Why doesn't Earth's axis change during the year? My understanding is that the Earth's axis points in the same direction in space during its entire orbit around the sun. And this is what causes our seasons.

My question is why doesn't the axis follow the orbital path (kind of like a car driving around a circle)?

I imagine the Moon plays a part in this.

As a secondary question, do the other planets in our solar system orbit the same as Earth (its axis points in the same direction in space for its entire orbit)?

My understanding is that the Earth's axis points in the same direction in space during its entire orbit around the sun. And this is what causes our seasons.

The second statement is correct. The axial tilt is the primary driver of the seasons.

The first statement is not exactly correct. There is a small but persistent change in the orientation of the Earth's axis. The change over the course of a year is small. The change over the course of 13,000 years is large. In 13,000 years the Earth's axis will be tilted by about 47 degrees compared to the current orientation. In yet another 13,000 years the Earth will be oriented very similarly to how it is oriented now. I'll write more about this later.

  My question is why doesn't the axis follow the orbital path (kind of like a car driving around a circle)?

That is not a good mental model. The only contact between your car and the ground are the bottoms of the four wheels on your car. The forces at these four small contact areas are what change your car's orientation and change your car's velocity vector. This means that the torque on your car as you turn the steering wheel and the centripetal force on your car that makes your care negotiate a turn are intimately connected. Turn the steering wheel gently and the heavy front end of your car follows suit. Assuming a gentle turn, the light rear end of you car soon follows suit. If you turn too hard, your car starts to skid instead of turn. Turn even harder yet and you risk having your car flip and roll. The motion of your car depends on that intimate coupling between force and torque.

Gravitation doesn't work like that. It instead acts on each and every bit of the Earth, all together, all at once, all the time. There is a slight variation in the gravitationally-induced acceleration across the Earth due to variations in distance to the Moon (or Sun), but these variations are small. The net gravitational force on the Earth is very much decoupled from the small net gravity gradient torque on the Earth.

The principal result of the gravity gradient torque exerted by the Moon and the Sun on the Earth is a very slow but large precession of the Earth's rotation axis (for more, read this wikipedia article on axial precession). This precession makes the Earth's rotation axis rotate about the normal to the ecliptic at a rate of one revolution per 25772 years. There are lesser effects collectively called nutations and polar motion that result from gravity gradient torque. These lesser effects are much, much smaller in magnitude but have much, much shorter periods. The largest of these lesser effects results from the 18.61 year cyclical variations in the lunar node. Even this largest component of nutation is small, two orders of magnitude smaller than the large but slow precession.


Why does mass naturally move closer toward's the center of other masses? If general relativity is understood to mean that mass curves space such that objects are still traveling in a straight line, why does mass naturally move closer toward's the center of other masses?

Think at the surface of a ball. A straight line within this curved surface is a great circle.
Take some point on the ball. Draw two great circles (apparent straight lines in the surface) starting in this point, with some non-zero initial angle. The two apparent straight lines intersect at the opposite side of the ball.
Wrongly assuming the surface of the ball being flat would result in the conclusion, that the lines must be curved.

The same applies to our three-dimensional space. We intutively assume it as flat, such that two lines starting at an observer with some angle should never intersect elsewhere. But in some cases they do; so we intuitively think the lines must be curved. But according to general relativity, it's actually our space-time which is curved by masses.

As an example, a ballistic curve looking like a parabola is actually a straight line in the curved space-time. So, two objects moving in different directions in e.g. Earth's orbit can eventually hit each other despite lack of drag or propulsion. They both move on a straight line (geodesic) in a curved space-time. The curved space-time allows these to lines to intersect later. Our three-dimensional Euclidean intuition interpretes this as curved trajectories.


How does inflation justify the nonexistence of magnetic monopoles? It is said that inflation justifies that magnetic monopoles don't exist. Can anyone explain how inflation theory explains the non existence of magnetic monopoles?

Let me start off by stating that scientists have not conclusively proven that magnetic monopoles do or do not exist. There has never had a confirmed detection of one so the jury is still out on if a magnetic monopole can or does exist at all. That being said, if monopoles do exist, no one has a good explanation for why we haven't detected one yet. There are hypotheses, but no solid proof.

To get to your question now, I think it is an incorrect statement to say that inflation theory explains the non-existence of magnetic monopoles. What inflation theory does potentially explain, is the extremely low density of magnetic monopoles in the universe. The idea here is that magnetic monopoles, for some unknown reason, are extremely unlikely to form, even in the intensely energetic and hot crucible of the Big Bang. Some estimates say that for 1 magnetic monopole that formed, $10^{29}$ "normal" particles formed, making magnetic monopoles extremely sparse and rare.

Where inflation theory comes in is that it says the universe expanded exponentially fast. If the monopoles were already sparse before this extreme expansion, the expansion would have only served to cause the magnetic monopole density in the universe be practically zero. It is quite possible that, post-inflation, the magnetic monopole density in the universe is so low that only one exists within the observable universe. Good luck finding that single particle out of the $10^{80}$ particle haystack.

In my mind though, that doesn't really answer the question as to why we don't see magnetic monopoles. Inflation just makes them extremely sparse, but why did so few form to begin with (assuming they formed at all). As I said above, nobody knows. A potential answer (without any evidence) is that magnetic monopoles are extremely energetic particles (e.g., with energies greater than the LHC of 14 TeV). If they have such extreme energies, even the Big Bang wouldn't have formed many of them. But again, that's just a guess.

Only time will tell what the real answer is.


How to find Exomoons? How would we be able to detect Exomoons? We detect exoplanets by seeing if the light gets faint from a star, but could we really use the same method for determining if there is an exomoon around that exoplanet? Thanks for your thoughts and answers!

I'm going to try to take a stab at answering this. With our current technologies, detecting exomoons can prove hard however there are various techniques being used today such as:

Analyzing data from the Kepler Spacecraft
Dynamic effects – the exomoon tugs the planet, which causes deviations in the times and durations of the host planet’s transits. This is similar to the radial velocity technique for detecting exoplanets. Source: UniverseToday
Transit effects – the exomoon may transit the star immediately before or just after the planet does. This will cause an added dip in the observed light. See this video for a great demonstration. This is similar to the light curve technique for detecting exoplanets. Source: UniverseToday
Gravitational Microlensing - which is a technique used to detect exoplanets like you stated above however it may also reveal signs of an exomoon. Read this source

I found this information doing some simple googling, feel free to edit or add to.


Years, months, days, and ... weeks? Why do we divide time into weeks?  Is there any celestial reason why humans do this?

one year: earth revolution around the sun
one month: moon revolution around the earth
one week: 7 days = ???
one day: earth rotation about its axis

The synodic period of the moon is $29.53$ days, a little shorter than a calendar month, which is on average about $30.4$ days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision.

Conventionally, the moon's appearance is divided into four phases: first quarter, full, last quarter, new. That means that on average, each phase lasts about $7.4$ days. Since calendars count days in integer amounts, a $7$-day period seems to be a natural choice.

The social importance of the seven-day period in Western cultural probably has much more to do with its religious significance in Abrahamic religions than astronomy per se (although certainly not unique to it). But its ultimate origin probably does lie in the natural division of the moon's appearance into four phases, which correspond to an apparent geocentric celestial latitude difference between the Moon and Sun of $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$.

That, the explicit answer to your question is

1 week = 7 days = one lunar phase.


Venus' magnitude during inferior conjunction I was doing an assignment on Stellarium when I observed Venus to have an apparent magnitude that didn't become less negative than -3, even at inferior conjunctions when we faced the dark side of Venus.

As far as I understand, apparent magnitude is subjective to an observer on earth, and that observer does not see Venus at all during an inferior conjunctions, so the magnitude should rise above zero.

I checked for other inferior and exterior planets and the moon on Stellarium; they all supported my theory.
I thought it was some sort of bug, but apparently it's a scientific fact.
I couldn't find any explanation online.

Did I understand something wrong?

I found references which indicate that Venus usually goes dark.  Check out the graphs in this paper for one example.

The exceptions happen when Venus' orbit, which is slightly out of the ecliptic, brings it to inferior conjunction when projected onto our (Earth's) orbital plane, but Venus remains at a slight angle relative to our viewing lines.

From UniverseToday,

  Not all inferior conjunctions of Venus are created equal. The planet’s
  orbit is tilted 3 degrees with respect to our own and can thus pass a
  maximum of eight degrees north or south of the Sun. Venus last did
  this on inferior conjunction in 2009 and will once again pass a
  maximum distance north of the Sun in 2017. For the southern
  hemisphere, the red letter years are 2007, and next year in 2015.


Will Pluto ever clear its path? I am not quite sure if it would, considering i'v looked through countless websites and found nothing and would like to know if it could clear its path and join the other planets?

In addition to the answer above, Pluto is a Neptune crossing trans neptunium object or TNO as it's often abreviated.  Being a TNO doesn't exclude the possibility of an object being a planet but crossing the orbit of Neptune does.   So, even if Pluto cleared out it's entire orbital neighborhood, it would still cross Neptune and naming it a planet would be problematic.

If it's orbit changed to being entire outside of Neptune, it would still face the problem of being one of several minor planets in that general region of similar size.

If you really wanted to push the theoretical to the limit, if Pluto's orbit moved so it stopped crossing Neptune's and if it merged with all the other similarly sized TNOs, then you could make an argument that it stands out as unique for it's region and a strong argument could be made that it should under those conditions, be called a planet, but that's a stack of improbabilities so high that winning next weeks mega millions top prize seems likely by comparison.

Also, if all the large known TNOs were somehow combined, they would still have a mass less than Mercury, though perhaps a slightly larger diameter, but that would be large enough where it just might get planet status.   It won't happen though.  Space is too big and object's orbits are too far from each other.   Planet formation happened within the proto-planetary disk at distances much closer to the Sun.   Uranus and Neptune likely formed much closer to the Sun than they are now.   There's not enough material and too much space in between for a planet to coalesce at Pluto distance from our Sun.


