky.

This page shows an image captured by Hubble of such a disk:

Strictly speaking then, one cannot see directly the black hole, as its view will be covered by the bright emission of the disk.

However, the presence of the disk will allow to observe the black hole thanks to its gravitational effect.


Can a habitable planet be smaller than 0.58 Earth radii? According to the below image, the lowest escape velocity a planet can have in order to still be able to retain water on its surface and have a temperature above freezing is 6.5 km/s minimum.
With Earth density (which is already a pretty high density to have for a planet smaller than Earth), 0.58 Earth radii has an escape velocity of 6.488 km/s, 0.585 Earth radii has an escape velocity of 6.544 km/s, and 0.59 Earth radii has an escape velocity of 6.6 km/s.
More narrowly, 0.581 Earth radii has an escape velocity of 6.499 km/s, while 0.582 Earth radii has an escape velocity of 6.51 km/s.  Thus, hypothetically, the smallest habitable planet has between 0.581 and 0.582 Earth radii.
My question is, is this true?  Is it impossible for a planet smaller than ~0.58 Earth radii to maintain liquid water on its surface?
This chart implies that a planet smaller than this can only retain water on its surface in its frozen state, or otherwise a lower escape velocity even if the planet were in the habitable zone would cause the planet to lose its water over cosmological timescales.
The only alternative would be to have a higher density than Earth which would increase its mass, gravity, and escape velocity, but as I said, Earth density is already quite high for such a small planet.

(source)

Short Answer:
In part Two of the long answer below, it is stated that a planetary mass object needs to have a mass of at least 0.1, 0.12, 0.23, or 0.25 Earth mass to be habitable.  Worlds with those masses and with a radius of 0.58 Earth radius should have densities of 2.827, 3.392, 6.502, or 7.067 grams per cubic centimeter.  Such densities are possible, though it would be unusual for such small mass worlds to have such high densities.
If such a world is made more dense, it could have such a mass with a radius a bit smaller than 0.58 Earth radius, though a small world with such a high density would even rarer and more unlikely.
Taking planetary density to a theoretical extreme, if a planetary mass object was made out of osmium, rough calculations indicate it could have those minimum masses with radii of 0.29 to 0.3945 of the radius of the Earth.
And of course several solar system bodies with no atmosphere and surfaces of solid ice are believed to have liquid water in subsurface oceans, and thus possibly be habitable for alien life in their subsurface oceans.  The smallest of them of them is Enceladus, a moon of Saturn, with a radius of only about 252.1 kilometers, or 0.039 the radius of the Earth.
Long Answer:
When you say "habitable" do you mean "habitable for carbon based liquid water using organisms in general" or "habitable for human beings, other intelligent beings, and multicellular land animals that need an oxygen rich atmosphere in particular"?
Part One of Two: Habitable for multicellular land animals that need an oxygen rich atmosphere.
As far as I know, the only scientific study of what planets would be habitable for multicellular land animals that need an oxygen rich atmosphere in general, and for humans in particular, is Habitable Planets for Man Stephen H. Dole, 1964, 2007.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf[1]
Dole believed that there are formulas for calculating the relationships between
the mass of a terrestrial type planet and its radius, volume, density, surface gravity, and escape velocity.  Thus the mass of a terrestrial type planet would determine its radius, volume, density, surface gravity, and escape velocity. And that is largely true for the few terrestrial type worlds in our Solar System, which formed from the same protoplanetary disc with similar percentages of different elements with different densities.
But even in our Solar System Mercury is much more dense than such a small planet should be.  It is believed that billions of years ago Mercury collided with an even smaller planet, and most of the lighter elements in Mercury's crust were lost in the impact, leaving only the denser elements behind.
And in the outer Solar System, where the protoplanetary disc included a higher percentage of the lighter elements,  many of the large, planetary mass, moons and other objects have, are made of mixtures of ices of various liquids with rock, and so are less dense than the terrestrial planets.
Astronomers have discovered thousands of exoplanets orbiting other stars, and have measured or calculated the masses, radii, densities, surfaces gravities, and escape velocities of some of them. Planets formed out of the protoplanetary discs of other stars with other percentages of various elements are likely to have somewhat different densities than terrestrial planets in our solar system.
A few large exoplanets are believed to have very high densities, and it is theorized that they are the remaining dense rocky cores of giant planets that most most of their light elements in various cataclysms.
So it is possible for a science fiction writer to justify a terrestrial type planet with a density somewhat different than what would be calculated based on the terrestrial type planets in our Solar System.  But not too much different.
On page 53 Dole sets a rough maximum possible mass for a planet habitable for humans.

