Is the full moon full everywhere on the planet? When the moon is full, fully lit by the sun, will it appear full everywhere on the planet?

E.g. if I where to set the date for the next three full moons for say: Sweden, Bali and Seattle how different would those dates be?

[This is intended to supplement @userLTK's answer and reaches the same conclusion.]

By definition, the Moon is full when its ecliptic longitude is exactly 180 degrees greater than the Sun's ecliptic longitude for a theoretical geocentric observer (first paragraph of https://en.wikipedia.org/wiki/Full_moon and definition of "full moon" at http://www.ncsm.city.nagoya.jp/astro/astrotool/glossary.html but couldn't find a 100% reliable resource).

By this definition, your location on Earth doesn't matter, since the full moon is defined by a theoretical observer.

Realistically, the moon's position for an observer on the surface of the Earth can differ by as much as a degree from the geocentric position. (This is the arctangent of 1/60, since the moon's distance is about 60 times the Earth's radius.)

Since the moon circles the ecliptic once every 27.321662 days or so (its "sidereal period"), its ecliptic longitude increases by about 0.55 degrees per hour.

Thus, roughly speaking, the "full moon" at a given point on Earth can differ by as much as ±1.8 hours from the true full moon. By this token, it takes the "full moon" about 3.6 hours to cross the Earth.

Of course, all of this refers to the moon's geometric position. If you allow for refraction, the results could be slightly different, especially since both the sun and moon experience over 0.5 degree refraction when rising/setting and the full moon rises right about when the sun sets and vice versa.


What is the origin of the dust near the sun? note: Solar Probe+ is now officially Parker Solar Probe

In https://space.stackexchange.com/q/17498/12102 I asked about the Solar Probe Plus mission. Reading a 2008 report I saw that there is an elevated exposure to high velocity dust near the sun. I understand the high velocity, anything in orbit that close will be moving fast, but I don't understand why there is dust there.
Simply put - wouldn't the gravity pull it in or the solar wind blow it away?
What is the origin of the refractory carbon and silicates? Is this orbiting leftover material from the formation of the solar system, or is it falling in from farther out, or is it left-over carbon and silicon that fell into the early forming sun and is now being blown back out?

4.3.5 Micrometeoroid and Dust. Solar Probe+ will encounter dust particles ranging in diameter from submicron up to several hundred microns and consisting of highly refractory carbon and silicate species with a typical bulk density of ~2.5 g/cm2. The particles will be traveling at relative speeds as high as 350 km/s. To define the shielding requirements for Solar Probe+, a dust model was developed based primarily on the work of Mann et al. (2004). The model employs the following assumptions...

above: screen shot from SolarProbePlus2008

above: screen shot from NASA's Solar Probe Plus Fact Sheet

above: illustration of NASA's Solar Probe Plus from http://solarprobe.jhuapl.edu/spacecraft/ Sun is up - to state the obvious!

There are two primary dust populations near 1 AU, interplanetary dust (IPD) and interstellar dust (ISD) [Mann, 2010].  I also discussed dust observations in detail at https://physics.stackexchange.com/a/160627/59023.

Interplanetary Dust

IPD of ~1 $\mu$m size drift sunward due to Poynting-Robertson drag while following roughly Keplerian orbits [e.g., Malaspina et al., 2014].  Closer to the sun, these particles break up due to collisions, sublimation/ablation, and/or sputtering.

Dust grains of ~0.1 $\mu$m size are the so-called "$\beta$ meteorites", which travel away from the sun due to the imbalance of radiation pressure over gravity [Mann, 2010].

The smallest dust grains with $\ll$ 0.1 $\mu$m size, the so-called nanograins or nanodust, act like large pickup ions, which are carried anti-sunward by the frame-dependent convective electric field (i.e., just the Lorentz force) produced when the dust grain moves relative to the solar wind flow (i.e., $\mathbf{E}_{conv} = - \mathbf{V}_{sw} \times \mathbf{B}_{sw}$, where the subscript $conv$($sw$) stands for convective(solar wind), and $\mathbf{V}$ and $\mathbf{B}$ are the bulk flow velocity and quasi-static magnetic field).  These particles can reach speeds in excess of 100 km/s relative to the sun [Meyer-Vernet et al., 2009].

Interstellar Dust

ISD was first discovered by the Ulysses spacecraft, which is ~1 $\mu$m size and moves at ~26 km/s relative to the solar system barycenter.  More recent work [Malaspina et al., 2014] has found a relationship between dust impact count rates and ecliptic longitude.

The reason can be seen from the following.  The Earth's transverse speed about the sun is ~29 km/s.  Thus, when the Earth moves anti-parallel(parallel) to the ISD flow direction the relative dust-spacecraft speed is ~55(~3) km/s, which produced an enhanced(depressed) dust count rate.  This occurs because there is a threshold impact speed necessary to produce a large enough plasma cloud (i.e., $\gtrsim$5-10 km/s depending on dust size) [Meyer-Vernet et al., 2009; 2014].

  What is the origin of the dust near the sun?

The primary sources of ~1 $\mu$m size near 1 AU are cometary debris trails, asteroids, planets, moons, and ISD [Mann, 2010; Zaslavsky, 2015].

  Simply put - wouldn't the gravity pull it in or the solar wind blow it away?

Some are attracted by a combination of gravity and Poynting-Robertson drag while the smaller grains are either "pushed" out by radiation pressure (i.e., $\beta$ meteorites) or "picked-up" by the solar wind Lorentz force (i.e., nanodust).

References

D.M. Malaspina et al., "Interplanetary and interstellar dust observed by the Wind/WAVES electric field instrument," Geophys. Res. Lett. 41, pp. 266-272, doi:10.1002/2013GL058786, 2014.
Mann, I. "Interstellar Dust in the Solar System," Annual Review of Astronomy and Astrophysics 48, pp. 173-203, doi:10.1146/annurev-astro-081309-130846, 2010.
Meyer-Vernet, N., et al. "Dust detection by the Wave instrument on STEREO: Nanoparticles picked up by the solar wind?," Sol. Phys. 256, pp. 463-474, doi:10.1007/s11207-009-9349-2, 2009.
Meyer-Vernet, N., et al. "The importance of monopole antennas for dust observations: Why Wind/WAVES does not detect nanodust," Geophys. Res. Lett. 41, pp. 2716-2720, doi:10.1002/2014GL059988, 2014.
Zaslavsky, A. "Floating potential perturbations due to micrometeoroid impacts: Theory and application to S/WAVES data," J. Geophys. Res. 120, pp. 855-867, doi:10.1002/2014JA020635, 2015.


How does a galaxy in Abell 2261 exist without a black hole in the center? Here a source says that astronomers looked for a black hole in the brightest galaxy in the cluster Abell 2261, A2261-BCG, but found nothing:

However, X-ray observations from NASA's Chandra X-ray  Observatory and
Hubble Space Telescope have found nothing.

How is it possible for this galaxy to not have a black hole in the center?

The Brightest Cluster Galaxy (BCG) in Abell 2261 ("Abell 2261-BCG") is a massive elliptical; these almost always seem to have supermassive black holes in their centers. In addition, the center of the galaxy has a large region of relatively low stellar density (a "core"), something usually thought to be produced by the merger of two SMBHs into one, an event which follows the merger of two massive galaxies to form one massive elliptical. (As part of the merger, the two SMBHs form a binary; stars near the center of the merged galaxy can gravitationally interact with this binary, with the usual result being a shrinking of the binary and the ejection of the star to a larger radius; thus, stars get preferentially kicked out of the center into the intermediate or outer parts of the galaxy, and the density of stars in the center gets lower. Eventually, the SMBH binary becomes so small that gravitational radiation shrinks it even further, to the point that the two SMBHs merge to become one in the center of the galaxy.)
The actual paper (Gültekin et al.) behind that article is about looking for X-ray emission from an accretion disk around the possible SMBH in the center of the BCG (and also in a few other locations near the center). Since they don't detect any X-rays, they conclude "there is either no $10^{10} M_{\odot}$ black hole in the core of A2261-BCG, or it is accreting at a low level." So there are really two possibilities:

There is a SMBH in the galaxy center, but it's quiescent (not accreting enough gas to produce a lot of X-rays).

There isn't a SMBH in the galaxy center, which is more exciting because it's unexpected.

Now, there actually is a theoretical mechanism which might occasionally eject a SMBH from the center of a galaxy. When two SMBHs merge to form one SMBH (see above), there can be a "kick" which causes the merged SMBH to be ejected from the galaxy center (the magnitude of this kick depends on things like the mass ratio between the two SMBHs, how fast the individual SMBHs were spinning before the merged, and what the directionw of those spins were relative to the orbit of the binary before the merger). In most cases, the ejected SMBH will fall back into the center of the galaxy -- possibly oscillating in and out a few times before dynamical friction slows its motion down and it settles back in the center of the galaxy. In extreme cases, it might be ejected fast enough so that it escapes the galaxy entirely. This whole process will eject a number of stars from the center of the galaxy, too, which might help explain why its low-density core is so big (i.e., a combination of the SMBH-binary-merger process and a resulting SMBH ejection).
So it's possible something like this happened in Abell 2261-BCG, and the SMBH is currently somewhere outside the galaxy center. (Or it's sitting the center and just doesn't show up in X-rays.) Of course, they didn't find any evidence for an X-ray-emitting SMBH outside the galaxy nucleus, either, so we really don't know -- and I wouldn't be surprised if there is a (quiescent) SMBH in the center after all.
(Rory Alsop is correct that not all galaxies have SMBHs -- for example, the local small spiral galaxy Messier 33 has been studied with the Hubble Space Telescope (Gebhardt et al. 2001, and the modeling of stellar velocities in its center suggests an upper limit of about 1500 $M_{\odot}$ for any possible black hole there. But every massive elliptical that's been studied carefully enough seems to have one, so we would expect Abell 2261-BCG to have one, too.)


