Can planets inside the Earth's orbit around the Sun appear to undergo retrograde motion?

Both Mercury and Venus undergo retrograde motion, but are visually close to the Sun when this happens, so we don't notice it.

Below is Mercury's right ascension from May 15 to July 15 of this year. Note that it starts off increasing, then decreases (retrograde motion) and then increases again.

*******************************************************************************
Ephemeris / WWW_USER Thu Feb 19 08:02:41 2015 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: Mercury (199)                   {source: DE-0431LE-0431}
Center body name: Earth (399)                     {source: DE-0431LE-0431}
Center-site name: GEOCENTRIC
*******************************************************************************
Start time      : A.D. 2015-May-15 00:00:00.0000 UT
Stop  time      : A.D. 2015-Jul-15 00:00:00.0000 UT
Step-size       : 1440 minutes
*******************************************************************************
Target pole/equ : IAU_MERCURY                     {East-longitude -}
Target radii    : 2439.7 x 2439.7 x 2439.7 km     {Equator, meridian, pole}
Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}
Center pole/equ : High-precision EOP model        {East-longitude +}
Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}
Target primary  : Sun
Vis. interferer : MOON (R_eq= 1737.400) km        {source: DE-0431LE-0431}
Rel. light bend : Sun, EARTH                      {source: DE-0431LE-0431}
Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2
Atmos refraction: NO (AIRLESS)
RA format       : HMS
Time format     : CAL
EOP file        : eop.150218.p150512
EOP coverage    : DATA-BASED 1962-JAN-20 TO 2015-FEB-18. PREDICTS-> 2015-MAY-11
Units conversion: 1 au= 149597870.700 km, c= 299792.458 km/s, 1 day= 86400.0 s
Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (>38.000=NO), Daylight (NO )
Table cut-offs 2: Solar Elongation (  0.0,180.0=NO ),Local Hour Angle( 0.0=NO )
**************************************************************************************************************
 Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O
**************************************************************************************************************
$$SOE
     2015-May-15 00:00     04 42 01.78 +24 08 32.2   1.53   4.36 0.69142867651373 -28.9445515  18.6988 /T 129.3762
     2015-May-16 00:00     04 43 23.87 +24 00 54.9   1.74   4.49 0.67512627148213 -27.4926615  18.0144 /T 132.4768
     2015-May-17 00:00     04 44 24.75 +23 51 35.3   1.96   4.63 0.65967892310678 -25.9844651  17.2466 /T 135.5875
     2015-May-18 00:00     04 45 04.61 +23 40 37.1   2.21   4.77 0.64511881456032 -24.4211059  16.3963 /T 138.7127
     2015-May-19 00:00     04 45 23.80 +23 28 04.3   2.47   4.91 0.63147747327791 -22.8037380  15.4651 /T 141.8554
     2015-May-20 00:00     04 45 22.86 +23 14 01.5   2.75   5.06 0.61878570842679 -21.1337209  14.4555 /T 145.0181
     2015-May-21 00:00     04 45 02.49 +22 58 33.7   3.05   5.20 0.60707343838108 -19.4128031  13.3707 /T 148.2014
     2015-May-22 00:00     04 44 23.61 +22 41 46.8   3.37   5.33 0.59636941752453 -17.6432855  12.2149 /T 151.4048
     2015-May-23 00:00     04 43 27.36 +22 23 47.7   3.71   5.45 0.58670087720487 -15.8281545  10.9939 /T 154.6255
     2015-May-24 00:00     04 42 15.08 +22 04 44.1   4.07   5.56 0.57809309959244 -13.9711803   9.7148 /T 157.8583
     2015-May-25 00:00     04 40 48.34 +21 44 44.8   4.46   5.63 0.57056894619759 -12.0769753   8.3871 /T 161.0937
     2015-May-26 00:00     04 39 08.89 +21 23 59.8   4.87   5.66 0.56414836550867 -10.1510071   7.0239 /T 164.3154
     2015-May-27 00:00     04 37 18.70 +21 02 40.1   5.29   5.62 0.55884790663755  -8.1995645   5.6456 /T 167.4924
     2015-May-28 00:00     04 35 19.87 +20 40 57.9   n.a.   n.a. 0.55468026744915  -6.2296724   4.2893 /T 170.5572
     2015-May-29 00:00     04 33 14.63 +20 19 06.0   n.a.   n.a. 0.55165390565664  -4.2489596   3.0439 /T 173.3313
     2015-May-30 00:00     04 31 05.30 +19 57 18.0   n.a.   n.a. 0.54977273920552  -2.2654833   2.1722 /T 175.2574
     2015-May-31 00:00     04 28 54.24 +19 35 48.1   n.a.   n.a. 0.54903595776444  -0.2875219   2.2105 /L 175.1837
     2015-Jun-01 00:00     04 26 43.78 +19 14 50.1   n.a.   n.a. 0.54943796056976   1.6766529   3.1266 /L 173.1900
     2015-Jun-02 00:00     04 24 36.24 +18 54 38.3   n.a.   n.a. 0.55096842790710   3.6190086   4.3897 /L 170.4268
     2015-Jun-03 00:00     04 22 33.81 +18 35 25.8   5.29   5.65 0.55361252502737   5.5319758   5.7564 /L 167.4122
     2015-Jun-04 00:00     04 20 38.57 +18 17 25.2   4.88   5.70 0.55735122910170   7.4086481   7.1443 /L 164.3125
     2015-Jun-05 00:00     04 18 52.42 +18 00 48.1   4.49   5.68 0.56216176249744   9.2429487   8.5180 /L 161.1913
     2015-Jun-06 00:00     04 17 17.09 +17 45 44.4   4.12   5.62 0.56801810956371  11.0297524   9.8584 /L 158.0785
     2015-Jun-07 00:00     04 15 54.09 +17 32 22.7   3.77   5.53 0.57489158958468  12.7649573  11.1530 /L 154.9902
     2015-Jun-08 00:00     04 14 44.76 +17 20 50.1   3.45   5.42 0.58275145605987  14.4455044  12.3929 /L 151.9356
     2015-Jun-09 00:00     04 13 50.19 +17 11 11.8   3.15   5.31 0.59156549247635  16.0693469  13.5713 /L 148.9204
     2015-Jun-10 00:00     04 13 11.32 +17 03 31.6   2.87   5.18 0.60130057734526  17.6353760  14.6833 /L 145.9474
     2015-Jun-11 00:00     04 12 48.85 +16 57 51.6   2.61   5.06 0.61192319592983  19.1433125  15.7250 /L 143.0178
     2015-Jun-12 00:00     04 12 43.37 +16 54 12.5   2.37   4.94 0.62339988172426  20.5935736  16.6938 /L 140.1313
     2015-Jun-13 00:00     04 12 55.28 +16 52 33.7   2.15   4.81 0.63569757625371  21.9871244  17.5876 /L 137.2863
     2015-Jun-14 00:00     04 13 24.86 +16 52 53.2   1.94   4.69 0.64878390050511  23.3253217  18.4055 /L 134.4805
     2015-Jun-15 00:00     04 14 12.31 +16 55 08.3   1.75   4.58 0.66262733523404  24.6097559  19.1469 /L 131.7103
     2015-Jun-16 00:00     04 15 17.69 +16 59 15.2   1.57   4.46 0.67719731085661  25.8420965  19.8117 /L 128.9719
     2015-Jun-17 00:00     04 16 41.05 +17 05 09.2   1.40   4.35 0.69246421084326  27.0239479  20.4002 /L 126.2604
     2015-Jun-18 00:00     04 18 22.35 +17 12 45.4   1.25   4.24 0.70839929524826  28.1567176  20.9133 /L 123.5707
     2015-Jun-19 00:00     04 20 21.53 +17 21 57.8   1.10   4.14 0.72497455264057  29.2415000  21.3516 /L 120.8972
     2015-Jun-20 00:00     04 22 38.49 +17 32 40.5   0.96   4.04 0.74216248879516  30.2789728  21.7164 /L 118.2337
     2015-Jun-21 00:00     04 25 13.15 +17 44 46.7   0.83   3.94 0.75993585914434  31.2693054  22.0086 /L 115.5738
     2015-Jun-22 00:00     04 28 05.39 +17 58 09.6   0.71   3.84 0.77826734979511  32.2120717  22.2296 /L 112.9109
     2015-Jun-23 00:00     04 31 15.13 +18 12 42.2   0.59   3.75 0.79712920966069  33.1061670  22.3805 /L 110.2380
     2015-Jun-24 00:00     04 34 42.27 +18 28 17.0   0.48   3.65 0.81649283449812  33.9497246  22.4625 /L 107.5478
     2015-Jun-25 00:00     04 38 26.75 +18 44 46.2   0.37   3.56 0.83632830264123  34.7400316  22.4769 /L 104.8329
     2015-Jun-26 00:00     04 42 28.49 +19 02 02.2   0.27   3.47 0.85660386201288  35.4734443  22.4247 /L 102.0858
     2015-Jun-27 00:00     04 46 47.47 +19 19 56.7   0.17   3.38 0.87728536859778  36.1453054  22.3070 /L  99.2988
     2015-Jun-28 00:00     04 51 23.65 +19 38 21.2   0.07   3.30 0.89833567803594  36.7498651  22.1247 /L  96.4642
     2015-Jun-29 00:00     04 56 17.02 +19 57 07.2  -0.02   3.21 0.91971399456822  37.2802130  21.8789 /L  93.5742
     2015-Jun-30 00:00     05 01 27.57 +20 16 05.6  -0.12   3.13 0.94137518555319  37.7282290  21.5704 /L  90.6210
     2015-Jul-01 00:00     05 06 55.30 +20 35 07.1  -0.20   3.05 0.96326907552925  38.0845640  21.2001 /L  87.5974
     2015-Jul-02 00:00     05 12 40.17 +20 54 01.8  -0.29   2.96 0.98533974152847  38.3386660  20.7688 /L  84.4961
     2015-Jul-03 00:00     05 18 42.16 +21 12 39.8  -0.38   2.88 1.00752484088954  38.4788698  20.2775 /L  81.3104
     2015-Jul-04 00:00     05 25 01.19 +21 30 50.5  -0.46   2.80 1.02975501346900  38.4925685  19.7269 /L  78.0344
     2015-Jul-05 00:00     05 31 37.13 +21 48 23.0  -0.54   2.73 1.05195341081967  38.3664848  19.1182 /L  74.6631
     2015-Jul-06 00:00     05 38 29.80 +22 05 05.9  -0.63   2.65 1.07403541451894  38.0870574  18.4526 /L  71.1926
     2015-Jul-07 00:00     05 45 38.91 +22 20 47.8  -0.71   2.57 1.09590861358459  37.6409536  17.7315 /L  67.6206
     2015-Jul-08 00:00     05 53 04.08 +22 35 16.7  -0.79   2.49 1.11747311584451  37.0157141  16.9564 /L  63.9467
     2015-Jul-09 00:00     06 00 44.78 +22 48 20.6  -0.87   2.41 1.13862226817234  36.2005225  16.1294 /L  60.1725
     2015-Jul-10 00:00     06 08 40.36 +22 59 47.7  -0.95   2.34 1.15924385212199  35.1870766  15.2529 /L  56.3023
     2015-Jul-11 00:00     06 16 49.99 +23 09 26.1  -1.03   2.26 1.17922180051076  33.9705111  14.3295 /L  52.3431
     2015-Jul-12 00:00     06 25 12.69 +23 17 04.6  -1.11   2.17 1.19843844390623  32.5502953  13.3626 /L  48.3050
     2015-Jul-13 00:00     06 33 47.27 +23 22 32.8  -1.19   2.09 1.21677724378686  30.9310023  12.3560 /L  44.2011
     2015-Jul-14 00:00     06 42 32.40 +23 25 41.0  -1.28   2.00 1.23412590569533  29.1228358  11.3140 /L  40.0476
     2015-Jul-15 00:00     06 51 26.58 +23 26 21.1  -1.37   1.91 1.25037969990154  27.1418048  10.2415 /L  35.8640
    $$EOE
**************************************************************************************************************
Column meaning:

TIME

  Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in
the 1st-column denotes a B.C. date. First-column blank (" ") denotes an A.D.
date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.
Later calendar dates are in the Gregorian system.

