As noted in http://aa.quae.nl/en/antwoorden/zonpositie.html#14 the
length of sunrise/sunset varies from approximately 128/cos(latitude)
seconds at the equinoxes to approximately 142/cos(1.14*latitude) at
the solstices.

More specifically, here's the length of sunrise/sunset at various latitudes:

Beyond 65 degrees north or south latitude, the sun does not rise or
set daily, and the length of sunrise/sunset increases significantly.

The data plotted above is the length of sunrise, but the length of
sunset is very similar.

All calculations for this program were made with this program:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12824.c

The raw output of sunrise/sunset times:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/sun-rise-set-multiple-latitudes.txt.bz2

You can verify these results at:
http://aa.usno.navy.mil/data/docs/RS_OneYear.php

The longest sunrise I found for 2015 was at 89 degrees 51 minutes
south latitude, 125 degrees east longitude. There, the sun starts
rising 20 Sep 2015 at 2352, bobbles up and down a bit (but never quite
sets), and finally finishes rising 43 hours and 21 minutes later, at
22 Sep 2015 at 1913, but see caveat at the end of this answer.

You can "verify" this by first visiting
http://aa.usno.navy.mil/data/docs/RS_OneYear.php with these
parameters:

to get:

Sun or Moon Rise/Set Table for One Year
             o  ,    o  ,                                                                              Astronomical Applications Dept.
Location: E125 00, S89 51                          Rise and Set for the Sun for 2015                   U. S. Naval Observatory
                                                                                Washington, DC  20392-5420
                                                            Universal Time

       Jan.       Feb.       Mar.       Apr.       May        June       July       Aug.       Sept.      Oct.       Nov.       Dec.
Day Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set  Rise  Set
     h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m   h m  h m
01  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
02  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
03  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
04  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
05  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
06  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
07  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
08  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
09  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
10  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
11  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
12  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
13  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
14  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
15  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
16  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
17  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
18  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
19  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****
20  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  2352       **** ****  **** ****  **** ****
21  **** ****  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
22  **** ****  **** ****  1842 1614  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
23  **** ****  **** ****       0708  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
24  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
25  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
26  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
27  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
28  **** ****  **** ****  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
29  **** ****             ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
30  **** ****             ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  ---- ----  **** ****  **** ****  **** ****  **** ****
31  **** ****             ---- ----             ---- ----             ---- ----  ---- ----             **** ****             **** ****

(**** object continuously above horizon)                                                      (---- object continuously below horizon)

Note that the sun rises at 2352 on September 20th, and doesn't set for
the rest of the year, verifying the sunrise start time.

Verifying the end time is a little tricker. To do this, visit
http://ssd.jpl.nasa.gov/horizons.cgi with the following parameters:

to get:

Revised : Jul 31, 2013                  Sun                                 10

 PHYSICAL PROPERTIES (revised Jan 16, 2014):
  GM (10^11 km^3/s^2)   = 1.3271244004193938  Mass (10^30 kg)   ~ 1.988544
  Radius (photosphere)  = 6.963(10^5) km  Angular diam at 1 AU  = 1919.3"
  Solar Radius (IAU)    = 6.955(10^5) km  Mean density          = 1.408 g/cm^3
  Surface gravity       =  274.0 m/s^2    Moment of inertia     = 0.059
  Escape velocity       =  617.7 km/s     Adopted sidereal per  = 25.38 d
  Pole (RA,DEC in deg.) =  286.13,63.87   Obliquity to ecliptic = 7 deg 15'
  Solar constant (1 AU) = 1367.6 W/m^2    Solar lumin.(erg/s)   =  3.846(10^33)
  Mass-energy conv rate = 4.3(10^12 gm/s) Effective temp (K)    =  5778
  Surf. temp (photosphr)= 6600 K (bottom) Surf. temp (photosphr)=  4400 K (top)
  Photospheric depth    = ~400 km         Chromospheric depth   = ~2500 km
  Sunspot cycle         = 11.4 yr         Cycle 22 sunspot min. =  1991 A.D.

  Motn. rel to nrby strs= apex : RA=271 deg; DEC=+30 deg
                          speed: 19.4 km/s = 0.0112 AU/day
  Motn. rel to 2.73K BB = apex : l=264.7+-0.8; b=48.2+-0.5
                          speed: 369 +-11 km/s

Results

*******************************************************************************
Ephemeris / WWW_USER Fri Jan  1 21:49:19 2016 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: Sun (10)                        {source: DE431mx}
Center body name: Earth (399)                     {source: DE431mx}
Center-site name: (user defined site below)
*******************************************************************************
Start time      : A.D. 2015-Sep-22 19:00:00.0000 UT
Stop  time      : A.D. 2015-Sep-22 20:00:00.0000 UT
Step-size       : 1 minutes
*******************************************************************************
Target pole/equ : IAU_SUN                         {East-longitude +}
Target radii    : 696000.0 x 696000.0 x 696000.0 k{Equator, meridian, pole}
Center geodetic : 125.000000,-89.850000,7.057E-13 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 125.000000,16.7540774,-6356.730 {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: DE431mx}
Rel. light bend : Sun, EARTH                      {source: DE431mx}
Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2
Atmos refraction: NO (AIRLESS)
RA format       : HMS
Time format     : CAL
RTS-only print  : NO
EOP file        : eop.160101.p160324
EOP coverage    : DATA-BASED 1962-JAN-20 TO 2016-JAN-01. PREDICTS-> 2016-MAR-23
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     Azi_(a-appr)_Elev
****************************************
$$SOE
 2015-Sep-22 19:00 *m  128.1772  -0.3117
 2015-Sep-22 19:01 *m  127.9272  -0.3109
 2015-Sep-22 19:02 *m  127.6771  -0.3101
 2015-Sep-22 19:03 *m  127.4270  -0.3093
 2015-Sep-22 19:04 *m  127.1770  -0.3085
 2015-Sep-22 19:05 *m  126.9269  -0.3077
 2015-Sep-22 19:06 *m  126.6769  -0.3069
 2015-Sep-22 19:07 *m  126.4268  -0.3061
 2015-Sep-22 19:08 *m  126.1767  -0.3053
 2015-Sep-22 19:09 *m  125.9267  -0.3045
 2015-Sep-22 19:10 *m  125.6766  -0.3037
 2015-Sep-22 19:11 *m  125.4266  -0.3029
 2015-Sep-22 19:12 *m  125.1765  -0.3021
 2015-Sep-22 19:13 *m  124.9264  -0.3013
 2015-Sep-22 19:14 *m  124.6764  -0.3005
 2015-Sep-22 19:15 *m  124.4263  -0.2997
 2015-Sep-22 19:16 *m  124.1762  -0.2989
 2015-Sep-22 19:17 *m  123.9262  -0.2981
 2015-Sep-22 19:18 *m  123.6761  -0.2973
 2015-Sep-22 19:19 *m  123.4261  -0.2964
 2015-Sep-22 19:20 *m  123.1760  -0.2956
 2015-Sep-22 19:21 *m  122.9259  -0.2948
 2015-Sep-22 19:22 *m  122.6759  -0.2940
 2015-Sep-22 19:23 *m  122.4258  -0.2932
 2015-Sep-22 19:24 *m  122.1757  -0.2923
 2015-Sep-22 19:25 *m  121.9257  -0.2915
 2015-Sep-22 19:26 *m  121.6756  -0.2907
 2015-Sep-22 19:27 *m  121.4256  -0.2899
 2015-Sep-22 19:28 *m  121.1755  -0.2890
 2015-Sep-22 19:29 *m  120.9254  -0.2882
 2015-Sep-22 19:30 *m  120.6754  -0.2874
 2015-Sep-22 19:31 *m  120.4253  -0.2865
 2015-Sep-22 19:32 *m  120.1753  -0.2857
 2015-Sep-22 19:33 *m  119.9252  -0.2849
 2015-Sep-22 19:34 *m  119.6751  -0.2840
 2015-Sep-22 19:35 *m  119.4251  -0.2832
 2015-Sep-22 19:36 *m  119.1750  -0.2823
 2015-Sep-22 19:37 *m  118.9250  -0.2815
 2015-Sep-22 19:38 *m  118.6749  -0.2807
 2015-Sep-22 19:39 *m  118.4248  -0.2798
 2015-Sep-22 19:40 *m  118.1748  -0.2790
 2015-Sep-22 19:41 *m  117.9247  -0.2781
 2015-Sep-22 19:42 *m  117.6746  -0.2773
 2015-Sep-22 19:43 *m  117.4246  -0.2764
 2015-Sep-22 19:44 *m  117.1745  -0.2756
 2015-Sep-22 19:45 *m  116.9245  -0.2747
 2015-Sep-22 19:46 *m  116.6744  -0.2739
 2015-Sep-22 19:47 *m  116.4243  -0.2730
 2015-Sep-22 19:48 *m  116.1743  -0.2721
 2015-Sep-22 19:49 *m  115.9242  -0.2713
 2015-Sep-22 19:50 *m  115.6742  -0.2704
 2015-Sep-22 19:51 *m  115.4241  -0.2696
 2015-Sep-22 19:52 *m  115.1740  -0.2687
 2015-Sep-22 19:53 *m  114.9240  -0.2678
 2015-Sep-22 19:54 *m  114.6739  -0.2670
 2015-Sep-22 19:55 *m  114.4239  -0.2661
 2015-Sep-22 19:56 *m  114.1738  -0.2652
 2015-Sep-22 19:57 *m  113.9237  -0.2644
 2015-Sep-22 19:58 *m  113.6737  -0.2635
 2015-Sep-22 19:59 *m  113.4236  -0.2626
 2015-Sep-22 20:00 *m  113.1735  -0.2618
$$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.