Why haven't Earth and Venus got any tiny moons? Or have they? Why haven't some meteoroids gotten caught in Earth's or Venus' orbit?

AFAIK most meteors are tiny fragments from comets. Shouldn't some comet tail sometime have passed Earth orbit at velocities suitable for our planet to capture such fragements? And 100 000s of asteroids have been detected. Why haven't the inner planets gotten asteroid like moons like the outer planets have? The asteroid belt isn't too far away.

Earth's large Luna might clean them away, but that won't explain Venus' lack of tiny moons.

The strength of the Earth's gravitational field compared to the Moon and the Sun is not enough to capture and hold satellites - there are too many disruptive forces that would rip them away over time.

However there are some objects at the Lagrangian points - the points where the gravitational fields of the Earth and other objects are equal and so it is possible to have a (likely-meta) stable orbit.

This gives some details about what might be found at the various Lagrangian points: http://en.wikipedia.org/wiki/List_of_objects_at_Lagrangian_points


What's the reason that we have a different number of days each month? It always was interesting for me to understand the answer for the question:

  What's the reason that we have a different number of days each month?

If the month is fixed on the time that the moon circles the earth, then the time of the circling isn't equal each month? And if the month fixed by the time that the earth circles the sun and then divided by 12 (365:12=30.4), why one month has more days and the other month has less days?

Before desktop computers, how did astronomers handle calculations of dates and times accurately, for example two plates taken decades apart, how would an astronomer calculate the time difference precisely?

You make a great point. The reason behind the discrepancy between the dates is due to a complicated history behind it.

The calendar is based on the calendar created by ancient Romans, which is based on one Moon cycle. One lunar cycle is

  29.53 days.

    www.universetoday.com/20620/lunar-year/

which does not evenly divide into the 365.25 days of the year.

Note: The Romans borrowed the calendar from the ancient Greeks to develop the idea of a 10-month calendar that left approximately 60 days unaccounted for.

Consequently, the earliest ancient Roman calendars had months that were either 29 or 30 days to account for this discrepancy.

The Romans started using the 10-month calendar in 738 B.C. Their months were initially called:

Mensis Martius (March)
Mensis Aprilis (April)
Mensis Maius (May)
Mensis Iunius (June)
Mensis Quintilis (July, quin- meaning "five")
Mensis Sextilis (August, sex- meaning "six")
Mensis September ("seven")
Mensis October ("eight")
Mensis November ("nine")
Mensis December ("ten")

To account for the remaining $\sim60$ days, Mensis Ianuarius (January) was added to the beginning of the year and Mensis Februarius (February) to the end of the year during Numa's reign around 700 B.C. with a leap year every 4 years.

  Leap years were decreed by proclamation the pontifex maximus, in such years an extra month was added to the calendar.

    Quoted from PM 2Ring

The calendar stayed in that order until 452 B.C. when a small council of Romans, called the Decemvirs, moved February to follow January.

However,

  in the initial 36 years of its adoption leap days were added every 3 years due to a misunderstanding (Romans used inclusive counting),

    Quoted from PM 2Ring

so that caused a big problem. In fact, by the start of the reign of Julius Caesar, the previous calendar was off by a whole week!

Julius Caesar noticed this and thus modified the Roman calendar in 46 B.C. to make each month have either 30 or 31 days, with the exception of Februarius[February], which had 29 days, with the three year leap year fixed into a four year leap year. Quintilis[July] was later renamed Julius[July] in his honor. Likewise, Sextilis[August] later became Augustus[August] to honor Augustus (Gaius Octavius, the first Roman Emperor). Augustus[August] was also given an extra day (taken away from Februarius[February]), so that Augustus and Julius would have an equal number of days, representing their equal power. This is the Julian Calendar.

  This calendar was extremely accurate, but was still not accurate
  enough as it drifts by approximately 3 days every 400 years. So after
  a few hundred years, the drift becomes noticeable
  (i.e. the seasons don't properly match the calendar), which is
  important for farming, and therefore human survival.

    quoted from  CJ Dennis

Then a pope named Pope Gregory XIII modified the Julian calendar again in October 1582. This is the currently used Gregorian calendar. He made it so that

  Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400.

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

This calendar is the most accurate calendar currently used, and won't be a single day off until the year of $\pm 3200$ AD. It isn't perfect though. Compared to the tropical year, it is 27 seconds too long. This is...

  due to astronomical effects (the
  rotation of the Earth speeding up and slowing down, etc.) we can't be
  more precise than that.

    quoted from CJ Dennis

As for your other question, unless the plates taken had accurate dates of when recorded, it would be rather impossible to calculate precisely.


How much does the sky change in a few thousand years? The "fixed stars" are not actually fixed, the earth's tilt changes over time etc., but all that happens slowly on human timescales.
Imagine a Babylonian astronomer (or astrologist?) teleported to the present, e.g. to present-day Iraq. Would she be able to tell that the sky is different? If so, what about a Maya  astronomer? In other words, how long does it take until the changes in the sky are perceptible to the human eye?

Here's part of the sky in the year 1

It is part of the sky you may know well,  Orion and the dogs.  I've marked the current positions of Sirius, Procyon and Betelgeuse, with green markers so you can see how their positions have changed over 2000 years.  It's not a lot.
The first thing that the Babylonian astronomer might notice is that there is a pole star, one that is fixed. 2000 years ago, Polaris was more than 10 degrees from the True North. Now it is less than 1 degree.  It isn't immediately obvious from a casual glance at the sky, but it would be noticeable and surprising to Babylonian astronomer.
If they started looking at the position of stars at a particular time and a particular date, they would find that things are in the "wrong" position.  Of course "on a particular date" assumes a calendar, and the Babylonian calendar was not the Gregorian calendar. With more measurement they would find that the point of Aries is now in Pisces. Again, its not something you would notice at a glance, but it is within the ability of a Babylonian astronomer.
Now the stars have actually moved since year 1.  But it would not be casually noticeable.  It would be within the ability of our Babylonian astronomer to measure, if we asked them to.  The relative positions of Sirius and Betelgeuse have changed.  The Babylonians could measure angles in the sky precisely enough.  But we don't know if they did, and if they did we don't know if they wrote them down.
So our Babylonian would not look up and immediately say "something's wrong".  But if we 1. Prepared him by telling them to take careful measurements of Polaris, Sirius and so on. and 2. Gave them some prompts on what to look for in terms of proper motion and precession,  then it would be possible for them to observe some changes.


can we calculate the velocity of galaxies at the edge of the observable universe? I read this post

Does the edge of the Universe travel faster than the speed of light?

but there was no calculation, so I tried it thusly:

The observable universe is 28.5 Gpc (gigaparsecs)
= 28500 Mpc (megaparsec)

https://www.google.com/search?q=diameter+of+the+observable+universe&oq=diameter+of+the+observable+universe&aqs=chrome..69i57.6887j0j7&sourceid=chrome&ie=UTF-8

The Hubble constant is 67.4 or 70 km/s/megaparsec.

https://www.nasa.gov/feature/goddard/2019/new-hubble-constant-measurement-adds-to-mystery-of-universe-s-expansion-rate

I’ll use the older value 67.4

v=H x D = 28500 Mpc * 67.4 km/s Mpc = 1,920,000 km/s = 1.9 x 10^9 m/s

speed of light 2.998 x 10^8 m/s

So the velocity of expansion at the edge of the universe is >6x greater than the speed of light?

I read these posts

https://physics.stackexchange.com/questions/107748/how-are-galaxies-receding-faster-than-light-visible-to-observers

So where are these measurements of galaxies moving faster than light?

and this

https://arxiv.org/pdf/astro-ph/0011070v2.pdf

From the arxiv paper, I get that galaxies can be receding at 3c based on GR, but my simple calculation comes up with 6c.

Can someone help me reconcile my curiosity with the facts?

I think the main reason the answer doesn't come out correct is that the simple Hubble's law equation:

$v = dH_{0}$

Is only applicable in the local universe (Hence $H_{0}$, the value of the Hubble constant today). The rate of expansion of the Universe (the Hubble parameter) is determined by the composition of its energy density. Today, the composition is about 30% matter and 70% dark energy. In the recent past the energy density was dominated by matter. In the more distant past the energy density was dominated by relativistic particles (photons and neutrinos). The rate of expansion of the universe is different in each of these "epochs". In general the universe used to expand more rapidly (expansion has slowed thanks to the effects of matter), but distances also used to be smaller (the universe itself was smaller). Combine all of this together and you have a very complicated relationship between distance and the recessional velocity we observe today. This relationship does not have a simple functional form.

I think the easiest question to ask is: "How far has a photon traveled such that if it were emitted at the beginning of the universe it would reach me today?". The answer to this question is the "cosmological horizon", which we commonly talk about as being the size of the observable universe. To answer this question you need to use the spacetime metric which applies for our universe (Called the Friedmann-Robertson-Walker-Lemaitre metric). This is basically the relationship between speed, time, and distance in the realm of general relativity. If our universe is flat (which we think it must be) this equation is actually quite simple, but unfortunately it requires that you know how the universe has been expanding over its lifetime. As I mentioned this relationship has no closed form equation or number, which is why you will never see a simple equation for the solution you are looking for. You need to integrate the solution numerically. Wikipedia has a good little article on the horizon.

If you're interested, you'd write down the FLRW metric for a photon travelling in a "straight path":

$ cdt = \frac{a(t) dr}{ \sqrt{1-kr^2} } $

Where $a(t)$ is the scale factor of the universe, which tells you about its expansion history. This is the complicated function. $k$ is the curvature constant, which for a flat universe is 0. You then get:

$ \int^{t_{0}}_{0} \frac{c dt}{a(t)} = \int^{r_{H}}_{0} dr = r_{H} $

Here $t_{0}$ is the age of the universe and $r_{H}$ is the particle horizon you're interested in. To figure out $a(t)$ you need the Friedmann equations, which are functions of the different energy densities (dark energy, matter, relativistic particles, curvature), and the rates at which they change in time. I used astropy to do the integration and got $\approx 3.35 c t_{0}$ which gives about 14 Mpc, so close enough to the answer that you could attribute any differences to a difference in measured parameters that go into solving the integral.

