Where is the exomoon? (Possible discovery of a companion to Kepler-1625b) The new paper in Science Advances Evidence for a large exomoon orbiting Kepler-1625b (open access!) uses a sophisticated combination of occultation light curve simulation, signal processing and statistical analysis to show that there is a significant chance that they have identified an exomoon, a moon orbiting an exoplanet.

But it's too sophisticated for me to understand how this Hubble data indicates a potential moon. Terms like "detrending",  "Markov chain Monte Carlo", and "Bayesian evidences" are beyond me.

Is it possible to choose one of those plots and add an arrow to say "this blip here, this is Kepler-1625b potential moon"? If not, is it possible to at least explain in a simple way what it is from this analysis that has led them to believe there may be a moon there?

Abstract:

  Exomoons are the natural satellites of planets orbiting stars outside our solar system, of which there are currently no confirmed examples. We present new observations of a candidate exomoon associated with Kepler-1625b using the Hubble Space Telescope to validate or refute the moon’s presence. We find evidence in favor of the moon hypothesis, based on timing deviations and a flux decrement from the star consistent with a large transiting exomoon. Self-consistent photodynamical modeling suggests that the planet is likely several Jupiter masses, while the exomoon has a mass and radius similar to Neptune. Since our inference is dominated by a single but highly precise Hubble epoch, we advocate for future monitoring of the system to check model predictions and confirm repetition of the moon-like signal.

"Detrending" refers to removing systematic effects that are caused by the spacecraft, detector or the environment and which are not part of the particular star. The systematic trends can be caused by a large variety of things; the main ones affecting Kepler are drifts in position and focus, thruster firings to compensate for angular momentum build up in the reaction wheels and the roll of the spacecraft every quarter. Because these trends are systematic and affect all the stars that Kepler sees in a similar way, the effects can be modeled and the trends removed - detrending. There are some examples in slide 15 and 16 of this Kepler workshop presentation which shows the "raw" brightness measurements (labeled 'SAP') and the detrended data on the right (labeled 'PDC').

You can see examples of this in the Supplementary Materials. On page 47 in Figure S2, you can see the data before detrending and you can see the dips caused by the transiting exoplanet (Kepler-1625b) but these are superimposed on long-term upwards or downwards slopes or curves. Detrending removes these effects (which will be common to all the stars, not just to Kepler-1625b's star) allowing you to get corrected flat light curves, enabling the search for the much smaller exomoon signal.

The evidence for the exomoon is in two parts. The timing of the planet transit that was observed by HST in October 2017 was 77.8 minutes early compared to the predicted time from the Kepler transits. This is shown (but not very well) in Figure S12 in the Supplementary Materials. If there was no exomoon or anything else tugging on Kepler-1625b, we would have expected the HST transit (the data point at transit epoch 7) to be on the extended black prediction line at an O-C value (y axis) of +77.8 minutes, ( off the top of their plot as it scaled). The second piece of evidence is that is if the transit is early, then they should expect to see the shallower/smaller transit of the exomoon after the transit as it would be on the other side of the orbit with the exoplanet. This is what they show in the bottom half of Figure 4; no matter which detrending method they use for the HST data, they see a small shallow dip after the main transit is over, centered around a time (x value) of 3056.25. This gives more confidence that the exomoon interpretation is real and not an artifact of the detrending process.

What they are saying after Table 1 is that the Δχ2 (the measure of the change in the goodness of fit of the model as you add extra things to the model) is better with the moon models (models Z and M) over one that just accounts for the transit timing variations (model T). This is because the presence of the exomoon is predicted to cause changes in the length and depth of the exoplanet transit but these are harder to measure than the change in timing.


Can magnetism escape a black hole? I know light, and practically nothing but gravity can escape a black hole.  My question is: can magnetism escape a black hole?

A couple things that convince me it can are:

Jupiter's magnetic field shape compared with the jets that can come from near black holes (I would think this might push material falling toward a black hole into or out of the poles):

Apparently black holes have very strong magnetic fields:

http://www.iflscience.com/space/magnetic-fields-can-be-strong-black-holes-gravity

Nothing "escapes" a BH - in the sense that a signal originating inside the event horizon remains forever inside. If something is observed moving away from the BH, then it was generated outside the event horizon. If it was generated inside, it would never be observed at all, forever and ever.

Gravity itself does not "escape" a BH - and neither does "not escape". Gravity is simply a characteristic of the metric of spacetime. If spacetime is warped in a certain way, gravity can be measured to exist. A BH is simply a very powerful distortion of spacetime, nothing more, nothing less. It is generated by a concentration of mass/energy, which warps spacetime, and then that concentration becomes trapped by this distortion that it has produced.

In that sense, gravity is simply part of the BH, because gravity is spacetime being warped, and because a BH is essentially just that - warped spacetime. A BH's gravitational field is part of the BH itself, extending to infinity (but getting weaker with distance). It doesn't "escape" because there's nothing there in the process of escaping.

It's like having a plastic bag tied into a knot to keep water inside, and someone asks "so how does the plastic escape the knot?" The plastic does not "escape" the knot, the knot is part of the plastic.

This all becomes easier to understand when you realize that gravity is not a thing, it's just an effect of spacetime being distorted.

EDIT: I think what you were really asking was - can a BH have its own magnetic field? The answer is yes.

A BH can have 3 characteristics: mass, spin (rotation), and electric charge (a.k.a. the no-hair theorem). All other characteristics of the matter falling in it are lost, except these three. If you drop a proton into a neutral BH, then the BH acquires a charge equal to one proton, and that's a measurable electric field.

Now consider a spinning BH with an electric charge, the Kerr-Newman metric. You have a charge, and you have spin. That means you have magnetism. So, yes, a BH can have a magnetic dipole. However, the rotation axis and the magnetic dipole axis must be aligned - a BH cannot be seen as "pulsing". Again, no signal from inside the event horizon can be observed outside.

However, you should not imagine the electric (or magnetic, same thing) field as "escaping" the BH. It does not escape. What happens is, when the charges were swallowed by the BH, the lines of electric field remain "glued" to the BH, which then acquires a charge. Those lines of electric field have existed forever, they don't "escape" anything, and continue to exist after the charge is trapped by the BH.

Note: electric fields and magnetic fields are one and the same. One could appear to be the other, depending on the motion of the observer.