A question about calculating the radius of a comet A Comet is orbiting around the sun with period of $T= 5.5$ years. The comet has radius $R$. After 5.5 years (A period) we understand that its radius has decreased about $20cm$. How can we calculate the R. We know that comet is approximately a sphere and it is made of Ice. We have the $G = 6.67 * 10^{-11}$ SI_Unit, $M_{sun} = 1.99 * 10^{30} Kg$, $\sigma = 5.67 *10^{-8}$ SI_Unit (Stefan-Boltzmann constant) , $L_{sun}=3.85 * 10^{26} w$, $l_f = 333 KJ/Kg$ , $l_v = 2260 KJ/Kg$. It would be fine that if the solution doesn't need the numerical value of density of ice but if it can't be solved without density of ice, you are allowed to use it.

I can calculate the distance to sun ($a$) with formula:
$T^2 = 4 *\pi^2 a^3 /(GM_{Sun})$.

$a= 4.66* 10^{11} m$

I also can calculate $F= L_{sun} / (4 * \pi * a^2)$.

I also approximated the power that the comet gives is: $P = F * Area$ (think ice don't reflect any light)

But I don't know how to calculate the R and formulate how it decreases every moment (because it affects the area).

Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness.
Orbital Distance
We're told the period is $T=5.5\:\mathrm{years}$. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, $a$). Since we're talking about a comet orbiting the Sun, we can simply use:
$$T^2 = a^3$$
where $T$ is the orbital period in units of $\mathrm{years}$ and $a$ is the semi-major axis in units of $\mathrm{AU}$. I find that $a = 3.116\:\mathrm{AU} = 4.66\times 10^{11}\:\mathrm{m}$. Good so far.
Volume of Evaporated Shell
We're told that the radius of the comet decreases by $\Delta R = 20\:\mathrm{cm}$. We can assume the comet is perfectly spherical and calculate the volume of the evaporated shell which should depend on both $R$, the radius of the comet, and $\Delta R$. This will be necessary since we need to know the total amount of ice that was evaporated. The volume of this shell is given by
$$V = \frac{4}{3}\pi(R^3-(R-\Delta R)^3)$$
$$V = \frac{4}{3}\pi(R^3-R^3+3R^2\Delta R-3R\Delta R^2 + \Delta R^3)$$
$$V = \frac{4}{3}\pi(3R^2\Delta R-3R\Delta R^2)$$
$$V = 4\pi R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$
Note that I've made a specific choice here. I've dropped the $\Delta R^3$ term from the second line. The reason being that it is a third order term and $\Delta R^3 \ll R\Delta R^2$. You could argue that I could drop the second order term, $R\Delta R^2$ since $R\Delta R^2 \ll R^2\Delta R$, but I'm choosing to keep this second order term around so we wind up with an $R$ to solve for in the final answer.
Power Input
The next step is to find the total energy input per second to this comet, e.g., the power. You've basically already defined this part. The flux on this comet is defined as
$$F = \frac{L_\odot}{4\pi a^2}$$
The power input is simply the flux times the area of the comet.
$$P_{\mathrm{in}} = FA = \frac{L_\odot}{4\pi a^2} \pi R^2 = \frac{1}{4}L_\odot \frac{R^2}{a^2}$$
Note that this assumes the comet is in a circular orbit and thus always at the orbital radius of $a$. If the comet's orbit had any sort of eccentricity, then $F$ would be a function of radius and you'd have a much harder time.
Evaporation Energy
Now we need to calculate the total energy necessary to evaporate the evaporated shell of volume $V$ from above. In order to evaporate a solid ice, you have to go through four stages of heating. First, you raise the ice's temperature to the melting point. The energy input for this is defined by the specific heat capacity for ice, $c_{\mathrm{ice}}$. Then you add energy to convert the ice into water. The energy input for this is defined by the latent heat of fusion, $\ell_f$. Now you can raise the temperature of the water until it reaches the next stage. This is defined by the specific heat of water, $c_{\mathrm{water}}$. Finally, you add energy to convert the water to gas, defined by the latent heat of vaporization, $\ell_v$.
All of these can be put together into a single equation.
$$E_{\mathrm{evap}} = c_{\mathrm{ice}}(m_{\mathrm{shell}}\Delta T_1) + \ell_fm_{\mathrm{shell}}+c_{\mathrm{water}}(m_{\mathrm{shell}}\Delta T_2) + \ell_vm_{\mathrm{shell}}$$
$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)m_{\mathrm{shell}}$$
Each one of the terms in this equation accounts for the energy input of one of the stages I described above. Note that $\Delta T_1$ is the temperature change from the initial temperature to the melting point ($273.15\:\mathrm{K}$). A reasonable temperature change might be $73.15\:\mathrm{K}$ (assuming an initial temperature of $200\:\mathrm{K}$), based on the temperature of comet 67P as determined by Rosetta. The $\Delta T_2$ is the temperature change from the melting point to the boiling point which is necessarily $100\:\mathrm{K}$.
You can look in a table somewhere and find that $c_{\mathrm{ice}} = 2.108\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$ and $c_{\mathrm{water}} = 4.187\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$.
Lastly, you need to define the mass of the evaporated shell, $m_{\mathrm{shell}}$. This is simply the volume already determined, multiplied by the density of ice/water. Technically those densities will be different, but they're close enough that we can just use $\rho_{shell} = 1000\:\mathrm{kg\:m^3}$. So ultimately we have:
$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)V \rho_{\mathrm{shell}}$$
For simplicity's sake, I'm going to define
$$\eta \equiv (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)$$
so that
$$E_{\mathrm{evap}} = 4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$
Putting it all together
We now know the total energy necessary to evaporate the comet's $20\:\mathrm{cm}$ shell as well as the energy input per second. We know that it receives this energy input per second for one orbital period of $5.5\:\mathrm{years}$ which means we can say:
$$T = \frac{E_{\mathrm{evap}}}{P_{\mathrm{in}}}$$
$$T = \frac{4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)}{\frac{1}{4}L_\odot \frac{R^2}{a^2}}$$
$$T = 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot} \left(1-\frac{\Delta R}{R}\right)$$
Again, for simplicity, I'll define
$$\xi \equiv 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot}$$
so that
$$T = \xi \left(1-\frac{\Delta R}{R}\right)$$
It should be pretty easy to see now that
$$\boxed{R = \frac{\Delta R}{1-T/\xi}}$$
The rest is just plugging everything in.


How do we know that a natural satellite cannot be a hollow object? In their 1966 book, Intelligent Life in the Universe, Carl Sagan and Iosif Shklovsky state that "a natural satellite cannot be a hollow object." How do we know this is absolutely true?

I'd caution that absolute statements are not good things in science. When Sagan says a natural satellite cannot be hallow, it is an absolute statement, but there's no way to 100% prove it true.

That being said, everything we know about planet/satellite formation tells us that the chances of a satellite forming as a hollow structure (or becoming hollow afterwards) is pretty much nil.

There is still much debate on the specifics of planetary formation, but the general process seems to be that small rocks and particles start to form, then they coalesce into bigger chunks and slowly that planetesimal grows in size. This type of formation process does not allow for the center to be hollow since it is the center which forms first! You cannot form the "edges" of your hollow planet and have those edges come together.

Moons and satellites can potentially have a different formation path. Look at our own moon for example. It is hypothesized that a Mars-sized proto-planet crashed into the Earth, flinging out a large amount of material. Much of this material then coalesced into our current Moon. Again, this process doesn't really allow for making something which is hollow.

Aside from the difficulty in forming a hollow satellite, you also have the challenge of maintaining the structure. Gravity constantly pulls everything to the center of the body. If there's nothing at the center, there's nothing preventing the outer shell from collapsing. It is technically possible that the satellite could not collapse, but that would require it to be fairly small and be perfectly structured such that the structural integrity of the rock forming the shell could withstand the crushing gravitational forces. Really though, such an object would have to be intelligently designed as the chances of it forming perfectly like that are non-existent. There's no possible way an object like Phobos could maintain itself without collapsing if it were hollow.


September 26 twelve hours sunrise to sunset? I noticed in today's newspaper (Boston Globe, September 26) that sunrise and sunset were both at 6:25. That's the twelve hour day I'd have expected at the equinox.

(Please retag if there's a better one)

A day would be twelve hours long at the equinoxes if the Earth had no atmosphere and if the Sun was a point rather than a sphere. Sunrise and sunset are defined as the time at which the upper limb of the Sun appears to rise above or set below an ideal oblate spheroid Earth and assuming average atmospheric conditions. While this calculation ignores terrain and variations in atmospheric condition, it does not ignore that the basic facts that the Earth does have an atmosphere and that Sun is a sphere. The standard atmosphere conditions means that sunrise (sunset) occur when the upper limb of the Sun first reaches (first falls below) 34 arc minutes below the horizon. That the Sun is not a point adds another 16 arc minutes to this, making sunrise/sunset occur when the center of the Sun is 50 arc minutes below the horizon.

This means that at the equinoxes, a "day" is at least six minutes longer than the twelve hours we are naively taught (and incorrectly reported twice a year by naive weathermen). As an aside, days are always longer than nights at the equator.


Does the sun have a vertical movment? As the Sun moves along with the planets does the sun exhibit vertical movement? That is, does it traverse in different planes?

Does the sun have a vertical movment?

Sure it does!

People who follow solar system bodies with the highest accuracy, including NASA's JPL, use the solar system barycenter (center of mass) for the origin of the plane of the ecliptic.

Since the big planets (Jupiter et al.) have some inclination, they pull the Sun up and down out of the plane of the Ecliptic by tens of thousands of kilometers. It's a small fraction of it's 600,000 kilometer radius, but it certainly happens.

You can see the data from the JPL Horizons website, or if you like Python, you can get virtually the same numbers by running the Skyfield package.

From Horizions I downloaded the position of the Sun and the four major planets of the solar system in 10 day steps from the years 1900 to 2100. Below I plot the three components X, Y and Z in the J2000.0 Solar system barycentric coordinates using the ecliptic as the reference frame. The black line is the motion of the Sun, and the color lines are for the four planets, but I have multiplied each by the the mass of the planet divided by the Mass of the Sun.

