Temperature of Earth if tidal locked with the Sun? Assume the Earth somehow stopped spinning once every ~24 hours, and instead started rotating at ~365 days per rotation, so it effectively would end up being tidal locked with the Sun.  In this scenario, the same side of the Earth would always be facing the Sun, and the opposite side of the Earth would always be in darkness.  Disregard all sorts of cataclysmic events that would transpire if something like this happened.

I think it would be safe to assume that life as we know it would quickly cease to exist except on the very edges of the day/night, but what would the approximate temperature stabilize at both on the point on the Earth directly facing the Sun, and a point on the surface of the Earth exactly away from the Sun?

That's very, very difficult to answer with a lot of if's. If the earth had no atmosphere it would be 255K-277K (-18-3°C) on the day side and a lot colder on the nightside. But if we include the atmosphere into our calculations we can get widely differing results.

For the sake of simplicity let's assume that earth tidal locks and nothing else happens. No change of atmosphere because many plants die or human population and therefore CO2 emmissions go down or something like this.

In this case I guess that the temperature would stay about the same. While one side of the earth enjoys permanent sunlight, the air on this side heats up and therefore rise to higher layers of the atmosphere. Cooler air from the other side of the earth streams in and the hot air in exchange streams to the cooler side where it cools down and returns to the bright side. So while one side of the earth is permanently sunny it is also permanently cooled down by air from the other side.

The dark side would not cool out because it is permanently heated by the hot air from the sunny side.

On the whole we would live on a very windy planet. But thats it.


Is it possible that all dark matter is made of rogue planets (free-floating planet)? Is it possible that all dark matter is made of rogue planets (free-floating planet)? (and other stuff like asteroids or meteoroids)

First of all I'll start with a few ideas:

Baryonic Matter: Baryons are elementary particles made up of 3 quarks. This includes protons and neutrons, and the term baryonic matter refers to matter made of baryons, such as atoms. Examples of non-baryonic matter includes neutrinos, free electrons and other exotic matter.
Things like planets, stars, dust, etc. are all made of atoms, and so are classified as baryonic matter.

Now, how do we know that dark matter is present in the universe?

Astronomers measure the gravitational pull of galaxies and galaxy groups/clusters based on how objects behave when interacting with these objects. Some examples of this include tidal gas/dust stripping, the orbit of stars in a galaxy and gravitational lensing of distant light from a large cluster. Using this they determine the mass of the galaxy (or galaxy group).
We can also determine the mass of a galaxy or group by looking at it and adding up the mass of all the objects (like stars, dust, gas, black holes, and other baryonic matter). While these methods both give us approximations, it is clear that the gravitational mass of galaxies and groups exceeds the baryonic mass by a factor of 10-100.

When astrophysicists first found this phenomenon they had to come up with a plausible explanation, so they suggested that there is some new, invisible matter called dark matter. (Aside: some astrophysicists also came up with other explanations like modified gravity, but so far dark matter does the best job at explaining observations).

Okay, so now how do we know dark matter is not any sort of baryonic matter?

There are a few reasons astrophysicists know that it is extremely unlikely that dark matter is baryonic. First of all if all the stars in a galaxy shine on an object it heats up, this heat causes the release of radiation, called thermal radiation, and every (baryonic) object above zero kelvin (or -273.14 deg celcius) emits this radiation. However, dark matter does not emit any radiation at all (hence the name dark!)

If dark matter were baryonic it would also mean that it could become light emitting. If we got a clump of baryonic matter* and put it in space it would gravitationally contract, and would eventually form a star or black hole** - both of which we would be able to see.

So, because of these reasons the dark matter in galaxies and in galaxy groups/clusters cannot be baryonic, and so cannot be planets, dead stars, asteroids, etc. It would definetely not be planets as there is no way 10-100 times the mass of the stars in a galaxy would be planets, as the mechanism for making planets relies on supernovae, and the number of supernovae needed for the that many planets would be far too high to match our observations. I hope that this answered your question!

*provided the clump of baryonic matter was large, and the amount there is in galaxies definitely is!

** we don't observe black holes directly, but can see radiation from their accretion disks.


Why does the Chandrasekhar limit affect white dwarfs differently? The Chandrasekhar limit is the maximum mass of a stable white dwarf star. Beyond this, a carbon-oxygen white dwarf will typically explode in a type 1a supernova, due to the nuclear reactions at those temperatures.

I've heard that oxygen-neon-magnesium white dwarfs, on the other hand, will not ignite. Rather, electron capture becomes energetically favorable and they become neutron stars. If I understand correctly, this also happens for carbon-oxygen white dwarfs in binaries, if the white dwarf has most of the mass in the system.

Why is this? Can anyone walk me through, step by step, what happens if a white dwarf (of both compositions) exceeds the Chandrasekhar limit, and how each differs?

Source: https://arxiv.org/abs/astro-ph/9701225

There are a variety of white dwarfs with various compositions, and analysing how they detonate in a supernova (or not) is an topic under investigation. A simple model, described in "How is the first detonation in Supernove type Ia triggered?" is of a helium shell initially igniting and that setting off the carbon in the core.

In this type of Type 1a supernova there is no neutron star formed, as the star is completely destroyed. The mass at which a white dwarf will undergo a type 1a supernova is just below the chandrasekhar limit, so neutronisation won't occur.

However in some white dwarfs with an atypical composition (as you note a Mg-Ne-O white dwarf) it is possible for the star to avoid detonation, and reach the chandrasekhar limit, and so for electron capture to occur, and a neutron star to form. It is worth noting that there is not definite observation of a white dwarf collapsing to a neutron star (where as there a lots of observations of type 1a supernovae) however these "accretion induced collapse" scenarios may explain some magnetars and short gamma ray burst

Progenitors of the Accretion-Induced
Collapse

So the two scenarios are,

Accretion occurs onto a Carbon-oxygen WD. The pressure and temperature in the core of a white dwarf increases until thermonuclear reactions begin (at about 1.38 solar masses). Since the white dwarf is degenerate it can't expand to reduce the rate of thermonuclear reactions, and the entire star detonates and is destroyed.
Accretion occurs onto a ONeMg WD. The star reaches 1.44 solar masses, electron degeneracy is no longer sufficient to prevent collapse. Electron capture occurs and the star collapses to neutron star.


Loading contour file in ds9 I have an ASCII file (.con) with the coordinates of vertices of different contours. This file is obtained from Matlab. Now after every set of contour ends there needs to be a blank row in the contour file. In place of this, I have a row with Nan Nan in it when Matlab writes the file. What I do is simply find and Replace the Nan with nothing. But when I load the contour file in ds9 it gives me weird contours. I'm also posting the result I expect (fig 1) and the result I'm getting (fig 2). The problem here that I've understood till now is that if I copy and paste every contour block of coordinates separately into the .con file the result is as expected but I want the whole process of writing the contours from Matlab into the file such that it gives me the expected result in ds9 to be automated (basically I don't want to copy and paste every block). Can somebody help with this?

This is what I expect:

Link for the .con file that gives the right result: https://drive.google.com/open?id=1tDMXtmywJnSjpi_CzCI5-F85oFpebnl4

Link for the .con file that gives the wrong result: https://drive.google.com/file/d/18rgN1UVtvcyfJSuh1mmbqOVQCna_aAUM/view

The solution as pointed out by samcarter in the above comment was to remove the tab in conto.con so that it had linebreak + linebreak instead of linebreak + tab + linebreak


Are there any planets or objects orbiting in Goldilocks zone? I was wondering whether there were any objects that orbit the "goldilocks zone" besides our home planet earth, and how the goldilocks zone of our solar system is defined.

From the Wikipedia article Circumstellar habitable zone, which is just another name for the Goldilocks zone:

  Estimates for the habitable zone within the Solar System range from 0.38 to 10.0 astronomical units, though arriving at these estimates has been challenging for a variety of reasons. Numerous planetary mass objects orbit within, or close to, this range and as such receive sufficient sunlight to raise temperatures above the freezing point of water. However their atmospheric conditions vary substantially. The aphelion of Venus, for example, touches the inner edge of the zone and while atmospheric pressure at the surface is sufficient for liquid water, a strong greenhouse effect raises surface temperatures to 462 °C (864 °F) at which water can only exist as vapour. The entire orbits of the Moon, Mars, and numerous asteroids also lie within various estimates of the habitable zone.

See also these questions:
 - What is the Habitable zone all about?
 - Why the Circumstellar Habitable Zone is defined as it is, if life could be possible outside of it?


Can looking at the sun through solar viewing eclipse shades everyday harm your eyes? I have a pair of Solar Viewing Eclipse Shades that say "Safe for Direct Solar Viewing" and would like to know if my eyes would be harmed if I looked at the sun everyday for about a minute wearing them.

It depends on the quality of glasses and intensity, nature and degree of harmfulness of sunrays which you are watching to. Every eclipse has different intensity so as it can harm you.

When there was a solar eclipse around 3-4 years ago, it was a full solar eclipse. That time I watched it from my school with total naked eyes. My eyes were not harmed from that, but my school principle did.

But I still recommend you to not to watch it directly, use projectors which would allow you to get the view without directly watching the sun. Only use original products which are being certified.


Are there any known objects in a "dual" tidally locked orbit? Is there any known pair of objects tidally locked with each other?  I know the moon is tidally locked with the earth (e.g. one side always facing earth), but I don't know of any pair of objects that are both tidally locked with each other.  For example, the planet would be tidally locked with the moon, and the moon would be tidally locked with the planet.

Pluto and its largest moon Charon are tidally locked to each other.

  Charon and Pluto revolve about each other every 6.387 days. The two objects are both gravitationally locked to the other, so each keeps the same face towards the other. This is a mutual case of tidal locking . . .

Because of Charon's large size compared to Pluto, and because its barycenter lies above the primary's surface, some astronomers have referred to the Pluto/Charon system as a dwarf double planet.