  Time tags refer to the same instant throughout the universe, regardless of
where the observer is located.

  The dynamical Coordinate Time scale is used internally. It is equivalent to
the current IAU definition of "TDB". Conversion between CT and the selected
non-uniform UT output scale has not been determined for UTC times after the
next July or January 1st.  The last known leap-second is used over any future
interval.

  NOTE: "n.a." in output means quantity "not available" at the print-time.

 R.A._(ICRF/J2000.0)_DEC =
   J2000.0 astrometric right ascension and declination of target center.
Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)

 APmag S-brt =
   Target's approximate apparent visual magnitude & surface brightness. For
planets and satellites, values are available only for solar phase angles in the
range generally visible from Earth. This is to avoid extrapolation of models
beyond their valid (data-based) limits.
   Units: MAGNITUDE & VISUAL MAGNITUDES PER SQUARE ARCSECOND

 delta  deldot =
   Range ("delta") and range-rate ("delta-dot") of target center with respect
to the observer at the instant light seen by the observer at print-time would
have left the target center (print-time minus down-leg light-time); the
distance traveled by a light ray emanating from the center of the target and
recorded by the observer at print-time. "deldot" is a projection of the
velocity vector along this ray, the light-time-corrected line-of-sight from the
coordinate center, and indicates relative motion. A positive "deldot" means the
target center is moving away from the observer (coordinate center). A negative
"deldot" means the target center is moving toward the observer.
Units: AU and KM/S

 S-O-T /r =
    Sun-Observer-Target angle; target's apparent solar elongation seen from
observer location at print-time. If negative, the target center is behind
the Sun. Angular units: DEGREES.

    The '/r' column is a Sun-relative code, output for observing sites
with defined rotation models only.

         /T indicates target trails Sun (evening sky)
         /L indicates target leads Sun  (morning sky)

    NOTE: The S-O-T solar elongation angle is the total separation in any
direction. It does not indicate the angle of Sun leading or trailing.

 S-T-O =
   "S-T-O" is the Sun->Target->Observer angle; the interior vertex angle at
target center formed by a vector to the apparent center of the Sun at
reflection time on the target and the apparent vector to the observer at
print-time. Slightly different from true PHASE ANGLE (requestable separately)
at the few arcsecond level in that it includes stellar aberration on the
down-leg from target to observer.  Units: DEGREES

 Computations by ...
     Solar System Dynamics Group, Horizons On-Line Ephemeris System
     4800 Oak Grove Drive, Jet Propulsion Laboratory
     Pasadena, CA  91109   USA
     Information: http://ssd.jpl.nasa.gov/
     Connect    : telnet://ssd.jpl.nasa.gov:6775  (via browser)
                  telnet ssd.jpl.nasa.gov 6775    (via command-line)
     Author     : Jon.Giorgini@jpl.nasa.gov

**************************************************************************************************************


Assuming the major heat input is from the star (the gas giant may also radiate significant IR) then you could seed clouds or freeze the water in order to increase the albedo.

Alternatively, you could increase the radiative efficiency by somehow painting the side of the moon facing away from the star black. Tricky, as no doubt this hemisphere changes with time quite rapidly.

If tidal heating from the gas giant is significant then you need to get the moon into a wider orbit!

The sunshade at the L1 point idea is widely discussed. Usually it is lenses that are considered, to avoid big problems with radiation pressure. I'm sure you could calculate some appropriate numbers to block a certain percentage of light in your system.


What does "Normalized to Si 10 to the power 6" mean?

The table you referenced is normalized so that the abundance of Silicon (Si) is exactly $10^6$, and all the other numbers are related to that.

"Abundance" in this table refers to the number of atoms.

Please note that this table is for the entire Solar System, taking into account the huge amounts of Hydrogen and Helium in the Sun. It will not be appropriate for creating Earth-like planets.


What (the heck) is a Super Worm Equinox Moon?

All those adjectives being smooshed together signify an uncommon event. That's why you've never seen them together like that before.

All 3 conditions have to hold true:

It's a supermoon, which means the full moon coincides with the moons perigee or nearest approach. That can make it appear up to 30% brighter than one at apogee (farthest away). These happen about every 13 months and it doesn't have to be exact so usually, we get 2 of them in a row, like we did this year. February had a supermoon and so did March.
It's a worm moon, which means it is occurring in the month of March (see @astrosnapper's answer for a better explanation of that).
It's during an equinox, basically the first day of spring (or autumn).

If any one of those isn't happening then it can't be called a Super Worm Equinox Moon.

Apparently, the term supermoon (all one word, by the way) is a relatively recent thing. I tried to view it on Google N Gram viewer, but...

It is a particularly bright full moon and it does deserve to have its own terminology, IMO.

Update: The rarity of the event is certainly relative. I saw a tweet from National Geographic that we also had a super worm equinox moon 19 years ago.


The aperture of your 4.5" telescope is one thing, it's also important what focal length you have. Is it a f/5 or rather a f/8? The f/8 would be suitable for viewing the Moon, Jupiter, Saturn, maybe even Mars and Venus. You can also buy a good solar filter, attach it in the front of the optical tube assembly, and view the sun. But be careful with that, and inform yourself beforehand! Your eyesight might be in danger otherwise. For the Moon it is advisable to buy a neutral grey filter, because it is also very, very bright.

As for the Messier objects: you can probably view the larger ones. The Ring nebula (M57) will be tough, because it is very small and faint. But open clusters, globular clusters and Andromeda's core will be possible. A few bright nebulas will also be possible, like M42 (Orion Nebula).

There is a good list of all Messier objects over at Astropixels. You should try objects of magnitude 4 or better, and that have a size of more than 5 arcminutes. Otherwise they may be too dim or small to find.


Why is there so little nitrogen in the Martian and Venusian atmospheres?

Nitrogen, with a molecular mass of 28 atomic mass units, is too light to have remained in Mars's atmosphere. Carbon dioxide, with a molecular mass of 44 amu, could (and does) exist on Mars, but it is rather sparse.
Venus's atmosphere appears to contain a small amount of nitrogen when viewed on a percentage basis. "Only" 3.5% of all of the gases in Venus's atmosphere is nitrogen. But that's not a particularly good metric. Venus has so very, very much carbon dioxide in its atmosphere that the CO2 dwarfs every other constituent. When viewed in terms of mass, Venus has a good deal of nitrogen in its atmosphere, four times more nitrogen than the Earth has in its atmosphere.


What would cause large jumps in brightness during a solar eclipse?

Eric Pauer, who produced the chart, has an interesting web site that covers this eclipse.  It is evident that the weather was not great at the time of the eclipse, with intermittent cloud, and it will have been passing clouds that caused the variations in intensity.

You can see the evidence from this page
http://www.pauerhome.com/eclipse/eclipse99/1contact/1contact.htm


Distance to Sun at Perihilion and Aphelion at max and min Eccentricity

I like the math of ellipses cause there's enough parts to keep it interesting, but it's not super complex either.   The answer to your question as you put it, is remarkably simple.  I've bolded the letters that represent distances for easier reading.

Eccentricity or $e = c/a$, where $a$ is the semi major axis and $c$ is the distance between the focal point (sun's location) and the center of the ellipse. $ea$ is the distance between the center point and the focal point.

So, the answer to your question is simply $a$ plus $ea$ and $a$ minus $ea$.   For the Earth and sun, the semi major axis or $a$, is $149,600,000\:\mathrm{km}$, then simply add and subtract $ea$ to both.

When $e = .0679$, $ea = 10.15\:\mathrm{million\:km}$. So $149,600,000 \pm 10,150,000$. $159,750,000$ and $139,450,000\:\mathrm{km}$ at furthest and closest.

When $e = .000055$, $ea = .008\:\mathrm{million\:km}$, so $149,680,000$ and $149,520,000$ at most circular.

Simple picture

More detailed picture

Maths and sources of pictures, here and here

footnote:  I hope this wasn't a homework question.


Are there known objects rotating at close-to-light speeds?

A particle travelling in a circle of radius $R$ at velocity $v$ has an acceleration of $v^2/R$. So either $R$ is very large, or a very strong force is acting to stop the particles at the surface of the object flying away. We see no very large objects (galactic clusters, for instance) rotating this quickly, so we need to look for strong forces.

One option is gravity. We know of sub-millisecond pulsars, which we believe are neutron stars with a radius of perhaps 10 km, rotating 1000+ times per second. This gives a surface velocity at the equator of about $60\,000 km/s$ which is $0.2c$. So those meet your conditions.

Defining how fast the surface of a black hole is rotating is difficult on several levels, but black holes do have angular momentum, and there is a dimensionless number relating that to their mass (or equivalently radius) which can be loosely thought of as the fraction of light-speed at which the event horizon is moving. This paper measures a value of about 0.44 for that parameter for the black hole in the centre of the milky way.

At the other extreme, this stackexchange answer suggests that the electrons in a hydrogen atom rotate about the nucleus at something a bit less than $0.01c$. This scales up linearly with the charge on the nucleus, so the inner electrons in a lead or uranium atom are in some sense rotating at close to light-speed.

Finally, the particles in a large particle accelerator, rotate around the centre of the accelerator at velocities very close to that of light.


