Expression for length of sunrise/sunset as function of latitude and day-of-year I would like to know how the length of sunrise/sunset (and also time between civil/nautical/astronomical twilight) varies with latitude and with day of year. Ideally what I am looking for is an expression that takes latitude and day-of-year as arguments and outputs the time taken for the sun to rise or set at that latitude on that day.

The equations necessary for a correct answer are somewhat lengthy, for a web interface that accepts the location and prints a calendar try: https://sunrise-sunset.org/.

For a lengthy explanation of the mathematics see Wikipedia's Calculating the equation of time and Daytime webpages for information about the duration of rising and setting, along with information about the differences in solar noon and the three types of twilight.

This is a QBasic implementation by Keith Burnett for twilight times:

'********************************************************************
'   Sun rise/set /twilight
'   From Explanatory Supplement
'   definitions and functions
DEFDBL A-Z
pi = 4 * ATN(1)
tpi = 2 * pi
twopi = tpi
degs = 180 / pi
rads = pi / 180
'
'   the function below returns the true integer part,
'   even for negative numbers
'
DEF FNipart (x) = SGN(x) * INT(ABS(x))
'
'   FNday only works between 1901 to 2099 - see Meeus chapter 7
'
DEF FNday (y, m, d, h) = 367 * y - 7 * (y + (m + 9) \ 12) \ 4 + 275 * m \ 9 + d - 730531.5 + h / 24
'
'   define some arc cos and arc sin functions
'
DEF FNacos (x)
    s = SQR(1 - x * x)
    FNacos = ATN(s / x)
END DEF
DEF FNasin (x)
    c = SQR(1 - x * x)
    FNasin = ATN(x / c)
END DEF
'
'   the function below returns an angle in the range
'   0 to two pi
'
DEF FNrange (x)
    b = x / tpi
    a = tpi * (b - FNipart(b))
    IF a < 0 THEN a = tpi + a
    FNrange = a
END DEF
'
'       the function below returns the time in 24 hour
'   notation - hhmm - given a decimal hours figure.
DEF FNdegmin$ (d)
c = ABS(d)
a = INT(c)
b = 60 * (c - a)
d$ = STR$(INT(a * 100 + b + .5))
d$ = RIGHT$(d$, LEN(d$) - 1)
WHILE LEN(d$) < 4
    d$ = "0" + d$
WEND
FNdegmin$ = d$
END DEF
CLS
'
'    get the date and geographical position from user
'
INPUT "     lat : ", glat
INPUT "    long : ", glong
INPUT "    zone : ", zone
INPUT "rise/set : ", riset
INPUT " Sun alt : ", altitude
INPUT "   year  : ", y
INPUT "   month : ", m
INPUT "   day   : ", d
day = FNday(y, m, d, 0)
PRINT "  day no : "; day
utold = pi
utnew = 0
sinalt = SIN(altitude * rads) 'go for the sunrise/sunset altitude first
sinphi = SIN(glat * rads)
cosphi = COS(glat * rads)
glong = glong * rads
DO WHILE ABS(utold - utnew) > .001
  utold = utnew
  days = day + utold / tpi
  t = days / 36525
  '     get arguments of Sun's orbit
  L = FNrange(4.8949504201433# + 628.331969753199# * t)
  G = FNrange(6.2400408# + 628.3019501# * t)
  ec = .033423 * SIN(G) + .00034907# * SIN(2 * G)
  lambda = L + ec
  E = -1 * ec + .0430398# * SIN(2 * lambda) - .00092502# * SIN(4 * lambda)
  obl = .409093 - .0002269# * t
  delta = FNasin(SIN(obl) * SIN(lambda))
  GHA = utold - pi + E
  cosc = (sinalt - sinphi * SIN(delta)) / (cosphi * COS(delta))
  SELECT CASE cosc
  CASE cosc > 1
    correction = 0
  CASE cosc < -1
    correction = pi
  CASE ELSE
    correction = FNacos(cosc)
  END SELECT
  utnew = FNrange(utold - (GHA + glong + riset * correction))
LOOP
PRINT "    UT   : "; FNdegmin$(utnew * degs / 15)
PRINT "   zone  : "; FNdegmin$(utnew * degs / 15 + zone)
END
'******************************************************************

This is an implementation by Barry Carter of rising/setting duration (see his GitHub for associated files):

// http://astronomy.stackexchange.com/questions/12824/how-long-does-a-sunrise-or-sunset-take

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "SpiceUsr.h"
#include "SpiceZfc.h"
#include "/home/barrycarter/BCGIT/ASTRO/bclib.h"

// since the sun travels no more than 362 degrees per day, and has an
// angular diameter of no less than 31 minutes, the fastest possible
// sunrise is 31 minutes/362 degrees or 2 minutes; thus, 120 below in
// bc_between below is valid

int main(int argc, char **argv) {

  SpiceDouble radii[3];
  SpiceInt n;
  char direction[10];

  furnsh_c("/home/barrycarter/BCGIT/ASTRO/standard.tm");

  // the year 2015
  double stime = 1419984000, etime = 1451692800;

  // test
  // double stime =  1388559600, etime = 1419984000;

  // lat/lon from argv
  double lat = atof(argv[1]), lon = atof(argv[2]);

  // the sun's radii
  bodvrd_c("SUN", "RADII", 3, &n, radii);

  double *results = bc_between(9, lat*rpd_c(), lon*rpd_c(), 0., stime, etime,
                   "Sun", -34/60.*rpd_c(), 120., radii[0]);

  // NOTE: can NOT ignore result 0 and 1, but must check if they are borders
  for (int i=0; i<1000; i++) {

    double rstart = results[2*i], rend = results[2*i+1];

    // if we start seeing 0s, we are out of true answers
    // TODO: this may break near 1970
    if (abs(rstart) < .001) {break;}

    // if the end result is too close to etime, result is inaccurate
    // same if start result is too close to stime
    if (abs(rend-etime)<1 || abs(rstart-stime)<1) {continue;}

    // TODO: this really inefficient (and not even always correct?)
    if (bc_sky_elev(6, lat, lon, 0., rstart, "Sun", 0.) >
    bc_sky_elev(6, lat, lon, 0., rend, "Sun", 0.)) {
      strcpy(direction,"SET");
    } else {
      strcpy(direction,"RISE");
    }

    // the "day" and length of sunrise/sunset
    printf("%f %f %s %f %f %f\n", lat, lon, direction, rstart, rend, rend-rstart);
  }
  return 0;
}

For an explanation see Wikipedia's Sunrise Equation webpage.

This is the algorithm for rising/setting times:

Sunrise/Sunset Algorithm

Source:
    Almanac for Computers, 1990
    published by Nautical Almanac Office
    United States Naval Observatory
    Washington, DC 20392

Inputs:
    day, month, year:      date of sunrise/sunset
    latitude, longitude:   location for sunrise/sunset
    zenith:                Sun's zenith for sunrise/sunset
      offical      = 90 degrees 50'
      civil        = 96 degrees
      nautical     = 102 degrees
      astronomical = 108 degrees

    NOTE: longitude is positive for East and negative for West
        NOTE: the algorithm assumes the use of a calculator with the
        trig functions in "degree" (rather than "radian") mode. Most
        programming languages assume radian arguments, requiring back
        and forth convertions. The factor is 180/pi. So, for instance,
        the equation RA = atan(0.91764 * tan(L)) would be coded as RA
        = (180/pi)*atan(0.91764 * tan((pi/180)*L)) to give a degree
        answer with a degree input for L.

1. first calculate the day of the year

    N1 = floor(275 * month / 9)
    N2 = floor((month + 9) / 12)
    N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))
    N = N1 - (N2 * N3) + day - 30

2. convert the longitude to hour value and calculate an approximate time

    lngHour = longitude / 15

    if rising time is desired:
      t = N + ((6 - lngHour) / 24)
    if setting time is desired:
      t = N + ((18 - lngHour) / 24)

3. calculate the Sun's mean anomaly

    M = (0.9856 * t) - 3.289

4. calculate the Sun's true longitude

    L = M + (1.916 * sin(M)) + (0.020 * sin(2 * M)) + 282.634
    NOTE: L potentially needs to be adjusted into the range [0,360) by adding/subtracting 360

5a. calculate the Sun's right ascension

    RA = atan(0.91764 * tan(L))
    NOTE: RA potentially needs to be adjusted into the range [0,360) by adding/subtracting 360

5b. right ascension value needs to be in the same quadrant as L

    Lquadrant  = (floor( L/90)) * 90
    RAquadrant = (floor(RA/90)) * 90
    RA = RA + (Lquadrant - RAquadrant)

5c. right ascension value needs to be converted into hours

    RA = RA / 15

6. calculate the Sun's declination

    sinDec = 0.39782 * sin(L)
    cosDec = cos(asin(sinDec))

7a. calculate the Sun's local hour angle

    cosH = (cos(zenith) - (sinDec * sin(latitude))) / (cosDec * cos(latitude))

    if (cosH >  1)
      the sun never rises on this location (on the specified date)
    if (cosH < -1)
      the sun never sets on this location (on the specified date)

7b. finish calculating H and convert into hours

    if if rising time is desired:
      H = 360 - acos(cosH)
    if setting time is desired:
      H = acos(cosH)

    H = H / 15

8. calculate local mean time of rising/setting

    T = H + RA - (0.06571 * t) - 6.622

9. adjust back to UTC

    UT = T - lngHour
    NOTE: UT potentially needs to be adjusted into the range [0,24) by adding/subtracting 24

10. convert UT value to local time zone of latitude/longitude

    localT = UT + localOffset

Here is an implementation of rising/setting times in Basic:

