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

  The apparent motion of the planets when they appear to move backward (from east to west) from the direction that they move ordinarily with respect to the stars.

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

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

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

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

TIME

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

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

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

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

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

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

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

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

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

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

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

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

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

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


Why would the interplanetary medium atomic hydrogen Lyman‐α emission fall as 1/r rather than1/r^2? The Live Science article NASA Spotted a Vast, Glowing 'Hydrogen Wall' at the Edge of Our Solar System links to Geophysical Research Letters The Lyman‐α Sky Background as Observed by New Horizons https://doi.org/10.1029/2018GL078808 the abstract of which is

  Recent observations of interplanetary medium (IPM) atomic hydrogen Lyman‐α (Lyα) emission in the outer solar system, made with the Alice ultraviolet spectrograph on New Horizons (NH), are presented. The observations include regularly spaced great‐circle scans of the sky and pointed observations near the downstream and upstream flow directions of interstellar H atoms. The NH Alice data agree very well with the much earlier Voyager UVS results, after these are reduced by a factor of 2.4 in brightness, in accordance with recent re‐analyses. In particular, the falloff of IPM Lyα brightness in the upstream‐looking direction as a function of spacecraft distance from the Sun is well‐matched by an expected 1/r dependence, but with an added constant brightness of ~40 Rayleighs. This additional brightness is a possible signature of the hydrogen wall at the heliopause or of a more distant background. Ongoing observations are planned at a cadence of roughly twice per year.

The constant brightness (indicator of "the wall") makes sense to me, a large resolved surface of constant emission intensity (W/m^2) will have a constant brightness, independent of distance.

Conversely, a point source's brightness will obey a $\frac{1}{r^2}$ (inverse square) law.

If absorption were the reason for the fall off, it would be Beer-Lambert as $e^{-\alpha r}$.

But the abstract describes a 1/r fall off of the interplanetary medium (IPM) atomic hydrogen Lyman‐α (Lyα) emission.

Question: How does a $\frac{1}{r}$ falloff arise in this case?

The brightness considered here is the brightness of scattered Lyman $\alpha$ as viewed in the upstream direction, i.e. away from the Sun.

The authors assume that the density of the atoms responsible for scattering the Ly$\alpha$ — the neutral hydrogen atoms — is 1) roughly constant throughout the Solar System and 2) is optically thin, meaning that the radiation field is not attenuated considerably with distance from the Sun.

Since the intensity of a point source falls off as $1/r^2$, the scattering emissivity in a given point should then also fall off roughly as $1/r^2$.

But when viewed from inside the Solar System and in the upstream direction, the brightness is the integral of all scattering events from that point to infinity, and since $\int 1/r^2 \propto 1/r$, you have your quoted brightness.


Could any known, living organisms on Earth survive on Mars? No life has been discovered outside of Earth (yet?), but do we know if anything that would be considered "living" on Earth could conquer Mars? (or maybe Venus?)

With the Mars One project on the way, I was wondering if it was possible to transplant some of Earth's life to Mars, and if something could survive there (naturally).

Many kinds of life are based on air and sun. Animals have to breathe oxygen, plants need the sun, and both need water. But there are many other kinds of life on Earth, like:

Very deep seawater life that lives under high pressure and without sunlight,
extremophile organisms that live in very hot water,
weird "arsenic life" which was debunked as being fully arsenic-based, but turned out to be something different...

Have there been any theories or studies about unusual forms of life on Earth that might survive on Mars?

The most likely candidate would be the Tardigrade. These little guys handle vacuum and radiation just fine. So long as water is provided, according to tests done in LEO the Tardigrade would survive on Mars. Even if they do dehydrate, they spring back to life once water is provided again.


(Thought Question) What would a giant far away mirror look like to a telescope? If there was a theoretically perfect mirror the size of say, our solar system, somewhere out in space that had a focal point of literally earth. What would that look like to a space or earth telescope? Could it be distinguished from other parts of space? Could you identify earth in it? (A la By putting a mirror in space, would we be able to see into the past?)

If there was a theoretically perfect mirror the size of say, our solar system, somewhere out in space...

Great thought experiment so far!

...that had a focal point of literally earth...

Rats! I was hoping for a flat mirror so the answer would have been slightly simpler.
It is true that concave mirrors are "magnifying mirrors" and if we stay closer to the mirror than twice its focal point (see below) we can see a magnified image. I don't believe that it provides better optical resolution than a normal mirror, but since the construction of our eyes is fixed (we can't change the optical system nor "pixel density") we use the mirror to "blow up" the image in the same way that a photographic enlarger blows up the image on a negative without improving the resolution of the image.
 source
In optics image information is encoded in the wavefront at any point whether in focus at that point or not. If we know the mirror's diameter and distance from Earth, we can apply the principle of diffraction in a simple way no matter how complicated the rest of the optical system.
For a hard-edge circular mirror (top-hat shaped apodization) we know that the Airy disk is the right principle to apply, and a simple definition of resolution gives us
$$\theta \approx \text{1.22} \frac{\lambda}{d}$$
for the angular resolution. Let's use 500 nm green light for $\lambda$ and twice Neptune's orbital radius (60 AU) for $d$. That gives us $4 \times 10^{-18}$ radians.
If the mirror were at the distance of Proxima Centauri or only $4 \times 10^{+16}$ meters, then the impact on resolving a wavefront from the perfect mirror based only on diffraction of a circular aperture would be of the order of centimeters!
The focal length of a concave mirror is the parallel-to-point distance, so we would really need the mirror to be at roughly the point-to-point distance of twice the focal length.
If the concave mirror brought our image back to us, say on a sheet of paper, it would be incredibly dim i.e. pretty much no photons except from the Sun itself. But if we ignore that then we would see ourselves blurred by about 1 centimeter.
If the image was kilometers in front of us then we could focus a telescope on that image in space and re-image it at the entrance pupil of an eyepiece or on to a sensor.
If the telescope were 1 meter in diameter, we could see roughly a 1 meter wide swath of Earth.
Of course this doesn't work so simply because the mirror would have to be pointed such that the location of Earth 8.5 years ago would be imaged where we were today.
Conclusions
Yes this is kinda possible from a Gedankenexperiment point of view; a 60 AU mirror out at Proxima Centauri with a focal length of twice that distance could produce an image of Earth nearby Earth (light time considerations in force) and we could look at that location in space with a large diameter telescope and see a patch of Earth 8.5 years ago roughly vignetted by the size of the telescope's aperture.


What celestial body (inside the solar system) has the highest flattening ratio? As a planet, Saturn has the highest flattening(ellipticity) which is 0.09796. So, it makes Saturn with the largest equatorial bulge as a planet and as such Saturn is the flattest planet.
However, I have encountered this statement from here:

Earth's Moon is even less elliptical, with a flattening of less than
1/825, while Jupiter is visibly oblate at about 1/15 and one of
Saturn's triaxial moons, Telesto, is highly flattened, with f between
1/3 to 1/2 (meaning that the polar diameter is between 50% and 67% of
the equatorial).

So, Telesto has a flattening value even higher than Saturn itself. I couldn't find any other body that has higher flattening value than Telesto. Does that mean Telesto has the highest recorded 'f' value?
Question: What celestial body inside the solar system has the highest recorded flattening value? Is it Telesto?
Note: Most of the asteroids, meteors, comets and small moons have irregular shapes and as such measuring the 'f' value is impossible and may have no useful value. I want to know which body has the highest recorded 'f' value.

I believe besides Telesto the dwarf planet Haumea in the Kuiper belt could also be in the mix for the highest flattening of objects in our solar system. Haumea is a triaxial ellipsoid where the 'f' value would around 1/2 between the largest and smallest major axis.


At what gravity does a body's gravitational lensing become observable? I wonder whether we know at how much gravitation a gravitational lensing around a celestial body becomes observable (visible in the sense of that we see stars on the wrong places, stars behind the body etc, similar like around this neutron star, but of course by far not as strong, I'd like to know where the boundary lies). Such visible gravitational lensing occurs around neutron stars, black holes and white dwarfs, right? What about red and brown dwarfs? They're also quite dense and have a high surface gravity (though not as high as the other mentioned bodies of course). Do they curve the space around them visibly?

The article 'A practical relativistic model for microarcsecond astrometry in space' (Klioner 2003; caution, very heavy duty maths...) describes the framework which was developed to support the data processing for the ESA Gaia astrometric space mission. This mission has the goal to measure the position (astrometry) of stars and other objects to approx. 20 microarcseconds (one millionth of an arcsecond, which is 1/3600th of a degree). In order to do this, you need to account for all the sources of error and deflection, such as light bending near massive objects, to a level that is well below you expected accuracy (this level of systematic error was chosen to be 1 microarcsecond, hence that part of title).