  The system is also unusual among planetary systems in that each is tidally locked to the other: Charon always presents the same face to Pluto, and Pluto always presents the same face to Charon: from any position on either body, the other is always at the same position in the sky, or always obscured.(110)


What exactly is the "paradox" in Olber's Paradox? To the extent of my understanding, Olber’s paradox states that if the universe was static and homogeneous, we should see a star at every point in the night sky and therefore the night sky should be equally as bright as day.

However, since the night sky is dark and non-uniform, it can be said that the universe is not static and not homogeneous. However, if this was already known, what exactly is the “paradox”? Why isn’t it called Olber’s Observation or something else?

Olber's Paradox was created at a time before the idea of a finite universe was accepted.  (It was thought of in the 1600's).  In order to resolve Olber's Paradox, you have to introduce the idea that either the universe had a beginning or it is of finite size.  (Note: the solution does not require an expanding universe).  So, at the time, it was a paradox.  Pretty much all astronomers considered the universe to be static and infinite.  Therefore, the fact that their observations didn't fit with what they expected made it a paradox.


Are the bright dots in this image individual stars? I saw this picture of M51 (Whirlpool galaxy) and noticed many, many bright dots around and in the galaxy. I have freehand-circled a few as examples. Are these individual stars (probably supergiants) in the Whirlpool Galaxy, or are they clusters, stars in the Milky Way, or something else?

Source: ESA/Hubble

A query to the Gaia Archive within 6 arc min of target name M51 finds 20 sources with phot_g_mean_mag <= 17.
Aladin Lite helps to locate them visually.
Five are circled in the lower half of the image in question:

Gaia source_id
ra °
dec °
parallax (mas)
magg

1551894661845524736
202.4034
47.1522
8.854 ± 0.017
14.3

1551988532650838016
202.4205
47.1790
2.387 ± 0.041
16.4

1551988670089792256
202.4236
47.1979
0.632 ± 0.025
15.1

1551988425277200640
202.4323
47.1537
0.813 ± 0.025
15.2

1551988567010595840
202.4417
47.1800
1.536 ± 0.012
13.3

These parallaxes indicate distances between 113 and 1600 parsecs, well within the Milky Way.
Of the three sources circled in the upper half, two seem to be listed in Gaia EDR3 but do not have good parallaxes there.
Aladin Lite shows SIMBAD matches near all three in
Hwang and Lee 2008, a catalog of 2200 star clusters in M51.

HL2008 id
ra °
dec °
magV
Gaia source_id

89770
202.494
47.216
20.0
1551988944968489984

13162
202.505
47.154
21.1
--

28066
202.531
47.167
19.8
1551987909882986368


In geocentric models, did the Earth rotate? In geocentric models, did the Earth rotate around its axis?

I assume that if the Earth rotated around its axis, then the Sun would not have to orbit around the Earth at a very fast rate, whereas if the Earth didn't rotate, then the Sun would have to orbit the Earth every day.

In the historic geocentric models, and in particular in the Ptolemaic model, the earth is immobile in the centre of the cosmos, and the sphere of the fixed stars rotates around it once a day, carrying the sun, moon and planets with it. There were astronomers who realised that this apparent motion could also be explained, from a purely mathematical point of view, by assuming that the earth rotates on its axis, but they rejected this option. By the way, neither Ptolemy nor any other serious astronomer believed that the earth was flat.


How much of the ambient radiation on Mars is arriving line of sight from the Sun? I read a lot about the amount of radiation bodies like the Moon and Mars receive, but it's hard to find information about where the radiation is coming from. It seems to me that it's cosmic radiation plus whatever is being radiating from the Sun. Would it be useful to build a habitat somewhere that's in perpetual shadow (e.g. deep crater), or to seek shelter in the shadow of a crater wall or mountain during a solar storm. Mostly I'm thinking about a habitat design for Mars, where the habitat will always be out of view of the sun. Sunlight could light the habitat via something like huge Mylar mirrors, that would not reflect much harmful radiation. But I'd like to know if the applies to the Moon.

Besides solar energetic particle (SEP) events, it's mostly (almost isotropic) galactic cosmic rays (GCR). GCRs may cause secondary radiation by interaction with the atmosphere or with rock. The smaller the solid angle of the visible sky, the smaller the mean galactic radiation dose. On Mars you get additional shielding by the atmosphere, hence dependence of the atmospheric pressure. More detail in this paper.
Shieldings from SEP events should be rich in protons, e.g. water or food, or you need to go deeper into underground, as soon as a SEP event is detected.
GCRs are partially shielded by the heliosphere. Hence solar maxima protect partially from GCRs, but SEP events get more likely during solar maxima.
Since GCRs are unpredictable, and SEP events can be detected, for solar maxima, protection strategies resultng in an overall reduced radiation dose can be elaborated.


Why could Quasi-stars ("black hole stars") have only existed when everything was hydrogen and helium? (no metal "contamination") This informative answer to What was the absolute limit to the possible sizes of the first stars formed from “primordial material with no metals”? led me to Wikipedia's Quasi-star; Formation and properties which says:

A quasi-star would result from the core of a large protostar collapsing into a black hole, where the outer layers of the protostar are massive enough to absorb the resulting burst of energy without being blown away or falling into the black hole, as occurs with modern supernovae. [...] Formation of quasi-stars could only happen early in the development of the Universe, before hydrogen and helium were contaminated by heavier elements; thus, they may have been very massive Population III stars.

But neither the answer nor Wikipedia plainly state just what it is about the presence of metals that would make Quasi-stars impossible, such that if they ever did exist it would only have been in the early universe.
Question: Why could Quasi-stars ("black hole stars") have only existed when everything was hydrogen and helium? (no metal "contamination")

The presence of heavier elements makes the medium absorb more radiation. This means a nascent star that would have been able to collapse in the absence of metals, will lose its outer layers to radiation pressure as the core starts to heat up.


Will James Webb see Population III stars? I have heard that James Webb will see the first stars that our universe produced. Can I assume that we may see galaxies that are so young that all of the stars in them are population III?

According to Rydberg et al. (2013) "Detection of isolated Population III stars with the James Webb Space Telescope", the answer for non-lensed population III stars appears to be no. For lensed stars, the detection would be difficult but the expected number of detections is very small:

  To detect even one 60 M⊙ Population III star when pointing JWST through the galaxy cluster MACS J0717.5+3745, the lensing cluster with the largest Einstein radius detected so far, the cosmic star formation rate of Population III stars would need to be approximately an order of magnitude higher than predicted by the most optimistic current models.

Prospects may be somewhat better for detecting population III galaxies. These would likely be small (dwarf) galaxies in low-density regions, as larger galaxies likely became rapidly polluted by supernova ejecta. From Zackrisson et al. (2012) "Detecting gravitationally lensed Population III galaxies with the Hubble Space Telescope and the James Webb Space Telescope":

  We find that the ongoing HST survey Cluster Lensing And Supernova survey with Hubble (CLASH), targeting a total of 25 galaxy clusters including J0717, potentially could detect a small number of pop III galaxies if ∼1 per cent of the baryons in these systems have been converted into pop III stars. Using JWST exposures of J0717, this limit can be pushed to ∼0.1 per cent of the baryons. Ultradeep JWST observations of unlensed fields are predicted to do somewhat worse, but will be able to probe pop III galaxies with luminosities intermediate between those detectable in HST/CLASH and in JWST observations of J0717.

(The designation J0717 is an abbreviated form of MACS J0717.5+3745 referred to in the Rydberg et al. paper.)


Astrophysical Jet Size? I am wondering about what is the estimated size of astrophysical jet that shoots out from a black hole, I don’t know if it is determined by the size of a black hole.

What would classify a black hole that is able to have at least 10 planets orbit it? Is it considered stellar?

If so, what would the estimated astrophysical jet or quasar be?

There are multiple categories of Black Hole powered astrophysical jets:

Microquasars: As their name implies, they are somewhat like a highly scaled-downed version of quasars, but unlike quasars, they are powered by accretion from a donor star. They contain stellar-mass black holes.

Quasars: They are huge systems of accretion disk(s) around supermassive black holes.  They are found in the centers of galaxies, and are very powerful, enough to outshine their host galaxies.

The size of jets depends on the type: microquasars have jets only a few light years long, whereas quasars have jets millions of light years long.
The size of the jets depends on the mass of the Black hole and the magnetic field of the black hole and the magnetic field of the accretion disk.


How can you determine the distance of a star using the magnitude equation? In high school we are undergoing a short course about astronomy. It is stated that you can use the apparent and absolute magnitude (m, M) to get a distance (D) of a star, with the formula
$$m - M = 5 \times \log(D) - 5$$
From which you free D:
$$D = 10^{\frac{m-M+5}{5}}$$
But how do you get both the m and M in practice, without knowing anything else? I can't imagine it.
I have asked my teacher but his explanation did not make sense to me, and as such I already forgot it.
Scenario:
you have a satellite which can measure the incoming photons  of a star. You now calculate the apparent magnitude using the formula:
$$m = m_\text{ref} - 2.5 \times \log(\frac{I}{I_\text{ref}})$$
Let's say you have a stable reference star and so you get $m=2$.
This is a nice start, but to determine the distance of the star you still need the absolute magnitude. How can you precisely acquire that using a HR-diagram? Because all the HR diagrams I have seen have a very fuzzy curve, and one temperature may map to multiple M.

This is what I'm struggling with: let's say you are somehow able to precisely measure that the star is 7500K. That means M can range from +4 to 0 (I guess we're also assuming it's a main sequence star, which we may not actually know), which gives us a possible D of about 4 to 25 (parsec?), which is an absolutely humongous range.
This can't be right, can it?