10 '         Sunrise-Sunset
20 GOSUB 300
30 INPUT "Lat, Long (deg)";B5,L5
40 INPUT "Time zone (hrs)";H
50 L5=L5/360: Z0=H/24
60 GOSUB 1170: T=(J-2451545)+F
70 TT=T/36525+1: ' TT = centuries
80 '               from 1900.0
90 GOSUB 410: T=T+Z0
100 '
110 '       Get Sun's Position
120 GOSUB 910: A(1)=A5: D(1)=D5
130 T=T+1
140 GOSUB 910: A(2)=A5: D(2)=D5
150 IF A(2)<A(1) THEN A(2)=A(2)+P2
160 Z1=DR*90.833: ' Zenith dist.
170 S=SIN(B5*DR): C=COS(B5*DR)
180 Z=COS(Z1): M8=0: W8=0: PRINT
190 A0=A(1): D0=D(1)
200 DA=A(2)-A(1): DD=D(2)-D(1)
210 FOR C0=0 TO 23
220 P=(C0+1)/24
230 A2=A(1)+P*DA: D2=D(1)+P*DD
240 GOSUB 490
250 A0=A2: D0=D2: V0=V2
260 NEXT
270 GOSUB 820: '  Special msg?
280 END
290 '
300 '        Constants
310 DIM A(2),D(2)
320 P1=3.14159265: P2=2*P1
330 DR=P1/180: K1=15*DR*1.0027379
340 S$="Sunset at  "
350 R$="Sunrise at "
360 M1$="No sunrise this date"
370 M2$="No sunset this date"
380 M3$="Sun down all day"
390 M4$="Sun up all day"
400 RETURN
410 '     LST at 0h zone time
420 T0=T/36525
430 S=24110.5+8640184.813*T0
440 S=S+86636.6*Z0+86400*L5
450 S=S/86400: S=S-INT(S)
460 T0=S*360*DR
470 RETURN
480 '
490 '  Test an hour for an event
500 L0=T0+C0*K1: L2=L0+K1
510 H0=L0-A0: H2=L2-A2
520 H1=(H2+H0)/2: '  Hour angle,
530 D1=(D2+D0)/2: '  declination,
540 '                at half hour
550 IF C0>0 THEN 570
560 V0=S*SIN(D0)+C*COS(D0)*COS(H0)-Z
570 V2=S*SIN(D2)+C*COS(D2)*COS(H2)-Z
580 IF SGN(V0)=SGN(V2) THEN 800
590 V1=S*SIN(D1)+C*COS(D1)*COS(H1)-Z
600 A=2*V2-4*V1+2*V0: B=4*V1-3*V0-V2
610 D=B*B-4*A*V0: IF D<0 THEN 800
620 D=SQR(D)
630 IF V0<0 AND V2>0 THEN PRINT R$;
640 IF V0<0 AND V2>0 THEN M8=1
650 IF V0>0 AND V2<0 THEN PRINT S$;
660 IF V0>0 AND V2<0 THEN W8=1
670 E=(-B+D)/(2*A)
680 IF E>1 OR E<0 THEN E=(-B-D)/(2*A)
690 T3=C0+E+1/120: ' Round off
700 H3=INT(T3): M3=INT((T3-H3)*60)
710 PRINT USING "##:##";H3;M3;
720 H7=H0+E*(H2-H0)
730 N7=-COS(D1)*SIN(H7)
740 D7=C*SIN(D1)-S*COS(D1)*COS(H7)
750 AZ=ATN(N7/D7)/DR
760 IF D7<0 THEN AZ=AZ+180
770 IF AZ<0 THEN AZ=AZ+360
780 IF AZ>360 THEN AZ=AZ-360
790 PRINT USING ",  azimuth ###.#";AZ
800 RETURN
810 '
820 '   Special-message routine
830 IF M8=0 AND W8=0 THEN 870
840 IF M8=0 THEN PRINT M1$
850 IF W8=0 THEN PRINT M2$
860 GOTO 890
870 IF V2<0 THEN PRINT M3$
880 IF V2>0 THEN PRINT M4$
890 RETURN
900 '
910 '   Fundamental arguments
920 '     (Van Flandern &
930 '     Pulkkinen, 1979)
940 L=.779072+.00273790931*T
950 G=.993126+.0027377785*T
960 L=L-INT(L): G=G-INT(G)
970 L=L*P2: G=G*P2
980 V=.39785*SIN(L)
990 V=V-.01000*SIN(L-G)
1000 V=V+.00333*SIN(L+G)
1010 V=V-.00021*TT*SIN(L)
1020 U=1-.03349*COS(G)
1030 U=U-.00014*COS(2*L)
1040 U=U+.00008*COS(L)
1050 W=-.00010-.04129*SIN(2*L)
1060 W=W+.03211*SIN(G)
1070 W=W+.00104*SIN(2*L-G)
1080 W=W-.00035*SIN(2*L+G)
1090 W=W-.00008*TT*SIN(G)
1100 '
1110 '    Compute Sun's RA and Dec
1120 S=W/SQR(U-V*V)
1130 A5=L+ATN(S/SQR(1-S*S))
1140 S=V/SQR(U):D5=ATN(S/SQR(1-S*S))
1150 R5=1.00021*SQR(U)
1160 RETURN
1165 '
1170 '     Calendar --> JD
1180 INPUT "Year, Month, Day";Y,M,D
1190 G=1: IF Y<1583 THEN G=0
1200 D1=INT(D): F=D-D1-.5
1210 J=-INT(7*(INT((M+9)/12)+Y)/4)
1220 IF G=0 THEN 1260
1230 S=SGN(M-9): A=ABS(M-9)
1240 J3=INT(Y+S*INT(A/7))
1250 J3=-INT((INT(J3/100)+1)*3/4)
1260 J=J+INT(275*M/9)+D1+G*J3
1270 J=J+1721027+2*G+367*Y
1280 IF F>=0 THEN 1300
1290 F=F+1: J=J-1
1300 RETURN
1310 '
1320 '   This program by Roger W. Sinnott calculates the times of sunrise
1330 '   and sunset on any date, accurate to the minute within several
1340 '   centuries of the present.  It correctly describes what happens in the
1350 '   arctic and antarctic regions, where the Sun may not rise or set on
1360 '   a given date.  Enter north latitudes positive, west longitudes
1370 '   negative.  For the time zone, enter the number of hours west of
1380 '   Greenwich (e.g., 5 for EST, 4 for EDT).  The calculation is
1390 '   discussed in Sky & Telescope for August 1994, page 84.


What is the origin of the OBAFGKM classification system? Once a body has been identified as a star it is classified into a subsection of either O,B,A,F,G,K or M (I think there are also 3 more letters recently added)

What is the origin of this classification system? Do each of the letters stand for anything?

The Wikipedia article on stellar classification has the full list of letters:

Class
Surface Temperature
Notes

O
> 33,000 K

B
10,000 - 33,000 K

A
7500 - 10,000 K

F
6000 - 7500 K

G
5200 - 6000 K

K
3700 - 5200 K

M
2000 - 3700 K

L
1300 - 2000 K
L T and Y are those new letters

T
700 - 1300 K

Y
< 700 K

History
This Cornell doc explains much of the history behind the system:

In the 1890s, many scientists were interested in developing a classification scheme for the stars. Edward C. Pickering at Harvard University, together with his assistant Williamina P. Fleming, assigned stars a letter according to how much Hydrogen could be observed in their spectra: stars labeled A had the most Hydrogen, B the next most, and so on through the alphabet. There were 22 types in all: this scheme was rather cumbersome, and it wasn't clear what its physical significance was.

In 1901, another of Pickering's assistants, Annie Jump Cannon, also begun to work on the classification sequence. Her meticulous observations led her to simplify the 22-type scheme into a simple sequence of temperature: OBAFGKM. Not only was the scheme much simpler than the previous, but it related the amount of hydrogen observed to a physical property of the stars for the first time.


How well can the Sun's surface be resolved by an observatory? There's no lack of photons, how well could a telescope resolve Sun spots and other surface features of the Sun? How would a high resolution Solar (space) telescope differ from for example Hubble which avoids the Sun by a 50° angle or so?

Solar telescopes typically have filters to reduce the amount of light reaching their detector. Depending on their purpose they may use, for example, neutral density filters to reduce the amount of light to a level where they can resolve details without blocking any of the visible spectrum, hydrogen alpha filters to observe the Sun at a wavelength of 656 nanometres for observing features in the Sun's atmosphere such as solar prominences and the chromosphere, or blocking the solar disc to observe the Sun's corona, flares and other ejecta.

Hubble, on the other hand, is designed with a very different purpose in mind such as taking long-exposure images to obtain the deep- and ultra-deep field full spectrum images. It avoids the Sun as even a very short exposure to the Sun would damage its instruments.


Relationship between absolute magnitude of a star and its luminosity? Why does this relationship involves the absolute magnitude of the sun and its luminosity?
How to derive an expression relating the absolute magnitude of a star to its luminosity?

So according to the magnitude equation, $$m - M = 2.5\log\left(\frac{d^2}{d_0^2}\right)$$
$$\Rightarrow M = m - 2.5\log\left(\frac{d^2}{d_0^2}\right)\ \ (eq1)$$

and Luminosity is $$L = 4\pi(d^2) \times f$$
 $$\Rightarrow d^2 = \frac{L}{4\pi \times f}\ \ (eq2)$$

Plugging eq2 to eq1 would have seem reasonable, but how is absolute magnitude of the sun and its luminosity used?

You are specifically talking about absolute visual magnitude; how bright an object would appear to our eyes if the object were 10 parsecs away from us. However, you must remember that stars emit a wide spectrum of light — much of it isn't visible to the naked eye.

Luminosity is a measure of the total amount of energy given off by a star (usually as light) in a certain amount of time. Thus, luminosity includes both visible light and invisible light emitted by a star. So there isn't a precise conversion between luminosity and absolute visual magnitude, although there is an approximation we can do.

Absolute bolometric magnitude is an analogous magnitude system considering all the energy emitted by the star. The formula relating absolute bolometric magnitude with luminosity is as follows:
$$L_\text{star} = L_0 10^{-0.4 M_\text{Bol}}$$

where $L_\text{star}$ is the star's luminosity, $M_{\text{Bol}}$ is the bolometric magnitude of the star, and $L_{0}$ is the zero-point luminosity (the luminosity of a star with $M_{\text{Bol}} = 0$) arbitrarily defined as $L_{0}=3.0128×10^{28}$ watts by the IAU. We can approximately convert between the two magnitudes via a bolometric correction:
$$M_{\text{bol}}\approx M_V + BC$$
The bolometric correction term $BC$ is empirically determined for the spectral class and evolutionary stage (although it can be modeled; see Torres (2010)). A table of values is given on the Wikipedia page.

The idea is that stars of the same spectral class and evolutionary stage have a similar distribution of wavelengths, so the discrepancy between all emitted wavelengths ($M_{\text{bol}}$) and visible wavelengths ($M_V$) should be similar.

Thus, the conversion between luminosity and absolute visual magnitude is approximately
$$L_\text{star} \approx L_0 10^{-0.4 (M_V + BC)}$$
Of course, this is only an approximation, but it can give us a pretty decent idea of what to expect. Trying it on 9 Pegasi, I predicted it would be around $1000 L_\odot$, which is still in the ballpark of the actual value $1950 L_\odot$. Mind you, the approximation would be better suited for less extreme stars than 9 Pegasi.


Why there are three equations governing blackbody emission? I learnt that the spectrum of a black body can be explained using

Planck distribution
Rayleigh -Jeans equation and
Weins displacement law

Can anyone tell me why there are three different equations that explain the same physical phenomena and when should I consider which equation?

Historically, two people (or groups of people) independently came up with different equations to model the blackbody equations in different parts of the spectrum. Rayleigh-Jeans law (classically derived) is valid for longer wavelengths and Wien's law (not Wien's displacement law) is valid for shorter wavelengths.