I believe when most scientists refer to the mass of a blackhole - i.e. a blackhole having the mass of a billion Suns, what they really mean is that the calculated gravitational pull of the blackhole is equivalent to that of such a mass would it exists at the location of the blackhole.

PS: Perhaps you could also bring some reference with respect to where exactly have you heard that 'It is said that in Blackhole 100% part of the mass is getting converted into energy'


How many things are wrong in this "artist view" of the TRAPPIST-1 system?

The 2 big planets are probably f & g and they don't look up to scale to me. While f is the size of Earth, or almost 4 times bigger than the Moon, but it's also further away, ~ 1.3 million kilometers, as opposed to 400,000km for the Moon. So I would say f will be somewhat larger than the moon but not anywhere near what the poster shows. G will appear even smaller than f because it's further away.
I haven't looked at the orbits in details but I think that sort of event should be possible. Trappist-1e being intermediate planet makes no difference.
As for "where is the sun" - it appears to be somewhere to the right, probably below the horizon. That makes sense because any base would probably be along the terminator.


What are the advantages of refractor telescopes over reflector telescopes?

reflector telescopes are newer and better then refractor telescopes

It's true that the first examples of reflectors we know of were made shortly after the first refractors we know of. But it's not correct to say that they're (always) better. First, each telescope model you might buy has its own specifications and build quality, which have to be carefully considered in terms of what you're trying to view. Second, certain refractors can outperform certain reflectors for certain tasks: for instance, many refractors can offer better views of planetary/lunar/cluster details than similarly priced reflectors could. Reflectors designed such that eyepieces have to be used at a standing height have the disadvantage of being difficult to use for things (e.g. planets) near the horizon. Refractors could also have the advantage of being potentially smaller and more portable than, for instance, something like a 10" dob. Reflectors have their advantages too: lack of chromatic aberration, for instance (although they do suffer from other forms of aberration). I hope this gives you a better idea of the pros and cons of different kinds of telescopes.


Difference between zone time and local sidereal time?

Universal Time (UT) and Greenwich Mean Time (GMT) are based on the average (mean) position of the sun. The mean position of the Sun crosses the meridian every 24 hours. There are subtle differences between UT and GMT, but they are the same for most practical purposes. (The Wikipedia article for Greenwich Mean Time points out that the start of the day under GMT changed from noon to midnight sometime in the past. The Time and Date.com article GMT versus UTC describes GMT as a time zone and UTC as a time standard.)

Since our clocks are also based on the mean position of the Sun, the time in your time zone is a fixed number of hours before or after UT.

Local sidereal time (LST) is the time according to the stars at your longitude. Let's assume that star X is on the meridian at 0 hours (midnight). The same star will return to the meridian after 23 hours 56 minutes 4 seconds (approximately) of clock time. In other words, the star time runs almost 4 minutes fast each day. Therefore, there are two key differences between sidereal time and the clock time:

Sidereal time is different at each longitude. Clock time is the same for all longitudes within a time zone.
To convert from clock time to sidereal time, you need to know the time and date.


How do I get flux density when I have mJy/beam , beam size and pixel size in an astronomical image

In essence read of the peak flux value and if a point source this will be the same as your integrated flux.

to elaborate on this:

In this question your source appears to be comparable to your beam. so I would treat as a point source. - note you might also want to consider that many people apply a "match filter" to their data if dealing with faint point source emission - this smooths the map - increasing the effective beam size within the map and also there may be some flux loss (see for example: http://adsabs.harvard.edu/cgi-bin/bib_query?arXiv:1707.00990)

You will see this (mJy/beam to mJy) is also discussed here: http://www.eaobservatory.org/jcmt/help/

For a real point source, a peak brightness value reported in units of mJy is the same as a peak brightness value reported in mJy/beam.

But what happens if we have a map in mJy/beam and we want to obtain an integrated intensity value, a total flux value? We first sum up a number of pixels and now we want to get our units correct from mJy/beam to mJy…

Total Flux = flux summed over a number of pixels/(number of pixels in a beam)

Then your units are:

[mJy*pixels/beam] / [pixels/beam] = [mJy].

Now we know that For a Gaussian:

Beam Area =  2 × π × σ² [arcsec]

where the σ of the Gaussian beam can be calculated from the JCMT FWHM values at 850 and 450 microns (reminder the beam components are provided in Dempsey’s 2013 paper).

FWHM = 2 σ √(2 ln 2)   [arcsec]

So we can use the FWHM to obtain σ to calculate the Beam Area and report the beam area in terms of pixels:

number of pixels in a beam = Beam Area [arcsec] / (pixel length)²

Finally I should add you will also want to do an aperture correction - see Dempsey et al 2013's calibration paper if you are using an aperture of a different size than 60".


This question has two parts:

Surface Temperatures

A very useful diagram which shows surface temperatures, and also gives you the temperature of any star you can observe is the Herzsprung-Russell Diagram, this one from le.ac.uk.

As you can see, the yellow of our own sun places it in the 4.5 kKelvin to 6 kKelvin, as noted in the question. This temperature is down towards the lower end of average. The main sequence, where most stars are, tops out about 20 kKelvin, and there are some up towards the 40 kKelvin region - they aren't shown here as they are much rarer.

White dwarves are a little hotter than our sun - between 6 kKelvin and 10 kKelvin.

Neutron stars are way off the main sequence - young ones can be over 1 MKelvin!

Core temperature:

Internally, the core temperatures are dependent on the mass of the star. In our sun, energy is delivered via the proton-proton chain mechanism, which occurs up to about 20 MKelvins, whereas more massive stars can use the Carbon-Nitrogen-Oxygen cycle - which happens from about 15 MKelvins upwards.

The differences are mainly down to convection and radiation differences - this extract from Wikipedia's Main Sequence page describes this in some detail:

  Because there is a temperature difference between the core and the surface, or photosphere, energy is transported outward via radiation and convection. A radiation
  zone, where energy is transported by radiation, is stable against
  convection and there is very little mixing of the plasma. By contrast,
  in a convection zone the energy is transported by bulk movement of
  plasma, with hotter material rising and cooler material descending.
  Convection is a more efficient mode for carrying energy than
  radiation, but it will only occur under conditions that create a steep
  temperature gradient. In massive stars (above 10 solar
  masses) the rate of energy generation by the CNO cycle is very
  sensitive to temperature, so the fusion is highly concentrated at the
  core. Consequently, there is a high temperature gradient in the core
  region, which results in a convection zone for more efficient energy
  transport. This mixing of material around the core removes the
  helium ash from the hydrogen-burning region, allowing more of the
  hydrogen in the star to be consumed during the main-sequence lifetime.
  The outer regions of a massive star transport energy by radiation,
  with little or no convection. Intermediate mass stars such as
  Sirius may transport energy primarily by radiation, with a small core
  convection region. Medium-sized, low mass stars like the Sun have
  a core region that is stable against convection, with a convection
  zone near the surface that mixes the outer layers. This results in a
  steady buildup of a helium-rich core, surrounded by a hydrogen-rich
  outer region. By contrast, cool, very low-mass stars (below 0.4 solar
  masses) are convective throughout. Thus the helium produced at the
  core is distributed across the star, producing a relatively uniform
  atmosphere and a proportionately longer main sequence lifespan.


How does one find the electric conductivy of a planet's atmosphere?

The conductivity of neutral gas is negligible, you need to estimate the free-electron conductivity in your gas.
For this, one could first consult a phase diagram, like referenced in this answer. There you'd see that for the atmospheric pressures and temperature for all gas giants, you will stay in the $H_2$-gas regime, so no exotic equation of state is necessary to describe the state of the visible atmosphere of the gas giants.

The updated models with and without Juno-data show that adjustments to structure models in the interior of the planet need to be made but the atmosphere, which you are interested in, is still an ideal gas.

To compute the amount of free electrons in the atmosphere, you need to compute the possible ions, and their ionization fraction. This gets really nasty for cold atmospheres, because molecules form. Mole fractions of those have to be computed by a chemical model, then ionized (possibly by a Saha-type ionization model) and translated into a conductivity (possibly by a Lorentz-Drude-type of model).

For your purposes that could get really involved, but I guess what you could do that is much simpler, is look at some published literature, e.g. here, just take and cite their conductivity profile.
However even in this source, the two models they use for the conductivity differ by 9 orders of magnitude in the neutral atmosphere, so you want to read carefully about their assumptions.


Are there any ways in which the light time difference between the center and the edge of the solar disk is noticeable?

@Astrosnapper's answer to Helioseismology of the Sun, what is actually measured? begins with:

  The two main methods to detect the solar p-mode oscillations, which have a period of about 5 minutes (so frequency of ~3 milliHertz) are line of sight Doppler shift and change in irradiance.

3 mHz is the peak of the power spectrum, but a 2D histogram of shows significant power at frequencies beyond 8 mHz, corresponding to a period of only 128 seconds.

What does this mean? A dopplergram is collected over an extended period of time, maps of line-of-sight velocity determined via doppler shift of a given spectral line, and fit to a model based on spherical harmonics. Below is an example of a dopplergram and a histogram of power as a function of frequency and spherical harmonic. The shading left-to-right is the doppler shift associated with the rotation of the Sun, I'm not sure if the orbital motion of the Earth and other kinematic effects have been subtracted.

I went with 8 mHz and calculated the time and phase delay due to the different distances of parts of the 695,700 km radius Sun at a given instant of time. (script):

I found some older data and a nice description Measurements of Frequencies of Solar Oscillations from the MDI medium-l Program by E.J. Rhodes, Jr., A.G. Kosovichev, P.H. Scherrer, J. Schou & J. Reiter which comes from The Michelson Doppler Imager aboard SOHO.

above: Fig. 1c: Full disk Dopplergram. (full-scale version). Full-disk Dopplergram obtained with MDI on 9 July 1996. The large left-right variation is the signal from solar rotation. The "texture" is a combination of the motion from the sound waves that permeate the Sun and the large convective cells called "super-granulation". The bright feature in the lower right quadrant is an active region. Source

above: Fig. 1b: Power spectrum obtained from 144 days the MDI Medium-l data for the modes averaged over the azimuthal order m. The power concentrates in ridges corresponding to solar acoustic (p) modes. The lowest weak ridge corresponds to the fundamental (f) mode. Source


No, the orbiting rocket was not visible to the naked eye.
Orbiter plus lem were about 65 square meters in size, about 25 X 25 feet.
At lunar distance they subtended about 0.000001 square degrees, and be at most 5 to 10 times as bright as the lunar surface; albedo 0.5 to 1 vs 0.1.
A speck 1/1000th of a degree on a side, and a little brighter than the moon is not visible in daylight. Likely it's well below the noise level in a daylight adapted eye.


Is the moon's revolution speed increasing or decreasing

From the cited article,

  The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth.