"Now it will be recalled that, to be considered habitable, a planet must have a surface gravity of less than 1.5 g. From figure 9, it may be seen that this corresponds to a planet having a mass of 2.35 Earth masses, a radius of 1.25 Earth radii, and an escape velocity of 15.3 kilometers per second.

Of course an exoplanet which formed from a protoplanetary disc around a different star, might be made of matter with a higher or lower average density, and so might have a slightly greater mass and radius than Dole calculated, while still having a surface gravity of 1.5 g or less.
And of course lifeforms evolving on an alien exoplanet could be capable in surviving in a much higher surface gravity than 1.5 g.
So what did Dole calculate for the minimum mass of a planet with an oxygen rich atmosphere habitable for humans, and would the minimum mass of a planet with liquid water on its surface be any higher or lower?
On page 54 Dole says:

However, if we take as a rough approximation that maximum exosphere temperatures as low as 1000 degree K are not incompatible with the required surface conditions of a habitable planet, then the escape velocity of the smallest planet capable of retaining atomic oxygen may be as low as 6.5 kilometers per second (5 X 1.25).  Going back to figure 9, this may be seen to correspond to a planet having a mass of 0.195 Earth mass, a radius of 0.63 Earth radius, and a surface gravity of 0.49 g.

Dole believed that a planet that small could retain an oxygen rich atmosphere, but didn't think that it produce an oxygen rich atmosphere in the first place.  Thus a planet with 0.195 of Earth Mass would probably not form a natural oxygen rich atmosphere but could be terraformed and given an oxygen rich atmosphere by an advanced civilization.
Dole used 2 different possible rules to calculate 2 different possible minimum masses for a planet with an oxygen rich atmosphere.  Those calculations produced a minimum mass of 0.25 and 0.59 Earth mass respectively.  So on pages 56 and 57 Dole decided that the true value would be intermediate and chose a value of about 0.4 Earth mass.

This corresponds to a planet having a radius of 0.78 Earth radius and a surface gravity of 0.68 g.

So a planet with the mass of 0.4 Earth mass would have a radius of 0.78 earth, which is 1.34 times a radius of 0.58 Earth radius, and even a planet with a mass of 0.195 Earth would have a radius of 0.63 Earth radius, which is about 1.08 times a radius of 0.58 Earth radius.
And I guess that 0.63 Earth radius is close enough to 0.58 Earth radius, that a planet with a lesser density and more light elements could possibly produce an oxygen rich atmosphere and replace as fast as it loses oxygen from the exosphere.  Or maybe a planet made of denser materials might have a high enough escape velocity and produce enough oxygen to retain an oxygen rich atmosphere.  and possibly that might stretch the minimum radius for habitability down to 0.58 Earth radius.
Part Two of Two: Habitable for carbon based liquid water using lifeforms.
Such a planet would not need to have an oxygen rich atmosphere, just a dense enough atmosphere to have sufficient pressure for water to be liquid at most temperatures on the planet's surface.  The lower the atmospheric pressure of a planet is the lower the boiling temperature of water is, until with a low enough atmospheric pressure ice will turn directly into water vapor without ever being a liquid.
"Exomoon Habitability Constrained by Illumination and Tidal Heating", Rene Heller and Roy Barnes, Astrobiology, Volume 13, number 1, 2013, is a comparatively recent discussion of the possible habitability of worlds in other solar systems, in this case as yet undiscovered exomoons orbiting exoplanets in other solar systems.  Heller and Barnes do not mention any reason to suppose that exomoons could be habitable if they were outside the mass range for an exoplanet to be habitable.  Therefore they summarize scientific opinions about the mass range for potentially habitable worlds.
https://faculty.washington.edu/rkb9/publications/hb13.pdf[2]
In section 2. Habitability of Exomoons, the last paragraph before section 2.1 2.1. Formation of Massive Exomoons, on page 20, discusses the mass range for habitable worlds, including exoplanets and exomoons.