I'd recommend Introduction to Cosmology by Ryden as a textbook if you really want to get into it.


How to create 3D animation of particles position vs time in python? I am trying to make a 3-D animation of stars' position vs time. I have the information of the stars' x, y and z co-ordinates over time. I was looking at the example of Matplotlib and did understood what I need to do but I don't want the connecting lines between different times. All I need is to show the position of the stars at different time and make a movie of it. This is the example I was looking at

https://matplotlib.org/gallery/animation/simple_3danim.html

After that I want to make it a bit complicated. I have the stars size and mass as well. I was wondering if I can change the size of the circles in the plot according to the radius and use different Colors for different mass range?

After that I want to make it a bit complicated. I have the stars size and mass as well. I was wondering if I can change the size of the circles in the plot according to the radius and use different Colors for different mass range?

That part has been addressed in previous questions and answers here, I remember reading maybe a year ago (+/- 0.9 years).

Here is something you may find helpful reprinted from my answer in Stack Overflow. It's based on astronomer and data scientist Jake VanderPlas' blogpost from 2013.

For your needs keep the dots but not the lines. What I recommend is that you start from this and get as far as you can, then ask a question in Stack Overflow showing what you've tried and explaining what you still need.

  Here is the Lorenz attractor both in 3D and animated. The script is in the following link (along with many goodies) in Jake VanderPlas' Pythonic Perambulations. You can learn a lot by going line-by-line through the script - it's an elegant use of matplotlib objects.

  https://jakevdp.github.io/blog/2013/02/16/animating-the-lorentz-system-in-3d/

  I added these two lines just before return in the animate function, and then used ImageJ to import the "image stack" and save the "animated GIF":

fname = "Astro_Jake_" + str(i+10000)[1:]
fig.savefig(fname)

  Note: For OSX it seems to be necessary to set blit = False in animation.FuncAnimation.

  Here is a minimal, simplified example of plotting lines in 3D based on the above:

def lorentz_deriv((x, y, z), t0, sigma=10., beta=8./3, rho=28.0):
    """Compute the time-derivative of a Lorentz system."""
    return [sigma * (y - x), x * (rho - z) - y, x * y - beta * z]

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.integrate import odeint as ODEint

x = np.linspace(0, 20, 1000)
y, z = 10.*np.cos(x), 10.*np.sin(x) # something simple

fig = plt.figure()
ax = fig.add_subplot(1,2,1,projection='3d')
ax.plot(x, y, z)

# now Lorentz
times = np.linspace(0, 4, 1000)

start_pts = 30. - 15.*np.random.random((20,3))  # 20 random xyz starting values

trajectories = []
for start_pt in start_pts:
    trajectory = ODEint(lorentz_deriv, start_pt, times)
    trajectories.append(trajectory)

ax = fig.add_subplot(1,2,2,projection='3d')
for trajectory in trajectories:
    x, y, z = trajectory.T  # transpose and unpack
    # x, y, z = zip(*trajectory)  # this also works!
    ax.plot(x, y, z)

plt.show()

I wrote this before I knew of PIL. Apparently you can use PIL to generate your GIFs instead of using an external program like I did.

See

Creating GIFs in Python Using Pillow (PIL Fork)
https://pythonprogramming.altervista.org/create-an-animated-gif-with-pil/
https://pythonprogramming.altervista.org/png-to-gif/

and have fun!

An example of a GIF with stars in it from the first blogpost (Nyan Cat):


Does terrestrial planets conduct heat from near to far side? Suppose a terrestrial planet with no atmosphere is tidally locked and very near the star like a hot Jupiter.

Is the far side still very cold or would crust would conduct heat from the near side to far side?

The transfer of heat would be insignificant. Consider the Earth, the core is as hot as the surface of the sun. Some of this heat does get transferred to the surface, but it is only 0.03% of the energy received from the sun. Rock is a good insulator, most of the heat is transferred by convection.

The far side of a tidally locked planet would be very cold. It would receive very little heating from the planet's interior or the near side.


Why is the observable Universe larger than its age would suggest? The age of the Universe is estimated at 13.8 billion years, and current theory states nothing can exceed the speed of light, which can lead to the incorrect conclusion that the universe can't have a radius of more than 13.8 billion light years.

Wikipedia deals with this misconception as follows:

  This reasoning would only make sense if the flat, static Minkowski spacetime conception under special relativity were correct. In the real Universe, spacetime is curved in a way that corresponds to the expansion of space, as evidenced by Hubble's law. Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance. → Ned Wright, "Why the Light Travel Time Distance should not be used in Press Releases"

That doesn't clear the matter up for me, and having no science or maths background beyond high school, further reading into Hubble's law isn't helping much either.

One layman's explanation I've seen offers explanation that the Universe itself isn't bound by the same laws as things within it. That would make sense – insofar as these things can – but the above quote ("Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance") seems more general than that.

Can anyone offer (or direct me to) a good layman's explanation?

The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static.

Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time.

A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is currently made of:

$$ H(a) = H_{0} \sqrt{\frac{\Omega_{m,0}}{a^{3}} + \frac{\Omega_{\gamma,0}}{a^{4}} + \frac{\Omega_{k,0}}{a^{2}} + \Omega_{\Lambda,0}} $$
where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a constant regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured today.

The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment.

Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression:

$$H(a) = H_{0} \frac{\sqrt{\Omega_{\gamma,0}}}{a^{2}}$$

At around:

$$ \frac{\Omega_{m,0}}{a^{3}} = \frac{\Omega_{\gamma,0}}{a^{4}} $$
$$ a = \frac{\Omega_{\gamma,0}}{\Omega_{m,0}}  $$
we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an acceleration of the coordinates of universe - something which has been confirmed (see: 2011 Nobel Prize in Physics).

So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral:

$$ D_{h} = \frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega_{m,0}(1+z)^{3} + \Omega_{\Lambda}}} $$

where the emission redshift $z_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the Friedmann–Lemaître–Robertson–Walker metric, a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.


Do all planets have a molten core? As we know, according to Wikipedia on Earth's inner core:

  The Earth's inner core is the Earth's innermost part and according to seismological studies, it is primarily a solid ball with a radius of about 1220 kilometers, or 760 miles (about 70% of the Moon's radius). It is believed to consist primarily of an iron–nickel alloy and to be approximately the same temperature as the surface of the Sun: approximately 5700 K (5400 °C).

Now the question is, do all planets have molten inner cores?

The short answer is no. Take Mercury for example in this comparison of Earth Mercury core. Mercury is thought to have a liquid outer core and solid inner core. The gas giants like Jupiter are thought to have a relatively tiny rocky core but the convective motion in the metallic hydrogen is what gives them the strong magnetic fields.

See also: Is Mercury's core liquid? for more on Mercury's core.


The formation of new stars as depicted in Feynman's lectures I've stumbled upon this curious passage addressing the formation of stars in Feynman's lectures on physics:

  Whether we have ever seen a star form or not is still debatable.
  Figure 7–12 shows the one piece of evidence which suggests that we
  have. At the left is a picture of a region of gas with some stars in
  it taken in 1947, and at the right is another picture, taken only 7
  years later, which shows two new bright spots.

While I fully agree with Feynman's later note† that it's extremely unlikely that this picture shows stars forming, what does it actually show?

There is no reference to it, and no mention of filter used, the physical scale or even of the celestial object shown, and I've gotten rather curious as to what the seven-year change represents. Variable stars? Or is the displacement due to proper motions? Film artifact? Something else entirely?

Figure in question (Source):

† Feynman concludes the paragraph with "[has the star formation somehow occurred?] Perhaps, and perhaps not. It is unreasonable that in only seven years we should be so lucky as to see a star change itself into visible form; it is much less probable that we should see two!" - indeed, the star formation rate of the whole Galaxy would be of this order!

imo, all the 5 stars are already present and with a little separation in the left image.
Their distance to Earth should have been investigated on the image of the right. The ones nearer could travel, proper motion, much faster than the ones on the background.
In short: they could be at large distances one from the others and it is a coincidence the fact that in the left image they are aligned. With zillions of stars out there the probability is not null.
There are very fast stars (VFS) for ex. US 708 (1200 km/s on WP e 708 km/s heliocentric radial velocity - Hirsch et al. 2005) not known at Feynman time.
The time of discovery of VFS was 2005 and 10 years later, Warren R, Brown et al, reported on the PROPER MOTIONS AND TRAJECTORIES FOR 16 EXTREME RUNAWAY AND HYPERVELOCITY STARS (2015).

EDIT ADD:
One thing that can be done is to use software enabled with Point Spread Function (PSF) to discriminate the number of sources and decide how many stars are in each of the images.
I will not do it because it takes time that I do not have.

There are several packages available, fi, DS9 (video: find a moving star), MAXIM DL, SExtractor. The images have to be scaned, eventually formated as FITS (python code) and if using SExtractor look at the procedure used by E. Bertin and M. Dennefeld (1997).
Search on youtube 'Detecting asteroids'


How does using Tidal energy affect the moon? Reading another question Where do we have it from that the Moon is migrating away from Earth?
I makes me ask what effect does harvesting tidal energy for electricity production have on the rate the moon gains angular momentum?
Presuming the most common harvesting methods involve underwater turbines in tidal zones with water flowing regularly in both directions (twice per day).

It is minute and inconsequential.  Without tidal energy harvesting, the motion of the water is ultimately converted into heat, it has a very small heating effect (that is swamped by the heating caused by the sun). With tidal energy harvesting, instead of heating the water, a little of the energy is converted to electricity (which is then converted into light or heat,  ultimately the energy becomes heat)
The tidal dams can change local tidal patterns, which could have a minute effect on the moon's motion, but this effect is minuscule.


