Is there a way to factor age into the mass-luminosity relationship for stars? I'm wondering if there's a way to formulate the luminosity of a star not only as a function of mass, but of age, too, and if so, what the formula for luminosity would look like. In the case of the Sun I'm aware it gets 10% brighter every billion or so years, but I doubt this relationship holds true for all mass ranges of stars, just like you have to adjust the formula for the luminosity mass relationship for stars of different masses.

If anyone could "shed some light" on this matter, it would be very much appreciated.

What you are looking seems to be the vertical path of a star in the Hertzsprung-Russel diagram (HRD).

The only problem is that stellar evolution is pretty complicated. See here a few simulated trajectories for different masses and chemical compositions from  Bertelli et al. 2008

Especially before and after the main sequence (i.e., regular boring hydrogen burning) the luminosity evolution becomes extremely variable.

From the simulations above one can also derive respective age-luminosity relations (taken from  Danchi 2013):

Again, you can see that the relation is very complicate for young or old stars and is fairly constant during the main sequence. The problem is, that it changes too little during most of the main sequence. So for most stars even with good mass and metallic measurement you won't be able to accurately estimate age just from luminosity. In addition to that, there are still uncertainties in our models.

Another thing one could do is trying to improve the mass-luminosity-relation by including age. I think this is also what the title of your question implies. The problem here it simply that we generally don't know the age of a star. But if we do, as Rob Jeffries explained in his answer, it is in principle possible.

One think that is done to get age from luminosity is to go all in and calculate so called isochrones in the HRD. These are lines with stars of same age but varying masses and metallicities and can be derived from simulations. If one then measures the luminosity, temperature and metallic one can look on which isochrone the star falls in the HRD (and  therefore what age it has). This however is still pretty inaccurate, especially on the main sequence, and is mostly done with whole star cluster, where statistics makes things easier.

This is however not my field of expertise so I would be happy it an actual expert could chime in. :)


Dropping a pebble into a black hole If one were to drop, say, a pebble, from a sufficient distance, into a black hole. The pebble would continue to accelerate until it were just short of the speed of light. In the next moment it's speed should exceed the speed of  light. Why not?

Gravity is no different from any other force. Doesn't matter it comes from a black hole, a star, a planet, or a lump of cheese floating in outer space. It's just a force, like any other.

Put a very powerful rocket engine on the pebble. In a short time it would accelerate the pebble near the speed of light. In the next moment it should exceed the speed of light. Why not?

Because that's not how it works. As you accelerate more and more, time dilation and space contraction also get stronger, so it gets harder and harder to get even closer to the speed of light, no matter what's pushing or pulling the pebble, be that a giant rocket engine or the enormous gravity of a large black hole. Same thing. The faster the pebble goes, the more warped space and time become, the harder it is to increase the speed.

It cannot even reach the speed of light because then time dilation, etc., would become infinite, effectively preventing it from reaching speed of light.

You cannot cheat relativity.

Note: This is not a rigorous explanation but it's close enough and should provide some intuitive understanding of what's going on.


Is it possible for two stars to be in a horseshoe orbit around a much larger star? I was reading about how Saturn has two moons, Janus and Epimetheus, that swap orbits once every four years. Could something like this happen on a much larger scale, but with stars instead of a planet and moons? Let's say you had a large, massive star with two red dwarfs orbiting around it. Could those two red dwarf stars fall into a horseshoe orbit, regularly swapping orbits every few years or so?

more about horseshoe orbits

Yes, it is possible, however such orbits are likely to be unstable.

For example in the paper

  Ćuk, Matija, Douglas P. Hamilton, and Matthew J. Holman. Long-term stability of horseshoe orbits. Monthly Notices of the Royal Astronomical Society 426.4 (2012): 3051-3056. doi:10.1111/j.1365-2966.2012.21964.x

it was found that if the mass of second body becomes larger than 1/1200 of the first body, horseshoe orbits become unstable. (Third body was assumed to have negligible mass).

Nonzero mass of the third body was investigated recently in a thesis:

  Balaji, Bhaskaran. Mass ratios in stellar triple systems that admit horseshoe orbits. Diss. Massachusetts Institute of Technology, 2016. abstract, pdf

and it was found that in addition to constraint of Ćuk et al. nonzero mass of the third body  further decreases stability of such systems, so that the lifetime of horseshoe orbit $\tau \propto m_3^{-1}$.

For stars only coupling of more massive giants with lighter  red dwarfs would fulfill  first constraint and then the third star would never have 'negligible' mass, one could conclude that horseshoe orbits when all three bodies are stars are always unstable. And for larger masses of smaller stars the lifetime of such orbits would be too short to expect such orbits in our vicinity.

However it would probably be possible to have stable horseshoe orbits of two stars (say a blue giant and red dwarf) and a planet.


Planetary observations: what to expect from the Powerseeker 50AZ? I bought the Powerseeker 50AZ, which has D=50mm; F=600mm F/12 which comes with 3 eyepieces: 20mm, 12.5mm and 4mm.

As far as I can tell it's not possible to colimate this telescope. As suggested here this is a very important part to see sharp images.

I have tried to observe jupiter and mars with it and jupiter I could see a dim small blurry image and mars a very tiny dot with absolutely no way of seeing details on any of those in what seemed the best focus I could get from them.
The only thing I can see details is the moon really. Also I have observed in what seemed very clear sky conditions.

So my question is:
How much can I expect to see from planets with this telescope?
With this current setup should I be seeing jupiter and mars in more details?

UPDATE:
Equipment link

https://www.amazon.com/Celestron-PowerSeeker-50-Refractor-Telescope/dp/B0000UMLYI/ref=sr_1_1?ie=UTF8&qid=1515177254&sr=8-1&keywords=powerseeker%2B50az%2Btelescope&th=1

Currently, Mars is very far from the Earth, so it appears to be very small. I doubt that any amateur telescope will show any details on Mars in this condition. Perhaps when it comes to opposition and is close to Earth, you might be able to see some faint markings with your scope.

Jupiter should show a sharp disk and nearly pinpoint moons with the 12.5 mm eyepiece (48 power). You should see two prominent bands on Jupiter.

The 4 mm eyepiece is probably poor quality, and since it provides 150 power, is too much power for a 50 mm scope. In ideal conditions, the general rule is 100 power for a 50 mm scope (2 times the scope size in mm). The clarity of the sky is not what makes a sharp image. There is a condition named "seeing" that indicates how sharp the viewing is. On some nights, the seeing is poor, so the images will be soft (not sharp) and perhaps shimmering, as if you are viewing the object through a stream of water. (Essentially, you are, except it is miles or kilometers of air!)


2SB mixer in radio astronomy? In literature about the instrumentation of radio astronomy, I frequently come across a type of mixer described as "2SB".  I'm familiar with single-sideband (SSB) and double-sideband (DSB) mixers, but I have never studied a 2SB mixer.

What is a 2SB mixer and how does it compare to a DSB mixer?  Ideally, I would also like to know of formal references that directly address the topic for further reading.

2SB is Dual Sideband, as opposed to DSB - Double Sideband. Here are a couple of papers you might find relevant and interesting:
a 2SB upgrade replacing a DSB
a paper that mentions the advantages of 2SB over DSB
Answers to What is the functional difference between DSB and 2SB mixers, and why exactly would 2SB be better for radio astronomy? in Signal Processing SE go into further detail.


If galaxies are moving away from each other then why are the Milky Way and Andromeda galaxy coming towards each other? The Andromeda Galaxy is approaching the Milky Way at about 684000 mi/hours, making it one of the few blueshifted galaxies. The Andromeda Galaxy and the Milky Way are thus expected to collide in about 3.75 or 4.5 billion years.

Why are some galaxies moving away and why are our galaxy and Andromeda coming towards each other?

The farther away, the faster the galaxies move away from us. But that's only the overall expansion of the universe. Locally velocities can differ and are to be added to the overall expansion. The Andromeda Galaxy is close enough to Milky Way to be able overcome the overall expansion of space. The Andromeda Galaxy and the Milky Way attract each other by gravity.

Hubble's law applied to Andromeda Galaxy/Milky Way returns an expansion of space of less than 70 km/s, since Hubble's constant is about 70 km/s/Mpc, and the distance is about 2.5 million light-years, less than 1 Mpc (= 3.26 light-years, see Parsec). That speed is slower than the about -300 km/s helio-radial velocity (space expansion already subtracted) of the Andromeda galaxy, hence feasible to overcome.


Aligning the equinoxes to the cardinal points on a circular calendar I want to draw a calendar as a circle, with the equinoxes at the top and bottom, and the solstices at the left and right. I am prepared to accept a little inaccuracy by having exactly 365 or 366 days in the circle depending on if it is a leap year.
If I just divide the circle into 365 equal segments, then it does not line up. I can get the solstices to line up very closely, but the spring equinox always comes in 2 days late, and the autumnal equinox 2 days early.
I think this is to do with the elliptical orbit of the Earth around the Sun, and the relationship of the axes of the ellipse to the axial tilt of the Earth.
Can anyone help me with the mathematics required to adjust the angle taken up by each day in order to skew the days such that the equinoxes and solstices fit to the cardinal points of a circle?

As Mike G said, the Sun's ecliptic longitude is ideal for this application.
I see from your profile that you're a coder, with some knowledge of Python. Here's a short Python script that prints solar ecliptic longitudes for each day for any given year. (It actually prints 367 days, I figured an extra day or two may be useful). The code mostly avoids features peculiar to Python, so it should be fairly easy to translate into other languages.
My code uses the algorithm that Mike G has mentioned from Wikipedia. (I also tried another algorithm from that page, but it's not sufficiently accurate).
"""
 Approx Sun ecliptic longitude

 Prints daily positions for a year.
 For dates between 1950 and 2050,
 it should be accurate to within
 0.01 degrees. For details, see
 https://en.wikipedia.org/wiki/Position_of_the_Sun#Ecliptic_coordinates

 Written by PM 2Ring 2020.12.6
"""

from math import sin, radians
from datetime import timedelta, date

# Approximate solar ecliptic longitude
# Uses day zero = Noon UTC, 1 Jan 2000
# so we subtract half a day
def eclon_approx(n):
    n -= 0.5
    L = 280.46 + 0.9856474 * n
    g = 357.528 + 0.9856003 * n
    gr = radians(g)
    el = L + 1.915 * sin(gr) + 0.020 * sin(2 * gr)
    return el % 360

oneday = timedelta(days=1)
day_zero = date(year=2000, month=1, day=1)

@interact
def main(year=2000):
    day = date(year=year, month=1, day=1)
    d = (day - day_zero).days
    for i in range(367):
        angle = eclon_approx(d)
        print('{:03d}, "{:%b-%d}", {:6.2f}'.format(i, day, angle))
        day += oneday
        d += 1