Section 6 of the paper discusses gravitational light bending and Table 1 gives the size of the effect for various bodies in the Solar System. The column headed by $\delta_{pN}$ gives the size of the effect in microarcseconds ($\mu$as). As expected the Sun has the largest effect at $1.75 \times10^6\,\mu$as or 1.75", with Jupiter second etc. The next two columns gives the angular separation a light ray has to pass to the body to cause a $1 \mu as$ or $10 \mu as$ effect. So for example, any light ray from a distant object passing within $11.3^\circ$ of Jupiter will cause at least a  $10 \mu as$ effect. The smallest bodies that have an effect larger than $1 \mu as$ are the dwarf planet Ceres (mass $8.958 \times 10^{20}$ kg), and the moons Dione ($1.05 \times 10^{21}$ kg) and Umbriel ($1.27 \times 10^{21}$ kg) - these masses are several thousand times smaller than the Earth for example.

Now the level of precision needed by Gaia is far more precise than and the deflections are tens of thousands times smaller than you would see in an optical image with a regular telescope e.g. Eddington's confirmation of general relativity during the 1919 total solar eclipse (space.com story). The Klioner (2003) paper includes a formula to estimate the size of the effect of the light deflection as a function of the density but you need to know what level of deflection you care about to get an answer. The formula (number 35 on page 11 of the paper) for the radius $L$ of a body producing a light deflection than $\delta$ is:
$$
L\geq \left(\frac{\rho}{1\,\textrm{g cm}^{-3}}\right)^{-1/2}\left(\frac{\delta}{1\,\mu\textrm{as}}\right)^{1/2}\times 624\,\textrm{km}
$$
where $\rho$ is the mean density of the body. Once you have chosen a level of deflection/lensing you want ($\delta$), you can chose suitable densities (e.g. asteroids are $\sim2\,\textrm{g cm}^{-3}$, Earth and other rocky planets are  $\sim5.5\,\textrm{g cm}^{-3}$)


Why are planetary orbits not all circular? Why are the orbits of the planets elliptical ? The circular orbit also satisfies all the conditions of gravitational force. Then why did nature choose the elliptical orbits ?

For a given amount of angular momentum, a circular orbit is the orbit with lowest energy.

Some orbits are elliptical (actually, all orbits approximate to ellipses) with a non-zero eccentricity, because they have yet to reach their lowest energy state, or because they have been excited from their minimum energy state circular orbit.


How is stellar mass of a galaxy obtained? I believe there is a difference between the mass calculated using the orbital speed and the stellar mass of the galaxy. So how is the stellar mass calculated?

Thank you :)

In short, you count stars. Of course you can't count or even see all the stars, so you have to count certain types of (bright) stars and then assume that the ratio of these to all the stars in their neighbourhood is similar to the ratio in a local part of the Galaxy where we can do a more complete census.

Once we have an idea of a number for the stars in any region, we can use an assumed mass distribution for the stars to calculate the total stellar mass.

There are details and nuances (e.g. the form of the stellar mass distribution depends on the age of the population), but that's the basic process in our own Galaxy.

In other galaxies, where you don't resolve individual stars, you work on the basis of mass to luminosity ratios for typical populations to convert a luminosity to a stellar mass.


Should I focus more on Aperture or Focal Length for a telescope? Let's assume there are three types of telescopes with the following specifications:

    Telescope                  Aperture              Focal Length
    --------------------------------------------------------------
    Example 1                     70 mm                    400 mm
    Example 2                     60 mm                    700 mm
    Example 3                     60 mm                    900 mm

As a beginner, should I look for a higher value in the aperture or the focal length?

There is one rule that is generally true for all deep sky objects (nebulae, stars, galaxies,...): Aperture matters!

For solar system objects, aperture is not that important.

The second most important thing is: What size are the objects you want to look at: Small objects need long focal lengths and high magnifications, large objects need short aperture for low magnifications.

With 400mm you could watch objects like:

Andromeda galaxy core
Orion nebula, other large emission or reflective nebulae (e.g. Pleiades)
large star clusters
low magnification lunar observations

With 900mm you could watch objects like

Planets (Saturn, Jupiter, Mars, ...)
high magnification lunar observations
planetary nebulae (e.g. ring nebula)

Note that 60 and 70mm aperture are still very small for telescopes! The aperture influences two things:

Light sensitvity: The larger the aperture, the more light you can collect. Very important if you live in a city!
Maximum resolution: Rule of thumb is that you can do aperture in mm times two as maximum magnification. I.e. for 60mm a 120x magnification is the absolute maximum which still is feasible.

The magnification is created by the eyepiece. E.g. when you have a 400mm focal length telescope and use a 10mm eyepiece, you get 400mm/10mm = 40x magnification.

Note: the shorter the eyepiece focal length, the more difficult it is to build. Good 5mm eyepieces can cost 100 USD and up. I personally started with a 750mm Newtonian with 150mm aperture and 25mm and 10mm eyepieces. That's a good allrounder, even though planets will appear rather small with the 10mm eyepiece. But you can later invest more money in good eyepieces, which you can re-use on better telescopes which you may buy later on.

Edit: One more thing -- the telescope mount is equally important as are the eyepieces and the telescope itself. A mount that fits the telescope easily is as expensive as the optical tube assembly itself. Hence many beginners start out with a Dobson telescope, which uses a very, very simple yet sturdy mount.


Are gas giants supported by thermal pressure? I've heard gas giants are supported because there is an equilibrium between thermal pressure and gravity. That is, if Jupiter were to suddenly begin compressing, temperatures would increase to the point that it would expand to its original size. Since it is in equilibrium, both the compression and thermal pressure counter one another, so its size never fluctuates.

However, Jupiter's average temperature is extremely low, so much so that its gases are electron degenerate. Is this a contradiction, or has the information above been just hearsay?

I am not sure what you mean by "thermal" pressure. Jupiter is supported by pressure, just like all objects that are in (approximate) hydrostatic equilibrium.

That pressure is provided by your everyday, temperature-dependent Maxwell-Boltzmann ideal gas pressure in the outer parts, but the free electrons in the interior become degenerate and so in these regions I suppose you would more accurately describe the pressure as being due to (partial) electron degeneracy pressure.

At the very centre there may be a liquid or even a solid core. In the metallic hydrogen model then this would still be degenerate electrons contributing the pressure. For a solid, rocky core, well solids are rather incompressible.

You can only describe Jupiter as being in approximate hydrostatic equlibrium. It is losing energy from its surface; this energy is supplied from gravitational potential energy and Jupiter is shrinking at a rate calculated to be about 2 cm per year. As it does so, the interior gas becomes denser and more and more degenerate and the pressure will become more independent of temperature. As a result, the rate of contraction will slow down and Jupiter will tend towards the radius of a "cold" body of that composition.


When was the parallax of a star first measured? Telescopes like Gaia measure the parallax of stars with a great precision. But for stars that are beyond 11 kpc, their parallax is still too small to be measured. With Earth-bound telescopes, only the nearest stars have a visible parallax. Was the parallax of any stars measured before the telescope era?
When was the parallax of a star measured for the first time?

Telescopes were apparently invented in 1609, but didn't become advanced enough to measure stellar parallax until the 1830s.
Observation of stellar parallax would be a big step in proving the heliocentric theory, and I think that the lack of detectable stellar parallax was used as an argument against the heliocentric theory in ancient times.  It was certainly used as an argument against the heliocentric theory in early modern times.

Stellar parallax is so small that it was unobservable until the 19th century, and its apparent absence was used as a scientific argument against heliocentrism during the early modern age. It is clear from Euclid's geometry that the effect would be undetectable if the stars were far enough away, but for various reasons, such gigantic distances involved seemed entirely implausible: it was one of Tycho Brahe's principal objections to Copernican heliocentrism that for it to be compatible with the lack of observable stellar parallax, there would have to be an enormous and unlikely void between the orbit of Saturn and the eighth sphere (the fixed stars).1

Source
After the Copernican theory gained popularity astronomers made many attempts to measure stellar parallax.

In astronomy, aberration (also referred to as astronomical aberration, stellar aberration, or velocity aberration) is a phenomenon which produces an apparent motion of celestial objects about their true positions, dependent on the velocity of the observer. It causes objects to appear to be displaced towards the direction of motion of the observer compared to when the observer is stationary. The change in angle is of the order of v/c where c is the speed of light and v the velocity of the observer. In the case of "stellar" or "annual" aberration, the apparent position of a star to an observer on Earth varies periodically over the course of a year as the Earth's velocity changes as it revolves around the Sun, by a maximum angle of approximately 20 arcseconds in right ascension or declination.