Standardized "constellation" regions? The current system of constellations is historical and has kind of arbitrary boundaries.

This has a number of obvious downsides:

Difficult to define boundaries need complex tables to express
Each constellation subtends a different solid angle (area of sky)
etc

Has anyone ever proposed, or does anyone use, a system of cutting the sky into regular chunks, that are either simple to define or similar in area?   Do any of these systems (if they exist) attempt to preserve the most important/famous constellations (as least the part people normally see) ?

Modern astronomers are not instructed on the constellations and pay little attention to them except as a naming convention for stars.  Positional information is nearly always either equatorial (ra,dec), or galactic (gl,gb), or supergalactic (SGL, SGB).  To improve the speed in which information on a region can be extracted from a database, a scheme called Hierarchical Triangular Mesh (HTM), is often used.  The sky is divided into three sections (0,1,2) and each of those is divided into 3 sections (00,01,02,10,11...) and this iterated until the level of finest size is reached.  One can address the finest region by using all of the digits or a larger region by dropping some of the final digits.


A beginner project in plotting celestial bodies in 3D coordinate space As a data visualization task, I'd like to plot some celestial bodies in 3D coordinate space. I assumed there would be a lot of databases where celestial bodies are recorded along with an XYZ coordinate relative to our sun. As I ponder more on it I begin to realize other features, for example how is the axis of alignment achieved and agreed upon?

Given the below data set, is it clear what system is in use and how I may interpret it in a way that gives me an XYZ I can begin plotting?

https://www.google.com/fusiontables/DataSource?docid=1TZ3eoWstpR8d0O2-eqoz3VFDsLwSSzNA5wX4LxE

Here you can choose your target body and center of observation. If you want cartesian coordiantes use "Vector Table".


What's that very bright star (south-oriented, appearing at 5 p.m.)? From my home I can see a very bright star (or a planet ?), appearing before 5pm local time, when I look toward South. It's really much more brighter than any other star, it's very often the only one you can see because of ambient light in the neighbourhood. It's even visible when it's a little bit cloudy.

Details:

My Location: N 50° 43.640' E 005° 41.430' (East of Belgium)
Time : 5pm (4pm UTC), December 21st, 2016
Direction : Almost straight on South (190°)
Altitude : 35°

Note that angles are rough estimates.

So, what's that star ?

That star is probably the planet Venus.

It is always close to the Sun, and Stellarium shows it to be in the south and well above the horizon at around 5pm local time.

To check, right above it should be Mars (also a planet).

In general, if you see a very bright star, it is very likely that it is a planet, and most likely to be Venus, Mars, or Jupiter. The latter two can appear at any time during the night, but Venus is always close to the Sun, so it can be seen just after sunset or right before sunrise. All planets appear close to the track of the Sun across the sky.


What are the main differences between solar flares and coronal mass ejections? Specifically, what causes them, what sort of damage can they cause, and what are the differences in their composition?

The following definitions are according to NASA.

"Solar Flares" are gigantic explosions from the surface of the Sun, occurring near sunspots along the dividing/neutral line between opposite magnetic field areas.

"Coronal Mass Ejections" (CME's) are:

  huge bubbles of gas threaded with magnetic field lines that are ejected from the Sun over the course of several hours.

Although the Earth's magnetic field protects the biosphere from the worst of these forms of 'space weather', according to the website "Space Weather: What impact do solar flares have on human activities?", the effects that do affect us are:

Disturbance of the ionosphere, disrupting radio communications.
heating, hence expansion of the atmosphere would cause drag on satellites, degrading their orbits, also affecting the accuracy of GPS.

From the Solar Weather site, an important distinction between CME's and solar flares in how they affect the Earth:

  Coronal mass ejections are more likely to have a significant effect on our activities than flares because they carry more material into a larger volume of interplanetary space, increasing the likelihood that they will interact with the Earth. While a flare alone produces high-energy particles near the Sun, some of which escape into interplanetary space, a CME drives a shock wave which can continuously produce energetic particles as it propagates through interplanetary space. When a CME reaches the Earth, its impact disturbs the Earth's magnetosphere, setting off a geomagnetic storm.

Very severe CME's can also potentially disrupt power grids and communications.


Which techniques are used to convert radio signals received by antenna to images? I realized that after answering the question How do astronomers convert radio signals received by their antenna to images? I had missed the point that the OP was only asking about software.

I've written this question about experimental techniques as a companion question in order to move my answer from there to here.

I can think of at least four very different ways that "astronomers convert radio signals received by their antenna to images."

1. Scanning

This is the oldest and easiest to understand way to image. The answer to the question How did single dish (or single receiver) radio telescopes originally generate images? sums it up concisely:

  They scan the object, if you point the dish a a point in the sky as the Earth rotates the dish scans across astronomical objects, then move the dish to point at a slightly different position and let it scan across the object again, and again. After a while you can re-construct an image from the scan lines in a similar way to analogue TV.

Here is an early 1D projection of scanning the plane of our Milky Way galaxy from The galactic system as a spiral nebula Oort, J. H.; Kerr, F. J.; Westerhout, G.  MNRAS 118, (1958) p. 379. This image is reversed tone for better clarity.

For more information see answer(s) to Why the blank wedges in this very early 21 cm map of the Milky Way? (Oort et al. 1958)

2. Interferometry (most common technique today)

If you receive an incoming wavefront from two or more different locations, which in this case means two or more antennas, you can use math to reconstruct an image. This is called interferometric imaging and these days it's done by feeding the signals from each antenna to a large, fast computer known as a correlator. However in the early days of interferometric imaging the interference was actually done electronically, with amplified signals from each antenna combined in a mixer and other electronics.

In fact, one of the earliest techniques of radio interferometry used a single antenna which collected both a direct signal from the source and a signal taking a longer path by reflecting off of water! This is called Sea interferometry and in this case the water acts as a Lloyd's Mirror. This is one dimensional imaging only.

Read more about this in Radio Astronomy at Dover Heights as well as in the paper Galactic Radiation at Radio Requencies V. The Sea Interferometer J. G. Bolton and O. B. Slee, Australian Journal of Physics, vol. 6, p.420, December 1953.

above: Very Large Array (Source)

You may have seen images of (material around) a black hole recently. In this case the signals from each antenna were recorded on to hard drives, which were then shipped to a central location before being put into a computer correlator, but mathematically it is the same process. See answer(s) to How does the Event Horizon Telescope implement the interferometry?

3. Direct imaging using focal plane arrays

The question What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE? shows the first image below, of the Parkes 21cm Multibeam Receiver and seems to be dated 1997. In this answer links to First image with Apertif: a new life for the Westerbork radio telescope which is shown in the second image.

4. Delay-Doppler radar

This is the coolest method! It is based on radar, which means that from one dish astronomers beam a carefully modulated radio signal towards a planet or asteroid, and receive the reflected beam in a separate dish. This kind of radar is called delay-doppler because it extracts two kinds of information from the returned signal, the total delay time giving distances, and the doppler shift caused by the velocity of the surface doing the reflecting.

If the body is rotating, different areas will be moving towards or away from us depending on their location, and different parts will be closer or farther away from us, also depending on location. If they receive enough signals over a long enough period of time, they can reconstruct a hypothetical image of the shape of the object.

See The Planetary Society's Senior Editor and Planetary Evangelist Emily Lakdawalla's How radio telescopes get "images" of asteroids and this answer for more on this technique.

Left: A very slow GIF Source.

Right: A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. Here is an animation that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top." From here. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?

Note: These two images do not necessarily have the same orientation.


How does an eclipse differ from an occultation? A comment in response to this question suggests that an eclipse differs from an occultation in that the former casts a shadow while the latter doesn't.

This isn't particularly satisfactory since blocking the light from one object with a second will always cast a shadow even if it's not visible to the human eye.

Similarly, an annular eclipse of the sun by the moon could be referred to as a transit.

Is there a more rigorous definition of these three terms that distinguishes an eclipse from an occultation or a transit?

There are a number of terms for these things:

conjunction:
When two objects have either the same right ascension or the same ecliptic longitude (usually as observed from Earth). That is, they lie in (almost) a straight line from the point of view of the observer. In the case of the Solar System, such as when the Sun and Venus line up from the point of view of an observer on Earth, this is a conjunction. (It is also a conjunction when Earth lines up between e.g. Mars and the Sun - it is an inferior conjunction from the point of view of Mars (but we can't observe from Mars)).

Furthermore, when Venus lines up between the Earth and the Sun this is an inferior conjunction, and when Venus lines up with the Sun but on the opposite side of the Sun then this is a superior conjunction. Thus the inner planets (Mercury, Venus) can have both inferior conjunctions and superior conjunctions with Earth while the outer planets (Mars, Jupiter, etc.) can only have superior conjunctions with Earth.

The "almost" in "almost a straight line" is important because the definition is by longitude. Although the planets are said to lie in the ecliptic plane, this is not completely flat - they do not all have the exact same orbit inclinations. Thus not every conjunction is a transit, occultation or eclipse - they do not pass directly in front or behind.

transit:
When an object passes in front of a larger object*, this is a transit. These are rarer than conjunctions because the objects need to be directly in line so that one can be seen crossing the other.

In the case of recent transit of Venus, when Venus passed in front of the Sun, the sunlight passing through Venus' atmosphere gave us the opportunity to collect data on its composition. When one of the moons of Jupiter pass in front of Jupiter itself* this is also a transit.

occultation:
When an object passes in front of a smaller object*, this is an occultation. The farther object is hidden by the nearer object. These are also rare** because, like transits, they also need to be directly in line so that the nearer object can be seen crossing in front of, and blocking out, the farther object.