The Planck Distribution approaches the two laws in its limits, as in, for shorter wavelengths it is approximately equivalent to Wien's law and for longer wavelengths, it is approximately equivalent to Rayleigh-Jeans law. However, the quantum mechanically formulated Planck's law is accurate at all wavelengths, so it is the one that should always be used. One can use the other laws, for example, for brightness temperature definitions in radio astronomy, they use the RJ law for convenience, but that's because the wavelengths are long enough, and the approximation of the Planck's law gives the same result. (https://en.wikipedia.org/wiki/Planck%27s_law#Approximations)

Wien's displacement law only relates the peak wavelength to the temperature, which is a different law completely, though Planck's law (differentiating the function to find the maximum) also gives the same result. Though in this case, the displacement law is not approximate. It is accurate and can be used whenever you want.


Plot SuperWasp Lightcurves I am trying to plot superwasp lightcurves in python from the fits files, using TMID and FLUX2, getting this:

There is a way to transform the information so I can plot the lightcurves with the kepler format, a example from kepler is something like this:

Any help is useful.

Not sure what you are expecting to see from the two datasets. Both datasets are examples of light curves, flux against time with different arbitrary origins for the time axis; SuperWASP TMID is integer seconds from Julian Date 2453005.5, Kepler uses BKJD = Barycentric Kepler Julian Date, but offset by 2454833.0. i.e., BKJD = BJD - 2454833.0. The SuperWASP datasets typically span 4 years (it's a little difficult to read the plot scales of the example) and are from a ground-based set of small (11.1 cm diameter) lenses with a CCD and so will have day gaps and weather. Kepler/K2 data are from a space-based 95cm telescope which was able to stare at the same patch of sky continuously for about 90 days before a break (and of course had no weather). Consequently there will be a considerable difference in appearance and photometric precision of the lightcurves.

This page at the NASA Exoplanet Archive gives details on the SuperWASP public data and this Python Jupyter notebook has info on plotting K2 light curves.

(Disclaimer: I worked on the SuperWASP project for several years)


Would an extremely bright star produce same kind of shadow band effect as seen in solar eclipses just before totality? This answer explains that the shimmering patterns seen on surfaces at the moment before totality of a solar eclipse are called shadow bands, and that the phenomenon is closely related to the twinkling appearance of stars (but not angularly larger planets) through the Earth's atmosphere.

If for some reason there happened to be a bright star close enough to Earth that its nighttime illumination visible on surfaces, but small enough in angular width that it twinkled, would we see shadow bands at night, every night when the star was out?

As there's a good chance the answer is "yes" to the hypothetical, I'd like to ask if this situation is in fact possible, or if a star were close enough to be bright enough to visibly illuminate a white surface at night, would it then be large enough in angular extent that it wouldn't twinkle any more?

The pattern from the Eclipse can be seen in the Smarter Every Day video Space Station Transiting 2017 ECLIPSE, My Brain Stopped Working - Smarter Every Day 175 at about 04:30.

Yes, the nonuniform illumination pattern would be visible for extremely bright stars.

Venus (mag -4.6) is bright enough to cast visible shadows.
Let's move Sirius, the brightest star in the northern hemisphere closer, to match the Venus brightness:

currently: mag -1.46, 8.6 light years away, 0.006" angular diameter
relative brightness difference: $\sqrt[5]{100}^{4.6-1.46} \approx 18$
required distance factor: $\sqrt{18} \approx 4.24$
at $8.6/4.24 \approx 2$ light years away, Sirius apparent diameter would be $0.006'' \cdot 4.24 \approx 0.026''$ which is still much smaller than planets

Moreover, we can even see planets twinkling sometimes (depending on the seeing), so the "point light source" is not a strict requirement (well, strictly speaking, point light sources do not even exist ;))

Interestingly, there are no mentions of shadowband-like phenomena in the supernova observation history, even though some of them were bright enough to be visible in daylight.

I guess we have to wait for the next Milky Way supernova to be sure...


How would a celestial map for Callisto depict the paths of Jupiter, the Sun, etc.? Let's say we're mapping Callisto's northern celestial hemisphere in a way similar to this map of Earth's.

Along with the stars (which let's just ignore for this question), this map will identify the paths of four objects:

The Sun
Jupiter
Ganymede (closer to Jupiter)
Himalia (farther from Jupiter)

Where on the map would these objects' paths be? Or at least how would I find the exact answers with math/software/etc.?

My assumptions/guesses:

Jupiter's path is relatively similar to the ecliptic but Callisto's orbital inclination or maybe Jupiter's might separate the two
Jupiter's moons are within a similar orbital plane to Callisto and would follow near Jupiter's path
the distance Callisto travels away from the sun in its own orbit is insignificant in determining the shape of the ecliptic

I have no idea if the axial tilt of Callisto or Jupiter would affect anything but it probably does.

If it helps, I've made this incredibly stylish and oh-so-fabulous template in case anyone wants to sketch it in Paint or something. The black dot at the center is the celestial north pole.

Thank you for your time!

I'm not sure this answers your question, but I used Stellarium (free) to generate this image from Callisto's north pole (so its celestial north pole, in Draco, is overhead) with orbit trails turned on:

Unfortunately, the image is not only messy, but may be inaccurate, since Stellarium doesn't do a great job of plotting orbits long-term: Stellarium 0.10.4: planet orbits change over time?

So, I wrote https://github.com/barrycarter/bcapps/blob/master/ASTRO/STELLARIUM/callisto.ssc to create a video of the view from Callisto's north pole, but the results are fairly uninteresting: https://youtu.be/O2GHJM54uhQ and come with several caveats:

I assumed that Stellarium is accurate, not just in where it displays the planet, but also in how it determines Callisto's rotational orientation, and any other details required to depict Callisto's celestial northern hemisphere.
The video is in "starchart mode", which means the stars don't move. In reality, the stars would make a complete rotation about every 16.69 days. More specifically, in the video, Jupiter moves through the constellations. In reality, Jupiter would remain fixed (Callisto is tidally locked), and the constellations would move.
The one year (+ about 2 months) period I chose is not "representative" and does not repeat. There may be quite a few more interesting things happening after the period the video represents. In particular, note that the Sun takes about 12 years (Jupiter's orbital period) to travel Callisto's ecliptic, and only a fraction of that travel is show in the video.


Discontinuity in space-time What I understood about it till now is that space-time fabric is kind of broken there. Please tell me whether I am correct.

If it's true then does it mean that perception of time will stop there?
I really can't get what broken space fabric is like. If it's a hole then how is matter supposed to be present inside that?
Is a black hole a discontinuity?

A singularity in physics is basically a big signpost saying "the theory you are using has probably ceased to be a good approximation of reality by the time you get here". An example is the infinite amount of short-wavelength radiation produced by a blackbody under the Rayleigh-Jeans law. This is obviously an incorrect description of reality and the issue ended up being resolved by the development of quantum theory: taking into account the corrections gives Planck's law, and the infinity goes away.

Inside a black hole, general relativity does predict singularities, and it seems to be fairly common to talk of them as though they are physical objects. However we should be cautious: even though general relativity is our best description of gravity so far, we already know that it does not describe the actual universe. The issue here is that general relativity is incompatible with the Standard Model of quantum field theory, which is our best theory of how physics works excluding gravity.

By the time you get to the region near where general relativity predicts a singularity, you're working with a very small region with an intense gravitational field. Quantum corrections are almost certainly going to be relevant here, and unfortunately we do not yet have a self-consistent theory of "quantum gravity" to describe what should happen in such a region and so far experiments (such as the ones being performed at the Large Hadron Collider) haven't given much of a clue as to how to proceed with developing one.

In a hypothetical universe where general relativity is correct then there would be a singularity (which is fortunately hidden from the rest of the universe by the event horizon), but we do not live in such a universe.


How to make a star chart from entries in a Bright Star Catalogue? I know very little in astronomy, but I want to use the data in a Bright Star Catalogue to create star map projections for a specific time and place.

For example one line is

{ "Dec": "-33° 31′ 46″", "HR": "13", "K": "4850", "RA": "00h 08m 03.5s", "V": "5.68" },

How would I convert this to an $x, y$ point on a printed star chart that could be used for viewing the sky at a specific time and place?

Here is a tentative answer based on your comment that you want to make a star chart. I'm not an expert in this but I used answers to RA/dec to Alt/Az program or method which link to http://www.stargazing.net/kepler/altaz.html

You have the star's position on the celestial sphere, convert it to decimal degrees:

     item              original             conversion                  decimal degrees
Declination:      -33°  31′   46″    -33 - 31/60. -  46/3600.         =   -33.529444
Right Ascension   00h  08m  03.5s   (  0 +  8/60. + 3.5/3600.) x 15   =     2.014583

You have the time $UTC$ and longitude $LON$, convert it to $LST$ (local sidereal time). From the question Local Sidereal Time and @DavidHammen's answer:

$$LST = 100.46 + 0.985647 d + LON + 15 UT $$

where

$LST$ is local sidereal time in degrees
$d$ is the number of days from J2000, including the fraction of a day
$UT$ is the universal time in decimal hours
long is your longitude in decimal degrees, East positive.

You have $RA$ and $LST$, get $HA$ (Hour Angle)

$$HA = LST - RA$$

You have $DEC$ and $HA$, get altitude and azimuth

$$ALT = \arcsin\left( \sin(DEC) \sin(LAT) + \cos(DEC) \cos(LAT) \ cos(HA) \right)$$

$$AZ = \arccos \left( \frac{\sin(DEC) - \sin(ALT) \sin(LAT)}{\cos(ALT) cos(LAT)} \right)$$

From there you have to decide how you want to plot altitude and azimuth on your map.

If you want to plot it on a circle of radius $R$ then use

$$X = R (1 - ALT/90) \cos(AZ)$$

$$Y = R (1 - ALT/90) \sin(AZ)$$

assuming $ALT$ is in degrees. In this map "top" or $X, Y = 0, R$ will be north and $X, Y = R, 0$ will be East. You will have to decide if that's the best way to plot it or if you want to mirror East-West.


Size of a black hole If there is no matter inside any of the black holes and all of them have a singularity in the center of them, which is infinitely small and infinitely dense, then how can they differ in size and weight?

The size of a black hole is determined by its energy and angular momentum. For a non-spinning black hole, there is a simple linear relationship between the black hole's mass and its Schwarzschild radius:

$$r_s=\frac{2GM}{c^2}$$

where $r_s$ is the Schwarzschild radius, $G$ is the universal gravitational constant, $c$ is the speed of light, and $M$ is the black hole's mass. The mass of the black hole comes from all the matter and energy that went into creating the black hole, as well as any additional matter & energy that's fallen into it afterwards.

Strictly speaking, the Schwarzschild solution is for a universe containing nothing but a single eternal black hole. But although it's unphysical it's still a useful solution when we want to model spacetime with spherical spatial symmetry (including planets & stars, not just black holes).

For a rotating black hole, the situation is more complex, and there are several horizons. Please see Wikipedia's article on the Kerr metric for details.

We cannot say exactly what happens at the core of a black hole. As discussed in this answer by Florin, we need a theory that unites General Relativity and Quantum mechanics to answer such questions.