A minimum mass of an exomoon is required to drive a
magnetic shield on a billion-year timescale (MsT0.1M4;
Tachinami et al., 2011); to sustain a substantial, long-lived
atmosphere (MsT0.12M4; Williams et al., 1997; Kaltenegger,
2000); and to drive tectonic activity (MsT0.23M4; Williams
et al., 1997), which is necessary to maintain plate tectonics
and to support the carbon-silicate cycle. Weak internal dynamos have been detected in Mercury and Ganymede
(Gurnett et al., 1996; Kivelson et al., 1996), suggesting that
satellite masses > 0.25M4 will be adequate for considerations
of exomoon habitability. This lower limit, however, is not a
fixed number. Further sources of energy—such as radiogenic
and tidal heating, and the effect of a moon’s composition and
structure—can alter the limit in either direction.   An upper mass limit is given by the fact that increasing mass leads to high pressures in the planet’s interior, which will increase the mantle viscosity and depress heat transfer throughout the mantle as well as in the core. Above a critical mass, the dynamo is strongly suppressed and becomes too weak to generate a magnetic field or sustain plate tectonics. This maximum mass can be placed around 2M4 (Gaidos et al., 2010; Noack and Breuer, 2011; Stamenkovic´ et al., 2011).

Their source for a minimum mass of 0.1 Earth Mass for a magnetic shield is

Tachinami, C., Senshu, H., and Ida, S. (2011) Thermal evolution
and lifetime of intrinsic magnetic fields of super-Earths in
habitable zones. Astrophys J 726, doi:10.1088/0004-637X/726/
2/70.

Their source for a minimum mass of 0.12 Earth mass for a substantial, long lived atmosphere is

Williams, D.M., Kasting, J.F., and Wade, R.A. (1997) Habitable
moons around extrasolar giant planets. Nature 385:234–236.

Their source for a minimum mass of 0.23 Earth mass for plate tectonics is

Williams, D.M., Kasting, J.F., and Wade, R.A. (1997) Habitable
moons around extrasolar giant planets. Nature 385:234–236.

Their sources for the weak magnetic fields of Mercury and Ganymede are:

Gurnett, D.A., Kurth, W.S., Roux, A., Bolton, S.J., and Kennel,
C.F. (1996) Evidence for a magnetosphere at Ganymede from
plasma-wave observations by the Galileo spacecraft. Nature
384:535–537.
Kivelson, M.G., Khurana, K.K., Russell, C.T., Walker, R.J.,
Warnecke, J., Coroniti, F.V., Polanskey, C., Southwood, D.J.,
and Schubert, G. (1996) Discovery of Ganymede’s magnetic
field by the Galileo spacecraft. Nature 384:537–541.

Their source for the importance of plate tectonics for habitability is:

Williams D.M. Kasting J.F. Wade R.A. Habitable moons around extrasolar giant planets. Nature. 1997;385:234–236. [PubMed] [Google Scholar]

Their sources for that maximium mass limit of 2 times the mass of Earth are:

Gaidos, E., Conrad, C.P., Manga, M., and Hernlund, J. (2010)
Thermodynamics limits on magnetodynamos in rocky exoplanets. Astrophys J 718:596–609.

Noack, L. and Breuer, D. (2011) Plate tectonics on Earth-like
planets [EPSC-DPS2011-890]. In EPSC-DPS Joint Meeting 2011,
European Planetary Science Congress and Division for Planetary Sciences of the American Astronomical Society. Available
online at http://meetings.copernicus.org/epsc-dps2011.

Stamenkovic´, V., Breuer, D., and Spohn, T. (2011) Thermal and
transport properties of mantle rock at high pressure: applications to super-Earths. Icarus 216:572–596.