I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether "speed of rotation" means orbital velocity or angular velocity, both are decreasing as the Moon slowly retreats from the Earth.

The Moon is retreating very slowly from the Earth, currently about 3.78 cm per year. Dividing by 385000 km (the Earth-Moon mean orbital radius) yields about 10-10 parts per year. That qualifies as "very slowly". For simplicity, I'll assume the Moon's orbit is essentially circular, plus this very slow spiraling out. For a circular orbit, the relation between orbital radius and angular velocity is $$r^3\omega^2 = G(M+m)$$
This essentially is Kepler's third law. Assuming $G$, $M$, and $m$ are constant, differentiating both sides with respect to time yields
$$r^2\omega (3\dot r \omega + 2 r \dot \omega) = 0$$
meaning
$$r \dot \omega = -\frac32 \dot r \omega$$
Suppose some quantity $q$ is of the form to $q=r^n\omega^m$. Differentiating with respect to time yields $\dot q = r^{n-1}\omega^{m-1}\left(n\dot r\omega + m r \dot\omega\right)$. Since $r\dot\omega = -\frac32 \dot r \omega$, this can be rewritten as $$\dot q = \left(n - \frac 3 2 m\right) \dot r r^{n-1}\omega^m$$
Quantities of this form include angular velocity ($q=\omega$, so $n=0, m=1$ in this case), orbital velocity $(q=r\omega$, so $n=m=1$), and angular momentum (q$=r^2\omega$, so $n=2, m=1$).  Note that $(n-\frac 32m)$ is negative for angular and orbital velocity but positive for orbit angular momentum. Orbit angular momentum thus has the same sign as does $\dot r$ while angular velocity and orbital velocity have the opposite sign as does $\dot r$.

As a sanity check, look to centripetal acceleration. This is given by $a = r\omega^2$. In terms of the above, $n=1$ and $m=2$. The term $n-\frac32m$ is clearly negative in this case, so centripetal acceleration decreases as the Moon retreats from the Earth. Centripetal acceleration is also give by Newton's law of gravitation: It is proportional to the inverse of the square of the distance between the Earth and the Moon, so this does indeed decrease as the Moon retreats from the Earth.


Are the orbits of all triple star systems at least technically unstable?

(Ignoring that all orbits are technically unstable due to the emission of gravitational waves.)

There are known solutions to the gravitational three body problem that can be shown to be stable. Lagrange found a three body solution for general masses where all three orbit the common center of mass in an equilateral triangular formation. Gascheau proved in 1843 that this solution is stable if the component masses satisfy

$$ \frac{m_1 m_2+ m_1 m_3 + m_2 m_3}{(m_1+m_2+m_3)^2} < 1/27$$

More recently, it was shown by Kei Yamada and collaborators (1 2 3) that if one includes the first order Post-Newtonian correction, this solution is modified by a (small) correction to the  triangle legs based on the component masses, i.e. it is only an equilateral triangle if all masses are equal. The general effect of the 1PN interactions is to decrease the region of parameter space where this solution is stable, but for suitably far separated systems there will still be masses for which the system is stable. Moreover, they also proved that (4) these solutions are stable under the emission of gravitational radiation (i.e. a triangular system will adiabatically evolve into another triangular solution).

The effects of 2PN interactions (and beyond) on the stability of these triangular solutions is currently unknown (as far as I am aware).


Should the sun appear very low in the horizon if its path is always between the two tropics?

Your explanation is correct, so you are not remembering correctly. The farther north you go, the lower the sun is to the horizon at noon.

Of course it depends on how far north you go. If you only travel 5° farther north in latitude, then the sun is only 5° lower in the sky. 5° is a "long" ways to travel on the earth, but in the sky 5° lower is not that noticeable.


Computing the local orbital frame vector directions in a geocentric reference frame

I would interpret this as the Local Vertical, Local Horizontal (LVLH) frame (see, e.g., Vallado and McClain, Fundamentals of Astrodynamics and Applications, 4th ed., p. 156). This is also known as the RIC frame (radial, in-track, cross-track) or the RTN frame (radial, transverse, normal).

The radial or local vertical vector $R$ is just the position vector from the center of the Earth to the spacecraft.
The in-track, transverse, or local horizontal vector $I$ is the vector in the orbital plane that is both prograde and perinduclar to the radial vector. It is not, in general, the velocity vector. (The in-track and velocity vectors are coincident if and only if the flight path angle is zero, i.e., you have one of a) a circular orbit, b) the spacecraft is at apoapsis, or c) the spacecraft is at periapsis). You can construct it by applying the Graham-Schmidt Process to the position and velocity vectors; $I = v - \frac{R\cdot v}{R \cdot R}R$.
The cross-track or normal vector $C = R \times I$ completes the right-handed set.


Determining sunrise and sunset times based on azimuth and elevation

$
\cos ^{-1}\left(-\frac{\tan (\lambda ) (\cos (\lambda ) \cos (\phi ) \cos
(Z)+\sin (\lambda ) \sin (Z))}{\sqrt{(\cos (\lambda ) \sin (Z)-\sin (\lambda)
\cos (\phi ) \cos (Z))^2+\sin ^2(\phi ) \cos ^2(Z)}}\right)-\tan^{-1}
(\cos (\lambda ) \sin (Z)-\sin (\lambda ) \cos (\phi ) \cos (Z),\sin
(\phi ) (-\cos (Z)))
$

is the amount of time from now (see important notes below) a celestial object will set, where:

$\phi$ is the azimuth of the object
$Z$ is the altitude of the object above the horizon
$\lambda$ is the latitude of the observer

Note the two-argument form of arctangent is required for accuracy: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent

Important notes (MUST READ!):

The result is in radians. To convert to sidereal hours, multiple by $\frac{12}{\pi }$
If the object has a fixed right ascension and declination, divide sidereal hours by 1.002737909350795 to get clock hours.
For the Sun (which doesn't have fixed right ascension/declination), do NOT divide as above. By not dividing, you compensate for the Sun's change in right ascension.
For the Moon, see "The Moon" section below.
The time computed above is for geometric midpoint setting. In reality, refraction near the horizon means the object will set later. Additionally, for objects that have an angular diameter (eg, the Sun), setting usually means when the top edge disappears over the horizon, which will make the set time even later. Accounting for both effects should be possible, but might require numerical methods instead of a closed form formula as above.
If the quantity inside the arccosine is greater than 1, the object is always in the sky and never sets or rises.
If the quantity inside the arccosine is less than -1, the object is never in the sky and thus also never sets or rises.
I did only minimal testing. As always, do not rely on my answers for anything important.

The Moon:

The Moon's right ascension and declination change rapidly, so this calculation does not work well for the moon.
You could compensate for the change in right ascension (and thus hour angle) by approximating the increase as 24 hours every 27.32158 days (its sidereal period) and do an iterative calculation.
An even better compensation for the change in right ascension would be to approximate the moon's movement in ecliptic longitude (which is more constant than its movement its right ascension) as 360 degrees per 27.32158 days and then project the ecliptic longitude back to right ascension, and then iterate.
Compensating for the moon's change in declination is more difficult. The moon's ecliptic latitude (which can be converted to declination) varies sinusoidally, but the equation $\sin (x)=a$ normally has two solutions. Unless you know whether the moon's ecliptic latitude is increasing or decreasing (ie, whether it's between ascending and descending nodes or vice versa), you won't know the direction of declination change.

Less important notes (optional):

See Need Simple equation for Rise, Transit, and Set time for more general equations on when an object rises/sets/etc.
Calculations for this answer at: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-object-riset-from-az-elt.m
Related calculations: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m
Mathematica was unable to find a simpler form for the "time to set" above, though I sense a simpler form does exist (I could be wrong).
If you know the Sun's declination (which you can get from its azimuth and elevation as above), you can almost determine the date. However, the sun reaches a given declination twice a year (example: it reaches 0 degrees declination on both equinoxes, by definition), so you can only know it's one of two days.

Solution notes:

I learned quite a bit answering this question, and thought it was only solvable numerically until I figured out the shortcut:

To convert from the azimuth/elevation sphere to the hour angle/declination sphere, you just rotate around the y axis by $\frac{\pi }{2}-\lambda$ (90 degrees minus the latitude) and then rotate $pi$ (180 degrees) around the z axis.
Once you have the declination, computing the hour angle when an object sets is easy.
You then subtract the setting time from the current hour angle to get the answer.

TODO: I was hoping to add some graphs/visualizations to this answer but couldn't quite get things working.


How would the solar system look in a Geocentric model?

Only for Sun, Earth and Mars, but here's an interactive Flash that illustrates it graphically. Third link from the top "Part I, Equivalence of Hypotheses (Flash)" Note that the planets don't move as you move the mouse over the image, only the circles (the theory) change.
http://science.larouchepac.com/kepler/newastronomy/part1/MegaEquivalence.swf for the animation. More generally: http://science.larouchepac.com/kepler/newastronomy/
(And I don't know or care who that LaRouge Pacman is, some rich guy for some reason financed good explanations of what Kepler did, that's what I know about it and that's all very good)

If you move the mouse pointer to the upper left you get the heliocentric system. To the upper right you get the geocentric system. The point is to illustrate how Kepler proved that Ptolemy, Copernicus and Tycho all had the same mathematical model for the solar system. They just used different frames of reference to it, just choosing different fixpoints from where to view it.

Some dots in the illustration represent important points in that ancient model. Some of them are stand-ins for the foci of an ellipse in a model made out of perfect circles. And the Sun was actually not the center of the Copernican model, a point near the Sun was. According to Copernicus the Sun was just a decoration which randomly happened to hang around near the center. Kepler was the first to come up with the thought that maybe there is a relationship between the geometry and the physical existence of stuff. An idea later developed to the theory of gravity by Newton and Einstein, who just added to Kepler's initial formula for describing the movements of the planets.


To clearify: In your question you distinguish between a star forming a neutron star when its core exceeds the Chandrasekhar limit and a star undergoing a Type II supernova.
These are, however, the same events.

To summarize: Generally an isolated star can die in one of basically four ways. With increasing probability with increasing mass these outcomes are:

1. Formation of a white dwarf: This is the outcome for stars with a initial mass below ~10 solar masses and includes our sun. This includes about 97% of all stars, as small stars are the most abundant.
At the end of its lifetime, when it has burned though all the hydrogen accessible to its core, it will start to burn gradually heavier elements in its contracting core. Up to which element this can go depends on the initial mass of the star. During this process the outer layers of the star begin to expands and it enters the red giant phase. At the end of which they shed their outer layers which then form a planetary nebular. All that is left after that is the cooling core of heavier elements. The white dwarf. This white dwarf is held up against gravity by degenerate electron pressure. This means the electrons are so close together that, to contract further, they have to be pushed in an higher energy state to not violate the Pauli exclusion principle.
An example would be the beautiful Ring Nebular where you can easily see the white dwarf in the middle.