A pure GR black hole has a mathematical singularity at its core, but most astrophysicists believe that's unphysical, and that a proper quantum gravity theory will eliminate that singularity. However, it's likely that the core of a black hole is still very small, since the quantum gravity corrections probably don't kick in until the size gets smaller than an atom. And of course even with a quantum gravity we still won't ever be able to observe black hole cores to validate the theory.

So we cannot currently say exactly what happens to energy and matter once it reaches the core of a black hole. But we do know that once anything crosses the event horizon of a black hole it can no longer affect the universe outside the event horizon, and that anything inside the event horizon must rapidly (in its own proper time) fall towards the core of the black hole, and that no known force can prevent that process.

The gravitational field of a black hole is sometimes described as a "fossil field". All matter & energy falling into the black hole modifies the spacetime curvature as it approaches the event horizon. And once it crosses the event horizon it can no longer change the spacetime curvature outside the horizon, so those curvature changes are preserved (until something else comes along to add its own curvature changes).

So it doesn't really matter what happens at the core of the black hole, since the exact nature of the black hole core doesn't affect what's happening outside the event horizon.


Why does Pluto's Orbit underlap Neptune's Orbit Why does Pluto's Orbit underlap inside of Neptunes Orbit. Clearly Neptunes Orbit is not overlapping on Pluto's. Yet, they fail to maintain the symmetric banding appearance all the other Planets maintain, similar to the quarter mile track you often see around football fields.

There are a few peculiarities of Pluto's orbit. These are:

Its high orbital eccentricity (e = 0.25) causes Pluto's perihelion to be ever so slightly smaller than Neptune's perihelion.
Pluto has an orbital resonance with Neptune. Its orbital period is exactly 3/2 of Neptune's. This orbital resonance is the cause of its orbital eccentricity that you ask about.
Pluto has an orbital inclination of 17 degrees to the ecliptic. No one knows why, we can only guess.

  The origin of Pluto's eccentricity explained:

During the later stages of planet formation, in this early Solar system, the planets were still forming. Neptune exchanged angular momentum with the remaining planetesimals, and its orbit expanded outwards. If Pluto were in a near-circular orbit larger than Neptune's (at about 33 AU), there is a high chance that Neptune could capture Pluto and lock it an orbital resonance (this probably happened when Neptune was at about 25 AU). As Neptune's orbit continued to expand outwards, this expansion drew out the eccentricity of Pluto's orbit.

Source and further reading: The Origin of Pluto's Orbit: Implications for the Solar System Beyond Neptune


What is the current accepted theory as to why Mercury, despite its size, has a similar density to Earth? According to the NASA web page overview about Mercury, despite the planet being just a bit larger than our moon, it's density is about 98.4% of Earth's.  This high density suggests a comparatively large iron core (compared to the rest of the planet).

Recent studies reported in "NASA Antenna Cuts Mercury to Core, Solves 30 Year Mystery" (NASA, 2007) have confirmed the presence of a molten core, surrounded by a silicate mantle (depicted in the illustration below (from the article):

The question is then, what is the current accepted theory as to Mercury has a still-molten core and a density like that of Earth?

The most-widely accepted hypothesis at the moment is that Mercury was struck by a large impactor that removed a significant fraction of its mantle (I believe this theory was originally proposed by Cameron & Benz in 1987, and the qualitative theory hasn't changed very much). For planets that are close to their parent stars (such as Mercury), the collision with the secondary body likely occurs at high velocity due to the rapid orbit speeds there (> 40 km/s). As this is much larger than a typical rocky planet's escape velocity (~10 km/s), material is lost from the planet in these collisions. This collision is also more likely to be "grazing," as "direct hits" are rare, and thus the outer parts of the planet are preferentially removed.

Because the mantles of rocky planets are primarily composed of silicates, which have a lower density than iron that resides in their cores, the collision results in a surviving planet with a higher average density.


Why don't (or can't) stars be more than 325 or so times the mass of the sun? What limits their size? Is there a particular reason why stars cannot grow as massive as they want to?
And why doesn't this limit apply to supermassive black holes?

There is basically an upper limit to the mass of a star because their luminosity is so great that the radiation pressure prevents the accretion of further mass.
However, the upper limit depends on the composition of the accreting material. This is because the effect of the radiation depends on the opacity of the material - stuff that is more metal-rich is more opaque. The upper limit you refer to is a bit too high for compositions similar to the Sun (I think more like $200 M_\odot$ would be the limit for that). For stars born in the distant past that were metal-poor, or even born from primordial material with no metals, the upper limit could be much higher.
For primordial material, with no metals at all, the physics that determines the upper limit is likely to be fragmentation. Primordial gas is quite transparent to radiation, but this also means it is poor at radiating away the heat from gravitational contraction. In principle, much larger bodies can only begin to contract towards being very massive stars if some of the gas is ionised or if there is hydrogen in molecular form. However, this cooling also reduces the Jeans mass, allowing the collapsing cloud to fragment into a cluster of smaller objects before a supermassive $(>10^6 M_\odot)$ star/black hole could form.
Nobody knows for sure how supermassive black holes formed. It seems likely that they grew from seed intermediate black holes that formed from very massive primordial stars - perhaps thousands of solar masses - or even by direct collapse to black holes of mass $10^4-10^6 M_\odot$ without ever forming a stable star.


Is it possible for galaxy clusters to interact? Galaxies commonly interact and collide with other galaxies. Is it possible that clusters of galaxies similarly interact and collide? Have people tried to study this phenomenon previously? I couldn't find anything specific in the literature.

Yes.

Theoretically, structure is expected to form first on small scales (stars and stellar clusters), and later on increasingly larger scales — galaxies, groups, and eventually galaxy clusters (see e.g. Longair 2006). This is confirmed, at least to some extent, observationally. For instance, galaxies have been detected out to a redshift of $z=11.2$ (400 million years after Big Bang; Oesch et al. 2016), while cluster have only been detected out to $z=2.5$ (2.6 billion years after Big Bang; Wang et al. 2016).

Thus, when clusters formed, the Universe had already expanded so much that interaction between them is rather rare.

It does happen, however. One of the most important examples is the Bullet Cluster, which consists of two colliding galaxy clusters. The reason I mention this is that it really beautifully confirmed the existence of dark matter. The image below (from NASA's Astronomy Picture of the Day) shows the two clusters after the collision. The stars and galaxies are so far apart that collisions between galaxies are rare, and collisions between stars virtually never happen. Thus, they have just passed right through each other, as seen in the image. The gas between the galaxies, however, collides, slows down, and is separated from the galaxies. This heats the gas to millions of degrees, emitting X-rays (seen in red). The blue stuff is a map of the mass distribution, made using gravitational lensing. This mass is clearly separate from the gas, but coincides with the galaxies, and gives a much higher mass than the visible mass, roughly 5.5 times more, which is exactly what is found for the ratio between dark and normal matter using other methods.


Portion of universe visible if gathering image from inflationary epoch Apologies in advance for my over-exposure to pop science.  I want to propose a telescope powerful enough to collect imagery from 10^-32 seconds after the “Big Bang”, or as soon as the end of the inflationary epoch.  With light coming from the universe when roughly 10 ly in diameter, along with the expansion to our current size, I would only see the tiniest fraction of light from so long ago with a telescope looking at that far of a distance.

Assuming my telescope has an identical view field as the Hubble Space Telescope, what fraction of the total volume of the universe at end of the inflationary epoch would I actually be viewing?

tl;dr: Your field of view would cover roughly one square centimeter of the sky at that time, and you would observe roughly 50 billionths of the observable Universe.

You can't really…

With photons, you will never be able to see further back than to recombination, when the Universe was 380,000 yr old, because until then, the free electrons made is opaque to radiation.

With neutrinos, you can (in principle) see back to 1 sec after Big Bang, but to "see" all the way back to inflation, you will probably need gravitational waves.

…but let's assume you can anyway

Anyway, lets's do a thought experiment and assume that you have a RLT (Ridiculously Large Telescope) with a field of view (FoV) of the HST. What fraction of the Universe can you see? The answer might surprise you:

The farther an object is from you, the smaller an angle it spans on the sky, i.e. the smaller it looks. This is true for birds and planets and even for the nearest galaxies. However, for distant galaxies a strange effect counteracts this: Because of 1) the finite speed of light and 2) the expansion of the Universe, distant galaxies were closer to you when they emitted the light you see today, and therefore spanned a larger angle. Hence, if you compare galaxies of the same physical size, they look smaller and smaller until around 15 Glyr (giga-lightyears), after which they'll look larger and larger.

The Hubble Ultra Deep Field has FoV of 2.8 by 2.5 arcminutes. The region of the Universe that this FoV spans is thus largest at a distance of 15 Glyr, where it span $4.8\times4.3\,\mathrm{Mlyr}^2$ (i.e. square-mega-lightyears). But then your FoV begins to span an increasingly smaller region. At recombination, it spans only $34\times30\,\mathrm{klyr}^2$ — in other words, if a Milky Way-sized galaxy were present there (it weren't), it would be larger than the FoV.

It you could look all the way back to 4 hours after Big Bang, your FoV would span roughly a square-lightyear. Three minutes after BB, when all the Universe's hydrogen and helium had just been created, it would span $\sim0.1\times0.1\,\mathrm{lyr}^2$. A few microseconds after BB, it would span roughly a square-AU (the distance from Earth to Sun). Approximately $10^{-22}\,\mathrm{s}$ after BB, it would span a square kilometer.

And at $10^{-32}\,\mathrm{s}$ after BB, your FoV would span $0.8\times0.7\,\mathrm{cm}^2$, i.e. roughly one square-centimeter!

Fraction of the Universe observed

The part of the observable Universe that you would observe is the volume "behind" your FoV. That can be obtained by integrating the area along the distance, but an easier way is to simply note the fact that, if your FoV were the whole sky, you'd observe the whole Universe. Since your FoV covers a fraction of the sky of
$$
\begin{array}\\
f & = & \frac{2.8'\times2.5'}{\mathrm{full\,sky}}\\
  & = & \frac{8\times10^{-4}\,\mathrm{rad}\,\,\times\,\,7\times10^{-4}\,\mathrm{rad}}{4\pi\,\mathrm{rad}^2}\\
  & = & 5\times10^{-8},
\end{array}
$$
this is the fraction of the Universe that you would observe.

At $t=10^{-32}\,\mathrm{s}$, the part of the Universe later became our observable Universe today, was about ten meters in radius. Another way to calculate the fraction is to realize that the above mentioned observed square-centimeter roughly comprises a fraction $f$ of the surface area of a sphere with a radius of ten meters.


Why does matter spiral into blackholes? Narrators of astrophysics documentaries constantly state that matter spirals into black holes as if it were intuitive that this would happen because of gravity. But the moon isn't spiralling into the earth. Newtonian physics suggest that an object would either fall past the event horizon in less than half a revolution, or it would orbit indefinitely until acted upon by a third party.

Why would an orbit gradually degrade?