In the second plot, I added the four together, and you can see that together they mirror the motion of the Sun.

The last two show the movement of the Sun in 3D (two views). Note that the vertical scale is much smaller than the horizontal scale.

Horizons setup:

below: Black thick line: motion of the Sun in barycentric coordinates in the J2000.0 X, Y and Z directions, relative to the ecliptic. Four thinner color lines: motions of the four largest planets, each weighted by its mass relative to the Sun's mass. Short period lines are Jupiter and Saturn with Jupiter the largest amplitude, long periods are Uranus and Neptune, with Neptune the largest amplitude.

below: Black thick line: motion of the Sun in barycentric coordinates in the J2000.0 X, Y and Z directions, relative to the ecliptic. Thin red line: sum of the motions of the four largest planets, each weighted by its mass relative to the Sun's mass.

below: Two views  of the 3D motion of the Sun in barycentric coordinates between the years 1900 and 2100. Units are kilometers, and note that the scales are different for each axis.

class Body(object):
    def __init__(self, name):
        self.name = name

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

names = ('Sun', 'Jupiter', 'Saturn', 'Uranus', 'Neptune')

bodies = []
for name in names:
    fname = 'sun motion ' + name + ' horizons_results.txt'
    body  = Body(name)
    bodies.append(body)

    with open(fname, "r") as infile:
        lines = infile.read().splitlines()

    iSOE = [i for i, line in enumerate(lines) if "$$SOE" in line][0]
    iEOE = [i for i, line in enumerate(lines) if "$$EOE" in line][0]

    print (name, iSOE, iEOE, lines[iSOE], lines[iEOE])

    lines = [line.split(',') for line in lines[iSOE+1:iEOE]]
    JD  = np.array([float(line[0]) for line in lines])
    pos = np.array([[float(item) for item in line[2:5]] for line in lines])
    body.lines = lines
    body.JD    = JD
    body.pos   = pos.T.copy()

years = 2000 + (JD - JD[3652])/365.2564
Sun, Jupiter, Saturn, Uranus, Neptune = bodies

for body in bodies:
    body.rsun = np.sqrt(((body.pos - Sun.pos)**2).sum(axis=0))

planets = [b for b in bodies if 'sun' not in b.name.lower()]

massdict = {'Jupiter':1.898E+27, 'Saturn':5.683E+26, 'Uranus':8.681E+25,
            'Neptune':1.024E+26, 'Sun':1.989E+30}

for b in bodies:
    b.mass = massdict[b.name]

if True:
    plt.figure()
    for i, thing in enumerate(Sun.pos):
        plt.subplot(3, 1, i+1)
        plt.plot(years, thing, '-k', linewidth=1.5)
    for p in planets:
        for i, thing in enumerate(p.pos):
            plt.subplot(3, 1, i+1)
            thing = thing * (p.mass/Sun.mass)
            plt.plot(years, thing)
    plt.show()

all_four = Body('all_four')
all_four.pos = np.array([p.pos*(p.mass/Sun.mass) for p in planets]).sum(axis=0)

if True:
    plt.figure()
    for i, thing in enumerate(Sun.pos):
        plt.subplot(3, 1, i+1)
        plt.plot(years, thing, '-k', linewidth=1.5)
    for i, thing in enumerate(all_four.pos):
        plt.subplot(3, 1, i+1)
        plt.plot(years, thing, '-r')
    plt.show()

if True:
    fig = plt.figure(figsize=[12, 10])  # [12, 10]
    ax  = fig.add_subplot(1, 1, 1, projection='3d')

    x, y, z = Sun.pos
    ax.plot(x, y, z)

    plt.show()


Can close binaries have a very eccentric orbit? For example, are there binaries with an orbital period < 5 day  and that are very eccentric?

What kind of factors could slow down orbital circularisation?

Are there examples of binaries with p<5day and e>0.1?

There are surprisingly many examples of short period, eccentric binry systems.

A very incomplete list would include

KIC4544587 P=2.19 d  e=0.288 +/- 0.026 Hambleton et al. (2013)

HD174884 P=3.66 d  e=0.2939 +/- 0.0005 Maceroni et al. (2010)

CoRoT 102918586 P=4.39 d  e=0.249 +/- 0.005 Maceroni et al. (2013)

HD313926 P=2.27d  e=0.209 +/-0.001 Rucinski et al. (2007)

In each of these cases, the stars are relatively warm, with thin (if any) subphotospheric convection zones. Tidal circularisation is much more effective in stars with thick convection zones (giants, or cooler main sequence stars). Therefore it is possible that all of these objects (and, as I say, there are many other examples) are too young to have circularised.

Some theory is presented by Zahn et al. (2005). Indeed, circularisation times are very short - basically you expect everything with a solar mass or below to be circularised if its period is less than 6 days, but the process is less efficient for higher mass stars. Circularisation will essentilly take place during pre-main-sequence evolution. Since a circular orbit represents the minimum energy configuration, the only way you are going to get non-circular orbits in solar-type and lower mass stars is if they are very young (less than some tens of millions of years), the binary system formed after their pre-main-sequence evolution or if there is an injection of energy via a dynamical interaction or an eccentric third body in a much wider orbit.


Why planet's orbit is not perpendicular or random ? Why planet's orbit is not perpendicular or random ? It always seems each planet is revolving on the same geometric-plane around the star.

Short answer: conservation of angular momentum.

Long answer:

The origin of almost any planetary system is a sparse cloud. That cloud starts to contract due (typically) to a pressure wave crossing it.

The cloud fragments as it contracts, and each fragment is what we know as a pre-star cloud.

Since almost always there is some movement in the matter in each cloud, the cloud as a whole starts to rotate, very slowly. Contraction helps because, due to angular momentum conservation law, when the cloud contracts, the rotation accelerates.

Soon we get a protostar with the most contracted matter, surrounded by a protoplanetary disk composed with the less contracted matter. The rotation of the whole system is in the same plane, due to conservation of angular momentum.

The protostar becomes a star, and the protoplanetary disk becomes a bunch of planets. Each planet, in turn, orbits the star and rotates on itself, all in the same direction, based on which point of the protoplanetary disk started accreting mass.

Later on, interaction among massive bodies disturbs this shiny process by changing some orbits and by having some collisions, but that's minor.


How could lithium burning take place in a quasi-star? According to Begelman et al. (2008), one of the most distinguishing features of the hypothetical quasi-star is that it's supported by radiation pressure from the accretion disk of the black hole in its core. The paper suggests that these stars form from Population III stars.

What's more, while almost all of the star's luminosity is from the accretion disk (which is far more luminous than thermonuclear reactions), temperatures around the central black hole are hot enough for fusion:

  Although we do not model the non-hydrostatic region of quasi-stars in any detail, in our case high temperatures are attained in the immediate vicinity of the black hole. However, even in this region the neglect of nuclear reactions is justified, first because black hole accretion is energetically much more efficient than fusion, and secondly because on-going accretion limits the time-scale over which inflowing gas is exposed to high $T$.

That's all fine, but what surprised me is the footnote on the third page:

  In the most massive quasi-stars, the central temperature may be high enough (a few million K) to initiate lithium burning. This is energetically negligible, and although the presence or absence of lithium does affect the opacity, the effect is small for the photospheric temperatures and densities of interest here.

Even though this part dismisses lithium burning as an important factor, since it's energy output doesn't compare to that of the accretion disk's, it still mentions that lithium burning can take place. Considering that these form from Population III stars, how is it that much lithium burning can happen in the first place?

Although lithium is a relatively light atom, it is considered a metal and lithium burning depletes the amount of lithium in a star. Since Population III stars are supposed to be nearly metal-free, shouldn't only hydrogen or helium fusion be a factor here, even if they only have a small effect?

Lithium, along with Hydrogen and Helium, was one of the 3 elements created in the Big Bang. Thus, it should exist to some part in any star that hasn't burnt all of it out, and as mentioned, it's not an easy thing to do.

Population III stars are expected to contain Lithium, and Beryllium as well. The amount, however, is not particularly high.


Fits file re-alignment I would like to re-align an image using a reference image (taken from GAIA). I am aware of the software Koords developed by the Karma team: https://www.atnf.csiro.au/computing/software/karma/

Unfortunately, the software crashes (with segmentation fault) everytime I try to add a source in either image. Is there a more up-to-date version of this tool? Or simply a different tool?

Update: Thank you for the responses! I will try to better explain exactly what I am trying to do :)

I have an X-ray image taken from Chandra and a GAIA-Aligned HST image. I know that the pointing for the X-ray image is off by ~.5 arcsec or so. I want  to re-align the X-ray image so that any X-ray point sources match their optical counter parts using the GAIA-Aligned HST image as my reference.  I hope this helps!

I ended up making my own small code that you can find here:
     https://github.com/crhea93/AstronomyTools/blob/master/Astrometry/Correction.py

Essentially, after inputting at least 3 matching target (found by eye in DS9 for example), the program will calculate the mean offset and apply it to the target image. The output is a new fits file of the target image with the proper correction.


Star - Black Hole binary. Star core Collapses. Does the black hole immediately grow from neutrinos? I was reading on how Type II Supernovae release as much as 99% of their energy as neutrinos. So I was thinking: suppose there was a large star and black hole binary system, but the two were not close enough for the BH to draw mass out of the star, so the star aged naturally instead.

If one were close enough to observe when the star's core collapses - but before the supernova bursts through the star's surface - would it be possible to see the black hole immediately grow in size for no apparent reason as it eats up the neutrinos from the companion star's collapse?

Yes. Neutrinos which hit the BH will add to its mass.

But you may be thinking that the neutrinos pass through the nascent supernova without interaction. In a dense enough environment, neutrinos do interact, and absorption of neutrinos by the layers beyond the core is a major source of the energy that disrupts the star.

