Emulation of an Orrery Orrery is a clockwork model of the solar system. I am trying to emulate one in 2D. Now, to emulate, I need to know what goes on inside. Can someone please explain the basic principle behind the clockwork? Or direct me to a resource that will explain all the machinery inside a simple Orrery.

The principle is almost exactly the same as a watch or clock, but instead of three concentric axles, you need 9 for the planets.

Have a google for Orrery kit - there are loads available. It is really all simple maths - you just need to know relative orbital periods in order to calculate cog sizes.

(picture from curiousminds.co.uk)
For moons, you do add a little complexity in the form of a transmission to support the Earth, so that the difference in movement can move the Moon.


Need help finding a star I've tried utilizing "Google Sky", however, once I've found the location of said star and am in the process of zooming in, the site craps out.
Anyway... if there's someone out there with a little more experience in this field, here are the coordinates of the star i'm searching for:
10h 10m 11.1s, +4 3' 12.3"

I'd appreciate the effort, and thank's in advance.

As far as I can see, this is just a faint star. Here is some data from the USNO B1.0 catalogue:

USNO B1.0 star

B1.0 ID: 0940-0181715

RA: 10h10m11.13s +/- 69 milliarcseconds

declination: + 4 03' 12.3" +/- 195 milliarcseconds

Epoch: 1977.9

Proper motion in RA:  12 +/-       2 milliarcseconds/year

Proper motion in dec: 6 +/-       6 milliarcseconds/year

Photometry:

Band mag   Cal Survey  Emulsi field  S/G   xi    eta

B1: 14.36   1  POSS-I  103a-O 550     4   -0.05    0.03

R1: 13.33   1  POSS-I  103a-E 550     1    0.05    0.01

B2: 14.07   2  POSS-II IIIa-J 782     9    0.09    0.26

R2: 12.79   2  POSS-II IIIa-F 782     8   -0.07   -0.02

I:  12.37   2  POSS-II IV-N   781         -0.04   -0.30

"Cal" = calibration: 0 means calibrated via bright standard stars on
the plate; 1 = via faint standards on the plate; 2 = faint standards on
adjacent plate; 3 = faint standards two plates away,  etc.
  "S/G" = star/galaxy index. 11=object looked starlike, to 0=fuzzy/galaxy-like.
  xi, eta = displacement of the star's position on this plate,  in
arcseconds,  relative to the mean solution.

This USNO-B1.0 data was downloaded from a VizieR server.


Day Length on Double Planets Assume two similarly sized bodies tide-locked to one another orbiting a barycenter between the two. That barycenter orbits a star. Since the two are tide-locked, their sidereal rotational period and sidereal orbital period (about one another) are the same (right?). There's no axial tilts or eccentricities to screw with things either for the sake of simplicity.

What would I base the length of a conventional day on for a society living on these worlds?

My gut's telling me to calculate the sidereal period based of the barycentral semi-major axes (or would I use the separation between the two worlds still?) for each and then compute the synodic periods and just use those, but my brain's too stupid to tell whether that's going to actually be anything close to a "solar day". I don't need perfect - just close enough.

Yes, if the planets are tide locked to one another, their orbital period and sidereal day are the same. That's assuming their orbital plane about each other isn't much inclined to their orbital plane about the star.

From the Wikipedia article on orbital period:

$T_d=2\pi*\sqrt{a_p^3/(G*(M_1+M_2))}$

Where G is gravitational constant, $M_1$ is mass of 1st body and $M_2$ mass of 2nd body. $a_p$ is semi major axis of elliptical orbit of the planets about each other.

For the planet's year you'd use a very similar equation.

$T_y=2\pi*\sqrt{a_s^3/(G*(M_0+M_1+M_2))}$

Where $a_s$ is the radius of their orbit about the sun and $M_0$ is the mass of the sun. If you like, you can leave off $M_1$ and $M_2$ as their masses are probably neglible in comparison to the sun's mass.

I am guessing by "conventional day" you want the time an inhabitant sitting on his front porch would measure between one high noon and the next.

Solar day = sidereal day * (1 + (sidereal day/year))

For example our solar day is about 1/365 longer than a sidereal day.


Do parallel universes exist? Do other universes like ours exist?
If they exist, how do we know that they exist when we have even not seen the ends of our own universe?

The simple answer is (as with so much in astronomy):
We Don't Know

Parallel universes may or may not exist. There is no definitive way to prove that these universes do or don't exist.

A parallel universe is a separate existence to ours. The Theories that suggest that there may be parallel universes are classified as theories of multiverses. There are many theories of multiverse, all of which propose different ideas about what could exist beyond the limits of our universe. There are also theories that suggest that the multiverse doesn't exist, although the theories with most support are by far the multiverse theories.

For a nice reference in book form, see Steinhardt and Turok's "Endless Universe: Beyond the Big Bang". Also, see Max Tegmark's work on multiverses levels I-IV (Max Tegmark -> See his Scientific American article entitled Parallel Universes).


The size of the radius of the event horizon of a black hole created by the merger of a black hole binary system Let's assume that you had a black hole binary system and everything said here is possible. Their large masses would lead to a large emission of gravitational waves. The loss of orbital energy and angular momentum to this gravitational radiation should ultimately cause the two black holes to merge into a single black hole containing the combined mass of the merged black holes.

With this said, does a greater combined mass of the newly formed black hole mean that the radius of the new event horizon is greater than the radius of the event horizons of the singular black holes?
I don't know if there is a formula connecting mass to radius, or if this is essentially a conceptually hypothetical concept, and the answer is simply that the larger the mass of the black hole, the larger the radius of its event horizon

The "radius" (there is no physical surface) of the event horizon of a rotating black hole, depends on both it's mass $M$ and angular momentum $J$, and is given by the equation
$$ r = \frac{GM}{c^2} +\sqrt{\left(\frac{GM}{c^2}\right)^2- \left(\frac{J}{Mc}\right)^2}.$$

It is thus difficult (for me anyway) to give a straightforward answer to your question. When two black holes merge, they will each have their own masses and angular momenta, plus there will be angular momentum in the orbit. The gravitational waves emitted during the merger can take away mass from the system as a whole (e.g. the final mass of the first observed black hole merger was three solar masses less than the summed masses of the merging components).

So, in general yes, for non-spinning black holes the event horizon grows as the total mass. But if the merger results in a black hole with maximal spin, where $J = GM^2/c$, then the final event horizon could be half the size given by the Schwarzschild radius ($2GM/c^2$), even though the total mass is larger than the black holes that contributed to it.

However, I think what is certainly true, is that it is not possible to add mass (even in the form of another black hole) to a given black hole and for the event horizon to get smaller, whatever the angular momentum contributed by that mass (section 4.2 of "Black Holes", by Raine & Thomas, 2015, Imperial College Press).


Plotting Astronomical Data Using Python I have been trying to create a graph which shows the distances between the stars and the earth, yet I could not obtain the desired graph as seen below (click to zoom in):

My graph:

As @uhoh suggested, I overlaid one graph on top of the other.

I apologize for posting this confusing graph

The combined graph denotes that the positions of the curves in my graph totally differ from those in the expected graph.

I followed this guide to calculate the position functions of the stars.

My Question: Did I made any mistakes that resulted in the anomalies?

Thanks in advance.

Annotations for the Code:

Name - Name of the star
RA - Right Ascension in degrees, ICRS coord. (J2000)
Dec - Declination in degrees, ICRS coord. (J2000)
pm_mRA - Proper motion in right ascension, in miliarcseconds per year
pm_mdec - Proper motion in declination, in miliarcseconds per year
vr - radial velocity in kilometers per second, a positive value means that the star is moving away from us
mparallax - parallax of the star in miliarcseconds
d - distance between the star and the earth

My Code:

def parseTextFile(file_name, delimiter=",", header=0):
    """ Parse a text file to a list. The file contents are delimited and have a header. """

    with open(file_name) as f:

        # Skip the header
        for i in range(header):
            next(f)

        data = []

        # Parse file contents
        for line in f:

            # Remove the newline char
            line = line.replace('\n', '').replace('\r', '')

            # Split the line by the delimiter
            line = line.split(delimiter)

            # Strip whitespaces from individual entries in the line
            for i, entry in enumerate(line):
                line[i] = entry.strip()

            # Add the contents of the line to the data list
            data.append(line)

        return data

fig = plt.figure()
ax = fig.add_subplot()

#time span
time = np.arange(-60000,100000,10)
count = 1

xdic = {}
ydic = {}
zdic = {}

#multiple lines of data
for star in parseTextFile(file_name, header=1):

    name = str(star[0])
    RA = float(star[1])
    Dec = float(star[2])
    pm_mRA = float(star[3])
    pm_mDec = float(star[4])
    vr = float(star[5])
    mparallax = float(star[6])

    pm_RA = pm_mRA * 0.001
    pm_Dec = pm_mDec * 0.001
    d = 1 / (mparallax * 0.001)

    #Transverse velocities
    vta = pm_RA * d * 4.740
    vtd = pm_Dec * d * 4.740

    #Linear velocities
    vx = vr * np.cos(Dec) * np.cos(RA) - vta * np.sin(RA) - vtd * np.sin(Dec) * np.cos(RA)
    vy = vr * np.cos(Dec) * np.sin(RA) + vta * np.cos(RA) - vtd * np.sin(Dec) * np.sin(RA)
    vz = vr * np.sin(Dec) + vtd * np.cos(Dec)

    #unit conversion from km/s to pc/year
    vx_pcyr = vx / 977780
    vy_pcyr = vy / 977780
    vz_pcyr = vz / 977780

    #initial positions
    xi = d * np.cos(Dec) * np.cos(RA)
    yi = d * np.cos(Dec) * np.sin(RA)
    zi = d * np.sin(Dec)

    #position functions
    x = xi + vx_pcyr * time
    y = yi + vy_pcyr * time
    z = zi + vz_pcyr * time

    distance = np.sqrt(x ** 2 + y ** 2 + z ** 2)

    ax.plot(time,distance,label=name)

ax.set_xlabel('Time (Year)')
ax.set_ylabel('Distance (pc)')
ax.legend()
plt.show()

I forgot changing the unit of the distance from parsec to light year.
This is a simple unit conversion error which should have been avoided.