There are probably four different processes, any or all of which may be described as "spiralling into a black hole"

The effects of general relativity. as @AtmosphericPrisonEscape has indicated, once you get very close to a Black hole Newtonian physics ceases to be a good approximation and must be replaced by general relativity (GR). When you do the calculations, you find that all circular orbits within 3 Schwarzschild radii of the black hole are unstable. Any object orbiting closer than that will definitely spiral in.
Gravitational radiation GR tells us that any two orbiting objects will emit gravitational waves, which carry away energy and angular momentum, slowly reducing the separation between the objects. This effect is, however, incredibly small, unless the objects are very massive and orbiting very close to one another (which means they must also be very dense). Thus this is mostly relevant for black holes and neutron stars in close orbits.
Turbulence in an accretion disk When a lot of gas or dust (which may be the remnant of larger objects pulled apart by tides) is orbiting a black hole, it tends to form a disk, called an "accretion disk". Basically the disk is oriented according to the average angular momentum of the stuff falling in, and anything orbiting in a different plane will collide with lots of the material in the disk and quickly join the disk. This is the same mechanism that caused the early proto solar system to form a disk, and indeed Saturn's rings (I know I'm oversimplifying somewhat).  Anyway, to remain in orbit, the inner parts of such a disk must be rotating faster than the outer parts, and this "shear" in the flow of gas tends to cause turbulence, which, it turns out, tends to transfer angular momentum outwards through the disk. This results in material at the inner edge of the disk getting slowly closer to the Black Hole until it reaches the Innermost Stable Circular orbit, at which point effect number 1 takes over.
Dynamical "friction"  If there are a lot of larger bodies (stars, asteroids, etc.) orbiting a large central mass, they will, from time to time, pass near each other, and  change each others orbits, transferring angular momentum between them. It turns out that averaged over many encounters, the effect of this is to drive larger bodies inwards and smaller ones outwards (possibly ejecting the smaller bodies from the system completely) so that it acts as something like friction on the larger bodies.


How does the evolution of a solar system not break the second law of thermodynamics? Please forgive: I am a layman when it comes to physics and cosmology, and have tried finding an answer to this that I can understand, with no luck.

As I understand it, the solar system evolved from a massive molecular cloud. To me, this seems to break the second law of thermodynamics, as I think it suggests order from disorder.

I know there must be something wrong with my logic, but am really stuck.

Can anyone explain this one in layman's terms?

(Posting to both "Astronomy" and "Physics", as it seems to overlap these subjects)

The total entropy actually increases, as the molecular cloud shrinks under gravity.

It may seem that as the molecules are getting closer, they are more ordered, which means less entropy. That is however only one part of the process. The second (important) part is: when the molecules are closer, they also have higher kinetic energy (since they descended into lower gravitational potential). So the gas is getting hotter, as it shrinks.

The increase in temperature of the gas is increasing its entropy, because the molecules occupy more momentum space. This increase in entropy via temperature is bigger than the decrease in entropy via the shrinking itself.

Later on, the hot condensed gas (or hot planet) radiates the heat into the space and cools down. You end up with a cold planet which indeed has lower entropy than the original gas cloud, because it is no longer hot. But the entropy increase was carried away by the radiated photons. So in total - entropy of the universe increased (the radiated photons are out there somewhere).

You can find some more detailed discussion of this topic on the excellent web page by John Baez or here.


Has J been used by itself as a spectral type? While answering the question "Why have brown dwarf classes been dubbed L, T and Y?" I noticed that in the paper Kirkpatrick et al. (1999) "Dwarfs Cooler than 'M': The Definition of Spectral Type 'L' Using Discoveries from the 2 Micron All-Sky Survey (2MASS)", the letter "J" was ruled out as being in use as a standard carbon star type.

So far I have not seen a source that uses "J" as a carbon star type by itself, I've only seen it in combination with "C", e.g. "C3,3J" or "C-J4 C2 5" (both from XHIP). This contrasts with the other carbon star types "R" and "N" which have definitely been used as separate identifiers.

Has "J" ever been used as a stand-alone spectral type, e.g. "J3"?

As far as I can tell, the J-type was introduced by Bouigue (1954), and have subsequently often called "J-type stars" or "J-stars", e.g. Abia & Isern (2000). This likely led to the J-type designation being put as "in use", though the spectral types themselves appear to always have been used in combination with "C".


Can cosmic rays alone produce noticeable aurorae in bodies far away from the Sun? Aurorae are caused by energetic particles interacting with planetary atmospheres. Part of this flux comes from the Sun (solar wind) while the rest is cosmic radiation. My main question is: could a planetary body located far away from the Sun (or any other star), with a suitable atmosphere and magnetosphere, have noticeable* aurorae, just from being exposed to cosmic rays? A related question would be: do we know how much of the aurora is caused by the solar wind, and how much by cosmic rays? And thirdly: is there any other way that a planetary body could produce aurorae, such as its own electromagnetic emissions?

*I know "noticeable" doesn't really work as a limit, but for the time being, I'm talking of electromagnetic radiation that can be distinguished clearly from the background of a planet's disc using instruments not more advanced than our current telescopes at interplanetary distances.

You may be interested to learn that Jupiter experiences aurora on its poles nearly (if not totally) non-stop. These aurora never cease, because they do not rely entirely on the sun as their source of charged particles, instead receiving particles from the volcanic ejections of its moon Io, as well as from cosmic rays. Granted, these aurora are occurring in the ultra-violet, but they are still 'noticeable'.

In short, yes aurora can be caused by non-solar sources, and conceivably cosmic rays at that. As to your other question concerning 'how much' of the aurora is due to which sources... I couldn't tell you how much of the auroras visibility comes from the sun, and I doubt there is much (or any) information on that subject, since determining the source of such convoluted light is likely outside the realms of both interest and empiricism. Perhaps you could pioneer that research? In any case, an intriguing subject to be sure.


Nodal Precession of Planets It is obvious that the longitude of ascending node ($ \Omega $) of a planet gradually changes with time. For example, the VSOP planetary theory accounts for this. In Astronomical Algorithms (Meeus 1998), one finds for example that, for Mercury, $ \Omega = 48.330893 - 0.1254227 T - 0.00008833 T^2 - 0.000000200 T^3 $ (referred to the standard equinox J2000.0).
While formulas to calculate the precession of the perihelion of a planet are easy to find (e.g., in “Calculation of apsidal precession via perturbation theory” by Barbieri and Talamucci), I am having a hard time finding sets of formulas pertaining to the advance (or regression!) of the ascending node.
Can someone please give me a few references of where I could find them? Thanks in advance!
(Reason: I’d be interested in calculating precise planetary orbits for an imaginary planetary system, or maybe even for known planetary systems such as TRAPPIST-1.)
EDIT: Most of the formulas found online (including those given in answers and comments) are either restricted to the Solar System, or restricted to one body orbiting another one. I’m looking for something that would be in the form of $ \dot \Omega = \sum_{i = 1}^{\infty} k m_i m_0^x $ where $ m_i $ is the mass of the perturbing body, $ m_0 $ is the mass of the perturbed body, and k and x are, well, what I’m looking for! Their determination probably includes G (the constant of gravitation) and R (the distance between the two bodies) and so on…

Perhaps you are looking for Lagrange’s planetary equations? Given some perturbation $H_1$ to the Hamiltonian, they give six equations to calculate the change of the orbital elements over time; a thorough discussion is given in Goldstein’s textbook on classical mechanics.
The equation for $\Omega$ is $$\frac{d\Omega}{dt}=\frac{1}{\sqrt{GMm^2(1-e^2)\sin i}}\frac{\partial H_1}{\partial i}$$ where $i$, $e$ are orbital elements.


How do one identify something they see in the sky? I've always wondered how someone looking at the sky into the enormous amount of stars and other things out there figures out that whether it is an already identified object or is he looking at something new? Is there some sort of tools like latitude and longitude to pin point something in the sky or some catalog where you can see if this is an identified object and what it is called, how it looks like and where it is located and all?

As you might have guessed, I don't have any science background, just asking out of a common man's curiosity, so please be gentle with the jargon you use.

There exist many astronomical catalogs, reaching from rather simple catalogs like the Bayer catalog for stars, or the Messier catalog for galaxies, nebulae, etc., to huge catalogs with millions of entries. A catalog with about a billion objects is scheduled for 2022.

There exist also printed editions of some catalogs.

Objects in the sky have several properties, by which they can be described and mostly identified.

One of the most frequently used properties is the position in the sky. From the local geographic position and the time, the observed position of the object can be translated into a celestial coordinate system, e.g. the equatorial coordinate system. The position of an object is described in this system by declination and right ascension.

Two other frequently used properties are magnitude and color.

Also frequently used are patterns of stars, meaning their relative position to each other, beginning with star constellations.

The analysis of steller spectra allows a more detailed identification.

Longer term observations may uncover variability and proper motion as additional properties.

Here is a description of parameters (including properties) provided by the BSC5p catalog.


Is a black hole a three-dimensional or four-dimensional object? Is a black hole a three dimensional or four-dimensional hole?

It would help to clarify your question on what exactly you mean by a 4th dimension.  Space-time for example is 4 dimensional everywhere, but it's 3 spatial dimensions and 1 time dimension.

Quantum theory allows for extra dimensions, but they're very small, smaller than an atom and impossible to measure by normal means.  CERN has looked for those folded dimensions but has found no evidence of them, and string theory defines 10 or 11 dimensions, most of them we can't see or measure, if they exist at all.

If you mean 4 spatial dimensions, then the answer is no, with a capital N.   There's no evidence that an extra spatial dimension would form in a black hole.  There's nothing in relativity that suggests that might happen.  An extra time dimension . . . probably not either, but it's fun to think about what 2 time dimensions would look like.

What happens inside a black hole is that one of the dimensions becomes like time.  It only moves in 1 direction, but that doesn't make it a 4th dimension, it just changes the nature of 1 of the dimensions.

If you actually had a 4th dimension (and the book flatland discusses this in pretty good detail), but if you actually had that, very strange things can happen, for example, you can be flipped over, like flipping a card over, which can represent 2 dimensions, from face down to face up.  Your left side would become your right side.  Your heart would be on the other side of your chest and all the left handed molecules in your body would become right handed molecules, which would make your ability to absorb and use nutrients that you digest problematic.   It might even make things taste and smell different.

You can't tie your shoes in the 4th dimension, but even if you could, your shoes could slide off your feet, no matter how well you tied them.   Food could fall out of your stomach, not through one of the two normal exits, but it could just slip out, like a ball rolling off a table.   Anything not attached could slip apart.   Sex would be much more difficult.  Bolts and nuts could just move sideways and detach, unless there was some electrostatic charge keeping them in place, in fact, the pressure that forms natural between a screw and a nut might cause all nuts and bolts to fly apart at considerable velocity once you entered 4th dimensional space.

Two people could also stand in exactly the same place, just in a different 4th dimensional space.   That brings a whole new aspect to the how many people can you fit in this Volkswagen game, as you could fit theoretically any number of people in the same Volkswagen.