SOLAR PRESENCE (OBSERVING SITE)
  Time tag is followed by a blank, then a solar-presence symbol:

        '*'  Daylight (refracted solar upper-limb on or above apparent horizon)
        'C'  Civil twilight/dawn
        'N'  Nautical twilight/dawn
        'A'  Astronomical twilight/dawn
        ' '  Night OR geocentric ephemeris

LUNAR PRESENCE WITH TARGET RISE/TRANSIT/SET MARKER (OBSERVING SITE)
  The solar-presence symbol is immediately followed by another marker symbol:

        'm'  Refracted upper-limb of Moon on or above apparent horizon
        ' '  Refracted upper-limb of Moon below apparent horizon OR geocentric
        'r'  Rise    (target body on or above cut-off RTS elevation)
        't'  Transit (target body at or past local maximum RTS elevation)
        's'  Set     (target body on or below cut-off RTS elevation)

RTS MARKERS (TVH)
  Rise and set are with respect to the reference ellipsoid true visual horizon
defined by the elevation cut-off angle. Horizon dip and yellow-light refraction
(Earth only) are considered. Accuracy is < or = to twice the requested search
step-size.

 Azi_(a-appr)_Elev =
   Airless apparent azimuth and elevation of target center. Adjusted for
light-time, the gravitational deflection of light, stellar aberration,
precession and nutation. Azimuth measured North(0) -> East(90) -> South(180) ->
West(270) -> North (360). Elevation is with respect to plane perpendicular
to local zenith direction.  TOPOCENTRIC ONLY. 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

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

The sun's angular diameter is about 32 arcminutes, so the sun's lower
limb is 16 arcminutes below the sun's center. When the center of the
sun has geometric elevation -18 arcminutes (-0.3 degrees), the lower
limb has geometric elevation -34 arcminutes. Since refraction near the
horizon is also 34 arcminutes, the sun's lower limb rises when the
sun's geometric elevation is -0.3 degrees.

In the table above, this occurs between 1914 and 1915, but my program
uses slightly more accurate data for the sun's angular diameter, and
the sun actually finishes rising between 1913 and 1914 (and closer to
1913).

You can then fly almost halfway across the world to latitude 89
degrees 51 minutes and longitude -19 degrees to see the
one-minute-shorter longest sunset, which starts at 23 Sep 2015 at 2128
and ends at 25 Sep 2015 at 1648, a length of 43 hours and 20 minutes.

In this case, you would use
http://aa.usno.navy.mil/data/docs/RS_OneYear.php to verify the ending
time of the sunset, and HORIZONS to verify the start time of the
sunset.

Polar sunrises and sunsets are considerably shorter:

At the North Pole, the sun starts rising at 18 Mar 2015 at 2015,
and finishes rising at 20 Mar 2015 at 0441, a length of 32 hours and
26 minutes.
At the South Pole, the sun starts setting at 21 Mar 2015 at 1650,
and finishes setting at 23 Mar 2015 at 0117, a length of 32 hours
and 27 minutes.
At the South Pole, the sun starts rising at 21 Sep 2015 at 0508,
and finishes rising at 22 Sep 2015 at 1400, a length of 32 hours and
52 minutes.
At the North Pole, the sun starts setting at 24 Sep 2015 at 0243,
and finishes setting at 25 Sep 2015 at 1131, a length of 32 hours
and 48 minutes.

Main caveat: Like HORIZONS and the sunrise/sunset tables above, I
assume 34 arcminutes of refraction at the horizon. That's reasonable
for most locations, but may be unreasonable close the pole, where the
longest sunrises and sunsets occur. In particular, refraction can
change rapidly at these latitudes, allowing for potentially much
longer sunrises and sunsets.

I now believe that http://what-if.xkcd.com/42/ is inaccurate, and will
ping the author to let him know.


Do/did the asteroids contain enough water to create Earth's oceans?

This is far too soon to draw sweeping conclusions. The Philae probe has only measured deuterium for one comet. But comets are diverse, of diverse origins. We need to take a lot more samples, from many different comets, before we can conclude firmly that water has, or has not, come from comets.

The only conclusion we could draw so far from this study is that earthly water has not come exclusively from comets similar to 67P/C-G.

What you're seeing now is the usual distortion that occurs in the media after some hugely popular experiment is completed - its conclusions are blown out of proportion. But that's not how science works.

Let's write this one down, and wait for further science to be put forth. We don't know enough yet about these objects.


Could different views of the universe simply be different points in time?

Could this occur if the distant body is moving laterally at a fast enough velocity relative to our position

Nope. I mean if the universe is flat it doesnt matter how fast the object goes you cannot see it twice in the sky.

  or if the shape of the universe itself loops around?

Yes. If the universe has a positive curvature. When we look at the sky, we can see the same object in two opposite directions.

In our universe this is not possible since if the universe is positively curved then the radius of curvature is much larger then the size of the observable universe. So its highly unlikely that we will see the same object in two different positions in the sky.


K-type dwarfs behave like a solar mass star, and will leave behind a (slightly) lower mass white dwarf.

F-type dwarfs will behave like a solar mass star and leave behind a (slightly) higher mass white dwarf.

Higher mass (hotter) stars will produce higher mass white dwarfs, up until the initial main sequence star is of about $8M_{\odot}$ (a spectral type of about B3). The corresponding remnant white dwarf mass increase up to about $1.2 M_{\odot}$. All these stars go through a giant phase, He core burning and asymptotic giant branch phases.

Above this then carbon burning will be initiated in the core and it is likely that burning will continue through to iron, followed by core-collapse and a supernova.


how to connect two eyepieces together with the x3 Barlow lens

You will get the maximum magnification by using the 10mm eyepiece with the barlow lens.  You cannot connect 2 eyepieces in the way you suggest, but even if you could add the 10 and the 25 to get a 35mm, that would give lower magnification, as I explained in a previous answer.


It doesn't matter if the body is made of gas, rocks, liquid or plasma, the four states of matter all have mass. So, as we know, mass create a gravitational field, and the more mass the stronger the gravity - and Jupiter has 317x Earth mass.


This is hardly an astronomy question, but I like drawing, so here you are:

Longitudes are measured from the Greenwich meridian, so the angle between Kampala and Quito is
$$\theta_\mathrm{Q} + \theta_\mathrm{K} = 82.5^\circ + 37.5^\circ = 110^\circ.$$
(remember that $0^\circ30' = 0.5^\circ$).
The shortest surface path is along Equator. Since $110^\circ$ is $\frac{110^\circ}{360^\circ} \simeq 0.3$ times the circumference of Earth at Equator, the length of path A (the dashed line) is
$$\mathrm{A}:\,\,d = 0.3 \times 40,075\,\mathrm{km} = 12,245\,\mathrm{km}$$

For path B (the solid line), you need a bit more  trigonometry. The radius of Earth is $R = 40,075\,\mathrm{km}\,/\,2\pi = 6,378\,\mathrm{km}$. The rest will be left as an exercise.


Is it certain that dark matter consists of particles? (And not just curved space)

Also asked at https://physics.stackexchange.com/questions/174080/how-do-we-know-dark-matter-isnt-curved-spacetime

Basically no. Or at least you can't have this idea and General Relativity. GR demands that you have something (matter/energy density) to cause the curvature. Curvature without cause is not part of the model. That's not to say that what you propose couldn't be true, but one would then have to ditch GR, which is a reasonably successful model (though some would argue that the dark matter problem is one of its biggest failures!).

I think a lot of parameter space has been ruled out for dark matter. We know it can't be baryonic from our understanding of primordial element abundances and also from the way it interacts (or doesn't interact) with normal matter in forming the basic structures in our universe on the right timescales. The microlensing surveys that have been conducted rule out large (planetary sized and above) dark matter bodies (and black holes, brown dwarfs, cold white dwarfs etc.), since to account for all the mass, these would produce lots of microlensing events that simply aren't observed.

An excellent primer on these topics has been produced by Garrett & Duda (2011).


Is it possible for a planet's axis to precess or wobble in such a way that its season cycle is of lesser duration than its year?

For this, definitely yes, if you tweak some parameters. On the Wikipedia page for Axial Precision, you'll find the set of equations and parameters that allow you to calculate the precessional rate for Earth's axis. Mainly, the Earth's precession is caused by a variety of factors but primarily due to the Moon and Sun (and thus related orbital/body parameters). If you look at those equations, and start tweaking values in them, you'll find you can force the axial precession period to be less than a "year". I'll go through and construct just such a system.

First though, I need to figure out the orbital parameters of the planet such that it has a 100-year long revolution period. This is acheived by using Kepler's Third Law, relating orbital period, orbital distance, and mass of the Sun (and Earth, but that's negligible). The relevant form of the equation, for our purposes is:

$$P^2 = \frac{a^3}{M_{\mathrm{Sun}}}$$

where, $P$ is the orbital period in years, $a$ is the orbital distance in AU (1 AU is the distance between Earth and the Sun), and $M_{\mathrm{Sun}}$ is the mass of the Sun in units of solar masses. If you're dictating that $P=100\ \mathrm{years}$, then I find I can set $a = 10\ \mathrm{AU}$ and $M_{\mathrm{star}} = 0.1\ M_{\odot}$ and make this equation true. What that means is that this new planet orbits a star which is 10 times less massive than our own star (making it a red dwarf) and orbits 10 times farther than Earth. My choice of parameters is all serendipitous and you could certainly find other parameters that work as well.