The density of the planet Earth is 5.513 grams per cubic centimeter.  A sphere with a radius of 0.58 Earth radius would have a volume of 0.195 of the volume of Earth.
So putting a mass of 0.1 Earth mass within a volume of 0.195 Earth volume would result in a density of 0.512 of Earth's density or 2.827 grams per cubic centimeter.
So putting a mass of 0.12 Earth mass within a volume of 0.195 Earth volume would result in a density of 0.615 of Earth's density or 3.392 grams per cubic centimeter.
So putting a mass of 0.23 Earth mass within a volume of 0.195 Earth volume would result in a density of 1.179 of Earth's density or 6.502 grams per cubic centimeter.
So putting a mass of 0.25 Earth mass within a volume of 0.195 Earth volume would result in a density of 1.282 of Earth's density or 7.067 grams per cubic centimeter.
In a comment, Xi-k says that since the relatively common element iron, a major component of the cores of most planets, has a normal density of 7.874 grams per cubic centimeter, higher than the figure of 7.067 grams per cubic centimeter quoted above, and the iron in the cores of planetary bodies is compressed to greater densities than 7.874 grams per cubic centimeter, a mostly iron planet could be habitable with a small radius.  Thus a planet 80 to 90 percent iron with 10 to 20 percent silicate crust could have an average density of about 7.067 grams per cubic centimeters.
This seems like a plausible way to get the radius of a habitable planet down to about 0.58 Earth radius. But an expert on planetary development could probably give a better estimate about the probability of planets being that small and dense.
At at the extreme of physical possibility, and very, very, very, very improbable, is a small planetary object made mostly of Osmium, a very, very rare element.  Such a planet might have to be created by a civilization collecting osmium from several solar systems and building a planet from it.

Osmium is the densest naturally occurring element, with an experimentally measured (using x-ray crystallography) density of 22.59 g/cm3.

https://en.wikipedia.org/wiki/Osmium[3]
So a sphere of solid osmium, with a thin layer of lighter materials necessary for life, could possibly exist - without discussing the probability of such a sphere forming.
Ignoring the compression and increased density of a planetary sized osmium sphere, and assuming that the hypothetical planet has an average density 4.0915875 that of Earth, It could have mass of 0.1 to 0.25 Mass earth with a radius of about 0.29 to 0.3945 Earth.


How would vehicles travel through the interstellar medium with its such low density? I've heard from a number of different places that the density in the interstellar medium can have an average of 1 atom per centimeter cubed. Perhaps I have a wrong understanding of what 'nothing' is (I am very knew to astronomy/astrophysics), but how would a vehicle, like the Voyager for example, be able to travel through it?

My (perhaps very flawed) thinking behind this is that if there are so few atoms not in close proximity, there couldn't be any (meaningful) forces interacting between them to provide any sort of way to travel through it.

The answer is in your question.

You said

  there couldn't be any (meaningful) forces interacting between them to provide any sort of way to travel through it.

That's the great thing about outer space. There normally aren't many particles around that could slow down a spacecraft via drag or friction. If you give something a push in outer space - somewhere that unrealistically is free from any gravitational interaction with other objects - it will travel forever. Objects travel through the ISM via Newton's first law - an object in motion tends to stay in motion.


Relationship between galactic rotation and stellar lifetime I have read that the lifetime of stars is much less than the period of rotation for a spiral galaxy. For example https://www.coursera.org/learn/evolvinguniverse/lecture/wlYhz/the-origin-of-spiral-arms the professor uses the existence of young blue stars whose creation is triggered by a persistent density wave to claim that the stars lifetime is less than the rotation period. I am not satisfied with this. Given the enormous timescale, how do we know this?

We know how long stars take to orbit in a spiral galaxy, because we know how big galaxies are (we can measure the size by knowing the distance using various methods such as Cepheid variables, and type Ia supernovae) and we know how fast the stars are moving. (We can measure the velocity of stars using the doppler method)

We know how long stars last for because we have a good model of how stars work. We can use this model to predict that large, hot stars will burn themselves out in a short time. We can verify this model by looking at the numbers of blue stars, red supergiants and supernovae.

From the first part we know that stars take about 250 million years to complete an orbit of the Milky way (which is typical for similar sized galaxies) From the second part, we know that giant blue stars live for only tens of million years, and so could not complete a full orbit of the milky way.

On the other hand, smaller stars like the sun live for billions of years and so make multiple orbits.