Here's the output for 2021:
000, "Jan-01", 280.79
001, "Jan-02", 281.80
002, "Jan-03", 282.82
003, "Jan-04", 283.84
004, "Jan-05", 284.86
005, "Jan-06", 285.88
006, "Jan-07", 286.90
007, "Jan-08", 287.92
008, "Jan-09", 288.94
009, "Jan-10", 289.96
010, "Jan-11", 290.98
011, "Jan-12", 292.00
012, "Jan-13", 293.02
013, "Jan-14", 294.03
014, "Jan-15", 295.05
015, "Jan-16", 296.07
016, "Jan-17", 297.09
017, "Jan-18", 298.11
018, "Jan-19", 299.13
019, "Jan-20", 300.14
020, "Jan-21", 301.16
021, "Jan-22", 302.18
022, "Jan-23", 303.20
023, "Jan-24", 304.21
024, "Jan-25", 305.23
025, "Jan-26", 306.25
026, "Jan-27", 307.26
027, "Jan-28", 308.28
028, "Jan-29", 309.30
029, "Jan-30", 310.31
030, "Jan-31", 311.33
031, "Feb-01", 312.34
032, "Feb-02", 313.36
033, "Feb-03", 314.37
034, "Feb-04", 315.39
035, "Feb-05", 316.40
036, "Feb-06", 317.41
037, "Feb-07", 318.43
038, "Feb-08", 319.44
039, "Feb-09", 320.45
040, "Feb-10", 321.47
041, "Feb-11", 322.48
042, "Feb-12", 323.49
043, "Feb-13", 324.50
044, "Feb-14", 325.51
045, "Feb-15", 326.52
046, "Feb-16", 327.53
047, "Feb-17", 328.54
048, "Feb-18", 329.55
049, "Feb-19", 330.56
050, "Feb-20", 331.57
051, "Feb-21", 332.58
052, "Feb-22", 333.58
053, "Feb-23", 334.59
054, "Feb-24", 335.60
055, "Feb-25", 336.60
056, "Feb-26", 337.61
057, "Feb-27", 338.61
058, "Feb-28", 339.62
059, "Mar-01", 340.62
060, "Mar-02", 341.63
061, "Mar-03", 342.63
062, "Mar-04", 343.63
063, "Mar-05", 344.63
064, "Mar-06", 345.64
065, "Mar-07", 346.64
066, "Mar-08", 347.64
067, "Mar-09", 348.64
068, "Mar-10", 349.64
069, "Mar-11", 350.64
070, "Mar-12", 351.64
071, "Mar-13", 352.63
072, "Mar-14", 353.63
073, "Mar-15", 354.63
074, "Mar-16", 355.62
075, "Mar-17", 356.62
076, "Mar-18", 357.62
077, "Mar-19", 358.61
078, "Mar-20", 359.60
079, "Mar-21",   0.60
080, "Mar-22",   1.59
081, "Mar-23",   2.58
082, "Mar-24",   3.58
083, "Mar-25",   4.57
084, "Mar-26",   5.56
085, "Mar-27",   6.55
086, "Mar-28",   7.54
087, "Mar-29",   8.53
088, "Mar-30",   9.51
089, "Mar-31",  10.50
090, "Apr-01",  11.49
091, "Apr-02",  12.48
092, "Apr-03",  13.46
093, "Apr-04",  14.45
094, "Apr-05",  15.43
095, "Apr-06",  16.42
096, "Apr-07",  17.40
097, "Apr-08",  18.39
098, "Apr-09",  19.37
099, "Apr-10",  20.35
100, "Apr-11",  21.33
101, "Apr-12",  22.31
102, "Apr-13",  23.29
103, "Apr-14",  24.27
104, "Apr-15",  25.25
105, "Apr-16",  26.23
106, "Apr-17",  27.21
107, "Apr-18",  28.19
108, "Apr-19",  29.17
109, "Apr-20",  30.14
110, "Apr-21",  31.12
111, "Apr-22",  32.10
112, "Apr-23",  33.07
113, "Apr-24",  34.05
114, "Apr-25",  35.02
115, "Apr-26",  35.99
116, "Apr-27",  36.97
117, "Apr-28",  37.94
118, "Apr-29",  38.91
119, "Apr-30",  39.88
120, "May-01",  40.85
121, "May-02",  41.82
122, "May-03",  42.79
123, "May-04",  43.76
124, "May-05",  44.73
125, "May-06",  45.70
126, "May-07",  46.67
127, "May-08",  47.64
128, "May-09",  48.61
129, "May-10",  49.57
130, "May-11",  50.54
131, "May-12",  51.50
132, "May-13",  52.47
133, "May-14",  53.44
134, "May-15",  54.40
135, "May-16",  55.36
136, "May-17",  56.33
137, "May-18",  57.29
138, "May-19",  58.25
139, "May-20",  59.22
140, "May-21",  60.18
141, "May-22",  61.14
142, "May-23",  62.10
143, "May-24",  63.06
144, "May-25",  64.03
145, "May-26",  64.99
146, "May-27",  65.95
147, "May-28",  66.91
148, "May-29",  67.87
149, "May-30",  68.82
150, "May-31",  69.78
151, "Jun-01",  70.74
152, "Jun-02",  71.70
153, "Jun-03",  72.66
154, "Jun-04",  73.62
155, "Jun-05",  74.57
156, "Jun-06",  75.53
157, "Jun-07",  76.49
158, "Jun-08",  77.45
159, "Jun-09",  78.40
160, "Jun-10",  79.36
161, "Jun-11",  80.31
162, "Jun-12",  81.27
163, "Jun-13",  82.23
164, "Jun-14",  83.18
165, "Jun-15",  84.14
166, "Jun-16",  85.09
167, "Jun-17",  86.05
168, "Jun-18",  87.00
169, "Jun-19",  87.96
170, "Jun-20",  88.91
171, "Jun-21",  89.86
172, "Jun-22",  90.82
173, "Jun-23",  91.77
174, "Jun-24",  92.73
175, "Jun-25",  93.68
176, "Jun-26",  94.63
177, "Jun-27",  95.59
178, "Jun-28",  96.54
179, "Jun-29",  97.50
180, "Jun-30",  98.45
181, "Jul-01",  99.40
182, "Jul-02", 100.36
183, "Jul-03", 101.31
184, "Jul-04", 102.26
185, "Jul-05", 103.22
186, "Jul-06", 104.17
187, "Jul-07", 105.12
188, "Jul-08", 106.08
189, "Jul-09", 107.03
190, "Jul-10", 107.98
191, "Jul-11", 108.94
192, "Jul-12", 109.89
193, "Jul-13", 110.84
194, "Jul-14", 111.80
195, "Jul-15", 112.75
196, "Jul-16", 113.71
197, "Jul-17", 114.66
198, "Jul-18", 115.61
199, "Jul-19", 116.57
200, "Jul-20", 117.52
201, "Jul-21", 118.48
202, "Jul-22", 119.43
203, "Jul-23", 120.39
204, "Jul-24", 121.34
205, "Jul-25", 122.30
206, "Jul-26", 123.25
207, "Jul-27", 124.21
208, "Jul-28", 125.16
209, "Jul-29", 126.12
210, "Jul-30", 127.07
211, "Jul-31", 128.03
212, "Aug-01", 128.99
213, "Aug-02", 129.94
214, "Aug-03", 130.90
215, "Aug-04", 131.86
216, "Aug-05", 132.81
217, "Aug-06", 133.77
218, "Aug-07", 134.73
219, "Aug-08", 135.69
220, "Aug-09", 136.65
221, "Aug-10", 137.61
222, "Aug-11", 138.57
223, "Aug-12", 139.52
224, "Aug-13", 140.48
225, "Aug-14", 141.44
226, "Aug-15", 142.41
227, "Aug-16", 143.37
228, "Aug-17", 144.33
229, "Aug-18", 145.29
230, "Aug-19", 146.25
231, "Aug-20", 147.21
232, "Aug-21", 148.18
233, "Aug-22", 149.14
234, "Aug-23", 150.10
235, "Aug-24", 151.07
236, "Aug-25", 152.03
237, "Aug-26", 152.99
238, "Aug-27", 153.96
239, "Aug-28", 154.93
240, "Aug-29", 155.89
241, "Aug-30", 156.86
242, "Aug-31", 157.82
243, "Sep-01", 158.79
244, "Sep-02", 159.76
245, "Sep-03", 160.73
246, "Sep-04", 161.70
247, "Sep-05", 162.67
248, "Sep-06", 163.63
249, "Sep-07", 164.60
250, "Sep-08", 165.58
251, "Sep-09", 166.55
252, "Sep-10", 167.52
253, "Sep-11", 168.49
254, "Sep-12", 169.46
255, "Sep-13", 170.44
256, "Sep-14", 171.41
257, "Sep-15", 172.38
258, "Sep-16", 173.36
259, "Sep-17", 174.33
260, "Sep-18", 175.31
261, "Sep-19", 176.28
262, "Sep-20", 177.26
263, "Sep-21", 178.24
264, "Sep-22", 179.22
265, "Sep-23", 180.19
266, "Sep-24", 181.17
267, "Sep-25", 182.15
268, "Sep-26", 183.13
269, "Sep-27", 184.11
270, "Sep-28", 185.09
271, "Sep-29", 186.08
272, "Sep-30", 187.06
273, "Oct-01", 188.04
274, "Oct-02", 189.02
275, "Oct-03", 190.01
276, "Oct-04", 190.99
277, "Oct-05", 191.98
278, "Oct-06", 192.96
279, "Oct-07", 193.95
280, "Oct-08", 194.94
281, "Oct-09", 195.92
282, "Oct-10", 196.91
283, "Oct-11", 197.90
284, "Oct-12", 198.89
285, "Oct-13", 199.88
286, "Oct-14", 200.87
287, "Oct-15", 201.86
288, "Oct-16", 202.85
289, "Oct-17", 203.84
290, "Oct-18", 204.83
291, "Oct-19", 205.83
292, "Oct-20", 206.82
293, "Oct-21", 207.81
294, "Oct-22", 208.81
295, "Oct-23", 209.80
296, "Oct-24", 210.80
297, "Oct-25", 211.80
298, "Oct-26", 212.79
299, "Oct-27", 213.79
300, "Oct-28", 214.79
301, "Oct-29", 215.79
302, "Oct-30", 216.79
303, "Oct-31", 217.79
304, "Nov-01", 218.79
305, "Nov-02", 219.79
306, "Nov-03", 220.79
307, "Nov-04", 221.79
308, "Nov-05", 222.79
309, "Nov-06", 223.79
310, "Nov-07", 224.80
311, "Nov-08", 225.80
312, "Nov-09", 226.80
313, "Nov-10", 227.81
314, "Nov-11", 228.81
315, "Nov-12", 229.82
316, "Nov-13", 230.83
317, "Nov-14", 231.83
318, "Nov-15", 232.84
319, "Nov-16", 233.85
320, "Nov-17", 234.85
321, "Nov-18", 235.86
322, "Nov-19", 236.87
323, "Nov-20", 237.88
324, "Nov-21", 238.89
325, "Nov-22", 239.90
326, "Nov-23", 240.91
327, "Nov-24", 241.92
328, "Nov-25", 242.93
329, "Nov-26", 243.94
330, "Nov-27", 244.96
331, "Nov-28", 245.97
332, "Nov-29", 246.98
333, "Nov-30", 247.99
334, "Dec-01", 249.01
335, "Dec-02", 250.02
336, "Dec-03", 251.03
337, "Dec-04", 252.05
338, "Dec-05", 253.06
339, "Dec-06", 254.08
340, "Dec-07", 255.09
341, "Dec-08", 256.11
342, "Dec-09", 257.12
343, "Dec-10", 258.14
344, "Dec-11", 259.16
345, "Dec-12", 260.17
346, "Dec-13", 261.19
347, "Dec-14", 262.21
348, "Dec-15", 263.22
349, "Dec-16", 264.24
350, "Dec-17", 265.26
351, "Dec-18", 266.28
352, "Dec-19", 267.29
353, "Dec-20", 268.31
354, "Dec-21", 269.33
355, "Dec-22", 270.35
356, "Dec-23", 271.37
357, "Dec-24", 272.39
358, "Dec-25", 273.40
359, "Dec-26", 274.42
360, "Dec-27", 275.44
361, "Dec-28", 276.46
362, "Dec-29", 277.48
363, "Dec-30", 278.50
364, "Dec-31", 279.52
365, "Jan-01", 280.54
366, "Jan-02", 281.56

That @interact decorator before the main function isn't standard Python, it's a feature of Sage. You can play with an interactive version of this script on a SageMath server here.