This is all absolutely true (pun intended). Finding absolute magnitudes is hard.  For many types of star, we don't really know their absolute magnitude and so we don't really know their distance.
For close-by stars we can get the distance by measuring the parallax (how far the star appears to move over a year due to the orbit of the Earth. Nearer stars appear to move more) From this we can work out absolute magnitude.
For other stars we can get their absolute magnitude in other ways. For example, a type of large bright star called a "Cepheid variable" will have a pulse rate P that is connected to it's absolute magnitude $M_\mathrm{v}$ by the following (empirically derived) formula:
$$M_\mathrm{v} = -2.43(\log_{10}P - 1) - 4.05$$
With this you can easily measure the period in days, and use that to find the absolute magnitude.  The apparent magnitude is easy to measure, it is how bright the star appears to be on Earth.
And with apparent and absolute magnitudes you can find the distance of the star.
Stars like Cepheid variables are called "standard candles", since their brightness can be determined exactly


What causes the dimensions of a star increase when its hydrogen fuel is exhausted? What causes the dimensions of a star increase when its hydrogen fuel is exhausted? For example, the Sun is expected to increase its radius 250 times. What causes this if its temperature is expected to fall? How gas can expand if the temperature falls?

The Sun will never run out of Hydrogen. This is a common misconception.

At this moment the Sun is fusing Hydrogen into Helium. This fused Helium remains at the core until it will reach a critical mass. At this point the core will begin to collapse. This collapse increases the temperature and pressure around the core where Hydrogen is being fused causing the increase the Hydrogen fusion which also causes an increase of the radiation pressure making the outer layers to expand and cool.

Once the core reaches a temperature of about 100 million Kelvin Helium fusion starts dramatically (Helium Flash) with high radiation pressure in the core. Here is where the Sun will reach 250 times it's current radius.

When the Helium fusion stabilizes the radiation pressure at the cores decreases making the Sun's radius to shrink.

The temperature will only drop at the surface because as it's expands it's further from the core.


How do astronomers detect the 'metals' in a star? If the atoms are presumably completely ionized? Atoms and molecules usually emit their characteristic wavelengths because of the electrons' energy levels....
Do the completely ionized ('naked') nuclei absorb and/or emit EM radiation?
If so, at what type of wavelength?  Mostly visible, IR and UV like the rest of the star?

You are correct that the characteristic emission and absorption lines we see in stars' spectra are from electrons that are bound to atoms making transitions between different energy levels.  That is possible because the elements in a star's photosphere are not fully ionized.  Hydrogen - the easiest element to fully ionize because its nucleus only has a charge of +1 - is pretty highly ionized in the hottest stars (O stars), which is why those stars have weak hydrogen lines in their spectra.
Completely ionized atoms (bare nuclei) do not absorb or emit electromagnetic radiation in the parts of the spectrum where stars emit most of their light.  They can absorb very high energy photons (gamma rays with energies of millions of electron volts) that have enough energy to excite their nuclear energy levels, but the photospheres of stars aren't hot enough to emit this kind of radiation (and as previously stated, the atoms there aren't fully ionized anyway).
Many stars have elements in their photospheres that are partially ionized, but the spectral lines we see from those atoms are still from the remaining bound electrons.  For example, Ca II (singly-ionized calcium, with 19 electrons instead of 20) has strong lines in the violet part of the spectrum of the Sun.


How would a small TCO (temporarily captured orbiter) or other natural Earth satellite most likely be detected? A sentence from the abstract of The population of natural Earth satellites states:

  At any given time there should be at least one NES of 1-meter diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286± 18) d.

As far as I know, the only known, well documented temporarily captured Earth satellite is 2006 RH120, which was in within 1 million km of the Earth for about 1 year. It has an estimated size of 2 to 3 meters, a rotation period (about it's axis) of about 3 minutes (presumably determined by photometry). According to Wikipedia:

  2006 RH120 was discovered on 14 September 2006 by the 27-inch (690 mm) Schmidt camera of the Catalina Sky Survey in Arizona.

Using data from the JPL Horizons database, it seems 2006 RH120 was only about 850,000 km from Earth at the time it was discovered.

Deep sky surveys are extremely prolific at detecting small solar system bodies. There was also a recent answer addressing the possibility that the Gaia spacecraft's could detect trojan asteroids associated with Earth's orbit, linking to the paper Detection of inner Solar System Trojan Asteroids by Gaia.

The question asked specifically about TCO's being detected by Gaia, but my question is more general: How would a small TCO (temporarily captured orbiter) or other natural Earth satellite most likely be detected?

There are Earth-based sky surveys, space based visible and near-visible, and possibly even infrared/thermal detections possible, even Earth-based radar, and possibly others.

Are there any searches specifically targeting TCOs?

above: a simple plot of the orbit of 2006 RH120 from JPL Horizions transformed to an Earth-centered Earth-Sun-synodic frame - red line with dots every 10 days, for the period when it was within a +/- 1.6 million km cube centered on Earth in this frame. The green thick band around Earth is the moons orbit during this 448 day period.

Cruithne and J002E3 were detected using ground based astrophotography.
J002E3 is thought to be "the S-IVB third stage of the Apollo 12 Saturn V rocket ... It is thought that J002E3 left Earth orbit in June 2003, and that it may return to orbit the Earth in the mid-2040s.1"
Ground telescope also found 6Q0B44E which orbits beyond the moon.
2002 AA29 was found via automatic sky survey.
Several of these objects are in horseshoe orbits.

Regardless, the discovery of these objects usually comes via conventional astrophotography.


Physical size in different redshift I have only recently started to study astronomy. I would like to ask if I want to get the angle of Mpc under different redshifts, but the result I calculated seems a bit strange. Did I make a mistake while converting?
from astropy.cosmology import FlatLambdaCDM
import numpy as np
from astropy import units as u
import math
cosmo = FlatLambdaCDM(H0=70, Om0=0.3)

d_A = cosmo.angular_diameter_distance(z=0.3)
print(d_A) #  Mpc

x = 3.29*u.Mpc

theta = (x/d_A *(180/math.pi) *(1/3600))
theta = theta*u.arcsec

print(theta)

You're correct, except that you should multiply, not divide, by 3600 (you're converting from degrees to arcsec, not the other way round). You get $5.7\times10^{-5}{''}$, but the correct result is 36002 times higher, i.e. $738.6''\!\!\!$.
But why not just let astropy do the job:
z     = 0.3
x     = 3.29 * u.Mpc
theta = x * cosmo.arcsec_per_kpc_proper(z)

print(theta.to(u.arcsec))
---> <Quantity 738.61092463 arcsec>


Are galaxies moving away from us faster than before? Are all galaxies moving away from us at constant speed even those that may be moving in our direction as the space is being formed? How does nothingness appear to push matter? A black hole sucks as space is compressing (being stretched without stopping) or compressed (like how our Sun pulling all planets) and is it the opposite phenomenon that show universe is expanding?

The recessional velocity $v$ of an object depends on two things: firstly it depends on how faraway an object is in terms of proper distance $D$, and secondly on the rate of the Universe's expansion as a function of cosmological time $t$, which is best expressed as the Hubble parameter $H(t)$. Specifically:

$$v = D \times H(t)$$

It's worth noting that this equation is slightly vacuous as it is merely the definition of recessional velocity, which is not something that can be directly measured.

As recessional velocity depends not just on a function of $t$, but also on $D$, the question as to whether objects are receding from us faster than ever before could be answered in several ways. Before I look at the different ways we could answer your question, I will note a few things. Firstly the definition of the Hubble parameter is:

$$H^2(t) = \bigg( \frac{\dot{a}(t)}{a(t)}\bigg)^2$$

where $a(t)$ is the scale factor which describes how the scale of the Universe changes with $t$, and $\dot{a}(t)$ is the first derivative of the scale factor with respect to $t$. Due to cosmological observations, the Universe is said to contain dark energy which causes the Universe's expansion to accelerate. What is meant by this is that at the current time $\ddot{a}(t) > 0$ where $\ddot{a}(t)$ is the second derivative of the scale factor with respect to time.

The first way we could look at your question is we could ask whether galaxies currently at a distance $D_0$ are receding faster than other galaxies that were previously at distance $D_0$.

From the definition of accelerating expansion and the Hubble parameter we can see that accelerating expansion does not imply that the answer to this question is "yes" and in fact if we assume dark energy takes the form of a cosmological constant (ignoring cosmic inflation), and we delve into the dynamics of the Universe, we find that galaxies currently at $D_0$ must be receding from us slower than the galaxies that were previously at $D_0$ were receding when they were at $D_0$. So in this particular sense the Universe's expansion is slowing down, even though we usually describe it as accelerated.

The second way we could answer your question is to ask whether the recessional velocity of any given galaxy is larger now than it has ever been in the past.

The answer to this question is more difficult as accelerated expansion does imply that the recessional velocity of a given galaxy increase with time, but the Universe's rate of expansion in previous epochs was decelerating. However, again taking dark energy as taking the form of a cosmological constant, we see that the answer is that galaxies achieve their highest recessional velocities twice: firstly at the Big Bang, and secondly in the infinite future. So the answer to this question is that galaxies are not currently receding from us faster than they have been at all previous times.

Recession velocity is different from peculiar velocity (i.e. the local velocity with respect to the CMB). We could add the two to find the 'real velocity', but as I've noted recession velocity doesn't have a direct physical meaning so what this 'real velocity' actually means is not straightforward.

Expansion is homogeneous, whereas the vacuum around a black hole is not homogeneous, so in this sense the 'sucking' of a black hole is not the opposite of expansion.


What percentage of a lunar meteor strike is blown back into space? Reading a discussion on another website, it was proposed that if a large meteor struck the moon, that a few days later, Earth would receive a meteor shower.

I thought about that and it seemed to me that one of the big unknowns would be what percentage of the impact would reflect back into space, away from the lunar gravity well.

The moon, having no atmosphere, then velocity out, could be pretty close to the velocity in, minus whatever's lost in heat and pressure waves and perhaps phase transition, though that could increase the explosive rebound and maybe other energy changes I'm not considering.

My question is simply, is there a fairly simple formula or estimate of what percentage would blow back into space following a lunar impact.  I'd suspect that it would depend on both the velocity and mass of the meteor and perhaps, composition.  If it was large enough and a direct hit, it could blow the entire moon into space, for example, but I'm thinking more standard asteroid size impacts not giant impacts.

Say, for example, a fairly large impactor like 99942 Apophis, about 27 billion kg, 370 meters in diameter.