Now, given those parameters above, we can try to calculate the precessional rate for this new planet. I'll assume this new planet has an orbiting moon, just like Earth does. The two important equations for calculating the Earth's precessional rate are to account for the contributions from the Sun and the Moon. The equations are given by:

$$P_{\mathrm{Sun}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Sun}}}{a_{\mathrm{Earth}}(1-e_{\mathrm{Earth}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$

$$P_{\mathrm{Moon}} = \frac{3}{2}\left(\frac{GM_{\mathrm{Moon}}(1-1.5\sin^2 i_{\mathrm{Moon}})}{a_{\mathrm{Moon}}(1-e_{\mathrm{Moon}})^{3/2}}\right)\left(\frac{(C-A)}{C}\frac{\cos \epsilon}{\omega}\right)$$

In these equations, $G$ is the Gravitational Constant, $M$ is mass in kg, $a$ is the orbital distance in meters, $e$ is the orbital eccentricity, $i$ is the inclination of the Moon's orbital plane wrt to the ecliptic, $(C-A)/C = 0.003273763$ is a measurement of the Earth's shape, $\epsilon$ is the Earth's axis' angle, nominally $23.43928^\mathrm{o}$, and finally $\omega$ is the rotation rate in rad/s.

Plugging in all the values for the Earth-Moon-Sun system (which you can find on the wiki page), you find that $P_{\mathrm{Sun}} + P_{\mathrm{Moon}} = 7.789\times 10^{-12}\ \mathrm{rad/s}$ and to convert to a time scale, you can multiply by  $2\times10^{-7}$. Doing so gives you the nominal Earth precessional period of about 26,000 years.

Given all of that, we can start playing with parameters to get it such that the precessional period for our new system is less than 100 years. I already said that $a_{\mathrm{Earth}}$ is 10 times larger and that $M_{\mathrm{Sun}}$ is 10 times smaller so we could get the period of 100 years. I found that I could make the following additional changes.

The moon is now 5.8 times more massive than our Moon.
The moon is now a quarter of the distance of our Moon.
The planet is spinning twice as fast.

With those changes, I find that the precessional period falls down to about 50 years, or half an orbit. You can continue to play with the numbers to get a different result.

An important consideration here is whether or not such a system is likely to even form. I think, given the system I've created, it is not outside the realm of possibilities. To have such a large, almost binary planetary system be so far away from a small star is unlikely, but I don't think impossible. The real difficulty here is that you've set the orbital period to such a long timescale. If you were to relax that a bit to something much shorter (say 10 years) you'd get a more realistic (and likely habitable) system.

  Earth's precession is very much longer than it's revolution period, but is it possible that these could be reversed and a planet still be habitable?

Certainly there's nothing to say a planet with a precessional period shorter than the revolution period is uninhabitable. What will make it uninhabitable is being too far from the star. The extreme orbital time you've provided necessitated having a planet that is likely too far from the star to be in the "habitable zone". You just have to play with all the parameters and find something that fits all your needs.

Another thing to consider is that if a planet has such a large and close moon as I've constructed, it will likely cause the planet to slow its rotation rate over short (astronomical) timescales. The planets rotation rate has a significant affect on the precessional rate and slowing it down over short timescales may mean your system doesn't keep the precessional rate under the rotation period for too long.


Math behind a uv plot in interferometry?

Rick Perley gives a lecture about interferometry to a biannual NRAO Synthesis Imaging Workshop.
Part 2 of his 2016 talk discusses u,v plane coverage.
The 2016 IRAM Millimeter Interferometry School lectures by J. Pety and F. Gueth may also help.

Let $\hat{u}, \hat{v}, \hat{w}$ be orthogonal unit vectors with $\hat{w}$ pointing from Earth to the radio source, $\hat{v}$ oriented northward, and $\hat{u}$ oriented eastward.
Let $B_{ij}$ be the baseline vector between antennas $i$ and $j$ in the array.
Then $(B_{ij} \cdot \hat{u}, B_{ij} \cdot \hat{v})$
and $(-B_{ij} \cdot \hat{u}, -B_{ij} \cdot \hat{v})$
are points in the u,v plot, and
$B_{ij} \cdot \hat{w} / c$ is a time offset between the signals at $i$ and $j$.

If both antennas observe the source for several hours, $B_{ij}$ rotates with the Earth, and the two points in the u,v plot grow into elliptical arcs.
The shape of those ellipses depends on the declination of the source, so
the same array will have different u,v coverage for sources at different declinations.

The interfering signals from the array are combined in a visibility function $V(u, v)$.
The image of the source is an inverse Fourier transform of $V$, typically deconvolved using an algorithm called CLEAN.
To minimize image artifacts, a diverse set of baseline lengths and orientations, without large gaps in u,v coverage, is desirable.


How many stars are in the Sculptor Dwarf Galaxy?

Estimates of star numbers are difficult to obtain. Red dwarfs are too dim to be seen as individual points so star counts cannot be obtained. The best guesses come from obtaining the mass of the galaxy and then guessing how much of that mass is made of stars. It is usually a low proportion, between 1 and 3%. You then guess how big the stars are on average to obtain a rough estimated of the star count.

Sources

Dwarf galaxies tend to have a high mass to light ratio meaning that more of their mass is in the form of dust, gas and "dark matter". Sculptor has particularly high ratio, of about 160$\Upsilon_\odot$, and a mass of about about 200 million times the sun.

If 1% of that mass is in the form of stars, but the stars are on average half the mass of the sun then Sculptor would have about 4 million stars.  The error bars on that number are very large: The mass could be as much as 320 million suns, and the other values are guessed by me (about 3% of the milky way is in stars, but the milky way has a much lower mass to light ratio)

Perhaps the final reason that this number is not easily available is that it is of relatively little use. While the size of a galaxy can be measured spatially or by mass, and the nature of the stars is of interest, the number of them is of less use.


Is the resulting light from a supernova a product of photons bouncing around in the Sun?

First, the Sun will not end up as a supernova - only a star $>8$ times the mass of the Sun will end its life in that way.

You also have the wrong idea about "trapped light" (photons bouncing around and gradually working their way to the surface). Photons are constantly emitted and absorbed again and don't travel very far (compared to the radius of the Sun), except at the photosphere, where by definition, they can escape.

Thus there is no storage of photons. What we see in a supernova is just a photosphere that is much hotter and much larger than the Sun. That is why it is many orders of magnitude more luminous than the Sun.

The heating (and explosive expansion) of the supernova envelope is ultimately derived from a small fraction of the gravitational potential energy released by the collapse of the stellar core.

Edit: The photosphere of a core collapse supernova is in fact about the same temperature as the Sun during most of the decay phase. Type Ia supernovae are a bit hotter. The huge luminosity is because the photosphere is huge.


The early sun was surrounded by a disc of dust and gas.  Such a disc is not stable, it will develop parts that are are more dense and parts that are less dense.  The combination of friction (between bodies of gas and dust) and gravity (pulling together) will tend to make the denser lumps grow denser, eventually coalessing into a ball of metal, rock and gas. This "proto-Earth" was comparable in size to the Earth now.

In the process, most of the rest of the dust and gas either fell onto the ball, or was pushed out of the way. It's not true that meteorites don't become part of the planet. Meteorites do add to the mass of the planet, but this is a slow process.

After this the early Earth collided with another planet. The other planet was destroyed and the remanents formed the Earth and Moon.  By now the Earth was roughly the same size as it is now.

There was then a period when the meteorite rate increased greatly. This added a little to the mass of the Earth.  The meteorites often were at least partly icy, and this water formed the seas.  The rocky an metallic parts of these meteorites were incorporated into the crust of the Earth.

By the time the "Late Heavy Bombardment" finished 3.8 billion years ago, the Earth was almost exactly the same size as now.  There has been some gain from later meteorite impacts, and some loss due from the top of the atmosphere, but these are minor.


22° Sun halo today in Zimbabwe. But there's a moon like thing inside the halo beneath the sun that cannot be seen with the naked eye, what is that?

It's a lens flare, caused by scattered sunlight within the camera's lens system. It looks very much like one of the examples on the Wikipedia page:

By Hustvedt - Own work


Gravity of a gaseous planet without a core

The gravitational force on a small mass on the outside of a planet is always the Newtonian $$F_{G}=-\frac{GM}{r^2},$$ so any planet, and particularly, any mass in the universe produces a gravitational field acting on everything else. So if, for example, the mass is $M=2\times 10^{27}\rm kg$ (i.e. one Jovian mass), then the gravity field outside the planet will always be the same (apart from Tides, higher order moments), no matter whether the mass is in Hydrogen or solids.
For the gas giants Jupiter and Saturn in our solar system, the mass in heavy refractories (i.e. everything heavier than Helium) is about $M_{\rm ref}\approx 15-20 \rm m_{\oplus}$, where $\rm m_{\oplus}$ is an Earth mass.
The rest of $M$ is hydrogen/helium. For Jupiter this is 300 $\rm m_{\oplus}$, Saturn about $75 \rm m_{\oplus}$.
This is a relatively large number of refractories in those gas giants, compared to solar composition, which is why we think that they have been formed via core-accretion, see Pollack (1996).
However there is another idea of how to form gas giants out there, which is that of gravitational disc instability, see Boss (2002). This idea posits that at very massive protostellar discs, which form planets, can become unstable and fragment into large clumps, which form gas giants directly. Those disc instability giant planets would have solar metallicity, i.e. a Jupiter-mass planet would have a refractory mass of only $M_{\rm ref} \approx 3 \rm m_{\oplus}$.
Those refractories would presumably sink to the planetary center and form a small core. Exoplanets that were found at large semi-major axis distances (hundreds of AU, compared to the Jovian 5 AU) from their stars, such as YSES 2b, are candidates for such disc instability models, and hence would host such small core. But that is as small of a core as it gets, you cannot have a core much much less massive than this.


Is the smaller companion of Pulsar J0453+1559 still the smallest known neutron star? Was it further confirmed to be a NS after Martinez et al. (2015)?