What does designation VY, NML, UU in star names stand for? I've read the a IAU page on the subject but i am no closer to understanding how this works. Also did search the already asked question, but didn't see anything on the subject.

What does bolded item in the names below stand for? How do they choose them?

VY Canis Majoris
NML Cygni
UY Scuti

VY Canis Majoris and UY Scuti are variable star designations. The first discovered variable star in a constellation is called R, the second, S, and then unto Z. After Z comes RR and so on. For a full description see this wikipedia page. NML Cygni is also a variable star, but its variable star designation is V1489 Cygni. The letters NML come from its discoverers: Neugebauer, Martz, and Leighton.


Why does squinting make hard-to-see objects clearer? So I, like most other astronomy enthusiasts, have, at some point, not been able to see something clearly, such as a star, and have had to squint to be able to see it better. When this happens, the object seems to become clearer, but the light seems to gain a cross, or sometimes a hexagon, or an irregular line shape when doing so. I've noticed a similar effect on different models of telescopes, especially the newtonian reflector series. More on this here: (http://www.youtube.com/watch?v=ipe3NN1yPzM). Is this a similar effect on the eye? But more importantly how does squinting improve the quality of the picture?

Squinting works the same way as a pinhole camera.

Ideally, light from a single point source entering your eye anywhere on your pupil will be focused on a single spot on your retina. But this works perfectly only if you have perfect vision; otherwise light entering near the top of your pupil may be directed to a slightly different spot on your retina than light entering near the bottom.

By squinting, you block out some of the light from the edges, effectively making your pupil narrower, creating a sharper but dimmer image on your retina. (You may find that it improves the vertical resolution more than the horizontal resolution.)

If you happen to be nearsighted (as I am), you can see a similar effect by looking through a small pinhole, or through a small aperture made with your fingers. If the light is bright enough, you'll see a dimmer but sharper image.

The irregular shapes may be interference from your eyelashes.


Is there any known moon of a moon? As far as I understand, a moon is an object in permanent orbit around a planet, dwarf-planet, asteroid, etc. If there was another object permanently orbiting this moon, would that be a moon-moon?

Is there any natural object of such kind that we know of?

If no, why are such objects so unlikely?

Edit: this is question is different from Do moons have moons? because I explicitly ask for a reason. Also, the terminology is included here.

No. One explanation for the equatorial ridge on Iapetus involves a now-lost subsatellite, so perhaps such objects existed in our Solar System's past.

Subsatellites tend to get rapidly removed by tidal forces causing changes to the orbit. Depending on the parameters of the system, the orbit may decay until the subsatellite collides with the moon or breaks up at its Roche limit. Alternatively it may migrate outwards until the subsatellite becomes unbound: at this point it would become a moon of the planet, though likely on an unstable orbit that would lead to scattering or a collision. The orbital evolution tends to be faster for more massive subsatellites. See Kollmeier & Raymond (2018) for details, who note that any subsatellites orbiting the candidate Neptune-sized exomoon in the Kepler-1625 system would need to be smaller than Ceres to survive.


How was the first exoplanet detected? How was the first exoplanet discovered? What method was used to detect it?

This question is a bit tricky because it's uncertain as to what counts as "discovering" an exoplanet. Today, we count an exoplanet as having been discovered if it can be detected either by two separate techniques or three separate transits (if found via the transit method). Some of the very very early hunts for exoplanets did indeed find them, but many of their initial results were dubious or questionable (and sometimes wrong) and weren't really confirmed until much later. So do you count a planet as being discovered when it was first potentially detected or do you not count it until its been completely confirmed?

First Exoplanet Detection

The very first time someone could claim they discovered an exoplanet was in 1988. The astronomers Bruce Campbell, G. A. H. Walker, and Stephenson Yang announced that they had discovered a planet around the star Gamma Cephei using the radial velocity method. However, their results were not that great since, at the time, the detection was at the threshold of their technological capabilities and their results weren't entirely believable. It wasn't until years later that this planet was actually confirmed as existing. Whether you count this as the first real exoplanet discovery or not depends on what you define as a discovery, given that the actual confirmation didn't come until years after the first detection.

First Confirmed Exoplanet Detection

In 1992, Aleksander Wolszczan and Dale Frail managed to find and confirm a planet around a the pulsar PSR B1257+12, again with the radial velocity method. At this time, their measurements were good enough to be believed and so considered "confirmed".

First Confirmed Exoplanet Detection Around a Main-Sequence Star

This case is particularly called out because it was the first time an exoplanet was detected and confirmed to exist around a normal main-sequence star. This occurred in 1995 around the star 51 Pegasi. This discovery was made by Michel Mayor and Didier Queloz. Many people would consider this the first "true" exoplanet discovery and will cite this case if you ask when the first exoplanet discovery was made.


What would happen when the speed of the Sun's spin changes? Does the Sun's spin have effect on the orbit of the Earth?

Does the Sun's spin have affect on the orbit of the Earth?

It has a very, very small effect. The Sun would have a larger equatorial bulge if it rotated faster, and a smaller equatorial bulge if it rotated slower. An equatorial bulge increases gravitational acceleration near the plane of the body's equator and decreases gravitational acceleration above  below the body's poles.
Since the Sun is already rotating rather slowly, it doesn't have much of an equatorial bulge. Moreover, the effects of oblateness drop off as 1/R^4, as opposed to the 1/R^2 drop-off in Newton's law of gravitation for a point mass.
The combination of small oblateness, 1/R^4 drop-off, and large distances makes the effects of the Sun's oblateness on the planet's orbits very small.
The effect is very, very small, about three orders of magnitude smaller than the general relativistic effect on Mercury's precession (which is already very small, about 43 arc seconds per century). The primary effect of the Sun's oblateness is to make Mercury precess by 0.0286 arc seconds per century. That said, the extremely precise observations of Mercury's orbit about the Sun do make that tiny precession statistically important because the uncertainty in Mercury's precession rate is down to
0.0015 arc seconds per century, according to Park, Ryan S., et al. "Precession of Mercury’s Perihelion from Ranging to the MESSENGER Spacecraft." The Astronomical Journal 153.3 (2017): 121.


How can I calculate the time between sunset and actual darkness? I would like to know the time it takes to go from sunset to dark based on GPS location.

In addition, ideally I would also like to have a formula that determines the percentage of sunlight at each minute past sunset. For example, after 15 minutes, there is only 50% sunlight left.

One may define three forms of twilight on Earth.  Although the actual amount of light depends on weather, topography, and land cover, they are defined as:

Civil twilight: Solar angle > -6°
Nautical twilight: Solar angle > -12°
Astronomical twilight: Solar angle > -18°

Human eyes see logarithmically so % of sunlight left is not a very useful measure, unless you are interested in incoming shortwave radiation for solar panel or other energy calculations.

To answer your question, you need to calculate at what time the Sun will be at those elevations below the horizon; this is calculated in a similar fashion to sunrise and sunset times.  However, on top of that you also need to consider atmospheric refraction.  See position of the Sun and sunrise equation.  The formulae are quite complicated, but you may be able to simplify them if you're only interested in the duration, not the absolute times.

The time after sunset at which those occur are a function of latitude and time of year.  On this carpet plot from Wikipedia you can read the length of different forms of twilight at 70°N (north of the Arctic circle):


How can the Sun burn without oxygen? How does the Sun burn without oxygen?

It might not be burning but a big part of the society speaks of it as burning.

So how does it work?

As you are suspecting, the sun burns in a different sense, not by chemical reaction with oxygen.

Atoms consist of a tiny, heavy nucleus, surrounded by an almost empty space, populated by electrons. Burning by chemical reaction with oxygen doesn't change the nucleus of atoms, but takes place in the hull of atoms: Atoms may assemble to form molecules; electrons change their orbitals (the way they surround the nucleus), and release some energy as heat.

Atomic nuclei are (positively) electrically charged, and repell each other.
But if small nuclei, like those of hydrogen atoms, come close together, they can fuse and form a larger nucleus. This nuclear fusion of hydrogen to helium (in this case) releases much energy, more even than fission of uranium in a nuclear power plant. The notion "burning" is used sometimes for reactions of atomic nuclei, too, if they release energy as heat.

To overcome the electrostatic repulsion of hydrogen nuclei, high pressure and temperature are needed. These conditions occur in the core of the Sun.


The equation for calculating how many times Earth (or our solar system) has revolved around the Milky Way I have seen it said on Quora and Reddit that it's about every 220-250 million years to go around the Milky Way from Earth. Wondering what the rough equation is for calculating it. That is, I guess, the angular momentum and distance from the center, so I could visualize it on a sped up animation and otherwise derive something like those numbers 220 to 250.

I'm not 100% sure what equation you would like, but I'll try at an answer to see how it goes.

The observed speed of the Sun orbiting around the center of the Milky way is about 220 km/sec. The distance to the center is about 8.1 kiloparsec, which is 2.5E+17 kilometers.

The time it takes to travel around a circle is the circumference divided by the speed,

$$\frac{2 \pi R}{v} \approx \frac{6.28 \times 2.5\times 10^{17}}{220} \approx 7.17 \times 10^{15} \text{seconds} \approx 227 \text{ million years}$$

The speed is a measured quantity. There isn't a simple way to calculate the speed with an equation, because the mass distribution in the galaxy is complicated and dominated by dark matter, whose presence we can only infer from the orbital motion of our own star as well as a large number of other stars in our galaxy.