In a bit more detail, the core collapse happens through inverse beta decay, where a proton and an electron react to form a neutron and a neutrino. This basically turns the star's iron core to neutrons. The energy is carried away by the neutrinos and the absorption of this enormous flux of neutrinos by the outer layers drives the SN.  The Wikipedia article has a good (but IMO somewhat incoherent) longer discussion.

Still the many neutrinos which do escape do so before the shock wave blows the star apart and the energy they carry will bulk up a nearby BH.

I doubt that it would easily be observable, though. Stellar-mass BHs are small in cross-section and would only absorb neutrinos from a patch of the sky proportional to their area. If a 10km BH was orbiting a million km star right at its surface, it would only absorb one part in 1010 of the neutrinos, producing roughly a 10-9 fractional increase in size.  Very difficult to measure!

If you're using this to provide an early warning of the core-collapse (perhaps so you can start you FTL drive and get outta there) it's fairly useless. First, the neutrino flux is probably high enough to fry people who are close enough to need the early warning.  And second, if you're far enough away not to be fried by the neutrinos, you have plenty of time to see the star rapidly swelling  and get moving before the comparatively slow shock wave gets near you.


What keeps galaxies together? The super massive black hole at the centre of Milky Way has a mass of about 4 million times that of our sun. Is this enough to keep the entire galaxy together, is it this black hole which keeps the galaxy together? I did read that there are also galaxies that don't have a super massive blackhole, what keeps those galaxies together?
I did ask this question at space stackexchange but it got closed because it was posted at wrong place.

The galaxy is kept together by the combined mass of the matter in the galaxy, of which the supermassive black hole is a negligible part. There are galaxies that don't have a central black hole (such as the Triangulum galaxy), and they are also held together by their combined mass.
In particular, the dark matter of the galaxy is what provides most of the mass that holds the visible matter together. Galaxies usually have far more dark matter than visible matter. This mass is distributed in a roughly spherical shape, with the greatest density towards the centre of the galaxy, and dropping off further from the centre. This "dark matter halo" is substantially larger than the visible disc.


Why the moon is blinding me right now? I'm on the most northern part of Germany (Schleswig-Holstein) it is 11.20pm right now and the moon is here as bright as I never saw it before I was just outside and it is as bright as if it was dim daylight. When the clouds moved away from the moon and I looked into the moon for a little moment it was so bright, that it actually glared me for a little moment. Is there anything specific going on that is just favored this event? Or is it something that simply happens sometimes and I just never noticed it before?

Other than it being a Full Moon and the Moon is high in the sky, there is nothing unusual going on.


Was it a space station? I just looked at the sky and saw a star, not so bright but moving very fast and in almost a minute it disappeared, as it was so fast.
Timing was in between 9.35-9.45 PM in Nuremberg, Germany.
My location is 49.45, 11.06.

So was it International Space station?

Anyway, I'm going to verify it after 92 minutes if i am able see it again.

Heavens-Above shows two fairly bright objects passing over Nuremberg at that time on 2020-04-14:

a Resurs-O1 rocket moving south to north
an Iridium-4 rocket moving north to south

The next visible ISS passes at your location are between April 28 and May 3, all before dawn.


What are the natural life sustaining resources available in the Jovian moon system? Hard science fiction writing question.  Premise: what natural resources are present in the Jovian system that would be necessary and desireable to an autonomous, self-sustaining human community?  Water on Europa, for example.  What else? Would like to keep this plausible without a lot of speculation.  Thank you.

The moons Europa, Ganymede, and Callisto all have substantial amounts of water ice on their surfaces, and for the latter two, also silicate rocks. There is also some CO2 in the ice as well as ammonia and methane. These give potential sources of carbon and nitrogen, and with enough energy you could make glucose and amino acids - Food.

Io is a crazy place, with lava flows, and sulphur covering the surface. Its not an easy place to mine

One potential source of energy is ${}^3\mathrm{He}$ This is a potentially useful fuel for a fusion reactor. It is found (in very low concentrations) in shaded parts of the moon. It may also be found on Jupiter's moons.

The minor moon Carme may have originated in the Kuiper belt (long ago) and have an interesting combination of carbon based substances. Amalthea has an interesting red colour - maybe carbon compounds again. There are quite a few moons that began as captured asteroids, but none seem to have been metallic.

There are lots of problems with life on Jupiter's moons: Cold - No air - radiation - etc


Why are nearby stars like Proxima Centauri and Barnard's star not visible to the naked eye? Proxima Centauri is the closest star to the earth other than the sun yet it isn't visible to the naked eye, but stars that are further away are easily found in the night sky. Why is that?

John's answer is correct.   For a few more details:

Stars brightness works out to roughly the 4th power of their relative mass.  This falls off a bit for very large stars, but for smaller stars it's in the ballpark.   (Red-giant phases not included.)

Proxima Centauri is about 12% the mass of our sun, and about 14% the sun's diameter.   At 12% its mass, its brightness, using the 4th power rule, is about 0.02% or 1/5,000th as bright.   That's total light.  Because it's smaller and colder, much of that light is in the infra-red range that we can't see.   In terms of visible light or visual magnitude, it's about 1/20,000th as bright as our sun.   That means, for it to be as bright as our sun from Earth in visible light, it would need to be about 1 million KM away.   If it was as far away as Pluto, some 40 AU on average, it wouldn't look very different than Venus at Venus' peak brightness, with a distinct red color.

Larger stars can have a brightness over 100,000 times brighter than our Sun.   So, because the mass of the star affects the brightness to about the 4th power, what it really comes down to is size matters.

Distance matters too, but only to the square of the distance, so mass of the star matters more.   Here's an article with a chart of the visually brightest stars in the sky.  Negative numbers and smaller numbers correspond to greater brightness.

It might go without saying, but virtually none of the stars that can be seen with the naked eye are red dwarf stars even though they're the most common type of star.  They're too dim for us to see at multiple light-years distance.


What is the current angle of motion of the sun with respect to the Galactic Plane? I understand that the solar system oscillates in its motion around the Milky Way. However at any given time, there must be a specific angle between the current direction of the solar apex and the galactic plane. This is the angle - in degrees - I am looking for.

To be specific. I'm not looking for the angle of the solar system with respect to the galactic plane, nor the angle between the celestial pole and the galactic plane, nor the angle between the plane of the ecliptic and the galactic plane. I am only looking for the current angle of motion between the solar apex and the galactic plane.

Also, is there more clear terminology for this? I've spent a few hours searching and it seems that I can't find an answer to what I would imagine is a relatively obvious question. Am I searching with the wrong terminology?

Thanks.

In an answer to this question, you can see that I describe the trajectory of the Sun around the Galaxy. Its current motion can be defined in a number of ways, according to your preferred frame of reference. From your question I judge that you want it in a "Galactocentric" frame, so we take the velocity of the Sun with respect to the "Local Standard of Rest" (LSR), which is the Sun's motion with respect to an average of local stars, and add something to it to account for the fact that local stars are orbiting the Galactic centre every 220 million years or so.

The velocity of the LSR is 10 km/s towards the Galactic centre, 5 km/s faster than an average star at the Sun's location tangentially to a line towards the Galactic centre, and 7 km/s upwards out of the Galactic plane (see for example here).

Thus, if we take this frame of reference, the component of the velocity of the Sun in the Galactic plane is $\sqrt{10^2 + 5^2} = 11.2$ km/s amd the component out of the Galactic plane is just 7 km/s. The angle this trajectory makes with the Galactic plane is $\arctan(7/11.2) = 32$ degrees. Note that there are uncertainties/disagreements about the exact value of the Sun's motion with respect to the LSR at the level of 1-2 km/s, so this final angle is also somewhat uncertain.

However, this rather steep angle does not take into account that the LSR is moving around the Galaxy at rather high speed. This speed is around 250 km/s in the Galactic plane and at right angles to the Galactic centre (e.g. Reid et al. 2014). There may be a more modern figure, but I cannot immediately locate it, but it is probably safe to assume an uncertainty of at least 5-10 km/s in this. With this we now have velocity components of 11.2 km/s out of the plane and 255 km/s in the plane. The angle you are seeking is then $\arctan(11.2/255) = 2.5$ degrees, with an uncertainty of only $\sim 0.3$ degrees. So if you were trying to make a drawing of the solar system motion with respect to the Galaxy, that is the angle you would use. Note that the Sun is currently above the Galactic plane by about 20 pc.


A black hole that doesn't take in matter? According to List of Common Misconceptions by Wikipedia, under the heading of astronomy, this line can be found:

  A black hole can act like a "cosmic vacuum cleaner" and pull a
  substantial inflow of matter, but only if the star it forms from is
  already having a similar effect on surrounding matter.

This is referenced to a paper written in Yale university:  "Frontiers And Controversies In Astrophysics Transcript 9", although that's a dead link. By linking so, I believe that it is a reliable source.

This got me thinking that is there a black hole that doesn't suck in surrounding matter? And is there a star that don't take in surrounding matter?

This is the full quote

  If, for example, the Sun were replaced by a black hole of equal mass,
  the orbits of the planets would be essentially unaffected. A black
  hole can act like a "cosmic vacuum cleaner" and pull a substantial
  inflow of matter, but only if the star it forms from is already having
  a similar effect on surrounding matter

It's badly written, but what it's basically saying is that it doesn't matter the form of the massive object that's being orbited by a planet or other object.  A stable orbit would remain stable, so, if our sun was to collapse into a black hole and nothing else changes, Earth's orbit would remain the same.  It would get dark and cold, but the orbit would be unchanged.

The misconception is, if the sun became a black hole than Earth would be sucked into the sun.   That's 100% false and that's all they're saying.

Gravity is a function of mass and distance.   Black holes have very high gravity, but a big part of the reason for that is because when they form out of dying stars, they become very small, a few solar masses squeezed into only about 10 miles across so the distance to the center of mass gets very small.

For a planet to be in any danger from a black hole, it would probably need to be inside the Roche Limit, perhaps only a few million miles away, which is several times closer than Mercury is to the sun, for example.   The safe distance, of-course, varies with the density and solidity of the orbiting object.