Are the planets in the TRAPPIST-1 system so close that inhabitants on one planet could see surface details on the other planets? In a discussion below the Space Exploration question Are there any to-scale diagrams of the TRAPPIST-1 system I was invited to ask a question like this, so I have.

Question: Are the planets in the TRAPPIST-1 system so close that inhabitants on one planet could see surface details on the other planets? How big would each planet appear from each of the other planets during conjunction or opposition? Bigger than Earth's moon?

Are the planets in the TRAPPIST-1 system so close that inhabitants on one planet could see surface details on the other planets?

Yes!

How big would each planet appear from each of the other planets during conjunction or opposition? Bigger than Earth's moon?

For some combinations, definitely!
Here's a diagram of the size of each of the planets as seen from each of the other planets. I've taken the semi-major axes and planet radii from a table in Wikipedia and used the Python script below to show the apparent sizes at conjunction or opposition. The axes are labeled in degrees and the largest case is 1.28 degrees or 2.5 times the apparent size of the Moon from Earth!
   a (Gm)        R (km)
b:  1.73          7150
c:  2.37          6984
d:  3.33          5000
e:  4.38          5804
f:  5.76          6671
g:  7.01          7322
h:  9.27          4930

For example, the top row shows the largest possible size of planets c through h from planet b. The black dots indicate the planet with itself.

import numpy as np
import matplotlib.pyplot as plt

data = (('b', 1.73, 7150), ('c', 2.37, 6984), ('d', 3.33, 5000),
        ('e', 4.38, 5804), ('f', 5.76, 6671), ('g', 7.01, 7322),
        ('h', 9.27, 4930)) # https://en.wikipedia.org/wiki/TRAPPIST-1#Planetary_system_data_charts

class Planet(object):
    def __init__(self, name, a_Gm, r_km):
        self.name = name
        self.a    = 1E+09 * a_Gm
        self.r    = 1E+03 * r_km

halfpi, degs = 0.5*np.pi, 180/np.pi

planets = []
for thing in data:
    planets.append(Planet(*thing))

arc      = np.vstack([f(np.linspace(halfpi, -halfpi, 201)) for f in (np.cos, np.sin)])
f        = np.array([-1.0, 1.0])[:, None]
c        = np.array([+0.8, 1.0])[:, None]
full     = np.hstack((arc, f*arc[:, -2::-1]))
crescent = np.hstack((arc, c*arc[:, -2::-1]))

all_sizes = []
for p in planets:
    p.arcs = []
    for pp in planets:
        if pp == p:
            p.arcs.append(None)
        else:
            size = 2*pp.r/(pp.a-p.a)
            all_sizes.append(abs(size))
            if size > 0:
                p.arcs.append(size*full)
            else:
                p.arcs.append(size*crescent)

max_size = max(all_sizes)
print('max_size: ', max_size)

if True:
    plt.figure()
    for (i, p) in enumerate(planets):
        y0 = -max_size * degs * i
        for (j, arc) in enumerate(p.arcs):
            x0 = max_size * degs * j
            if type(arc) is type(None):
                plt.plot([x0], [y0], 'ok', markersize=6)
            else:
                x, y = degs*arc
                plt.plot(x+x0, y+y0)
    plt.gca().set_aspect('equal')
    plt.show()


Energy loss from Doppler effect The energy of electromagnetic radiation is related to frequency; higher the frequency, higher the energy level. If electromagnetic waves have lower frequency when they arrive on Earth than originally emitted due to the Doppler effect, where, according to the conservation of energy, does the excess energy go?

It's spread out in time. If a source emits a 1W pule of energy lasting 1 second and the receiver is receding so fast that it is Doppler shifted to a frequency which means the power is just 0.5W, then the pulse will take 2seconds to arrive (since the end of it had further to travel).


Why weren't the Hubble light-echo images of V838 Monocerotis supplemented by ground telescopes? Drawing from this answer

  The V838 Monocerotis expansion (not a supernova) and the observation of the subsequent "spectacular" light echo was quite a notable event! From Nature 422, 405-408 (27 March 2003)

The Hubble images are quite spectacular (discussed further in that answer). I've included them again below. The color images are 83x83 arcsecconds.

A later conference paper laments the fact that no HST images could be obtained the following year (2003):

  "We received HST observing time at five epochs in 2002 through DD allocations: April, May, September, October, and December. All of the observations were made with the Advanced Camera for Surveys (ACS), which had been installed in HST during SM3b in March 2002. I need not emphasize to this audience how extraordinarily unfortunate it is that no HST observations were obtained during 2003—the loss of this opportunity is truly incalculable. However, the echoes were imaged twice in 2004 through the Hubble Heritage program, in February and October. More happily, the HST Cycle 14 allocation committee did award our team observing time for an intensive HST imaging campaign from October 2005 to January 2006, and we also have two more epochs of observations scheduled in Cycle 15 for late 2006 and early 2007".

My question is: Why could't observations from one or more of the ground telescopes that are using Adaptive Optics been used to at least supplement the data in 2003? I understand they may not provide the same quality, but not even close? Not even for the polarization mapping? Is it related to the extended area of the target? Is a field of 80 arcsec too large to correct all at once? Is the pattern too complex, or is it the lack of an isolated bright "guide star" the problem?

note: I need a reasoned and hopefully quantitative answer, not your "best guess" or speculation. Thanks!

Figure 2 of the Nature paper describes the preservation of the actual light curve (history) within the structure of the light-echo shell:

  "FIGURE 2. HST images of the light echoes
  The apparently superluminal expansion of the echoes as light from the outburst propagates outward into surrounding dust is shown dramatically. Images were taken in 2002 on 30 April (a), 20 May (b), 2 September (c) and 28 October (d). Each frame is 83" times 83"; north is up and east to the left. Imaging on 30 April was obtained only in the B filter, but B, V and I were used on the other three dates, allowing us to make full-colour renditions. The time evolution of the stellar outburst (Fig. 1) is reflected by structures visible in these colour images. In b, for example, note the series of rings and filamentary structures, especially in the upper right quadrant. Close examination shows that each set of rings has a sharp, blue outer edge, a dip in intensity nearer the star, and then a rebrightening to a redder plateau. Similar replicas of the outburst light curve are seen propagating outwards throughout all of the colour images."