There are some invariances that actually lend themselves well to estimating an answer to this question.  From the Deep Impact mission white paper:

A 1 mm projectile striking a target at 10 km/s will yield the same
result as a 1 m projectile striking at the same speed, provided all
distances are scaled by the same ratio of 1,000 = 1 m/1mm and all
times are multiplied by the same factor. Thus, if the 1mm projectile
makes a crater 6 cm in diameter in 300 ms, the 1 m projectile will
create a 60m diameter crater in 300 s.

The same paper gives us a set of three equations to estimate crater diameter. These equations are the result of experimental research combined with modeling the Navier Stokes differential equations governing the conservation of energy and momentum in materials. The first equation is:
$$\pi_2=\frac{1.61gL}{v_i^2}$$
Where $\pi_2$ is the inverse of the Froude number, $g$ is the local surface gravity, $L$ is the projectile diameter, and $v_i$ is the impact velocity. We will use an impact velocity of 12.6 kms taken from this paper. The second equation is:
$$\pi_D=C_p\pi_2^{-\beta}$$
Where $\pi_D$ is a dimensionless measure of crater diameter, and $C_p$ and $\beta$ are material constants. The third equation is:
$$\pi_D=D(\frac{\rho_t}{m})^{-1/3}$$
Where $m$ is the mass of the asteroid and $\rho_t$ is the density of the lunar surface.  The top 10 or so meters of the lunar surface are regolith at about 1.5g/cm³, but the next several km are mixture of bigger, more dense ejecta from billions of years of impacts, combined with finer grained regolith, according to this paper.  Given the overall moon density of 3.34 g/cm³, we'll estimate the density of this top layer as 2.5 g/cm³. Solving for $D$ and then plugging in all the values of our variables gives us a crater radius $R$ of approximately 2.5 km.
In an asteroid impact, the fastest moving ejecta comes from near the center of the crater.  The ejecta closer to the edge of the crater has much lower velocity:

The equation governing the velocity of the ejecta $V_{ej}$ is therefore partly a function of $r$, the distance from the center of the crater.
$$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$
Where $\epsilon$ is a material coefficient of 1.8.
We can solve the above equation for $r$ and solve using $V_{ej}$ as 2.38 km/s, the lunar escape velocity.  This yields a value of 275m for $r$.  That means that from the crater with a radius of 2500m, the only ejecta to reach escape velocity comes from the interior most part of the crater with a radius of 275m.
The depth of the crater can be estimated as $d=L\sqrt{\rho_p/\rho_t}$ where $\rho_p$ is the density of the projectile (asteroid).  Since the density of the projectile (2.6g/cc) and the estimated density of the lunar surface (2.5g/cc) are similar, then the depth would be approximately $L$, or 370m.
If we estimate the volume of the crater as a cylinder, then the total volume of lunar ejecta will be $\pi*R^2d$ is about 7.21e9 cubic meters, and the lunar ejecta to reach escape velocity will be $\pi*r^2d$ or about 8.79e7 cubic meters or a little more than 1% of the total ejecta.  This may not seem like much, but the total volume of Apophis is only 2.65e7 cubic meters.  That means that the ejecta reaching escape velocity will be more than 3 times the original volume of Apophis!
Notes:

The ejecta plume is in an inverted cone, so even given ideal geometry, most won't make it to earth.
The ejecta that are just under escape velocity will obtain an orbit that leads them to crashing back into the lunar surface.
The field of asteroid impact modeling is still in quite a bit of flux. Also, the models tend to perform less well for very large asteroid impacts since we have no ability to perform experiments on that scale.  It wouldn't surprise me if the total escape velocity ejecta calculated above differed by an order of magnitude in either direction to a similarly sized lunar asteroid impact.
My codes is available on request.


Will Jupiter eventually stop shrinking? Jupiter is currently shrinking due to the Kelvin-Helmholtz mechanism. Will this mechanism eventually hold or getting slower? If so, at what size of Jupiter will it stop and why? If not, what will eventually happen to Jupiter?

The rate of shrinkage is decreasing. The reason is that the core of Jupiter is becoming increasingly electron degenerate. Electron degeneracy pressure is almost independent of temperature. Once degeneracy sets in, an object can continue to cool with almost no change in its interior pressure, and hence no radius change. A cosmic example is a white dwarf.
This is not an on/off process, so at no point will you be able to say "Jupiter has now stopped shrinking", but Jupiter's radius will not reduce by more than about 10-20% from what it is now. The ultimate fate of Jupiter is to become a cold, ball of metallic hydrogen (with impurities).


How is it possible for a collision to be responsible for Uranus's axial tilt? I realize that the collision explanation is in any case highly (purely?) speculative, but I'm curious how it would work.  If Uranus is a ball of gas, why wouldn't any colliding object just pass through (or perhaps be destroyed/significantly damaged by the heat of friction)?  What would there be for the colliding object to hit?  Did Uranus have a significant solid mass at one time?  Is this a function of the speeds that would have been involved (sort of akin to skipping rocks on a pond, if that's a relevant analogy)?

For objects on the scale of a planet, the state of matter doesn't really matter much.  A colliding planet or planetesimal would not "just pass through"  The amount and density of the gas would prevent that.
Some easy experiments:  Put your hand outside a car as you are driving along, you'll feel that "gas" (ie air) has real substance and pushes against your hand quite hard.  Now imagine that you are driving at 20 km/s.... What happens when a piece of rock from space hits the atmosphere? It doesn't just pass through, it burns up in the air as a meteor.
Moreover the pressure and density in Uranus is huge.  Not so far below the surface are layers of water,ice, ammonia, methane.  And the pressure is so great that the density is more that that of water.
The collision of two gas planets could result in a combined planet that had a rotation axis inclinded at 90 degrees or more, and tidal effects and equatorial bulge effects would pull the moons into orbits around the equator.


DIY tracking motor - gear mating I have a Skywatcher 130/900 and I found that if I do an accurate polar alignment I'll need only one motor to track a celestial body, so I decided to build my motor. Right now I'm facing a problem with the mating of the spur gear that drive the worm gear

Using a caliper I found that the gear has the outer diameter of 50mm, and a tooth is 1mm in height, so the reference diameter is 49mm; the gear has 140 teeth, and so the module (reference_diameter / n_of_teeth) is 0.35.
After this I used Solidworks to model a spur gear that could match it, and printed this

When I tried to check the mating, I found this

The teeth seems right, but they're too short and can't mate.
I used a caliper and counted the teeth to obtain the gear specs, and I think is kinda stupid as approach, but I could not think of something better to mate my gear. Is there a better approach to this?
Then, the idea was to try and remove the gear (removing the screw) and replace it with a 3d printed gear, but it does not seems to work (and I think it could break the telescope mount). Did someone try to do it?
Searching online I found that (obviously) there are some tracking motors and another question came to my mind: how can they mate the gear? Is there some kind of standard? Does it mean that every telescope's mount has some kind of standard gear? If yes, where can I found them?
EDIT: That's the situation: on the right there is the spur gear that I'm trying to mate (or eventually remove, if possible) and in the center there is the worm gear that move the telescope on the RA

EDIT2: Here's the project of an involute gear with Solidworks. (I followed this tutorial) but as you can see, the teeth overlap.

As a reminder, I have the following parameters

Pitch diameter (PD): 49mm (caliper measurement)
Number of teeth (Z): 140
Module (PD/Z): 0.35

Am I doing something wrong (or am I missing something)?

Gear teeth usually have a special shape which is either an epicycloid or involute curve. Making worm gears is even more complicated because the teeth have a helical curve in addition to having an involute profile.

Making good-quality gears requires specialized tools, knowledge and equipment.


Why smaller black holes are considered to be more dangerous than bigger black holes? More massive the body is more is the gravity.Still often smaller black holes are considered to be very dangerous than the bigger ones,why?

All black holes are dangerous if you get close enough, but smaller ones are more dangerous because they cause stronger tidal effects, so they can cause spaghettification well before you reach the event horizon, whereas you can cross the event horizon of a supermassive black hole without the tidal effects becoming noticeable.

Very small black holes would also be dangerous due to the large amount of Hawking radiation they'd emit. E.g., a 1 million metric ton hole has a luminosity of over 356 terrawatts. However, there's no known mechanism to produce black holes that small, apart from waiting for a large hole to get small by evaporating via Hawking radiation, and that won't even start to happen for a very long time, since the cosmic microwave background is currently around a billion times hotter than even the smallest stellar black holes.

Actually, the main danger in the neighborhood of a black hole is if the hole is active, that is, if it is currently surrounded by matter that it's absorbing. That matter gets accelerated to very high speeds due to the gravitational acceleration and when it collides with other matter large amounts of dangerous radiation is released.


How to measure mass of planets' core from orbit I am told in an astrophysics lecture the following.
The mass of Saturn's core was measured by Cassini when it completed its final flyby between the rings and the planet itself. It was also found that its core has a sharp boundary.
Juno measured the mass of Jupiter's core & found that its boundary is a bit fuzzy.
How were these measurements done in principle?
(Ie what was measured from which we can deduce core mass?)

I certainly don't know the details of these kinds of calculations, but as my thought is a bit too long for a comment I'll write it up as an answer.
If you measure the flattening of a planet due to rotation (e.g. by measuring its rotation period) and the gravity (measured while keeping the orbiter at constant height above the planet), then you have everything you need to calculate the moment of inertia I/(MR2) using Radau's approximation (not the most informative link, but I couldn't find a better one). After that, it's a matter of taking a model for a planet, e.g. a two-layer model with a core and a mantle, which has free parameters x, the fractional core radius, and f the difference in density between both layers of the planet. Compare your measured moment of inertia with the model and you've got x and f for that planet, which allows you to calculate the mass of the core (if you also have the total mass of the planet from gravity measurements). The figure below shows the two-layer model and the measurements for some solar system bodies. The value 0.4 on the y-axis is the value for a homogeneous planet. From that you can deduce that Ganymede is a very differentiated planet. Hope that helps a little bit.