The Copernican heliocentric theory of the Solar System had received confirmation by the observations of Galileo and Tycho Brahe and the mathematical investigations of Kepler and Newton.[10] As early as 1573, Thomas Digges had suggested that parallactic shifting of the stars should occur according to the heliocentric model, and consequently if stellar parallax could be observed it would help confirm this theory. Many observers claimed to have determined such parallaxes, but Tycho Brahe and Giovanni Battista Riccioli concluded that they existed only in the minds of the observers, and were due to instrumental and personal errors. However, in 1680 Jean Picard, in his Voyage d’Uranibourg, stated, as a result of ten years' observations, that Polaris, the Pole Star, exhibited variations in its position amounting to 40″ annually. Some astronomers endeavoured to explain this by parallax, but these attempts failed because the motion differed from that which parallax would produce. John Flamsteed, from measurements made in 1689 and succeeding years with his mural quadrant, similarly concluded that the declination of Polaris was 40″ less in July than in September. Robert Hooke, in 1674, published his observations of γ Draconis, a star of magnitude 2m which passes practically overhead at the latitude of London (hence its observations are largely free from the complex corrections due to atmospheric refraction), and concluded that this star was 23″ more northerly in July than in October.[10]

Consequently, when Bradley and Samuel Molyneux entered this sphere of research in 1725, there was still considerable uncertainty as to whether stellar parallaxes had been observed or not, and it was with the intention of definitely answering this question that they erected a large telescope at Molyneux's house at Kew.3 They decided to reinvestigate the motion of γ Draconis with a telescope constructed by George Graham (1675–1751), a celebrated instrument-maker. This was fixed to a vertical chimney stack in such manner as to permit a small oscillation of the eyepiece, the amount of which (i.e. the deviation from the vertical) was regulated and measured by the introduction of a screw and a plumb line.[10]

The instrument was set up in November 1725, and observations on γ Draconis were made starting in December. The star was observed to move 40″ southwards between September and March, and then reversed its course from March to September. [10] At the same time, 35 Camelopardalis, a star with a right ascension nearly exactly opposite to that of γ Draconis, was 19" more northerly at the beginning of March than in September.[11] These results were completely unexpected and inexplicable by existing theories.

Source
So as a result of searches for stellar parallax, the aberration of light was discovered by James Bradley.
Bradley continued to research the aberration of light, and made another unexpected discovery, the nutation of the Earth's axis.

Nutation was discovered by James Bradley from a series of observations of stars conducted between 1727 and 1747. These observations were originally intended to demonstrate conclusively the existence of the annual aberration of light, a phenomenon that Bradley had unexpectedly discovered in 1725-6. However, there were some residual discrepancies in the stars' positions that were not explained by aberration, and Bradley suspected that they were caused by nutation taking place over the 18.6 year period of the revolution of the nodes of the Moon's orbit. This was confirmed by his 20-year series of observations, in which he discovered that the celestial pole moved in a slightly flattened ellipse of 18 by 16 arcseconds about its mean position.3
Although Bradley's observations proved the existence of nutation and he intuitively understood that it was caused by the action of the Moon on the rotating Earth, it was left to later mathematicians, d'Alembert and Euler, to develop a more detailed theoretical explanation of the phenomenon.5

https://en.wikipedia.org/wiki/Astronomical_nutation[1]
As it turned out, the changes in the apparent positions of stars due to the aberration of light and nutation of the Earth's axis are many times larger and easier to detect than even the largest stellar parallax of even the nearest star.
It was not until the 1830s that astronomical instruments became advanced enough that the first stellar parallaxes were detected and measured, after centuries of failed attempts.  And the parallaxes of only three stars were measured during the the 1830s.

The large proper motion of Alpha Centauri AB was discovered by Manuel John Johnson, observing from Saint Helena, who informed Thomas Henderson at the Royal Observatory, Cape of Good Hope of it. The parallax of Alpha Centauri was subsequently determined by Henderson from many exacting positional observations of the AB system between April 1832 and May 1833. He withheld his results, however, because he suspected they were too large to be true, but eventually published them in 1839 after Friedrich Wilhelm Bessel released his own accurately determined parallax for 61 Cygni in 1838.[62] For this reason, Alpha Centauri is sometimes considered as the second star to have its distance measured because Henderson's work was not fully acknowledged at first.[62] (The distance of Alpha Centauri from the Earth is now reckoned at 4.396 ly or 41.59 trillion km.)

Source
In 1804 Piazzi reported that 61 Cygni had a very large proper motion and was probably one of the closest stars to Earth, and thus a good candidate for parallax observations.  There were many unsuccessful attempts to measure the parallax of 61 Cygni.

When Joseph von Fraunhofer invented a new type of heliometer, Bessel carried out another set of measurements using this device in 1837 and 1838 at Königsberg. He published his findings in 1838[31][32] with a value of 369.0±19.1 mas to A and 260.5±18.8 to B, and estimated the center point to be at 313.6±13.6. This corresponds to a distance of about 600,000 astronomical units, or about 10.4 light-years. This was the first direct and reliable measurement of the distance to a star other than the Sun.[27][33] His measurement was published only shortly before similar parallax measurements of Vega by Friedrich Georg Wilhelm von Struve and Alpha Centauri by Thomas Henderson that same year.[34] Bessel continued to make additional measurements at Königsberg, publishing a total of four complete observational runs, the last in 1868. The best of these placed the center point at 360.2 ±12.1 mas, made during observations in 1849.[27] This is close to the currently accepted value of 287.18 mas (yielding 11.36 light-years).[35]

Source

The distance to Vega can be determined by measuring its parallax shift against the background stars as the Earth orbits the Sun. The first person to publish a star's parallax was Friedrich G. W. von Struve, when he announced a value of 0.125 arcseconds (0.125″) for Vega.[37] Friedrich Bessel was skeptical about Struve's data, and, when Bessel published a parallax of 0.314″ for the star system 61 Cygni, Struve revised his value for Vega's parallax to nearly double the original estimate. This change cast further doubt on Struve's data. Thus most astronomers at the time, including Struve, credited Bessel with the first published parallax result. However, Struve's initial result was actually close to the currently accepted value of 0.129″,[38][39] as determined by the Hipparcos astrometry satellite.[4][40][41]

Source
So the first three measurements of stellar parallax are almost tied for first place.


Are there recent estimates of Kuiper belt mass? What is the mass of the Kuiper belt? Obviously estimates will be highly uncertain, but for my application I would like to find a decent current estimate.

The papers I found so far: (Gladman et al. 2001) estimates it to 0.1 Earth-masses, (Trujillo, Lewitt & Luu 2001) gets the mass of bodies between 100 km and 2000 km to 0.03 Earth-masses, (Kenyon & Bromley 2004) estimate on the order of 0.5 Earth-masses (assuming a starting mass of 10 and then losing 95%), (Fraser et al 2014) estimate just $\sim$0.01 Earth-masses.

These papers are a bit old, and I suspect there have been more recent observations improving on the estimates. There is a fair number of papers discussing how to resolve this discrepancy between disk models predicting 10-30 Earth-masses and these low numbers, but I am more interested in empirical estimates.

So, what is the current consensus on the Kuiper belt mass and its uncertainty?

Surprisingly, I also have not been able to find any generally recent (5 years or less ago) information about this topic.

I did find a research paper from 2007 (Here). You may want to check out page 11 in Section 4.3. It states that the maximum mass of the Kuiper Belt is 0.1 Earth masses. They also talk about how it was estimated and such.

I haven't been able to find anything else besides Wikipedia articles (which you probably aren't looking for).

Very strange how there hasn't been a lot of research on this. An intriguing topic, I must say.

That's all I could find... Hope that helps...?


Solar series 142 and 147 started same year - series 147 and 152 will be ending same year. Why? Coincidence? This question was left over from the last question I had, so I am asking it separately here:

I have been looking into two solar eclipses (Saros series 147 and 152) that will both be occurring in 2021. Each of these eclipses will be happening near the opposite poles of the earth.  Both are part of young saros series. Saros 152 is comprised of 70 eclipses, while saros 147 contains a total of 80 eclipses. These series are 10 eclipses difference in length and began 180 years apart. So, I happened to check out their ending dates and found that both series are projected to end during the exact same year - 177 day apart – together in the year 3049. I thought that both of these series ending together one semester apart in the very same year was a pretty amazing coincidence.

So, checking further – it seems new saros series start from as few as 7 years apart to as many as 100 years apart (this is extremely variable spacing, but it averages to about 31 years apart over the usual 1260+ year active series lifespan). I wanted to know why series 147 and 152 would both be ENDING at the same time.  I went back to the NASA website, and checked the 5000 year listing of all solar saros series numbering 0 through 180. I only found one other such instance of two solar saros series ENDING in the same year – series 49 and 54 both ENDED together in 32 CE – so it seems like a pretty rare event.