Asteroids occulting stars are studied because while we can calculate the asteroid's orbit and its distance from us using its speed, knowing how gravity and orbits work, etc., we can't see clearly enough to determine the size (think few kilometres in diameter but millions of kilometres away). But when an asteroid passes in front of (occults) a distant star, the star is effectively a point source (looks tiny), from the amount of time the occultation lasts and the asteroid's speed we can calculate its size***.

eclipse:
The Moon passing in front of the Sun is, depending on the conditions, a partial eclipse, total eclipse or annular eclipse.

Clearly, by the definition, when the Moon fully blocks the Sun it is an occultation and when it only partially blocks the Sun it is a transit, but
this is a special case and is called a solar eclipse rather than a transit or occultation. (Similarly when the Moon passes through the Earth's shadow - when the Earth is directly between the Sun and the Moon - it is called a lunar eclipse.)

The word eclipse comes to us via Latin and French from ancient Greek ekleipsis, from ekleipein 'fail to appear, be eclipsed', from ek 'out' + leipein 'to leave'. (see eclipse on Wikipedia)

This alignment is also known as a syzygy****.

* Smaller or larger as they appear to the observer, an asteroid is most definitely smaller than a star but because it is so much nearer it looks bigger.
** A lookup of occultations can return a long list so that it may not seem rare, but when you compare how many asteroids there are, and how many stars, in relative terms it actually does not happen very often.
*** We can set limits on its size, given that the "average" asteroid looks like a lumpy potato rather than being some nice, regular shape that would make it easy to work these things out.
**** We could keep adding terms and googling them all day... ;-)
***** There is no footnote with five *


What star system/Galaxy did LIGO observe the gravity wave? Are there any coordinates for the Black Hole system, or galaxy for which the LIGO observations, observed the gravity wave? Any coordinates of the black hole system for which the gravity wave signal has been observed?

The location of GW150914 was rather poorly constrained by the 6.9 ms time difference between the event when it was observed by the two separated LIGO detectors at Livingston and Hanford.

Further details are here
https://www.ligo.caltech.edu/image/ligo20160211b and this site provides a visualisation of where the waves came from.

The location OF GW150914 can only be constrained to be somewhere within 600 square degrees with 90% confidence. There are millions of possible host galaxies within this region.


Open access table of visible stars with magnitude, coordinates, and possibly color? I'm looking for a link to a table of visible stars that is open and available to everyone. It should have magnitude, RA, dec, and possibly some indication of color. This will be used to produce somewhat realistic night skies as a backdrop for showing the motion of the planets in the sky.

I'm not so interested in names, constellations, etc. These are of course useful to know and possibly to plot, but what I'm primarily after is the information necessary to illustrate star position, brightness, and some color information.

edit: to reiterate, "...table of visible stars that is open and available to eveyone." I'm assuming that in 2016 the visible stars are not behind a paywall, am I wrong? Approx RA, dec, mag - are these available for open access and usage?

The Hipparcos catalogue by van Leeuwen (2007) contains all the information you require, plus estimates of distance from parallax. It is open and free to use for scientific purposes.

http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I%2F311

The direct page that describes the catalogue contents and ftp site is http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/311

The tables themselves are at ftp://cdsarc.u-strasbg.fr/pub/cats/I/311


Black Hole, Object or Portal? Some say that a black hole is a rip in space-time, a portal linking two points in the universe as one. How sound is this? I've read that when a giant star collapses, what is left behind (a massive neutron star?) is so dense that light cannot even escape its gravitational pull. Is this phenomenon caused by an actual object? If one were to travel into a black hole, should not one find only really dense matter?

There should be nothing special about crossing the "event horizon" - the line beyond which light cannot escape and so, in theory at least, one might "travel into a black hole" and not be aware of any physical changes at all.

However, I suspect what you mean is what would happen if one travelled to the centre of a black hole. Here we may speculate but our current theories of physics breakdown at this "singularity" and so we cannot know - nor is it easy to think of how we might design an experiment, given our currently available technologies, to test any theory we have.


The M-theory to understand black holes I am a layman in astronomy. Can someone please explain it to me whether the M-theory can be used to explain what is happening in the centre of the black hole where the singularity exists? Or is it still necessary to peek inside a black hole to gain that information?

Can someone please explain it to me whether the M-theory can be used
  to explain what is happening in the centre of the black hole where the
  singularity exists?

In Layman astronomy - yes, M-Theory (and probably a few string theories) do explain in detail what happens at the singularity, but it's a mathematical construct, not a testable hypothesis, so what it tells you might be wrong and little more than creative but irrelevant mathematics.

Looking inside a black hole, well, that's not a good idea either.  Even if you were able to see what happened, you wouldn't be able to send a message back telling anyone, so that's also no help.

Quantum physics is all a model anyway, but it's a good model because it provides accurate answers and it predicted the existence of particles that were later discovered.   Relativity is a good model too that stood up to testing.   Good models are useful and highly accurate and provide answers.

M-theory is a mathematical construct.  That's a different thing entirely.   It could still end up being right, but until it's testable, it's really just a mathematical exercise.   They really shouldn't call it a theory, it should be M-Hypothesis and String-Hypothesis.  They aren't real theories.

If I may quote Shakespeare from Henry IV, just for fun:

  GLENDOWER

I can call spirits from the vasty deep.

  HOTSPUR

Why, so can I, or so can any man;
But will they come when you do call for them?

I'm personally much more comfortable with the "both accepted theories, relativity and quantum theory, break down at the singularity, so we don't know what happens".

If you prefer that M-Theory says what happens, preferring that answer is entirely optional, but I think the majority of people in this field are more comfortable with the we don't know answer.


When was Hubble tension first noticed? When was this term first used? When was Hubble tension first noticed? When was this term first used? Who used this term for the first time?

Hubble tension refers to the incompatibility between different measurements of the value of the Hubble constant. These measurements are incompatible up to more than $5 \sigma$. This incompatibility arises between what we measure "nearby" and what we measure further away, and indicates that there might be some physics we don't understand yet.

Now there has always been some disagreement between the different values for $H_0$ -- in the early 20th century, estimates ranges from 50 to 550 km/s/Mpc -- but the error bars on those measurements were very large. So in the early days of the Hubble constant, although the proposed values varied wildly, there was no tension (well, there were quite a few arguments, but no Hubble tension yet), since it was accepted that the constant was the same on all scales.

Values of the Hubble constant from 1920 to 1962

The most likely value for $H_0$ mostly shrank in the middle of the 20th century, as bias in previous measurements were discovered. By the 60's, most agreed that $60 < H_0 < 130$. But around 1975, there started to be a divide between those who though that the constant was around $55$ km/s/Mpc and those who thought it was around $100$ km/s/Mpc. For instance, de Vaucouleurs found $H_0 = 50$  km/s/Mpc in 1970 and $H_0 = 100\pm 10$ km/s/Mpc in 1977. But error bars were still large, and it was expected that the values for $H_0$ would converge.

Values of the Hubble constant from 1975

Up until very recently, the difference between measurements could reasonably be explained by the size of the error bars. But as measurement errors decreased, the range of possible values did not. This is what led to what we call the Hubble tension.

What really made the Hubble tension clear were the measurements of the Cosmic Microwave Background (CMB) done by Planck in 2013. These were the best observations of the CMB that indicated that $H_0 = 67.74±0.46$ km/s/Mpc, in complete disagreement with the value around $73$ or $74$ km/s/Mpc derived from type Ia supernova.
The Planck observations made the discrepancy between these values statistically significant.
Around the same time, measurements of the Baryonic Acoustic Oscillations (BAO) confirmed the tension between the values obtained at low redshift (with the standard distance ladder) and the values obtained at high redshift.

The Hubble tension was noticed before it was formally called that way. See the following articles that describe this tension without using the expression "Hubble tension":

Planck and the local Universe: Quantifying the tension, 2013
THE 1% CONCORDANCE HUBBLE CONSTANT, 2014
Cosmological implications of baryon acoustic oscillation (BAO) measurements, 2014
The Hubble Constant, 2015

In fact, a search on arxiv, ads or google scholar for "Hubble tension", or "Hubble-parameter tension" before 2014 yields no results. It seems that in 2015, this exact phrase started to really be used (for instance here).

In the last five years, more measurements of $H_0$ have been made, confirming the tension between the "high-redshift" values and the "low-redshift" values, and solving the Hubble tension is a hot topic of research.


How close would the sun have to get to a supermassive black hole to have Jupiter stripped and slingshot into interstellar space? Please make whatever assumptions are needed.  Assume sun approaches black hole on some sort of hyperbolic trajectory.  Assume size of supermassive black hole is same as the one at center of our galaxy.  How close would sun have to get in order for Jupiter to be “slingshot” into interstellar space.

Closer than about 10000AU, 1.5 trillion km or about 2000 times the orbital radius of Jupiter.

Three body systems are peculiar beasts, and but I'm going to change the question to "Would Jupiter be in a stable solar orbit?" This makes it a question of Hall radius.

The Hall radius is $r_H = a\sqrt[3]{m/3M}$ where $m/M=1/4000000$ is the ratio of solar mass to black hole mass (the mass of Jupiter is treated as negligible!) and $a$ is the orbital distance of the sun about the black hole

If we set $r_H$ to 5 (AU) and solve for $a$ we get a value of about 10000  (I'm only working at 1 s.f. accuracy here. In other words, if the sun is orbiting the black hole at 10000AU then Jupiter would be in a marginally stable orbit.  If the sun got closer than 10000AU then Jupiter would tend to orbit the black hole and not the sun.

However this is only an upper bound. On a hyperbolic orbit the sun would spend little time near the black hole and so even if Jupiter had drifted in the black hole's gravity as the sun passed by, as the sun and jupiter moved away, it would tend to fall back into orbit around the sun. To model whether Jupiter would actually be lost would require knowledge of the initial conditions and numerically solving the differential equations of motion.