Ultimate Graph:

Data Used:

  Name,RA(deg),Dec(deg),pm_RA(mas/yr),pm_Dec(mas/yr),Vr(km/s),parallax(mas)
  Wolf 359,164.120271,+07.014658,-3842.0,-2725.0,19.321,418.3
  Proxima Centauri,217.42895219,-62.67948975,-3775.75,765.54,-22.40,768.13
  Alpha Centauri,219.900850,-60.835619,-3608,686,-22.3,742
  Barnard's star,269.45207511,+04.69339088,-798.58,10328.12,-110.51,548.31
  Gliese 445,176.92240640,+78.69116300,743.61,481.40,-111.65,186.86
  Luhman 16A,150.8218675,-53.319405556,-2754.77,358.72,23.1,500.51
  Sirius,101.28715533,-16.71611586,-546.01,-1223.07,-5.50,379.21
  Lalande 21185,165.83414166,+35.96988004,-580.27,-4765.85,-84.69,392.64
  Ross 248,355.479122,+44.177994,115.10,-1592.77,-77.715,316.7
  Gliese 65,24.756054,-17.950569,3321,562,29,373.70

Code:

%matplotlib qt
import numpy as np
from mpl_toolkits.mplot3d import axes3d
from matplotlib import pyplot as plt
from matplotlib import patches

def parseTextFile(file_name, delimiter=",", header=0):
    """ Parse a text file to a list. The file contents are delimited and have a header. """

    with open(file_name) as f:

        # Skip the header
        for i in range(header):
            next(f)

        data = []

        # Parse file contents
        for line in f:

            # Remove the newline char
            line = line.replace('\n', '').replace('\r', '')

            # Split the line by the delimiter
            line = line.split(delimiter)

            # Strip whitespaces from individual entries in the line
            for i, entry in enumerate(line):
                line[i] = entry.strip()

            # Add the contents of the line to the data list
            data.append(line)

        return data

if __name__ == "__main__":

    file_name = 'C:\\Users\\The Wings of Dream\\Desktop\\UWO-PA-Python-Course\\Lecture 5\\Homework 2\\star_data.txt'

#Program Begin:

fig = plt.figure()
ax1 = fig.add_subplot(211, projection='3d')
ax2 = fig.add_subplot(212)

time = np.arange(-60000,100000,10)
count = 1

xdic = {}
ydic = {}
zdic = {}

for star in parseTextFile(file_name, header=1):

    name = str(star[0])
    RA = float(star[1])
    Dec = float(star[2])
    pm_mRA = float(star[3])
    pm_mDec = float(star[4])
    vr = float(star[5])
    mparallax = float(star[6])

    pm_RA = pm_mRA * 0.001
    pm_Dec = pm_mDec * 0.001
    d = 1 / (mparallax * 0.001)

    vta = pm_RA * d * 4.740
    vtd = pm_Dec * d * 4.740

    vx = vr * np.cos(Dec) * np.cos(RA) - vta * np.sin(RA) - vtd * np.sin(Dec) * np.cos(RA)
    vy = vr * np.cos(Dec) * np.sin(RA) + vta * np.cos(RA) - vtd * np.sin(Dec) * np.sin(RA)
    vz = vr * np.sin(Dec) + vtd * np.cos(Dec)

    vx_pcyr = vx / 977780
    vy_pcyr = vy / 977780
    vz_pcyr = vz / 977780

    xi = d * np.cos(Dec) * np.cos(RA)
    yi = d * np.cos(Dec) * np.sin(RA)
    zi = d * np.sin(Dec)

    x = xi + vx_pcyr * time
    y = yi + vy_pcyr * time
    z = zi + vz_pcyr * time

    xdic['x'+str(count)] = x
    ydic['y'+str(count)] = y
    zdic['z'+str(count)] = z

    distance = np.sqrt(x ** 2 + y ** 2 + z ** 2) * 3.26156

    ax1.plot(xdic['x'+str(count)],ydic['y'+str(count)],zdic['z'+str(count)])
    ax2.plot(time,distance,label=name)

    count = count + 1

w_oort, h_oort = 160000, 3.2
ax2.add_patch(patches.Rectangle((-60000, 0.03), w_oort,h_oort,color='slateblue',alpha=0.2))
ax2.annotate('Oort Cloud', xy=(15000,1.6), size=12)

plt.axvline(x=0,color='gray',linestyle='--',linewidth='0.5')

#plotting constraints
ax2.set_ylim(0.0, 10.0)
ax2.set_xlim(-60000, 100000)

ax1.set_xlabel('x axis')
ax1.set_ylabel('y axis')
ax1.set_zlabel('z axis')
ax1.title.set_text('Motion of Stars in Space')
ax2.title.set_text('Distance-Time')
ax2.set_xlabel('Time (Year)')
ax2.set_ylabel('Distance (Light Year)')
ax2.legend()
plt.show()

ax1 subplot will give you a 3d parametricplot showing the motion of the stars.


Solar system data API? I'm prototyping a real time solar system / milky way discovery app.

Could you please help me on:
-What is the best orbital data public API I can use?
-What is the standard numerical way to define orbit?

I'm looking for something like JPL Small-Body Database Browser, but for every celestial objcets, and / or missions.

Here is a short answer. @barrycarter's helpful comment links to a much larger list of resources. Also, you may find answers below [this question] helpful.

There are Spice kernels for many (but not all) exploratory spacecraft as well, you can view them through the Horizons interface or find them via ftp.

For most objects in Earth orbit, Two Line Elements (TLEs) from Celestrak or Space-Trakck are useful. See this answer as well.


How much lighter would someone be on the Jupiter-facing side of Io? Io is the Galilean moon closest to Jupiter. Its average surface gravity is 0.183g and it has a tidal lock with Jupiter. So if you stood on the near side of Io at the equator, how much lighter would you be compared to average (say you weigh 18.3 pounds on Io, how much would you weigh due to Jupiter's gravity on Io's near side)? Jupiter has 318 Earth masses, this is why I think you'd be quite lighter on Io's near side on the equator.

On the leading or trailing points of IO, you are orbiting Jupiter at the same speed as the center of mass of Io, so you experience its gravity only
On the equator, facing Jupiter, you are 1821.6 km closer to Jupiter, but still orbiting with the same period. So there is a bit of Jupiter's gravity that is not countered by Io's orbital speed.
How much?
Gravity from Jupiter is more by: Assume 1kg at Io core, vs 1kg at io-jupiterface
G(core) = 0.71813N
G(jupiterside) = 0.7244N
Differnece = 0.00627N
In addition, you have less centrifugal force from the orbit, to be being in an orbit that is "smaller" by 1821.6km but has the same period.
Centrufugal at Io(core) = .71813N (as expected, at Io's center of mass, its orbital centrifugal force matches Jupiter's attraction, hence an orbit)
Centrifugal at IO(jupiterface) = 0.7151N
Thus the difference in perceived weight of a 1kg mass is
(.71813-0.7151) + (0.7244 - 0.71813) = 0.0093N
Which is a difference of 1.295% , or 1 part in 77


Requirements to resolve position of Jovian Whistlers up to magnitude of Red Spot with amateur radio equipment? Can I sense a bright star pointing an eight foot antenna towards it? is a while ago, but the discussion about this question I found very interesting. Please forgive me that I am not an expert in Jovian Whistlers but I tend to understand it as massive EM events on Jupiter.
I am wondering: Could we use a Yagi-Uda antenna or an array of it to resolve the source of the radio signals on Jupiter? For instance, we could have the Jupiter-disk covered by a square of $40 \times 40$ pixels, and the goal would be to figure out from which pixel the signal originated.
The idea behind that is that a cool research question for an (advanced) science project in school would be a question like "Did the lightning occur inside the Red Spot?" and Yagi-Uda antennas are fun to build, and do not require as much space as other antenna types.
References

Strong whistler mode waves observed in the vicinity of Jupiter’s moons

tl;dr
No, not from Earth.
Because of the ionosphere.
Whistlers are VLF (very low frequency) hundreds to thousands of Hertz. Our ionosphere blocks electromagnetic radiation very reliably below say roughly 10 MHz depending on time of day and solar activity because it's a plasma.
The image below shows peak whistler emissions at about 1 kHz.
You can imagine that the electrons in the ionosphere look more like electrons in a lossy metal to a radio wave in the kHz range; some will bounce off like it's a shiny metal marble, others will be absorbed, but without question None shall pass!
LOFAR works from 10 MHz to 250 MHz.
See also

Why did it take five years to "figure out" how to use astrometric calibration sources to deblur LOFAR images?
What can be learned from low frequency radio astronomy available outside of Earth's ionosphere?

Update!
No, not even from cis-lunar space!
Because the interplanetary electron density is still too high.
@ProfRob's comment on the "What can be learned..." question reminds us:

How low is "low"? The plasma frequency in the interplanetary medium near the Earth is about 100 kHz. https://physics.stackexchange.com/a/519164

and that answer goes into substantial detail on how to calculate the plasma frequency of various media.
The Chinese lunar orbiter Queqiao had a low frequency receiver, and backing up ProfRob's point:

Additionally, this satellite hosts the Netherlands–China Low-Frequency Explorer (NCLE), an instrument performing astrophysical studies in the unexplored radio regime of 80 kilohertz to 80 megahertz. It was developed by the Radboud University in Netherlands and the Chinese Academy of Sciences. The NCLE on the orbiter and the LFS on the lander work in synergy performing low-frequency (0.1–80 MHz) radio astronomical observations.

There are further references in the article.
It seems you must build a spacecraft and fly it into close proximity to Jupiter -- within a few attenuation lengths -- to hear the whistlers.
More
Your reference Strong whistler mode waves observed in the vicinity of Jupiter’s moons links to the Open Access Nature paper Y. Y. Shprits et al. (2018) Strong whistler mode waves observed in the vicinity of Jupiter’s moons which includes Figure 2. shown below.
From Whistlers (radio):

A whistler is a very low frequency or VLF electromagnetic (radio) wave generated by lightning.[1] Frequencies of terrestrial whistlers are 1 kHz to 30 kHz, with a maximum amplitude usually at 3 kHz to 5 kHz. Although they are electromagnetic waves, they occur at audio frequencies, and can be converted to audio using a suitable receiver. They are produced by lightning strikes (mostly intracloud and return-path) where the impulse travels along the Earth's magnetic field lines from one hemisphere to the other. They undergo dispersion of several kHz due to the slower velocity of the lower frequencies through the plasma environments of the ionosphere and magnetosphere. Thus they are perceived as a descending tone which can last for a few seconds. The study of whistlers categorizes them into Pure Note, Diffuse, 2-Hop, and Echo Train types.
Voyager 1 and 2 spacecraft detected whistler-like activity in the vicinity of Jupiter known as "Jovian Whistlers", supporting the visual observations of lightning made by Voyager 1

Figure 2 from the Nature paper linked above.