But then, I checked to see if there were any occasions of two saros series BEGINNING together in the same year. Going back to the NASA website again with this question, I checked the 5000 year listing of all solar saros series numbering 0 through 180. I only found one single example of two saros series both BEGINNING in the same year. This time series 142 and 147 both BEGAN together in the year 1624 CE.

Not only is solar saros 147 involved in BOTH of these "same year" cases, but all three series eclipses will be presenting themselves during a single upcoming semester series in 2021 – solar series 142 (Dec 14, 2020), 147 (June 10, 2021),
and 152 (Dec 4, 2021).

  Can anyone explain to me the actual mechanics behind what causes saros series to begin and end? I am familiar with Saros/Inex
  Panorama – but unfortunately they do not answer this question, and I
  can’t even read them online anyway.

  Does anyone know why solar saros 147 may have STARTED in the same year as saros 142 or why saros 147 will be ENDING together in the same
  year as saros 152? Are these two occasions both very rare
  coincidences, while at the same time, both involve the same series 147?

The reason Saros Cycles begin and end is down to the positions of the orbits of the Sun, the Moon and the Earth. From the main Wikipedia page on Saros:
For an eclipse you need the 3 bodies to line up at a full or new moon,

and repeat occurrences of these lunar phases result from solar and lunar orbits producing the Moon's synodic period of 29.53059 days...however for the majority of these, the shadow of the Earth or Moon fall to the north or south of the other body.

It can only happen when a full or new moon is around the ecliptic.

The period of time for two successive lunar passes through the ecliptic plane (returning to the same node) is termed the draconic month, a 27.21222 day period. The three-dimensional geometry of an eclipse, when the new or full moon is near one of the nodes, occurs every 5 or 6 months when the Sun is in conjunction or opposition to the Moon and coincidentally also near a node of the Moon's orbit at that time, or twice per eclipse year.

Also:

Two Saros eclipses have the same appearance and duration due to the distance between the Earth and Moon being nearly the same for each event because the Saros eclipse cycle is also an integer of the anomalistic month, the period of the eccentricity of lunar orbit, 27.5545 days.
After one saros, the Moon will have completed roughly an integer number of synodic, draconic, and anomalistic periods (223, 242, and 239) and the Earth-Sun-Moon geometry will be nearly identical

The reason we get varying start and ends is because of this. Those roughly integer periods vary.
This diagram from that Wikipedia page illustrates perfectly why they can be different lengths, stop and start at changing intervals etc.


Distribution of spectral types of "giant" stars I'm looking for the distribution of spectral types in the "giant" star category. This category is defined as basically B and O main sequence stars and any star that is larger than main sequences outside of it.

My first thought was that the M, K, G, F, and A type distribution were the same as they are in the main sequence but then I realized M has to be dropped, and then i'm not quite sure how K, G, F, and A types evolve and usually only Red Giants are mentioned and I couldn't find numbers for the giants so I'm stuck.

To be clear, I want to know the percentage of giant stars that are in each of the spectral types. So if there are 1000 stars in a given area and 100 giant stars in that areas I only want to know the percentage of each spectral type of the stars in that 100.

OK, I see what you want.

What I did was take the revised Hipparcos catalogue and select all stars closer than 50pc. At this distance there should be a complete sample of the types of stars you were interested in. I can plot an absolute magnitude versus colour diagram for these 7096 stars (see below) and use this to select giants and OB main sequence stars.

I do this by using data from chapter 2 of Zombeck (1992), I define O and B main sequence stars to have $B-V<0$ and $M_V<5$ (to exclude hot white dwarfs). Evolved giants are selected (a bit arbitrarily) to lie significantly above the main sequence.

Then I can plot a frequency histogram of the B-V values of those stars and label them with spectral types, again according to tables in Zombeck (1992).
The result is seen below.

Of these 235 stars: None are O stars, 29 are B stars, about 120 are G giants, about 80 are K giants and only 2-3 are M giants


Does precession affect this measurement? I don't have the astronomy vocabulary to ask this question.   You can find a better picture at https://www.timeanddate.com/astronomy/axial-tilt-obliquity.html.
I want to know if the distance between points A & B on my graphic has a name, and if it would grow shorter or longer as the earth's rotation slows down.  Would this distance only reflect the 23° wobble in the rotation of earth or would it also reflect Precession of the Equinoxes.

No special name.
The problem that I see here is that the distance will vary over the year.  At the moment of solstice, you can easily find the distance by basic trigonometry:

If the center of the Earth is C, then $AC/AB = \tan(23.5^\circ)$ so $AB = 6370/\tan(23.5^\circ)=13000\, \mathrm{km}$

At any other time the distance will be different, and at the equinox the axis of the earth is parallel to the tangent plane of the Earth, so the distance is infinite. So the distance AB is not a fixed value, and varies over the year.
In the longer term the precession of the equinoxes will affect when solstice occurs, but not the angle 23.5.  This angle also varies in the long term between 22.1° and 24.5°, over a cycle of about 41,000 years.


How does the Hubble telescope capture long-exposure photos? I recently learned (if I understood correctly) that the famous Ultra Deep Field photo(s) required something like a 10-day exposure on a very tiny piece of the sky to achieve. How can this occur when the telescope is constantly in motion zipping around earth at ~17,000 mph? Wouldn't the earth and/or possibly the moon get in the way every 90 minutes or so?

I suppose my question applies to just about any photo it takes since it would be out of alignment of the target in less than a second.

The answer to this is that such images are not taken continuously. The HST did not stare at one part of the sky for 10 continuous days, but rather it stared at one part of the sky for short periods over a long time which, when added up, amounted to 10 total days of observations. To quote Wikipedia on the subject of how the Hubble Ultra Deep Field (HUDF) image was taken:

  The observations were done in two sessions, from September 23 to October 28, 2003, and December 4, 2003, to January 15, 2004. The total exposure time is just under 1 million seconds, from 400 orbits, with a typical exposure time of 1200 seconds. In total, 800 ACS exposures were taken over the course of 11.3 days, 2 every orbit, and NICMOS observed for 4.5 days. All the individual ACS exposures were processed and combined by Anton Koekemoer into a single set of scientifically useful images, each with a total exposure time ranging from 134,900 seconds to 347,100 seconds. To observe the whole sky to the same sensitivity, the HST would need to observe continuously for a million years.

In the above, ACS is the Advanced Camera for Surveys and NICMOS is the Near-Infrared Camera and Multi-Object Spectrometer.

That, I think, is the basic answer to your question, but there's a few interesting and related points I'd like to point out. First, the HUDF image was actually comprised of many individual images rather than a single 10-day exposure image (as the quote indicates). This would necessarily have to be the case, even if Hubble had the ability to stare at the HUDF region of the sky for 10 days without interruption. Cameras work by taking in light. The Charge-Coupled Device (CCD) which actually collects the light (and converts it to an electric charge) can only hold so much charge before it becomes saturated. At some point, you have to "stop recording" and read out the CCD which clears the accumulated charge. That way, you can resume again with a fresh slate. A 10-day continuous exposure would have easily saturated Hubble's CCD. So even if Hubble did spend 10 continuous days observing this patch of the sky, it would have been broken into many, many smaller exposures. This means, even if the Moon did get in the way during part of such a 10-day exposure, they could have just not made any observations during that time.

The other related fact I wanted to point out is that Hubble has something known as the Continuous Viewing Zone (CVZ). This is a region of the sky which is never obscured by the Earth, Moon, Sun, etc. as Hubble orbits. As the name implies, it allows for continuous viewing, without interruption. The Hubble Deep Field (HDF) (the precursor to the HUDF) was observed in this region. However, there are reasons why you wouldn't want to observe in this region such as it being polluted by "Earthshine".


How clearly would I be able to see a galaxy with the naked eye if viewed from a "close" distance? In images of other galaxies they look very bright and vivid against the blackness of space (example). If I was located a few tens of thousands of lightyears away from a galaxy, would I see something similar or would it be more like a slightly brighter smudge in space?

I'm wondering since it seems we can barely see the features of our own galaxy with the naked eye from earth.

Edit: I see that someone has downvoted the question, it would have been good to have some feedback! I realise the question might be a bit vague, but I thought it was interesting to know if galaxies would look as bright as we see them in the pictures.

The surface brightness of an extended source in independent of distance. So for the surface brightness of a large spiral go look at M31 (The Great Galaxy in Andromeda) . What will change with distance is the apparent size, but the surface brightness will remain the same.

The surface brightness of a near edge-on galaxy will be greater than one face-on due to the greater number of stars close to any sight-line, and the integrated brightness will of course vary with distance.