Here is a picture from the most recent review I could find - Horvath et al. (2020).
The companion of J0453+1559 is still the least massive of the double neutron stars with very precisely measured masses. I cannot find any citations to the Martinez et al. paper (here is a list) that either update the mass or shed any further light on its true identity. It appears to have been accepted in the literature as a neutron star - though there is still a school of thought that it could yet be a massive white dwarf (Tauris & Janka 2019).
As you can see from the picture there are several other objects with only slightly higher masses, and some of these are pulsars and are unambiguously neutron stars (e.g. a pulsar of mass $1.251 \pm 0.021 M_\odot$, McKee et al. 2020). There are some neutron stars in X-ray binaries and with white dwarf companions that may have a lower mass; but the uncertainties could push them above J0453+1559.


Orbital elements: Difference between longitude of perihelion and argument of perifocus

I know the answer now. Turns out that longitude of perihelion ($\overline\omega$) is defined as the sum of longitude of ascending node ($\Omega$) on the ecliptic plane and argument of perifocus ($\omega$) on the orbital plane: ($\Omega$ + $\omega$). It's strange to add up two non-coplanar angles, even when the inclination between the two planes is very small, but that's what it is.


Could Venus or Mercury have a moon that we haven't detected?

It's unlikely that either Mercury or Venus could have moons to begin with. Both of these planets are pretty close to the Sun — and in general, this prevents moons from finding stable orbits.

If a moon were too close to the planets, it would fall within the Roche limit and be torn apart by tidal forces. If a moon were too far from the planets, it would fall outside the Hill sphere and be pulled into the Sun.

The zones where moons around these planets could be stable over billions of years is probably so narrow that no body was ever captured into orbit when the planets were first being accreted.

Now, it gets even more complicated in Mercury's case. Its Hill sphere extends about 3000 miles (~4828 km) — that is, any satellite more than 3000 miles from the planet will be pulled away by the Sun's gravity.

However, its Roche limit extends to 3600 miles (~5794 km), so any satellite within this would be ripped apart. Venus lacks this excuse, but the region between where the Sun would pull away a satellite and the Roche limit is probably too small for a satellite to happen to have formed there.

Next, you have to realize that inner planets generally don't have moons. Solar wind pressure usually sends dust and debris flying away from the planets, so they don't have any spare material for moons.

Remember that Earth's Moon is thought to have formed from a collision with the Mars-sized planet Theia. This makes Earth an oddity in itself. Meanwhile, Mars is thought to have captured Phobos and Deimos from the asteroid belt — so Mars and its moons didn't condense from the same dust mass.

With all this, it'd be extremely surprising if we found moons on either Mercury or Venus.


How much water on the earth is made from hydrogen comes with solar wind?

Here is a way of getting a rough upper bound: the density $\rho$ of the solar wind is about 4 atoms per cm$^3$ (that we can assume are hydrogen), and it travels at $v=400-500$ km/s. That means that were the Earth to catch every atom directed at it it would gain $\pi R_\oplus^2 \rho v$ atoms per second, $2.0$ to $2.5\cdot10^{14}$. Which gives us a mass flow of $3.4$ to $4.3\cdot 10^{-13}$ kg/s. Earth has been around for 4.5 billion years, so over that time at most about 49 to 61 tons of hydrogen have arrived.
The total amount of liquid water on Earth is $V=1.386\cdot 10^9$ km$^3$ (which doesn't count the sizeable amount of water bound in rocks). About 11.19% of this mass is hydrogen, so the weight of hydrogen is $0.1119 V\rho_{water}=1.55\cdot 10^{20}$ kg. So the ratio between hydrogen from Earth and from the sun is about $2$ or $3\cdot 10^{15}$.
Still, in a half liter (about 27.7 mol water) there are $3.33\cdot 10^{25}$ hydrogen atoms, so there should be about 11 to 16 billion solar hydrogen atoms in the cup if this estimate is right.
Maybe Earth's magnetic field captures more atoms, which might increase the value. Likely most solar wind is instead pushed away and the true value is much lower. But I am pretty confident there are some solar atoms in the cup.


How bright is Comet PANSTARRS C/2017 S3 now?

The solar elongation (angle from the Sun to the object) is about 15 degrees right now and decreasing as the comet moves in towards perihelion (closest approach to the Sun) on August 15. As James K says this means that comet is lost in the Sun's glare. Anything below 30 degrees solar elongation is tricky to impossible to see.

Assuming it survives the close perihelion passage of 0.21 AU, it is predicted to enter the field of view of the SOHO LASCO Sun-observing coronographs on August 24:

  Assuming there is anything left of C/2017 S3, perihelion will occur this month on the 15th at a rather small perihelion distance of 0.21 AU. Being only 2 weeks from perihelion, it is already at a small elongation of 31 degrees on the 1st. Most observers will lose sight of it during the first week of August as its elongation drops below 25 degrees on the 4th and 20 degrees on the 6th. If the comet is still a going concern it will be visible in the SOHO LASCO C3 FOV between August 24 and September 13. The comet appear around the 7:30 position, passes to the east (left) of the Sun (halfway between the Sun and edge of FOV) and exits the FOV around the 12:30 position.

The above info is from the ALPO Comet Section. Latest LASCO image and movies are available here and many other places.


How do we know dark matter/dark energy exists?

Some additions to the answer of MBR:

In fact, we do not know that dark matter and dark energy do exist, but we have indirect clues. You will often see claims that dark matter and dark energy are two of the major problems of cosmology today, including by professional astronomers, but this is an epistemological misconception: you cannot call a hypothesis a problem.

Currently we have several observational anomalies (as defined per T. Kuhn): namely the rotation curve of galaxies, the luminosity of supernovae, the power spectrum of the CMB... (and some additional things like the difficulty to measure G or to detect directly gravitational waves).

According to the history of science, we have 3 ways to solve these anomalies:

ontological approaches (dark matter and dark energy)
legislatives approaches (extensions of General Relativity)
paradigmatic approaches (inhomogeneous Universes)

Dark matter and dark energy are two ontological approaches that seems to work pretty well: by adding two simple components we can fit a wide range of observations. This is why astronomers like them: their modelization are relatively easy and allow to fit everything well. But for the moment we do not know their nature.

Some say that with WMAP (Wilkinson Microwave Anisotropy Probe) and Planck space observatory we have entered in the era of precision cosmology: but this is only true if our paradigm of dark matter/dark energy approaches is correct. If this is not the case, we currently do not interpret correctly what we measure, but ultimately more accurate measures can lead to a better view of current cosmological anomalies.

So dark matter and dark energy are two elegant and minimalist approaches to several fundamental problems. This is a very hot topic in cosmology, but we may realize, one day, that we only had problems because our current paradigm was erroneous.


Does time also pass more slowly for a galaxy that is traveling at relativistic speeds, where the speed difference is due to Hubble expansion?

Yes, but it has little to do with the Special Relativistic time dilation of a moving body - in the sense that recession velocity is due to the expansion of space between galaxies rather than their relative motions.
The cosmological expansion leads to a redshift $z$ and also to time dilation by a factor $(1+z)$. i.e. Those distant clocks appear to run slower to us by a factor $(1+z)$. And vice versa - our clocks would appear to run slow from the perspective of an observer in a distant galaxy.
The time dilation effect has been observed in the light curves of high redshift supernovae (Blondin et al. 2008) and the durations of gamma ray bursts (Zhang et al. 2013) and is consistent with this prediction. This is an important observation since it is very difficult to account for this time dilation effect using a "tired light"  explanation.


App Stellarium and hourly wobbling motion of Neptune

This appears to me as just the consequence of having an altazimuthally mounted “camera” (the simulation). Objects are tilted towards the left when they rise, towards the right when they. set. You can verify that easily by looking at the Moon when it rises and sets from your own place. This is due to the horizon being “tilted” with respect to the Earth’s plane of rotation (unless you are exactly at the pole).
I’m not super familiar with Stellarium, but as far as I know, there’s a way for the observer to not be located on the Earth, or at least, you can change the latitude of the simulation. Try that, and if you set yourself as being at the Earth’s (north or south) pole, you shouldn’t see that anymore.


Wikipedia uses: mon, tue, wed, thu, sat, sun.

ISO 8601 defines an exchange standard for dates based on weeks and days, which looks like this: 2017-W01-3 or this 2017W011 (Year, ISO week number, number of day of week starting with Monday). Not so surprisingly it does not standardize names of days as far as I can tell.

So you could claim to use ISO-8601 conventions and simply use 1-7 (starting with Monday).

A comprehensive listing of abbreviations of day names for European countries is given by the European Medicines Agency for blistered drugs for the curious.


Where can I find thousands of actual images of stars in the galaxy?

The online version of the digitised sky survey will give you as many images of the sky as you want. Be aware that these are based on photographic plates. Also be aware that every imaging system has its own peculiarities in terms of the shape of the point spread function (i.e. the shape of a star in the image) and features such as diffraction spikes.
https://archive.eso.org/dss/dss


If someone was above the galaxy, how much of it can he see?

The Andromeda galaxy is not too different from ours, and yet you can see it with your naked eyes at 2.5 million light years distance. You do need to be under a very dark sky, far from cities, and it does look like a mere white smudge, but it is visible.

Now compare that with the fact that the Milky Way itself is quite visible to the naked eye if you're far from cities...

How do I see the milky way?

...and basically interpolate somewhere in between "little white smudge" and "river of stars filling up half the sky" to get an idea of how our galaxy would look like from its vicinity.

It would not be bright anyway, and you would have to adapt your eyes to complete darkness first. But you would definitely see its outline. Whether you would distinguish anything of the shape of its arms or not, is harder to say, and would probably depend a lot on the actual distance.


What is the speed of gravitational waves ?

Gravitational waves travel at the speed of light.

There are many possible sources of gravitational waves. The two confirmed detections so far (14 Sep 2016) are merging black hole binary systems, but sources of (detectable) gravitational waves may include merging neutron stars, short period binaries containing neutron stars or white dwarfs, gamma ray bursters or supernovae.