From Astronom. J. 135, 2, 2008 or ArXiv

  Figure 2. Images representing the degree of linear polarization, p, for each of the four epochs of data shown in Figure 1. Image scales and orientations are the same as in Figure 1. The image stretch is linear, ranging from black representing zero linear polarization to full white representing ~50% linear polarization. These images illustrate the apparent outward motion of a ring of highly polarized light in the light echo.

Adaptive optics does not work (or at least did not 10+ years ago) on ground based astronomical telescopes at visible wavelengths. No images remotely comparable to HST could be obtained for faint$^*$ objects from the ground in the BVI wavebands.

I checked for papers discussing adaptive optics at other wavelengths and couldn't find any. The field of view is a factor. Back in 2003 there wouldn't have been instruments that could have covered V838 Mon at IR wavelengths with NIR AO observations. It can just about be done now at Gemini South, but field of view is still an issue for AO visible wavelength observations.

$^*$ NB "Lucky imaging" or speckle imaging can work from the ground for resolving binaries etc, but is not so good for imaging faint, extended objects. Having said that, I have seen some images of things like planetary nebulae taken with Lucky Imaging, but that is now and that was then. Nobody was doing it in 2003.


Do comet orbits lie in the same plane as the other planets? Two typical comets that I know are Donati's and Halley's. So comets also have orbits, but are their orbits on the same plane (such as the ecliptic) as the other planets?

Those two comets happen to have large orbital inclinations to the Ecliptic: 117$^\circ$ for
C/1858 L1 (Donati)
and 162$^\circ$ for
1P/Halley.
Long-period comet orbits are oriented essentially at random, suggesting that they originate in the
Oort cloud,
a sphere several thousand AU from the Sun.
Many short-period
comet orbits are inclined only a few degrees to the Ecliptic,
suggesting origins in the
scattered disc,
a torus several tens to a few hundred AU from the Sun.


What would the view be list from TRAPPIST-1g, looking back towards TRAPPIST-1? There's a lot of interest with the recent discovery of seven earth sized planets around TRAPPIST-1. I saw this poster, and the NASA video says that in the system you could see a lot of other planets whipping around in their orbits.

It got me wondering:

Obviously that poster is exaggerated, but how exaggerated?
What would you actually see with the naked eye, standing on TRAPPIST-1g, and looking back toward the other planets?
How big would they appear?
What might TRAPPIST-1 look like?

Let's take a look at the star and the planet's characteristics first. We have the star TRAPPIST-1:

$M = 0.082 M_{\odot}$
$R = 0.117 R_{\odot}$

The planets are all roughly in the scale of $0.7 R_{\oplus}$ to $1.2 R_{\oplus}$ with their semi-major axes of:

b: $1.66 \times 10^6 \; \mathrm{km}$
c: $2.88 \times 10^6 \; \mathrm{km}$
d: $3.14 \times 10^6 \; \mathrm{km}$
e: $4.19 \times 10^6 \; \mathrm{km}$
f: $5.57 \times 10^6 \; \mathrm{km}$
g: $6.73 \times 10^6 \; \mathrm{km}$

So if we would stand on the surface of TRAPPIST-1g and look towards the star, we would be at $1/25$ the distance to the star compared to the sun. The star however is roughly $1/8 R_{\odot}$. That would make the star appear about at about $3$ times the diameter compared to our Sun from the Earth.

TRAPPIST-1f would every few days pass TRAPPIST-1g with a closest approach of $1.16 \times 10^6 \; \mathrm{km} \approx 3 \times r_{\mathrm{Earth-Moon}}$ which is roughly three times the distance to the Moon.

Since $R_{\mathrm{TRAPPIST-1f}} \approx R_{\oplus}$ and $R_{\oplus}/R_{\mathrm{Moon}} = 3.67$ it would seem that TRAPPIST-1f would appear roughly the size of the Moon on closest approach.

The other planets are similar in radius but their distances are $2$ to $5$ times further away and would thus appear smaller by that factor.


Is there any difference between M87's image and predictions? Today, the EHT released (somewhat blurry as compared to simulations) images of the M87 black hole.
Did the image contain anything which might lead to a correction to General Relativity or were all the observations according to what had been predicted?

The observation did indeed confirm general relativity, see e.g. Lisa Grossman wrote in October 2020 for ScienceNews: The first black hole image helped test general relativity in a new way

That iconic image, of the supermassive black hole at the center of the galaxy M87 about 55 million light-years away, showed that the shadow closely matched general relativity’s predictions of its size. [...]

Specifically, the researchers used the size of the black hole to perform what’s known as a “second-order” test of general relativity geared toward boosting confidence in the result. That “can’t really be done in the solar system” because the gravitational field is too weak, says EHT team member Lia Medeiros of the Institute for Advanced Study in Princeton, N.J.


Could a heavy exoplanet squeeze water into being liquid and 4° warm? Let's say there is an exoplanet orbiting its host star well outside its habitable zone. Suppose further that it has a lot of water, being perhaps comparable to earth when it comes to the volume ratio of water compared to the remainder of the planet.
Now water at its densest is 4° Celsius warm. My question is: Could the exoplanet's weight force the water into a liquid state by squeezing it to the temperature of lowest density?
My intuition would say no, because after all, temperature is nothing but molecular movement, and every kind of accelerated movement is supposed to lower the total energy of a system, which seems to be absurd.
Also, this seems to be an explanation of the so-called "faint young sun paradox", which is well-known in geology.
If the answer differs from that, I would be pleased to read an explanation of how my reasoning was flawed.

Hardly. Maybe. Probably. It depends.
Liquid water at atmospheric pressure is densest at 4°C - but that is a function of pressure and temperature.
The phase of a material (solid, liquid, gas, critical) depends on both, pressure and temperature - and so does the density which depends on both, linearily within one phase, discontinuous at phase boundaries. The phase diagramme of water which describes the phase you find for any combination of temperature and pressure is surprisingly complex for water and thus one of the least well-established ones.
Maybe one can answer your question already with a clear 'yes', if you consider a sub-surface ocean an acceptable answer. Enceladus' interior is somewhat heated by tidal interaction with Saturn. From Cassini observations we know that under about a 30km thick ice sheet on Saturn's moon Enceladus we have an ocean of liquid saline water - a source for the geysir activity on that moon.