 Source

From Wikipedia's Milky Way:

  Galaxy rotation curve for the Milky Way. Vertical axis is speed of rotation about the Galactic Center. Horizontal axis is distance from the Galactic Center in kpcs. The Sun is marked with a yellow ball. The observed curve of speed of rotation is blue. The predicted curve based upon stellar mass and gas in the Milky Way is red. Scatter in observations roughly indicated by gray bars. The difference is due to dark matter


Can tidally locked planets have liquid oceans on their night side, and would such a planet be habitable? Let's say we have a tidally locked planet orbiting a star. And let's say that the conditions on its surface are just right for water to exist on its surface. Conventional wisdom says that the water on the day side will be liquid, while the night side will be covered with a sheet of ice.

However, I'm wondering if this is a hard and fast rule. What I want to know is if it is possible for tidally locked worlds to have liquid water persist on its night side, and what conditions could bring this about. And would those conditions still leave the world habitable?

Short Answer:

With current knowledge nobody knows.  Some calculations suggest that it might be possible for a tidally locked planet to be habitable and to have life.

Long Answer:

The first stop searching for information about the habitability of tidally locked planets in red dwarf star systems might be the Wikipedia article Planetary Habitability.

  Astronomers for many years ruled out red dwarfs as potential abodes for life. Their small size (from 0.08 to 0.45 solar masses) means that their nuclear reactions proceed exceptionally slowly, and they emit very little light (from 3% of that produced by the Sun to as little as 0.01%). Any planet in orbit around a red dwarf would have to huddle very close to its parent star to attain Earth-like surface temperatures; from 0.3 AU (just inside the orbit of Mercury) for a star like Lacaille 8760, to as little as 0.032 AU for a star like Proxima Centauri[76] (such a world would have a year lasting just 6.3 days). At those distances, the star's gravity would cause tidal locking. One side of the planet would eternally face the star, while the other would always face away from it. The only ways in which potential life could avoid either an inferno or a deep freeze would be if the planet had an atmosphere thick enough to transfer the star's heat from the day side to the night side, or if there was a gas giant in the habitable zone, with a habitable moon, which would be locked to the planet instead of the star, allowing a more even distribution of radiation over the planet. It was long assumed that such a thick atmosphere would prevent sunlight from reaching the surface in the first place, preventing photosynthesis.

  This pessimism has been tempered by research. Studies by Robert Haberle and Manoj Joshi of NASA's Ames Research Center in California have shown that a planet's atmosphere (assuming it included greenhouse gases CO2 and H2O) need only be 100 millibars (0.10 atm), for the star's heat to be effectively carried to the night side.[77] This is well within the levels required for photosynthesis, though water would still remain frozen on the dark side in some of their models. Martin Heath of Greenwich Community College, has shown that seawater, too, could be effectively circulated without freezing solid if the ocean basins were deep enough to allow free flow beneath the night side's ice cap. Further research—including a consideration of the amount of photosynthetically active radiation—suggested that tidally locked planets in red dwarf systems might at least be habitable for higher plants.[78]

https://en.wikipedia.org/wiki/Planetary_habitability#Red_dwarf_systems1

This article links to a longer discussion of the habitability of red dwarf star systems.

https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems2

And of course the various sources listed in those articles are worth studying.


From original burst, fraction of stellar mass still surviving on Main sequence Suppose that all stars in this galaxy were born in a single major-merger burst event about 10
Gyr ago. From this original burst, I want to compute the fraction of stellar mass still surviving as stars in the
main sequence ? For this, I have got to use a Salpeter IMF, and a star formation range between 0.1 and 120 solar
masses.

What I have done is starting from Salpeter IMF : $$\Phi(m)\text{d}m=\Phi_{0}\,m^{-2.35}$$

with $$\Phi_{0}$$ a constant normalization.

From this, I integrate from $$m_{1}=0.1\,\text{M}_{\odot}$$ to $$m_{2}=120\,\text{M}_{\odot}$$

$$N(0.1<m<120) = \int_{0.1}^{120}\,\Phi(m)\,\text{d}m = \Phi_{0}\,\bigg[\dfrac{0.1^{-1.35}-120^{-1.35}}{1.35}\bigg]$$

This result depends on the valeur of $$\Phi_{0}$$ and I don't know how to deal with it in order to get $$N(0.1<m<120)$$ ?

Moreover, it seems that I have to take into account of the age of the major-merger burst event (10 Gyr).

From these 2 principles, how could I calculate the fraction of stars surviving in the main sequence ?

Any help is wlecome, Regards

When you are calculating fractions, rather than absolute numbers, the value of $\Phi_0$ does not matter, since it will be a multiplying factor in both the numerator and denominator.

You have (almost, see below)  successfully got an expression for the denominator of your fraction.

The numerator is found by calculating an equivalent integral from your lower limit to an upper limit that is instead defined by the most long-lived main sequence stars that are still "alive" - i.e. those with a lifetime equal to the age of our Galaxy.

Finally, you were asked to find the fraction of stellar mass surviving, not the fraction of stars. The stellar mass existing between two mass intervals is
$$M_* = \int_{m_1}^{m_2} m\Phi(m)\ dm$$


Does matter accumulate just outside the event horizon of a black hole? My understanding is that time slows and approaches stopping when approaching the event horizon of a black hole.  I have seen this explained several places, including a brief explanation in the last paragraph under: http://en.wikipedia.org/wiki/Black_hole#General_relativity, quoted below:

  Oppenheimer and his co-authors interpreted the singularity at the boundary of the Schwarzschild radius as indicating that this was the boundary of a bubble in which time stopped. This is a valid point of view for external observers, but not for infalling observers. Because of this property, the collapsed stars were called "frozen stars",[17] because an outside observer  would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius.

Does this mean then that no matter actually falls into a black hole (except possibly what was there at its formation)?  Would this also mean matter is accumulating just outside its event horizon?  As I understand it, this would be the perspective from outside the black hole.  If this is the case, I wonder if we would observe a tremendous amount of matter surrounding the event horizon, but it would be extremely red shifted?

Edit:

I noticed an answer to a different question, especially the end portion, provides some insight here as well:
https://astronomy.stackexchange.com/a/1009/1386

Edit:

These YouTube videos someone put together explain the concept very well, and seem to indicate this idea is gaining traction!

https://www.youtube.com/watch?v=yZvgeAbrjgc&list=PL57CC037B74307650&index=118
https://www.youtube.com/watch?v=b1s7omTe1HI

Edit:

This new YouTube video describes this idea very well, and describes it as the way black holes work!

https://youtu.be/mquEWFutlbs

What you're describing is basically the "collapsed star" (Eng) or "frozen star" (Rus) interpretation of black holes that was common prior to the late mid-1960s. It was a mistake.

Suppose you are distant and stationary relative to the black hole. You will observe infalling matter asymptotically approaching the horizon, growing ever fainter as it redshifts. Does it mean that matter "clumps" around the horizon? To find out, suppose you throw yourself towards the black hole to try to catch the matter that you see. What you will find is that it fell into the black hole long ago.

In other words, the most sensible way to answer whether or not infalling matter clumps on the horizon is to look at the situation from the frame of that infalling matter. And there, it is clear: no, it does not clump, as it crosses the horizon in finite proper time. (As an aside, for a Schwarzschild black hole, falling from rest is exactly Newtonian in Schwarzschild radial coordinate and proper time.)

The "comoving viewpoint" was recognized by Oppenheimer and Snyder in 1939, but it was not until the 1960s, with the work of Zel'dovich, Novikov, et al., that it was generally recognized as truly significant in the community. In 1965, Penrose introduced conformal diagrams based on the Eddington-Finkelstein coordinates (1924/1958) that showed quite clearly that the stellar collapse is not slowed, but instead continues to a singularity. For an overview of the history of this change of viewpoint, cf. Kip Thorne, et al., The Memberane Paradigm (1986). These topics are commonly covered in many relativity textbooks.

Ok, but since it still takes an infinite amount time in the frame adapted to a stationary distant observer, does that mean that the horizon never forms in that frame? It does form: the underlying assumption in the argument that it does not would be either that the infalling matter needs to reach the center for the horizon to form or cross a pre-existing horizon to make it expand. But that assumption is simply not true.

An event horizon is defined in terms of future lightlike infinity, roughly speaking in terms of whether or not light rays escape if one waits an infinite amount of time. That means the location of the horizon at any time depends on not just what has happened, but also what will happen in the future. In the frame of the distant stationary observer, as matter falls towards the event horizon, it does slow down to asymptotically approach... but the horizon also expands to meet it. Similarly, the initial collapsing matter does not need to collapse all the way to the center for the event horizon to form.

  How can the finite life-time of the Black hole due to Hawking radiation be made consistant with the infinite amount of time (future) needed for the expansion of the event horizon (in the outer time-frame)?

There's no need to: [edit]that a particular time coordinate doesn't cover the full manifold is a fault of the coordinate chart, not of spacetime[/edit]. From every event, send out an omnidirectional locus of idealized light rays. The event horizon is the boundary of the spacetime region from which none of these light rays escape to infinity. This question has an objective answer--for any given light ray, either it will escape or it won't.

An external observer would need to wait infinitely long to know for sure where the event horizon is exactly, but that's a completely different issue.  With Hawking radiation, the black hole shrinks, but it doesn't change the fact that light rays from some events will fail to escape, and thus that an event horizon will exist.

Here's a Penrose diagram of a spherically collapsing star forming a black hole that subsequently evaporates:

Light rays run diagonally at ±45° on the diagram. Note that there is a region from which outgoing light rays (running diagonally lower-left to upper-right) don't escape and instead meet the $r = 0$ singularity (the bolded, undashed horizontal line). The horizon itself is the the $r = 2m$ line marked on the diagram and its extension into the star: it should actually go from the (dashed, vertical) $r = 0$ line on the left, rather than extending from the star's collapsing surface. That's because some of the (idealized, noninteracting) light rays from inside the star will also fail to escape to infinity.

Now suppose that on this diagram you draw timelike curves that stubbornly stay away from the horizon, and you insist on using a parameter along them as a time coordinate. Does the fact that you've chosen coordinates that exclude the horizon needs to be made consistent with whether or not the event horizon actually exist? The resolution is simple: if you want to talk about the horizon, stop using coordinates that exclude it.


Were there any images of Sanduleak -69 202 (progenitor of SN1987A) before it exploded? We all know about SN1987A, the closest observed supernova since Kepler's time. Its progenitor was Sanduleak -69 202, a magnitude 12 blue supergiant, catalogued in 1970. Were there any images of this star before it exploded, and how did we know that this was the star that produced said supernova?