For a brief introduction see Gravitational Wave Sources at the ALIGO site.


What affects the evolution curve of a star's luminosity as a function of time?

I think this question is too broad, but I'll take a stab at it.

The Russell-Vogt (or sometimes Vogt-Russell) theorem is that the position of a star in the HR diagram is determined by its mass and composition.

The luminosity is determined mostly by its central temperature and composition. In turn, the central temperature depends on mass and radius and the radius depends on the luninosity and effective temperature.

Thus the question you ask fills textbooks. But to first order.
The time dependence of luminosity is set by the time dependence of mass - ie mass loss (or gain) and the rate of change of composition, particularly in the nuclear burning regions of a star.

In a star like the Sun, mass loss is relatively unimportant, so it is the rate at which hydrogen is turned into helium in the core that sets the timescale for luminosity evolution. Other processes that alter the core composition like mixing (due to convection, rotational mixing or diffusion) are thought to be second order effects.

Restricting myself to the main sequence as an example. The luminosity of the Sun is set by the core temperature. The the nuclear reactions become faster as the mean particle weight increases and the core contracts and the temperature rises to maintain hydrostatic equilibrium. The rate at which this occurs is given by the luminosity of the star divided by the amount of H available (proportional to the mass of the star).

Overall chemical composition drives this as a second order effect. A lower metallicity star is less opaque to radiation. The radiation escapes more efficiently and the star is smaller and hotter at the surface for the same luminosity. Smaller stars are hotter in the middle for the same mass and thus more luminous. Thus the luminosity evolution of low metallicity stars is faster than those of higher metallicity. See for example Figs 1 and 2 of Bazan & Mathews (1990).


The relation of the lit side of the moon and the sun

I'll assume you mean the lit side of the Moon (this wasn't clear from the original question, but I've edited it).

Suppose there's a crescent Moon visible in the sky during the day. Plot a straight line between the tips of the crescent, then plot a perpendicular straight line across its center across the sky. Intuitively, that line should pass through the Sun. You're saying it doesn't, and you're asking why.

In fact, that line does pass through the Sun, and what you're seeing is an optical illusion.

If you imagine the sky as a perfect hemisphere, with you at the center, the geometry will match up. But we we tend to imagine it as a flattened dome, probably because the horizon is obviously far away, but we have no reference points for the zenith so we imagine it being closer. This is (part of) the source of the "Moon illusion", where the Moon appears bigger when it's close to the horizon. In fact, it's very nearly the same angular size, but since we imagine it being farther away it must be "bigger".

The same mention distortion of the sky's geometry makes it look like the axis of the Moon's lit side doesn't point directly to the Sun.

Next time you see this, try holding up a straight rod intersecting both the Sun and the Moon. I think you'll find that the rod passes directly across the axis of the Moon's lit side.


Why are tides high only on a full moon day?

There are two main gravitational causes of tides: the Moon, and to a lesser extent the Sun.

When the moon is full or the moon is new, the Earth, Moon and Sun are roughly aligned, and the Lunar tide combines with the Solar tide to give a "Spring tide" that has a larger range.

When the moon is at first quarter, or third quarter (ie a half moon), the solar tide and the lunar tide are acting against each other, and you get a "neap tide" with a lower range than a spring tide.

The size of the tide on a particular day at a particular location is strongly affected by the local shape of the coast and sea floor, Some places only have one tide each day, or tides of different size, due to these effects.

Note the name "Spring tide" is not from the season. You get Spring tides every New and every Full moon. It comes from an old English word meaning "to bulge" The tide "bulges" during New and Full moon.

So you are not correct to say that tides are higher on the day of Full moon. The New Moon's gravity does create tides, and tides at New moon are as large as the tides at Full moon, and larger than the tides at half moon.


What is the orbit of Earth around the Sun?

According to wikipedia and other sources, a planet and a star always move in a circular orbit around the common center of mass of the both bodies ...

This is not true. In the absence of other gravitational sources, a planet and a star move in elliptical orbits about the common center of mass. Ancient scientists assumed circular orbits, but only because circles were somehow most pure. There were problems with this: It didn't match what those ancient scientists saw. So they modified these circular orbits by tacking on other circles. It was a mess.

Kepler cleaned this mess up by noting that the orbits of the planets are very close to elliptical rather than circular. This empirical result was later strengthened by Newton, who derived that elliptical orbits (not circular orbits) are a natural consequence of a force that follows an inverse square law. Gravitation is such a force.

The orbits of the planets about the solar system are not quite elliptical because planets are attracted to one another as well as to the Sun. Nonetheless, the orbits are still very close to elliptical. The Sun is more than 1000 times more massive than Jupiter. Jupiter, Saturn, and the other planets represent tiny perturbations on the nearly elliptical orbits of the planets about the Sun.

What about other perturbing effects, such as nearby stars and the Milky Way as a whole? These are immeasurably small. A passing star will perturb planetary orbits over the course of the tens of millions of years or more. The Milky Way as a whole has essentially no effect whatsoever. Theoretically, the effect most certainly does exist, but it is so small that it can be ignored, even over the lifespan of the solar system.


Will it be dangerous when close stars become Red Giants?

Life on Earth will not be adversely affected when nearby stars become red giants.

Certainly, things will get very messy inside those systems: red giants throw out a lot of gas and dust. Red giants are a lot more luminous than the Sun because they have such a huge surface area, but they don't emit large amounts of dangerous radiation, like X-rays or ultraviolet light. In fact, a red giant is cooler than the Sun, so it emits relatively little UV light.

According to Wikipedia, it's possible that a red giant system could even be habitable, so life in nearby systems should be quite safe.

  Although traditionally it has been suggested the evolution of a star into a red giant will render its planetary system, if present, uninhabitable, some research suggests that, during the evolution of a 1 M☉ star along the red-giant branch, it could harbor a habitable zone for several billion years at 2 AU out to around 100 million years at 9 AU out, giving perhaps enough time for life to develop on a suitable world. After the red-giant stage, there would for such a star be a habitable zone between 7 and 22 AU for an additional 109 years.
  Later studies have refined this scenario, showing how for a 1 M☉ star the habitable zone lasts from $10^8$ years for a planet with an orbit similar to that of Mars to $2.1×10^8$ yr for one that orbits at Saturn's distance to the Sun, the maximum time ($3.7×10^8$ yr) corresponding for planets orbiting at the distance of Jupiter. However, for planets orbiting a 0.5 M☉ star in equivalent orbits to those of Jupiter and Saturn they would be in the habitable zone for $5.8×10^9$ yr and $2.1×10^9$ yr respectively; for stars more massive than the Sun, the times are considerably shorter.

However, the Sirius and Procyon systems mentioned in the question are both binary systems, and both Sirius B and Procyon B are already white dwarfs. That makes the picture more complicated, and definitely more dangerous inside those systems, but neighboring stellar systems won't be harmed.

When Sirius A and Procyon A become red giants their white dwarf companions will get bombarded by a lot of material. If a white dwarf accretes enough hydrogen, it can lead to a runaway fusion reaction, in other words, a nova explosion.

  Hydrogen fusion may occur in a stable manner on the surface of the white dwarf for a narrow range of accretion rates, giving rise to a super soft X-ray source, but for most binary system parameters, the hydrogen burning is unstable thermally and rapidly converts a large amount of the hydrogen into other, heavier chemical elements in a runaway reaction, liberating an enormous amount of energy. This blows the remaining gases away from the surface of the white dwarf surface and produces an extremely bright outburst of light.

A nova is bright, but it only emits a tiny amount of energy compared to a supernova, and poses no danger to neighboring stellar systems.

Actually, it's very unlikely that either Sirius B or Procyon B will have nova events: they are simply too far from their companions to accrete enough hydrogen for that to occur. (Thanks to Peter Erwin for that info).


What happens to the shrinking universe in the presence of the pressure of Hawking radiation?

In a shrinking, roughly 3-spherical universe with only a black hole, Hawking radiation should follow a geodesic line and return to the black hole, without excerting radiation pressure to the universe as a whole.

Therefore it's hard to see, how Hawking radiation should establish an equilibrium with gravity.

More feasible seems, that the shrinking universe can prevent the black hole from further evaporation at some point, since evaporated particles and radiation are directed back into the black hole by the overall curvature of spacetime.


the more concentrated the mass, the more powerful is gravity?

To simplify things, I will assume that we have a spherical body and that we are considering Newtonian Physics.

If we have two bodies of the same mass but different radius, as long as we are outside of both bodies, and at the same distance from their centres, the force of gravity will be the same:
$$
F = G \frac{m M_i}{r_i^2}
$$
where $m$ is "our" mass, $M_i$ is the mass of body $i$ and $r_i$ is the distance between our centre and the centre of the body, and $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$ is the Gravitational Constant.

So, there is no dependence on the density of the body, only on its total mass.

However, if you are thinking about the gravitational acceleration $g$ at the surface of different planets, not only the masses may differ, but also the distance to their centre. Note that $g$ is given by:
$$
g = G \frac{M_i}{r_i^2}
$$

So, computing for Earth we have that $M_{Earth} = 5.972 \times 10^{24} kg$ and $r_{Earth} = 6.371 \times 10^6 m$, which gives, $g_{Earth} \approx 9.820 m s^{-2}$, slightly different from the standard gravity: $9.80665 m s^{-2}$.

The same computation for Mars, which has $M_{Mars} = 6.39 \times 10^{23} kg$ and $r_{Mars} = 3.3895 \times 10^6 m$, gives $g_{Mars} \approx 3.71 m s^{-2}$. So, since Mars has, roughly, $1/10$ of Earth's mass and $1/2$ of the radius, the acceleration of it's gravity is around $40\%$ of Earth's acceleration.


How do I know, mathematically rather than from observation, if a moon is full?