What is the oldest thing? I'm well aware of the celestial bodies considered as the oldest visible in our galaxy.

My question might feel a bit odd: What is the oldest thing in the universe? I am puzzled with this question as the only answer that I have found by myself sounds weird.

As the notion of "old" only makes sense in an observable and ellapsing timeline, the most obvious answer to me would be "Time is the oldest thing". There is no such thing as a measurement of "old" until there is actually mass to allow time, so which predates which? I'm pretty sure the answer will tightly be linked to Planck's constant, but how?

What am I doing wrong? How could I find relief in an answer that wouldn't be just philosophical, but also logical? :)

Pretty much every hydrogen atom that's in a glass of water has a proton that dates from 1 / 1000000 seconds after the big bang.  That's older than the cosmic microwave background, which dates from almost 400,000 years later.


Forms of stellar orbits around the galactic center Do all stars in orbit around Sagittarius A have approximately circular orbits or are there stars that orbit the Milky Way center very elliptically, like being in the Milky Way's outer rim at apoapsis and in the core worlds close to Sagittarius A at periapsis? Do we know what orbital eccentricity the Sun (respectively the Sun-Jupiter barycenter) has?

I'm not sure what the focus is on Sgr A*? Only the stars that are very close to the Galactic center can be said to be "orbiting Sgr A*", the rest of the stars in the Galaxy orbit in the the entire Galactic potential, to which Sgr A* is a minor contributor.
The stars near Sgr A* have a wide variety of orbital eccentricities. This is shown clearly in the picture below (credit: Cmglee, Wikimedia commons).

Stars in the Galaxy at large also show a variety of orbital eccentricities in their non-Keplerian orbits. The orbits are non-Keplerian because they are not orbiting a point-like source of gravity and the orbits do not form closed shapes.
Many, in fact most, of the stars near the Sun are in the Galactic disc population and have roughly circular orbits, confined close to the plane of the Galaxy. About 1% of nearby stars are part of the old halo population which have a wide dispersion of eccentricities and orbital inclinations.
The Sun is currently in a mildly "elliptical orbit" (it is not a closed ellipse as discussed above and it wobbles above and below the Galactic plane), since its motion is not entirely tangential to the Galactic centre - it has components of 5-10 km/s towards the Galactic centre and upwards out of the Galactic plane, compared with a tangential orbital velocity of around 240 km/s. See https://astronomy.stackexchange.com/a/12311/2531 and How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?


Determining line ratios in planetary nebula I know that the line strength ratio tells us how hot the electron plasma in a nebula is, and also give information about the electron densities in the nebula. But how do you compute the line ratios? Say we consider two emission lines (both forbidden) decaying to the ground state. If I have the Einstein A coefficients and the statistical weights, how do I compute the line ratios? Any explanation (preferably with examples) will be highly appreciated.

You need a model for the excitation of the line.  But since you say you have two forbidden lines, you can try assuming the density is well below the "critical density" (which is the density at which point the upper levels of the transitions get quenched by collisions instead of waiting as long as it takes for the "forbidden" radiative transition to occur).  Then if you are talking about two transitions from the same upper level, the line ratio is simply the ratio of the Einstein A values for the two transitions (note that the Einstein A value normally already includes the statistical weights of the lower levels).  If you are talking about two different upper levels, then you need a model for how those upper levels get populated, and the line ratios will be the ratios of the upper level populations times the Einstein A values.  It's not always easy to know how the upper levels get populated, so that's why it's hard to give you an answer to your question, it is very context dependent.  But the main simplification of using forbidden lines is that you don't have to worry about the photons getting reabsorbed before they get out.  (A detail is that line ratios are often given in terms of energy flux ratios, rather than photon ratios, so you'd need to multiply the Einstein A values by the photon energies if the transitions are at very different energies.)


Can matter fly directly into a black hole and avoid the accretion disc? From what my non-scientifically-trained mind understands, matter typically enters a black hole via getting caught up in the accretion disc going faster and closer to the black hole until it passes the event horizon, the point of no return. Using Earth as an analogy for a black hole, I assume the accretion disc would orbit around the equator, similar to Saturn's rings, and the matter might eventually crash into Brazil or somewhere along the equator?

My main question is, can matter simply do a straight shot into the black hole at a higher latitude, say straight into New York City, avoiding the disc altogether?

As you can probably tell by my simplistic analogy, I am a complete layperson. While fancy jargon and artistic looking equations might be expected (which is fine), please also dumb it down to an average intelligent business administration major's level of scientific understanding.

In your analogy: A satellite stays in orbit because it has angular momentum. This always balances the force of gravity, thus when being in the satellite (or in the ISS) it seems as if you're weightless, because force balance. This is very important to understand, because it is the reason that in space all orbital motions can -in principle- go on forever. Because angular momentum is conserved.

Now in order to crash into the Earth/BH you need to have very low angular momentum (which is equivalent to having only linear momentum straight down into the BH) or you have to loose the angular momentum somehow.
Near Earth, this happens to the ISS, because it has a low-earth-orbit between 350-450 km and at this height there are still enough atmospheric particles around to produce drag.

Around black holes it is a very similar thing.
I will define the Schwarzschild-radius $R_s = \frac{GM}{c^2}$ as a relevant length-scale. Then, as long as we are outside $3 R_s$ distance from the BH, we need, as on Earth, friction to dissipate our angular momentum. In the case of accretion discs that form around BH's this can be accomplished through turbulent dissipation as laid out in the famous paper by Shakura and Sunyaev.

Then, only very close to the BH, namely at a distance $< 3R_s$, space-time distortions help us to fall into the BH without the need for any additional source of friction. As you see, this is a region very close to the Black Hole and it is misrepresented in popular accounts of BHs that all matter orbiting it will magically fall into it.


Seeing units, arcsec VS cm I am very used to working with seeing in arcseconds, but sometimes I come across the seeing in cm, e.g. here.

I have looked online for a way to convert from one to the other, but I can't find any. How would this be done?

Seeing is typically measured as the FWHM of the seeing disk, but can also be expressed though the Fried parameter $r_0$, which measures the size, or strength, of the parcels of gas that cause the turbulence in the atmosphere. The relation between the two is (e.g. Vernin & Munoz-Tunon 1995)
$$
\text{seeing} = 0.98\frac{\lambda}{r_0},
$$
where $\lambda$ is the wavelength of the light.

$r_0$ can by calculated by integrating the "turbulence strength" along the line of sight$^\dagger\!\!$, and  scales as $\lambda^{6/5}\!$. Because of this dependence, the "usual" seeing has only a weak ($\lambda^{-1/5}$) dependence on wavelength.

From the Wikipedia article on seeing:

  At visible wavelengths, $r_0$ varies from 20 cm at the best locations to 5 cm at typical sea-level sites

$\dagger$Actually, $r_0$ refers to a distance corresponding to the line of sight towars zenith; observing at an angle $\zeta$ from zenith introduces a factor $\cos^{3/5}\!\zeta$.


Is it safer to be near a star or a black hole? If a star or black hole neared the solar system, which would do the least damage?  For the sake of a good comparison, lets assume the candidate star is an "average" 5 solar mass star, and the black hole is a 5 solar mass black hole.  My thoughts at first were that the star would be the least devastating, but now that I think about it more, since not even radiation, etc... is being emitted by the black hole, (except possibly from hawking radiation and from any in falling matter), the black hole might actually be safer!

the candidate star is an "average" 5 solar mass star, and the black
  hole is a 5 solar mass black hole

Then their gravity is identical. Black holes don't have magic powers. A 5 M☉ star and a 5 M☉ black hole exert the same attraction from the same distance. The only difference is that the black hole would be much, MUCH smaller (about 30 km diameter in this case), so you could get a lot closer to the center, which is where the extreme gravity happens. But at cosmic distances they are the same.

Hawking radiation from a 5 M☉ black hole is negligible, likely too small to measure - its temperature is about 10^-8 Kelvin.

If it has an accretion disk, that might be a problem, but usually it only generates two relativistic jets at the poles - if you're not hit by the jets, you're fine.

The overall thermal glow from the accretion disk of a 5 M☉ black hole might be intense, but it can't last very long. It's not a continuous burn like a star, unless it's a much bigger black hole with a huge accretion disk and plenty of source material nearby to feed it - which is not the case here.

Your conclusion is technically correct. Gravitationally they are the same, but the star has the added problem of generating extra heat. Not that it would matter anyway, because both would completely disrupt orbits in the solar system, and the Earth is going to be flung out into space and would freeze anyway. It's just that the star would bake us first, and THEN we would freeze solid.

Or we would settle into a highly elliptical orbit around either the Sun or the invader body, which will have us alternately baking to death and freezing to death.