Comparison of wave power and pitch angle distributions for the encounter on 6 September 1996 (orbital segment G2). a The distance between Ganymede and the Galileo spacecraft, b measurements of the magnetic field, c pitch angle distribution of 527–884 keV electron fluxes, d same as c but normalized by the value of fluxes at 60° local pitch angle, e dynamic spectrogram of electric field very low frequency (VLF) spectral density, f dynamic spectrogram of magnetic field VLF spectral density


How are the gaps filled in images captured by CCD arrays? Large telescopes don't use a single CCD; they use arrays of them. Since CCDs can't be seamlessly joined, this means that every captured image must have gaps. But, the resulting images almost never have gaps when they are presented to the public.

How are these gaps filled? Does the telescope capture a number of overlapping images and then merge the results? Wouldn't this result in differing exposure levels for differing areas of the image, which, even though it could be compensated for in the processing, would give different noise characteristics to the different areas?

In general, the common practice is to dither exposures and fill in the gaps. The dither serves multiple purposes:

provides some exposure in the gaps between the CCDs
smooths out pixel-to-pixel variations in the response.
prevents one bad pixel from ruining an entire observation

The dither pattern varies from telescope to telescope. For example, the following image shows the dither pattern of Chandra telescope.

Image from havard.edu

The effect of dither is removed during high-level ground processing of the data.


What is the size of the universe.is it infinite? What is the exact size of the universe.is it infinite?

"No one knows if the universe is infinitely large, or even if ours is the only universe that exists." - NASA

Still, the "observable universe" is a sphere roughly 92 billion light-years in diameter.


What are some night sky objects I could see with my Celestron UpClose 20x50 Porro Binocular? I bought this binocular because I've read countless times that one should start with a pair of binoculars before diving into telescopes. I've seen details of the moon and I can notice Venus round shape on some nights. I also saw Jupiter and it's two larger moons, as faint as the faintest stars, but this was on a beach trip so the sky was clearer than what I'm used to.

I live on a place with a 7 or 6 on the Bortle Scale, is there something else I should try to see?

Depending on the season you can try different objects. Try large star clusters and nebulae:

The Pleiades
Orion nebula
Andromeda galaxy
M13 globular cluster

Also double stars like epsilon Lyrae.

It helps to have a foto tripod and an adaptor for the binoculars. Gives much better view when the binoculars are resting on the tripod.


Why don't globular clusters flatten with a galactic disc? Globular clusters lie in the galactic halo, outside of the disc. However, galaxies are more or less a collection of material and objects — why is it, then, that most stars form a plane due to the angular momentum, but some patches of stars do not?

In fact, globular clusters often contain some of the earliest stars formed in a galaxy, right? How come they do not flatten, yet a lot of the older stars do?

Globular clusters formed whilst the gas of the proto Milky Way was still approximately spherically distributed.

The gas forms a dissipative system that loses energy and collapses (within the first billion years) to a disk whilst conserving angular momentum.

Formed stars and clusters are essentially collisionless so the halo stars continue to have a spherical distribution, whilst more recently formed stars formed from gas already in a disk and so continue to be in a disk.


Is Earth's Surface "In orbit"? I'm having trouble understanding relative angular/tangential speeds at increasing altitudes above Earth's surface. In particular, I find this comparison of tangential velocities on Wikipedia very confusing. According to it, the tangential speed of Earth's surface (465.1 m/s) is different from the tangential speed required to "orbit" at Earth's surface (7.9 km/s). Why are these different values? My understanding of the Earth was always that material on and within the Earth is orbiting the center of the Earth just like satellties do. Time for multiple questions in one post...

Is material on Earth's surface not in free fall around Earth's center?
How are geostationary orbits even a thing? Seems like the only orbit that could be geostationary would be standing on the Earth's surface.
What changes as you orbit further above the Earth's surface? Does your angular speed increase or decrease? Does your tangential speed increase or decrease?
Is magma near the center of the Earth not rotating faster than material in the crust, like in an accretion disk?
Can two objects be orbiting (circularly) at the same altitude but with different tangential speeds?

Thanks in advance!

1. Is material on Earth's surface not in free fall around Earth's center?

No. Material on the Earth's surface -- or inside it -- is not in orbit, and so is not in free fall. You can temporarily put yourself into an orbit (and thus into free fall) by jumping up into the air, or jumping off a higher surface. When you do this, you are briefly in a very eccentric orbit (one which would take you very close to the center of the Earth, if the Earth wasn't a solid body) -- but then you hit the ground and are no longer in orbit.

The Earth rotates in the same way that a spinning top rotates; this has nothing to do with orbits.

2. How are geostationary orbits even a thing? Seems like the only orbit that could be geostationary would be standing on the Earth's surface.

Again, the surface of the Earth is not orbiting. The Earth rotates as a rigid body, with (as AtmosphericPrisonEscape noted) residual angular momentum left over from its formation, like a spinning top.

Because your angular speed in an orbit decreases the further away you are from the Earth, there will be a point where it happens to match the Earth's spin rate. If you arrange the orbit so it is above the equator and in the same direction as the Earth's spin, then you will always be above the point on the equator: a geostationary orbit.

3. What changes as you orbit further above the Earth's surface? Does your angular speed increase or decrease? Does your tangential speed increase or decrease?

Both your angular speed and your tangential speed decrease the further away you get. (Your angular speed would decrease even if your tangential speed stayed the same, because the circumference of your orbit increases with altitude; but in fact the tangential speed decreases as well.)

4. Is magma near the center of the Earth not rotating faster than material in the crust, like in an accretion disk?

The Earth rotates approximately as a rigid body, so in general, no. The molten outer core (which is not magma) may rotate slightly slower, while the solid inner core might rotate a little faster, but we're talking about $\sim 0.1$ degrees per year differences, and this has nothing to do with orbits. (The Earth is nothing like an accretion disk.)

5. Can two objects be orbiting (circularly) at the same altitude but with different tangential speeds?

Ignoring minor deviations due to things like the non-spherical nature of the Earth, mass concentrations in the crust, etc., the orbital speed for a circular orbit is a function of the altitude only. So two objects in circular orbit at the same altitude must have the same tangential speed. (Note that they can have different velocities, because velocity is a vector quantity -- so you can have two object orbiting in different -- even opposite -- directions at the same altitude, at least until they run into each other.)


Will the 2016 Perseid meteor shower be visible in Sydney, Australia? As the title states, will viewers in Sydney, Australia be able to see the Perseid meteor shower in August, 2016? Assume the viewer is well outside of the city and light pollution is a minimal problem.

According to Australian Geographic, the event will be visible in the northern parts of the country – specifically, "head north for the best chance to witness the event" – but it is not clear whether it will be possible to see anything in the southern parts.

If any viewing is possible, what would be the recommended dates and time of night to see the most?

The International Meteor Organization calendar for 2016, page 12, gives the Perseid radiant declination as +58$^\circ$ and says "regrettably, it cannot be properly viewed from most of the southern hemisphere."
At latitude -34$^\circ$ the radiant never rises above the horizon, so there may be fewer Perseid meteors than sporadic non-Perseids.
This does not change year to year as the AG article seems to suggest.
What does vary is interference from the Moon; this year it's full on the 18th, so the article is correct to recommend the morning or two before the predicted peak on August 12-13.
Also check your local weather forecast.
The predawn hours, e.g. 2 hours before sunrise, are generally best for meteors because the morning side is forward as the Earth orbits the Sun.


Is there any plausible astronomical source of energy that could wipe out humanity without rendering the planet inhabitable? I'm aware of hard radiations from space can wipe out life but I believe that would also causes the planet to be inhabitable for a period of time. Is there any other kind of energetic astronomical event that could destroy life through a transient pulse of radiation (particles or photons) without any "cool down" needed for new life to develop?

The event could be known or theoretical or just plausible that spans entire galaxy if necessary.

The cockroach and the tardigrade can withstand just about anything.  Next, any radiation blast consisting of wavelengths reasonably attenuated with water wouldn't affect, directly at least, all the deep-sea creatures.


What would happen if a star passed through the barycenter of two binary stars? I was doing some research on Alpha Centauri A and B, and thought to myself: 'What would happen if something passed directly through the barycenter of these two stars?' I may theorise that a passing star will offset the barycenter with its gravity, sending the two stars in whatever direction it would go. I am not certain about this, nor am I certain that anyone has the answer. But if they do, please answer, I'd really like to know!

If we assume the binary pair have equal masses, then by symmetry a star passing through their barycenter at right angles to their rotation plane will not change their symmetry.

I did some simple simulations where a star with mass 0.1 flies through a binary in an originally circular orbit.

The first case is with an interloper moving at speed 20. It passes through, losing a bit of momentum to the binary, that now moves slowly in the same direction.

The second is an interloper moving more slowly, with speed 1. Now the binary moves along more, and the interaction causes a quasiperiodic oscillation in seimajor axis.

Just dropping the interloper with zero velocity makes it bound to the binary, oscillating back and forth, driving complex quasiperiodic or chaotic motion.

It is possible to make the pair scatter by throwing in a sufficiently heavy interloper with enough momentum that first drags the binaries close, transfers enough momentum to unbind the system, and then passes along.

Of course, if the direction is not exactly symmetric things can turn messy. Note that there are non-chaotic solutions too.

To sum up, interlopers can deposit energy and momentum, split binaries, or create chaotic tangles. But it is not the barycenter that is the important part.


Mars looks like a blur I took my telescope out tonight (I am new, just got the telescope a few days ago, it's my first one) and I was observing Mars, however all I see is just an bright orange circle, with no surface detail, regardless of magnification. The telescope I'm using is a 8in Newtonian, 1200mm focal length. Here's what I saw with a 8mm eyepiece and 2x Barlow:

Mars was about 32° above the horizon when I took this image.

Did I just do a poor job with collimation? Is the the observation angle? Please let me know!