Yes, it is clearly visible in photographs of the LMC from the 1980s.  It is shown in this comparison from the Australian Astronomical Observatory
We know it was this star in the most simple possible way: The supernova occurred exactly at the same location as this star, and when we look now, that star has gone (replaced by a little bipolar supernova remnant)


Why was the (small) Hubble better able to find KBO targets for New Horizons than large adaptive optics ground telescopes? When initial searches for a Kuiper Belt object for New Horizons to fly to after passing Pluto did not find good targets, the Hubble telescope was used, and it resulted in the current targeted flyby for 2019. Initial searches used ground based telescopes. When the search risked running out of time without a good target, the Hubble was brought in to help.

According to what you read about the current generation of large earth based telescopes that use adaptive optics, those telescopes have a much larger angular resolution and light gathering area than the Hubble. So why was the Hubble better capable of finding a good target?

edit
Part of the answer may be that the current generation of adaptive optics telescopes only do adaptive optics in infrared, at least if KBOs are best observed in visible light which I don't know but I've moved that to a separate question.

Regarding everyone suggesting that atmospheric absorbtion is to blame, how does that square with this: The 8.3m Subaru telescope (which was one of the telescopes used in the ground search) has a light collecting area of 53m2. The Hubble has a collecting area of 4.5m2. So atmospheric absorbtion would need to be 91.5% for them to collect the same amount of light. Sure atmospheric absorbtion is high for some infrared wavelengths, but surely not that high over all the relevant wavelengths.

I suspect it's a combination of two things:

Stable, guaranteed high-resolution imaging across the entire field of view, something not possible with ground-based adaptive optics;
Very low background in the optical for HST (Hubble), versus a very high background for ground-based AO in the near-infrared.

Most adaptive optics systems are only able to correct in a small region (the "isoplanatic patch") around a bright "guide star" (say, half an arc minute in radius at most); even with artificial laser guide stars, you still need a moderately bright guide star for so-called "tip-tilt" correction. This means you can only do adaptive optics searches in limited parts of the sky.

HST, on the other hand, delivers high-resolution imaging throughout its entire field of view (several arc minutes across), all the time, regardless of where it's pointed.

To make matters worse, the trajectory of New Horizons is close to the Galactic plane, so there are lots of faint background stars. This makes it harder to pick out possible Kuiper Belt objects, making a very accurate and stable point spread function (such as that of HST) even more important.

These searches are best done in the optical, to minimize the sky background. The absence of atmospheric sky glow (mostly scattered light from the Sun and Moon) for HST makes it easier to quickly detect faint sources like KBOs. The fact that adaptive optics systems, as you and Rob Jeffries note, work almost entirely in the near-infrared, where the atmospheric background is much higher, makes it even worse for them.


Massive degraded impact crater on Mercury - chemical evidence? The planet Mercury has a myriad of impact features - this has been known for some time. However, observe in the following images, a relatively large 'High-Mg region' detected on the planet (below):

Image source: Messenger website

From the group's press conference website, they state that:

  The unusual composition of the high-Mg region and the relatively thin crust in this region suggest that it may be the remains of an ancient and highly degraded impact basin.

How is the 'unusual composition' of this region suggestive of being the remains of an ancient, highly degraded, impact basin?

There is still some research to do to definitively answer this question. Having said that though, according to research published in the paper "Evidence for geochemical terranes on Mercury: Global mapping
of major elements with MESSENGER’s X-Ray Spectrometer (Weider et al. 2015), the high Mg region (HMR) could be due to the consequences of an impact event - the authors from the article suggest that:

  therefore, that the HMR may also be an impact related
  feature, and that its high Mg/Si ratio is evidence of
  mantle excavation during a large impact event early in Mercury’s
  history (>4.1 Ga).

and later, partly resurfaced by subsequent volcanism and smaller impact events.

The authors elaborate further that the mechanism behind it could be

  If an ancient large impact at the HMR site penetrated into a
  vertically differentiated mantle (with additional lateral variations
  possible) and reached a deep, otherwise unsampled layer (i.e., the
  lherzolitic source proposed by Charlier et al. 2013*), the HMR composition
  would be the signature of high-degree partial melting of
  that mantle source.

Additional resource cited:

Charlier et al. 2013, Phase equilibria of ultramafic compositions on Mercury and the origin of the compositional dichotomy, Earth and Planetary Science Letters


Polynesian wayfinding techniques - how does this work? In Disney's "Moana", we get a hint of a Polynesian wayfinding technique of holding your hand to "measure the stars" (note, you're not giving the sky a high-five). We get to see two different constellations measured: Crux (the Southern Cross), and Orion.

By matching the positions of the stars either in real life or on Stellarium, I was able to figure out something of how the Southern Cross is used. You place your fingers against the four brightest stars (two pointers plus the two brightest of the cross itself), and then the south celestial pole is right in the curve of your outstretched thumb. (Aside: It makes an awesome demo in Stellarium to place your hand on the screen, do the measurements, and then turn on the grid, highlighting the south pole beautifully.)

What I haven't managed to do, though, is match the corresponding measurement in Orion. The fingers are placed near Orion's belt and heel, but I can't figure out what is being measured. Can anyone provide either certain information, or plausible guesses?

(I can provide a screencap if there's an appropriate way to include it in the question.)

This is from an article in the Conversation, by Duane W. Hamacher, and Carla Bento Guedes:

"Voyagers measure the angles between stars and the horizon using their hands. The width of your pinkie finger at arm’s length is roughly one degree, or double the angular diameter of the Sun or Moon.

Hold your hand with the palm facing outward and thumb fully extended, touching the horizon. Each part of your hand is used to measure a particular altitude.

The hand method used by Nainoa Thompson to find the altitude of the Polaris. Journal of the Polynesian Society
In Hawai'i, the “North Star”, Polaris, is Hokupa'a, meaning “fixed star”. It lies close to the north celestial pole. The altitude of Hokupa'a indicates your northerly latitude.

In the film, we see Moana Waialiki using this technique to measure the altitude of a group of stars. Look closely and you can see that she’s measuring the stars in Orion’s Belt. The position of Moana’s hand indicates the star above her index finger has an altitude of 21º. Given that the movie takes place about 2,000 years ago near Samoa, the position of Orion indicates they are travelling exactly due East.

Moana measures altitude of Orion’s belt stars. Walt Disney Studios Motion Pictures
Later in the film, we see Moana navigating by following Maui’s fish hook. In the various Polynesian traditions, the hook was used to pull islands from the sea. It is represented by the constellation Scorpius, which rises at dusk in mid-May. This indicates southeasterly travel.

Looking at Scorpius - Maui’s Hook - in the same orientation as shown in the film. Stellarium
However, the positions of the stars are not fixed in time. Over the 3,500 years that Polynesians have been exploring the Pacific, the stars have gradually shifted due to precession of the equinoxes.

From the latitude of Samoa, the Southern Cross has lowered from 60º altitude in 1500 BCE to 41º today. Those navigating by the stars must gradually adjust their measurements as the positions of stars slowly shift over time."


How are galaxies moving so much and why are they moving? I don't understand and I haven't understood for a while.
How is it that all the galaxies in the universe are moving, how they are moving and how fast?

There are several sources of movement for galaxies.

Galaxies may rotate.  Galaxies are formed essentially from matter collapsing under gravity into "clumps".  This process isn't uniform due to gravity from other clumps of matter and so it is very rare that all the particles in a single clump "fall" directly into it's centre.  This creates rotation (Angular Momentum) and so the eventual galaxy will rotate.  Note that some galaxies show very little overall rotation, with the individual stars orbiting the centre of the galaxy in a multiple directions.

Galaxies interact with each other. Most galaxies reside within a group or cluster of galaxies, our galaxy is within a group called the Local Group with Andromeda and several ($\geq$52) dwarf galaxies.  These dwarf galaxies are usually part of a "satellite-subsystem" where they orbit a large galaxy under gravity.  These orbits are affected by the other local galaxies and occasionally some systems can be "expelled" due to the gravitational forces happening to speed the body up past the velocity required to escape the system.

The major galaxies in a group or cluster interact with each other, orbiting a centre of gravity.  This is due to the same mechanism as galaxy rotation.  This is usually a very dense area of galaxies, or a massive Brightest Cluster Galaxy (BCG).  In the local group, the two main galaxies (our Milky Way and the Andromeda Galaxy) are getting closer together and will collide in approximately 4 billion years.

The universe is expanding. The universe is expanding and so galaxies appear to be moving away from us in every direction.  We see this as redshift, a change in the wavelength of light as it reaches us due to the difference in speed between the origin an us.  This is not really anything to do with galaxies moving, and more to do with the fabric of the universe.  Whether this will continue, or whether the universe will stop expanding and collapse is a current active area of research.


Supposed SN 1006 was only 4 lightyears away what would the blasts impact be like on earth? One of the commenters on my answer to https://worldbuilding.stackexchange.com/a/86733/3460 questions that the blast would even have a noticeable impact if that supernova was in Alpha Centauri. And since my answer is just based on my intuition, I would like to figure that out.

So radiation put aside, what would the blast of SN 1006 be like for earth when arriving, supposed it went supernova just 4 light-years away?

Last week, a paper was posted on arxiv (https://arxiv.org/abs/1707.04253v1) which studies the "The Resilience of Life to Astrophysical Events", in particular it deals with the temperature change of oceans (as boiling oceans would evaporate the water of a planet which seems to be bad for biological life forms...). For a assumed SN explosion of Proxima Centauri the temperature change of Earth oceans would only be 0.1 degree Celsius - not enough to extinct life by boiling the oceans.

(extract from Sloan, Batista and Loeb, 2017)


Tidal locking of a moon When a moon is in orbit around it's planet but is so close that it will become tidally locked does it experience a large 'wobble' in between the phase when it still has its own rotation and the final phase when it is always showing the same face to the planet?  I can only imagine that it would 'wobble' very widely, back and forth like a pendulum until finally settling into a locked position.

There's no back and forth motion on the way towards tidal locking.   There's kind of two ways to look this kind of "settling", there's overshooting and correction, basically a back and forth, and there's a very gradual force that grows smaller, basically moving towards zero but not overshooting.    I realize that's just re-writing your question, but it helps to understand the 2nd option.

I'll use the Earth as an example, because the Earth is in the gradual process of becoming tidally locked to the Moon.  It's the same process, but easier to visualize.

Because the Earth rotates faster than the Moon orbits, the tidal bulge moves ahead of the moon.   This is called the tidal bulge offset.

Source.

It's the size of the bulge and the angle of the offset (and the mass and distance of the Moon) that causes the familiar gravitational effects, the Earth's rotation slowing down and the Moon moving away from the Earth.

Source

The current angle of offset for the Earth's tidal bulge to the Moon is about 3 degrees, often exaggerated in diagrams so it's easier to see.  (I should find a source for that - I've not been able to find a good one).

From the Moon's point of view (and if we assume a circular orbit), the tidal offset stays in the same place.  It's a factor of the relative rotation speed.   As Earth slows down, the angle of offset will move towards zero.    At tidal locking, the angle of offset becomes zero and there's no force to continue in the other direction.    In a simple mathematical model, tidal locking wouldn't ever happen completely, it would just gradually get closer and closer, like an object shot from the Earth at exactly escape velocity.