There is also the very tiny chance that, due to orbital disruptions, we would collide with something else, either with the invader, or with the Sun, or with another planet. This would mean a quicker death and might be overall "preferable".

Regardless, having a 5 M☉ body entering the system is not a good scenario for life on Earth.

http://xaonon.dyndns.org/hawking/


What would the night sky look like if the Milky Way were the only galaxy in the universe? I'm curious to know what the sky would look like without any other galaxies out there. How much do other galaxies factor into the stars we see? Does the Milky Way account for most of them? Would the night sky look normal? Or would it be very empty?

I feel sure this is a repeat, but couldn't immediately find it. The only things in the night sky we can see (with the naked eye) that are not part of our own Galaxy are (on a good night) the Andromeda galaxy and the Large and small Magellanic clouds. Every individual star brighter than $V=6$ and visible to the naked eye is in the Milky Way.

So it would hardly look any different.


What constellation does sun occupy by location and date? How can i find the location of the sun and which constellation it would occupy based on the location and date of year.

for example: an observer is in Des Moines, Iowa on April 10th, which constellation does the sun appear to occupy?

You can see for yourself for any date and location using the free Stellarium program (http://stellarium.org/).

The constellation that the Sun is in as seen from Earth does not depend noticeably on the precise location of the observer on Earth.  It depends mostly on the time of year.  Here are the approximate days of the year at which the Sun enters a new constellation:

Aries: April 18
Taurus: May 14
Gemini: June 21
Cancer: July 20
Leo: August 10
Virgo: September 16
Libra: October 31
Scorpius: November 23
Ophiuchus: November 29
Sagittarius: December 17
Capricornus: January 19
Aquarius: February 16
Pisces: March 12

The "approximate" part is that the Sun may be up to one day earlier or later than the quoted dates, just like the start of the seasons may be one day earlier or later than the average.  (Actually, it is the calendar that is early or late with respect to the phenomenon.)  Also, because of the precession of the equinoxes, these dates shift by about 1 day every 70 years.

See http://aa.quae.nl/en/antwoorden/sterrenbeelden.html#6.


Are there any to-scale diagrams of the TRAPPIST-1 system? Recommended to cross-post from SpaceEx: https://space.stackexchange.com/questions/38397/are-there-any-to-scale-diagrams-of-the-trappist-1-system

I was recently told that the planets in TRAPPIST-1 are close enough together that you can see them from each other, as visible, discernible planets "hanging in the sky". i.e. much like the moon.

This assertion is compatible with this question: How many things are wrong in this "artist view" of the TRAPPIST-1 system?

I'm trying to understand the implications of the claim, and started by trying to find a to-scale astronomic diagram of some, or all, of the TRAPPIST-1 system, showing the star, the planets, their sizes and orbits, to scale.

Obviously such a diagram of the (Sol-Terra) Solar System would be completely useless - a few pixels in the centre, surrounded by a 8 other single-pixel-dots around the diagram.

But if the claim is true for TRAPPIST-1, then I would expect such a diagram to be actually "usable".

But I couldn't find such a diagram. Does one exist?

As BMF already pointed out, you won't really see any planets in such overview diagrams for TRAPPIST-1, either. At least not with more than a few pixels per planet.

As I did not find a good model online, I would recommend looking at "Games" like Universe Sandbox or SpaceEngine. For the latter, older versions are even free.

Then you can generate screenshots like these:

Inner 3 planets to scale in front of a white background:

The whole system with the habitable zone:

And a double eclipse from TRAPPIST-1d:


What happens if I were to add angular momentum to a maximally spinning black hole? I am aware that a "naked singularity" is not possible. However, what would happen if I added angular momentum to a black hole near the maximal angular momentum limit? Does it asymptote?

How would you add angular momentum?

The only way to do that is to drop something with angular momentum (relative to the centre of mass of the BH and the thing, if you want to get picky) into the BH. Could be a spinning object, or an object fired in at a oblique angle, or circularly polarised light or even suitably polarised gravity waves.

Whatever you do, you increase both the mass and the angular momentum of the BH. Now the limit on the spin of a black hole is expressed in terms of a relation between these two quantities and the Schwarzschild radius of the black hole (which is determined by the mass) and it turns out that whatever you add, and however you add it (if you're firing something in obliquely, it has to actually hit the BH so it can't be too ar from the centre of mass) you cannot increase the angular momentum enough relative to the increase in mass to break that limit.

The equations you have to look at are on wikipedia see especially this section. One way of expressing the limit is given there as
$GM^2 < Jc$  where $J$ is the angular momentum. Suppose you have a black hole just under this limit and add an object of mass $m$ angular momentum just under $Gm^2/c$ velocoty just under $c$ and on a trajectory that just grazes the Schwarzchild radius of the hole at $\frac{2GM}{c^2}$. The angular moment of the result is just under $$GM^2/c + 2GMmc/c^2 + Gm^2/c$$ where the terms are respectively the original angular moment of the hole, the angular moment from the grazing impact and the angular momentum of the added body. This works out to be (just under) $$G(M+m)^2/c$$, the limit for the (now-enlarged) black hole.


How far away is this galaxy? Arp 299 is in the news, with most sources reporting it as 140 or 150 million light-years away. But what kind of distance is this?

The paper which is trhe source for the news item has Supplementary Materials which mention the distance but don't appear to say what kind or give details:

  The merging pair of galaxies Arp 299 (Fig. 1) at a 44.8 Mpc distance (33; corrected to the cosmic microwave background reference frame; H0 = 73 km s^-1 Mpc^-1) is one of the brightest nearby LIRGs with an IR luminosity LIR ~ 7x10^11 L_⊙ (34; L_⊙ = 3.833×10^26 W).

Is it obvious what method is used (and thus what type of distance this is) to measure the distance to this galaxy? Is there a standard reference/database/etc. giving basic information like distances of astronomical objects like galaxies? I'd like to know the answer for this galaxy, but answers that generalize will naturally be more interesting.

Edward L. Wright has a cosmology calculator that is useful in converting between different types of distances. His page on distances gives many methods for measuring distances; I feel that if I knew which type (if any) of these methods was used to find the distance I'd have a decent chance of figuring out what kind of distance it was (and could find the other distances by bisection).

The article references a paper by Huo et al. (2004; that's the reference number "33" in your quoted text). Huo et al. merely list a redshift of "0.0103" for Arp 299 in their Table 1 (without any apparent sources); combined with the fact that the authors of the Science paper say what Hubble constant they assumed (73 km/s/Mpc), this tells us that the distance is redshift-based. That is, they used the redshift plus the Hubble constant to derive the (comoving) distance.

If you look up Arp 299 in the NASA Extragalactic Database and scroll down the resulting page, you can see a list of possible redshifts ("Calculated and Corrected Velocities") for this galaxy, ranging from 3088 km/s (which is in fact $z = 0.0103$) to 3511 km/s. Immediately below that, you can see the derived distances ("Hubble Flow Distance"), assuming $H_0 = 73$, the same value the authors say they used. The closest number to their value is the 44.4 Mpc, assuming "3K CMB" correction (which seems to match the authors' description). Why this isn't exactly the same as the authors' value of 44.8 Mpc I don't know; possible a slightly different value of the redshift was used.

At this relatively small redshift, the differences between the different distances that pela mentions are pretty minor. E.g., plugging $z = 0.0108$ (NED's "3K CMB" corrected redshift) and $H_0 = 73$ into Ned Wright's cosmology calculator gives you a comoving distance of 44.2 Mpc, a luminosity distance of 44.7 Mpc, and an angular-diamter distance of 43.8 Mpc. The differences between these are all smaller than any plausible uncertainty based on using the redshift.[*]

So, to summarize, the distance in the paper is (as pela says) the comoving distance; the method used to derive it was the redshift of the galaxy (corrected to a CMB-relative value) plus a Hubble constant of 73 km/s/kpc.

[*] E.g., if we assume a redshift uncertainty of 300 km/s, based on possible peculiar motion of Arp 299, then the distances are uncertain by +/- 4 Mpc.


Where is the Perseids meteor shower coming from? Swift–Tuttle Comet has orbital period of 133.28 years. Yet, people say that debris and dust from this comet causes this meteor shower. This does not make any sense. Or do they want to say that from 1992, every year, for 28 years Earth picked up debris left behind so many years ago while passing the August part of orbit? It's logical to assume that all this debris would be exhausted by now...