There's another answer here claiming that an off-axis aperture mask will improve things. That is a fairly popular meme, but it's entirely incorrect, and makes misleading claims about the effects of the central obstruction. All that does is it hides existing problems with the instrument.

If the off-axis hole appears to improve things, something else is very wrong and you need to fix that first; it's not a solution, it's a symptom pointing at serious issues. More importantly, it reduces precious aperture, therefore capping the resolving power of the instrument in a situation (observing Mars) when you need all the resolving power you can get. Please don't do that. Ignore the off-axis mask altogether.

Now, what to do to get better results with Mars?

Collimation

Make sure your telescope is in perfect collimation. There is no one technique to rule them all. There are many ways to collimate a telescope. Whichever method you choose, make sure you actually use it before observing Mars. Small deviations from perfect collimation will affect results on this difficult target.

Your instrument's manual, or the manufacturer's website should provide some info. Here are some links to get you started:

https://garyseronik.com/a-beginners-guide-to-collimation/

https://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/

Some info about the importance of collimation:

http://www.astrophoto.fr/collim.html

Regarding laser collimators, which are very popular: A laser collimator is very quick and can be very good, but provided the laser itself is perfectly centered. Plug the laser into the focuser, watch the laser spot on the primary mirror, and slowly turn the laser like a knob - if the spot moves in a circle as you turn the laser, then it's not centered. Some lasers can be adjusted until they are centered (2 or 3 adjustment screws on the barrel), others can't. A laser that's not centered will not provide good collimation for your telescope.

There are many other collimation techniques, keep learning.

You don't need to be in perfect collimation every single time - a lot of faint fuzzies (galaxies, etc) don't require any careful collimation. But for Mars and the Moon you better do it right. Don't obsess over it, but try and do a good job. Personally, I at least check collimation every time no matter what, but with practice now it only takes me 5 minutes to fix any deviation, so it's not a chore.

Thermal equilibrium

If your telescope is not at the same temperature as the air, convection vortices will appear inside which will affect performance. Take the instrument outside at least 1 hour before the observation and let it vent out all the heat.

Some reading material:

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

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

Seeing

This is basically atmospheric turbulence. When it's bad, the image is distorted and fuzzy. When it's good, the image is clear and you can see small details. Basically, Mars is impossible to watch if seeing is not good or very good. Sometimes, spells of bad seeing can last for days or weeks. There's not much you can do except put the scope in the car and drive someplace else where seeing is good at that time.

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

Seeing is basically weather-related, so it can be predicted to some extent. Go on the Clear Dark Sky website, search for a location close to your home, and pull up the chart:

http://www.cleardarksky.com/csk/

Look at the row called Seeing. If it's dark blue for the time when you plan your observation, chances are good that seeing will not get in the way. If it's light blue or white, the forecast is not good.

Opposition

Mars is only worth trying around oppositions, when it's closest to Earth. This happens roughly every two years.

The previous opposition was in July 2018 when Mars had an apparent size of 24 arcsec.

Currently (November 2018) the apparent size of Mars is 11 arcsec, less than half the size at opposition.

The next opposition is in October 2020, when it will be about 22 arcsec.

A few weeks before and after opposition Mars is worth trying.

https://www.wolframalpha.com/input/?i=mars+apparent+size+in+october+13+2020

Focus

Make sure the scope is perfectly focused. Point it at a fixed star (Polaris is quite convenient since it does not move), and adjust the focuser until the image of the star is as small as it can be.

It's pointless to try fine focusing if the scope is not collimated (see above). Collimate first, then you'll be able to achieve a sharper focus.

Also, if there are seeing issues, or thermal issues, a sharp focus will be hard to achieve.

Other factors

Mars is a difficult object to observe. Good quality optics are important. Aperture is important (more is better - please disregard the memes about limiting the aperture). Experience is important - you'll get better in time.


Why solar eclipse paths are symmetrical? Reading upon the eclipse of March 20, 2015, I stumbled upon this page: http://www.timeanddate.com/eclipse/list-solar.html. What caught my eye is that for each year there are two eclipses whose paths are almost symmetrical relative to equator.

I'm just curious why that's always exactly so.

The orbit of the moon is inclined by 5.14° to the ecliptic.

As you may know, the ecliptic is the apparent path of the Sun across the sky, so it is also the plane of Earth's orbit around the Sun. The plane of the Moon's orbit is inclined by 5.14° to the plane of the Earth's orbit. The intersection of the two plans is a line bisecting both planes.

There is a nice picture of this at http://www.cnyo.org/tag/orbital-plane/. [I'll insert it here after checking the copyright]

So, it is only when the Earth is very near to the intersection line that an eclipse can occur, because it is only then that the Sun, Moon and Earth can be in a strait line. This occurs for slightly more than a month (called an eclipse season), twice a year.

Another angle comes into the picture too. The axis of the Earth is tilted, so the equator is at 23.4° to the plane of the ecliptic.

So, the pattern you have observed comes about because of that geometry, where mirror image eclipse paths occur about six months apart.

The linked pages explain a lot more of the complexities.


Does our universe have to be embedded in a higher dimensional space? I've heard that our universe may be open or closed.  If it's closed it might have a toroid shape.  If this is the case, would that imply that our 3 spatial dimensions have to be embedded in a higher dimensional (spatially) space?  For example, the old asteroids video game was apparently a torus mapped to a 2-dim video screen, and it stayed in 2 dimensions, but with a specific behavior of moving objects.

The universe being open, closed, or flat only determines the type of geometry one must use to describe distances (and time). For open and closed geometries, Euclidean Geometry is not what one should use. I would also agree that our universe being open, closed, or flat has nothing to do with the number of dimensions it contains.

At present, there is no direct evidence that there are any additional dimensions over 3+1 dimensions (this just means three spatial dimensions and one time dimension). However, many GUT theories do include additional spatial dimensions in an attempt to unify all of the forces.

Also at present, we have very good evidence that the universe is 'flat'. What this means is that the angles of a triangle have to add up to $180^{\circ}$ and distances are measured in the standard Euclidean way. When we talk about the universe being flat, this is a purely global statement. Locally, however, it is completely possible to live in curved space. We actually do live in curved space. The mass of the Earth is curving space and time in a way that General Relativity predicts, and therefore clocks run very slightly differently depending on where you are on the Earth's surface, and distances are very closely approximated by Euclidean distances, though they're not.

How do we actually determine that the universe is globally flat, you may ask? We use what's known as a standard ruler. Much like a standard candle, if we think we understand the physics, any deviation from what we would predict gives us new information about the universe (in the case of supernovae, it's that they can be used to measure distances, and therefore map out the expansion of the universe). We use the angular size of fluctuations in the Cosmic Microwave Background Radiation (we think we understand the physics behind the fluctuations fairly well) to test how much, if at all, the universe deviates from flatness. The latest results from Planck show very good agreement with the standard picture provided to us by the standard LCDM cosmological model.

Below is the power spectrum of the temperature fluctuations of the CMB. The location of the first peak is what cosmologists use to measure flatness.


Why can't we detect water on Mars in the same way as in exoplanets? Often I see news articles claiming astronomers found exoplanets with water. If we are capable of detecting water on such distant objects, why don't we employ similar methods to find water on Mars or Titan moon?

We can look up for water in the atmosphere of the planet, not at the planet itself. We can do that when a planet passes in front of a star. That tehnique is called transmission spectroscopy.


Radioactive decay is the main source of the earth's interior heat: fact or theory? It's believed the main reason that the interior of the Earth hasn't cooled down in 4 billion years is because of the ongoing decay of radioactive elements that keeps adding more heat than is lost to the various cooling processes. Is this a proven fact or are there other possibilities?

There are really only six ways you could heat the interior of a planet. Gravitational collapse and compression, fission, fusion, crystallization, a very large impactor, and tidal flexing.

Fusion is out because Earth would be a rock-star. (Been waiting to use that one.)

Of the remaining possibilities, nuclear fission explains the majority of Earth's current temperature gradient.

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


What is a "differential chemical abundance"? Probably this is a extremely basic question, but after a lot of time searching about it without results (in all the sites/publications i've found they only use this term but there is no specific definition)i came here with the hope of get an answer to my question: in the context of chemical abundance in stars, what is a differential abundance?

I already know what is a chemical abundance, the question is simply about the implication of the "differential" word.

Specifically, my doubt emerged when i was reading this publication.
In the second paragraph of the section 3.2 "Stellar Parameters" is the first encounter with this concept:

  We adopted the same line list as in Ramírez et al. (2014). The
  EW  measurements  were  used  to  obtain  elemental  abundances
  with the
  abfind
  driver of MOOG (Sneden 1973), using Kurucz
  model atmospheres (Castelli & Kurucz 2004). Then, line-by-line
  differential abundances were obtained.

  The stellar parameters
  $(T_\text{eff}, \log g, {\rm[Fe/H]}, v_t)$
  of HIP 68468
  were determined by imposing a differential spectroscopic equilibrium  of  iron  lines  relative  to  the  Sun  (e.g.  Meléndez  et  al.
  2014a), using as initial parameters the values given in Ramírez
  et al. (2014). The solar values were kept fixed at
  $(T_\text{eff}, \log g, v_t)
= \rm
(5777\,K, 4.44\,dex, 1.0\,km\,s^{-1})$. Our solution for HIP 68468
  gives consistent differential abundances for neutral and ionised
  species of different elements, and also for atomic and molecular
  lines, as discussed in more detail in the next section.

If the line emission is optically thin, then the measured line flux is proportional to the abundance $F_{\nu} \sim n$ of a species along the line of sight (if it's only one species).

The problem here is that $F_{\nu}$ needs to be measured relative to a baseline. If that fails, e.e. when the stellar signal is too noisy, one doesn't know what this should be. But a measurment at two different wavelengths $\nu$ and $\nu'$ that lie reasonably close together, can at least yield a difference in abundances $F_{\nu}-F_{\nu'} \approx n-n'$ relative to this unknown base level.
This allows then to construct an atmospheric model, up to the unknown base level, which one can try to fix with other physical arguments.


Determination of orbital elements for Trans-Neptunian Objects, how? How are orbits calculated for Trans-Neptunian Objects?  I understand there are six standard parameters for defining an orbit (link) but I am not sure how telescope observations translate into these parameters.  Telescope observations tell us where the object is on the Celestial Sphere but can't even tell us (without a good spectroscopic line to do a redshift analysis with) how fast the object is moving towards or away from us.