Will just a glimpse (less than a second) of sun during partial solar eclipse damage eyes? This doubt arose when I thought that on Normal Days, at least I have observed bright sun while going outside, going to office, in park etc. for less than a second. It happens almost daily. Not deliberately but that's how things work I guess. Similar example is playing cricket. Many times sun comes in the way of ball in the air and again I have a glimpse of sun. To me, it is very common as I've seen in international cricket matches on TV.
On the other hand, in case of the days of Solar Eclipse, on one or two occasions, when the sun wasn't very bright due to clouds, I've deliberately observed the partial eclipse just for a glimpse, just to confirm it's actually cut. And I confirmed it. (How fast it was? It was not like I stood still and observed the sun for less than a second and then closed eyes. It was like this: I started moving my head and sight from earth surface and in a faster way moved my head towards sun and without any break, stopped at safe location on earth again where there was no sun). And it did literally no harm to me.
So my doubt is:
Why there's so much theories that you must not stare sun even for less than a second or even a glimpse by mistake? If you see news on eclipse day, you'll see full news about this. But on any other normal day, no such news. Is there something special, something more bright on solar eclipse day? Will it damage eyes?

Glancing at a partial solar eclipse is about as dangerous as glancing at the Sun on any other day. If you look at the Sun a few minutes after sunrise or a few minutes before sunset, when the Sun's altitude is low, the light is filtered through a lot of air, and most of the ultraviolet is scattered, so it's a lot less dangerous than looking at the Sun in the middle of the day. However, during an eclipse, people tend to look at the Sun for more than just a few minutes, and that's not healthy for the eyes.
The big danger of looking at a solar eclipse without proper protection occurs when it's a total (or nearly total) eclipse. When the eclipse is near totality, the sky is dark, so your pupils dilate, letting in more light. But the light rays coming from that small sliver of Sun are just as bright as normal, and have the same proportion of UV (ultraviolet) as full sunlight. Those UV rays can cause a lot of damage to the retina very quickly, and that damage can be permanent.


Why do nebulae turn into accretion disks but planets do not? What are the factors that prevent planets from also turning into disks, like the stellar dust in a nebula does?

The Earth is not a perfect sphere, but rather it's squished at the poles and it bulges at the Equator, owing to its rotation working to squish it into a disk.

I'm tempted to say that the Earth couldn't become a disk because of the density of the material in it fighting back against being pressured into a disk in a way that in stellar gas you can't? But then, why isn't the gas around the gas giants a disk? Is it because of the solid nucleus of those planets fighting against that?

The circumstellar disk is perhaps about 2%-3% the mass of the central star.   Large planets may form in similar ways.   The mass ratio of the disk to what it orbits depends a lot on the angular momentum and I would think that only large planets that clear out a considerably larger volume of space are likely to have enough angular momentum to form disks around them.    Jupiter's Galilean moons and most formation moons probably form planetary debris disks during their planet's formation.

This ring system around a brown dwarf that orbits a larger star could be compared to a circumstellar disk.

Disks of this nature, both stellar or theoretical-planetary are temporary, so only very young solar systems might have them and we probably don't have telescopes good enough to get a good look at young planets during their formation period.   A large part of the reason we don't hear about planetary disks is that we might not have good enough telescopes to observe them.

(Note, if someone with better knowledge than me answers this, I can remove my answer).


Can air pollution affect the observed positions of stars? I live in a big suburban area where pollution is quite high, and I have been testing a sextant by measuring the angle between pairs of stars. I compared the measured angle with the one obtained from ephemerides, where the latter takes account of the alteration of the body altitude due to refraction in the atmosphere.
For instance, I measured the angle between Castor and Pollux, both at apparent altitudes of ~ 39 degrees.
The comparison works pretty well, but the sextant appears to systematically overestimate the angle by ~ 1'. Can the air pollution in my area (not light pollution) alter the refraction index of the surrounding air mass and thus be responsible for this discrepancy?
Thanks!

The air pollution has a negligible effect on the refraction, because pollution barely affects the air density.
However, the refraction is proportional to the air pressure and inversely proportional to the air temperature in Kelvin. So different weather conditions or ground elevations will result in different refraction values.


How to Calculate Planetary Equilibrium Temperature in Binary systems I'm working on a program that generates the basics stats of a terrestrial planets in binary star systems. I'm not the best at this kind of math, so I'm having trouble calculating the planetary equilibrium temperature.

The formula I'm currently referencing was posted in a thread from almost 4 years ago: How to calculate the expected surface temperature of a planet.

It seems reasonably accurate since (if memory serves me) using it yields a good answer for Earth and Mars. However this is for a system with one radiating body not a system with two radiating bodies (binary stars).

My instinct is to treat the two stars as one big star and add the wattage $(w/m^2)$ the planet receives from the various stars. So total watts = watts star 1 + watts star 2. That seems reasonable to me, at least for the purposes of calculating planetary temperatures.

I've already done work to calculate the watts received by the planets from the various sources. I've used this equation, subbing in values for both stars.

$$
F(L/S^2)
$$

L = Luminosity (in fraction of Solar), stars 1 & 2 (respectively): 0.0159 & 0.7758
S = Semimajor Axis (in AU), stars 1 & 2 (respectively): 0.1228 & 3.6000
F = Earth Mean Solar Flux $(w/m^2)$ 1362

So the above calculation yields a Mean Solar Flux for the example planet of $1344w/m^2$ from Star 1 and $92w/m^2$ from Star 2. Adding those together I get a mean global wattage of $1436w/m^2$.

However, I'm unsure how to modify the original equation to utilize the above calculate solar flux (ideally without having to reference the original luminosity and semimajor axis variables). What is the correct way to alter the formula?

Any help would be very much appreciated!