Can two heavy objects circling around their C.M. be separated because of the speed of gravity? Imagine two massive objects, with the same mass (M) circling around their center of mass (C.M.). Let's assume that the distance between them is 1 light hour. Don´t the two bodies get accelerated and move away from each other because they feel the gravity of each other as it was 1 hour ago, because of which a force tangent to the direction of the speed develops?
Can it be that gravity acts instantaneously? I've read about experiments proving the speed of gravity to be equal to the speed of light, but these were disputed.
Isn't there both in Newtonian mechanics and GR there a time delay? The difference being though that space and time in Newtonian mechanics are separate and absolute (implying instantaneity?), while in GR the both are a whole and absolute.

Can two heavy objects circling around their C.M. be separated because of the speed of gravity?

No. Newtonian mechanics does quite well assuming instantaneous gravity. You can get relativistically correct orbits either by doing a complete GR calculation, or using a simple approximation as already eluded alluded to in @antispinward's answer.
This excellent answer to Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch? in Space Exploration SE explains that we get the wrong answer if purely Newtonian mechanics is used except for slowing down the speed of gravity, but that's because one is neither treating gravity correctly nor using Newtonian mechanics correctly.
Many answers to How to calculate the planets and moons beyond Newtons's gravitational force? include a way to treat this problem using a well-accepted approximate treatment of general relativity.
From this answer:

The acceleration of a body in the gravitation field of another body of standard gravitational parameter $GM$ can be written:
$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$
where $r$ is the vector from the body $M$ to the body who's acceleration is being calculated. Remember that in Newtonian mechanics the acceleration of each body depends only on the mass of the other body, even though the force depends on both masses, because the first mass cancels out by $a=F/m$.

and later:

The following approximation should be added to the Newtonian term:
$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$


3D array in FITS data I am an amateur astronomer currently working on Spitzer telescope dataset with python (astropy and aplpy). From the examples in astropy's website, I understood that the FITS data is a way to represent image data with human-readable headers.  However, I now have a fits image ( SPITZER_I2_46469632_0000_0000_2_bcd.fits ) which is a 3D array of dimensions (64, 32, 32). I believe the second and third coordinates to be the length and height of the image in pixels. What about the first dimension? How do I interpret such images and NaNs in the data?

It appears to me that the full Spizter frame should normally be 256x256 but, if observing a very bright source or when requiring high temporal resolution, a different observing mode is performed wherein only 32x32 pixels are read from the CCD. However, what happens is that the exposure times are short and a total of 64 frames are observed in total, creating a data cube of 64x32x32. If you want, you can treat each individial 32x32 frame as its own image, or else stack them all together for the full image.

Full details can be found here.


What is the galaxy M87 doing these days? The massive elliptical galaxy M87 in the Virgo cluster is 53,490,000 light years away. It also contains one of the largest, heaviest supermassive black holes in the known Universe. But it's also my understanding that the stellar population of this galaxy is mostly old, red supergiants. Is there any evidence for new star formation in M87? If there isn't, what is the fate of this galaxy and what is it like in modern times? That may be impossible to know for sure but what I wonder is if the remaining stars and matter in the galaxy have been swallowed up by central black hole which may be thousands of lightyears across by now.

The answer to your question is relatively straightforward: Not much. I'll delve into it in two parts: Star formation and the activity of the supermassive black hole at the center.
Star formation
M87 home to many Population II stars. They have little hydrogen and helium in them, and the clouds of gas and dust in the galaxy don't have much, either. In fact, many dust clouds can be destroyed by the intense radiation emitted by the accretion disk of the central black hole. This combines to make it seem unlikely that there will be much star formation in the galaxy.
That said, just because a galaxy doesn't have a lot of materials for star formation doesn't mean it can't form lots of new stars - at least, no under certain special circumstances. Galaxy collisions can induce high rates of star formation - starburst galaxies are good examples of this. While M87 might not be at a high risk for a collision, there is still the possibility of one. M84 is nearby, and may have had an encounter with M87 in the past. While it is a long shot (and M84 has a low star-forming rate), an interaction with the two could spawn new stars.
The supermassive black hole
You seemed particularly interested in the black hole at M87's center. It is a supermassive black hole, with a mass billions of times that of the Sun. Quite a monster, by many standards. Yet it is not doomed to swallow the galaxy whole anytime soon. It is sucking in gas and dust at a rate of $\sim10^{-1}M_{\odot}$ each year (see Di Matteo et al. (2002)) - and there are trillions of solar masses of gas, dust, and stars in the galaxy. For comparison, Sagittarius A* accretes matter at a rate of between $\sim10^{-8}M_{\odot}$ and $\sim10^{-4}M_{\odot}$ per year, depending on the exact model (see e.g. Quataert et al. (1999) and Shcherbakov et al. (2012). Still, the supermassive black hole in M87 couldn't accrete all of the matter in the galaxy in the age of the universe, if it continued at this rate.
In summary: Star formation in M87 is at a low rate, and while a galactic interaction could create a new wave of star formation, it is unlikely. The central black hole is sucking in lots of gas and dust, but it will take billions upon billions of years before it eats up a substantial amount of matter.