Additionally, the motion of these objects on the sky is dominated by Earth's motion relative to the objects rather than the motion of the objects themselves.  This seems to me to offer additional complications.

My question is: how do 2-D observations get fit with a reasonable degree of accuracy to a 6-D parameter space with the additional complication of Earth's motion relative to the object?

Don't be afraid to be mathematical in your response.

You've asked a big question, too big perhaps for a Q&A forum such as this. Your question is the sole subject of graduate level aerospace engineering classes, e.g., University of Colorado ASEN 5070, Introduction to Statistical Orbit Determination, and is the subject of multiple graduate level texts, e.g., Statistical Orbit Determination by Bob Schutz, Byron Tapley and George H. Born. To have a chance in that class, you'll need to already be well-versed in multivariate calculus, linear algebra, probability and statistics, numerical methods, and computer programming.

That said, you might want to look at Bernstein and Khushalani (2000), "Orbit fitting and uncertainties for Kuiper belt objects," The Astronomical Journal, 120.6:3323.

The preferred approach in statistical orbit determination is to collect data over multiple orbits. That's a luxury that is not possible with Trans-Neptunian Objects that have only been seen a handful of times, and only over a small arc of the multi-hundred year long orbits of those objects.

One thing Bernstein and Khushalani did to overcome this was to realize that TNOs are nearly inertial (Newton's first law) objects. Gravitation is but a small perturbation of inertial behavior at such distances. Another thing they did was to take advantage of the fact that for observations separated by a short span of time (e.g., a day or two), almost all of the apparent motion is due to the Earth rather than proper motion of the target object. This gives a good estimate of the distance to the target.

Their approach involves doing some of the regression in Cartesian space, with gravitation being a small perturbation, and then switching to orbital element space to complete the regression. Along the way, they worry about whether they have enough information to do the full orbital element space regression.


A real light source, or an artifact? I was taking shots of the 2017 eclipse using my D7200 camera with a 1000 ND filter. As I was going through the pictures after downloading them to a pc, I noticed that there's what appears to be a tiny light source present off to the side of the sun. I have circled it in the picture below. It is also present in the pictures during totality, which I took without the ND filter.

It doesn't seem like a camera or a lens artifact, since it showed up pretty consistently in different shot configurations, but I did not know that a 300 mm zoom with an 1000 ND filter could capture a light source besides the sun and the moon in full day light. If it is not an artifact, what could be the source of this light?

I agree with the comment by StephenG. This is a stuck pixel. As the link suggests, they are a very common flaw in digital cameras, and not an astronomical feature.

I think this because it appears, on zooming in, to be a single pixel defect, most light sources even with the best focus, cover more than 1 pixel. The filter you are using is very powerful removing nearly all light. An object would have to be exceptionally bright to be seen through such a filter. If it was real then it would be so bright that when you removed the filter, that one spot would have totally overexposed your image. Since stuck pixels are pretty common, and you might not have noticed them in a picture taken at day, or of a star-field, it is a safe bet that that is the problem.


What is the Inner Horizon of a rotating black hole? My understanding is that any object with non-zero angular momentum which undergoes gravitational collapse to a black hole will become a rotating black hole. In reality all black holes will be rotating black holes because all objects undergoing gravitational collapse will have some angular momentum (even if it is relatively small).
I understand that the inner structure of a rotating back hole contains:its boundary i.e the event horizon, the “inner horizon” and the singularity which is ring-shaped. Once an object crosses the event horizon it has passed the point of no return and the singularity is a one-dimensional ring of infinite density. However,  I’ve been unable to locate  an article explaining exactly what the inner horizon is which is understandable to someone who has some mathematical background, but without a degree-level knowledge. For example, entering “Inner Horizon” into Wikipedia redirects to the following:
https://en.wikipedia.org/wiki/Cauchy_surface#Cauchy_horizon
Which with my lack of detailed mathematical background is incomprehensible!
I have found that many of the popular science articles about black holes make statements which are unsatisfactory and  and raise a lot of other questions .
E.g. https://www.discovermagazine.com/the-sciences/peering-inside-realistic-black-holes states
“To visualize the inner horizon, consider this thought experiment: Imagine you could enter a spinning black hole and cross the event horizon while holding a flashlight that’s pointed outward. That light would eventually freeze along, or within, a spherical surface. That’s the inner horizon, ….. the point where the black hole’s gravity is strong enough to keep light from leaving, but not strong enough to drag it all the way into the center. The trapped light becomes so energized that it turns into a firewall of radiation, incinerating anything that passes through it.”
Does anyone know of an article explaining exactly what is the inner horizon is to the educated layman? Or would be able to explain themselves  by answering this question? I would like to know what happens when an object crosses the inner horizon and WHY?

This is an attempt at a completely non-mathematical description although I'll use the proper terms for things.  Caveat: it may be wrong: my memory of this is rather distant now and I haven't spent much time trying to sort out the details.  I would welcome corrections (and/or a better answer!).
This answer also started with the intention of being a brief sketch and has turned into something much bigger.  It is as a result very thinly referenced: sorry for that.  I will try to make it better.
Some definitions
The first thing to think about is what sort of paths through spacetime do massive objects take?  The answer is that they travel on what are called 'timelike curves', and, more precisely future-directed timelike curves.  A future-directed timelike curve is simply a trajectory that some massive object, like a person, say, could take through spacetime (you actually need to be fussy about this and insist that the curve is smooth and probably a few other things).  Things which are massless, like light, can travel on 'null curves' and there are also 'spacelike curves' which nothing travels on: null curves represent the boundary between timelike and spacelike curves in some sense.
Future-directed timelike curves are often called 'worldlines', because they are the curves corresponding to the paths objects like people and rockets take through spacetime.
So you can think of spacetime as being full of these timelike curves: each curve represents some possible future for an object starting from a given point in spacetime (and also, going the other way, some possible past for an object which ended up at this point in spacetime).  A 'point in spacetime' is an event: its both a position in space and time.  Spacetime is made of events.
There are special subsets of these families of curves called geodesics, which are the curves that things which experience no acceleration travel on.  A timelike geodesic is the curve a stone would follow (if you threw it in a vacuum and there were no other forces on it...).  Geodesics don't really matter below.
Timelike curves have lengths, and the length of them is the time experienced by an object following them, which is called the 'proper time'.  Null curves have zero length (yes, this is fine).  Geodesics are extrema of length: the geodesic (or geodesics) between two points are the longest curves between those two points.  This is like the definition of a straight line in ordinary geometry but longest rather than shortest.  It's also how the twin 'paradox' works, of course.
Schwarzschild
So first of all, let's look at a Schwarzschild black hole: one that is not rotating.  It has one special surface, which is the event horizon.  The important thing about the event horizon is that any future-directed timelike curve which passes through it will meet the singularity, and will do so after only finite proper time.  What this means is that, once you've passed the horizon there is no way back through it, and at some time after passing through it (time from your point of view, so proper time) you will hit the singularity.  (Caveat: in GR, in real life you presumably meet whatever quantum thing actually replaces the singularity).  This holds for all timelike curves, not just geodesics: you can have a rocket which is as powerful as you like and it doesn't help (although the acceleration from it will kill you).
Kerr
OK, the Kerr solution is more complicated: it has three special surfaces, two of which are horizons.
The static limit
The first special surface is what's called the 'static limit'.  To see what this is you need to know about a weird thing in GR called 'frame dragging'.  Frame dragging is a feature by which rotating objects in GR (such as planets, stars, anything) tend to 'drag' the spacetime around them so that objects near them 'want' to rotate with the large rotating thing.
Well, for a Kerr BH there's a region where, in order not to rotate with the object, you would need to be going at the speed of light.  Or, in other words, all timelike curves within this region get wrapped around the BH.  Within the static limit you can't stay still: you must rotate with the BH.
But the static limit is not a horizon (it does touch the outer horizon at the poles): you can pass into and out of it.  The region between the static limit and the outer horizon (next) is called the 'ergosphere' of the BH.  (Alternatively, the static limit may be called the ergosphere and the region between it and the horizon the 'ergoregion'.)
The outer horizon
The next special place is the outer horizon.  This is just an event horizon as we commonly know it: once you've passed through it, there's no going back.  Any future-directed timelike curve which passes the outer horizon will meet ... something ... in finite proper time.  Once you've passed the outer horizon then you're gone from the point of view of any observer outside the BH (and I think that, as with Schwazschild, it takes an unbounded amount of their proper time for this to happen although it makes no difference in practice).
The outer horizon is, I think, always called the outer horizon.  It is common to label regions of spacetime:

region I is outside the outer horizon
region II is between the outer and inner horizons
region III is inside the inner horizon

So once you're inside the outer horizon you are travelling through region II, and you spend only a finite time in it.  After that, the 'something' you meed is not, this time, the singularity, it is ...
The inner horizon
The inner horizon is, in a way, the nearest thing to the Schwarzschild singularity for the Kerr solution: it's the place that you will get to, in finite proper time, once you pass the outer horizon.  But it's not very like the singularity because it's not a singularity: it's just another one-way door through which you pass.
The inner horizon is a 'Cauchy horizon' (see below).
Once you're inside the inner horizon, you're in region III.  And as before there's no going back.  But region III is very strange indeed.  If you take it seriously, it has two extremely strange features:

not all future-directed timelike curves hit the singularity (I think perhaps almost none do);
there are closed future-directed timelike curves in this region.

Both of these lead to very awkward situations: a closed timelike curve (CTC) is a time machine: it is a worldline which you can move along into your own past.  So that's a bit awkward.  Fortunately these CTCs are all inside not one, but two horizons.
Except: what happens to the worldlines which miss the singularity?  Well, if you take this seriously, what you have to do is to do what's called 'extending' the spacetime of the Kerr solution, and once you do this you have two possible fates for these things:

they can pass through the singularity, which is ring shaped, and get into some very strange, asymptotically-flat spacetime, which I don't understand at all;
they can miss the singularity altogether and then pass out, through a horizon, into a new copy of region II from which they inevitably pass, though another horizon, into a new copy of region I;

And now, in this new copy of region I there is another copy of the Kerr solution (or, really, the same solution again), and they can pass in through another horizon into yet another copy of region II, and thence into another region III and so on.  There is an infinite chain of these things, in both directions.
A Penrose diagram of Kerr
One way of seeing this is to use a Penrose diagram, which is a picture of spacetime which has been conformally transformed to make it compact.  What that means is two things: a conformal transformation is one which can change lengths but doesn't change angles, and making spacetime compact means that points at infinity are brought in to a finite distance, so you can have a picture of an entire, infinite, spacetime on a finite bit of paper (well, OK, you have to suppress two space dimensions in the usual way).
Here is a diagram I drew of the Kerr solution (any errors in this are mine).  This only includes two copies of it: you need to remember that there is an infinite chain of these things.  Time goes upwards.