A 4th spatial dimension is probably impossible, because it breaks too many rules and laws as we know them, and it's never been observed.

Black holes, by comparison, can do strange things like slow down time, and they aren't fully understood, but a 4th spatial dimension probably isn't possible, unless you mean at quantum distances, in which case . . . maybe.

Feel free to read this for a different approach to a somewhat similar question.


Hills' mechanism for making Hypervelocity Stars In Hill’s mechanism, the gravitational tidal force of
  a single MBH disrupts an approaching binary. One star is captured on an eccentric orbit around
  the MBH and, by conservation of energy, the other star escapes with a final velocity equal to the
  geometric mean of the ∼$10^4$ km/s
  infall velocity and the ∼$10^2$ km/s binary orbital velocity.
  Hypervelocity Stars and the Galactic Center,
  Warren R. Brown, p.3

I don't understand how the high energy/velocity can be transferred to the star.

(Hills' 1988 Nature paper is still paywalled.)

Consider two unit masses with velocity $V$. Give them a small push apart so they have velocities $V+v$ and $V-v$ with $v\ll V$ . Their kinetic energies are then $$\frac{(V+v)^2}{2} \approx \frac{(V)^2}{2} +vV$$ and $$\frac{(V-v)^2}{2} \approx \frac{(V)^2}{2} -vV$$
We have transferred energy $\approx vV$ from one mass to the other with the expenditure of a much smaller  energy $v^2$.

Now suppose the masses are initially moving with some tiny velocity and then travel into a potential well where they have velocity $V$. If we push while they are in the well then one will have enough kinetic energy to escape the well plus $vV$, while the other will be trapped in the well.

This is the same principle as the Oberth effect, where the ejected reaction mass plays the role of one of the masses.


Was that actually a shooting star during the supermoon eclipse on September 27th 2015? During the super-moon total lunar eclipse on September 27th a meteor-like streak passed very close within view of the event. It appeared to happen at the very middle, the maximum eclipse.

[CafferyPhoto][1]

At the time, I thought it must have been some organization shedding an unused satellite. This would be a pretty epic time to do so as so many eyes around the world are looking at the same thing. Maybe even NASA marking the halftime of the event. Can anyone confirm or shed any amount of light on what this was?

[1][https://www.facebook.com/cafferyphoto/photos/a.279223605465113.81307.120967987957343/915070541880413/?type=3&theater]
note - photo for proof only - shutter cut off leading edge

QUESTIONS:

People on the Caffery's Facebook page seem to have seen in in California and other locations as well, I happened to be in St Thomas, Virgin Island at the time of the event.  CafferyPhoto quotes. "I was only exposing for half a second so the shutter cut off the latter half of the trail, darn it!" - Seems in was taken in the NorthWestern US sometime around 7:30pm.

There are about 5 sporadic meteors visible per hour most nights, and since thousands of people were taking photographs of the moon that night, it is not impossible that someone captured a meteor in the same shot as the moon.

A couple of things it was not

It was not an unused satellite. When they re-enter you see a fireball as it breaks up, not a single streak, as here.
It was not "Nasa marking half-time". The object would have been visible from quite wide area (but not world wide, not even countrywide) But the association with the eclipsed moon would only have been visible from a particular location. Nasa isn't going to be interested in entertaining the  good folks of Yakima, at the expense of everyone else.

What it could be:

It might be a meteor. Normally you would see a brighter head as the meteor changes brightness as it falls, but if this was a 1/2 second exposure, that could have been cut off.
It could be a satellite, which appear to be a steadily moving star, and would look like a streak in a timed exposure. However, looking at Heavens above doesn't show any satellites near the moon at that time.
It could be a deliberate forgery.

I reserve judgement. One would need exact locations and timings.


Why was the July 27th 2018 lunar eclipse not visible in North America? I was just wondering if it had anything to do with the fact that last year we were blessed with the opportunity to see the total solar eclipse in our neck of the woods. So this is the universe's way of making things 'even'. Or is it pure chance and luck?

Jokes aside, there is probably a physics explanation regarding tidal forces or angular momentum, so I am curious how these factors affect which parts of the world get to witness the event.

Nothing so romantic as the universe being fair.

The path of the moon is tilted by about 5 degrees from the path of the sun. but twice a month the moon must cross the path of the sun.  When this moment coincides exactly with full moon there is a lunar eclipse. Since the sun is on the opposite side of the Earth, lunar eclipses are only visible during the night, and this one happened to occur during the day for the USA. When it is day in the USA it is night in China, and the eclipse was visible from the other side of the world.

Of course, these things aren't random, but there is no simple formula like "if you see a solar eclipse, then the next lunar eclipse won't be visible."  You can just go to any of the eclipse prediction websites to find when the next one will be visible in your area.


Electrical potential difference between planets In dealing with electricity, we usually refer to Earth as the electrical neutral or ground. Is there any evidence that other planets' ground is at the same electrical potential level as the Earth's?

If somehow, there was a lossless conductive connection between two planets, would current flow accros It? Would this be theoretically a source of energy?

I dunno how or why a planet would have a significant net charge, but supposing it did, here's some initial calculations.

If the planets have a different net charge, you can treat them as capacitor plates.
Capacitance in its simplest form is

$C = \frac{\epsilon *A}{d}$
The permittivity $\epsilon$ of vacuum space is 8.85 x 10-12 farad per meter .
Let's choose a nearby (nonexistent) planet at 5 LY = 5 * 9.461e+15 m .

Looks like the capacitance is pretty close to zero.  So the energy available,

$E = \frac{CV^2}{2}$  is going to be nonexistent unless there's a seriously large voltage.


Do black holes have energy? So black holes are created by certain dying stars and when the star runs out of nuclear energy gravity wins out and the star implodes. An entire star’s mass collapses down into a smaller and smaller volume of space. Which then creates a black hole, so my question is, do black holes actually have energy to create that type of vacuum to suck light in?

If a black hole was created by a huge star running out of fuel and then collapsing in on it self, would the black hole have energy and if so where would it come from, or is it just like the vacuum in space with no energy at all?

(Please correct me if I said anything wrong about the black holes as I am very young and don't quiet grasp the concept of it yet.)

In outer space, a vacuum isn't what sucks in matter to a black hole. That may be the case for your household vacuum cleaner, but everywhere in outer space is essentially a vacuum. Instead, it's the black hole's powerful gravitational pull.

However, black hole can have energy. One type that comes to mind is angular kinetic energy. This is the energy that a spinning object has - in this case, a rotating black hole (see also Kerr black hole and Kerr-Newman black hole). Black holes can also have translational kinetic energy - the energy of an object in motion. Also black holes have gravitational potential energy - potential energy due to gravity - and black holes with electric charge can have electric potential energy.

So black holes do have energy, although perhaps not in the form you thought.

Keep asking good questions!

I hope this helps.


Accuracy of calculating the vernal equinox? Calculating the exact time of the vernal equinox is essential for many ephemeris calculations. NASA Goddard Institute for Space Studies uses (at least for some purposes) the following, rather short source code:
Function VERNAL (IYEAR)
      Implicit Real*8 (A-H,O-Z)
      Parameter (EDAYzY=365.2425d0, VE2000=79.3125)
      VERNAL = VE2000 + (IYEAR-2000)*EDAYzY
      Return
      End

What that means becomes clear when reading the comment inside the code:

For a given year, VERNAL calculates an approximate time of vernal equinox in days measured from 2000 January 1, hour 0.
VERNAL assumes that vernal equinoxes from one year to the next are separated by exactly 365.2425 days, a tropical year [Explanatory Supplement to The Astronomical Ephemeris].  If the tropical year is 365.2422 days, as indicated by other references, then the time of the vernal equinox will be off by 2.88 hours in 400 years.
Time of vernal equinox for year 2000 A.D. is March 20, 7:36 GMT [NASA Reference Publication 1349, Oct. 1994].  VERNAL assumes that vernal equinox for year 2000 will be on March 20, 7:30, or 79.3125 days from 2000 January 1, hour 0.  Vernal equinoxes for other years returned by VERNAL are also measured in days from 2000 January 1, hour 0.  79.3125 = 31 + 29 + 19 + 7.5/24.

I highlighted the essential part of the quote. We know for sure that this algorithm will give the wrong date of the vernal equinox in the year 5333 when it will be off by more than 24 hours, but I suppose it can already be incorrect earlier if the time stamp for the vernal equinox is close enough to midnight. When would be the first time that this algorithm picks an incorrect day?
My question: Is the above quoted algorithm the most accurate way of calculating the vernal equinox for times far in the future or past? How would you do it accurately for time scales of 1000s of years, say with e.g. for an application for archeology?
PS: WolframAlpha calculates vernal equinoxes in the distant future with giving hours and minutes and without any error bars which seems too good to be true.

Calculating the exact time of the vernal equinox is essential for many astronomical calculations.

I dispute that claim. What is true is that calculating the exact time of the vernal equinox is essential for some astrological and religious calculations.

My question: Is the above quoted algorithm the most accurate way of calculating the vernal equinox for times far in the future or past?

Of course not. The value of 365.2425 days is an exact value; it is the average number of days per year per the Gregorian calendar. The Gregorian calendar repeats over a 400 span. In any 400 span, there will be 97 leap years (96 non-century leap years, plus one century leap year) with 366 days and 303 years with 365 days. That results in 146097/400 days in a year on average, or exactly 365.2425 days.
The Gregorian calendar was developed as a replacement for the Julian calendar because scholars attached to the Catholic Church were finding it increasingly more difficult to correctly calculate the date on which Easter should fall. The Julian calendar had a leap year every four years. That was a bit too much, and it resulted in an ever increasing number of days between the Jewish Passover and the Christian Easter. The switch to the Gregorian calendar, which is where that 365.2425 values comes from, was motivated by religion rather than astronomy.


Are there any galaxies visible in the night sky around the Orion constellation? I'm writing a story for NaNoWriMo, and in it there is an extraterrestrial race whose home galaxy is visible from Earth. This home galaxy is visible (or even partially visible or barely visible) in the night sky as a close neighbor to the Orion constellation. "Close neighbor", as in "within or around" the Orion constellation.

Is there any such galaxy?

The short answer is there are no galaxies in or around Orion that are visible to the naked eye.

The International Astronomical Union (IAU) has some simplified star charts for each constellation. The charts show galaxies the brightest galaxies that are visible in amateur telescopes as red ovals. The chart for Orion and the surrounding constellations show no galaxies. Of course, larger telescopes are able to show fainter objects, so you are bound to find a galaxy in every constellation with a large enough telescope.

The galaxies that are easily visible to the naked eye are as follows: (There are a few other that are visible to the naked eye, but they are not as obvious.)

Large Magellanic Cloud in the constellations of Dorado and Mensa
Small Magellanic Cloud in the constellations of Tucana and Hydrus
Andromeda Galaxy in the constellation of Andromeda


Where's the Prime Meridian of celestial coordinates ...or what's the 'direction of vernal equinox'?