The Moon phases can be defined by the phase angle between the Sun, Moon and Earth; for example, at 0°, the Moon is defined as full, and at 180° it is defined as new. If you want to know how bright the Moon is at a given angle, we would use the phase angle to find the apparent and absolute magnitudes of the Moon.

Absolute magnitude, when referring to illuminated objects (objects that do not produce their own visible light), simply means their apparent magnitude if viewed from 1 AU away. This means it is almost entirely dependent on the object's phase angle. Right now, you're asking about how bright the Moon would seem to a person on Earth, so we'll find the apparent magnitude. The formula to find an illuminated object's apparent magnitude (in the Solar System), if we know its absolute magnitude $H$, is:

$$m = H + 2.5 \log_{10}{\left(\frac{d_{BS}^2 d_{BO}^2}{p(\chi) d_0^4}\right)}\!\,$$

Where $d_0$ is 1 AU, $\chi$ is the phase angle (in radians) and $p(\chi)$ is the phase integral (integration of reflected light). $d_{BO}$ is the distance between the observer and the body, $d_{BS}$ is the distance between the Sun and the body, and $d_{OS}$ is the distance between the observer and the Sun. This formula probably looks pretty scary, but it can be simplified with some approximations. First, we can approximate the phase integral as this:
$$p(\chi) = \frac{2}{3} \left( \left(1 - \frac{\chi}{\pi}\right) \cos{\chi} + \frac{1}{\pi} \sin{\chi} \right)$$
Where $\chi$ is the phase angle, in radians. In the Moon's case, we can set $H_{Moon} = +0.25$ (this is the absolute magnitude during a full moon), $d_{OS} = d_{BS} = 1$ AU and $d_{BO} = 0.00257$ AU. Now we get the formula:

$$m_{Moon} = 0.25 + 2.5 \log_{10}{\left(\frac{0.00257^2}{p(\chi)}\right)}$$

So now, we've got a formula that approximates the apparent magnitude of the Moon at any given phase angle. However, even though this gives a close approximation, it is not 100% accurate. Astronomers use empirically derived relationships to predict apparent magnitudes when accuracy is required.

Here's a quick script I wrote to calculate the apparent magnitude, given any phase angle:
https://jsfiddle.net/fNPvf/33429/


@called2voyage's comment points you to an in depth discussion. I'll add to that a simple summary of some things.

If you imagine that the astroid is in a simple, circular orbit around the Sun that's defined only by gravity, let's call that the nominal orbit.

Now consider the effects of sunlight.

If the asteroid were white or reflective, then a lot of the incident sunlight will reflect back towards the Sun, pushing the asteroid away from the Sun. Since both gravitational attraction and the Sun's brightness vary as $1/r^2$, so to the asteroid it feels like gravity is just a little weaker or the Sun is a little lighter, but otherwise still feels like gravity. If the asteroid became reflective slowly, it would just move to a slightly larger radius and remain in circular orbit.

Remember that the incident photons, and the reflected photons both contribute to the push!
If the asteroid were black and absorbing, then the absorption of the photons would give a push, and then they would warm the asteroid. Warm things radiate heat, and once equilibrium were reached, the radiation would also be a source of pressure.

If the astroid didn't rotate fast, but instead turned slowly so that it had a warm side always facing the Sun and a cold side always facing away, then the radiated infrared photons would also result in a push away from the Sun! So a dark asteroid that turned so slow that it had a permanent hot and permanent cold side would also be at a larger radius than nominal. The difference between the two depends on the functional form of the diffusivity and roughness of the surfaces.
However if the dark asteroid were rotating fast, say every few hours or tens of hours, then the radiation would be a little stronger in the forward or backward direction with respect to the orbit. If it was rotating the same sense as its orbit, the hot side would turn backwards, radiate backwards, and push the asteroid forward. That would slowly, but continuously raise the orbit of the asteroid. The longer this happened, the larger and larger its radius would become.

This is called the Yarkovsky effect and over long periods of time (millions of years) can have a big impact. For an asteroid to just miss the Earth, maybe a few years would be enough.

It turns out that @ BobJacobsen's answer to Why paint only one-half of Bennu? is, in a highly condensed and stylized form; "because the Yarkovsky effect is strong enough that there's no point in wasting good 'space paint' on the other half."
If you are a tiny dust particle near the Sun, there is yet another thing that happens,  called the Poynting–Robertson effect.

Tiny bits of dust have a higher area to mass ratio, so even though they may be in thermal equilibrium and not have a hot and cold side, tiny forces make a big difference. And if you are close to the Sun, then the forces are less tiny i.e. larger. @honeste_vivere's excellent answer points out that the Poynting–Robertson effect can cause the radius of the dust particles' orbits to decrease over time. An effect similar to (identical to) astronomical aberration will cause more pressure at the front of the dust particle, pushing it backwards and causing it to spiral closer to the Sun over time.


Aren't there more naked-eye-visible stars in the Milky Way plane?

You can tell a lot about Galactic structure by just looking. The ~5000 stars that can be seen with the naked eye have a roughly "lognormal" distribution of distance. I show plots below which were generated from the most recent version of the Hipparcos parallax catalogue. Fig.1 shows results for all stars with $5.5<V<6.5$  (i.e. very faint naked eye stars). The Hipparcos catalogue is almost complete for these stars, though some stars are so far away that the distance is highly uncertain - nevertheless, the general picture should be ok.

The median distance is about 440 light years. But the premise of your question is I think, that you dispute that this is far enough for the non-spherical distribution of stars in our Galaxy to become apparent. The answer is actually that it is, but only just. The Sun is very close to the plane of the disk of our Galaxy. The scale height of stars above this disk varies depending on stellar age and mass. Very approximately, the exponential scale height is 300-500 light years for most of the stars in our Galaxy.

This is just small enough that if we look at the demographics of stars in two Galactic latitude regions we do see a difference. Fig. 2 below shows stars with $5.5<V<6.5$ at low Galactic latitudes (within 15 degrees of the plane in green) and more than 45 degrees from the plane (in blue). There is a clear and significant difference. More stars can be seen at greater distances towards the plane, betraying the non spherically-symmetric nature of stars in our Galaxy.

In other words the structure of the Galaxy is not too small to affect the distribution of naked eye stars.

When we look at the Milky Way we are seeing millions of unresolved stars that are in general more distant still. The effects of the disk-like nature of the Galaxy become more apparent as one goes to greater distances. In particular, at high Galactic latitudes one simply runs out of stars and hence there is no uniformly illuminated sky - leading to Olber's paradox, as correctly stated in another answer. But at low latitudes, there are sufficient stars that in general, within the resolution of our eyes there are many stars whose light sums to provide a visual stimulus. This picture is interrupted by dust. Dust in the Galaxy is even more concentrated towards the plane than the stars. It is for this reason that beyond a few thousand light years, dust plays a major role in shaping the Milky Way structures that we can see, effectively blocking light at very low Galactic latitudes.

Fig 1: Probability distribution of distance for naked eye stars in the Hipparcos catalogue

Fig 2: Probability distribution of distance for naked eye stars divided into low and high Galactic latitude regions (i.e. towards and away from the Galactic plane).

IMPORTANT EDIT:

After the answer was accepted I did a bit of double checking. I split the Hipparcos sample into bright ($V<6$, 5000 stars) and very bright ($V<3$, 173 stars) and looked at the distribution of stars per unit area on the sky as a function of Galactic latitude (Fig. 3). The results are shown below and more obviously make the point. It turns out that the asymmetry between in an out of the plane is clearly seen, even in the brighter sample. The reason is that the very bright stars are not all that much closer than the bright stars. Perhaps a median of 200 light years vs 350 light years. Thus the scale height of the Galactic disc must be small enough that the difference from spherical symmetry is already appearent at these kinds of distances.

Fig. 3: Normalised plot of number of stars per unit area versus Galactic latitude for a sample of bright and very bright stars. Note the concentration towards the Galactic plane.

FURTHER EDIT: You can even see from the plot above that there is also a slight concentration towards negative latitudes; the peak density is around -5 to -10 degrees in both cases. This is likely because the Sun is above the Galactic plane at present (though I also wonder if dust plays a role). The Sun is currently 60 light years above the plane and heading upwards (see How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? ). The combination of the results I have shown may be sufficient to conclude that the Galactic disc has a "characteristic thickness* not more than a few hundred light years, but more than 60 light years!


Did the Moon ever approximate a geosynchronous orbit above the early Earth?

Maybe. The period of a 20,000 km orbit is about 7h 50m. Wikipedia says:

  [The Earth] would have experienced a day some five hours long after
  the impact, and the Earth's equator and the Moon's orbit would have
  become coplanar.

In the present era, the Moon's orbital plane is tilted by
5.145° to the ecliptic (the Earth's orbital plane); the obliquity of the ecliptic (relative to Earth's equatorial plane) is about 23.4°. However, as Wikipedia says:

  The Moon's orbit around Earth has many irregularities (perturbations),
  the study of which (lunar theory) has a long history.

The post-impact Earth was very hot, and had a magma ocean for a while. However, it soon had liquid water again because the high atmospheric pressure pushed the boiling point of water way up.

From Wikipedia:

  [From the Theia impact] a fair fraction of material should have been
  vaporized by this impact, creating a rock vapor atmosphere around the
  young planet. The rock vapor would have condensed within two thousand
  years, leaving behind hot volatiles which probably resulted in a
  heavy $\mathrm{CO_2}$ atmosphere with hydrogen and water vapor. Liquid water
  oceans existed despite the surface temperature of 230 °C (446 °F)
  because at an atmospheric pressure of above 27 atmospheres, caused by
  the heavy $\mathrm{CO_2}$ atmosphere, water is still liquid. As cooling
  continued, subduction and dissolving in ocean water removed most $\mathrm{CO_2}$
  from the atmosphere but levels oscillated wildly as new surface and
  mantle cycles appeared.