The power radiated from the planet, assuming it is at the same temperature across its entire surface:
$$P_{\mathrm{rad}}=\epsilon\sigma T_{\mathrm{eq}}^4 \cdot 4\pi R_{\mathrm{pl}}^2$$
Where $\epsilon$ is the emissivity (to match the formula in the question, set this to 1), $\sigma$ is the Stefan-Boltzmann constant, $T_{\mathrm{eq}}$ is the equilibrium temperature, and $R_{\mathrm{pl}}$ is the radius of the planet.
The power absorbed by the planet from the stars, assuming it is sufficiently far away from each star that there is negligible illumination from that star beyond 90° from the substellar point:
$$P_{\mathrm{abs}}=\sum_i{f_i\left(1 - A_i\right) \cdot \pi R_{\mathrm{pl}}^2}$$
Where $f_i$ is the flux from the $i$th star, and $A_i$ is the planetary albedo with respect to the $i$th star (this may be different because the reflectivity will be wavelength-dependent, and the stellar temperature affects which wavelengths are being emitted).
So equate radiated and absorbed power, and rearrange:
$$T_{\mathrm{eq}} = \left[ \frac{\sum_i{f_i\left(1-A_i\right)}}{4\epsilon\sigma} \right]^{1/4}$$
The limitation here is that the fluxes are going to be time-variable due to the planet's distance changes from the stars (same applies to a planet in an eccentric orbit around a single star), and it is non-trivial to figure out how to average this out. The real planet will take time to heat up and cool down, and in that time the distances and fluxes will change.
So take the numbers you get from this "equilibrium temperature" calculation with a pinch of salt.

As requested, here are some example calculations. First off, the sanity check: Earth. Taking 1361 W/m2 as the value of the solar constant and using an albedo of 0.3 and an emissivity of 1, the temperature works out as
$$T_{\mathrm{eq}} = \left[\frac{1361\ \mathrm{W\!\cdot\!m^{-2}\,\times \left(1-0.3\right)}}{4 \times 1 \times \left(5.6704 \times 10^{-8}\ \mathrm{W\!\cdot m^{-2}\!\cdot\!K^{-4}} \right)}\right]^{1/4} \approx 255\ \mathrm{K} $$
A bit chilly, but not too bad: the greenhouse effect keeps the actual Earth warmer, which could be taken into account by using an emissivity $\epsilon < 1$.
Now for the example in your question, with fluxes of 1344 W/m2 and 92 W/m2. Let's also suppose that while the first star is Sunlike (so I will keep the 0.3 value for the albedo), the second star is cooler than the Sun. This star emits more of its light in the red where the planet is less reflective. I will use an albedo of 0.25 for this star.
The calculation works out as follows (dropping the units in the calculation step for the sake of space):
$$T_\mathrm{eq}=\left[\frac{1344\times\left(1-0.3\right) + 92\times\left(1-0.25\right)}{4\times1\times\left(5.6704\times10^{-8}\right)}\right]^{1/4} \approx 258\ \mathrm{K}$$
Hope that helps!


Why are Hubble images "Cut" Whenever you take a look at a photo that is taken by hubble,
Lets say M87 or Eagle Nebula or maybe Ultra Deep space field,

The Image is always "Cut"

Like this :

or this

I want to know reason for this ,

Any ideas?

I think it's during photo capture or processing error... or no details about elements to color?

The "stair-step" pattern visible in the first image is characteristic of the the Hubble's Wide Field and Planetary Camera 2 (WFPC2). This pattern is created by the arrangement of the four CCD chips. This image shows how the three larger "wide field" chips and the smaller, single "planetary camera" chip are arranged to create the pattern.

The bottom picture appears to be a slice of the original Hubble Deep Field, which was also captured with WFPC2.

So the answer, for these two images, is that the shape of the image is an artifact of image capture.

However, in general, finished images such as these could take up any arbitrary shape. Many of the modern images contain data from multiple views, multiple cameras, even multiple telescopes all drizzled together using specialized photo processing software.


Energy carried by solar wind What are the velocity, mass, and charge distribution of the solar wind.

Near the earth within the magnetosphere in the ecliptic
Near the earth but outside the magnetosphere in the ecliptic
Outside the ecliptic at 1 AU

I would like to understand the energy content of the solar wind, and how it compares with the solar radiation(solar constant is $1360\, Wm^{-2}$)?

What are the velocity, mass, and charge distribution of the solar wind.

Velocity
The solar wind speed has a large range of variation, between ~250–820 km/s [e.g., Chen et al., 2014; Gopalswamy, 2006; Jian et al., 2011, 2014; Kasper et al., 2012; Maksimovic et al., 1998; Marsch, 1983; McComas et al., 2013; Schwenn, 1983; Stverak et al., 2008, 2009] near the ecliptic plane.  These values are not including interplanetary shocks, which can have speeds exceeding 2000 km/s.

The speed is generally higher at higher latitudes out of the ecliptic plane, tending to be over 650 km/s [e.g., McComas et al., 2008; 2013].

Number Density
The number density also has a large range of values, from ~2–90 $cm^{-3}$ [e.g., Chen et al., 2014; Gopalswamy, 2006; Jian et al., 2011, 2014; Kasper et al., 2012; Maksimovic et al., 1998; Marsch, 1983; McComas et al., 2013; Schwenn, 1983; Stverak et al., 2008, 2009].  Again, these do not include interplanetary shocks or coronal mass ejections (CMEs).

Charge State
The alpha particle to proton number density ratio varies between ~1-5%, depending on solar cycle and solar wind speed [e.g., Kasper et al., 2012; Schwadron et al., 2014].

We have also measured the ratio of $O^{7+}/O^{6+}$ and $C^{6+}/C^{5+}$, finding ~1-30% and ~20-200%, respectively [e.g., Schwadron et al., 2014].

  Near the earth within the magnetosphere in the ecliptic

The properties of the terrestrial magnetosphere vary so widely, you would need to narrow down this question.  For instance, the charge states are completely different (e.g., we observe $O^{1+}$ but not $O^{7+}$) but the number densities range from ~$10^{-2}-10^{3} \ cm^{-3}$.

  Near the earth but outside the magnetosphere in the ecliptic

See responses above to first part.

  Outside the ecliptic at 1 AU

We don't have any measurements near 1 AU that are at high latitudes.  Some spacecraft have done out of ecliptic polar orbits with high apogees, but the heliocentric latitudes were still within ~$10^{\circ}$ of the ecliptic plane.  The notes above discuss our only real measurements out of the ecliptic by the Ulysses spacecraft.

  I would like to understand the energy content of the solar wind, and how it compares with the solar radiation(solar constant is 1360 W $m^{-2}$?