Here's a Q on Roche Limits and Black holes if interested.


What would happen if those gravitational waves were much stronger? What if those gravitational waves detected would be actually so strong that the distance of those mirrors moved would be larger? Not 1/1000th of proton but rather one millimeter, or one meter or even more? What would happen to us? How big would these black holes had to be and how much energy would be released in such a process?

Based on a post by an Astrophysicist my understanding is as follows:

Gravitational waves would still be faint to detect irrespective of the size of the Black Holes. BTW aren't black holes meant to be more concentrated rather than bloated?
The distance from ground zero is what matters rather than the size of the colliding black holes and even if they were a billion times closer they would still not measure up to 1 mm.

There wouldn't be significant impact from the waves themselves rather the Black Holes are the ones that would do the damage.
For the shift to be 1 mm we would require a force enormously stronger than that of a gravitational wave.

Here's the link to the article: http://www.forbes.com/sites/briankoberlein/2016/02/13/could-gravitational-waves-ever-be-strong-enough-to-feel/#1605e7fa4aac

Disclaimer: All my points are a result of my learning from the above link rather than any native knowledge or original research that I profess to have done.


Could the trajectories of non-periodic comets be used to infer properties of the 'ninth planet'? Non-periodic comets are comets which have very long orbital periods (>200 years or more), spending most of their time in the outer solar system.

Planet X, recently revived by researchers at Caltech, is a proposed planet orbiting in the farthest reaches of the outer solar system. The Caltech researchers infer its presence from to its effect on the orbits of Kuiper belt objects.

If such a planet exists, and is the cause of the Kuiper belt object alignments, could it not also knock such objects in to the inner solar system, creating non-periodic comets with huge aphelion orbital distances? Could our current observations of non-periodic comets be used to infer properties of the planet?

The idea is basically to look for comets coming from the same direction. Look for a pair of comets that at some time earlier in their orbit was at the same place at the same time. That is a data point, but an extremely inaccurate one. It does not have to mean anything, and you are going to collect a large amount of statistical noise. For this purpose, normal periodic comets are close to useless, as they are in the same region for decades or even centuries as they fall slowly towards the sun. Hyperbolic (non-periodic) comets on the other hand, have a shorter stay in each region, allowing us to better pin-point a possible encounter. The problem is, there are not actually a lot of them observed, and the length of the time horizon for accurate enough observations is not that long.

The properties you can obtain about the ninth planet, if it exists, from large amount of data on non-periodic comets are:

Current distance within a reasonably large error margin. (+-10 AU)
Orbital period; not so accurate, but you may get an OK estimate.
Inclination; not accurate, but together with the orbital period, estimates will grow more accurate over a longer time horizon of observations.


Best approximation for Sun's trajectory around galactic center? What is the current best approximation for the path the Sun takes around the center of the Milky Way?

I have found some information on the approximate position of the Milky Way's center, the speed of the Sun's orbit, and the inclination/declination of the Sun's orbit. I have also seen models suggesting the Solar System oscillates up and down as it revolves around the galactic center.

I am wondering if there is any agreed upon approximation (a 3d plot of position vs time) that includes all of these factors.

The Sun orbits in the Galactic potential. The motion is complex; it takes about 230 million years to make a circuit with an orbital speed of around 220 km/s, but at the same time it oscillates up and down with respect to the Galactic plane every $\sim 70$ million years and also wobbles in and out every $\sim 150$ million years (this is called epicyclic motion). The spatial amplitudes of these oscillations are around 100 pc vertically and 300 pc in the radial direction inwards and outwards around an average orbital radius (I am unable to locate a precise figure for the latter).

This is established by measuring how the Sun moves with respect to the average motion of the stars in the solar vicinity - the so-called local standard of rest. Estimates of the Sun's motion with respect to the LSR vary a little. According to Dehnen & Binney (1998) the Sun currently moves at 10 km/s inwards, at about 5 km/s faster than the average star tangentially to the Galactic centre and at about 7 km/s upwards out of the Galactic plane. Though as I mentioned previously, these motions are of an oscillatory nature around the mean value. The Sun is currently about 8.5 kpc from the Galactic centre and about 20 pc "north" of the Galactic plane.

I leave it as an exercise for the reader to visualise this orbit. From a side-view it would look like a pseudo-sinusoidal oscillation. From above the disk it would look like a slightly eccentric ellipse with a very large precession.

All of the above assumes that the Sun moves in a smooth Galactic potential. In fact the presence of spiral arms, a Galactic bar and the clumpy presence of giant molecular clouds can and will perturb the Sun's Galactic orbit around the regular behaviour described above, so predicting the Galactic location of the Sun in a billion years or so is not necessarily going to be very accurate.


How do I know, mathematically rather than from observation, if a moon is full? I know about the equations to describe the orbit of a moon around a planet. I know the moon's semi-major axis and eccentricity, and the same for its host world with the star they orbit.

Is there any equation that tells me how much of the moon is illuminated at night, and possibly how bright, as seen from the planet?

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

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

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

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

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

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

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


Is there a way to calculate the mass of of a stellar remnant given the initial mass of the star? Assume there is a star in a completely empty region of space, away from any outside influence.
Is there a way to calculate the mass of a resulting stellar remnant, a white dwarf, neutron star, or black hole(variations is accuracy are completely fine) that will result after a star dies, only given the star's mass around the time in which will die? (like an equation or something)
And, is there a way to calculate the resulting remnant's mass only given the star's initial mass during its formation?
It would be appreciated if one were to provide a concrete method or graph for determining such.
Thanks in advance!

For massive stars and their remnants, one can only be guided by theoretical measurements at present, since we do not have any masses for isolated neutron stars and black holes and no measurements of the masses of the progenitors of any neutron stars or black holes (only limits). Here are two plots taken from the extensive review by Limongi (2017). These show the relationships you want (the red lines) between the final remnant mass (on the y-axis) and the initial progenitor mass on the x-axis. The calculations are highly dependent on metallicity, so plots are given here (there are more in the paper) for solar metallicity ([Fe/H]=0) and a thousandth of solar metallicity ([Fe/H]=-3). There is also a dependence on rotation, which may be important.

You have to appreciate that these red curves are highly uncertain and model-dependent.
The situation is much clearer at lower masses. The masses of white dwarfs, even isolated white dwarfs, can be estimated quite well and their presence in binary systems and clusters of stars gives the means to estimate their ages. Combining this with their cooling ages gives a good estimate of the time that their progenitor spent as a "normal" star and then stellar evolution models give an estimate of the progenitor mass. Both the modelling and measurements are much more secure and so the initial-final mass relation for white dwarfs is in quite good shape.
The plot below is from a summary by Kalirai (2013). It is likely to be superseded in the next year or two by the wealth of data coming in on white dwarfs from the Gaia satellite. The names on the plot are the clusters in which the calibrating white dwarfs were found. You will see on the plot that Kalirai gives a simple formula that represents this relationship and might be what you are looking for. I believe the dashed line is from some stellar evolution model.


What are some applicable problems with the correct usage of G? In some fields of astrophysics, where distances are measured in parsecs (pc), velocities in kilometers per second (km/s) and masses in solar units (), it is useful to express G as approximately 4.302 x 10^-3 pc M^-1 (km/s)^2 ...
Is G not approximately 6.67 x 10^-11 N(m/kg)^2? Any reasonable explanation would be appreciated.

I think one of the reasons for using different units is how you take the measurements for the other variables in the equation.

For example, let's say you have two marbles, and you want to calculate the gravitational force between them (not a lot!). You place them a certain distance apart, and measure it with your meter stick. You find the distance between them to be $0.146 \text{ meters}$. Now you weigh them, and find their masses to be $0.00156 \text{ kilograms}$ apiece. You decide to use the value of $G$ with units including meters and kilograms.

Now let's say you have two stars far, far away. You take measurements based on parallax and find that they're roughly $35,400 \text{ light-years}$ away from each other. Your instruments can't measure this to any greater accuracy. You measure their masses by calculating how much they deflect light, and come up with a figure of $10 \text { solar masses}$ apiece. You therefore decide to use some form of $G$ that, for some reason, has solar masses and light-years as some of its units.

Basically, you choose different forms of a constant depending on the units you're able to measure with your instruments. You won't be able to figure out exactly how many meters your stars are apart from each other, but you can figure out just about how many light-years apart they are, with less than a light-year of uncertainty.


When a neutron star accretes matter, will its mass increase? When a nova gets matter, the outcome is that it may lose more matter because of eruptions. Only some kind of nova's masses can increase stably.

A white dwarf may become a supernova after it exceeds the Chandrasekhar limit, and we study that much. A neutron star can get mass too, why do not we talk about their end much?

How about a neutron star when it gets matter from accretion? It is necessary that its mass will increase?
When will it collapse?

The neutron stars have about 1.1 to 3 solar masses. If the neutron star accretes mass, its mass will increase and once it crosses the Tolman–Oppenheimer–Volkoff limit, the star will collapse to form a black hole.

On way of this happening is in the case where one of the binary stars becomes a neutron star as in X-ray pulsars. The neutron star receives gas supply from its companion, which increases its mass,for example CenX3.

Source: nasa.gov

Though this method could lead tothe formation of black hole, no observational evidence exists as of today.


What happens to a star after it has become a red giant? When our sun reaches the end of its lifetime , it will turn into a red giant. How long will it be a red giant, and what happens after a star has been a red giant? Does it turn into a planet after a long enough time spent, when it has used up all its helium?

It will become a planetary nebula like e.g. the Cat's Eye nebula that was formed by the death of a star with about the same mass as the Sun ~ $1\ M_\odot $:

              Composite image using optical images from the HST and X-ray data from the Chandra X-ray Observatory

The Sun, and any red dwarfs above about 0.25 solar masses, will expand into what's called a red giant, a late stage of stellar evolution. At this stage, the star starts to fuse different elements, and eventually throws off its layers as a planetary nebula, leaving behind a white dwarf made of carbon and oxygen.