Why is the speed of light 299,792,458 meters/sec? Ok, I am majoring in physics (4th year) and I never understood this fundamental (kinda) question. Maybe I haven't explored it enough.

For example, why does it take 8min20sec for the light from the sun to get to us?
I know the answer to this question on a 'surface' scale. The sun is 1AU away, c=3E8 m/s, and d=v/t to get approx 8 min 20 sec.

my question is on a deeper level.

Say you could "ride a photon" I KNOW THIS IS IMPOSSIBLE, but just say you could. Or a better question: what does a photon experience? The photon, as per my understanding, would leave the sun and (if it is on the right trajectory) hit the Earth instantaneously. A photon leaving Alpha Centauri would see the universe all at once, in a infinitesimal small unit of time (if directed out to space).

If a photon sees everything all at once, why do we perceive it to have a speed? I am sure this has something to do with frames of reference, special relatively, Lorentz transforms? but just seems strange. why is the speed of light finite to us... if it was infinite would this be problematic?

Speed of light being finite is one of the fundamentals of our Universe.

If it were infinite, this would have a major implication in causality.

Besides, in non-quantum physics, light is just an electromagnetic wave. Eelectromagnetic field is described by Maxwell's equations, which predict that the speed $c$ of electromagnetic waves propagating through the vacuum depends on the dielectric permittivity $ε_0$ and the magnetic permeability $μ_0$ by the equation $c = {1\over\sqrt{ε_{0}μ_{0}}}$ so you can not have an infinite speed of light unless electric permittivity or magnetic permeability were zero, which in turn would cause all sorts of odd things to electromagnetical attraction (and thus, to matter existence beyond elemental particles).


Number density of stars on the Galactic plane The number density of stars (e.g. the number of stars per cubic parsec) is given by
$$N(z) = N(0) e^{-|z|/h}$$
where $z$ is the height above the Galactic plane, $h$ is the scale height for a specific component of the Galaxy, and $N(z)$ is the number density at different heights above the plane.

The scale heights for the thin disk and thick disks are:

Halo stars: $h=3\textrm{kpc}$
Thick disk: $h=1.5\textrm{kpc}$
Thin disk stars: $h=300\textrm{pc}$
Thin disk gas: $h=100\textrm{pc}$

Unfortunately I cannot find the values for the number density at the galactic plane: $N(0)$

Do you have any references in which I can find these numbers?

The number density of stars is not well constrained, because it hinges critically on the behaviour of the stellar mass function at low stellar masses. Low-mass stars tend to be dim and hard to detect. What is reasonably well constrained, instead is the mass density of stars, where low-mass objects contribute little, even if their number is large.


density of dark matter halo I was surfing the net for knowing the actual density of dark matter halo for a particular Mh(Mass of the halo) and finally, I got two equations:
 for pseudo-isothermal halo
and
 by NFW-profile
So, with which density distribution of dark matter halo should I go corresponding to any Mh?