2. Supernova and the formation of a Neutron star:
The core of heavier stars ($9-100 M_{\odot}$) are, too, mostly held up by degenerate electron. During the lifetime of the star more and more heavy elements collect in a growing core. With increasing core mass/density, however, the electrons reach higher and higher energies. At some point these energies become high enough for electrons to be "pushed" into protons, becoming neutrons and releasing lots neutrinos in the process. This reduces the degenerate pressure until it can no longer support the core against the gravitational pressure of the outer layers.
This happens when the core exceeds the mentioned Chandrasekhar limit. This is what initializes the core-collapse of heavier stars.

Even heavier stars with hotter cores are increasing held up not only by degenerate pressure but regular thermal pressure. For those stars the collapse is additionally promoted when photodisintegration of heavy elements saps the core of more and more thermal energy and leads to its further contraction.
(See the first pages here for a more detailed description of how the SN is triggered)

Independent from how the collapse started, it continues until at some point the degenerate neutron pressure and most importantly the repulsion of the strong nuclear force kicks in.
At this point the core-collapse suddenly halts and the proto-neutron star becomes virtually incompressible. This all happens so fast, that the outer shells basically free-fall onto the proto-NS and "bounce off". Together with the energy provided by the intense neutrino flux, this drives away the outer shells and leads to the explosion of the star. This is a core-collapse supernova. All that is left is a expanding supernova remnant and a central neutron star, a pulsar. The composition of the remaining outer layers at the time of the explosion determines the type of the SN.
Most famous example: The Crab nebular.

3. Core-collapse SN with black hole formation:
At higher masses it can happen that not all of the outer layers can escape the gravitational well of the proto-NS. Some part of it can fall back onto it. During this process the NS can gain mass again and eventually exceed the Oppenheimer-Volkoff-Limit. When this happens even the pressure from the degenerate neutron gas is not sufficient anymore. At this point there is nothing that can stop the gravitational collapse to a black hole. The accretion of the additional material again releases huge amounts of energy and can lead to the formation of jets. This is what is suspected to be a Hypernova and can be accompanied by a gamma ray burst (GRB).
See below for and animation of a jet braking though the expanding supernova shell and producing a GRB:

4. Core-collapse with direct BH formation:
An alternative, less violent way for forming a BH is a direct collapse. This mostly happens for very heavy stars.
In this scenario the proto-neutron star is unable to initialize the ejection of the outer layers and just collapses to a BH. No huge explosion, no supernova remnant.
This is sometimes called a failed supernova. And we have actually seen such a star just disappearing:

5. Pair-instability supernova and complete disruption:
For stars with masses between about 140 and 260 $M_{\odot}$ (see respective gap in the figure below) temperatures in the core can become hot enough that thermal radiation reaches gamma ray energies. Specifically photons can reach energies high enough to produce e$^-$/e$^+$ pairs. This process transfers thermal energy into the mass of the formed particles. This leads to a contraction of the core. At that stage the core has not jet fused many heavy elements and consists mostly of Helium, which now undergoes rapid fusion as the core contracts. The released energy complete disrupts the stars leaving no compact core and only an expanding shell. Such heavy stars mostly formed in the early universe as the first generation of stars.

Which one of the scenarios 2 - 4 will happen to a star with a zero-age main-sequence (ZAMS) mass of more than 15 solar masses is hard to tell and an active field of research. It all depends on ZAMS mass, binarity, metallicity, angular momentum, and magnetic fields, among others properties of the progenitor star. Current computational capabilities are just not enough to produce reliable 3D magneto-hydrodynamic simulations with good temporal and spacial resolution and scale.

But from here there is a rough estimate of what we think is the most probable outcome for different masses and metallicities:

And from here there is a set of actual 1D simulations of the supernova 1987A showing all kinds of different outcomes for different masses and initial models of the star:
(Although 1D simulations tend to have a hard time to initialize explosions)

How complex these simulations are can also be seen in nice 3D videos like this one.

So you see, to determine which stars go through which final fate we have to torture huge supercomputers and even then we still have a hard time coming up with definitive answers.


Why would a tidally-locked rocky planet have a first-order spherical harmonic surface temperature distribution?

It probably wouldn't. But when studying these things, you don't want to go in assuming you know more than you do or you might bias the analysis.

I am thinking that the incident flux at a given point on the tidally-locked planet would be
$$I∼\max(0,\cos(\theta))$$
where θ is the static zenith angle at a given point, and so the temperature would be something like
$$T∼I^{1/4}∼\max(0,\cos(\theta))^{1/4}.$$

Unfortunately this assumes that the planet is a blackbody with zero heat redistribution over the surface, e.g. by winds or currents in a magma ocean, and no sources of heat on the nightside such as tidally-driven hyperactive volcanism which is something you certainly do not know starting out.
Spherical harmonics are a generic set of basis functions over a sphere, so it does make sense as a fit which does not assume any physical processes operating or their relative importance. In fact, this is noted in Louden & Kreidberg (2018) "SPIDERMAN: an open-source code to model phase curves and secondary eclipses" (reference 15), which is referenced when they talk about using a spherical harmonic model. Conveniently, the paper shares the lead author with the LHS 3844 b paper, so presumably this does reflect some of the thought that went into the LHS 3844 analysis. A relevant quote from that paper:

A useful and physics-independent model is a sum of spherical harmonics. This method was used for the case of the phase curve of HD 189733b by Majeau et al. (2012). An example map generated by SPIDERMAN is displayed in Figure 4. The main observational features of a phase curve, including the offset hotspot, can typically be recovered with a only the first spherical harmonic, with the centre offset from the substellar point. (Cowan et al. 2017) explore the effects of odd harmonics in phase curve data, and find that these can correspond to weather features in the planet atmosphere.

(emphasis mine)
Kreidberg et al. (2019) do note that a simple sinusoidal fit produces unphysical negative temperatures on the nightside which can be corrected using odd harmonics, hence their choice of the first-order model.
Physics-based models probably would start to introduce a whole bunch of poorly-constrained parameters, probably overkill for the first analysis: this is mapping the terrain, figuring out how the terrain ended up like that is another stage entirely. This is a Nature paper, so brevity is par for the course.


What kind of reddish wood is it exactly that was historically used to line the inside of observatory cupolas (domes), in order to absorb moisture?

Probably western red cedar, thuja plicata, especially if the observatory is in North America. Western red cedar is easily obtainable in North America. It has a light, red, wood that is particularly resistant to decay, and ideal for use in exposed positions.

  The soft red-brown timber has a tight, straight grain and few knots. It is valued for its distinct appearance, aroma, and its high natural resistance to decay, being extensively used for outdoor construction in the form of posts, decking, shingles, and siding.[28] It is commonly used for the framing and longwood in lightweight sail boats and kayaks. In larger boats it is often used in sandwich construction between two layers of epoxy resin and/or fibreglass or similar products. Due to its light weight—390 to 400 kg/m3 (24 to 25 lb/cu ft) dried—it is about 30% lighter than common boat building woods, such as mahogany. For its weight it is quite strong but can be brittle.

Wikipedia: Thuja plicata

It may be worth pointing out that while "cedar" refers to a group of Cedar woods, trees of the proper Cedrus genus are rare and localized:

  Cedrus (common English name cedar) is a genus of coniferous trees in the plant family Pinaceae. They are native to the mountains of the western Himalayas and the Mediterranean region, occurring at altitudes of 1,500–3,200 m in the Himalayas and 1,000–2,200 m in the Mediterranean.

Wikipedia: Cedrus


I might be misunderstanding your question, but using HORIZONS (http://ssd.jpl.nasa.gov/horizons.cgi) with these settings should give you the data you need:


This is an interesting question!
Let's first ask; What does "200x" mean exactly?
If we set a photocopy machine to 1.5x or 2x we get a paper copy that we can set next to the original and compare them directly. With a ruler we can confirm that something that's 1 inch on the original is 1.5 inches or 2.0 inches on the copy.
But how big or how much closer something looks is a more subjective because our vision system is not flat. Our vision works with angles rather than distances.
For small angles like several degrees or so, we can use the small angle approximation fairly well, so let's go with that.
When a telescope is focused far away, near infinity, perhaps 100 times farther away than the length of the telescope itself, we can treat 200x as the angular amplification of 200.
Something in the distance (e.g. 4 miles away) that's 0.01 degrees wide will appear 200 x 0.01 = 2 degrees wide through the telescope.
If we move that object 200 times closer, it will also appear almost exactly 2 degrees wide.
So as long as we are considering things far away and small, we can say that something that appears 200x larger through the telescope would appear to be the same size as if it were 200x closer.
4 miles times 5280 = 21,120 feet. Divide that by 200 and we get 105.6 feet.
At 21,120 feet I would appear about 0.015 degrees tall. Viewed through a 200x telescope I would appear about 3 degrees tall. If I moved to 105.6 feet away, I would also appear about 3 degrees tall.
This works as long as we use the small angle approximation and our telescope is focused on things far away, say 100 times the length of the telescope or farther. The same 200x telescope would not necessarily provide exactly 200x angular amplification if they eyepiece were pulled out so far that one could focus on something across the room.


How do you find the altitude of the Sun if you are on the Moon?

@DavidHammen is essentially correct. According to stellarium, the Sun gets as high as 89 58' 48":


How can the universe be expanding faster than speed of light?

It was a loud explosion

First, we have to clarify what the Big Bang was. The name is a misnomer, as it was neither loud nor an explosion. All we know is that the Universe is currently expanding — that is, space is literally being created between all matter. We have plenty of answers here explaining this more intuitively, such as mine (which also answers your question about its implications on special relativity).

Anyway, if we were to look further back in time, we'd expect there to be less and less space between matter. Our equations tell us that at 13.8 billion years in the past, the amount of space in the Universe should have been zero.

This is a problem. In general relativity, you can't have a metric with zero space. Thus, we know general relativity is incomplete. It cannot explain what happened at the moment of the Big Bang.

  [the] universe came to existence

No such claim is made. All we know is that something must have happened back then. Our current physical models do not accurately represent the extremely early Universe.

  Hence my question is that everything is traveling faster than the speed of light?

The expansion of the Universe is not happening at the same rate as it was back then. It is slow enough that everything within our Hubble sphere is receding at a lower velocity.


Would things be exactly the same if the Sun was orbiting the Earth and the Earth was in the center of the Solar System?

It depends on what you mean by "in the center" or "switch places."  If you physically moved the Earth to where the Sun is, and the Sun to where the Earth is, and gave each the speed that the other had while keeping the same reference frame, this would do tremendous violence to all the other planets in the solar system (though the Earth would continue to orbit more or less the same).