A red dwarf that is too small to become a red giant will not turn into a planetary nebula: its fusion processes will eventually cease and it will probably produce a white dwarf made mostly of helium. But the main-sequence lifetime of these very small stars is longer than the age of the universe, so this has never actually happened yet. The helium white dwarfs that do exist were formed in binary star systems (and were formed during their complex dynamics).

Red dwarfs are small main sequence stars - the smallest and dimmest that are still able to fuse hydrogen in their core. Brown dwarfs are even smaller (in terms of mass), but cannot even fuse hydrogen - they are thought to be able to fuse deuterium and lithium. Stars smaller than a brown dwarf turn out to be gas giants like our Jupiter, Saturn, Uranus and Neptune.


Why would the side of the moon that faces earth be as dark as the far side of the moon? Somebody suggested that the side of the moon that always faces the earth is as dark as the far side of the moon, but is that really the case? Doesn’t earth-shine make the earth-facing side of the moon overall brighter? Or does the absence of a moon atmosphere negate any possible gain from earth-shine?

Somebody suggested that the side of the moon that always faces the earth is as dark as the far side of the moon, but is that really the case?

That is not the case.
I'll look at two different wavelengths: visible and radio.

Visible
Doing astronomy from the surface of the Earth when the Moon is full is much more difficult than is doing astronomy from the surface of the Earth when the Moon is new. While much of this difficulty arises due to the Earth's atmosphere, some is the direct result of the presence of a largish, well-lit object that has same angular size as does the Sun. The full Earth as seen from the Moon is, on average, over 40 times brighter than is the full Moon as seen from the Earth. Reading a newspaper on the Moon at night under a full Earth would be a piece of cake compared to reading a newspaper on the Earth at night under a full Moon.

Radio (and microwave)
These are the wavelengths where an observatory on the far side of the Moon would truly shine. The Earth radiates significant amounts of electromagnetic radiation in the radio and microwave wavelengths. There are a few places on the Earth where radio towers and microwave towers are absolutely forbidden, such as the area around the Murchison Radio-Astronomy Observatory in Australia, the area around the Itapeting Radio Observatory in Brazil, and the area around the National Radio Astronomy Observatory (Green Banks, West Virginia) in the US. These Earth-bound radio quiet zones are still plagued with radio and microwave radiation reflected by the atmosphere and by radiation from satellites orbiting the Earth.
Radio telescopes on the far side of the Moon would be shielded from the large amounts of electromagnetic radiation in those wavelengths emitted by the Earth, and possibly even from the electromagnetic radiation in those wavelengths emitted by satellites orbiting the Earth. Moreover, radio telescopes on the far side of the Moon would not be hindered by the Earth's atmosphere. The transparency of the Earth's atmosphere at radio wavelengths ends at the very long wavelengths that would be useful in studying the early universe, and it ends at the upper microwave that would be useful in studying molecular gas clouds.


Available energy for potential life on Titan There is an interesting discussion about whether there could be life on Saturn's moon Titan. (For example here.) The life could use reaction of hydrogen and acetylene which are being produced by photolysis in Titan's atmosphere. I would like to know what total energy is available for such life per square meter per day, or at least how much hydrogen is produced in Titan's atmosphere per day.

There are some estimates in article McKay and Smith, Icarus, 2005, but I have problem translating the quantities they use for the estimate to total amount of hydrogen produced per unit of time or energy produced per unit of time.

I can give a rough answer to your question and some brief thoughts on the paper, and I invite correction from anyone smarter than me.   Your article talks about energy per mole, but energy of production on a daily basis would be a product of solar energy.

To produce hydrogen in Titan's atmosphere requires UV light, from this article, UV light of 1600 Angstroms (160 NM) (Source)

Quote:

  Methane is a carbon atom surrounded by four hydrogen atoms, and it can
  be broken apart by ultraviolet light at wavelengths of about 1600
  Angstroms. The fragments, or radicals, that are produced from this
  process are very chemically reactive. They’re things like CH, CH2, and
  in some cases CH3.

Only a small percentage of the sun's light is in that range.   8%-10% of the sun's light is in the UV range, but maybe only 1%-2% in the upper UV3 range of 160 NM.  (Source)

The same Wiki article, using Saturn as a guide, Saturn receives between 13.4 and 16.7 Watts per square meter (Source), so figure an estimate of 15 watts per square meter total solar energy, 1%-2% of that strong enough to pull a hydrogen atom off Methane, so 0.15 - 0.3 watts per square meter.   That might not sound like much, but on a moon the size of titan, that's a lot of square meters so the low wattage isn't a big deal.   That's a peak estimate though as some of the hydrogen would recombine with the CH3 it split from, some energy would likely be converted to heat, and perhaps the biggest problem, mentioned in the article above, a share of the hydrogen, perhaps the lion's share would simply be lost off the planet.

  In the atmospheres of the giant planets, the hydrogen stays around
  because of the high gravity of those planets. Their atmospheres are
  primarily hydrogen anyway, and after the fragmentation occurs, the
  products sink into the deeper atmosphere, and methane is
  reconstituted. It’s a complete chemical cycle.

  On Titan, as far as we understand, that does not occur. Because the
  gravity is low, hydrogen should escape. The Voyager ultraviolet
  spectrometer saw a corona of hydrogen around Titan, which is a good
  indication that hydrogen is escaping. Now, if hydrogen is going away,
  the products that can be made from the methane are going to have a
  higher carbon to hydrogen ratio than methane itself, and you’re not
  going to be able to remake methane. So as far as we understand the
  photochemistry, Titan should be destroying methane and making more
  carbon-rich products, the simplest of which are acetylene – C2H2 – and
  ethane – C2H6. These are made directly from methane

This is my problem with the article.  I don't see any way the hydrogen finds it's way down to the planet's surface to be used in the way the article suggests, life in the lakes of Titan.    It's an interesting idea and the people who wrote it are probably PHD educated and I'm not, but I suspect the majority of hydrogen produced high in Titan's atmosphere would remain high in it's atmosphere and over time, lost to the solar wind.

It does raise the question, because Titan should lose it's Methane over time, why does it still have so much Methane - a very good question.

Lets look at the timing.   Huygens landed on Titan January 14th, 2005.

Your article was written  14 January 2005; revised 18 April 2005

and here's an article from a few months later, November 2005.

http://www.spaceref.com/news/viewpr.html?pid=18410

  The origin of methane in Titan's atmosphere is a mystery because it
  gets broken down by sunlight and particle radiation from space in the
  upper atmosphere. If surface lakes and pools were the only source, all
  of Titan's methane would be lost by this mechanism in less than a
  hundred million years, a short time for a moon that's been around
  since the formation of the solar system 4.5 billion years ago.�
  Components of the methane molecules react with each other and
  atmospheric nitrogen. As they descend, they form larger and heavier
  molecules that comprise the orange haze that blankets the moon.
  Because Titan is very cold (292 degrees below zero F, or minus 180
  degrees Celsius) these heavy compounds condense and rain out on the
  surface.

  "We have determined that Titan's methane is not of biological origin,
  so it must be replenished by geologic processes on Titan, perhaps
  venting from a supply in the interior that could have been trapped
  there as the moon formed,"

So, I'd chalk this one up to timing.   January-April 2005, life on Titan was a viable theory, which has since been de-theoried, or, whatever the correct term is.


Are there any natural satellites in retrograde Sun orbit? As I watch the video of the asteroids being discovered, all I see is prograde orbits. Their prevalence is fully understandable and justified, but the Sun gets interstellar visitors occasionally and I see no specific reason why it couldn't catch one into a retrograde orbit. They would obviously be a minority and likely not nearly as long-lasting as these in prograde orbits, but they should exist.

Did we observe any? If so, what sort of percentage of the whole are they? If not, why?

Yes, there are, but they are very rare.

Wikipedia has a list of retrograde asteroids, and mentions:

  As of September 2016, of the more than 700,000 minor planets known, there are only 95 known retrograde minor planets.

That means only 0.014% of known minor planets are in a retrograde orbit.

The asteroid with the highest known inclination is 2013 LA2, with an inclination of reportedly (175.189±0.007)° according to the JPL small-body database browser.


River-like artifact in perseverance rover's photo of Mars Good to know that perseverance rover has touched the surface of Mars finally. We will have nice new opportunity window for exploring our closest "friendly" planet of solar system.
Here's one of Mars photos perseverance has made :

Question is, What are this object roots marked with rose color ? I suspect it can be remains of past river in Mars or is it just an artifact of wind+sierra effect ?

The image in question appears to come down from the rover "down-look" camera.  Here is a raw image of that, while this is a raw image taken next to it (you can line up the crater part in the left of the first with the main crater in the second).
While both list Sol 4 as the acquisition date, they list Feb. 23 and Feb. 24 as the Earth dates, while local time is separated by just 14–15 minutes.  Despite those dates, which seem pretty incompatible to me, I have a feeling that that's when the images were downlinked to Earth from the rover.  The "down-look" camera is "mounted beneath the rover, looking downward at the surface during landing."  And, those do not look like images of the surface to me that I would expect if you were just a few inches away, instead they look like meter-scale pixels, not millimeter-scale pixels.  So, I think these were images taken while it was landing, a few minutes apart.
Given that premise, I looked at a close-up of the HiRISE image of the landing site, which is nominally 25 cm/pix, and that crater in the second image matches the crater in the HiRISE image on the left, just rotated about 90° clockwise.  So, this was a descent image, not an image from the surface.
With that knowledge, those are sand ripples, which are basically small dunes.  These are all over Mars, so it's nothing unusual and does not have anything to do with it being near or in a lakebed, but just have to do with the wind and particle size of the martian regolith.