In reality, because objects aren't perfectly symmetrical and because orbits aren't perfectly circular, the "take forever to get to zero", doesn't happen.  But it's still a gradual approach to zero and no overshooting and returning because the force on the rotation approaches zero as the angle of offset approaches zero and the force needed to change a planet's rotation is enormous and there's essentially zero force to make that happen.

With more eccentric orbits, like Mercury and it's 3:2 rotation to orbit resonance, that might be different.   Not sure how eccentricity might play into this.   That's more complicated.


Hazy patch with huge comet tail Tonight between 4:30 am & 5 am, we were star gazing on a skyscraper. We saw hazy illuminated triangle patch. We took out star app but didn't find any comet passing through that region & I was sure it wasn't Andromeda Galaxy. I'm not sure what I watched.

Please note we thought it could be light of aeroplane, but we were sure it wasn't as it didn't move quickly. The second thought was some form of satellite. I think there isn't any satellite with that much light.

Please can anyone help me identify hazy triangle patch in the sky. I am attaching star map of my sky. The region I saw the hazy patch is marked with red circle on the map.

I'm based out of Mumbai. I don't want my curiosity to die without knowing what I saw, I'm happy to be wrong if it isn't any stellar object. Thank you.

As you see from the app, there are no hazy objects there, and in the light polluted Mumbai skies, deep sky objects would not be easily visible. There are no bright naked eye comets visible at the moment.

Satellites appear like moving stars, not hazy. The most likely hazy object would be a high cloud or part of a contrail, perhaps lit by the light of the city, or by the first light of dawn.


How to get the axial tilt vector(x,y,z) relative to ecliptic According to the orbital plane XYZ, I understand that the axial tilt is the angle between the planet axis rotation and any normal of the plane (actually Z in right handed system). BUT, this is not sufficient to define a vector direction, I need another angle into the XY plane. Where do I find this information?  Actually I believe it's relative to the direction of the periapsis. Am I right?

Following all the nice conversation on the answers, I maybe need to ask

How to get the axial tilt vector(x,y,z) relative to ecliptic or orbital plane, given ONLY ONE angle, i.e. 23.439281 deg for the Earth, 25.19 for Mars, etc.

Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit.
Earth's north pole vector in
equatorial coordinates is
$$\vec N_{\oplus,eq} = (0, 0, 1)$$
To transform this to
ecliptic coordinates, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get
$$\vec N_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$
In a spherical coordinate system, two angles define a unique direction.
In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$.
This IAU report, table 1,
gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years.
To convert these to rectangular form
$$(x_{eq}, y_{eq}, z_{eq}) =
(\cos \alpha \cos \delta,
\sin \alpha \cos \delta,
\sin \delta)$$
Then if you want J2000 ecliptic coordinates
$$(x_{ecl}, y_{ecl}, z_{ecl}) =
(x_{eq},
y_{eq} \cos \varepsilon + z_{eq} \sin \varepsilon,
z_{eq} \cos \varepsilon - y_{eq} \sin \varepsilon)$$
But if you want another planet's orbital plane to be the $xy$-plane,
then you also need the orbit's inclination and longitude of ascending node;
I leave that transformation as an exercise for the motivated reader.


Obtaining TESS input catalog (TIC) IDs for Gaia `source_ids`? The TESS input catalog (TIC) "uses the GAIA DR2 catalog as a base" according to the catalog's website. The Gaia DR2 database provides source_ids for each observation of a target. I want to get the TIC number for a target based on the Gaia source_id. Is there an accepted way of doing this that I'm missing?

To be more specific, I can get a list of Cepheids from Gaia DR2 using their archive and the ADQL command select * from gaiadr2.vari_cepheid. The first result has the Gaia source_id as 4052922352453886976. How can I find the TIC number for this target star? Is the best way of doing this using the RA and DEC of the Gaia observation and finding the closest match in the TIC? Since the TIC is based on the Gaia DR2, I would assume there would be a direct corresponding target and finding the closest RA and DEC match would be unnecessary.

You can query TIC by Gaia ID. Using the MAST Portal, MAST Catalogs, TESS Input v7, Advanced Search form, I specified:

GAIA ID: 4052922352453886976

and got a single result:

TIC ID (ID):            50559830
RA (ra):                275.945292 (18:23:46.870)
Dec (dec):              -25.672353 (-25:40:20.47)
TESS Mag. (Tmag):       13.579
Version (version):      20180610
UCAC ID (UCAC):         322-168973
2MASS ID (TWOMASS):     18234686-2540204
GAIA ID (GAIA):         4052922352453886976
GAIA G Mag. (GAIAmag):  14.5068

Since the MAST portal currently provides TICv7, the Gaia ID is probably from Gaia DR1.
The TIC release notes say Gaia DR2 data are new in TICv8.

  9  GAIA  The matched Gaia DR2 identifier has been updated in TICv8.
  Gaia IDs in TICv8 are included for stars that are found in the Gaia-provided Gaia-2MASS look-up table.
  These identifiers have been updated by the Gaia team from DR1 to DR2, and may not be consistent between TICv7 and TICv8.


Moving from computerised to normal amateur telescope? A few years ago I purchased a Celestron Nexstar 130 SLT computerised telescope thinking that it would be a perfect to get me into astronomy. The issue is the computer doesn't work at all and manually finding objects was a pain so I haven't used it in years. Now I want to get back into the hobby the 'proper' way by tracking objects with star maps and I want to get it right since photography and spectroscopy really excite me and that's what I want to end up doing. Despite all of this I can't seem to find a dovetail Equatorial mount that will fit my optical tube so now I feel I am stuck with a motorised mount that doesn't allow me to point my tube precisely and is really no fun. Is there any way for me to get into the hobby with my current optical tube using a different mount, use my computerised mount accurately enough for decent observation, or do I really have to buy a whole new optical tube just for the mount?

A Celestron Nexstar 130 SLT is not suitable for spectroscopy and the only photography that it will be mush use for would be afocal or DSLR imaging of the Moon (or Sun with a suitable filter).

For more serious work you will need a equatorial mounted scope with a better focuser.

Exactly which one to choose will depend of your budget and what you want to do.

When you say the computer does not work can you be more specific.


Why is dark energy preferred to the cosmological constant? Dark energy and the cosmological constant are often identified, but Peebles and Ratra explain that

  "Einstein did not consider the cosmological constant to be part of the stress-energy term... a new constant of nature,
  $\Lambda$, appears in the addition to Einstein’s original field equation. One can equally well put Einstein’s new term on the right hand side of the equation, and count [...] as part of the source term in the stress-energy tensor. The distinction becomes interesting when $\rho_\Lambda$ takes part in the dynamics, and the field equation is properly written with $\rho_\Lambda$, or its generalization, as part of the stress-energy tensor."

In other words, Einstein treated $\Lambda$ term as a modification of GR with a "new constant of nature", while dark energy may potentially manifest itself in some other ways. Later they write about dark energy:

  "The idea that the universe contains close to homogeneous dark energy that approximates a time-variable cosmological “constant” arose in particle physics, ...in cosmology..., and on both sides by the thought that $\Lambda$ might be very small now because it has been rolling toward zero for a very long time. The idea that the dark energy is decaying by emission of matter or radiation is now strongly constrained by the condition that the decay energy must not significantly disturb the spectrum of the 3K cosmic microwave background radiation."

So if I understand this correctly the idea of a dynamical effect came from suspecting that $\Lambda$ may be time dependent, but so far that was not detected. And since "one can equally well put Einstein’s new term" on either side of the equation mathematically both interpretations are equivalent. Still, the dynamical interpretation seems to be overwhelmingly preferred.

  Are there empirical reasons for conjecturing dark energy, some spatial non-uniformity in expansion perhaps? If not, what are the theoretical reasons for preferring it to the cosmological constant of modified GR?

A cosmological constant should be considered a special case of dark energy. The effective stress-energy tensor for a cosmological constant is proportional to the metric $g_{\mu\nu}$, so in a local inertial frame will be proportional $\mathrm{diag}(-1,+1,+1,+1)$. This is equivalent to perfect fluid with energy density and pressure directly opposite one another, but more importantly, it is the only possible form for the stress-energy that would give the exact same energy density and pressure in all local inertial frames.

If by 'dark energy', we understand it to mean all the contributions to stress-energy in the above form, then there is no reason for this to be constant, and plenty of reasons why it might not be, as this situation is not exceptional in fundamental physics. For example, there could be false vacua with various different energy densities, and they must be invariant across inertial frames.

In particular, the basic idea of inflation considers a flat FRW universe with expansion driven by a scalar field $\phi$ at a local extremum of its potential, $V'(\phi_0) = 0$, which yields an exponential expansion with constant energy density $T^0{}_0 = V(\phi_0)$. More refined models, such as slow-roll inflation, could therefore be directly interpreted as a time-varying dark energy density, while eternal inflation would also include spatial variability. There's plenty of other inflationary models besides.

One the interpretational flip-side, one could always have a $\Lambda$ that corresponds to the energy density of the true vacuum, and the rest as separate contributions on top of that. It's just not as useful in a cosmological context compared to grouping all 'dark energies' together, as all stress-energy gravitates equally.


Is the Sun really burning and getting smaller? Is the Sun really burning and getting smaller? If so, because there is no air in space, what can be keeping it burning and what materials get burnt? If no air is required, what can speed up or can stop the burning process?

I won't bother answering how the Sun "burns" its fuel — that is already answered in this question. Your second question is whether the Sun is getting smaller; the answer depends on what you mean by "smaller". We'll analyze this in two ways:

Radius

If we define the radius of the Sun by distance between the core and photosphere, then the Sun is not getting smaller. In fact, it is growing over time, as demonstrated in this graph:

Image courtesy of Ribas (2009), colored by user RJHall on Wikipedia under the Creative Commons Attribution-Share Alike 3.0 Unported license

The height of the photosphere is increasing ever so slightly, even in the Sun's main sequence stage; specifically, Table 3 of Sackmann et al. (1993) states that at the Sun's zero-age main sequence, it was $0.9R_\odot$ in radius, and at its terminal age main sequence it will be $1.6R_\odot$. That means that, over the course of about 10.9 billion years, the Sun's radius will have increased by $0.7R_\odot$ — about 70% its current radius.

If we do the math $\left (0.7R_\odot \over 10.9 \ \text{Gyr} \right )$, we could find that the Sun's average increase in radius (from beginning to end of main sequence) is about 4.5 centimeters each year (1.77 inches). This calculation shouldn't be taken too seriously, since the growth rate is not linear, but it demonstrates how slowly the Sun has been growing.

The Sun will continue to grow at a very slow rate until it leaves its main sequence stage. Once it enters the red giant branch, it will grow many times its current size, to approximately 1 AU.