Can a telescope ever increase the apparent luminance of an extended object? From what I know about common telescope designs, telescopes don't increase the apparent luminance of extended objects compared to the luminance seen with the naked eye. In this sense extended objects don't appear "brighter" (per unit solid angle of object/image), although the total light flux received from the object (the illuminance) can increase due to the increased magnification (the object looks bigger through the telescope). The best that can be done is keep the luminance the same, which requires eliminating transmission losses. In addition to transmission losses, the light received by the eye per unit solid angle of image is further reduced at high magnifications when the exit pupil is smaller than the viewer's pupil.
Can there be no telescope design (purely optical, e.g. without using electronic eyepieces) that makes extended objects appear "brighter" (in the above sense) than with the naked eye, by somehow overcoming the limitation of the observer's pupil? If so, can this be proven? If not, what would such a design look like?

Can there be no telescope design... that makes extended objects appear "brighter"... than with the naked eye, by somehow overcoming the limitation of the observer's pupil?

I've left out all of the parenthetical qualifications and I'll answer this, let's see if it gets at the heart of the question.
tl;dr: No, because etendue; the same reason that a wall doesn't get brighter when we walk towards it and why we can't go outside with a magnifying glass and concentrate blue sky.

I thought I'd leave the other answer in place for contrast and as a teaching moment, but the down voters had a go at it, so now you can't see it unless you have 10k reputation.
I'd claimed that a pair of 7x50 binoculars would make the object 49 times larger in solid angle but collect $(50/6)^2$ or about 69 times more light.

If my fully dark-adapted pupil is 6mm in diameter then the aperture is 8.3 times larger in diameter than my pupil, but the image is only 7 times larger. We square the ratio to get the ratio of surface brightnesses, so it will appear to be
$$\left( \frac{50/6}{7/1} \right)^2 \approx 1.42$$

However the OP pointed out in a comment that this would produce an exit pupil larger than the 6 mm entrance pupil of the eye.

Thanks for the answer. I'm not sure about the conclusion though. With 7x50 binoculars, the exit pupil is 7.14 mm in diameter, meaning not all of the incident light falls on the retina since the eye pupil is smaller. Specifically, the fraction of of light that does enter the eye is $(6/7.14)^2=0.705$, which is precisely the reciprocal of the factor 1.42 that you calculated. So in the absence of transmission losses I think the luminance remains the same as that seen with the naked eye.

That was an Aha! moment, nature is smart, or at least I'm not.
I replied:

omg I think that I have I failed to recognize something as fundamental as conservation of etendue. Now it looks like my answer is wrong. :-( ...

This failed for the same reason that a wall doesn't get brighter when we walk towards it and why we can't go outside with a magnifying glass and concentrate blue sky on a sheet of paper. In classical mechanics the analogy is conservation of phase space and Liouville's theorem


Our Solar System’s interstellar movement in relation to the Alpha Centauri System Our solar system is moving at approx. 52,000 mph, so where is the Alpha Centauri system, which is assumed to be also moving through interstellar space at a similar velocity since it is part of our Galaxy. Is the A. Centauri system moving in front of us at 12 o’clock, parallel to us at 3 o’clock, or some other configuration?

If you were looking down on the galaxy from above with 12 o'clock pointing from the Sun to the Galactic center, then using SIMBAD we can get the galactic co-ordinates as approx (315,-0.6) in longitude, latitude respectively. Longitude is measured anti-clockwise from the line from the Sun to the galactic center, so Alpha Cen would be approximately between 1 and 2 o'clock. ESO have a map here showing the nearby stars within 12.5 light years.


Did nobody in the Astronomy community think 12,000 new satellites in LEO might be a problem? This answer to Can / should the Starlink satellites be blackened? mentions Space News' SpaceX working on fix for Starlink satellites so they don’t disrupt astronomy where SpaceX chief operating officer Gwynne Shotwell is quoted as saying the following:

Shotwell admitted that nobody in the company anticipated the problem when the satellites were first designed.
“No one thought of this,” she said. “We didn’t think of it. The astronomy community didn’t think of it.”

Question: Did nobody in the Astronomy community think ahead of time that 12,000 new satellites in LEO might be a problem?
Did SpaceX sneak this in under the astronomical observer community's nose by starting "small" (SpaceX's 4,425 satellite constellation - what's the method to the madness?) and then "drifting" up to 12,000 satellites without anybody realizing it? Or did the community actually have some good idea that this was likely going to be a problem?

fyi it may not stop at 12,000:
Space News: SpaceX submits paperwork for 30,000 more Starlink satellites

WASHINGTON — SpaceX has asked the International Telecommunication Union to arrange spectrum for 30,000 additional Starlink satellites.
SpaceX, which is already planning the world’s largest low-Earth-orbit broadband constellation by far, filed paperwork in recent weeks for up to 30,000 additional Starlink satellites on top of the 12,000 already approved by the U.S. Federal Communications Commission.

SpaceX could one day operate 42,000 Starlink satellites if it builds and launches 30,000 in addition to the 12,000 for which it already has FCC approval. Credit: SpaceX

“No one thought of this,” she said. “We didn’t think of it. The astronomy community didn’t think of it.”

What utter nonsense. SpaceX are either lying or they didn't bother to go looking for any other opinions.

Astrophotographers have been complaining about satellites messing up their photos for years. When Starlink was announced astronomers complained even louder.

Go to any popular astronomy forum and search for "starlink" and see for yourself #1 #2.


What is a "21cm/mm absorption system" in the context of measuring old photons from quasars? This answer links to Wikipedia's Time-variation of fundamental constants which cites Further Evidence for Cosmological Evolution of the Fine Structure Constant (J. K. Webb et al. 2001), the abstract of which says:

  We describe the results of a search for time variability of the fine structure constant, alpha, using absorption systems in the spectra of distant quasars. Three large optical datasets and two 21cm/mm absorption systems provide four independent samples, spanning 23% to 87% of the age of the universe. Each sample yields a smaller alpha in the past and the optical sample shows a 4-sigma deviation: da/a = -0.72 +/- 0.18 x 10^{-5} over the redshift range 0.5 < z < 3.5. We find no systematic effects which can explain our results. The only potentially significant systematic effects push da/a towards positive values, i.e. our results would become more significant were we to correct for them.

The paper discusses the "21cm/mm absorption systems" further but I don't understand yet because I'm simply not familiar with the kind of work described here.

The text does refer to:

  HI 21cm absorption lines can be compared with molecular transitions detected at mm wavelengths to constrain $g_p \alpha^2 $ ($g_p$ is the proton g-factor).

so I am thinking that "system" refers to a system of transitions rather than a piece of observational equipment. Is it possible to describe the basics and/or point to a source that does?

They have both 21cm and mm-wave observations for those systems.

Caption to Fig.1

  The hollow squares correspond to two 21 cm and molecular absorption systems.

By "system" they are referring to sets of absorption features caused by multiple clouds of material along the line of sight to a single quasar. Each cloud imposes its own set of absorption features. The totality of these absorption features is referred to as a system.


If quasars are powered by black holes, why are they so bright? If a black hole can swallow everything, even light, then how can a black hole power anything, particularly something as bright as a quasar. Why doesn't the black hole swallow the light of the quasar?

Are quasars formed when two black holes combine?

I assumed it might be like an orange when we press it from middle then its juice comes out from both ends.

Is my assumption right...or is there another reason?

The light emitted isn't actually from the black hole itself but the swirling matter that surrounds it called the accretion disk. The light emitted from the accretion disk is far enough to escape from the pull of the quasar. A link from wikipedia sufficiently explains this:

  Quasars are believed to be powered by accretion of material into supermassive black holes in the nuclei of distant galaxies, making these luminous versions of the general class of objects known as active galaxies. Since light cannot escape the black holes, the escaping energy is actually generated outside the event horizon by gravitational stresses and immense friction on the incoming material.

I am supposing that part of the question, you are asking why quasars are brighter than black holes. That's because of the larger amount of in falling matter into the black hole creates greater friction thus more light.


How to visually chart HORIZON ephemeris I'm trying to build a 3D model of the Sun, Earth and Moon in in Javascript (using Three.JS, which relies on x,y,z coordinates for all objects) in order to simulate the Aug. 21 eclipse. I've successfully figured this out using the Kepler method that NASA describes, but would like to use the HORIZON ephemeris data for greater accuracy, given how sensitive the timing of an eclipse can be.

I've read up on celestial coordinates, but still can't quite figure out how to take the output of the HORIZON tool and convert RA, Dec, etc into 3D coordinates, the way the Kepler document outlines. Can anyone point me to an authoritative primer one how to get from celestial coordinates to geometric coordinates?

Coordinates based on AU are just fine -- I've already figured out how to scale that in the simulator.

Thank you!

I just went through this by myself.  The principal thing you have to do is choose the vector format for your ephemeris, and then download the ephemeris as a plain text (ascii) table. This is way easier than using the Spice tools.


What is the direction of a comet's dust tail before and after perihelion? I recently created a visualisation of comet ISON's orbit. However, I received a critique that the dust tail of the comet should be pointing away from the Sun after perihelion.

As far as I know, the ion tail of the comet definitely points away from the Sun as it is blown away by the solar wind.

I want to know what is the direction of the dust-tail of the comet, before and after the perihelion. Is it always trailing the path of the comet (to a fair degree of approximation) or are there other significant effects?

Comet tail always aims away from the sun, as it's torn from the comet's gas cloud by solar wind. It trails slightly behind, as the comet moves along its orbit while the gas travels directly away, but that's relatively minor - the speed at which the gas and particles are pulled away forming the tail is much higher than the comet's orbital velocity.

Your visualisation shows it trailing behind the comet as if it was moving in some gaseous/liquid medium that stops the lighter particles while letting the heavy comet head travel ahead. This is not the case - as the comet is far away from the sun, the gas cloud just forms its atmosphere and travels with the comet. As it approaches though, the intensity of solar wind increases and it pulls away the atmosphere which can't be protected against it by magnetic field or strong gravity, as the comet has neither.