It is difficult to estimate the effects of the interaction between the early Moon and that primordial ocean.


Approximate formula to find velocity from cosmological redshift

This formula is exact if the expansion is linear ($a(t) = H_0 t$) and all peculiar velocities are zero. In that case, the comoving distance to the object is $$\int_{t_\text{then}}^{t_\text{now}} \frac{c\,\mathrm dt}{a(t)} = \frac{c}{H_0} \ln \frac{t_\text{now}}{t_\text{then}} = \frac{c}{H_0} \ln (1{+}z)$$ and the present recessional velocity is $H_0$ times that.
In the real world, the expansion has been not too far from linear, and recessional velocities are not too large, so it's reasonably accurate.


Why can't we feel the Earth's revolution?

You do, but it's too small to really notice

First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.

Think about driving a car, or riding in an airplane.  Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".

However, When you take a sharp turn, or take off from the runway, you definitely feel something.  That's the acceleration.  It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.

More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.

Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.

Acceleration is quite noticeable, depending on its magnitude.  Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.

So how large is the acceleration keeping Earth in orbit?  About 0.0059 m/s².
What about the acceleration of Earth's rotation?  A ever so slightly larger 0.0339 m/s².

Small wonder that it seems like you can't feel these forces!


What are the most popular computer programming languages in observational astronomy?

Python is gaining popularity and replacing MATLAB in many fields of science as the tool for fast prototyping and writing research code.
For example, have a look at:

https://www.astropy.org/
http://www.astropython.org/
https://python4astronomers.github.io/
https://www.astroml.org/
https://asteca.github.io/

(I'm not working in astronomy as a researcher but in some other field.)


Polar satellites for Global Navigation Systems

IRNSS
is intended to serve an area spanning 100 degrees in longitude by 80 degrees in latitude.
Anywhere in that service area, at any time, a receiver needs to compare distances to four of the satellites.
To determine latitude, the satellites can't all be above the equator, so some of the orbits are inclined 29$^\circ$. Those satellites oscillate north-south in pairs. Polar orbits aren't needed; global systems serve the poles with orbits inclined only 55$^\circ$.


Which is the closest exoplanet to Earth?

Earlier this year (2016), scientists used the radial velocity method to discover a planet orbiting Proxima Centauri: Proxima Centauri b. It was announced in Anglada-Escudé et al. (2016). Here are some of its basic properties, as reported by the authors of the paper and known as of August 2016:

Mass: At least 1.3 Earth masses
Semi-major axis: 0.05 AU
Maximum eccentricity: 0.35
Orbital period: 11.2 days
Surface temperature: 234 K (see Table 1)

Now, as the authors note, the planet may not be habitable:

  The habitability of planets like Proxima [Centauri] b—in the sense of sustaining an atmosphere and liquid water on its surface—is a matter of intense debate. The most common arguments against habitability are tidal locking, strong stellar magnetic fields, strong flares and high ultraviolet and X-ray fluxes; but none of these have been proved definitive.

However, further observations might help us learn more about the planet and determine if it could be habitable for any type of life.

It should be noted that the controversy about planets around Alpha Centauri continued after this question was initially answered by user8. Alpha Centauri Bb was shown to be nonexistent by Rajpaul et al. (2015). However, in another twist, observations by Demory et al. (2015) seemed to show the existence of Alpha Centauri Bc (at the time, Alpha Centauri Bb was still a possibility, so now "Alpha Centauri Bc" would actually be "Alpha Centauri Bb" if it existed), an Earth-like planet. However, there has been no more evidence to confirm the existence of this planet - and at any rate, Proxima Centauri is closer to the Solar System than Alpha Centauri, so Proxima Centauri b wins no matter what.


Having some experience with the only logarithmic unit in regular use (the dB, for sound intensity), I have to say: please, no.

A logarithmic representation can be handy in some cases. This chart is an excellent example of what can be done if you choose the scale on your image correctly.

Note that this has been done without introducing a new unit for distance, it just plots log(x) instead of x on the vertical scale.

But if you were to use a logarithm in the fundamental unit for distance, you'd introduce some problems:

cosmological distances no longer match other distance units and any conversion between them involves using a log function.
calculations involving cosmological distances get more complicated because you've now got to use log() everywhere.
you lose the linear relationship between numbers: object X is twice as far away as object Y, but the difference in your unit becomes log(2) making distances far less intuitive.


Units of time are actually quite arbitrary.

Their relationship to astronomical features is historical:

it's what people used when counting rotations and revolutions of the planet were the best tools to hand. Nowadays, they are legacy values: the "true" units are other things that happen to be feasible to measure using modern tools (wavelengths of light, resonances of atoms), with particular values chosen and scaled to match the old values for continuity.

And other units derived from them were purely matters of arithmetical convenience. Factors of 12 and 60 were common because they make it easy to do arithmetic in thirds and quarters, and when doing fractions, they reduce easily. So a value like a "second" is absolutely and utterly arbitrary; it applies to outer space exactly as well as it does here.

The second's arbitrary value is the fundamental unit of the metric system, that that's what scientists use to communicate with each other. The numbers can be scaled for their convenience: as long as they're labeled, one system is as good as any other. (Especially with computers to do the work.)

Astronomers try to measure rotations and revolutions of distant bodies, and when speaking of them in public they might use "day" or "year" to help people visualize it. But when speaking to each other, they know what the technical terms mean, and avoid the ambiguity of layperson speak.

  So in the end, it's all just "seconds", scaled up or down. It may be
  converted to something else for public consumption or even with each
  other, but when it comes down to brass tacks they find it's easiest to
  all use the same unit, even if it's completely arbitrary.

(It is worth noting that some physicists use a rather unfamiliar set of units call "natural units", under which important constants like the speed of light and the pull of gravity work out to precisely 1. These are convenient when you're pushing around equations, but require a fair bit of effort to translate into clocks and rulers when you want to build an actual experiment.)

  Another approach can be Through an atomic clock.

Modern atomic clocks use atoms to determine the passing of time. So you would measure it using that.

  For example 1 second is defined as --- the duration of 9192631770
  periods of the radiation corresponding to the transition between the
  two hyperfine levels of the ground state of the caesium 133 atom This
  mean that if you can make a laser shine on a ceasium 133 atom and by
  counting the periods of the light emitted you will now how many
  seconds have passed.

As a interesting side note sins our orbit around the sun is not constant it is sometimes necessary by the scientific community to prolong a year by a second or so every now and then to make clocks match the moment of the earth. There called leap seconds


How to compensate the effect of tectonic activity in devices like LIGO?

There are four aspects to this:

You suspend the mirrors and other components using the best suspension mechanism that has ever been built. This is very effective at isolating the experiment.
There are two detectors thousands of km apart. To claim a genuine signal it should be seen identically in both in less than the light travel time between them.
Many of the astrophysical GW sources (e.g. merging black holes) produce characteristic signals that look nothing like random seismic activity - for instance a correlated increase in frequency and amplitude terminating in a "chirp".
Lastly, LIGO is not immune to seismic noise - it is this that partly limits its sensitivity below frequencies of around 30 cycles per second. The seismic noise component at higher frequencies is much smaller.

The way that LIGO isolates the astrophysical signal from the various noise components that are present is described in this paper by the LIGO and VIRGO collaborations. A crude characterisation of the process is that you cross-correlate your signal against a library of expected signatures from astrophysical GW sources and look for a high degree of correlation (which you wouldn't expect in random noise).

The characterisation of the noise (including seismic noise), a description of how it is assessed and monitored (including how data is rejected when deemed too noisy) is found in a further LIGO publication.


The two coincided about 800 years ago. The December solstice and perihelion date coincided in 1246, 773 years ago.

There are many different concepts of what qualifies as a "year". Three of them are the sidereal, tropical, and anomalistic years. The sidereal year measures how long it takes for the Earth to complete one orbit about the Sun with respect to the remote stars.

The tropical year measures how long it takes for the Earth to pass from one March equinox to the next. The tropical year is about 20 minutes shorter than the sidereal year due to the slow changes in the Earth's rotation axis caused by the Moon and the Sun.

The anomalistic year measures how long it takes for the Earth to pass from one perihelion to the next (or one aphelion to the next). The anomalistic year is about 7.8 minutes longer than the sidereal year due to gravitational perturbations from the other planets, with Jupiter and Venus dominating over all the others. The tropical year is thus shorter than the anomalistic year by about 28 minutes. That the tropical year and anomalistic year have slightly different frequencies means that there is a very slow beat frequency with a beat period of about 26000 years.

Human civilization arose about half of this period ago (about 12000 years ago) when the last glaciation ("ice age") ended. At that time, the anomalistic and tropical years were almost 180 degrees out of phase with respect to where they are now. The end of a glaciation typically occurs when northern hemisphere summers coincide with perihelion. That makes summers in far northern latitudes short in duration but rather hot.

One of the key proxies regarding whether glaciations start or end is the mean July temperature at far northern latitudes. When perihelion occurs a few weeks after the June solstice, the short but hot summers that result makes snow from the previous winter melt rather than build up. The melting that results from such short but hot summers triggers a positive feedback that makes the end of a glaciation very rapid, in a geological context of "rapid".

To clarify things,

  So, there is no particular reason in orbital mechanics; it just happens that we live in an era when they are close together?

Exactly.

Or maybe not. The end of the last glaciation, which occurred about 12000 years ago, massively spurred human development. The cycle then was almost exactly opposite what it is now. It took about 10000 years for humanity to develop to a somewhat advanced level (e.g., Roman Empire level), and then only another couple of thousand to develop to the stage where we are now. Whether humanity can survive for another 13000 years without wiping itself out will determine whether this was happenstance.