However, if what you are talking about is simply the language of what you call the center, then you are talking about changing to a different set of coordinates where the Earth is at the stationary center of your coordinate system.  Doing that presents no problems, and nothing is different except the language you are using to talk about what is going on.  One of the great things Einstein did was to reframe the laws of physics in a form that was independent of coordinates, allowing us to forever separate what the laws say is happening from the language we use to talk about what is happening (i.e., our chosen coordinates).  So it all comes down to whether you think "being in the center" is a real thing, or just the language of some coordinate system.  If the former, then no, things would not be at all the same, if the latter, then yes, they'd be exactly the same, if you are careful to use coordinate-independent laws like general relativity.


Complex life in binary black hole - Sun(s) system

The Roche limit for a solar-mass (or 10-solar-mass) black hole is much larger than the Schwarzschild radius, or even the last stable orbit (which is a few times the Schwarzschild radius). So the Roche radius is a good limit (and, yes, you're right about the limit for an Earth-like planet and a 10-solar-mass BH.)

You needn't assume the BH has an accretion disk; I suspect most of the ones wandering around in our galaxy don't at the moment, just because accretion disks accrete (go into the black hole) over time, and so you only see them when there's some relatively recent source of gas to create the disk. (Most of the Milky Way black holes we know about are in close binaries, where the companion star is close to the Roche limit and is a regular source of gas.) Now, in principle the stellar wind(s) from the star(s) in your system could contribute to a very weak, tenuous accretion disk, but in practice it's unlikely to produce enough radiation to be of any concern, especially as you're assuming a planet with an atmosphere.

So there's no need to worry about whether or not the planet is tidally locked to the black hole, unless you want to specify the length of the day-night period. Of course, if the planet is only a few tenths of an AU away from the BH, it probably will be tidally locked before very long.

The variation in distance from the star(s) you suggest should be fine in "habitable zone" terms. In fact, given that the planet orbits the BH quite rapidly (about 1.3 days for a 0.05 AU orbit around a 10 solar-mass BH), it won't spend much time at either extreme, so you could probably have it go inside and outside of the formal habitable zone and still be habitable, as long as the average insolation is in the habitable range.  This would create some interesting day/night/seasonal effects, particularly in the tidally locked case: e.g., at a certain time of "year", the "eastern" hemisphere would experience noon when the stars were only 1.95 AU away, while for the "western" hemisphere noon would happen when the stars were 2.05 AU away. Half a year later, the situation would be reversed.

One last curiosity: assuming the black hole has an "average" spin, would it be realistic if the two Stars had a tilted orbit with respect to the black hole spinning axis?

That really depends on how the system formed in the first place. If the initial pre-BH star and the other star(s) formed from the same collapsing cloud, they would probably have similar spin and orbital angular momenta; if the spin of the BH is entirely inherited from its progenitor (which goes supernova to form the BH), then the orbital plane might be similar to the BH's spin. But if the system got put together later (maybe an encounter between a binary consisting of the BH and another star and the current binary pair, in which the BH's original companion star gets ejected), then the two stars could have a tilted orbit.


Your question is valid and is a common operation. When computing when an object will be visible, many observers want to be able to draw a globe or map that is marked with the position from which an object like Pluto is directly overhead. Such a position on the globe at a given instant (and you do say “at the current time”) is, all other things being equal, the best-situated place from which to observe the object.

Unfortunately, the “libastro” C library that PyEphem wraps seems to only provide one instance of this concept: Earth satellites have .sublat and .sublong attributes, because it is so common to want to draw the path of a satellite on a globe or map.

But libastro does not generalize the concept. From what I can see, there is no way to generate the point on the Earth that is directly below any other Solar System object. And since I have not been in the habit of trying to extend libastro, it is likely that PyEphem will not gain this ability.

However, I have been developing a replacement for PyEphem that is written in pure Python and that I will be free to extend, called Skyfield. I will hopefully have this concept working there soon, and when I do so I plan to make it work for all objects, and not limit it to Earth satellites!

(In the meantime, as pointed out in the comments: you could try asking PyEphem for the geocentric RA and dec and, if you could adjust RA by the Earth's current hour angle, then you could turn those into an approximate latitude and longitude.)


Does a continuous deformable mirror cause diffuse reflection?

Given your reply it seems your definition of diffusion is slightly different from my sense. But the phenomenon is true that at segmentation locations wavefronts are being deflected not in a uniform manner, causing the diffusion issue posted in this question.
Before addressing this issue, let me briefly describe how adaptive optics works in practice. In a typical AO system, a wavefront sensor measures the incoming wavefront distortion (usually it's slopes), given which a deformable mirror corrects for. The deformable mirrors consist of multiple actuators, which enable the mirror deformation. These actuators are controlled by commands.
The input/output are:

Input: wavefront slopes $\mathbf{y}$
Output: actuator control commands $\mathbf{x}$

These two are usually related through a so-called interaction matrix $\mathbf{A}$:
$$
\mathbf{y} = \mathbf{Ax}
$$
This matrix models all system artifacts, including the diffusion issue in this question. Since we would like to solve $\mathbf{x}$ from $\mathbf{y}$, we would also like the pesudo-inverse of $\mathbf{A}$, denoted as $\mathbf{A}^{-1}$, so in real-time the AO system can solve the actual commands as:
$$
\mathbf{x} = \mathbf{A}^{-1} \mathbf{y}
$$
$\mathbf{A}$ and $\mathbf{A}^{-1}$ are usually calibrated before running the AO system.
Now consider this question:

Yes, there are diffusion reflection from the mirror. But this issue contributes in part to the interaction matrix $\mathbf{A}$. As long as people calibrate it carefully it does not really affect AO performance.

For normal AO systems both measurement and correction wavefront lateral resolutions are low: typically at thousands of measurements, i.e. $\mathbf{x}\in\mathbb{R}^{1000}$ or so. In such a situation, I would conjure that the diffusion reflection at segmentation locations is not significant enough compared to the overall sub-mirror reflection, i.e. the fill factor of the sub-mirror is high enough to ignore this effect.


Did we discover how nebulae hold their shapes?

"Nebula" is quite an imprecise word; over the years, it's been applied to everything from interstellar dust clouds to supernova remnants to the birthplaces of stars to even galaxies. Wikipedia separates true nebulae into four categories: H II regions, planetary nebulae, supernova remnants, and dark nebulae. In this case, it might be better to make the distinction between nebulae that primarily surround on star/star system and nebulae that exist in interstellar space, potentially encompassing many stars.
Circumstellar nebulae
I'll classify planetary nebulae and supernova remnants as circumstellar nebulae. They're both the result of a star passing into the final phase of its life.
Planetary nebulae (which include the Cat's Eye Nebula) are the remains of the outer layers of gas given off by relatively low-mass stars, like the Sun. Balick & Frank (2002) subdivide these based on morphology, into four categories:

Round
Elliptical
Bipolar
Irregular

The names describe the nebulae's characteristics fairly well. However, they show observation bias; for instance, a bipolar nebula may appear to be found if its lobes are lined up perfectly with Earth. The exact reasons for the distinctions aren't well-know; Wikipedia cites Jordan et al. (2002) (which I think you may have already read) in saying that magnetic fields could be the cause, but there are obviously other reasons for a lack of spherical symmetry: intense rotation (possible in an AGB star), violent mass loss (unlikely in stars of this low mass), a strong stellar wind, interactions with other bodies, etc. The latter may have caused the dramatic lobes of M2-9, where interactions with a binary partner produce extreme distended streams of gas:

Image courtesy of the ESA under the Creative Commons Attribution 3.0 Unported license.
Supernova remnants fall into three basic categories:

Shell-like, which are simply spherical shells
Composite, characterized by radio emission from a pulsar (with potential shell-like structure)
Mixed (also sometimes termed "composite"), which has clear characteristics of both of the above, with X-ray and radio emissions

NASA has three pages on the different types, though it uses "composite" in place of "mixed", designating "composite" nebulae "Crab-like", after the Crab nebula. The shape of a remnant depends on which type it is; for instance, composite nebulae may be shaped by the stellar wind of the central pulsar.
Interstellar nebulae
H II regions (which include the Eagle Nebula, and of course the Pillars of Creation) and dark nebulae fall into this category. Both may be part of or interact with molecular clouds, which are the star-forming regions of the galaxy. H II regions, like molecular clouds, are shaped by the stars inside them. Radiation ionizes and even pushes away the gas surrounding young, hot stars, breaking up the nebulae. However, variations in dust density mean that different parts may "evaporate" at different rates, leading to some weird shapes - including the Pillars of Creation.
Evolution and dissipation
When I first read your question, I focused on the phrase (emphasis mine)

how nebulae hold their shapes

because nebulae are actually often extremely short-lived, on astronomical timescales. While the exact lifetimes vary, they're typically quite short when compared to stellar lifetimes:

Planetary nebulae: On the order of $\sim10^4$ years, although this may be influenced by the progenitor and the nebula's morphology (see Badenes et al. (2015)).
Supernova remnants: On the order of $\sim10^5$ or $\sim10^6$ years, although this depends on how you define the lifetime of a remnant. It can differ if you define it on X-ray emission, radio emission, or something else entirely.
H II regions: On the order of $\sim10^6$ or $\sim10^7$ years, depending on the evolution of the star(s) inside the region. Their radiation pressure will dissipate the gas cloud, so its lifetime depends on their characteristics. As Stars, Nebulae and the Interstellar Medium explains, H II regions would live longer if they weren't dissipated by the stars. Otherwise, they could survive for orders of magnitude longer.


If we draw a circle and draw a 315 degree angle on that circle, in the counter clockwise direction, we would roughly get the same point as when we draw an angle of 44.8 degrees in the opposite direction (the equivalent of drawing -44.8 in the same direction) because 315+45=360. There therefore does not seem to be an error in your results... Visually the two angles are the same. As for the slight inaccuracy, it may be due to floating point precision errors or measurement errors.
If you want positive right ascension, all you need to do is add 360 if it is negative. Before doing that (this is general advice, it may not be important to your code), you ought to subtract the decimal part, get the remainder on dividing by 360, and re-add the decimal part.  Edit: The difference between GST and GMST is called the equation of time (for Grinich), and it does not grow beyond 16 minutes, so I think you can use it for estimates, but then there would be no point of rounding your results to the tenth decimal or so. I honestly don't know if there is a "GST for dummies" (normally I wouldn't post an incomplete answer, but the question was edited), but using GMST should suffice if the 16 minute error margin is acceptable.


Can electromagnetic waves be affected by gravity?