Newtons law of Gravitation The Force of attraction between two bodies is proportional to the product of their masses. And considering the equation $F=ma$. Acceleration due to gravity will be more for objects lower mass. This contradicts the established theory that objects with different masses take the same time to reach the ground if dropped from the same height. Please explain

For gravity the force is :

$$F=\frac {G M_1 M_2} {r^2}$$

But the acceleration :

$$a=\frac F {M_2} =\frac {G M_1 } {r^2}$$

and this is the same for all bodies as it depends only on the primary's mass ($M_1$).


What is the tidal radius of the Sirius system? The Fomalhaut system is a trinary system, with Fomalhaut A (1.9 M), Fomalhaut B (0.7M) and Fomalhaut C (0.2M). Fomalhaut C is 2.5 light years from Fomalhaut A, and 3.2 light years away from Fomalhaut B. The tidal radius of the Fomalhaut system is 6.2 light years.

The Sirius system (Sirius A, 2M, Sirius B, 0.9M) is roughly 1.1x more massive than Fomalhaut A and B. What would be some ways to calculate a tidal radius of the Sirius sytem, to get an estimate of at what distance other stars could orbit the Sirius system?

The tidal radius for a star or collection of stars near the Sun and on a circular orbit (matches Sirius reasonably well) is given by (e.g. Pinfield et al. 1998)
$$R_T \simeq \left( \frac{GM}{2(A-B)^2} \right)^{1/3} ,$$
where $A$ and $B$ are the Oort constants and $M$ is the total mass.

Using the value of $A-B$ from Feast et al. (1997), this reduces to
$$R_T = 1.43 \left( \frac{M}{M_{\odot}} \right)^{1/3}\ \ {\rm pc}.$$

Using this formula, I get that the tidal radius for Fomalhaut is 2.01 pc (in reasonable agreement with whatever source you are using). As the mass of the Sirius system is only marginally larger then the tidal radius will be almost the same, since it depends on the cube root of mass.


Why is a black hole's Hawking radiation temperature inversely proportional to its mass/size? As a black hole shrinks in volume and mass, shouldn't its temperature get lower?  Shouldn't it evaporate more slowly?
Naively, (very naively), I think that with a smaller surface area (as per its event horizon), it would have fewer and fewer random quantum foam particles popping up just outside its surface....

In a sense, you're right.  Vacuum energy over a surface area is proportional to the surface area so larger area means more of the particle-antiparticle occurrences and that's the source of Hawking radiation, that is, if you don't mind the virtual particle explanation, which is the only explanation I sort of understand.  I look at the equations and my brain goes blank.
But surface area isn't the only factor.  The more important factor is how quickly the gravitational field drops off, which, for stellar mass black holes or larger, the drop off over particle-anti particle distance is tiny, given a radius of at least a few miles and a very small particle-anti particle distance or high probability of a small distance.  That small drop-off in gravitational field leaves little chance for the outermost of the two virtual particles to escape.
That's why the only black holes that are thought to be hot are theoretical micro black holes, which are tiny.  The smaller the theoretical black hole, the faster it's gravitation falls with distance, so, even small distances like those between a particle and an anti-particle, make it much easier for one of the particles to escape the gravity, effectively stealing energy from the black hole and radiating outwards.


How to convert from Black body intensity to MJy/sr? How does one convert from  ${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}}$ to ${\rm MJy \cdot sr^{-1}}$
and from ${\rm ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}}$ to ${\rm MJy \cdot sr^{-1}}$

Part 1
On the one hand, ${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}} = 10^{-3} {\rm J}/({\rm m^2} \cdot {\rm sr})$ has basic dimensions ${\rm mass} \cdot {\rm time}^{-2} \cdot {\rm angle}^{-2}$ and is a radio brightness.
On the other hand, there is ${\rm 1 MJy}\cdot{\rm sr}^{-1} = 10^{-20} {\rm J}/({\rm m^2} \cdot {\rm sr})$ which has indeed the same basic units. A possible difficulties could be the conversion from Jansky to SI units: $1\, {\rm Jy} = 10^{-26} {\rm W} \cdot {\rm m}^{-2} \cdot {\rm Hz}^{-1}$ and (less likely) the fact that ${\rm M}=10^6$. This said, there is a factor of $10^{-17}$ between the two values.
Part 2
We start with ${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot cm^{-1}} = 0.1 {\rm W}/({\rm sr \cdot m^2})/{\rm m} $ which is easily identified by Wolframalpha as unit for a volume scattering emission coefficient. This is not the same as the expression in part 1. However, since you state that it is about Black body intensity, you can follow the argumentation of uhoh in the comments to convert from a wavenumber (given in ${\rm cm}^{-1}$) to a frequency (in ${\rm Hz}$).

However if you'd like to try it yourself, start with $f= c \lambda$, take the derivative to get
$$\frac{{\rm d}f}{{\rm d}\lambda} = -c \frac{1}{\lambda^2} $$
and rearrange to get
$${{\rm d}\lambda} = -c \frac{1}{\lambda^2} {\rm d}\lambda$$
derivative to get
$${\rm d}f = -c \frac{1}{\lambda^2} {\rm d}\lambda$$
and use $c \approx 2.9979\cdot10^{10} {\rm cm}\cdot{\rm s^{−1}}$. Use that to convert your number in ${\rm cm}^{−1}$ to ${\rm Hz}^{−1}$ [...]


What contributes the most to the seasonal temperature variation? The seasonal temperature is ultimately due to the precession of the Earth around the axis. But what I'm curious about is...  is it due more to the side experiencing winter being farther from the sun or is it more due to the fact that the days are shorter and the nights are longer?

[ The earth is actually closer to the sun in the Northern hemisphere's winter. The seasonal temperature variation is predominantly due to the angle the earth makes with the sun. In the northern hemisphere in winter the angle is such that the earth is tilted with north pole away from the sun and the sunlight hitting the earth is spread over a much larger area than if it was pointed towards the sun. Due to this tilt the sun is also lower in the sky and has the effect shown in the 2nd figure.


What is the reason the Jovian planets are in descending order of size? Is there any scientific reason for the jovian planets to be in descending order of size or is it purely coincidental? In terms of radii it goes Jupiter, Saturn, Uranus and Neptune.

Since the initial formation of the solar system and after the initial mass formation of each of the planets let's see what what we can derive from the laws of physics.

Any mass that could further accumulate on any planet would have to come from 3 places:

In the nearby vicinity of the planet's orbit.
From outside the solar system and extra Pluto Kuiper belt objects.
From the Sun as solar wind.

We can surmise that in the first 200 million to half billion years in the life of the solar system, the planets would have grown in the most mass by (1) sweeping up the asteroids and debris in their orbit and vicinity.  In fact, Jupiter with its huge mass continues to pull away stray asteroids to a lesser extent to this day.

The occasional encounters with objects coming from (2) outside the solar system are too rare and far in between to be of significance.  The cosmic rays from external to the solar system are also not significant for mass gain.

That leaves (3) the solar wind.  Whether a planet gains or loses mass from the solar wind is a balance between the

(a)  Gravitational escape velocity of the planet, which is directly proportional to the mass of the planet

(b)  Average temperature of planet's surface and atmosphere

(c)  Average speed or kinetic energy of the particles from the solar wind

(d)  Intensity of the planet's magnetic field.

(e)  Intensity of the solar wind, which is inversely proportional to the square of the distance from the sun.

The (e) factor was and still is much too strong for the inner 4 planets to gain mass from the solar wind.  Being that the escape velocity would be too low to keep the atmosphere with the ongoing assault of the solar wind as is the case with Mercury and our moon.  We can thank the earth's magneto sphere for saving our atmosphere from the solar wind.

Now when we get to where Jupiter's orbit is, about 10AU, the intensity of the solar wind is one hundredth of what it is on the earth.  The solar wind that does reach Jupiter is decelerated and deflected towards the poles by Jupiter's very strong magnetic field .  And since the escape velocity is so high due to Jupiter's mass, most of the solar wind is absorbed into Jupiter resulting in mass gain from solar wind over time. Not even hydrogen can escape Jupiter's  gravitation.

The same net gain from solar wind applies to the next 3 gas giants Saturn, Uranus, and Neptune.  Since these are in order further away, the solar wind intensity correspondingly less.  All three still have a high gravitational escape velocity due to their size and the ones further out have a very low surface temperature so the gases don't have enough average energy to escape.

As we go further out from Jupiter the same dynamic of mass gain through captured solar wind plays out, albeit less and less due to the distance from the sun.

While the solar wind for purposes of planetary  mass gain may seem insignificant now, we must understand that this was not the case earlier on.
And over 5 billion years much mass can be gained in this manner.


Is the rate of gravitational wave detection roughly what was expected? LIGO's first light was back in 2002 but it has only recently started regular observation. With roughy 50 events confirmed, is it possible to compare its detection rate to predicted rates?
There were probably some wide ranges bracketed, and predictions may have evolved after construction began as the science evolved, so there can't be a precise answer. But is it possible to at least say if the rate is roughly in the same ballpark as what was anticipated?
Are there any glaring deficiencies or surprisingly high rates for specific types of events at all?

Observations are made in "runs". As of December 2019, LIGO has made 3 runs, and made 50 detections of gravitational waves. Maintenance and upgrades of the detectors are made between runs. The first run, O1, which ran from 12 September 2015 to 19 January 2016, made the first 3 detections, all black hole mergers. The second run, O2, which ran from 30 November 2016 to 25 August 2017, made 8 detections, 7 black hole mergers, and the first neutron star merger. The third run, O3 began on 1 April 2019; it is divided (so far) into O3a, from 1 April to 30 September 2019, and O3b, from 1 November 2019 until it was suspended in March 2020 due to COVID-19.

I’d have to dig more to find some references, but the existence of $M > 30\ M_{\rm Sun}$ black holes was a big surprise. So a bunch of the events that have been detected are from a class of objects that wasn’t predicted at all.  Originally I believe that neutron star - neutron star mergers were expected to be the main class of events.   I don’t know how that NS-NS merger rate has stacked up against expectations.