The solar wind ram(dynamic) pressure is typically only ~1 nPa or $10^{-9} \ J \ m^{-3}$.  This is highly variable and can change in milliseconds (e.g., interplanetary shocks), but that would still only be $10^{-6} \ W \ m^{-3}$.  If we make a hand-wavy argument that this drops to zero in ~3 $R_{E}$ (i.e., upper bound on thickness of magnetosheath), then the power per unit area can be up to ~20 W $m^{-2}$.  However, I would not read too much into that number as the actual power dissipated per unit area is different for numerous reasons.

References

C.H.K. Chen et al., Geophys. Res. Lett. 41, pp. 8081, 2014.
N. Gopalswamy, Space Sci. Rev. 124, pp. 145, 2006.
L.K. Jian et al., Solar Phys. 274, pp. 321, 2011.
L.K. Jian et al., Astrophys. J. 786, pp. 123, 2014.
J. C. Kasper et al., Astrophys. J. 745, pp. 162, 2012.
M. Maksimovic et al., Geophys. Res. Lett. 25, pp. 1265, 1998.
E. Marsch, Fifth International Solar Wind Conference 228, pp. 355, 1983.
D.J. McComas et al., Geophys. Res. Lett. 35, pp. L18103, 2008.
D.J. McComas et al., Astrophys. J. 779, pp. 2, 2013.
N.A. Schwadron et al., J. Geophys. Res. 119, pp. 1486-1492, 2014.
R. Schwenn, Fifth International Solar Wind Conference 228, pp. 489, 1983.
v. Stverak et al., J. Geophys. Res. 113, pp. 3103, 2008.
v. Stverak et al., J. Geophys. Res. 114, pp. 5104, 2009.


Birthdate differences calculation relative to earths orbit duration I have got a question as I am trying to figure out what the real difference between the birth date of my friend and me is.

I am born on 1983 6th February 7 am. My Friend is born on 1984 7th February 1 am.
1982 was a leap year as the real duration of earths orbit around sun is ca. 365 1/4 days. Okay... But I am struggling to figure out if need to add/subtract 8 hours to my or my friends birthday to figure out what the real difference in our birth dates is, in relation to how long earth orbits around the sun. That leap year is at end of February makes it even harder to me as I don't know how that fits in into the calculation.

The sidereal year is 6 hours and 9 minutes longer than a 365 day calendar year. To simplify your question, assuming you an your friend were born on the same date and time in 1983 and 1984 you would be 365 days, 6 hours, and 9 minutes older than your friend.

Since there is no leap day between the two times you are considering, it doesn't affect the calculation.


Do we live in a galactic bubble? I have heard a certain statement, which can be summarized succinctly into "we live in a local bubble", implying that solar system is located in an intestrellar dust underdensity region.

This region is allegedly produced by a supernova explosion, which happened near the Sun in the distant past. The bubble size was claimed to be of order of few tens of parsec, and the fact of its existence allows one to easier deal with dust extinction for nearby stars.

Another claim I have heard about this bubble, is that this concept is obsolete nowadays, for if even if there were a supernova taking place near the Sun, gas diffusion timescales are short enough, so that no vestige would have remained from this past event by now.

I would like to ask you, therefore, to comment on the text above and in particular indicate, whether such an underdensity really exists, and what do we know about it.

As I have discovered afterwards, the described bubble exists and is in fact a known phenomenon. For those, who want to read more, here is a wiki link http://en.wikipedia.org/wiki/Local_Bubble and the links therein.

Otherwise, it is an underdensity in ISM (not only dust) in the region of Orion Arm (which is a minor spiral arm). The underdensity is up to ten times comparing to normal density in the Milky way, depending on the component of ISM in consideration. The shape of the cavity is complex, typical sizes being of order $100 \textrm{pc}$, but when it comes to practical observations, one should be wary when neglecting dust extinction at distances over $50 \textrm{pc}$.


Does blackhole merging break their event horizon segregation? So far as I know, if we have 2 blackholes A and B, both having their own event horizons, then what's inside the event horizon of A remains completely unknown to us, and same for what's inside the event horizon of B.

Now, suppose A and B coalesce (merge). So we now have a single blackhole C, with its own new event horizon.

This would mean that what was inside the event horizon of A now "knows" what is inside the event horizon of B.  Would this mean that all event horizons of all blackholes are actually "linkable" together since they might all merge at some point?

Edit: After watching the suggested video, I can try rewording the question:

Suppose Alice and Bob are outside blackholes A and B.

Alice jumps into blackhole A, and Bob jumps into blackhole B.

Bob never sees Alice crossing the event horizon (according to the video below; it's the same Bob than in the video: he cannot see Alice crossing the event horizon until "t=infinity"), but Bob crosses the event horizon of blackhole B (not part of the video).

Let's say that the two blackholes are also falling to one another so let's say they are "now" (that concept is probably misplaced) merging. So the two event horizons now become one.

Are Alice and Bob reunited despite Bob never seeing Alice cross the event horizon of A? How would such situation (2 blackholes) be drawn using the diagram in the video in answer below?

This might be related to Do black hole singularities actually merge? and What happens to the information on the event horizons of two merging black holes? but I'm unable to get my answer from there... but I'm only a computer engineer, I'm not a physicist, so the answer might be lying there and I don't see it.

We're also not physicians, not even Astrophysicians, but we're physicists.

One of the more famous physicists, Leonard Susskind, discusses in his excellent youtube lecture series on general relativity (I believe it was episode 7, which I've linked) that during a black hole merger the Schwarzschild surface (not a sphere anymore) deforms and 'comes out to get you'.

Otherwise one would run into the problem that early research into black holes had, when they were still called Collapsars. This problem was that a Collapsar would eternally collapse due to the gravitational time dilation and never actually form a Schwarzschild radius that the static solutions indicate should exist. Then also, black holes could never merge.

But this problem has been solved with the realization that the Horizon deforms during a merger. The Horizons themselves then merge, and A and B come into causal contact.


How long does it take a star to transition from main sequence to red giant? I have read that it takes billions of years for stars of approximately the size of the Sun to get through the main sequence, after which it will enter the red giant phase. When that happens, how long does it take for the star to balloon outward when it becomes a red giant?