The comet sheds material each time it get close to the sun. The sun heats this dirty snow ball, the ice evaporates and tears dust and smaller rocks with it.
These dust particles then follow a similar orbit around the sun as the comet. But as they are ejected with some velocity and react differently to the solar wind & radiation, their orbit is slightly different from their parent comet.
Different orbit means generally different orbital velocity and, over time, the particles spread out along the comets orbital path. They form a steam of material that constantly dissipates and is replenished each time the comets approaches perihelion.
And each time the earth passes through the part of the solar system this stream occupies, we see meteor.
Here is a nice interactive animation:
https://www.meteorshowers.org/view/Perseids


Axial wobble or binary star system Is our solar system affected by a body unobserved as of yet that causes our solar system to curve through space in an orbit? Is this idea outlandish or us there any evidence to support it? Consider this information:

  "In 2007 Voyager 2 unexpectedly reached the edge of the heliosphere at a distance approximately 1.5 billion km less than the distance traveled by Voyager 1, indicating the SS is bullet shaped in the direction of the interstellar magnetic field."

Another bit of information is that a binary system with a moving SS was suggested as far back in modern times as 1894 -- with predictions 40x more accurate than an earth wobble to calculate precession of equinoxes (over 100 years).

Also unknown to me was precession was observed millennia ago, for which we have some records, but today's accepted notion was formed in the 1850's -- before it was accepted that the solar system even moved. Our current theory of precession takes no account for SS movement. Zero.

Lasly local bodies do not precess at all, only distant stars and -- from what I understand -- all of them.

Is a binary star system just bunk? Can it be tested for?
Is the axial wobble theory tried tested and come up true every time or Does this theory have predictability issues that have to be tweeked constantly?
Why is it that the moons orbit around the earth, as well as the earths orbit around the sun, are increasing and the precessional cycle is speeding up? Shouldn't that have the opposite effect?

There is absolutely no possibility that our Sun is part of a "binary star system". Even the least luminous star (with a mass of about 8% that of the Sun) would have an apparent brightness and proper motion that would easily have led to its discovery. See for example Is Sun a part of a binary system? and  Is it possible that the Sun has a binary partner (the Nemesis Theory) that has eluded detection?


why hasn't Nasa gone back to our Moon? why has not Nasa gone back to our Moon since the first landing on the moon

We didn't only visit the Moon once! Neil Armstrong, Buzz Aldrin and Michael Collins were the first manned mission to land on the Moon with the famous Apollo 11 landing, but 5 later Apollo missions also landed safely on the Moon and did various exciting things.
This Wikipedia article gives a brief list of all the manned and unmanned missions that have investigated the Moon, and this NASA article gives a nice history of the Apollo program with all the manned missions. (So far the US are the only country to land astronauts on the Moon.)

As for why there haven't been more visits since Apollo, there are a couple of points:

1) We have continued investigating the Moon, through remote observation, close encounters and landings, just not with humans on board. The future of human space flight is quite a debated question, since humans are very fragile, have lots of requirements in terms of food, space, health etc. and are not really very efficient at many tasks. Automated investigation of space is by far cheaper and easier. The main advantages humans have are intuition and adaptability. We can change plans, make snap decisions and respond much quicker than a robot controlled by people far away on Earth.

2) Turn the question around: why would we go back to the Moon? Why spend money and time investigating this rather boring lump of cold rock when we could go to places we have absolutely no data on, or very shaky theories, or places that have environments hugely different to our own? If we want to test scientific theories, we often want extreme environments, not just something  a little bit different to home. And if, as your comment says, we are looking for other life or future homes, these things are very rare so we will have to look a lot further away than the Moon to find something that interesting!


What would change if Venus rotated at a speed and direction similar to Earth? I am interested in two aspects of this question. Imagine that Venus had a similar rotation speed and the same direction of rotation that we have here on Earth.

What would change for Venus itself? I'm interested about changes in the core and mantle behaviour, a potential magnetosphere, changes in the atmosphere and tectonic plates.

What would be different for the other planets near the Sun? I'm interested about changes in gravitational resonance and potential orbit changes.

I'm interested about changes in the core and mantle behavior, a
  potential magnetosphere, changes in the atmosphere and tectonic
  plates.

A lot of this is very theoretical.   Venus' atmosphere is nearly 100 times as massive Earth's atmosphere.  It's also nearly all CO2, a greenhouse gas, which means it's good at trapping heat.   That wouldn't change if Venus' rotation rate was faster.

"Core and mantle behavior" is a bit of an unusual statement, cores rotate with the planet and they slowly cool, there's some convection vs conduction going on, but it's difficult to study the cores of other planets.  It's hard enough studying the core of Earth.   But what we know of the core of Earth, or think we know is that Earth's core and Earth's crust rotate at slightly different rates.  This may be due to Earth's crust slowing down over time and the core, not slowing down as fast.   There's a slight variation in rotation rate.   Also, Earth's inner core is solid, Earth's outer core is somewhat liquid (though extremely viscus), and as it cools, the inner core slowly grows, the outer core shrinks and in the process heat is released.   This process creates a flow, which may be partially influenced by the Moons orbit, and together the circulations are thought to create Earth's magnetic field.   This isn't certain, but the flow of the outer core is thought to be the cause and Earth's rotation may play a role in that.

So if Venus rotated much faster, in combination with it's proximity to the Sun and the solar tidal forces, it's possible a faster rotating Venus would have a permanent magnetic field.   It's far from certain, but that's a possible outcome.

Plate tectonics is more tricky.  Various studies suggest that planetary size has a lot to do with plate tectonics, and abundant water plays a role too, oceans making plate tectonics easier.   Venus is on the small size and it has almost no water, so it's not a good candidate for plate tectonics.   Venus also has very little granite.    This is geology more than astronomy, but abundant granite (IMHO) plays a role in plate tectonics.   The rocky bodies in our solar-system.   Mercury, Venus, Mars, the Moon, are all, basically basalt like on their surfaces.   Earth formed large granite land masses that float above the more dense basalt like layers under the oceans.   Without granite land masses and without oceans, I don't think Venus is a good candidate for plate tectonics.

Did Venus ever have plate tectonics, like billions of years ago?   Maybe.   I have no idea, but maybe.   Would it have plate tectonics if it rotated faster?   I doubt it.  A magnetic field - maybe.  Plate tectonics, I don't think so.

This question is probably too general for stack exchange and it may be closed, but I don't believe that just spinning Venus faster would change all that much apart from maybe giving it a magnetic field, and that wouldn't change all that much on Venus. It has an induced magnetic field anyway.

If you could give Venus an ocean, maybe make it more massive and move it further away from the sun, then some interesting things might happen like plate tectonics, which could have lead the planet down a different path, perhaps more similar to Earth.

Earth was able to absorb CO2 into it's vast oceans and take in CO2 by photosynthesis and rocks were able to absorb CO2 by plate tectonics, and Earth was far enough from the sun to not have it's surface water evaporated and lost like what probably happened to Venus.

I think planetary formation, based on size, type of star, temperature, initial materials, etc, will be a very interesting field of study when/(if) big enough telescopes are set up to actually get a look at distant planets.   It's very difficult to even observe, much less, study a planet that's light-years away.   The so far twice or three ties delayed James Webb Space Telescope might get the ball rolling on this subject with better observations of exo-planet atmospheres, but it's a long ways away from being well understood.  The James Webb ST is an enormously difficult undertaking, so we should be patient.  With luck, it'll be up and taking in images within a few years.

Finally, would Venus' rotation, if it was faster, affect other planets?    Basically not at all.   Jupiter's rotation doesn't affect Earth.   Rotation rate doesn't affect other objects at great distances.


Could dark matter be considered a medium? Long ago, it was theorized that light had to travel through a medium, until the particle/wave duality of light was discovered.  Then it changed to the idea that light can travel through the vacuum of space as a particle.

Light travels at different speeds depending on the medium that the light is traveling through.  It travels slower in water than it does through air.

However, is there any possibility that dark matter is a medium that light travels through?  Could it be conceivable that what we recognize as light speed is really just the speed at which light travels through this medium?  If we were to remove dark matter, would light go faster?

I don't see any reason why dark matter couldn't be considered a medium just as air or water are, but this does not mean that your conclusion is correct.

Dark matter does not interact with photons, however the reason we know dark matter exists is because of the measurable effect of its gravity.  Galaxies have a tendency to clump around it.

Light travels unimpeded at its standard rate between galaxies where we don't detect any dark matter, so interesting question though it is, I'm afraid the answer is no.


How to choose initial state vectors for a resonant orbit numerical simulation? How do higher numbers affect orbital resonance? is a really interesting question, and I thought I would try to do a simple numerical simulation of various resonance ratios to see if some are inherently stable or inherently unstable.
I understand that numerical simulation is inexact and so it can introduce its own instabilities so attention will have to be paid to the choice of integrators and results taken with a large grain of salt.
In order to get something to happen within say a hundred or a thousand orbits the masses of the orbiting bodies have to be large enough to have a significant impact on each other1, and so the state vectors I choose to initialize the calculation will not necessarily be those of circular orbits each body neglecting the other.
Are there any guidelines or rules of thumbs for initializing resonant orbits?