A note on how this graph was created: first of all, this isn't my graph, I took it from my lecture notes. The moment of inertia factor for a homogeneous body is calculated as
$$I = \frac{8\pi}{3}\int_0^R \rho r_0^4 dr_0 = \frac{8\pi}{15}R^5 $$
$r_0$ is the mean radius of a shell within the planet (because planets generally aren't round).
Then you take a simple two-layer model for a planet where $\rho(r) = f\rho_0$ for $0\leq r \leq xR$ and $\rho(r) = \rho_0$ for $xR < r \leq R$,  where $R$ is the radius of the planet.
The moment of inertia factor is then
$$I/MR^2 = 2[1 + (f - 1)x^5]/5[1 + (f - 1)x^3] $$
which is the curve that you see in the figure.


What would happen if you jumped out of a moving spacecraft? I've watched a film where one of the characters claimed to have rapidly disembarked a moving spacecraft during a 5g burn (?!) and immediately wondered how he survived.

For example, you're on a spacecraft that's travelling at 100mph (or any other random speed) and you go to the nearest airlock and jump out (I.E: leave the spacecraft).

Assuming you have a strong enough protective suit (I.E: this isn't a murder/suicide):

What effect does this have on your body?
Does the vacuum of space mitigate or exacerbate the negative effects?
Will you just "float out" into space or will you actually be moving with/in the general direction of the spacecraft?
Do you slow down or speed up relative to the spacecraft?
How do the effects change as the speed changes?

I apologise for the multi-part question, but this question can't really be summarised in a single sentence.

UPDATE: Also, has this ever happened before? Has anyone ever "fallen out" of a spacecraft and made it back safely?

Thank you.

If the spacecraft is not accelerating, the astronaut leaves the airlock and just floats outside, not moving relative to it. If the rocket is accelerating they "fall" down towards the stern from the perspective of people onboard; the astronaut would instead say the rocket is accelerating "upwards" and they are just floating.

In space, away from gravitational fields and in vacuum, only relative motion matters. You cannot tell whether the astronaut and spacecraft are moving at a high or low velocity compared to the rest of the universe just by looking at the two of them. So one can flip between the perspective that the craft is fixed and by turning on the engine it makes the astronaut (and everything in the craft) is accelerated, or the the perspective that the astronaut is floating in place and the rocket is speeding away. They are equivalent.

Astronauts leave spacecraft like the ISS all the time, and only experience free-fall despite the station moving at high speed around the Earth. As far as I know nobody has ever done a spacewalk when a craft is accelerating, for obvious safety and sanity reasons.


Could the solar system survive in a void? If the solar system were suddenly placed in the middle of a large void... would everything still function as normal?

Would we even notice any difference here on Earth apart from the night sky?

...and speaking of the night sky, would that even look much different to how it is already? I imagine just less colourful and and more sparse, but seeing whole galaxies instead of individual stars?

Or would moving the solar system into a void just mean the end of the world as we know it?

If you took away the Milky Way except for the Solar system (but kept the MW's satellite galaxies), the  "naked eye" night sky would feature the Andromeda Galaxy, 2.5  million light-years away, the Large Magellanic Cloud (LMC), 160,000 light-years away and the
Small Magellanic Cloud (SMC), 200,000 light years away.


How to prove to my friend that aliens have never visited earth I come from a physics background where I don't believe something unless the evidence is shown to me, however I have a friend who believes Aliens have visited earth , which he believes based on conspiracy theories.
My question is: is there any solid scientific reasoning as to why the conspiracy theories aren't true. As for him I need to give him evidence as to why they don't exist , and not the other way around .

I don't think you can prove that aliens have never visited Earth. It's theoretically provable if one had a time machine, but without one, we can't really prove that aliens didn't visit Earth (say) 2 billion years ago. The physical evidence would be long gone (especially with plate tectonics).

What you can say is that the burden of proof is on the person making the assertion. The null hypothesis is that aliens have never visited Earth. If your friend asserts that aliens have visited then the onus is on him to provide proof. If you can refute all the proofs he brings up, you can argue the null hypothesis remains the one to believe.


Converting Earth Moon Barycenter coordinate to Geocentric coordinate I have Earth Moon Barycenter coordinate (J2000) generated from VSOP2013 in Astronomical Units,

JD=2415020.5
X=-0.1968857640    Y=0.9633224602     Z=0.0002147057    AU
X'=-0.0171376156   Y'=-0.0035078892   Z'=-0.0000011481  AU/day

How to convert it to Geocentric?

How to convert it to Geocentric?

You don't. That's the position and velocity of the Earth-Moon barycenter with respect to the center of the Sun. Converting that to geocentric (a) doesn't make sense, and (b) even if it did, can't be done without knowing the location of the Earth-Moon barycenter in geocentric coordinates.

The position of the Earth-Moon barycenter plus the geocentric inertial position of the Moon expressed in the J2000 ecliptic frame plus the Earth-Moon mass ratio lets you compute the position of the Earth with respect to the Sun (or the position of the Sun with respect to the Earth). Add information about the positions of other planets with respect to the Sun and you can compute the positions of those other planets with respect to the Earth.

You cannot do the above with VSOP2013 alone because VSOP2013 does not contain information about the position of the Moon with respect to the Earth. You'll need a separate lunar ephemeris for that.


Are white dwarfs the ones always absorbing mass from a companion? It seems to me every time I see stellar remnant binary systems where there is a white dwarf and another star or something, it appears that the white dwarf always pulls mass from the star.

What if the white dwarf is with a neutron star? Will the white dwarf draw mass from the neutron star? What if the white dwarf was orbiting a supermassive star(not giant stage yet)?

A white dwarf only draws mass because it is denser that it's companion star. If a nearby object was denser than the white dwarf, and exerts enough gravity on the white dwarf, then it will draw mass from the white dwarf.

If the white dwarf is with a neutron star, and the neutron star is close enough to exert a gravitational pull that can distort a white dwarf, then the neutron star will draw mass from the white dwarf.

Now, you see, stars are way less dense that stellar remnants. No matter how massive, it's the density that counts. So if the white dwarf orbits any star, it will draw mass from the star.


How many degrees does the sidereal month orbit? It's written everywhere that anomalistic month orbits 360 degrees completely, then how many degrees does the sidereal month orbit?

Here is written that the sidereal month orbits 360 degrees too, but it doesn't make sense that they have the same orbit degrees since the anomalistic month is longer.

According to my logic, if 360 degrees equal to 27.55455 days, then 27.321661 days (=sidereal month) equal to 356.9 degrees, rather than 360 degrees. Is it correct?

The Moon does move 3.1° in the 0.23 day difference between those month lengths, but it goes 360° in a sidereal month and 363.1° in an anomalistic month.

The anomalistic month, from perigee to perigee, is slightly longer because the perigee moves.
The lunar apsides take 8.85 years to go all the way around the circle.

Similiarly, the draconic month is slightly shorter because the nodes move the other way in a cycle of 18.6 years.


How do we know that the discovered M87 black hole isn't just a star surrounded by a dust disk? To an untrained eye like mine, pictures of stars surrounded by dust discs look very much similar to the picture of the M87 black hole.
Here are some pictures of these dust discs:

And here is a picture of the M87 Black Hole:

I find it quite difficult to differentiate between the two in a meaningful, non-hand-wavy way. What is the scientific explanation for the distinction? I'm not doubting the scientific explanation, I just don't understand it.

Black holes are often studied (and discovered!) by observing their effects on objects around them. Stellar-mass black holes, for example, can be found by determining the orbit of any luminous companion. Supermassive black holes, by comparison, affect the motion of numerous stars and clouds of gas in their immediate vicinity. By fitting the motions of those stars, astronomers can determine that there must be an extremely massive object in that location, and typically a supermassive black hole is the only possibility.
In the case of M87, these measurements were first conducted in the late 1970s (Sargent et al., Young et al.). Both groups noted that the velocity dispersions near the nucleus required a central mass on the order of $\sim5\times10^9M_{\odot}$. Mass/luminosity ratio profiles were also calculated based on photometry, and both sets of observations noted a steep rise in $M/L$ near the center. Neither group was able to rule out other possible explanations, like a compact star cluster, but a supermassive black hole was - to quote Young et al. - "the most attractive of the models considered". Further observations over the last four decades have ruled out those other options.
The image produced by the Event Horizon Telescope is consistent with a supermassive black hole, as the EHT Collaboration wrote in the first of their papers on the observations:

It is also straightforward to reject some alternative astrophysical interpretations. For instance, the image is unlikely to be produced by a jet-feature as multi-epoch VLBI observations of the plasma jet in M87 (Walker et al. 2018) on scales outside the horizon do not show circular rings. The same is typically true for AGN jets in large VLBI surveys (Lister et al. 2018). Similarly, were the apparent ring a random alignment of emission blobs, they should also have moved away at relativistic speeds, i.e., at ~5 μas day−1 (Kim et al. 2018b), leading to measurable structural changes and sizes. GRMHD models of hollow jet cones could show under extreme conditions stable ring features (Pu et al. 2017), but this effect is included to a certain extent in our Simulation Library for models with Rhigh > 10. Finally, an Einstein ring formed by gravitational lensing of a bright region in the counter-jet would require a fine-tuned alignment and a size larger than that measured in 2012 and 2009.

There are other arguments you can test yourself. For example, the photon ring matches calculations from general relativity, assuming the now-accepted mass of the black hole.
In short: Stellar and gas dynamics require the presence of a large mass in the center of M87, and the image rules out many non-compact objects.


Why is the flux density and amplitude different for galaxies than stars? This might be a simple question, but I'm having a hard time answering it myself - or at least answer it correctly, I think.
I'm sitting with different spectrals; many for stars and one for a galaxy. The one for the galaxy has a flux density of 1e-13, while the stars range from 1e-7 to 1e-12.
My question is: why is galaxy's flux density lower than even the lowest stars? The galaxy also have very characteristic emission peaks. How is that reflected in the spectrum of the stars, or is that another phenomena?
To clarify, I am talking about a plot like this. I had to scale the stars spectrum down by a lot in order to match the galaxy's spectrum. Why is that?
Also why can't the stars produce the same spiky amplitudes as the galaxy? Is it because of the galaxy emission lines from billions of stars and vast amounts of dust and gas?