It takes a star like the Sun about a billion years to go from the end of core hydrogen burning to the beginning of helium core burning.  One might not call that entire phase a red giant phase, however, because the puffing out process takes quite a while to get going.  In the mean time, the star is what is called a "subgiant," rather than a red giant, and physically the difference is that in a full-blown red giant, the electrons in the core have lost so much heat that they are approaching close to their quantum mechanical ground state.  This implies the core is very small, about the size of the Earth, so a lot of heat must be lost before the star gets to that state.

After that, we can say the star is a red giant, but it won't be all that large until the core builds up more mass.  A ball of electrons near their ground state will shrink as more mass is added to it, and as the core shrinks, the electrons gain kinetic energy, and importantly, so do the helium ions.  When the core mass gets large enough, the helium gets enough kinetic energy to start fusion, and that's the end of the red giant phase.  The star ceases to expand and instead contracts to the "horizontal branch."  Most of the huge expansion occurs near to the end of that process, so the Sun will look like a red giant for less than perhaps 100 million years, depending on how big and red you want it to be in order to count.  A reasonable source is https://www.space.com/22471-red-giant-stars.html.


What is the axial tilt of a planet measured relative to? I am very much a beginner on the astronomy front but I understand about planets having different axial tilts, hence why Venus turns the opposite direction from the Earth and Uranus turns sideways.

However, I am confused as to what the axial tilt bearing is measured from. For example, if all planets were in a perfect line from the sun, would they spin perfect in alignment on their own various axial tilts? Would some still be 'off-set' by a few degrees to one side or the other?

Do we actually know please?

If you look at the Astronomical Almanac for the year 2011 as an example, in the table at the top of page E3, you find two measures of axial tilt. The third column is the declination of the planet's north pole for the mean equinox and equator of date 2011 January 0, 0 hours terrestrial time. So this is measured with respect to the Earth's orbit. The last column is the inclination of the planet's equator to the planet's orbit.

If you look at the right ascension column in the same table, you see that on that particular date the right ascensions are all different.


Is the moon inside earth’s atmosphere? If so, what are the consequences? Recently, a paper was published claiming that the Earth’s atmosphere extends far beyond the Moon. This has been reported on by a number of news sources and websites, including the ESA, Business Insider, Space.com, and EarthSky.org.

If these findings are true, how does this affect the current understanding related to the measured atmospheric pressure, the currently accepted pressure in space between the Earth and the Moon, the pressure on the Moon, the effect of Earth's gravity on the atmosphere, and Earth's ability to maintain this extensive atmosphere while traveling through the Solar System? Is the atmosphere gradually lost or replenished?

Image credit: European Space Agency

It's long been known that Earth's atmosphere gradually reduces in density, and is detectable to very high altitude.

A 1976 atmospheric model has data up to 1000 km of altitude. At that altitude, atmospheric density is on the order of 10-15 kg/m3. This is already better than the best vacuum we can create on Earth's surface. So we can call this an atmosphere (because it contains gases also found on Earth at ground level), but its consequences are limited.

It's sparse enough that at this altitude, atmospheric drag on satellites is negligible for the lifetime of a satellite.

Earth's atmosphere loses some mass to space: at high altitude some atoms achieve escape velocity. This is replenished by evaporation and outgassing on Earth. For example, helium gas starts its life as alpha particles from uranium decay, the gas seeps through the ground, rises through the atmosphere (because it's much lighter than air) and eventually escapes.

We already know with very high accuracy that the Moon slowly moves away from us as it's slowed down. Most of that slowdown is caused by tidal effects, but there will also be an atmospheric component. Due to the very low density, this component will be small.


Why aren't secondary mirrors offset to get rid of diffraction spikes due to the support vanes? Some kind of ellipsoidal shape mirror could reflect to a secondary mirror which is not in the way of the infalling light. Two advantages are immediately obvious. The primary mirror would not be partly shadowed by the secondary. And the support structure of the secondary would not cause diffraction spikes in point sources.

What are the reasons for why this is not only not the standard, but quite uncommon? Especially on the greatest telescopes, one would think that the additional production difficulties, if any, are not a relevant consideration given the budget and clever engineers behind them.

There is a 1.6 meter off-axis telescope for solar observation at Big Bear Observatory called the New Solar Telescope (NST).  Off-axis is particularly helpful in solar observations because it reduces stray light and eliminates a rotating diffraction pattern which can be severe when the sun is the target.  However, for most astronomical work the additional costs are just not worth it, yet. The extra mechanical costs may be greater than the extra costs for the optics.  The total weight of optics and supporting structures of professional telescopes can be many tons.  There is additional strain required to rotate and support an asymmetric structure like this, i.e. bigger motors are needed.  Galaxies are not bright enough to produce diffraction patterns and the issue of stray light from nearby bright objects is far less of a problem than the foreground stars, zodiacal light, and night sky light actually in the line of sight.  If a nearby star is producing a bothersome diffraction spike, one can usually rotate the field to avoid this. I am not saying that stray light or diffraction effects are never a problem, just that this is not a top problem and, if we need to trade down in size at all for an off-axis design, most astronomers would not accept it. Perhaps if relative costs for these telescopes come down in the future, they will be more common.


How are the newly discovered Janus/Epimetheus rings different from the other rings of Saturn? What are the newly discovered ring systems of Saturn, and the circumstances relating to the discovery? Is there something that makes them different from the old well-known rings, like their formation?

The ring in question is the Janus/Epimetheus ring system:

http://saturn.jpl.nasa.gov/science/moons/epimetheus/

Scientists believe the ring formation was due to the break up of one moon. Cassini has taken photographs of the moons and Voyager was the satellite to allow scientists to discover the moons and the rings which follow their orbit. What is seemingly special about these rings is the structure.

http://saturn.jpl.nasa.gov/photos/imagedetails/index.cfm?imageId=2277


What is the relationship between the focal length and f number of a telescope compared to a guide scope? How do you determine the optimal length of a guide scope for a given imaging telescope?

I imagine it would have to take into consideration the pixel size of the guide scope, but would the aperture matter as well?