tl;dr The answer is opinion-based, but choosing an NFW profile probably won't offend anyone.
Pseudo-isothermal profile
There are many more or less physically motivated functional forms for the density profile of a dark matter halo.
The pseudo-isothermal sphere is a less unphysical profile than the simple, isothermal profile (which is unphysical in the center and has a diverging integrated mass for $r\rightarrow\infty$). It gives good fit to the rotation curves in the optical disk of galaxies, but not to the light distribution.
NFW profile
Arguably, the most popular profile is the NFW profile (Navarro, Frenk, & White 1996;1997) — with a total of ~13,000 citations, these two papers are among the most cited papers in astrophysics ever. Its popularity is probably partly due to its simplicity. It isn't really physically motivated, but simply turned out to give good fits to numerical simulations.
Einasto profile
However, subsequent N-body simulations with higher resolution
(
Navarro et al. 2004;
Hayashi & White 2008;
Gao et al. 2008;
Springel et al. 2008
)
showed small, but systematic, deviations from the NFW profile which are more accurately modeled with an Einasto (1965) profile:
$$
\rho(r)=\rho_{-2} \exp \left[\frac{-2}{\alpha}\left\{\left(\frac{r}{r_{-2}}\right)^{\alpha}-1\right\}\right],
$$
where $r_{-2}$ is the radius at which the slope of the profile is $-2$, $\rho_{-2} = \rho(r_{-2})$, and $\alpha$ is a fitting parameter with best-fit values that typically lie in the range $[0.12,0.25]$, increasing with halo mass.
However, since there isn't much physics here either, and this form is more complicated, the NFW profile seems to remain the profile of choice for most people.
Other profiles
Many profiles can be written as double power-laws
$$
\rho(r)=\rho_{0}\left(\frac{r}{r_{0}}\right)^{-\gamma}\left[1+\left(\frac{r}{r_{0}}\right)^{\alpha}\right]^{(\gamma-\beta) / \alpha},
$$
where the parameters $\alpha$, $\beta$, and $\gamma$ can take a variety of values (Tab. 5.1 from Mo et al. 2010):
$$
\begin{array}{lll}
\hline
\hline (\alpha, \beta, \gamma)  & \text{Name} & \text{Reference} \\
\hline(1,3,1) & \text{NFW profile} & \text{Navarro et al. (1997)} \\
(1,4, \gamma) & \text{Dehnen profile} & \text{Dehnen (1993)} \\
(1,4,1) & \text{Hernquist profile} & \text{Hernquist (1990)} \\
(1,4,2) & \text{Jaffe profile} & \text{Jaffe (1983)} \\
(2,2,0) & \text{Modified isothermal sphere} & \text{Sackett & Sparke (1990)} \\
(2,3,0) & \text{Modified Hubble profile} & \text{Binney & Tremaine (1987) }\\
(2,4,0) & \text{Perfect sphere} & \text{de Zeeuw (1985)} \\
(2,5,0) & \text{Plummer sphere} & \text{Plummer (1911)} \\
\hline \hline
\end{array}
$$
At small and large radii $\rho$ is proportional to $r-\gamma$ and $r-\beta$, respectively, while $\alpha$ gives the sharpness of the break.
Some profiles are better at describing the inner structure, some are better farther out, some are better when you include gas, etc. What you choose depends on your particular problem, and your mood.


Could the Earth use gravitational lensing / bending of light to see it's own bottom? Specifically, I have often wondered whether gravitational lensing might be able to (theoretically) be used in order to see light reflecting off the Earth that has zoomed off into space and been bent by one object after another until it comes full circle back to Earth again. Given the distance the light would have travelled, could we not use this technique to peer back through time at our own planet?

Is it reasonable that this could happen given the size of the universe and the mass and distribution of objects within it?
Given the restraints above, would be be able to estimate how far back in time we are looking?
Given the above, how bright would something have to be in order for its light to travel this distance and still be detectable?

Whimsical question. Similarly, you could reason that someone on a planet far, far away would be holding up a mirror, such that we could see our own reflected past again.

Unfortunately, both options are practically infeasible, even if the light would be undistorted and redirected perfectly towards us. Remember that (1) Earth is small, (2) the reflected signal is weak as it is, (3) any object capable of reflecting this back towards us would be a black hole, the closest of which is further away than most known exoplanets, and the signal falls of with $r^{2}$ with $r$ twice (!) the distance to the object doing this lensing.

Now, if we somehow were able to observe it anyway, it would be straightforward enough to estimate the look back time - the distance to the lens would presumably be fairly well known.

How bright it would have to be is not really a sensible question - there are too many variables in play for that (which (hypothetical future) telescope are we using, what are the observing conditions, how far away is the source, what kind of source are we talking about...). And at the end of the day, the brightness of your source is not even the biggest problem; finding a suitable lens is.