Mass

The Sun is losing mass for two reasons: the nuclear fusion in its core converts some mass into energy, and high energy protons and electrons are constantly ejected from the Sun's corona as solar winds. Wood et al. (2002) shows that the Sun's mass-loss rate $\dot{M}$ is (approximately) inversely proportional to the square of its age $t$:

$$\dot{M} \propto t^{-2.00 \pm 0.52}$$

As of now, $\dot{M} = 10^{-14}\frac{M_\odot}{\mathrm{yr}}$. After the Sun leaves the main sequence, it will lose its mass at a much faster rate. As stated in Sackermann et al., the rate at which it a post-main sequence star loses mass can be generalized with Reimers' law:

$$\dot{M} = -\eta (4 \times 10^{-13}) \frac{LR}{M}$$

where $\eta$ is a value depending on the star (derived to be ~0.6 for the Sun), $M$, $L$ and $R$ are in solar units, and $\dot{M}$ is in solar masses per year. This is simplified, but gives us an idea of how the Sun's phase affects its mass-loss rate.

By the time the Sun enters the red giant branch, it will have only 72.5% its current mass, and its luminosity and radius will reach $2300L_\odot$ and $170R_\odot$, respectively. Plugging those values in, we find that the rate of mass loss will be $1.3 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$. It will lose mass even faster in the asymptotic giant branch, when its mass will become ~54.1% of what it is now, meaning $\dot{M}$ will be $2.5 \times 10^{-7} \frac{M_\odot}{\mathrm{yr}}$.


Local Standard Time Meridian - LSTM (Units) I was just wondering what units I could scale down to (or or shouldn't) for working out values like Local Solar Times, or Local Time etc...

From the papers I'm looking at, I can't see any mention of SI's or ISO's so I'm a little lost as to what to use as a reference.

Thanks

Ade

In a time, the word "standard" normally means the civil time, recognized by the government in the place of interest. I'm not sure about other countries, but in general conversation in the USA, "standard" also means that daylight saving time is not in effect. "Daylight" time (as in "Eastern Daylight Time") means the civil time, recognized by the government, and daylight saving time is in effect.

These can be expressed in the ISO 8601 format; the standard gives the example "15:27:46+01:00" meaning the time zone is one hour ahead of (east of) Greenwich.

In astronomy, time can also be the local time of the observer, ignoring the existence of time zones. This is local mean time, and is obtained from Universal Time by adjusting for the longitude of the observer. Usually the exact form of Universal Time that would be used would be UT1.

Local apparent solar time is based on the actual position of the sun; it is the time given by a sundial. It can be about 15 minutes ahead of or behind local mean time.


How long was an early Earth year? Roughly how long did it use to take for the early Earth 4.5~ billion years ago to complete one revolution around the Sun?

(I know the Earth was spinning much faster, so a day would only take 2-3 hours, so I'd like to have the answer in terms of current Earth days for better understanding - if possible.)

(I Google'd the heck out of this question and couldn't find an answer.)

There are two components to take into account for this question

the earth rotation rate on itself
the orbiting period around the sun

A day back then was shorter than today's earth day. The earth was rotating quicker in the past, but its rotating period around the sun didn't change much because of Kepler's law.
I cannot say how it was 5 billion years ago (anyway, did the earth had something you could call a surface to track the day from then?), but you could find data derived from fossils analysis for the last billion years, e.g. here from Nasa's SpaceMath, on how long a day was back then - I realize now the figures below assume the orbiting period of the earth around the sun (~8766 hours) stayed constant over this period of time. This assumption is wrong as explained here - however stating it "has been close to its current value for the last 2-3 billion years", so below figures are an approximation

Period
Age(years)
*Days/year
*Hours/day
ratio

Current
0
365
24.0
1.000

Upper Cretaceous
70   million
370
23.7
0.988

Upper Triassic
220  million
372
23.5
0.979

Pennsylvanian
290  million
383
22.9
0.954

Mississippian
340  million
398
22.0
0.917

Upper Devonian
380  million
399
22.0
0.917

Middle Devonian
395  million
405
21.6
0.900

Lower Devonian
410  million
410
21.4
0.892

Upper Silurian
420  million
400
21.9
0.913

Middle Silurian
430  million
413
21.2
0.883

Lower Silurian
440  million
421
20.8
0.867

Upper Ordovician
450  million
414
21.2
0.883

Middle Cambrian
510  million
424
20.7
0.863

Ediacarin
600  million
417
21.0
0.875

Cryogenian
900  million
486
18.0
0.750

From which I would roughly interpolate: 600 days earth self-rotation per year full rotation around the sun 2 billion years ago

Edit
We are saying here that earth was rotating quicker in the past:

Detailed  studies  of  fossil  shells,  and  the  banded  deposits  in  certain  sandstones,  reveal a much different length of day in past eras! These bands in sedimentation and  shell-growth  follow  the  lunar  month  and  have  individual  bands  representing  the  number  of  days  in  a  lunar  month.
By  counting  the  number  of  bands,  geologists can work out the number of days in a year, and from this the number of hours  in  a  day  when  the  shell  was  grown,  or  the  deposits  put  down.  The  table  above shows the results of one of these studies.


Why does Jupiter break comets apart? I understand that there are magnificent tidal forces that come with getting too close to Jupiter, but what is causing comets like Shoemaker-Levy to get torn apart?

I wouldn't expect that the gravitational forces are so strong that spaghetification occurs, is it?

It is.  Tidal disruption occurs when $\rho_{Jup}(R_{Jup}/R)^3$ > $\rho_{comet}$ where R is the separation and $\rho$ is the density.  The density of Jupiter is 1.33 g/$cm^3$ and the comet was estimated to have a density of only 0.5 g/$cm^3$ based on the radius at which it disrupted.  That is, disruption took place at separation R ~ $(1.33/0.5)^{1/3} R_{Jup}$.


What will happen when landing on Jupiter? Jupiter is a gas giant, so landing on it will not be like landing on Earth, our Moon or Mars etc., as it does not have a solid surface like these.

If we have a hypothetical spaceship or probe landing on Jupiter, and it can withstand the enormous pressure, what or how will be the sequence of events?

I already read this:

  A major problem in sending space probes to Jupiter is that the planet
  has no solid surface on which to land, as there is a smooth transition
  between the planet's atmosphere and its fluid interior. Any probes
  descending into the atmosphere are eventually crushed by the immense
  pressures within Jupiter.

What I do not understand:

Is this enormous or immense pressure of atmosphere or Jupiter's surface?
While descending, will we know that we have transitioned from the atmosphere to the surface?
Why does animated footage (real or conceptual) of Jupiter appear like a high viscous liquid (e.g. lava) rotating in an opposite direction, rather that gas(es)?
Does Jupiter have an inner solid core?

I've already gone through these:

Is Jupiter made entirely out of gas?
Could we fly/drive through Jupiter?
If Jupiter is a gas-giant then why don't its features change?

Jupiter does not have a "surface" and nor is there anything but an arbitrary division between interplanetary space and where its atmosphere begins.

The crushing pressure is its atmospheric pressure. The deeper into the atmosphere you go, the greater the column of gas that lies above you. It is the weight of this column of gas that is responsible for the rapid increase in pressure with depth.

The answer to your last question is most definitely addressed in the duplicate question about whether Jupiter is entirely made of gas. There is quite likely to be a liquid phase nearer the centre and there may be a solid core of order ten times the mass of the Earth. It is not a settled question.

The gas motions you talk about are essentially belts of weather systems in the upper layers of Jupiter's atmosphere. It is all most definitely gas that you can see.


Would Europa be an ocean planet if it were in the habitable zone? If a Europa-like body were in the Sun's habitable zone, let's say in an orbit between Earth and Mars, would the body become and remain a water ocean planet? In the habitable zone, the Sun would warm up and melt Europa's icy surface which will first turn into water vapor because Europa has no atmosphere, but the water vapor might form an atmosphere around Europa (which would contain oxygen as well) and eventually Europa might get a liquid water surface. Would that be possible? And would the atmosphere remain or couldn't it remain due to Europa's low gravity? But what if the Europa-like body had a Mars-like gravity and/or an Earth-like magnetosphere?

According to Arnscheidt et al. (2019) "Atmospheric Evolution on Low-Gravity Waterworlds", the transition between "planet-like" and "comet-like" (escaping) atmospheres occurs at a surface gravity of around 1.48 m/s2. As a caveat, the waterworlds being considered in the paper are objects which have water reservoirs of 40% of the total mass, which is a much higher fraction than Mars or Europa, such worlds would be more like scaled versions of Ganymede.
The surface gravity on Europa is 1.315 m/s2, putting it on the comet-like side of the transition. According to this, it seems unlikely that Europa would be able to maintain long-term habitable conditions in the face of atmospheric escape.
Martian gravity is 3.71 m/s2, putting it on the planet-like side of the transition, so a "super-Ganymede" with Mars-like gravity would likely be able to maintain long-lived habitable conditions.
They also note that the high albedo of ice means that deglaciating ice worlds tend to skip the long-lived habitable state entirely, going straight from frozen to runaway greenhouse/atmospheric escape:

The ice-albedo feedback can hamper transitions from snowball states to temperate states: this has
already been demonstrated for terrestrial-mass worlds
(Yang et al. 2017). We can incorporate the ice-albedo
feedback into our model using the simple albedo step
function
$$
A(T_s) = \begin{cases}
\alpha_I & T_s < 273\,\mathrm{K} \\
\alpha_L & T_s \ge 273\,\mathrm{K}
\end{cases}
$$
where $\alpha_I$ is the albedo of the icy (snowball) state, and
$\alpha_L$ is the albedo when there is surface liquid water. Hysteresis plots for different choices of $\alpha_I$ are shown in Figure 5. We observe that a snowball state experiencing
a stellar flux-driven deglaciation generally bypasses the
long-lived state entirely, except for very low $\alpha_I$ values.
Although the mechanism setting the inner edge of the
habitable zone is different, the conclusion of the habitable state likely being bypassed upon stellar flux-driven
deglaciation is the same as that of Yang et al. (2017).

(emphasis mine)


Can a supernova or a close passing star create a transient sun gravity focus "habitable zone"? This question was inspired by the previous question Would a black hole passing next to a star create a deadly focal point due to gravitational lensing?. I understand, as gravity also acts on light, that any massive object acts as a sort of lens. In the case of our sun, the focal distance starts about 550au, and there are several proposals to send probes there, given we could use the sun as a lens for a huge "telescope". See for example, Hesperides: Solar/nuclear missions to the Sun׳s inner gravity focus and First Stop on the Interstellar Journey:The Solar Gravity Lens Focus.
So I wonder, could the intensity of this effect be enough to melt water and create a sort of habitable zone (perhaps more of a habitable spot) in the opposite direction of a supernova as seen from the sun, by concentrating its light rays. I put the expression habitable zone inside quotes in the title because I don't think that sort of environment would be actually be habitable, as ionizing radiation of the supernova would be concentrated as well, so that would be a extreme environment. Yet, I can think of some interesting observable effects. I think that "habitable spot" would be tiny, but if we were lucky enough for a cometary nucleus pass through it just at the right time, perhapts it could develop a coma that we would be able to observe from Earth. Countless supernovas happened since solar system formation, so I would not be surprised if outer solar system contains a couple remnants of objects roasted this way.
The light of a near passing star would be concentrated too, but I think the effect would be weaker when compared with what a supernova like, say, SN 1054 can achieve.