Determining the percentage area of the moon's disk shaded with respect to time I've looked online and I've found sources such as this paper on a lunar phase simulator as well as this article about lunar phases.
I'd like to be able to model the area of the lunar disk which is shaded with respect to the moon's angle or time of the lunar month and I'm not sure how to go about doing this. In essence, I would like to be able to find a function f(x) where I can feed in a value such as 0 (meaning 0 days since the start of the lunar month) and I'd be able to get an output like 0 since it's a new moon. Likewise, f(14ish) should be new moon or fig 5 on the image below and so I'd get a value close to 1.
I'm focused a bit more on the maths rather than real values, so I'm willing to make assumptions such as the moon and earth being perfect spheres for the sake of the model.

I know there's a lot of maths involved and I'm going to have to do some integration, but I'd appreciate it if someone could point me to sources to get started on this since the papers I'm finding are either children's lunar phase worksheets or not related to this topic.

If I interpret your question correctly, you want to calculate the area of the moon disk as seen from Earth which is shaded, correct? You might then be looking for a way to calculate the position of the moon's terminator line. For this, you might have to dive into celestrial mechanics, more specifically about the ephemerides of moon and Earth. This is necessary as you  have to calculate the position of Earth and moon in an appropriate sun-centered coordinate system.
In a basic course on celestial mechanics this does not necessarily occur since the priority there is to calculate the stability of the orbit of objects.
Related questions and answers

Get a table of Moon phase angle per day
How to calculate the moon phase angle?

References

Klaus Hempe, Jürgen Molt: Sterne im Computer. Berechnungsprogramme für den Hobby- Astronomen (1999). IMHO an accessible explanation of the numerics, but in German.

http://www.internetsv.info/MoonCalc.html seems helpful.

Wolframalpha's moon phase calculation always give you the illuminated percentage of the moon disk.


Do unattached objects contribute to a planet's gravity? If you could measure Earth's gravity with infinite precision, and did so both with Earth and its full population, and Earth with no people, would there be a direct difference in gravity? Is this even a valid way to think of gravity?

Yes, it would change the attractive force by 10^-14 if there was a world without people.

Depending on the distance, you can measure gravity of people and buildings, in fact they used gravity to measure gaps in the pyramids of Giza once.

Today gravity surveys can detect caves and reservoirs of 2m to 50m at 2m to 150m depth, so in the future, we will be able to discover many caves and fossils with nanotechnology gravimeters.
 http://physicsworld.com/cws/article/news/2016/mar/30/new-gravimeter-on-a-chip-is-tiny-yet-extremely-sensitive


Can SETI certify whether or not Proxima b is inhabited by beings using electromagnetic communication? Can SETI certify whether or not  Proxima b is inhabited by beings using electromagnetic communication?

By certify I mean by past, present or future observations using technologies available.

Using current technology (and by that I mean experiments and telescopes that are available now) we would probably be unable to detect life on Earth even if observed from a distance of 4 light years, which is the distance to Proxima Centauri.

A "blind" search could look for radio signatures and of course this is what SETI has been doing for lots of different stars, including, I imagine, Proxima Centauri. If we are talking about detecting "Earth-like" signals, then we must assume that we are not talking about deliberate beamed attempts at communication, and so must rely on detecting random radio "chatter" and accidental signals generated by our civilisation.

As I have remarked in the linked question: The SETI Phoenix project was the most advanced search for radio signals from other intelligent life. Quoting from Cullers et al. (2000): "Typical signals, as opposed to our strongest signals fall below the detection threshold of most surveys, even if the signal were to originate from the nearest star". Quoting from Tarter (2001): "At current levels of sensitivity, targeted microwave searches could detect the equivalent power of strong TV transmitters at a distance of 1 light year (within which there are no other stars)...". The equivocation in these statements is due to the fact that we do emit stronger beamed signals in certain well-defined directions, for example to conduct metrology in the solar system using radar. Such signals have been calculated to be observable over a thousand light years or more. But these signals are brief, beamed into an extremely narrow angle and unlikely to be repeated. You would have to be very lucky to be observing in the right direction at the right time if you were performing targeted searches.

It has been suggested that new radio telescope projects and technology like the Square Kilometre Array may be capable of serendipitously detecting radio "chatter" out to distances of 50 pc ($\sim 150$ light years) - see Loeb & Zaldarriaga (2007) - and so Proxima Centauri would be well within this range. This array, due to begin full operation some time after 2025 could also monitor a multitude of directions at once for beamed signals.


Solar System formation, considering its and the universe's age It is known that the Sun is 4.6 billion years old, and the complete Solar System is of a similar age. The class of stars to which the Sun belongs seems to be quite common. Stars of its class can live for 8 or 9 billion years before moving to the red giant stage. Also, the whole universe is less than 14 billion years old.

Thus, not many generations of Sun-like stars could have existed before the first stages of our Solar System. And heavy elements are created when stars collapse. While there could be a lot of star classes with much shorter lifespans, I'm wondering: where do the heavier elements came from? Not only for our Solar System, but many others. I think it is very likely that our Solar System is of average age; do the numbers match between dead stars and remote stellar systems? Are there studies about this?

Early in the life of the universe it is thought that star formation in "metal-free" gas probably favoured larger stars. These Population III stars had short lives and very quickly enriched the interstellar and intergalactic medium with nucleosynthesis products.

In fact the main enrichment of the interstellar medium (ISM) in our own Galaxy is thought to have taken place some billions of years before the Sun was born. Stars that are being formed now have basically the same composition as the Sun. The plot below shows the iron abundance (on a logarithmic scale) as a function of deduced age for stars in the Gaia-ESO spectroscopic survey (from Bergemann et al. 2014). As you can see, there has not been too much change in iron abundance over the last 10 billion years.

I think the key to this is that most of the heavy elements in the gas in our Galaxy come from stars more massive than the Sun, and which have much shorter lifetimes as a result. [Stellar lifetime is roughly proportional to $M^{-2.5}$]. So several "generations" of these have lived and died (and must have done so).

To add some more detail. Almost all the elements heavier than Helium (known as "metals") are produced inside stars. To get back into the ISM you have to get the "processed" material out of the centres of the stars. This occurs in basically three ways. (i) Supernovae - the explosions associated with the final collapse of very massive stars ($>10 M_{\odot}$). The supernova progenitors have short lives $<50$ million years, so many generations of such stars can have lived and died before the Sun was formed, and probably of order a few hundred million to a billion had done so. (ii) Giant star winds. These occur towards the ends of the lives of stars with mass of $<10M_{\odot}$. Some fraction of processed material gets dredged out of the centre and expelled into the ISM. The lifetimes of these stars is 50 million to $>10$ billion years. Most of the enrichment is caused by stars about in the middle of this range. These stars go on to fade as white dwarfs. (iii) Type Ia supernovae. Explosions triggered by mass transfer on to a white dwarf (probably in binary systems) causing a detonation and the scattering of metals into the ISM. Progenitors have lifetimes similar to the giant stars in (ii).

If you look carefully at the colour scheme of the plot above, you will notice that ancient stars have a higher ratio of magnesium to iron. This supports the ideas in the paragraph above. Magnesium is an alpha-capture element, produced mainly in short-lived massive stars and gets into the interstellar medium through type II supernovae. Conversely, although iron is also produced in massive stars, it is primarily produced and dispersed in type Ia supernovae that have much longer lived progenitors. Thus Mg enrichment occurs more rapidly than Fe enrichment in the early Milky Way.

The numbers of white dwarfs we see in the Galaxy and the deduced rate of supernova explosions that are occurring (e.g. from counting the number of neutron star remnants) also matches up with what is required to produce the metallicity of the ISM and the Sun. But there are plenty of variables and unknowns and not everything is solved. For instance it looks quite likely that star formation needed to be higher in the past; and some parts of the Galaxy (e.g. the bulge) have a higher metallicity, thought to be due to a period of very vigorous high mass star formation 10 billion years ago.


Not clear on how to calculate sub-radar latitude How to find sub-radar latitude in reference to radar telescopes hitting near earth asteroids?

In the common case asteroids will rotate around a nearly fixed axis, like the Earth. That axis of rotation defines a north pole and south pole for that asteroid. The north pole is the one where, when a viewer is seeing the asteroid from above it, the asteroid is rotating counter-clockwise. Once you have north and south poles, you can define an equator and, thus, a system of latitude. I have no idea what is done for asteroids that are undergoing chaotic tumbling.

The "sub-radar latitude" is the asteroid's latitude that is directly facing the radar viewer. A good example from the literature that uses the term many times, and includes a short definition, is arXiv:1101.3794. A short excerpt:

  [...] implying significant change in the object’s sub-radar latitude (the angle between the asteroid-Earth line and the object’s equatorial plane).


How loud would the Sun be? Sound can't travel through outer space. But if it could, how loud would the Sun be? Would the sound be dangerous to life on Earth, or would we barely hear it from this distance?

The Sun is immensely loud. The surface generates thousands to tens of thousands of watts of sound power for every square meter. That's something like 10x to 100x the power flux through the speakers at a rock concert, or out the front of a police siren. Except the "speaker surface" in this case is the entire surface of the Sun, some 10,000 times larger than the surface area of Earth.

Despite what "user10094" said, we do in fact know what the Sun "sounds" like -- instruments like SDO's HMI or SOHO's MDI or the ground-based GONG observatory measure the Doppler shift everywhere on the visible surface of the Sun, and we can actually see sound waves (well, infrasound waves) resonating in the Sun as a whole! Pretty cool, eh? Since the Sun is large, the sound waves resonate at very deep frequencies -- typical resonant modes have 5 minute periods, and there are about a million of them going all at once.

The resonant modes in the Sun are excited by something. That something is the tremendous broadband rushing of convective turbulence. Heat gets brought to the surface of the Sun by convection -- hot material rises through the outer layers, reaches the surface, cools off (by radiating sunlight), and sinks. The "typical" convection cell is about the size of Texas, and is called a "granule" because they look like little grains when viewed through a telescope. Each one (the size of Texas, remember) rises, disperses its light, and sinks in five minutes. That produces a heck of a racket. There are something like 10 million of those all over the surface of the Sun at any one time. Most of that sound energy just gets reflected right back down into the Sun, but some of it gets out into the solar chromosphere and corona. No one can be sure, yet, just how much of that sound energy gets out, but it's most likely between about 30 and about 300 watts per square meter of surface, on average. The uncertainty comes because the surface dynamics of the Sun are tricky. In the deep interior, we can pretend the solar magnetic field doesn't affect the physics much and use hydrodynamics, and in the exterior (corona) we can pretend the gas itself doesn't affect the physics much. At the boundary layers above the visible surface, neither approximation applies and the physics gets too tricky to be tractable (yet).