Yes (and somewhat No apparently too)
If you follow Einstein's theory of general relativity then gravity has no direct effect on electromagnetic waves itself.
Gravity however affects space time itself warping it towards the gravity source. Thus any electromagnetic wave passing close enough to the gravity source will bend towards that source when observed from the outside.
From the point of the wave however, it follows a straight path and is not affected by gravity at all.
If you mix in quantum theory, it's getting even more weird, because suddenly parts that work well for general relativity no longer work within the quantum model. This is part of current research in physics to bring general relativity and quantum theory into alignment.


WHY implies a reason, which is venturing in to the religious realm.

The circumstances under which an event such as the Big Bang occurs are guess work at best. Since the Big Bang is a singularity from which time itself started there is no real before the big bang. That's the first problem. The second is that because of the spatial singular nature of the big bang no information from 'before' it could ever pass through to our current universe and provide us with any hints.

There are a number of oddball theories that explore the idea of physics before the big bang. Some investigate what occurs on scales of time shorter than that of the Planck time, where quantum mechanical processes may give us hints in to the cause or mechanics of a big bang event. Proving or, indeed, disproving these theories is currently impossible experimentally.


Distinction between signal from terrestrial and signals from celestial sources

There's a few indicators where a source is located:
Direction: Radio waves are just like light waves only with longer wavelength: you have to look into the direction of the source in order to see it. Terretrial sources usually are not high up in the sky.
Apparent movement: Another important aspect is Earth's rotation: the sky rotates with respect to Earth's surface. Thus if the source moves with the rotation speed of Earth and remains stationary in a celestial reference frame, it is very likely a source outside of Earth - and vice versa.
Location / Parallax: If you have several receivers, they will disagree on the direction a signal comes from, but will agree much better the further away a source is. Celestial sources are far away enough that the direction will be (nearly) identical. The amount of disagreement, the so-called parallax gives you a geometric indicator about the distance to the source.


How to calculate the maximum and minimum solar azimuth at a given location?

Thanks to Ralf Kleberhoff for Pointing me in the right direction.

Step 1: Find Summer Solstice Date

For the accuracy needs of this question simply finding the hemisphere and picking the most common date will work. (June 21 in the Northern Hemisphere and December 22 for the Southern Hemisphere). If greater accuracy is required. There are many resources available.

Step 2: Find Sunrise and Sunset Times on Solstice Date

This is straightforward. However, for completeness. (Note: 90.833 is the solar elevation angle for sunrise and sunset.. However, 90 is sufficient for this accuracy and simplifies the equations.)

$ha = \arccos\left(\frac{\cos(90^o)}{\cos(\Xi^{rad})\cos(\delta^{rad})} - \tan(\Xi^{rad})\tan(\delta^{rad})\right)^o =  \arccos\left(-\tan(\Xi^{rad})\tan(\delta^{rad})\right)$

$solarnoon = \frac{720 - 4\Xi - \Gamma}{1440}$

$sunrisetime = solarnoon - \frac{4ha}{1440}$

$sunsettime = solarnoon + \frac{4ha}{1440}$

Step 3: Solve Azimuth Equation for Sunrise and Sunset Times

Simply plug the $sunrisetime$ and $sunsettime$ into the equations provided in the question.


How many (or percentage) satellites of Solar system planets have their orbits synchronized like the Moon (always facing earth)?

This is not really known; while the rotation period of most of the larger satellites is known, and they're mostly synchronized, for the smaller ones it's harder to determine and they're less likely to be synchronized:

  One conclusion is that, other things being equal ... a large moon will lock faster than a smaller moon at the same orbital distance from the planet. ... A possible example of this is in the Saturn system, where Hyperion is not tidally locked, whereas the larger Iapetus, which orbits at a greater distance, is.

Wikipedia has a list of synchronized satellites and 'suspects', if you want to investigate further.


Go to the Kepler/K2 MAST database https://archive.stsci.edu/kepler/data_search/search.php or https://archive.stsci.edu/k2/epic/search.php

Search on a temperature ($T_{\rm eff}$) range: e.g. for K-stars  4500 .. 5200

A (big) list of objects will be found and presented to you in a table. Mark the ones you want (all the Kepler objects should have available data, but not all the K2 ones will unless you limit your search to the earlier K2 campaigns) and then click on the "submit marked data for retrieval from STDADS"

It will then ask you to confirm that you want to download the data with various options to get at the data and a tick box list of what you want (including light curves).

Then once you've got the data you need to look at the FAQ and descriptions of data format etc. that can be found at the main Kepler archive pages https://archive.stsci.edu/kepler/

If on the other hand you just want to look at the light curves then you can do that without downloading the data by clicking on "plot marked light curves" on the search results page.


The photo shows a conjunction of the Moon and Jupiter.
A few hours later, observers in Europe
(APOD,
virtualtelescope.eu)
saw the Moon pass directly in front of Jupiter, an event called an occultation.

Stellarium shows Venus and Aldebaran below them, but in the photo they are probably behind a cloud.
Capella is visible at far left, slightly higher than the Moon.


One truly safe way of solar flare observations is to not directly observe them (with your eyes or imaging technology) at all, such is the case with radio-astronomy.  There are a number of examples of how this could be done, for example, the Sudden Ionosphere Disturbance Monitor, this device measures disturbances in the ionosphere that includes the effects of solar flares.  This operates in the VLF (3-30kHz) range.

From the Society of Amateur Radio Astronomers, the advice is

  For VLF solar flare observations you will need a strip chart recorder and a radio receiver capable of operating in the noisy 20 to 100 kilohertz radio band. These receivers are quite simple and may be home constructed.

(Plans for these can be obtained from that group)

A repository of further links (far too many to list here) can be found on the website Amateur Radio Astronomy on the Internet.


In general one can only measure the mass of a star if it is orbiting another body or is orbited by another body.

In essence, the method exploits Kepler's third law, though the exact technique depends on what observations are available, on what timescale and with what resolution (spectral and angular) is available.

Other techniques can be calibrated using such objects as exemplars - for instance there are reasonably well-calibrated mass-luminosity or mass-color relationships.


Any plans for TESS after it finishes northern sky survey?

NASA missions that have gone beyond their original Prime Mission lifetime, such as TESS, go into the NASA Senior Review process every 3 years. The most recent one of these was the 2019 Senior Review which reviewed Hubble, Chandra, Fermi, NuSTAR, NICER, Swift, Newton-XMM along with TESS. This review panel looks at the whole operation of the mission, whether it's doing good science, how the operations are conducted, how much lifetime there is left in the spacecraft and the cost of continuing operations.

The full panel report is available but all 8 missions were recommended for continued funding and TESS was rated as "best of the rest", just behind Hubble and Chandra (which have an enormous science legacy). The panel report is a recommendation to NASA but they normally follow the report as best they can and as the budget permits (Congress can, and occasionally does, intervene in this process by adding or removing funding for specific missions).

As detailed in the NASA response to the panel, the 8 missions were all extended for the Fiscal Year 2020-2022 but as noted, this is contingent on NASA's Astrophysics Division receiving the funding it asked for in the FY2020 Presidential Budget Request and the Congressional budgeting process. Extensions beyond 2022 are possible but depend on the results of the 2022 Senior Review. The TESS spacecraft is in a very stable orbit which allows it to maintain the orbit with minimal use of the propellant consumables on board (which normally sets the end of the mission as was the case for Kepler/K2 for example)

The panel noted:

  The expected planetary return for TESS’s extended mission is high, pushing well beyond the prime mission. Due to the longer time baseline, together with solid Kepler exoplanet demographic statistics, the team makes a strong case that the extended mission will nearly triple the detection of planets smaller than 4 Earth radii in the extended mission, compared to the prime mission, nearly triple the number of detected planets in the habitable zone, and more than triple the number of planets with periods beyond 20 days


If you peruse Google for Earth cooling rates, you might find what you're looking for. Here's one of many sources.  I strongly recommend reading the entire page,as it discusses some of the geology-based estimates as well.

  Kelvin's primary attack on geologic dating was that measurements of
  the rates of geologic processes were highly uncertain, if it was
  possible to measure them at all (Hallam, 110). Accordingly, Kelvin
  sought to apply the knowledge of physics to the problem. Given that
  temperature increases the further one descends below Earth's surface,
  Kelvin concluded that Earth was slowly cooling. He set out to
  calculate the time required for the Earth to cool, and thereby
  solidify, from an initially molten state. The idea that Earth had
  begun as an incredibly hot sphere of liquid dates back to Descartes
  and Leibnitz. This assumed initial condition was the linchpin for
  Kelvin's entire method (Hallam, 110). Bits of material at the surface
  would sink before solidifying, creating convection currents that kept
  the Earth at a uniform temperature until solidification began at the
  core (Hallam, 110; Knopf, 445). Kelvin needed to know: (1) the
  temperature at Earth's core, (2) the temperature gradient with regard
  to depth below the surface, and (3) the thermal conductivity of rocks.
  The gradient was established to be around one degree Fahrenheit for
  every fifty feet. Kelvin made his own measurements of conductivity.
  The problem was determining the temperature at the core. This is where
  Kelvin's theory of solidification enters the picture. Because the core
  was thought to be solid rock, its temperature could not exceed the
  melting point of rocks (Hallam, 110). He constructed the following
  equation:

  dθ / dx = S / h √(π t)

  The temperature gradient is expressed by dθ / dx; h is the thermal
  conductivity; x is the distance below the surface, and θ is the
  temperature (Holmes, 445). In 1862, Kelvin arrived at a likely age of
  100 million years. Because of uncertainties in the data, the lower and
  upper limits were 20 million and 400 million years (Dalyrmple, 26;
  Hallam, 111).

That said, there are plenty of more recent papers which estimate the cooling rates (with and without the help of the radioactive elements) based on the kinds of data now available as to the current and historical makeup of the core, mantle, and outer layer(s), heat transfer cycles, etc.


Could there be additional Neptunian satellites between Larissa and Hippocamp?

[My answer may be a bit off as astronomy is just my hobby]

First off, I would assume eccentricity has little function into the investigation because why would it be a moon when it can escape Neptune's gravity?[assuming you mean Neptunian eccentricity] I mean, unless you are thinking about a moon escaping?
Now, let's move onto the distances. You notice when you graph the moons that there is a considerable gap in distance between S/2004 N 1 and Larissa. I believe, after checking this on multiple sites, that there is indeed such a gap.
Neptune Moon Origin
First, let's take a look at one of the most accepted theories on Neptune and its moons. It is well known that Neptune is very close to the Kuiper Belt. In the Kuiper belt, there are objects like Pluto in orbit around the Sun. Well, Neptune having a very strong gravitational pull would actually yank objects out of orbit into its orbit. Most people believe that this is what happened with Triton. Making up more than 80% of the mass of all of Neptune's Moons,  if would have wreaked havoc on the then stable orbits of its original moons.
This consequently could have cause many moons that were large[relatively speaking] to gain/lose eccentricity which would have cause some to spiral inside of the Roche Limit of Neptune ultimately causing the formation of several rings around Neptune. However, in the case of Saturn, we see moons inside of the Roche limit, so we can also deduce that the Roche limit of a planet varies on how massive/large an object is.
We can now conclude that almost all of Neptune's current moons and their orbits are not its original ones. To answer your question, we look at the three body problem. I am not a physicist to I don't know much on the relative equations for the three body system to calculate the exact orbits of possible moons or calculate proof why there isn't a moon. However, I will give you a relative, not full equation blast, explanation.
Three body problem
The three body system states that in long term, the three body problem may lead to either and for the most part, stabilization of the orbits of moons around a planet or, for some period ratios, be the cause of the chaotic motions.
In Neptune's case, Triton introduction has cause for a short period, chaotic motions, which then stabilized into new stable orbits.