1I'll assume a central force, i.e. the orbiting bodies will not affect the position of the body they orbit, and so there won't be any indirect coupling through it.
Related:

Just how "locked" are resonant-chains of exoplanets thought to be? (e.g. K2-138 and TOI-178) (currently unanswered)
Why does the exoplanetary system TOI-178 challenges current theories of planet formation?

I don't know if there are rules of thumb for initializing resonant orbital simulations.  I might suggest:

Setting initial state vectors' Z-Components to zero.  Keeping the whole simulation in the X-Y plane can help with debugging and visualization.

Choosing initial state vectors that are equivalent to orbits with non-zero eccentricity. Orbital resonances tend to amplify or dampen eccentricities, but the effects can be multiplicative. So if you choose circular orbits, you may see less of an effect.

Choosing initial state vectors that line up apogees and perigees with closest contacts.

Running multiple simulations with slight perturbations to initial state vectors to test stability.

Some interesting initial experiments might be to see if you could replicate the instability of the L1 and L2 points of an orbit.  Or perhaps put some little masses in the Kirkwood Gaps and see if they get expelled.

Notes:

I tried writing some of my own numerical simulations to simulate natural planetary flybys. I used an algorithm outlined by Voesenek, but I was getting round-off errors that shot my simulation. Instead of investing more time to fix my simulation, I ended up using Universe Sandbox 2, that I paid for and downloaded from Steam.

Universe Sandbox probably doesn't have the fidelity you need for resonance stability simulations.  Gallardo wrote a pretty slick long term orbital integration software package called ORBE, that is free to the public for educational purposes.  I wish I had read Tabaré Gallardo's 2017 Exploring the orbital evolution of planetary systems before writing my own simulator, since he uses a neat trick by Encke to account for the Sun's gravity analytically via 2-body problem equations while separately treating the inter-planetary gravitational accelerations numerically. This allows larger time steps and prevents some kinds of round-off error.  Gallardo's software is designed to test long term stellar system stability.  Also, it uses orbital elements rather than state vectors, which are a bit less intuitive to me. Gallardo's used of ORBE to run Kirkwood Gap stability simulations is totally AWESOME!!  Here is a capture of one of the figures from Gallardo's 2017:


I saw 3 consecutive object disappearing in the sky at same nearby location I was roaming on my terrace and I saw star-like moving object that gradually became small, red colour and disappear. Within a few minutes I saw another similar-looking  object a little distance away coming from opposite of first one and disappearing with red colour at nearly the same location. Within 5-6 minutes, I saw another similar object that soon disappeared. The object was moving from a lower altitude to a higher one. What was it?

You may have seen some artificial satellites passing into the Earth's shadow.
If so, Heavens-Above could help you identify them.
The reddish color just before disappearance would be due to the atmosphere scattering shorter wavelengths of sunlight.


How long New Horizons will remain operational? As of now the New Horizons probe is well beyond the orbit of Pluto, moving towards the Kuiper belt. In an article I read that New Horizons will eventually follow the path of Voyager probes to the interstellar space. But how long it will remain functional and as further it will travel away from us it will also required very long time to receive data from the probe. Is there any pre-planned fate of the probe ?

The answer is in the other section:

New Horizons Mission Info

What I can say off the top of my head is that the mission has been extended and they have identified another object that the spacecraft can reach. However, after the next rendezvous it will be out of fuel and will not be able to maneuver.
BUT -- the probe is powered by a Plutonium-238 radioisotope thermal generator (RTG), and will likely to continue functioning for the foreseeable future. The relevant question is: how long will the government fund the NASA operations team that is 'talking' to the probe. That is anyone's guess.

New Horizons Spacecraft Components

PS - I'll update this answer later once I talk with some colleagues. I started, and worked for the company (KinetX, Inc) that did the navigation for New Horizons. I no nothing of the science, but my friends did the navigation and know the near-term details for the mission.


Have we Observed Continuum Emission from Neutron Stars? Have we detected continuum optical emission from any rotating neutron stars that do not have an accretion disk dominating the light? I ask because I know we have observed Doppler broadening of spectral lines from the rotation of even ordinary stars rotating at ordinary rates, but it seems to me that only the rapid rotation of a neutron star would be fast enough to make the Doppler broadening of the thermal continuum peak to be observable.

Yes - quite a few isolated neutron stars have been observed, where any magnetospheric emission or accretion-related emission is either negligible or has been otherwise separated.

As you suspect, this emission is thermal in nature. Neutron stars are roughly approximated by black bodies but, like "normal" stars, they do have atmospheres and strong magnetic fields that modify the black body spectrum considerably. If the neutron star is isolated then the only reason that you might be able to detect thermal emission from the surface is that it is very young (less than about a million years after formation) and still hot. The difficulty is in ruling out any residual, pulsed magnetospheric component or a contribution from accretion of the interstellar medium itself.

A couple of examples to look at in the optical part of the spectrum would be Kulkarni & van Kerwijk (1998) who detected optical radiation from the counterpart to RX0720.4-3125; and Walter & Matthews (1997) who claimed a detection of thermal optical emission detection from the isolated neutron star RXJ185635-3754 (and see picture below).

There are several more examples of thermal emission detected at X-ray wavelengths from young, isolated neutron stars - the so-called X-ray dim neutron stars (see review by Mereghetti 2010).

Of course if you are extending your definition of continuum emission to include non-thermal (but not accretion-related) processes, then continuum emission is detected from all isolated pulsars. Synchrotron radiation is a continuum process and has been detected right across the electromagnetic spectrum in many pulsars.

A visible band HST image of the isolated neutron star RXJ185635-3754, attributable to Walter & Matthews 1997.


How to calculate moon overhead and underfoot time? For hunting purposes, some hunters follow the moon phases to hunt accordingly. To give some context, here's some links:

Lunar cycle's effects on hunting: https://www.acurite.com/blog/weather-stations-with-moon-phase-for-hunting-fishing
Solunar theory: https://en.wikipedia.org/wiki/Solunar_theory

I would like to know how to calculate the time when the moon is directly overhead in the sky (moon overhead) and when the moon is directly underfoot in the opposite side of the earth's sky (moon underfoot). Those are the information that I have:

Moon rise time
Moon set time
Moon phase (first, full, last, new)

Those information come from AccuWeather API: https://apidev.accuweather.com/developers/astronomyAPIGuide#MoonPhasesRange. There are some other APIs that give the moon phase in degree format or between 0 and 1 but we chose AccuWeather.

I tried to calculate this with only the moon rise and set time. I thought that simply calculating halfway between rise and set time will suffice. However, I didn't always get the same results as the moon overhead and underfoot in this website: http://www.solunarforecast.com/solunarcalendar.aspx

Is there a formula to get the time of moon overhead and underfoot with the information that I have?

UPDATE

I used the definition

  when the moon is directly overhead in the sky

because that is what I found from my research. However, it's seems the correct definition is when the moon is at highest point in the sky.

Also, since it's for hunting purposes, I don't need the time to be accurate to the minute (but that could be nice).

At the accuracy you need, the calculation can be quite simple:  It takes 12 hours and 26 minutes for the moon to go from "overhead" to "underfoot", and 12 hours 26 minutes to go from "underfoot to overhead".

So if you notice that the moon is overhead at 20:00 one evening, it will be overhead at 20:52 the next evening,  then 21:44, 22:36 and so on.  (Or, even easier,  approximate to "50 min" and use "add an hour, minus 10 min"

The actual motion of the moon is rather complicated, as the moon's orbit is elliptical and perturbed by the sun. If you need more accurate values you need to use a complex method of calculation, which is what the websites are using.

If you only have rise and setting time, the culmination will be roughly in the middle

The Solunar calculator page contains some obvious errors (such as on 5th May)


What was this red deep sky object? I was looking through sky yesterday and saw something red in sky (naked eye ) like nebula or something then tried using stellarium and mobile observatory apps to find out what it was, But couldnt find anything like this in them. I also tried to capture it on my phone but nothing visible except brightest stars.

I just know it was in cygnus - vulpecula - sagitta or somewhere near it.

As i wear glasses and dont have access to dark sky, it was my first naked eye view of some sort of deep sky object. Ill become so happy if i find what that red object was.

I think there are no deep sky objects (or extended objects) that show any color to the naked eye. They are too faint to activate the color sensors, so they appear white. I think you saw something nonastronomical.

You will have a better idea the next time you see it; get a better estimate of its location so that you can find it on a star chart. Use binoculars if you have them.