For both Ecliptic and Equatorial coordinate system, the Wikipedia article gives "Primary direction (0° longitude)" - "Vernal Equinox", linking to article on the event of the Equinox which never mentions any directions associated with it.

I guess the direction will be associated with the axis connecting the Sun and Earth during vernal equinox, but which direction of the line? Outward, with the meridian running opposite of the Sun, or inward, the meridian crossing the Sun at vernal equinox? Or maybe my guess is entirely wrong?

Generally, how would one go about converting celestial (ecliptic or equatorial) latitude to Earth latitude or vice versa - e.g. finding where on Earth an object of given celestial coordinates is in zenith at given time? The linked article has several formulas of converting between various celestial coordinate systems but either no conversion to Earth coordinates or I failed to understand some of the conversions.

The equatorial plane and the ecliptic plane cross each other, like shown in this illustration:

The "Vernal Equinox" is simply the direction of the line where they cross.

You can not obtain the Earth latitude from the ecliptic latitude alone, as the planes are not parallel. You must know the ecliptic longitude as well.

From equatorial to Earth latitude it is much simpler, the latitude is equal to the declination.


What are the 4 candidates for Planet Nine reported in 2017? From Wikipedia Planet Nine#Citizen science:

  In April 2017, using data from the SkyMapper telescope at Siding Spring Observatory, citizen scientists on the Zooniverse platform reported four candidates for Planet Nine. These candidates will be followed up on by astronomers to determine their viability. The project, which started on 28 March 2017, completed their goals in less than three days with around five million classifications by more than 60,000 individuals.

But it has not been updated for nearly 3 years. Is there any follow up study that confirm what these 4 candidates actually are?

A frequently cited study about Planet Nine is arXiv:1601.05438: Evidence for a Distant Giant Planet in the Solar System, but I can confirm that after 2016 there are not too many studies on the subject, at least I could not find many. There are two interesting studies I would like to share, but none of them spoke of four candidates.

Spitzer Follow-up of Extremely Cold Brown Dwarfs Discovered by the Backyard Worlds: Planet 9 Citizen Science Project. The abstracts reads

We present Spitzer follow-up imaging of 95 candidate extremely cold brown dwarfs discovered by the Backyard Worlds: Planet 9 citizen science project, which uses visually perceived motion in multiepoch Wide-field Infrared Survey Explorer (WISE) images to identify previously unrecognized substellar neighbors to the Sun. [...]

Even more interesting is What If Planet 9 Is a Primordial Black Hole? by Jakub Scholtz and James Unwin from 2020:

We highlight that the anomalous orbits of trans-Neptunian objects (TNOs) and an excess in microlensing events in the 5-year Optical Gravitational Lensing Experiment data set can be simultaneously explained by a new population of astrophysical bodies with mass several times that of the Earth (M⊕). We take these objects to be primordial black holes (PBHs) and point out the orbits of TNOs would be altered if one of these PBHs was captured by the Solar System, inline with the Planet 9 hypothesis. Capture of a free floating planet is a leading explanation for the origin of Planet 9, and we show that the probability of capturing a PBH instead is comparable. The observational constraints on a PBH in the outer Solar System significantly differ from the case of a new ninth planet. This scenario could be confirmed through annihilation signals from the dark matter microhalo around the PBH.

I hope that this is helpful for you.


the length of second, minute or hour, what defines the time of exo planetary bodies we measure time based on earth and earth mass\gravity plays a role in the measurement ? so how and what would be time if not earth time on other exo planets since they will be having different mass/gravity and probably more variables affecting time measurement.

I can say Gliesa has a day of 30 hrs(assumed) ,

So will the length of second, minute and hour, likeewise day or month will fall under its own time system?

You have 2 separate concepts here. The length of a day and a year are unconnected with how you measure seconds, minutes and hours.

The length of a day is defined as how long it takes for the planet to rotate on its axis, and a year is how long it takes to orbit its primary.
Seconds, minutes and hours are a measurement chosen by us back in history, and chances are we will still use the same wherever we go in the universe. The way we measure it is designed to let us always measure seconds consistently.

Even on Earth, we say that a day is 24 hours, but it isn't really, as the two sets of concepts don't match up. We haven't defined a second as a fraction of the time it takes the planet to rotate - it is defined based on physical constants. So if you were to say that Gliesa has a day that is 30 units long - it would be astonishing if those units were exactly hours!


How can I know the source size using the spectrum variability? When reading some astrophysical papers I saw things like this:

The sources are variable on very short timescales, implying a compact emission region.

How can we conclude or measure the source size, or emission region size (or its compactness) from the variability of the spectrum?

A source cannot coherently vary on timescales shorter than the source size divided by the speed of light.
That is because there is no way that different points on the source could "communicate" with each other and arrange for a coordinated increase or decrease in brightness.
Therefore the shortest timescale of variability gives an upper limit to the size of the emission region.


How long was a lunar month in ancient times? As the Moon is getting farther and farther from Earth, it's natural that the time it takes to circle around our planet is going to increase, thus the month will be longer.
However, I read that the reason it's straying from us also is causing the Earth to revolve more slowly. As the day is getting longer, maybe a lunar month will remain more or less the same?
Fast rewind it back to billions of years ago, did that hold true too? How many 'days' did it take for the Moon to go around the Earth, say, 4bya? Are there any formulae to calculate this?
I'd appreciate more the answers that consider both the day length of specific ancient times and its conversion to modern day, for easy imagination. Any sources that list several values for different periods in the past would be great, too.

'Day' is the worst possible unit to measure the revolution of the Moon around Earth as it is the temporal unit which is directly influenced by the tidal friction and deceleration. We can reasonably assume that the length of the year is not impacted by lunar tides and subject to much less variation.
This is only a partial answer as I will have to leave the math to get an actual number for time reasons.
The only measurable quantity for the past is the length-of-day of Earth and the tidal rythm by means of their impact on biorythms and in turn its impact on sediment stratigraphy and growth markers e.g. in fossil shells and similar.
One summary with some values and further references on how the past rotation can be estimated is this paper by Deines & Williams. If you know the length-of-day at a time, you can calculate the yearly angular deceleration of Earth. That knowledge can be employed, using the conservation of angular momentum and calculate the orbit of the moon and thus length of month (thus one full orbit around Earth).


Any tip on maintaining a log of celestial objects? I am an amateur observer and Olympus 10x50 binocular is my tool. As I keep locating celestial objects (Planets, Stars etc), I would like to maintain a log of the observed objects. The purpose of the log is to list down the number of celestial objects I have observed/located. I need suggestion on what data points should I capture which helps in making some sense out of it and also encourage my friends to get into star gazing.

I suggest you to take a look at the Amateur Astronomy Observers Log Web Site, where everybody can share their astronomical logs.

The logs contain:

Instrumentation used
Sky condition (seeing, light pollution, ...)
Accurate date and time of the observation

Specific informations that you could add depend by the kind of object that you are watching. For example you could try to estimate the magnitude of a variable star, or you could describe colour variation of Jupiter bands.

Some software can give you more support. There are specific astronomical log software for smartphone (for example Stargazing Log for Android).
Morover in the Linux planetarium software KStars when you open the detail of an object you have a specific tab where you can register logs in form of simple text.


Calculating distance of comet C/2019 Y4 (ATLAS) SO, I decided I would take some pictures of the comet C/2019 Y4 last night. Afterwards, I thought I would try and do a rough estimation of distance based on my observations. I don't have any fancy measuring tools, so a few approximations were made, but my result was way off. I've clearly done something wrong and was hoping someone could point out where I have gone wrong and correct me.

So, I basically layered my first picture over the second in photoshop and marked the start and end position over the course of 1 hour.

I apologise for the poor photo, the exposures were short and I haven't done any processing as I was only interested in the position.

So,to me, it looks like it has moved approximately he same distance as the 2 stars to the right. These are TYC 4360-0254-1 and TYC 4360-1679-1 according to Sky Safari Pro.

I used the measurement tool in the app to measure the distance in arc-minutes, which gave me 1'55.1"

I converted this into degrees using Google, which gave me 0.0258333333 degrees.

I then decided to work out the approximate distance between my start and end points by using my latitude to work out the speed the earth was rotating (cos(latitude) x 1670) which I found HERE.

My latitude is approx 50.82N, so I got my speed at 1055.04km. I used that as a distance. I now have a triangle with a base of 1055.04km and a top angle of 0.02583 degrees.

I turned this into a right-angle triangle by putting a line down the middle. I can now work out all 3 angles. I have 90, 0.010129 (half the original angle) and 89.987 (90-0.010129) I also have an opposite, adjacent and hypotenuse. I only know the adjacent which is half my original distance (527.52km), and I want to find the Opposite.

Using SOHCAHTOA, I can say tan(89) = opp/adj, thus opp = 527.52 x tan(89.987) which gives me 2,324,974.545km.

Sky safari gives a distance of 153.1 million km.

So where did I go wrong? Was it because of poor approximations? I know that the linear distance for the adjacent side isn't quite accurate because of earth curve, and I know the comet was also moving, and I know my measurement of the start and end position wasn't too accurate, but I was hoping to at least get within 10%, so I think I must have gone wrong somewhere in my method, rather than poor approximations.

I do hope I have put enough effort in for this to be considered a good question. I have tried re-measuring and double checking, but if there is any other info people need, then by all means, ask and I'll do what I can

EDIT

Because of the comment by barrycarter, I realised I had forgotten the movement of earth itself. As this is approximately 107207km/h, I included this in my calculations and ended up overshooting and getting 238,577,507km.

So, I made tiny adjustments to my parallax angle, and if I change this slightly, then I get very different answers.

Based on that, is it possible my crude measurement of the parallax angle is to blame?

It appears that you are doing a simple parallax calculation, using a baseline obtained from the Earth's rotation over an hour. You appear to be neglecting the most significant change, the motion of the comet.

If you plot the position of C/2019 Y4 in SkySafari, you will see that around this date the comet's location against the star background shifts about 37 arc minutes per day, or a bit over 1.5 arc minutes per hour. The location is based on the orbital elements of the comet (MPEC 2020-C76), but also includes the effect of the Earth moving in its orbit, as mentioned in the comments.

There is no simple way of determining the distance of a solar system object from two position measurements alone. A rough estimate of the orbital elements requires at least three position and time measurements; the MPC orbital elements referenced above are based on many observations over several months.


Is there a threshold on distance/size for a tidal locking? I know that some systems tend to tidal locking (such as earth-moon), which occurs basically because the difference in the gravitational pull on one side is significantly different from the pull on the opposite side. (There's a lot of great answers here.) But what can be considered a significant different?
In other words, is there a threshold on distance or size in order for systems to tend to tidal lock?

Thanks!

You can see an example of tidal locking and atmosphere simulation for a planet closely orbiting a dim star. They show a simulation of the atmosphere and some interesting theories about the movement of gasses due to tidal locking (convection) that occurs between the bright and dark side of the planet.  The link goes directly to the discussion of tidal locking:

KEPLER 186F - LIFE AFTER EARTH

A similar question was posted in the physics forum and it looks like that could be close to the answer you are looking for:

Tidal Lock Radius in Habitable Zones

I am also interested in tidal locking and hope someone may post a more concise answer.