In the image here, it shows the relative orbits of stable moons and Neptune's rings. We can conclude that the respective Roche limit of Neptune would cause any possibly try of ring matter collapsing into a moon to be foiled if the mass was too high. This means that only small moons form between rings Adams and Galle, with Galle marking the "will tear anything apart no matter the size" Roche limit.
Moons between the gaps
There can be many moons there in the gaps between the rings, as long as the moon is small enough to fit, and its gravity is not influencing the rings so much that it is noticeable, most likely very small to be picked up by our telescopes. It brings up another problem: How small must a moon be to be no longer a moon?
Moon definition
There isn't a restriction. However, the most widely accepted one is that a moon is an object with a relatively clear orbit[little debris]. We can say that there may be many moons between the rings, but are too small to be detected.
Moons in the large gap you notice
Now moving on, we have a gap between Larissa and Proteus / S/2004 N 1. Clearly, no rings call this area a home. So this gap is somehow mysteriously "uninhabited".
I doubt that there isn't a "moon" there, even if it were really small. Hippocamp is a very small object. In fact, it is the smallest moon discovered around Neptune. The relative Roche limit is not here, so it probably is possible for ring material to recondense into a moon with a stable orbit, and many times, a moon might happen to be covered up when we search. As I said, there is no size limit to a moon, so, because this area is clear of debris, there might be quite a few that we can't find.
A possible scenario [saying that there are a lot of moons] might be that the moons will gravitationally interact with each other, such as kick each other out, part of the three body problem. However, this can be easily detected, as rings would be seemingly affected. This obviously isn't so, due to everything being in a stable orbit, because the "chaotic" period is already passed, and debris settles down and recondense whenever possible.
Note: There can be moons also at the lagrange point if the point is near the "gap" you are taking about.
Another Note: Similar to lagrange points, uhoh mentioned another very good concept of co-orbital configurations, which is the interaction between two celetrial objects. Places with extremely possible stable orbits are called Trojan Points. One example of this is the Trojan asteroid groups which share the orbit with Jupiter. There are two Trojan Points near a given object so those points might harbor many stable moon orbits there too.
Answer
So my guess is that yes, there are stable orbits within this gap, and there is a good chance there are additional moons out there.
We are discovering more and more moons today. Jupiter is one good example.

@uhoh I would think that a considerable gap must be necessary, because moons would kick each other out or collide if the space were too small, forming another chaotic period. Orbits must have settled for moons to reform and since there isn't another gravitational disturbance, there shouldn't be more chaos. Depending on the size of a given moon, the gap must be large enough so that the moon's gravity shouldn't disturb another moon's orbit. Such events would quickly settle if it would happen because objects tend to move to stable places.


Why does this graph for sunlight intensity on land has a steeper slope during sunrise as compared to sunset?

Too long for a comment, alas:

https://www.mesonet.org/index.php/site/about/other_measurements/#srad notes:

  Solar radiation measurements in the morning and evening are sensitive
  to obstruction (e.g., trees) on the east and west horizons. A delay of
  sunrise or early arrival of sunset at a particular station may be
  explained by examining panoramic site photos available at
  http://www.mesonet.org/index.php/site/sites/mesonet_sites.

Also dig around a bit starting at http://mesowest.utah.edu/cgi-bin/droman/meso_graph_climo_ndb.cgi?stn=SEAM5&unit=0&hours=24&day1=0&month1=&year1=2014&hour1=00&windred=&time=LOCAL&var=SOLR&lastyear=1&vnamev=Solar%20Radiation&stationname=SEAGULL&vlabel=%C2%A0W/m*m and you should find more data.


I suppose size limit of the comet is still under process but astronomers have found a few large ones though!
The comet Holmes which appeared in 2007 became brighter and also swelled up in size. Its nuclear size is only 3.6km (2.2 miles), but it released so much dust and gas so that it expanded to about 869,900 miles (1.4 million kilometers) i.e larger than the sun.
The largest comet in terms of nuclear size was discovered in the year 1729, comet Sarabat which had a nucleus of about $100 {\rm km}$.


How can I improve the Field of View of my home-made telescope?

There basically are three main properties which influence what you see with your telescope:

The aperture defines how much light you get, thus how sensitive your setup will get. The bigger, the more light you collect, thus the fainter objects you will be able to detect in principle. There's the somewhat sense-free saying that you can compensate (clear) aperture only by clear aperture.

The focal length is the primary factor which drives your field-of-view. You can somewhat vary it by the choice of your eye piece. The magnification is the ratio of the two focal length, of your primary optics and the eye piece: $m = \frac{f_{tube}}{f_{eye}}$. Thus the bigger this magnification is, the more details you can see, the easier you resolve things which are close to eachother. However, phyics strike here and the size of the aperture limit how much magnification makes sense due to the diffraction ocuring at it. The so-called useful magnification (also see here, thus maximum magnification) is by rule-of-thumb about twice the aperture in mm. You can increase or decrease your focal length on an existing telescope with either a Barlow lens or focal reducer respectively. Both, of course, result in some optical degradation, as any optical system does.

The actual field-of-view is also determined by the eye piece properties and its opening angle. This can vary in a limited range only and is usually around 50°. See also this telescope calculator by Sky & Telescope.

Taken aperture, focal length of tube and focal length of eye piece, you derive one additional property which immediately affects what you see: the so-called opening ratio or f-ratio which is the ratio of focal length of your tube and the clear aperture. This determines how bright you will perceive a certain object as that defines how much light from an object is projected per unit area. Telescopes with identical f-ratio will show identical objects at the same brightness - independent of their actual field-of-view. Thus this number really defines how well you can see faint objects like nebula or faint / distant stars.
If I should put it in three sentences, this is it in short: your choice of focal length determines how big an area you see of the sky. The longer, the smaller your FOV. The ratio of aperture and focal length defines how faint objects you see.
As to the inverted image: astronomical telescopes usually have that, it keeps the system simpler. There are terrestrial telescope which employ an additional lens. Taking the words of uhoh's comment: The nature of real images is that they are upside-down and reversed left-right. It is possible to use a 1:1 relay lens to re-image your focal plane again but it will make your system more complicated to build and to hold everything in the right place. You can also use a prism to flip it in one direction, or two prisms to make an image inverter. If you don't want to ask a new question about using the relay lens, you can just read about it in wikipedia.


Trouble understanding speed-dispersion in (elliptical) galaxies

The total light of a galaxy is the sum of perhaps billions of stars, each with their own line of sight velocity. You can characterise this distribution with a mean and a standard deviation (dispersion), and this will be reflected in the absorption line profiles of that galaxy.

Equally you might choose to measure this distribution as a function of position in a galaxy, if you are able to spatially resolve different regions. For instance one could put a spectrograph slit across a galaxy and get the velocity dispersion as a function of position along the slit.

When you are talking about nearby stars, you can measure the 3D velocities individually, providing you can measure the radial velocity, proper motion and distance of each star. You can then find the mean and standard deviation for this group of measurements in each velocity coordinate.


I took the liberty of looking a bit closer at your image and annotate several features.

If you look closely, you see that several of the brighter objects in your image do the exact same random walk. Those are e.e. the objects A, B and C. This already strongly hints to some shaking of the camera, rather than A being a UFO.
If you look closer at your own image (which you always should), you see those lines all over the image from other weaker stars as well.

This automatically begs the question: How long was your exposure? Or was it composed from several single frames? Because it looks like one of those options  for me.

To substantiate this, let's look at some other features: Feature D doesn't show any spots, thus must have been taken while a shake of the camera was happening, thus must have moved quickly, thus this must be a shooting star. Of course a shooting star would appear like this as well on a non-shaking camera.
For contrast, feature E is also linear, but shows spots, which means while this line was putting light into the camera it must have stopped shaking on regular time intervals. This is a bit weird. But because the angle of the line, just as D, doesn't match any of the streaks that we follow in the figure A, this must also be a self-moving object. Now a self moving object of much lower speed than a shooting star (D), can be a satellite or an airplane.
Then there's feature F which clearly is also slow-moving, but does a weird rectangle that I cannot match to the 'master-shake' of feature A. It could mean that it's also self-moving taken together with one of the corners of the master shake. But then I don't understand why F is doing the cornery thing, but E does not. Another possibility in my opinion is, that E and F are both satellites, which faded in and out of brightness at different moments when the picture was taken, thus a different 'shaking' was imprinted on them.

So yeah, summarizing
Your image does some weird things, most of them clearly caused by camera-shaking. I don't understand all the details in your picture for that matter, but that's far from a reliable UFO claim. In particular because essentially EVERY UFO claim is based on shaky/nonsharp images.

P.S.
Also note how the grass in the lower left is shifted from really dark, to slightly dark. There are two central features in the grass that can be brought to roughly, but not exactly match when moving them over each other in photoshop. This indicates that this is the same patch of grass, just being photographed at two different times, after the grass was moving around a bit.
I suggest a more stable mount for your camera, if you're sure you didn't bounce into it by yourself.


Convert Stellarium video capture to 360 for youtube

360° panoramic media conventionally use equirectangular projection
and a 2:1 aspect ratio.
I made a still panorama this way:

Configuration > Tools > Screenshots: Custom size 2048 x 1024
View > Sky > Projection: Cylinder (i.e. equirectangular)
View > Sky > Stars: Relative scale 0.70 (limits brightest stars to a reasonable size)
FOV 180° (Ctrl+Alt+1)
Other view options as desired (e.g. turn off ground, atmosphere, labels)

Then I saved a screenshot (Ctrl+S) and injected the appropriate metadata.
The result is viewable as a 360° panorama.

Constellations near the center of view look OK in any view direction,
but stars near the zenith look rather dim and rice-shaped.
I guess this could be remedied by something like the "Spheric mirror distortion" option (Configuration > Tools > Planetarium options) but within the "Cylinder" projection.
I don't know of a Stellarium plugin which does this yet.