A galaxy is obviously going to be much more distant than the stars. On the other hand the galaxy contains a lot of stars which will contribute to the overall spectrum.
Whether the galaxy or a star will be brighter will depend roughly on the ratio of $N/d^2$, where $N$ would be the number of stars and $d$ would be the distance to the galaxy.
What your plot has done is empirically remove this factor by linearly scaling the galaxy spectrum so it matches with the stellar spectrum over most of the wavelength band shown.
On the other hand, the light from a galaxy is not only made up from the stars that are within it. There are contributions to the light from (for example) hot gas near any active galactic nucleus and also relatively hot gas in the interstellar medium, possibly associated with bursts of star formation, supernova remnants and nebulae ionised by hot stars. These latter contributors are responsible for the emission lines seen in the galaxy spectrum.


If the moon is losing energy dragging the oceans around, why is it moving away from earth instead of falling like any other satelite would? If a satellite loses energy it starts to circle lower and finally hits the atmosphere. The moon is moving like an inch a year away from the earth even though it is losing energy pulling the oceans around.

No. The energy budget of the Moon is actually increasing. The Earth's spin is slowing down, and that energy gets transferred to the Moon.

Why is the Moon receding from the Earth due to tides? Is this typical for other moons?

Whatever else may be happening (like the losses you mention), this transfer from Earth's spin to Moon's orbital energy is the main process responsible for raising the Moon's orbit.


Question regarding the Milky Way when calculating galactic space velocities for galaxies I have been calculating galaxy space velocities (where proper motions are known) in order to measure their orbits of the Milky Way using the method proposed in the appendix of http://www.aanda.org/articles/aa/pdf/2011/01/aa13415-09.pdf

The question I have is: Do we put the Milky Way through the same calculation process to get an initial velocity, or do we have that it's (0,0,0) as it is essentially our origin point (ie the U velocity component would mean that the Milky Way is moving in the direction away from its own galactic centre which doesn't really make sense to me as this could be any direction)?

I have put it through the process and found an initial space velocity of the Milky Way of (-11.1,232.24,7.25) (my method differs slightly from the link since I have used an updated velocity vector for the motion of the sun with respect to the Local standard of Rest) which essentially just comes from correction solar and galaxy rotation motions.

So, is this calculated vector correct for the Milky Way or should it be (0,0,0)?

Thanks

v$_r$ and μ are radial velocities and proper motion as seen from the Sun (ie heliocentric)  and the equations in the appendix convert to v in the $S_o$ frame which is the center of the Galaxy.  So, the motion of the MW is (0,0,0).  If you want to, you can solve for v$_r$ and μ for the MW by plugging v=(0,0,0) into the equation, and find the heliocentric proper motion and radial velocity of the Galactic center. But it is not needed.


How bright are geostationary satellites due to reflected sunlight? We are all familiar with seeing low-Earth orbit satellites scooting across the sky in a couple of minutes. Many of them are bright and readily identifiable with the naked eye. Geostationary satellites on the other hand are much further away and would remain fixed in position (though moving very slowly with respect to the background stars).
How bright are geostationary satellites in reflected sunlight? Can any of them be spotted, even with a telescope?
Related questions:
How will Starlink affect observational astronomy?
How do satellites impede current telescopes?

Have a look at Is it possible to capture geostationary satellites with DSLR?

See the excellent advice in this answer to Are any geosynchronous satellites visible with the naked eye?

No, but they are easily seen with a small telescope on a sturdy mount. March and September are the best times. Use an app to help you. My favorite way is to keep M11, the Wild Duck Cluster, in view with a medium power eyepiece. Every few minutes, a "star" will slowly track through the southern edge!

It's a fun and instructive exercise to calculate the declination of an equatorial GEO orbit from a given latitude!

GEO satellites can be quite large, and while they are said to be +10 to +14 magnitude they can flare far brighter due to flat surfaces, both solar panels and thermal radiators. From http://www.satobs.org/geosats.html

Typically the satellite will be in the mag. +11 to +14 range (or dimmer), but brightening by several magnitudes when the geometry is favourable (around mag. +5 to +6 is not untypical). One satellite is reported to have briefly been visible to the naked eye at mag. +3!

See Richard Huziak's (Saskatoon Centre (contact info in link) excellent tutorial Observing Geostationary Satellites reprinted from Saskatoon Skies:

...I found three other satellites, all between magnitudes 13 and 13.5, each a degree or two from the last! Chances are that one of the satellites is Canada’s Anik C1, although I could not identify it for certain.

From Sky and Telescope's How to See and Photograph Geosynchronous Satellites

Many geosynchronous satellites shine between magnitudes 10–12, so you can spot them in telescopes as small as 4 inches. They're also easy to photograph. High ISOs and fast, low light lenses aren't necessary, just a camera capable of a several-minute-long time exposure — long enough for the stars to trail, so you can easily tell them apart from the satellites. Set your shutter speed to "B" and ISO at 400. You can hold the shutter button down with your finger, but a shutter release cable is much better and vibration-free. Use a 100–200-mm telephoto lens, focus sharply, and expose for 2–4 minutes. When you enlarge the image, you'll should see long trails and a line of pinpoint dots — satellites!

See and enjoy the videos of GEO satellites colliding with space junk linked in Are commercial communications satellites in GEO being constantly monitored by telescopes?
From this answer to Are there any satellites in geosynchronous but not geostationary orbits? we can see that while many objects in GEO have a well station-kept equatorial orbit, some are allowed a small inclination and others can have quite a large one!

Large communications satellites can have large flat surfaces that can dramatically increase their visibility at certain oriatiantations. These include solar panels that are often articulated to always point at the Sun, and large thermal radiators that have high thermal emissivity (are dark) in thermal IR yet highly reflective (mirror-like) in visible to avoid solar heating. For more on that see answers to What are these very large, square panels on Inmarsat 5? and What is the function of this array of what looks like mirrors on TESS? and some cool photos of spacecraft in question.

For some math, see answers to At what magnitude would adjacent large geostationary communications satellites see each other? and When would we detect a tiny meter size natural satellite in a geostationary orbits?  and Calculating the apparent magnitude of a satellite

For a truly weird looking communications satellite bound for GEO see Why do these satellite antennas look so weird? (strangely shaped with little white dots…)

From How do commercial broadcast satellites in GEO produce such carefully shaped signal footprints? you can see thermal radiators that are highly reflective in the visible (and near IR)

From How will GOES-R simultaneously point some instruments down at Earth and others sunward?


Why did the big bang produce a low entropy universe? In order for the universe to be as it is, containing stars and galaxies, it is a requirement that the early universe was at a relatively low entropy state.

Why did the big bang produce this low entropy universe?

The second law of thermodynamics states:

  the total entropy of an isolated system can only increase over time

If applied to our universe, one can conclude that the entropy of our universe must necessarily increase over time (or, improbably, stay constant), resulting in the final heat-death of our universe. Just as Georges Lemaître saw that the Universe was expanding and surmised that it must have been smaller in the past (implying a Big Bang), we can also surmise that if the entropy of our Universe is increasing to a maximum entropy wherein the Universe experiences a heat death, it must necessarily have been lower in the past. This begs the question, why was it lower? How did the universe come to be in a state with such low entropy such that stars and galaxies formed. Why did it not spring into existence, already in the heat death stage?

The answer to this is, I don't know. And I'm not sure anyone could know or give a solid answer, backed up by facts and observational evidence. At best, there are some guesses and hypotheses which I can go over a bit.

First off, let me point out that the second law of thermodynamics may not in fact apply to our Universe as a whole. The problem with the laws of physics is that they often apply in localized cases, say when we're talking about a science lab or a galaxy. But when you try to apply them to the universe as a whole, there's just too many unknowns and usually the rules change. For example, is our universe an isolated system? Nobody can answer that because nobody can view the entire Universe. We're limited to only observing the observable Universe so already we can't definitively say that the Universe adheres to one of the primary pre-conditions of the second law.

Not to mention, you have the difficulty that it may be impossible for all, absolute laws to apply to the Universe as a whole, when we already know some of them don't. The laws of physics all rely on each other. If you change one, you affect all the others. We know (or at least some believe) that the law of conservation of energy does not necessarily hold for the Universe as a whole. At any local point in space and for any localized event, it does, but not when applied to the entire Universe. So already the laws of physics are messed up and that will have unknown consequences for other laws of physics when applied to the Universe as a whole.

Energy in our universe is not conserved because it is expanding. This expansion is constantly dumping new energy into our universe, effectively making our universe not "isolated" even if it may be based on the dictionary definition of the word. As such, it's entirely possible for the entropy to be driven down by this energy input via expansion. The initial inflationary epoch, which likely violated energy conservation globally, could have resulted in our low, initial entropy state.

But as I said, a lot of this is guessing and hypotheses. I don't think there's a good answer out there to this good question, and if there is, I'd really like to hear it.


What name was voted for Tau Boötis b by the Name Exoworlds Project? In 2015, the exoplanet Tau Boötis b was included in the  Name Exoworlds Project, a project where various international astronomy organizations would propose names for a selection of known exoplanets that complied with a set of guidelines presented by the International Astronomical Union. Upon completion of submissions, voting was opened to the public, with the selected and approved names being published.
The IAU decided that the name voted for Tau Boötis b did not not meet the naming guidelines, and annulled the result of the vote.
I have no objection to them deciding to not use the name, and  I don't want this to be come an argument as to whether they should have annulled it. I just want to know: has it ever been revealed what the public-voted name was, and if non-obvious, the reason it was annulled?

The statistics for the 2015 vote are available at http://www.nameexoworlds.iau.org/statistics
This indicates that the winning names were "Shri Ram Matt" for the star, and "Bhagavatidevi" for the planet. The rationale given there is as follows:

One of the proposals for tau Bootis fails to comply with the IAU policy for naming exoplanets, explicitly forbidding the use of names of persons that are known to have been involved in political, religious or military activities. The IAU respects the proposed name, and the high number of votes cast in its favor, but decided to annul the vote for tau Bootis.