Is the expansion accelerating or Decelerating? According to Hubble's Law, the farther a galaxy is, the farther it is moving away. But do we take into account the fact that we are actually looking in the past?

For example, there are two galaxies A and B at distance of 5 and 10 billion years respectively. Now, when we observe A we are looking at how it was moving 5 billion years ago. The same applies for B. So, now we conclude that 5 billion years ago space was expanding at a slower rate while it was expanding comparatively faster 10 billion years ago. What's wrong with this conclusion?

The light we see now is not a direct indication of how the galaxy was moving at the time the light was emitted. In the cosmological frame, the galaxies aren't moving (on average, at least); rather, space between them is expanding. The rate at which it is expanding is the Hubble parameter.

If cosmic expansion is homogeneous and isotropic, then all distances on the cosmological scale should be affected in the same manner, proportionally, unless other forces are involved. Thus the distance between some galaxy and us is affected in the same proportion as the wavelength of the light:
$$\frac{D_\text{now}}{D_\text{then}} = \frac{\lambda_\text{now}}{\lambda_\text{then}}\text{.}$$
The light we see now is a direct indication of the scale factor, i.e. by what factor the distance has grown since the time of emission. We can also consider distance $D$ as a function of cosmological time $t$ and define a recession velocity as the rate at which it is changing:
$$v_\text{r} \equiv \frac{\mathrm{d}D}{\mathrm{d}t} = \underbrace{\left[{\dot{D}}/{D}\right]}_{H(t)}D\text{.}$$
Therefore, the recession velocity now is given by the Hubble parameter $H(t)$ now.


Why are so many sun-like stars found in multiple systems? How do planets form in such a situation? It has been reported that perhaps half of nearby sun-like stars are found in multiple systems.

Why is this so common, and how do planets form in such a situation?

Kaitlin M. Kratter, “Sibling Rivalry Begins at Birth,” Nature, Vol. 518, 12 February 2015, p. 173

You are correct - when we look at stars in the sky, about half of them are actually binary systems, with a range of separations. The frequency of binarity is strongly mass dependent; the 50% figure is appropriate for stars like the Sun, but it appears to be closer to 100% for high-mass O-stars and as small as 20% for the lowest mass M-dwarf stars.
There is also a change in the separation (and corresponding orbital period) distribution. The distributions of separations looks "log-normal" (like a Gaussian when you plot log separation on the x-axis). For solar type stars the peak is at around 50-100 au, but binaries become more compact (maybe 10-30 au) for low-mass stars. There is a tail of binaries with much smaller and larger separations.

These observational data are explained (or are in the process of being explained) as a combination of nature and nurture. The basic process of star formation results from the collapse and fragmentation of molecular clouds. If a collapsing fragment has significant angular momentum then the most energetically favourable way it can continue to collapse whilst conserving angular momentum is to sub-fragment into two pieces orbiting each other. This is thought to be the most basic binary formation process. Other processes include the formation of low mass companions through instabilities in cicumstellar disks or interactions between young stars in dense protostellar environments. Exactly how this leads to the mass dependence of the binary frequency and separation distribution is a topic of contemporary research; there is no definitive answer I can give.

After that I think there are two basic ways, or rather two basic locations, in which planets could form in a binary stem. Either around one or other of the stars or as circumbinary objects. The likelihood of each will depend how and when the binary formed, the binary component separation and mass ratio.

The planets would likely form in the same way that they do around single stars. If the binary star components are originally formed close together, there could be a circumbinary disk from which a stable planetary system formed. Both circumbinary disks and circumbinary planets have been observed.

Disks around individual components are truncated by tidal forces at a radius roughly corresponding to a third of the binary separation. Inside this radius it may be possible to form planets. Thus widely separated binaries could have planets orbiting the individual stars.


Was the whole Universe close to Big Bang very small, or just very dense? I am sorry this question is probably silly for professional astronomers, of which I am not one.

I often hear at lectures that immediately after the Big Bang, the universe was small, say, the size of grapefruit or something like that. But because of inflation which stretched space at super-light (but finite), speed, the observable universe now, may not be the whole universe, and the whole universe may be infinite. If it is infinite, then it would seem that it had to be infinite in infancy as well, just very dense. In fact, it seems "in the limit", it had to be infinite even at Big Bang.

So, questions:

Was early universe small, or only dense but still infinite?

How about at Big Bang, was it infinite?

When astronomers say early universe was small, do they simply mean "the part of the universe which corresponds to our observable universe, was small"?

I am sorry this question is probably silly for professional astronomers, of which I am not one.

This question is by no means silly. Your question is a common one about cosmology (the study of where the universe came from, how it is evolving and what its fate will be). The media often butchers these concepts horribly, resulting in a lot of confusion (out of all scientific information, they seem to have the hardest time reporting cosmology accurately). Your inquisition is definitely a good thing.

  When astronomers say early universe was small, do they simply mean "the part of the universe which corresponds to our observable universe, was small"?

Well, they're usually referring to the entire Universe. In my final paragraph, I explain what this implies for the observable universe.

  If it is infinite, then it would seem that it had to be infinite in infancy as well, just very dense. In fact, it seems "in the limit", it had to be infinite even at Big Bang.

You're closer to the truth. When we talk about the expansion of the Universe, we're really saying that space is being created between all matter.

As you mentioned, the Universe may be infinite. It is not like a ball, but rather like a flat grid, and its "expansion" just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them. Below is a gif I've made to demonstrate this:

A more useful way to describe this is to say the grid is expanding — that space itself, as a coordinate system, is growing. As an analogy, imagine are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.

So the same thing is happening with our universe. The grid is in fact growing, and objects are being swept away with it.

OK, now that we've gotten the core concepts down, I'll introduce one more bit of terminology. The "scale factor of the Universe" refers to how much the Universe has expanded, compared to now. For example, if in a billion years the scale factor is 3, that means that every object in the Universe is 3 times farther from each other compared to now. If the scale factor 700 million years ago was 0.8, then everything was closer by a factor of 0.8 at that time. By definition, the scale factor is 1 right now.

So, if the Universe is expanding now, we'd expect it to be smaller as we look further back in time — i.e. the scale factor would be less. General relativity predicts the scale factor to be zero at 13.8 billion years ago. This would mean that every object would be zero times its current distance from us — in other words, there would be no space.

If you think a Universe without space is impossible, you're correct. We apparently have a contradiction. In GR, you can't have a spacetime with zero space.

Our modern physical theories work fine up a few fractions of a second after the moment of contradiction, and our observations do agree with the idea of an extremely dense early universe. However, our theories break down as we try to model the Universe at earlier and earlier times, until they no longer prove accurate, preventing us from explaining the most interesting moment.

This is why the moment of the Big Bang is one of the biggest mysteries in cosmology. Theories like quantum gravity have arisen to try to explain the conditions near the Big Bang, but none are sufficient as of now.

  I often hear at lectures that immediately after the Big Bang, the universe was small, say, the size of grapefruit or something like that.

Indeed, the problem stems from the ambiguity when one says "universe". In this case, they're referring to the observable universe, which is actually spherical. The observable universe was indeed much smaller near the time of the Big Bang, compared to its radius now.

This is because its radius actually depends on our Universe's scale factor*, which means that at the moment GR predicts the scale factor to be zero, it also predicts the size of the observable universe to be zero.

Obviously that can't be the case, since as we've explained above, it shouldn't be possible for the scale factor to be zero. However, we can say with reasonable confidence that the observable universe was likely the size of a grapefruit at one point, if not smaller (although "grapefruit" seems an arbitrary choice for comparison. I can't actually find the paper that first uses this analogy, so what they originally meant is a bit unclear).

*Measuring distances is actually a bit tricky in cosmology; in some cases, we want to talk about distances or motion of objects while neglecting the Universe's expansion. To save you the need to learn a lot of terminology, I'm right now taking into account the expansion of the Universe when talking about the observable universe's size. The observable universe also grows due to factors besides the Universe's expansion, i.e. light from further and further galaxies reaching us.


What is this graph showing? This is from the NED

What do these three titles mean?

log fb (Jy)
MRK 1014
Log v (HZ)

Why is there LINE,LINE,LINE,LINE,LINE written along the bottom? What does that signify?

What are the dots and lines showing? A range of information? In that chase, what's the relevance of the dot?

Is it like this: The dot is the log v and then the horizontal line shows the range of MRK 1014 recorded for that log v and the vertical line showing the log fv recorded for it?

Why does the log v between 12 - 16.3 have small amounts of log fv and MRK 1014, but the log v between 15 - 19 have no log fv at all, but a large amount of MRK 1014. The same goes for the 2 log v points between 10.4 - 12, except these two have a large amount of log fv but no MRK 1014

Maybe if you could explain the units to me and give me links to further reading it would all become a lot clearer...

Update Okay MRK 1014 is the same of the object. (Doh!) However, what is that titless axis showing?

This is a spectral energy distribution (SED). Since astronomical objects don't emit light a single frequency an SED tells you how much emission you're getting across a range of frequencies.

"log" indicates it's on a logarithmic scale
$\nu$ represents frequency in Hz (e.g. $log~\nu=12$ is $10^{12}$ Hz)
Jy is a unit of specific flux density $f_{\nu}$ is the specific flux
MRK 1014 is the name of the object
Each of the points are observations there is a link on the page for "photometric data points" to see where everything comes from. All the positions have a point, which is measurement of the specific flux density at that given frequency. The vertical lines are the error bar for the flux measurement and the horizontal is the range of frequencies a measurement covers (e.g. a filter might have a central frequency but it filters light over a range of frequencies).
There are a few upper limits which are indicated by vertical lines with arrows pointing down.
LINE indicates that this is an estimate from a spectrum and it is the measurement of the flux of that line, not from a imaging filter.