Supernovas have a light curve going from normal to peak and back on the order of a month. Hence the spot would go from too cold to habitable and back very quickly.
The same is true for stars passing the focal point: their motion (typically tens of km per second) makes the heated point move equally fast.
Habitable zone denotes a region where a planet could have liquid water (and maybe other benign environmental properties). This implicitly assumes that this is a stable state, since the context is planetary orbits and the emergence of life, a process that takes long time. A ultra-brief habitable zone is missing the spirit of the concept, even if it may be correct to state that a certain location is habitable over a short span of time.
As the deadly focal point post discusses, the brightening is also not enormous: to get a habitable spot, you need a fairly nearby supernova or star, so the gain is not enormous.


What determines if a comet's debris will produce periodic meteor showers? Background
I was wondering if the debris from the comet NEOWISE that recently passed Earth would provide meteor showers in the future.
I looked up that the Perseid meteor showers happen when the Earth passes through the debris left behind by comet Swift-Tuttle which has a nucleus diameter of 26 km. In comparison, the comet NEOWISE has a nucleus diameter of 5 km. I am guessing this means it is too small to have a debris trail that produces periodic meteor showers? I was not able to find data about the length of these comets' tails.
The Question
What determines if the debris from a comet will produce periodic meteor showers in the sky?
After thinking about it a bit myself it seems like these variables should be important:

Diameter of the comet
Length of tail when comet reaches Earth's orbit (which probably depends on comet size and composition)
Orbit of comet and its relation to Earth's orbit
Orientation of comet / comet's tail in relation to Earth when it crosses the Earth's orbit
Whether or not tail extends to Earth as the comet moves farther away from Sun
Whether or not the debris persists after the comet has left

I was hoping someone could provide more insight into what determines if periodic showers will be produced by a comet. Specifically, what variables are involved (comet size, length of tail, comet's orbit in relation to Earth's orbit, etc.). I'm particularly interested in what determines the length of a comet's tail as well as whether a debris trail persists long after the comet has left.

Meteor showers tend to be formed from dust from short period comets.  As they swing repeatedly around the sun, they build up a stream of dust in orbits close to their own orbit. Some will be pushed slightly off the comet's orbit and end up orbiting faster or slower than the comet, and eventually the stream of dust will go all around the sun.
The other thing that there needs to be to get a meteor shower is a comet whose orbit is close to crossing the Earth's orbit. Comet Swift-Tuttle is one such body. Its orbit passes only 130,000 km from the Earths orbit (close enough that there was at one point some concern that it could collide with the Earth in the future. This has now been ruled out for the next two thousand years).
Neowise will release dust, but most of that dust will still be near the comet now and it will continue to orbit with the comet. So that dust will be leaving the inner solar system and will become too spread out to cause a significant meteor shower by the time it returns in several thousand years. Neowise doesn't approach the sun often enough to build up a significant stream of dust particles.
So to summarize it's not really the comet's tail intersecting the Earth's orbit that leads to showers but the debris stream built up along the comet's orbit as it repeatedly goes around the Sun. In long-period comets the debris stream is spread out too much along the orbit to produce a significant stream but short-period comets can produce a significant stream leading to showers.


Telescope and its construction I want to build a reflecting telescope in my home. I've attached a drawing of my plan that contains some explanations of what I understand so far, and highlights the parts of the design for which I have questions.

What type of mirror should i use, parabolic or spherical, and what are the factors considered when making that decision?

Does the glass used to make the mirror affect the incoming light rays? What about the secondary mirror in the middle of the tube?

Are there other things to consider that would help for higher magnification?

Some telescopes have a glass window at the front where light enters the tube, why is that? Would I need to add it here?

This is a very broad topic. You can literally write entire books on it. I'll stick a short answer here for future reference.

You have two main options: refractor or reflector.

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

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

Refractor is easier. You'll need a relatively large lens with a long focal length - this will be your primary optics, at the top of your telescope. The length of your scope will be dictated by the focal length of the primary lens.

You can measure the focal length by going outside when the Sun is up and trying to focus the Sun to a tiny pinpoint image - the distance then between lens and image is the focal length.

As an ocular or eyepiece you could use a smaller lens with a short focal length. Much better, you could use an eyepiece from an old microscope or binoculars.

The total distance between the objective (primary lens) and ocular (eyepiece) is the sum of their focal lengths. If the primary lens has focal length F1, and the eyepiece has focal length F2, then when looking at infinity the distance between them is F1 + F2. This will help you calculate the total length of the tube.

However, if the ocular (eyepiece) is a divergent lens (the kind of lens that makes things seem smaller) then the total distance is F1 - F2. By the way, it's hard to measure the focal length of a divergent lens, because it doesn't make an image you can project.

The magnification of the instrument will be:

M = F1 / F2

Do not worry too much about magnification. Make something that works first. More magnification is not always better. In the beginning, aim for something in the 10x ... 50x range. Refine it later.

Make sure the tube is rigid. A plastic pipe like the ones used in water installations should work well.

Make sure all lenses are centered and aligned with precision. You don't want your lenses to be leaning in any direction. Their main axis should coincide with the axis of the scope. This is important for performance.

If you get too many color fringes (little rainbows) around objects, try and reduce the diameter of the primary lens by placing a paper ring in front of it - this makes the image a bit darker but should diminish the pesky rainbows somewhat.

Mount the eyepiece on a smaller tube that's sliding into the main tube. Cut the main tube a little shorter than F1 + F2. This will allow you to move the eyepiece back and forth until the image is in good focus.

If you find a shop that makes prescription eyeglasses (for people with eye or vision problems), they have so-called "blank" lenses that they use to cut the glasses from. See if you can buy one of those for your primary optics. A focal length of about 1 meter should be fine (opticians call it a 1 dioptre lens, which you can find in pretty weak prescription glasses).

Sometimes you can pull a good lens out of a broken projector or something, but make sure the diameter is big enough (50 mm or bigger should be fine) and the focal length is big enough (I would say anywhere between 0.5 and 2 meters - that's between 2 and 0.5 dioptres - but keep it close to 1 m if possible).

The first telescope I made, many years ago, was very much like this. A blank lens from a prescription glasses maker, a plastic pipe, an eyepiece from a toy microscope. It worked reasonably well, I could see with it craters and mountains on the Moon.

Making a reflector is much harder, but the performance of the instrument is higher. Read the Stellafane guide, it's very comprehensive:

https://stellafane.org/tm/

What's described in there is a dobsonian telescope. That's a newtonian reflector on a simple alt-azimuth mount.


Struggling to use my new telescope I recently got my first telescope, a Bushnell reflector telescope. I spent a good while setting up my alignment. And I can see the moon perfectly! I went to look at another, Mars. I got it aligned in my finderscope. And i go to focus to see. And i see....basically the mirror in my scope with the lines. Its clearly getting light from the star....but i can't get it to focus on the star at all! Help!!

It looks to me like you have two problems.

You need to adjust the focus. All telescopes come with a knob that alters the focus. This is pretty easy.
the shadow you see is from the secondary mirror, and it is off centre. I don't know what the word for this is (off-collimation?), but I had it the other night and I fixed it by wiggling the secondary mirror and making sure it was flush with the hole it's supposed to fit in.

Try using the focus and if it's still bad, take another shot and post it here and maybe someone with more experience than me can help.


Why isn't the black hole white? I recently saw the first image of a black hole. As I understood, it is covered with bright, hot matter. In this case, how can we see the black disk (event horizon), instead of a bright disk due to the matter surrounding the black hole? Or is the matter around the black hole disposed in a disk shape and the astronomers got lucky that the disk was not oriented towards Earth?

What's going on here is that you have been misled into thinking the ring-like structure has anything to do with the accretion disk. It doesn't, or at least only indirectly.

The disk is referred to as geometrically thick, but optically thin (see sections 1 and 2 of paper V issued by the Event Horizon Telescope collaboration on 10 April 2019). This is actually the opposite of the disk visualised in the film "Interstellar".

Because the disk is geometrically thick, it covers the whole picture. Because it is optically thin, we can see through it to the black hole. That is the basic answer to your question.

In an optically thin plasma, the brightness you will see is proportional to the optical path length (physical length multiplied by an absorption coefficient) along that sightline. The photon ring marks radiation travelling towards us that has been bent around the black hole, or has even orbited it several times. Hence those sight lines have larger optical path lengths and that is why we see it as a bright ring.

Radiation travelling towards us from plasma in front of the black hole, has a small optical path length and is not very bright. In addition, the sight lines to plasma behind the black hole cannot travel through the black hole or even close to the event horizon. Hence the circular "shadow" inside the photon ring.


Is **Voyager I’s** reduced data transmission rate as described in [this article][1] because of the distance or because its transmitter getting slower? Is Voyager I’s reduced data transmission rate as described in this article because of the distance or because its transmitter getting older & slower?

Transmitters don't slow down as they age. Red shift isn't involved, at all. As a transmitter gets further away the signal at the receiver gets weaker, so it's deliberately slowed down. Data channel capacity depends on signal to noise ratio and bandwidth. As the signal gets weaker the data rate can be slowed to reduce the necessary bandwidth.  Noise doesn't increase: it's just a loss of signal strength at the receiver. Narrowing the receiver bandwidth reduces the noise received, but also limits the information bandwidth.


How to identify stars in photographs? I can look at a star chart and identify things like constellations in the sky.  But if I take a picture with my DSLR camera (35mm with a decent zoom lens, no astronomy specific optics), I run into difficulty. I find that the camera can see a lot more stars than my eye, and I find that I start to have great difficulty even identifying simply constellations.  For instance, when I look for the constellation Cassiopeia in the night sky it is very easy to spot: it looks like a sideways "W".  But when I take a picture I don't see a W at all in the picture, I see about 16 stars around where the "W" should be a weird angles to each other, and it's hard to figure out which stars to connect to form the actual constellation.

And it just goes downhill from there, as other constellations don't even have a shape as easily recognizable as Cassiopeia, and as a result I have a great deal of difficult even recognizing what I am looking at.

A photograph with a regular camera is not quite the same as looking through a telescope, because the camera has a much wider field of view.  And at the same time, there does not seem to be a linear relationship between the actual magnitude of the star and how bright it looks.  As I mentioned above, I see 16 stars which look somewhat comparable in brightness, but I know that they differ by several magnitudes.

Is there some kind of trick or technique that I can use to quickly identify stars in pictures I take?  In looking for clues, I have not seen anybody mentioning this problem.

Take another photograph of the same field that is less exposed. (Doesn't matter if you do this by shortening the exposure time, decreasing the ISO, etc.) This will give you an image with far fewer stars so you can easily pattern match your images.