Is Earth losing orbit? Is the Earth losing or gaining distance from the sun overall? Could asteroids over time add weight to the Earth and push it farther from the sun more than having a degrading orbit around the sun? Does asteroids and space dust increase the mass the the outer planets over sustaining or pulling the Earth from the Sun?

What determines the Earth's orbital separation from the Sun is the total mass of the Sun and Earth, as described by Kepler's third law.

The orbital radius of the Earth is increasing because the Sun is gradually losing mass both by emitting radiation and through the solar wind at a rate of around 5 million tonnes per second (e.g. Noerdlinger 2008).

Any gain of mass through the accretion of dust or asteroids (by the Earth or Sun) is negligible by comparison.

This mass loss rate is only sufficient to widen the Earth's orbit by about 1.5 cm per year.


How to accurately measure the temperature of red supergiants? I just (February 28th, 2021) heard the news on progress how to measure the temperature of super giants:

Red supergiants are a class of star that end their lives in supernova explosions.
Their lifecycles are not fully understood, partly due to difficulties in measuring their temperatures. For the first time, astronomers develop an accurate method to determine the surface temperatures of red supergiants.

The original article by Daisuke Taniguchi and team is called Effective temperatures of red supergiants estimated from line-depth ratios of iron lines in the YJ bands, 0.97-1.32 μm but is not available (yet).
Maybe somebody can help me understanding the following quote intuitively:

However, by looking at the ratio of two different but related lines - those of iron - we found the ratio itself related to temperature. And it did so in a consistent and predictable way.

How could the ratio between the intensity of spectral lines correlate with temperature? Furthermore: How could we actually show that correlation if we do not know the temperature in first place?
References

What are the possible solutions to the Red Supergiant problem?

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


Standard conditions for a heliacal rising I imagine there are several possible conditions for determining the time of a helical rising of a star, the elevation of the star at sunrise, the latitude of the observer, and the azimuth of the star.

Is there any kind of equation (or plot/graphic) that relates these variables and can be used to predict whether a star will be visible? (Ignoring weather, ie assuming some standard atmospheric condition).

What I do is compute the limiting magnitude for the position of the object I am interested in, and subtract it from the magnitude of the object (I call this the magnitude contrast).  If negative, it should be visible with the naked eye.  I use the function limmag_jd() in the Fortran library libTheSky for this.

In order to find the best moment in a given night, I compute the contrast for say every 15 minutes, and find the minimum.  You can do this for a number of nights, until the best contrast found is negative, which would be the heliacal rising.

I'm not sure whether there is an easy way using a simple equation - the limiting magnitude will depend on things like the positions of Sun, Moon and the object in question, the phase of the Moon, the magnitude of the object, etc.


How to calculate sunrise and sunset on on a given date and location? I have a given date at any time in future or past eg: 2016-04-23 or 2022-05-12
Now given at any location(lat,lng). How do i find the time of sunrise or sunset.

Assuming you mean this in practical sense:

Just use one of the numerous free websites that give this data:

Eg. https://keisan.casio.com/exec/system/1224686065

Assuming you want to know about the math:

Check out the sunrise equation


Calculating the Diameter of Jupiter through Image I was trying to calculate the diameter of Jupiter from a picture I took of it. Here's the information I was able to get that I needed to calculate the diameter:

  Focal Length of Telescope: 1.2 m

  Multiplier for Barlow Lens: 2x

  Camera CCD (Sensor) dimensions: 14.8 x 22.2 (mm)

I found the diameter to be 135 pixels in my image. Here's what I did:

\begin{align}
f_e &= \text{Focal Length $\cdot$ Barlow Multiplier}\\
 &= 1.2 \cdot 2\\
 &= 2.4
\end{align}

Then, to calculate the field of view:

$$\text{Field of View} = \frac{\text{Sensor Size}}{f_e} \cdot \frac{180}{\pi}$$

Where the sensor size is the length in millimetres of the size of the camera CCD (I used the horizontal axis, or the longer side, to get the 135 pixels, meaning that I'd use 22.2 mm), and the second fraction is there solely to convert radians into degrees. Substituting my values, I got:

\begin{align}
\text{Field of View} &= \frac{0.0222}{2.4} \cdot \frac{180}{\pi}\\
&= 0.53^\circ
\end{align}

As my entire image is 4271.8 pixels wide, and I know the angular field of view, I used ratios to get the field of view of the 135-pixelled Jupiter to be $\theta = 0.017^\circ$.

Then, using basic trigonometry, I was able to calculate the radius of jupiter:

$$\tan \left(\frac{\theta}{2}\right) = \frac{\text{Radius of Jupiter}}{\text{Distance to Jupiter}}$$

The photo was taken on March 22, 2016 - Wolfram Alpha gives 667.6 million as the distance to Jupiter on that day:

$$\tan \left(\frac{0.017^\circ}{2}\right) = \frac{\text{Radius of Jupiter}}{667.6 \cdot 10^6}$$

$$\therefore \text{Radius of Jupiter} = 97\ 577\ \mathrm{km}$$

Googling the radius of Jupiter gives $69\ 911\ \mathrm{km}$ as the result. Is there any reason why my value is so significantly off? Anything wrong with my procedure, anything I'm missing? I was not able to to upload this image I took as the file size was too large, however, the image does not lack clarity and I am completely stumped as to why my result is off by such a significant amount.

Any help will be greatly appreciated, thanks in advance.

Your barlow is marked "2x" but I'll bet it isn't.

How is your camera mounted on the barlow? More importantly, what distance is between the camera's focal plane and the barlow?

An experiment to try (or at least to think about).  Get an eyepiece extension tube (or improvise something) and put it between the camera and barlow, increasing the spacing. You will have to refocus a bit and you will get a different (I guess bigger) image size, even though the barlow is still "2x". To get an exact "2x" barlow you might need a particular spacing, and eyepieces are normally closer to the eyepiece than cameras because of the extra bulk of the camera.  (I think this is the most likely cause of your discrepancy; your camera is currently mounted further back than an eyepiece would be so projection is making the image larger.)

A second experiment to try: without the barlow, photograph the moon and see how close you get to its expected size.  If it's close, then our assumptions about the barlow are in question.

Third thing to try: are you absolutely sure about the distance from Earth to Jupiter? (It's probably OK but I'm always suspicious about online calculators.)


strange moonset around 10 pm (3.2 2014, NZ) - explanation? during a visit to New Zealand we observed (on 3.2 2014) a strange moonset in Te Anau.

Around 10 pm, shortly after sunset, the moon appeared, but then quickly began to go down, and about 1 hour or less it "touched" the mountains on horizon and shortly after it completely disappeared and then we did not see it anymore.

I did not bring a camera or camcorder, so I only have this bad photo from a mobile phone:

The whole process was pretty fast, so I only have one picture. I've never seen anything like it - usually (in Europe) the moon comes out after sunset rather than going down :-)

There was also one guy living there more than 10 years, but he was apparently surprised too.

Can anybody explain this? Is it something normal in this location?

thanks

February 3, 2014 has been short (three days) after new moon. The moon is then about 40 degrees or about 3 hours behind the sun. Hence, what you observed is quite normal, in Europe, too.

The only difference is, that the moon looks to the other side when seen from the southern hemisphere. Here, for 2014, a moon phase calendar of the northern hemisphere.


How the watches are showing perfect time? The actual time for one day is 23hrs, 56mins, 4.1secs right? Then how can the clocks and watches can show perfect time?
I mean, if I observe the sun rise at 6:00am this day, tomorrow I should observe it before 6:00 or I can say at 5:56 approximately as our measuring devices follow 12+12 i.e., 24hr day. But in practice again it will be 6:00am when the sun rises. How it is possible?

Those 23hrs, 56mins, 4.1secs are actually the sidereal rotation period of Earth. It is how much time the Earth needs to perform one rotation relative to the stars. But that is not exactly the same as solar time. By the time the Earth has completed one rotation, it has also moved a little forward in its orbit around the Sun. Thus, it needs a few more minutes of rotation to catch up so the Sun is at the same location of the sky.


If a photon were 1 Planck length away from the event horizon of a blackhole, would it be able to escape the blackhole's gravitational pull? I understand that one could theoretically replace the sun with a blackhole without affecting the orbits of the planets. This leads me to believe that a blackhole isn't a "cosmic vacuum cleaner" so much as it is a concentration of mass that has the strongest gravitational field known.

Intuitively, this must mean that there is a threshold beyond which escaping the blackhole's gravity becomes impossible even at light speed, which is the event horizon. Since a Planck length is the smallest unit known at which quantum gravitational effects become apparent, would it be reasonable to assume that a photon that is one Planck length away from the event horizon of a blackhole can escape its gravitational pull?

It depends on the direction the photon travels.

If the photon is directed straight away from the black hole, it will escape (but will be very redshifted).

Obviously all directions towards the event horizon will lead to capture. But since the spacetime is so curved, this also includes many directions that point away from the hole: the photon trajectory will curve around and hit the event horizon anyway. Closer to the event horizon the cone of directions where photons can escape gets smaller and smaller, until it is essentially just the direction straight away. (See slide 6 in this presentation)

See also  this interactive web page; try it for a fixed position looking along the "orbit". The higher up in the picture the closer you get to the horizon. I have found that one can get past the $1.5R_s$ limit (the photon sphere, where exactly half of the sky is directions leading into the hole) by moving to a different browser tab, scrolling, and returning; the effect between 1 and 1.5 $R_s$ is to make the sky look like an ever smaller ball - photons emitted in any other direction will be captured.


Distance parallax and cosmological distance in universe in expansion I know that we cannot measure cosmological distances with parallax, but this is more a philosophical question. Am I correct in assuming that if we could measure distances using parallax (like stellar parlax used for nearby stars) the distance we would measure would correspond to the "angular diameter distance" ? https://en.wikipedia.org/wiki/Angular_diameter_distance

Yes, exactly. The angular diameter distance is defined such that it is compatible with "regular" trigonometric rules, just as the luminosity distance is compatible with regular relations between flux, distance, and luminosity.