So, here is what is going on in this diagram.

At the bottom is region I, for which the two outer diagonal edges are past & future null infinity, $\mathcal{I}^-$ and $\mathcal{I}^+$ respectively (not marked).
The outer horizon is the line between I and II.
The inner horizon is the line between II & III, and you can see that once you've got into II you are going to hit this.
The singularity is the dotted line.  If you pass through it you end up in the unmarked region to the left or right of it, and this is the bit of the thing that I just don't understand at all.
If you otherwise miss it (which you almost always do) you end up passing through another horizon into a new II region.
And from that you pass out through another horizon into a new I region, which has its own  $\mathcal{I}^-$ and $\mathcal{I}^+$, also not marked.
And if that's not enough for you you could then travel into yet another II and do the whole thing over again.  Personally at this point I decide I've had enough and just go off to live my life in the new I region.

As I said above, I think this is (now more) correct, but any it may (still) not be.  Note also that the static limit, which is in region I, isn't shown here at all: it just has the horizons.
Probably a toy
But I think it's commonly accepted that this is pretty clearly all just a theoretical toy, because this only works for a solution which actually is a Kerr BH: what people call an 'eternal black hole'.   Real physical BHs are not like this, because they originate from collapsing objects: no BH can be older than the age of the universe.  Therefore I believe that the inner structure of real BHs is probably not like this at all.
In particular I mentioned above that the inner horizon is a Cauchy horizon.  A Cauchy horizon is something which, loosely, divides spacetime into regions where there are closed spacelike curves (no problem with these!) and closed timelike curves (big problem with these!), and serves to 'censor' the nasty region (the one with the CTCs) from the nice region (the one with only closed spacelike curves).  However, I think it is the case that Cauchy horizons have two unfortunate features:

they're not stable under plausible assumptions, so tiny perturbations turn them into something else (not sure what) which would make them unphysical;
if you cross one you experience an infinite flux of energy which is clearly also unphysical since this could only happen for eternal BHs.

Both of those things, in particular the lack of stability, make it seem unlikely that the internal structure of real spinning BHs is Kerr.  Least of all would this seem to be true for BHs which originate from collapse, which is probably all BHs, since whatever internal structure there may be does not yet exist from the perspective of an external observer.
However this all well beyond my competence, and I am simply not sure what the current state of things in this area is.
References

Some lecture notes on the Kerr solution which used to be at https://www.roma1.infn.it/teongrav/leonardo/bh/bhcap3.pdf but no longer are.  I believe these may be a chapter from this book which I found via Leonardo Gaultieri's home page: I have not read the book but the notes looked very good.
On Crossing the Cauchy Horizon of a Reissner-Nordstrom Black-Hole, S. Chandrasekhar and J. B. Hartle, Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences 384, no. 1787 (1982) – I think this is the thing you'd want to read to understand Cauchy horizons, and I may once have read it, but it's behind a paywall so I can't check now.  Certainly Chandrasekhar did a lot of interesting work on this sort of thing.
Wikipedia page on Kerr solution is probably reasonable.
Wikipedia page on Penrose diagrams likewise.

I wish I had some better references, sorry.


Which is the shape of the sky? I'm a software developer and I'm developing a planetarium and the first thing that I don't know is: What is the shape of the sky?

First, I though that it is a semi-sphere, but searching on Internet I have found a lot of youtube tutorials, and all of them draw a semi-sphere but flatten (you can search skydome on Google and find a lot of tutorials).

My question is, how can I represent sky on a virtual 3D world? How much do I need to flat a sphere to get a real sky representation?

For computer software, the easiest way to take a sphere (and/or hemisphere) and flatten it into a flat shape (usually a rectangle) is the equi-rectangular projection (also known as the plate carrée), because it has the simplest formula relating pixels and coordinates:

$$x = w \times \frac\lambda{360} + \frac w2$$

$$y = -h \times \frac\phi{180} + \frac h2$$

where $x$ and $y$ is the pixel point, $w$ and $h$ are the width and height (in pixels) of your map's rectangle, $\lambda$ is the longitude in degrees and $\phi$ is the latitude in degrees

This all assumes that your map puts the 0 longitude, 0 latitude point in the center. In other words, the prime meridian runs down the map's middle vertically and the equator runs down the map's middle horizontally. It also assumes that north and east are positive, while west and south are negative.

Notice that you don't even have to take the sine or cosine of anything. That's why the equi-rectangular projection is so simple. It basically just takes longitude and latitude and makes them $x$/$y$ coordinate with only an offset. The offset comes because, usually in computer programming, the origin is at the top left instead of the center of the screen.

Notice also that I negated the latitude in the $y$ formula, because most computer systems in 2D coordinates have the positive $\bf y$ direction pointing down. This is opposite of a standard graph in math, where positive $y$ goes up.

In your case, longitude and latitude will actually be right ascension and declination respectively. They are basically stellar coordinates for the celestial sphere as opposed to ground coordinates for the geode.

There are many, many other map projections possible, but equi-rectangular is the simplest.

However, your question is a little vague because you mentioned a "virtual 3D world", yet also asked how much you need to flatten it. If you already have a 3D model of a sphere in your computer graphics, you don't have to flatten it at all. You just draw the 3D model (sphere) with a texture on it. If you want to represent it as a map, then of course you have to flatten it, and equi-rectangular projection is the simplest way.


Orion Skyquest xt8i Intelliscope Dobsonian Telescope Lens I have the telescope in the title, which camps with 10mm and 25mm eyepieces. Both cannot really see anything in detail past the moon. All planets look like stars. What are good eyepieces for seeing all the other planets in detail, respectively?

The telescope you have is quite well regarded, and the eyepieces that come with it are of reasonable quality.  If you can see the moon clearly, including craters and other features, then you should be able to see that the planets, particularly Jupiter and Saturn,  are not like stars.  At the very least they will look like disks, not points of light, and saturn's rings will be unmistakable. The 10mm lens will give magnification of 120, which is more than adequate.  My guess is that you are not looking at the planets, but stars close to them.
Make sure sure the finder scope is properly aligned - use the moon for that - and then try again.


Can a tidally locked planet have their own habitable zone? As we know, if a planet is near a star and is tidally locked, then at near side it is very hot and at far side it is very cold. But between 2 sides there should be a gradient of temperature change and should have an area which the temperature is suitable for lives. Can we also say it has a habitable zone?

As you suggest, it might be possible for a habitable corridor to exist along the stationary terminator. But there are ideas around for more than that. The planet might be exposed to tidal forces the volcanism of which warms the far side. It might have a thick stormy ocean or atmosphere which evens out the surface temperature (All but the smallest planets have atmospheres). It might have habitable moons which when tidally locked to the planet rotate regularly versus the star. And if it, like Mercury, has an eccentric enough orbit it might rotate relative to its star although it is tidally locked. Planetary diversity is huge.


How would one calculate the synodic period between 3 planets? I don't know if this is the correct term, but how would one calculate the time it takes for 3 planets to lie on the same line?

Normally three planets would not line up along a simple line like that.  Planets have orbits with different inclinations, so at best they'd be in the same plane.

Working that out requires a detailed calculation of their motions.  The simplest approximation is to actually assume everything has the same inclination and treat the orbits as circles (which they're not precisely).

For this simple model, a planet's angular position $\theta$ at any time $t$ is given by :

$$\theta(t) = \theta_0+t\frac {2\pi} T$$

Where $T$ is the orbital period and $\theta_0$ is the initial angle.

What you want is the value of $t$ satisfying :

$$\theta_1(t) = \theta_2(t) + 2m\pi = \theta_3(t)+2n\pi$$

Where the $n$ and $m$ values are integers.

Now doing this for two planets is easy, we get :

$$t_{12}=\left(\frac {\theta_{10}-\theta_{20}}{2\pi} - m\right)\frac{T_1T_2}{T_2-T_1}$$

$$t_{13}=\left(\frac {\theta_{10}-\theta_{30}}{2\pi} - n\right)\frac{T_1T_3}{T_3-T_1}$$

Remember we have to work out those $m$ and $n$ values to get a result !

But the problem is that we need to get integer values and all the other values are real numbers.  This means that there may be no (exact) solution that produces integer values and hence no one time when they all line up.

Now in the real world their are some other issues :

The orbital inclinations mean we need to work in three dimensions
The orbits will be ellipses, not circles, and if we want even more accuracy they're not even convenient shapes like that (look up the three body problem to get an idea why).
The speed of light is finite.  So what "lined up" means is not as simple as it seems.  There might be only one observer who would say they were lined up and every other observer would consider them not lined up.
Planets have non-zero sizes, so we can allow them to be lined up within a range of values, not a simple set of numbers, but a whole set of ranges of values.
Again with time in general relativity, whose time are using as a reference.  That's a really complicated one in general relativity (more than it might seem), but fortunately to a good approximation we can ignore relativity a lot of the time or use small corrective factors.
All our measurements for the parameters of the orbit will have a finite accuracy and some uncertainty range.  So, particularly over very long periods of time, our calculations can become inaccurate.


Relation between black hole mass and radius, and our universe's Is there a graph of known black holes, with their estimated mass in the X axis and their estimated radius in the Y axis? If so, where can we find it? I would like to know if a black whole with the whole estimated mass of our universe would have the estimated radius of our universe (which means our universe could be a black hole, that's why light can't escape it and it looks "finite").

According to the standard ΛCDM cosmological model, the observable universe has a density of about $\rho = 2.5\!\times\!10^{-27}\;\mathrm{kg/m^3}$, with a cosmological consant of about $\Lambda = 1.3\!\times\!10^{-52}\;\mathrm{m^{-2}}$, is very close to spatially flat, and has a current proper radius of about $r =  14.3\,\mathrm{Gpc}$.