In terms of dBA, if all that leaked sound could somehow propagate to Earth, well let's see... Sunlight at Earth is attenuated about 10,000 times by distance (i.e. it's 10,000 times brighter at the surface of the Sun), so if 200 W/m2 of sound at the Sun could somehow propagate out to Earth it would yield a sound intensity of about 20 mW/m2. 0dB is about 1pW/m2 , so that's about 100dB. At Earth, some 150,000,000 kilometers from the sound source. Good thing sound doesn't travel through space, eh?

The good folks at the SOHO/MDI project created some sound files of resonant solar oscillations by speeding up the data from their instrument by 43,000 times. You can hear those here, at the Solar Center website. Someone else did the same thing with the SDO/HMI instrument, and superposed the sounds on first-light videos from SDO. Both of those sounds, which sound sort of like rubber bands twanging, are heavily filtered from the data -- a particular resonant spatial mode (shape of a resonant sound) is being extracted from the data, and so you hear mainly that particular resonant mode. The actual unfiltered sound is far more cacophonous, and to the ear would sound less like a resonant sound and more like noise.


Starting out in practical astronomy I want to start observing the night sky, what advice would you give to get started?

If you're dead broke (like most people) I would suggest you take look at a small refractor telescope, you could pick up a cheap-o Celestron Powerseeker 60/70 for something like 40 bucks. The useability of such a small telescope is very limited, but you'll be able to get a nice clear image of the rings of Saturn, four of Jupiter's moons, or nice views of the moon, maybe even a few Messier objects if you're patient enough.
-but-
If you're willing to drop a bit more coin, then I would suggest nothing smaller than an 8" reflector. For less than 200 bucks you can find something that will really impress you.

I used to take my crappy little telescope and drive up into the hills where light pollution was at a minimum, where I would just sit for hours trying to find Messier objects. Loads of fun!

As far as joining an amateur astronomy society, I would suggest you check with your local Community College's astronomy professors. At least here in California, one of the requirements of the astronomy classes is that there be opportunities to participate in actual observing. This means that there will more than likely be an astronomy group having "star parties" and other observing events, which will give you some good connections to competent groups.


What is Event Horizon of a Black Hole? The boundary of a black hole is said to be surrounded by event horizon - the point of no return! What is its significance in terms of general relativity?

It is exactly the point of no return for light. In other words, it is the point (actually more like a sphere) in which the escape velocity from the Black Hole gravity reaches $c$.


Are there sufficient observational data to measure non-Newtonian perihelion advances of any Asteroid and Comet orbits? Anomalous (i.e. not predicted from Newtonian theory) advances of the perihelion direction have been observed for many solar system planet orbits and have been accounted for by Einstein's General Relativity.

Basic Orbital elements have been obtained and published for many Asteroids and Comets.  But are the existing observational data on any Asteroids or Comets sufficient (in terms of accuracy and extent of observations) to determine/estimate rates of non-Newtonian perihelion advance and are any such determinations publically available?

I am by no means an authority on this area but prompted by user /u/called2voyage I will refer the 1994 Astronomical Journal paper by Shahid-Saless (Colorado) and Yeomans (JPL) Relativistic Effects on the Motion of Asteroids and Comets.

To paraphrase their abstract: They study the predicted effects arising from relativistic perturbations on the motions of asteroids and comets and show that for a number of such objects, inclusion of relativistic contributions in the equations of motion give rise to significant improvements in the orbital solutions.  They go on to argue that ignoring such corrections to the equations of motion gives incorrect solutions.  They point out how the use of masses derived from relativistic ephemerides together with purely Newtonian equations of motion results in an inconsistent hybridized non-Newtonian, non-relativistic model.

With respect to the posted question:-
(a) are the existing observational data on any Asteroids or Comets sufficient (in terms of accuracy and extent of observations) to determine/estimate rates of non-Newtonian perihelion advance?

...the authors short-list 15 asteroids with the largest predicted relativistic perihelion precession rates (from a long-list of 156 commonly-studied objects).  For example Icarus has the greatest rate (0.101 arcseconds/year) and had (in 1993) been observed over 43 years thereby producing a predicted cumulative relativistic precession over this period of 4.34 arcseconds. Observations referred to are a mix of optical and radar (no spacecraft telemetry) and are those used to produce the ephemerides model from which orbital elements are extracted.

The authors note that, assuming a (then) current observational accuracy of a few tenths of a second of arc, the relativistic contribution to precession should be detectable for Icarus and other asteroids near the top of the (ranked by precession-rate) list. As with the planets the majority of asteroid perihelion precession will be caused by the perturbing influences of other planets, but the authors do not present calculated values for these contributions).

(b) are any such determinations publically available?

...the authors selected six asteroids whose motions are significantly affected by general relativistic effects. They then calculated a set of orbital elements for each orbit using JPL's development ephemeris DE200 (with Earth and Moon perturbations treated separately).  They did this in two ways, firstly with Newtonian equations of motion alone and secondly with the addition of non-Newtonian General Relativisitic equations of motion.

The authors indicate that the residuals (differences between observed and predicted positions at various times) are improved (i.e. reduced) by application of non-Newtonian corrections to the equations of motion. For example the RMS residual for Icarus is improved by 30%. Thus it is technically possible to calculate differences in the orbital behavior (such as perihelion advance) predicted by (i) the pure-Newtonian model and  (ii) the Newtonian plus non-Newtonian model.

However the paper does not actually present explicit accounts of total observed perihelion precession or a breakdown into Newtonian and non-Newtonian components.  In any case, due to complexity, such a breakdown could only apply over the period of observations which is limited (maximum 61 years in 1993) compared to the timescale (a few hundreds of years) over which (predominantly Newtonian) planetary-induced perturbations might be expected to average out.

However it would be possible (assuming there are no major perturbing encounters with planets or the asteroid belt) to use the provided asteroid orbital element data in a numerical orbit simulator program to roughly model the perihelion precession rate of an asteroid orbiting the Sun in the presence of the other planets using Newtonian equations of motion (i) without and (ii) with relativistic modifications calculated from $\frac{F}{m} = \frac{GM}{r^2} \left(1+\frac{3V_t^2}{c^2}\right) = \frac{GM}{r^2} \left(1+\frac{3GM.SLR}{c^2.r^2}\right)$ (see Walter-Goldstein-Schwartzchild Formulation ).  That way the amount of asteroid precession resulting from each of the two different sources (Newtonian, non-Newtonian) could be modelled.

Note: As /u/UhOh/ commented, an alternative, possibly superior, equation for the extra, relativistic acceleration is given in the Shahid-Sales & Yeomans paper cited above as Eqtn 3.11, on page 1886.


Maritime telescopes: Stabilisation requirements for optical vs. radio telescopes? SOFIA stands for Stratospheric Observatory for Infrared Astronomy. She says:

My telescope stays stable with a spherical bearing, shock absorbers, and gyroscopes.

I suppose a similar system would work for an optical telescope on a maritime vessel as well. But what about a radio-telescope on a ship with a dish of around 10m? What are arguments (ideally mathematical ones) against a similar construction like for SOFIA?

References

Tom Nardi A Miniature Radio Telescope In Every Backyard. hackaday.com (2019)
Gregory Redfern: Cruise Ship Astronomy and Astrophotography - The go-to astronomy resource for cruise travelers

You asked for arguments against this, but I don't think there are any other than money.  Consider the AN/SPQ-11 passive radar that was aboard the US Navy ship Observation Island.  The actual receiver area is not as large (7M according to https://fas.org/irp/program/collect/cobra_judy.htm), but we could imagine a larger antenna being placed on a larger ship such as the USS Enterprise.  The beam (maximum width) of the Enterprise is 41 meters, which could certainly accommodate a 10-20m dish.
I think the only reason this would not be built would be lack of funds.  From an engineering perspective, it seems doable.


Is there an antonym for 'transit', i.e. the transit of a celestial body through the meridian I'm writing software to calculate the rising, transit, and setting of a body on the celestial sphere for a specific location and date. It seems to me that the opposite of transit, at least in the case of the Sun and for circumpolar stars, would also be of importance.
In the northern hemisphere this would be the time when a circumpolar star would pass through the line between the celestial north pole and true north, and be at its lowest height above the horizon.
Is there an antonym (opposite) for the word transit?

The transit of a star when it reaches its highest point is called the upper culmination and when it reaches its lowest point is called the lower culmination. I think both are referred to as meridian transits.


How to project an arbitrary elliptical orbit onto the night sky plane? An ellipse E is arbitrarily oriented in 3-D space with the origin at one of its foci. Use the standard elements provided in the figure. Assume the reference direction $\gamma$ or $\gamma'$ is the positive Y-axis direction. Find the projected position vector of the ellipse onto the plane of reference as a function of the angle $\nu$.
After some calculation, I obtained the position vector of the celestial body to be
$$r=\frac{a(1−e^2)}{1+e\cos(\nu)}$$
Where $a$ is the semi-major axis and $e$ is the eccentricity. Now, I begin with the $(r,ν)$ coordinates in the plane of the ellipse and project it onto the reference plane using the $\cos(i)$ factor. How, then, can I find the orthogonal components $(x,y)$ of this projected position vector on the reference plane? A parameterized solution would be appreciated as well.

The (x, y, z) position of an object on its orbit with reference to a reference plane are given by:
u = ω + ν
x = r (cos Ω cos u − sin Ω sin u cos i)
y = r (sin Ω cos u + cos Ω sin u cos i)
z = r sin i sin u
(MEEUS, Jean. Astronomical Algorithms, Second Edition, p. 233)
This should help you find what you need.