Only for at flat and non-expanding universe would these distances coincide with each other, and with the "true" distance, which is what you would measure if you lay out meter sticks (after freezing the Universe in time, so it doesn't expand more before you're done).


Why is T-Tauri So Dim? I'm looking at stellar objects and doing some math here and...
Now, I can't find these values directly listed, but from it's relative magnitude and distance, (10.27, 471ly,) I calculate T-Tauri to have an absolute magnitude of 4.47, and when I convert that to solar luminosities I get 1.39.
I've run that on objects for which I can find absolute magnitudes and solar luminosities listed, and I'm getting the correct values.
But every description of T-Tauri stars I find says they are brighter than main sequence stars of equivalent mass. A main-sequence star about twice the mass of the sun should be much more than 1.39 times the brightness of the sun.
Now I must turn to people who actually know something about astronomy:  Is the prototype T-Tauri star "weird" in some way, and not actually typical of the class of objects that were named after it, or do I have some bad math or bad data here? What else could be the problem?

According to Herczeg & Hillenbrand (2014) the intrinsic bolometric luminosity of T Tau is about $7L_\odot$ (for an assumed distance of 479 light years).
The main difference between their estimate and yours is that they have: (i) Taken account of extinction towards the object, which considerably dims its visible flux by about 1.25 magnitudes (a factor of 3.2). (ii) Taken account of the spectral energy distribution, which differs from the Sun.
You cannot in general calculate a ratio of luminosities from a difference in visual magnitudes, unless the spectral energy distributions of the two sources are the same. T Tau has a greater proportion of its flux emerging at near infrared wavelengths than the Sun.
T Tau is in a multiple system with a much lower mass companion (which contributes little to the luminosity), but this has led to a quite accurate mass determination of $(2.12 \pm 0.10)\ M_\odot$ (Kohler et al. 2015).
Given a main-sequence mass-luminosity trend of around $L \sim M^3$, then it is apparent that T Tau has around the luminosity it would be expected to have on the main sequence.
This would suggest that it has finished descending its Hayashi track in the HR diagram and is currently traversing, roughly horizontally, towards hotter temperatures and slightly rising luminosity, in excellent accordance with stellar evolutionary models (see below) if its age is $\sim 2$ Myr.
T Tau is only unsusual compared with most "T-Tauri stars" in that it has quite a high mass. Most "T-Tauri stars" are a solar mass or lower, and if found in young stellar clusters and associations with ages $<10$ Myr, you can see from the plot below that they would have higher luminosities than they end up with on the main sequence.

Pre main sequence evolutionary tracks and isochrones from Steven W. Stahler and Francesco Palla, "The Formation of Stars"; doi:10.1002/9783527618675


Is there a closest free-return trajectory to a black hole? The scenario I am imagining is a test particle falling from infinity with an initial off-center velocity. In Newtonian mechanics the path would be a hyperbola. But I assume that, near a black hole, that is no longer a good approximation.
I have read about the innermost stable circular orbit, and it seems like there might be an innermost free-return distance, where any object that falls within that distance will need to use rockets to get back out. I may have even come across this idea, but I am not finding it with a Google search.
I have an argument that there must be such a distance. Imagine that you could fly past a supermassive black hole just half a meter above the event horizon and continue on to deep space without using rockets. Tidal forces aren't so great at the event horizon for a SMBH, so there should be no reason you can't extend your arm just over half a meter down and not pull back a bloody stump. But you can't retrieve things from below the event horizon. This paradox is resolved if, in fact, the only way to escape from that close to the EH is to exert massive force using rockets (or some other method). Even though you are not yet inside the EH, your free fall path will take all of you there, and the force to prevent that would pulp you anyway. I think it is related enough to the main question to ask: Is this analysis correct?

The trajectory of a ballistic body, whether in Newtonian or Relativistic physics depends on the initial energy and angular momentum.
The difference is that in Newtonian physics, if the mass is compact enough, the infalling object (if given an initial kinetic energy) will never hit the central object, unless it is a direct hit, and will scatter off to infinity.
This can also happen in GR, but only if the specific angular momentum exceeds a threshold that increases with the initial energy of the particle, but can never be less $4GM/c$, and only as small as this for an object that falls from infinity with no initial kinetic energy. If the specific angular momentum is less than this then the infalling object will always end up in the black hole (for a freefalling trajectory).
For the limiting case, the closest approach to the black hole is $2r_s$, which is the local maximum of the effective potential.
However, the body can get closer if it has more angular momentum and more initial energy. If we allow the angular momentum and energy to get arbitrarily large then the body can get to $r=1.5 r_s$.
So I think this is the answer you are looking for. This is the value for a Schwarzschild black hole. For a spinning black hole, things are more complicated.
EDIT: Some details
The effective potential in the Schwarzschild metric is given by
$$ \frac{V_{\rm eff}}{mc^2} = -\frac{r_s}{2r} + \frac{L^2}{2m^2 r^2 c^2} - \frac{r_sL^2}{2m^2r^3c^2}\ , $$
where $r_s$ is the Schwarzschild radius and $L/m$ is the specific angular momentum of the free body.
The radial equation of motion in terms of the proper time $\tau$ of the body is
$$\frac{1}{2c^2}\left( \frac{dr}{d\tau}\right)^2 = \frac{1}{2}\left( \frac{E^2}{m^2 c^4} -1 \right) - \frac{V_{\rm eff}}{mc^2} \ .$$
To find the local maximum of the effective potential, differentiate with respect to $r$ and equate to zero. This gives you that the specific angular momentum at the maximum is
$$ \frac{L^2}{m^2} = \frac{c^2 r_s r^2}{2r - 3r_s}\ . $$
Allowing the energy of the body to be $E = \gamma mc^2$ (where $\gamma$ is the Lorentz factor) one can set $dr/d\tau = 0$ at the scattering minimum radius and then subsititute in the value for $L^2/m^2$ at the maximum of the effective potential. The rationale here is that the body will get to its closest point when it has an energy corresponding to the maximum in the effective potential. If it has any more energy than this it will just plunge into the black hole. Any less and it will scatter at a larger radius.
This condition gives you a quadratic equation in $r$ which can be solved in terms of $\gamma$. If $\gamma=1$ then the quadratic becomes a linear equation with $r=2r_s$. This corresponds to the case of a particle falling in from rest at infinity and meeting a local maximum of the potential of $V_{\rm eff}=0$. The body then scatters off to (just) infinity (i.e. the marginally bound case). This is illustrated below, where the horizontal purple line is a line corresponding to an energy $E/mc^2=1$. The peak at B is at $r=2r_s$ where $V_{\rm eff}=0$ (I've also shown the equivalent Newtonian potential for contrast).

If you allow $\gamma$ to be larger then the the viable solutions for the minimum $r$ lie in the bounds $1.5 r_s \leq r \leq 2r_s$. If $\gamma \rightarrow \infty$ then the scattering minimum radius $r \rightarrow 1.5r_s$, which also requires $L/m \rightarrow \infty$. The plot below shows the potential for $L/m = 10GM/c$, with a purple line representing a body with $E/mc^2=2$ which would scatter to infinity. The peak at B is just above $r=1.5r_s$.

If you want to play with this (my) Geogebra widget, you can find it at https://www.geogebra.org/m/u7y6huer


Does Oxygen-burning process produce Neon? In the book "Horizons_ exploring the universe-Cengage learning (2018)", p200, it states that:

Helium fusion produces carbon, and some of the carbon nuclei absorb helium nuclei to form oxygen. A few of the oxygen nuclei can absorb helium nuclei and form neon and then magnesium.

But when I check Wikipedia for Oxygen-burning process, there doesn't seem to have any steps that will produce Neon, but when I check for Carbon-burning process, two carbon-12 nuclei can fuse to a neon and helium nuclei, so is there an error in the book? Or there are other steps of oxygen burning that will produce neon?

The pages you’re looking at describe fusion of two similar nuclei with each other (e.g. oxygen with oxygen).  But fusion doesn’t have to involve identical nuclei, and reactions involving more abundant nuclei will in general occur at a higher rate (though this can vary based on other factors, too).
Long before oxygen-oxygen or neon-neon reactions become important, reactions involving helium nuclei (also called “alpha particles”) take place.  That is what the text in your book is referring to: C + He gives O, and O + He gives Ne. See the list in this Wikipedia article on the alpha process.


Storm with lightnings and no thunder Last evening I assisted in an interesting phenomenon in my region (Eastern Europe). It was a dark sky with heavy rain and blinding lightning (which struck every 5 seconds or so), but NO thunder at all.
I was thinking that the explanation might be:

The clouds are very high in the sky and and the sound does not reach the land
The wind deviated the sound somewhere else.
It was not a storm and aliens were conducting an experiment in the atmosphere

I'm not a meteorologist, so if you have a better explanation, please let me know it.

"No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning. However, it IS possible that you might see lightning and not hear the thunder because it was too far away. Sometimes this is called “heat lightning” because it occurs most often in the summer. "

here is more information regarding lightning:
http://www.nssl.noaa.gov/education/svrwx101/lightning/faq/


Expansion of Space in the Very Distant Future Given a nearly infinite timeline, would the expansion of space eventually cause individual particles to become "unobservable" from one another due to the vast distance and increasing rate of expansion with that distance.

Maybe someone can give chapter and verse on this, but my understanding is that bound objects are unaffected by the expansion for the case of dark energy represented by a simple cosmological constant.

But a "big rip" will occur in a finite time if the dark energy equation of state $P = w \rho$ is characterised by $w < -1$. In this scenario, the expansion accelerates so quickly that the scale factor becomes infinite in a finite time and structures must disintegrate when the size of the observable universe becomes smaller than that structure.

The present state of play is that $w$ appears to be very close to -1 (a cosmological constant). But given that dark energy is not understood, who can really say whether it might change in the future...