From this, we can conclude that the total mass of the observable universe is about
$$M = \frac{4}{3}\pi r^3\rho \sim 9.1\!\times\!10^{53}\,\mathrm{kg}\text{.}$$
Sine the universe at large is nonrotating and uncharged, it's natural to compare this to a Schwarzschild black hole. The Schwarzschild radius of such a black hole is
$$R_s = \frac{2GM}{c^2}\sim 44\,\mathrm{Gpc}.$$
Well! Larger that the observable universe.

But the Schwarzschild spacetime has zero cosmological constant, whereas ours is positive, so we should instead compare this to a Schwarzschild-de Sitter black hole. The SdS metric is related to the Schwarzchild one by
$$1-\frac{R_s}{r}\quad\mapsto\quad1 - \frac{R_s}{r} - \frac{1}{3}\Lambda r^2,$$
and for our values we have $9\Lambda(GM/c^2)^2 \sim 520$. This quantity is important because the black hole event horizon and the cosmological horizon become close in $r$-coordinate when it is close to $1$, a condition that creates a maximum possible mass for an SdS black hole for a given positive cosmological constant. For our $\Lambda$, that extremal limit gives $M_\text{Nariai} \sim 4\!\times\!10^{52}\,\mathrm{kg}$, smaller than the mass of the observable universe.

In conclusion, the mass of the observable universe cannot make a black hole.

  Well, we don't fully comprehend black matter, do we? And it was just "yesterday" that we discovered the "black energy", wasn't it?

If GTR with cosmological constant is right, we don't need to "fully comprehend" it to know its gravitational effect, which is what the calculation is based on. If GTR is wrong, which is of course quite possible, then we could be living in some analogue of a black hole. But then it's rather unclear what theory of gravity you wish for us to use to try to answer the question. There's no remotely competitive theory that's even approaching general acceptance.

  From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation.

Actually, the point of that calculation was to show that it's at least prima facie plausible. The Schwarzschild radius calculation doesn't rule out the black hole--quite the opposite. However, it's also not appropriate for reasons I explained above. The more relevant one actually does have mass more than one order of magnitude apart, and shows inconsistency. So if GTR with Λ is correct, it's unlikely because the ΛCDM error bars aren't that bad.

However, even if we still treat it as "close enough", that does not by itself imply what you want. The question of what kind of black hole all the mass of the observable universe would make, if any, is quite different from whether or not we're living in one. The black hypothetical needs to be larger still.

The biggest point of uncertainly, though, is the cosmological constant, even if GTR is otherwise correct. If we're allowed to have very different conditions outside our hypothetical black hole, then we could still have one, but then we get into very speculative physics at best, and just complete guesswork at worst.

So treat the above answer as conditional on the mainstream physics; if that's not what you want, then there can be no general answer besides "we don't know". And that's always a possibility, although not a very interesting one.


What decides the direction in which the accretion disk spins? Planets lie on the same plane because of the accretion disk formed during the Protostar stage, as I read in this question.

I also read about the collision of particles in the gas cloud causing the overall spin to be in just one direction. But what decides the direction in which the accretion disk spins relative to the direction of the core?

(I'm thinking that it might be because of some primary conditions - maybe the direction in which the core is spinning.)

Stellar systems are born from clouds of turbulent gas. Although "turbulence" means that different parcels of gas move in different directions, the cloud have some overall, net angular momentum. Usually a cloud gives birth to multiple stellar systems, but even the subregion forming a given system has a net, and non-vanishing (i.e. $\ne0$), angular momentum.

Parcels moving in opposite directions will collide, and friction will cause the gas to lose energy, such that the cloud contracts. Eventually subclouds moving in one direction will "win over" subclouds moving in other directions such that everything moves in the same direction, keeping the original angular momentum (minus what is ejected e.g. through jets).

This means that the central star will rotate in the same direction as the circumstellar disk and that, in general, the planets that form subsequently, will also not only orbit the star in the same direction, but also spin in the same direction around their own axes. This is called prograde rotation. Sometimes, however, collisions between bodies may cause a planet or asteroid to spin in the opposite direction. This is called retrograde rotation and is the case for Venus and Uranus.


Why there is no smoke around the Sun? Where there is fire, there is always smoke. So why there isn't any smoke near the Sun?

The sun isn't on fire.   Fire is actually very rare in the solar system.   It requires chemical potential energy, which happens on earth because of life.  Photosynthesis uses solar energy to build things out of carbon, oxygen, nitrogen and hydrogen (and some other elements).   It's these carbon chain structures that are ultimately flammable, and pretty much, only in an Oxygen atmosphere.   Oxygen atmosphere is produced by the same photosynthesis.

The sun's heat comes from nuclear fusion which doesn't make smoke.    Also, the nuclear fusion only happens deep the inside of the sun, the outside is just a thick blanket of molecules - mostly hydrogen.   The sun also isn't transparent.  The light we see from the sun is like the light we see from a red hot piece of metal or lava.   Red hot metal doesn't smoke (unless you drop water on it - which is actually steam.)   The sun, like red hot metal or lava, glows bright because of temperature.

In a certain sense, the tail of a comet is kind of like steam being made by the heat of the sun.


Relationship between Hubble paramater and absolute magnitude I have a set of data that is absolute magnitude of a star and its redshift. Is there any way to compute the value of the hubble parameter given that magnitude and redshift?

Maybe. To calculate the Hubble constant (H=v/d), you need the know the velocity (v≈c*z￼￼) and distance of a distant object (or rather as many objects as possible). The velocity of these objects must be dominated by the expansion of the Universe. So stars in our local group that are still bound to us via gravity wont work. So as long as the stars you use are not really bright or even supernovae in distant galaxies there is no way to get the Hubble constant.
If your stars are far enough away, however, you still have get their distance. This you can do with the distance modulus, i.e., difference between absolute and apparent magnitude:

m-M = 5 log(d)-5

So assuming the stars are distant stars for which you also know their apparent magnitude (e.g. Cepheid variables), than yes you can calculate H.
If they are just random stars, than not.


How would Alpha Centauri A appear from the surface of Alpha Centauri Bb? I'm trying to imagine what a hypothetical observer on the surface of Alpha Centauri Bb would see in the daytime sky. I know it would vary depending on the orbit of b around B, as well as the position of B and A in relation to each other.

What would the "sun" in the sky look like at different times on Bb? I know that at times B and A would overlap and appear as a single sun, but what about when they don't?

I'm trying to imagine what a hypothetical observer on the surface of
  Alpha Centauri Bb would see in the daytime sky. I know it would vary
  depending on the orbit of b around B, as well as the position of B and
  A in relation to each other.

My calculations are based on data from these wiki pages:

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

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

The planet Alpha Centauri Bb orbits the star Alpha Centauri B on an orbit with a semi-major axis of 0.04 AU. That's 25x closer to its star than Earth is to the Sun.

Alpha Centauri B has 86% the diameter of the Sun.

Therefore, ACB would appear in the sky of ACBb as a huge, extremely bright and hot disk, about 21x bigger than the Sun as seen from Earth (25 x 0.86). The heat would destroy and melt many materials.

ACB looks a bit more orange than the Sun, but the color difference is small.

ACA and ACB orbit each other, being separated by as little as 11 AU, or as much as 35 AU. ACA is about 22% bigger in diameter than the Sun.

Therefore, ACA as seen from ACBb would appear as a very small "sun" (about 1/10 the size of the Sun as seen from Earth) when at the closest approach, or basically an extremely bright dot, with no apparent size, so shiny it would hurt the eye, when at the maximal distance.

ACA would provide essentially no heat, and very little light, when compared to ACB (or even when compared to what we get from the Sun here on Earth). But it would be very similar in color to the Sun.

At the closest approach, ACA would look very similar to the Sun as seen from Saturn. At the maximal distance, it would look like the Sun seen from Neptune.

Summary:

ACB is a huge fireball of doom, 20x bigger than how we see the Sun, a bit of an orange hue, scorching hot, melting and igniting everything. ACA is a tiny disk, or even just a dot, yellow-white, very bright, providing essentially no heat and much less light than ACB.

ACA would grow bigger, then smaller again, following a cycle of 80 years.

ACA and ACB would get closer together in the sky, then drift further apart, on a cycle following ACBb's "year" - which is only 3.2 days in our terms. A very, very short "year".

It is not unlikely, although this is speculative, that ACBb is tidally locked to ACB, because they are so close together. If that's the case, ACB would only be visible from one side of ACBb, and would appear to remain "fixed" in the sky.

Here's an artist's impression. Keep in mind that ACB is a giant fireball, so this is more like a fisheye lens image (horizon to horizon in one snapshot). The size of ACA is overestimated. The Sun as seen from Earth would be the size of the letter B in that image.

  I know that at times B and A would overlap and appear as a single sun,
  but what about when they don't?

Well, do they? Do we know that for sure? For the overlap to happen, the orbit of Bb would have to be relatively coplanar to the mutual orbits of A and B. This seems likely, but I'm not sure if it has been established as a fact.


How far could be an object from the Sun and still be under the influence of its gravitational field? I'm trying to see how far can our star reaches with its gravity. I'm asking if anyone could give info as to what's our star's limit or the furthest object found in our solar system.

The Sun's gravity extends infinitely, but eventually solar objects would be unstable due to the influence of other stars. The minor planet "Sedna" has an orbit which takes it nearly 1000 AU (0.016 light years) from the sun at its furthest point (but now it is a lot closer)

It is also thought that billions of comets must orbit in the outer part of the solar system, out to 50000AU, or 0.8 light years, (or possibly further) forming the Oort Cloud. However, at such distances, they could not be directly observed. This marks the greatest distance at which orbiting solar system bodies can be found.


Is it possible for particles to be accelerated FTL inside BH travelling towards the singularity? Maybe just maybe it is possible for a particle to travel FLT inside of black hole  because of the singularity gravitational force

This could be due to the fabric of space it self falling towards the singularity FTL beyond the event horizon. So technically
A particle can go FTL moving with space itself not not moving through it, and this would break any laws of relativity

But if I'm wrong please tell me.

It's a little hard to follow the question.

The short answer is: No it doesn't move faster than light.