How was calculating the celestial latitude using an astrolable of any help to sailors? I'm sorry if this sounds like a silly question but I don't understand how measuring the location of a star would help in sea navigation.

Measuring the angle to a known star allows the navigator to calculate the latitude of the ship in the absence of any visible (known) landmarks.

Since they would know (or refer to an almanac) the celestial latitude (declination) or altitude of the star at their home port or some other known port, they can measure the position of the star and take the difference between the measurement and the reference to calculate their latitude at sea.

For example, if their home port were located at 54°N then the angle from the horizon to Polaris would be 54°. If they take a measurement of Polaris while at sea and find it to be 30° above the horizon then they know that they are 24° farther south, i.e. latitude 30°.

Similarly, if there were a star that is directly overhead (90° from the horizon) when it transits the celestial meridian at home, then they know that that star has a declination of 54°. If they take a measurement of it while at sea and find it to be 6° north of zenith (84° from the horizon) then they know that they are 6° south of home.

From all of this, knowing the date lets them calculate the sunrise and sunset times, and if they have an accurate clock they can work out their longitude. In the past, longitude was a bit more guesswork as clocks were not all that accurate at sea.


Precision in the measurement of the distance to the Sun How precisely can the distance to the Sun be measured? Wikipedia says the distance to the Moon can be measured upto millimeter precision. But Wikipedia article on distance to the Sun says only about Astronomical Unit and nothing on the precision of the measurement of the distance to the Sun. I am certainly aware that we now use radar to measure the distance to the Sun, and I remember reading the precision in its measurement somewhere on the Internet, but I can't find it anymore. When I try to find the relevant information on the Internet, all I find is educational articles on things such as parallax, which is certainly superceded by radar measurements, or articles on lunar distance, which I am not looking for. Relevant references will highly be appreciated.

Edit: I just found out on Internet that we measure the distance to the Sun through distance to the Venus or Mercury. Anyway I want to know the precisions or errors in these measurements..

This is a partial answer too long for a comment, but it may help to get the ball rolling...
Starting from this answer we can see that the standard gravitational parameter of the Sun used by JLP in their development ephemerides is 1.32712440040944E+20. While that doesn't mean we know the Earth's orbit to 1 part in $10^{14}$ it hints that it's believed to be known surprisingly well!
It's possible that something like this might shed some light on the subject, but it's not an easy question to answer. The Planetary and Lunar Ephemeris DE 430 and 431, IPN Progress Report 42-196 (February 2014)  Part of the problem is that we don't know exactly where the solar system barycenter is with respect to the Sun because there may be bodies far away we haven't detected yet. That's only a small effect on the Sun-Earth orbit. If I had to guess, I'd say it's somewhere between 1 and 100 meters uncertainty between the Earth and the center of the Sun. Where the edge of the Sun falls is a horse of a different color!


What is the minimum distance from city that allow to see Milky Way plane? The light pollution fades the starlight coming from the Sky and this dont let people to see the Milky Way disc. Is there any study about this?

The premise is certainly true. An exact distance isn't possible to determine since it would depend on how much light pollution is coming from the city, and how much haze (water droplets and particulates) is in the air. You could also be a "short" distance away but on a mountain with much greater elevation than the city.


What is it that I have been seeing in the sky? So I'm interested in space scientifically. I do have a firm belief that there could be intelligent civilizations in our universe. Just wanted to get that out of the way.

I've been spending the evenings in the garden due the weather. Past two days I've seen the same similar thing and I'm not sure what it is. I have observed many 'shooting stars', planets, satellites, and aircraft. It's just one solid bright white light. I actually checked it with binoculars I had in the kitchen an yeah all I can see is bright white light. It is much brighter and larger to my eye than satellites appear. It is about 22:00 when I see these it is pretty dark but I wouldn't say peak darkness. It moves steadily faster than large aircraft like commercial flights that I can see. It moves in a straight line I think and is fully visible but just as that bright light.

I could possibly go to lengths and get images of it if anyone wants I took video on both occasions. They are pretty clear and I tried adjusting stuff on the camera to improve the quality while filming.

Anyway, would be interested in what it could be obviously. Responses will be appreciated.

EDITS:

Okay, so the town I'm in is called Colchester, United Kingdom.
I was facing West, it came from above the cloud, was to the right side of Bootes. Carried on in a straight line across the whole sky travelling East until it was no longer visible. The light remained bright the entire time.

Oh, and 22:00, GMT.

I am uploading the videos and images, I will link to it.

The International Space Station can appear as bright as Venus, typically crosses the sky in about 6 minutes, and passes directly over the UK several evenings this week.
Heavens Above provides a table of ISS passes visible from a given location.
Their chart of a July 25 pass is consistent with your revised description.


Searching for stars with a given surface temperature and an absolute magnitude? Do you have in mind any database/sites that I can search for stars that matches a given surface temperature and an absolute magnitude?

For example, when I enter Temperature=9,500K & AbsoluteMagnitude=0.58 then the search engine returns the result Vega and some other stars.

Any suggestions are appreciated!

Though not much convenient, I found one.

https://github.com/astronexus/HYG-Database

This is a database that contains information of a lot of stars.

I may use excel to use this database and derive any needed results.


Why the forbidden lines of [OIII],[NII] and [SII],[OII] are sensitive to electron temperature and electron density, respectively? I am trying to understand the topic of "Metallicity estimates", either HII regions or planetary nebula. For this latter, the electron temperature (and density) must be known. I am studying the formalism of Osterbrock.

I know, for instance, that if you use the lines [OIII]4363 and [OIII]5007 the electron temperature can be computed. In the same way, the ratio between [OII]3729 and [OII]3726 provides an estimation of electronic density.

Actually, I also know that, for the case of [OIII], its metastable excited states are relatively separated. For the case of [OII], such metastable excited states are very close. Such separation, in the metastable excited states, is crucial to compute the temperature and density.

My point is that I do not understand why this separation is important, from a physical point of view. Why the ratio between these lines can estimate such properties?

You can check examples of what I am talking about in "Electronic Density" and "Electronic Temperature" sections of this link: http://www.astronomie-amateur.fr/feuilles/Spectroscopie/NGC2392.html

I hope to make me understood.

Thank you very much for your help.

Forbidden lines like these arise in thin gases, where collisional de-excitation is unlikely. Forbidden lines become "quenched" by collisional de-excitation when the densities reach levels that depend on the detailed quantum physics of each transition (i.e. different for every line/transition).

If you study the ratio of a pair of lines at densities well below their "quenching thresholds", the ratio will depend only on the relative occupation of their upper levels, which in turn depends on the temperature, but is insensitive to density.

If on the other hand you choose a pair of lines with very similar upper energy levels, the ratio will be insensitive to temperature. However, if one transition is around or even above its "quenching threshold", then the ratio becomes very sensitive to density.


Moving stars in the sky Today at 7:24 pm, I saw to stars moving parallel to each other for a few seconds in the sky. Then, both suddenly vanished. What phenomenon is this?

Generally bright points of light moving across the sky at a constant speed are satellites. Meteors are much faster and usually have a trail.
If you saw two or more close together and moving at the same speed they were probably some of the Starlink satellites launched by SpaceX.
The sudden vanishing is usually because they either moved into Earth's shadow, or the angle of the reflecting surfaces moved so that they didn't reflect light to you as an observer.


Do elliptical galaxies look similar to globular clusters? If we got close enough to an elliptical galaxy would it look similar to a globular cluster?

Your question is not very well defined since you don't say what observational properties you wish to compare. From your emphasis on distance, I assume that you want to compare the visual appearance of elliptical galaxies with globular clusters of the same angular extent on the sky.

In which case ellipticals and globulars are easily distinguishable with the aid of a telescope (I assume you do not mean naked eye objects, since the nature of globular clusters is pretty difficult to make out with the naked eye, but if you do mean that then of course most smudges on the sky are indistinguishable with the naked eye). The reason for this is that true elliptical galaxies contain factors of $\sim 10^4$ more stars than even the most massive globular clusters and are about 100 times the physical diameter. Thus, if they had the same angular extent, an elliptical galaxy be $\sim 100$ times further away, would look much smoother and less grainy than a globular cluster, because it contains $10^4$ times more stars per unit area, but would have around the same surface brightness.

Of course there are such things as dwarf spheroidal galaxies, but here the problem is the other way around. They have about the same number of stars as globular clusters, but have diameters a factor of 100 larger. Thus if they had the same angular extent they would look similar to a globular cluster in terms of "graininess", but since they would be 100 times further away, they would be $10^4$ times fainter!


Is there a simple, analytical formula for the lunar phase brightness curve? Brightness of the lunar disk varies as a function of phase angle. This phenomenon is very well established, as is the increase in brightness at low phase angles.

Is there a simple analytical formula that describes this function? It looks like brightness could be modeled as an exponential as a function of phase angle, modulo the opposition surge.

Thanks.

EDIT: I realize that these effects depend on the wavelength of the reflected light. I'm looking for an average curve, though.

The magnitude of the Moon without the opposition surge is given by a simple equation:
$m = -12.73 + 1.49 \cdot |\psi| + 0.043 \cdot \psi^4,$
where $\psi$ is the phase angle in radians (Allen 1976).  This can be converted to flux using $m \propto -2.5 \cdot ^{10}\!\log F$,
where $^{10}\!\log$ denotes the 10-base logarithm (as opposed to e.g. the natural log).
For the increase in flux due to the variable distance and opposition surge, I multiply F with the factor
$\left(\frac{\Delta_0}{\Delta}\right)^2  \cdot  \max\left(1, 1.35 - 2.865 \cdot |\psi| \right),$
where $\Delta_0$ and $\Delta$ are the mean and current distance to the Moon, respectively.  However, I can't find a reference for the last term at the moment, and I can't remember where I obtained it.  I'll post that information if I find it.


Strange trail left by shooting star I saw something really odd last night that I cannot understand.
Last night I got up around 3am to catch a glimpse of the upcoming Perseids meteor shower, now I know it's not peaking until next week but we have clear skies so I thought I'd take a look. I've seen hundreds of meteor showers, so know what to expect.
Whilst looking at another part of the sky a large shooting star radiated from the Perseus constellation, I just missed it! However, what was left in the sky was most odd, a really long dotted line of what looked like dust trail from the meteor. Usually, these trails stay where there are, and slowly fade away. But this trail didn't fade and very slowly moved across night the sky in an absolutely perfectly straight line until it faded into the horizon.
From the moment I first saw it to when it fully disappeared into the horizon was probably about 30 seconds or more. What did I see?
(Gloucestershire, UK).

Thanks for your replies. Yes it appears what I saw was Elon Musk's Starlink satellites. I didn't even know they existed, what a sight! Thank you. Chris