Estimating the angle covered by the star trails and deducing how long the exposure lasted

The answer is given here. One minute of time corresponds to 15 arcminutes (written as 15'). This is because in 24 h the Earth revolves 360º, so
$$\textrm{angle per time} = \frac{360º}{24 \textrm{ h}} =\frac{21,600'}{1440 \textrm{ min}} = 15'/\textrm{min}.$$
If you turn this fraction upside down, you see that 1' corresponds to 1/15 min, or 4 seconds.

That is, you measure the angle (let's call it $\theta$) of any of the star traces, as seen from the center (notice that the Northern Star is not exactly at the center, so that it itself traces a tiny arc instead of a dot). From the picture below, I get roughly $\theta = 135º$. The exposure time is thus:
$$t_\mathrm{exp} = \frac{\theta}{\textrm{angle per time}} = \frac{135º}{360º/24 \textrm{ h}} \sim 9\textrm{ h}.$$

By the way, if you mark the position of the ends of the trails, you can recover the stellar sky. I found Ursa Major, marked by the yellow dots.


How can professional astronomers keep amateurs engaged given the ever-increasing power of modern telescopes?

The role of amateurs in making discoveries will likely change. Telescopes like the LSST will make it even more unlikely that an amateur with a smaller telescope will detect an object first. But the LSST will produce a vast amount of data that will need to be processed, and amateurs may still play a role in this, as citizen scientists do with projects like Folding At Home.

Professional astronomers will continue the same outreach they currently do. Amateur astronomy is still rewarding without the vain hope of making new discoveries. It's enjoyable for the same reason people still visit scenic wilderness instead of just studying maps and photos of it on their computers.


How many new galaxies enter the observable universe each day/year/decade?

What we see at the outer reaches of the observable universe is a surface of the cosmic microwave background (CMB), which is at the time when the universe, still quite homogeneous (perturbations were a few parts in $10^5$), cooled enough for the electrons to combine with the protons and make hydrogen atoms.  Each day, we can see a light-day farther and so we slowly see a different (larger) surface of the CMB (but there were no galaxies at this epoch).  Therefore, as we observe the CMB over many years, we will always see the same time epoch, but at different spherical shells.  Will we see new galaxies?  Yes. Galaxies are forming continuously in view at high redshifts (2-20?).  In a few million years, the small perturbations that we see today in the CMB will develop into young galaxies within our view.  One cannot provide numbers for how many we might be seeing because small galaxies merge into bigger ones and those merge again.  At high redshifts there is a lot of merging going, so the rate is particularly ill defined there.

How is this changed by the acceleration from the cosmological constant (aka dark energy)? From the Friedmann equation one sees that the Hubble constant reaches a final constant value of $c\sqrt{\Lambda/3}$ or about 40% higher than its present value.  Eventually, in billions of years, the CMB will shift from $3^oK$ to much lower and then the CMB will fade (redshift) away.  After that, those new distant galaxies will also fade away.  And then more and more of the universe will go dim.


Are we still in the Solar Maximus stage? Will this winter 2013/14 be a good aurora-viewing season?

From this data here:

http://www.swpc.noaa.gov/SolarCycle/

http://solarscience.msfc.nasa.gov/SunspotCycle.shtml

We are in a local maximum but the actual values have been lower than predicted for this time.

As for getting an idea for the aurora, I would keep an eye on the sunspot numbers and the space weather.

http://www.swpc.noaa.gov/today.html

http://www.gi.alaska.edu/AuroraForecast


How do scientists know there are about 300 billion stars in a galaxy and there are about 100 billion galaxies?

The way it works is as follows. We do detailed studies of stars in the solar neighbourhood. This establishes the local density of stars and the mix of masses they possess (called the stellar mass function). We compare that with the mass function of clusters of stars and note that to first order it appears invariant.

We can then triangulate the problem in various ways: we can make a model for the stellar density of the Galaxy, assume it all has the same mass function and hence get a number of stars. The model may be based on crude light-to-mass conversions, but more often would be based on deep surveys of the sky - either narrow pencil beam surveys from HST, or broader surveys like SDSS, The key is to be able to count stars but also estimate how far away they are. This is highly uncertain and relies on some assumptions about symmetry to cover regions of our Galaxy we cannot probe.

Another method is to count up bright objects that might act as tracers of the underlying stellar population (eg red giants), compare that with the number of giants in our well-studied locale, and from this extrapolate to a total number of stars, again relying on symmetry arguments for those bits of the Galaxy that are distant or obscured by dust.

A third way is to ask, how many stars have lived and died in order to enrich the interstellar medium with heavy elements (a.k.a. metals). For example, it turns out there must have been about a billion core-collapse supernovae to create all the oxygen we see. If we assume the mass function is invariant with time and that supernovae arise from stars above 8 solar masses, then we also know how many long-lived low-mass stars were born with their high-mass siblings and hence estimate how many stars exist today.

The number, whether it be 100 billion or 300 billion is no more accurate than a factor of a few, but probably more accurate than an order of magnitude. The main issue is that the most common stars in the Galaxy are faint M dwarfs,that contribute very little light or mass to the Galaxy, so we really are relying on an extrapolation of our local knowledge of these objects.

The number of galaxies problem is easier, though the number is less well defined. We assume that on large scales the universe is homogeneous and isotropic. We count up how many galaxies we can see in a particular area, multiply it up to cover the whole sky. The number must then be corrected for distant faint galaxies that cannot be seen. The difficult here is that we are looking into the past and the number of galaxies may not be conserved, either through evolution or mergers. So we have to try and come up with a statement like "there are n galaxies in the observable universe today that are more luminous than L". I think this number is certainly only an order of magnitude estimate.


What is the average color of the universe?

According to astronomers at Johns Hopkins University, it's beige. Note, they previously announced it was turquoise, but they identified a bug in their calculations, redid the numbers, and now it was determined the average color is beige. Kind of boring, yes, but kind of what you'd expect.

In short, they collected "detailed light measurements from more than 200,000 galaxies. They then constructed a "cosmic spectrum," which represents all the energy in the local universe emitted at different optical wavelengths of light" and averaged it. More details at link below.

NPR : The Color of the Universe Is...


An asteroid that remains outside Earth's atmosphere, and therefore does not impact Earth, will have no effect on the planet.

Asteroids, even biggish ones like the dinosaur killer, are pretty small compared to the Earth. They do not have enough mass to graviationally affect earth's orbit noticeably. And if they miss earth, the pass quickly.

The concern would be that if an asteroid can pass close to Earth, that it may collide in a future orbit. The asteroid's path will be changed by the earth's gravity. However, a very close encounter will enable astronomers to get a very high quality position on the rock, and so be able to predict its future path with very great accuracy.

Comet tails aren't a worry. We passed through Halley's tail in 1910, the effect on the Earth was precisely nothing. ☄


How to accurately measure the temperature of red supergiants?

Stars behave like blackbodys. Not perfect idealized blackbodies, however, the spectrum of a star is close enough to the standard blackbody spectrum. Reason why you can use the Wien's Law to calculate an estimate of its surface temperature:
$\lambda_{\rm max} = (0.29 {\rm\, cm\, K}) / T$
Where $\lambda_{\rm max}$ is the frequency of maximum measured emission of light. Problem is, measuring a star’s spectrum is not always easy. The most common approach is to observe the star with two different filters, a blue one and a visual yellow one. Measuring the difference of the Blue bigthness and the Yellow one gives you the actual color of the star. In some cases more color filters are used (UV, Near Infrared, V, ...).
Now, as of why the used the iron lines: When a Star ages, its core gets more and more iron rich due to nuclear fussion (iron is the last element that produces more energy when getting fussioned than it takes for the reaction to happen). When the balance between outwards radiation and gravity gets unstable, the star collapses. In the case of Red giants a supernova most likley follows. The Red giant that was beign messured in the paper had probably a lot of iron, so the spectal lines of iron where the most "stable" ones that did not get altered by the stars upper atmosphere. They probably used two frequencys of different iron configurations to propperly estimate the $\lambda_{\rm max}$ of the star.
Hope this clarifys things a bit.


There is nothing wrong with your explanation.  You could add some explanatory notes like $a_{min} = l-c$ and $a_{max}=l+c$. Also, I suggest that you put the equation $l+\delta = \frac{\pi}2-a_{min}$ on a separate line.

It is very good to figure out these equations on your own.

$ $

Here is an alternative explanation with less algebra:

If you walk outside at night, you will notice that the north star has an elevation of $l$ degrees if your latitude is $l$.
The north star does not move.
The angular distance between stars does not change  (in the short term!).
In order for the a star to be circumpolar, it needs to be within $l$ degrees of the north star (otherwise it will dip below the horizon when its hour angle is 12 hours.)
So, the angle $\theta$ between the north star to a circumpolar star needs to be less than $l$.  That angle is $\theta=90^\circ-\delta = \pi/2 -\delta$ where $\delta$ is the declination of the star.
Finally,
$$
\begin{align} \theta &\leq l\\
\pi/2 -\delta &\leq l \\
\pi/2  &\leq  \delta + l \\
\pi/2  - l &\leq \delta.
\end{align}
$$


Could someone explain RA/dec in simple terms?

A system projected into the sky

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

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

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

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

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

So how do we use this system?

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

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

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

What's over my head right now?

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

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

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


Perhaps I have always been misunderstanding the concept of space weather, rain from space now that's a cool idea.

Seriously though, the solar wind particles are at such a a high temperature that molecules cannot form, instead it is an ionised gas containing mainly Hydrogen and Helium and only very tiny traces of other elements.


How to Distinguish between Temperature and Doppler effect using Black-body radiation?

Without any other information, you cannot distinguish between the two effects.

$$ T = T_0 (1 + z) $$

A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T_0$ with redshift $z$.

For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody.  For the CMB, we can use the fact that atomic combination happens only below a particular temperature.