First off from the perspective of the outside observer, the particle doesn't reach the Schwartzchild radius, due to gravitational time dilation. From the viewpoint of the distant observer (the best kind to be when black holes are involved) the particle's clock slows to a stop as it nears the event horizon.

Your velocity is a relative and a local property, if you want to talk about the velocity you need to say what it is relative, and when black holes are involved you need to be careful about which metric you are measuring the velocity in. From the point of view of the infalling particle, it enters and crosses the event horizon. It may then be falling faster than the speed of light relative to people on the outside, but it is behind the event horizon, and can not return to them. It is not moving faster than light relative to any point which it can return to. So this doesn't break causality, nor does it allow for high-speed transport.

All that is certain is that in a short time (as the particle sees it) it will reach a singularity and then time and space will cease to exist for it, and all you are left with is mass.

Hadrons might become some kind of quark gluon plasma on their way in, but this doesn't make it lose mass, nor does it have any effect on the whether it moves faster or slower than light.


Could binary gas giants have ring and moon systems? Would any moons/rings around either of the gas giants stay intact or get ripped apart by the tidal forces between the two giants? If any moons/rings are able to form around either of the giants, is there a limit to how large a ring system could be, or how far a moon could orbit?

It would depend how close the gas giants were too each other, but in theory it would be possible.   Similar to the binary star system (image below), provided you ignore the AUs and scale this down from stars to planets, the same principal applies to orbits around two gas giants orbiting each other.

Source article.

They would have relatively small hill spheres depending on how close they were and corresponding, even smaller true regions of stability. But planetary ring systems tend to be quite close to the planet anyway and larger ring systems need to be almost exclusively inside the Roche Limit.

The binary gas giant orbits would have to be outside each other's respective Roche Limits anyway and provided there was sufficient distance between then, like maybe 4-6 times the larger of the two Roche limits (ballpark guess), then it's possible to have both gas giant planets with their own ring systems.

What's not possible is to have two orbiting disks around two bodies where the orbiting disks intersect each other, like they had in the Startrek Discovery premier episode.   And that was forming star systems, not gas giants, but it's still impossible for any orbiting ring or debris disk to intersect the way they did it.   It was a cool looking image, but clearly impossible.

I also want to point out that a ring system around both planets like the top of the first image with the writing "disk around both stars", that's not possible with a typical ring system.  Ring systems need to be in tight orbits around their planet, though faint/disburse ring systems like Saturn's E ring can exist further out.   The further out a ring system, the more likely it would be disturbed by the binary planet.


Why does the angular size of the Moon change? I know that the orbit of the moon is elliptical.
But I cannot understand this graph:

The number of 'bumps' in the graph is around 13~14 in a year, so I concluded that the each 'bump' indicates one lunar month.
But why aren't these bumps symmetrical? Since the moon revolves around the Earth once in one lunar month, shouldn't they be symmetrical?

The moon's orbit is elliptical, and it moves faster when it is closest to the Earth.  This is the case for all elliptical orbits, and was discovered by Kepler, who gave a neat rule for predicting how fast an orbiting body moves. (https://kids.britannica.com/students/assembly/view/90830)

For the moon, this means that when it is largest, it is also moving fastest. So the peaks in the graph (where the moon is large) are narrow, as the moon quickly moves away from periapse.  When the moon is smallest, it is moving slowly, so the troughs in the graph are wide. It changes size slowly when it is far from the Earth.
The bumps are not all the same size because the orbit of the moon is perturbed by the sun.  This changes the amount that the moon changes size. https://eclipse.gsfc.nasa.gov/SEhelp/moonorbit.html
When the eccentricity is low, there is less difference in size.  When it is large there is a big difference.
This is the cause of "supermoons", when the full moon occurs at the same time as periapse.  When the eccentricity is at a maximum, the supermoon will appear especially large and bright.

The Moon's orbit is one of the most complex orbits in the solar system!


Question related to Sun rise and Sun set times I am not sure whether this question is right or not so bear with me.

I understand 'Sun rise' and 'Sun set' is caused by the Earth's rotation.

I read that 'Sun light' takes around 8 minutes to reach Earth.

Every city in the world has its own 'Sun rise and Sun set' timings.

Lets assume the 'Sun rises' at 6 AM in 'City A'.

Does this mean that 'City A' sees/feels the light of the Sun which was in the sky @ 5:52 AM ?

In other words, does 'Sun rise' and 'Sun set' timings change according to 'light years'?

I understand 'Sun rise' and 'Sun set' is caused by 'earth rotation'.
  I read 'Sun light' takes around 8 minutes to reach earth
  Every city in this world has its own 'Sun rise and sun set' timings.

This is all correct. The apparent motion of the sun in the sky is mostly due to the rotation of the Earth.

  Lets assume 'sun arises' at 6 AM in 'City A'.
  Does it mean that 'City A' sees / feels the light of Sun which was in the sky @ 5:52 AM ?

It means that at 6AM, light from the sun shines on city A. That light had left the sun at about 5:52. At 5:52, city A was in the shadow of the Earth. The time that the light left the sun is fairly irrelevant to the timing of sunrise. A 5:52 the light left the sun, in the 8 minutes that followed the Earth rotated by about 2 degrees and by 6 AM city A was just in the right position to catch a little of the light. We don't change the timings. We would not say that the sun rose at 5:52, because at 5:52 the sun appears to be below the horizon.


Is the sphere of influence of our milky way's center super massive black hole same as the Milky way's radius? The sphere of influence is a region around a supermassive black hole in which the gravitational potential of the black hole dominates the gravitational potential of the host galaxy(found it in wiki). So is it the same? If not, what holds the milky way together?

No. The black hole at the centre of the Milky way is only a few million solar masses. The same amount of mass, made up mostly of stars, would be enclosed within $\sim 10$ parsecs of the Milky Way centre. The bulge of the Milky Way, which is a few thousand parsecs across contains more than 20 billion solar masses.

Thus the dynamics of our Galaxy, outside of the central few parsecs are hardly influenced by the black hole at all.

What holds the Milky Way together? The gravity of everything else - stars, gas, dust, dark matter (the latter probably being the most important once you get to galactocentric radii beyond the Sun).


How fast is the universe expanding? I've hear several theories stating that the universe is expanding faster than the speed of light, others claim that the universe expands faster the further away you measure it. Which of this is correct and how do you prove it (mathematically)? Furthermore, does this correctly imply, then, that eventually galaxies will be so far away, and moving so fast, that we will never see them again?

Thank you in advance!

Both are correct, although the first can be further explained a bit. I won't give you a mathematical proof, though; Instead I'll play with characters.

Let's assume, for the sake of gedankenexperiment, that:

The speed of light is 5 characters/sec;
our universe is expanding at 1 character per 5 characters per sec.

This is our current universe, and we launch a photon from body A aiming at body E.
(Space generated each second is marked with a # symbol.)

T=0s A----B----C----D----E           Bodies
     *                               Photon     - 19 chars to E
T=1s A--#--B--#--C--#--D--#--E
      --#--*                                      17 chars to E
T=2s A--#---B-#----C#-----#D----#-E
      --#-----#-*                                 17 chars to E
T=3s A--#----B#-----#-C---#----D#-----#-E
      --#-----#-----#*                            18 chars to E
T=4s A--#-----#B----#----C#-----#---D-#-----#--E
      --#-----#-----#-----#*                      19 chars to E

I know, the graph isn't too granular, and the space generation isn't evenly distribute. I apologize for that, but it's for the sake of demonstration.

Notice that at T=2 some space is already generated between A and the photon. But that's irrelevant: E is sitting at the event horizon, and will never be reached by the photon, because the amount of space being generated between photon * and body E is equal, or superior, to the speed of light.

Given any positive expansion rate, there will be an event horizon - a point where the accumulated dilation of space is more than the amount of space a particle moving at the speed of light can travel.

A galaxy sitting initially at say, 1000 characters from A at T=0, will be at staggering 1200C at T=1 - that's 40 times our speed of light.

At T=16s, B (that was passed by the original photon at T=1) will be sitting exactly where E was relative to A, and at T=17 will fall out of our event horizon. A new photon emitted from A will never reach it.


What's the relationship between Moon phase and the Milky Way visibility? I've been using an IOS app to predict visibility of the Milky Way.

I'm planning a trip to capture the Milky Way on 9th March 2015, but the app shows the visibility will be very weak on that date.

The Moon doesn't even rise until 9:12am.

Does the Moon phase still affect the visibility of the Milky Way even if it is not above the horizon?

Moon Phase
With the moon phase, it's going to be in Waning Gibbous on March 9th, so it will be reflecting a lot of light back to earth. Any light that interferes with the sky will make dimmer objects harder to see.

The Moon's reflectivity figures taken from here:

Fraction of the moon: 0.61371425766053
Illumination of the moon disk: 88 %

Location
It will also depend a lot on your home location (hemisphere) as to the position of the Milky Way at different times. Your proximity to any light pollution will of course have an impact, as living on the outskirts of a city myself, it is impossible to view the Milky Way. Only when in very rural areas on clear nights have I had a good view of it.
New Moon
On the 20th March there is a New Moon, so this may be a more opportune time to try to capture it.


Mechanism for Brown Dwarf Fusion I've read (at here, among other places) that during the Degenerate Era, star formation will end and the last stars will go out. But it was noted that there is still the possibility of star birth, coming from the collision of two stellar remnants, such as brown dwarfs (although white dwarfs also have the potential for such a collision). When two brown dwarfs collide, if they have enough combined mass, fusion can begin.

This is much different from how normal stars form today, and how they fuse hydrogen and other elements. So my question is this:

Will there be any major differences between the fusion process in the new "star" and a star with the same mass that formed via traditional star formation (i.e. collapse of a gas cloud)? Also, how will this affect the new "star's" life cycle?

Your scenario is, of course, rather speculative and considers the very distant future. I don't like that article you're referencing, some stuff seems inaccurate (for example the fraction of baryons locked into stars is tiny and will always remain tiny).

If two brown dwarves collide (extremely unlikely, but if you wait long enough, anything with some tiny non-zero probability eventually will happen), they form a new star with their combined mass. Since the brown dwarves have not undergone any hydrogen burning, the new star has the same composition as a new-born star of the same mass would have had (except for lithium) and, if massive enough, will evolve just like an ordinary M dwarf (or so). In other words, it will fuse hydrogen into helium as all stars on the main sequence.