The aperture of the guide scope matters because it affects how faint the stars are which are available for guiding - the larger the aperture the fainter the stars it will pick up which gives you more choice when picking a good one as a guide star. This is especially important if you are imaging from an urban area where light pollution may otherwise limit the number of visible stars.
The length of the guide scope is not directly important, although generally larger diameter objective means longer anyway. The weight of the guide scope does however have to be also considered as your mount has to be able to deal with the extra load you are placing on it, lighter is better here of course.
When you talk about pixel size I presume you mean pixel size of the CCD camera you are attaching to the guide scope itself? I am not expert enough to go into great details there, if you need more in depth advice on that side of astrophotography I can thoroughly recommend the Cloudy Nights forum, there are some very experienced astrophotographers on there who are very willing to share their knowledge and experience with all no matter how much of a beginner you may be.


Could an object block out the Sun from Earth's view for 3 hours? how large, massive, far away (from earth), and fast would an object (planet-like) have to be to block out the Sun for 3 hours and not disturb Venus, Earth, or the Moon?

Is it possible?   Not really.  Not with your criteria.   Could some mad scientists make it happen, like if we make the Moon bigger?   Er, maybe.

Types of eclipses.

Nasa's eclipse page (scroll down for maps over 20 year periods).

2001-2020 map below.

The map has a notation for hybrid eclipses (see here), not all that relevant to the question.

Total solar eclipses (blue lines on the map), are quite small because the umbra or fully shaded region that the Moon casts on Earth is small, maybe 150 km in diameter on average, 267 km max.  All planets and objects that orbit the sun cast an Umbra that extends behind them.   The Moon (diameter 3,474 km), is always casting an umbra somewhere in space, being about 1/400th the diameter of the sun, it's umbra is about 1/399th the distance between the sun and the moon, or roughly 375,000 km on average.   Eclipses happen on Earth when the Earth is lined up below the Moon's Umbra.  Mostly they don't line up which is why eclipses are rare, but the Umbra is always somewhere in space, not all that far from the Earth, following the moon's path.

Rogue Planet

Your scenario mentioned planet-like.  Now, if it's a planet from our solar-system, that would imply disturbing the orbits of Earth/Venus/Moon, which you specified you didn't want.   There's also real danger if you were to have a planet like object in a near Earth orbit that passes from time to time, close enough to cast it's Umbra over Earth.   For safety reasons "just passing through" is the way to go.

A rogue planet on a hyperbolic trajectory past the sun that, by very rare coincidence, passes between the Earth and the Sun, but close enough to cast it's Umbra over Earth would be very rare to say the least, but a theoretically possible scenario to cast a total eclipse over Earth, but it would need to pass very close and at just the right angle to block the 1/2 of 1 degree of arc that the Sun takes up in the sky.

How close?

Mars is roughly twice the diameter of the Moon, so would need to pass within about 750,000 km to the earth, and line up just right, to cast a tiny umbra that moved across the Earth.

Venus, about 3.5 times the diameter, about 1.3 million km

Jupiter, about 41 times the diameter of the moon, would need to pass within about 15 million KM to cast an umbra on the Earth of similar size to the typical Moon's eclipse Umbra.

And those are the minimum distances.  You'd need the theoretical planet to be pretty well inside those limits to have an Umbra of significant size and for a 3 hour eclipse, bigger is better.

What would a rogue planet passing that close look like?

Consider that it's passing close enough that it looks larger than the Moon, at least for a 3-plus hours.   How close it would need to be depends on the size, as noted above.

The minimum velocity for an object on a hyperbolic trajectory just passing through the solar system but near Earth distance would be Earth's escape velocity, or the orbital velocity of the Earth (108,000 kph) times the square root of 2, and if it approached on the same plane and in the same direction as the Earth, the relative velocity between this Rogue planet and Earth, the minimum relative velocity would be about 45,000 kph, give or take.   That's absolute bare minimum as a theoretical rogue planet passing inside the Earth's orbit would usually be traveling a fair bit faster than that.

So, for a planet like that to remain between the Earth and the Sun for 3 hours, it would need to be about 3 times 45,000 km in diameter, or, nearly the size of Jupiter.

A small rogue planet couldn't cast an Umbra over Earth anywhere close to 3 hours because it would pass over the Earth much too quickly.   A Jupiter-sized Rogue planet could, theoretically, if it was closer than closer than 15 million km, cast it's umbra over Earth for 3 hours or so, but that would be close enough to change the Earth's orbit and, at the very least, give the Earth significant climate change by pushing the Earth a few million km either closer or further from the sun.    Not an ideal scenario.

So there's no scenario that does what you ask.  You'd need a planet the size of Jupiter, with the mass of Neptune or perhaps less to leave the Earth/Moon undisturbed.    A Neptune mass planet with a large ring system . . . just maybe.   But given how rare your scenario is to begin with, I'm going to say basically no.   You could do it, but it would change the Earth's orbit, or you could do it with a smaller planet but for much less than 3 hours.  In any case, it would be a preposterously rare scenario no matter what.

What if we move the Moon?

With rogue planets, I just looked at the relative velocity between the planets, which is close enough for a near-sighted umpire.  When you look at the Moon's umbra, the Moon's motion is much slower, so you also have to take into account the Earth's rotation too.

The Moon (Synodic period), takes 59 minutes to cross the sun, that's tip to full cover, and another 59 minutes back to tip.  (based on a 29.5 day synodic orbit, 29.5 days to cover 360 degrees, the sun, roughly 1/2 of 1 degree in the sky).   So, 158 minutes of partial eclipse, and a record, of 7 minutes total eclipse in one spot on Earth (7 minutes, the record for a total eclipse is possible when the Moon is slightly larger than the sun, close to it's perigee).

So if our mad scientists pull the moon closer to the Earth, the period of time for a total eclipse increases but the rate that the Moon moves across the sun also increases, so you're effectively increasing the 7 minutes of total but decreasing the 158 minutes of partial.   If you pull the Moon all the way into a geostationary orbit, some 10.5 times closer, the Moon would span an enormous 5.25% of the sky, but it wouldn't move in the sky while the sun would and the sun, which crosses about 1 degree of sky in 4 minutes would pass behind the enormous geo-stationary moon in just 21 minutes.   You can't create a 3 hour eclipse by pulling the Moon closer.

This is where the Earth' rotation speed comes in.  The Moon moves over the Earth in it's orbit at about 3,683 kph, and relative to the sun, but because the Earth is in orbit to, the effective velocity relative to the sun (Synodic not sidereal) is about 8% less.

The Moon's umbra effectively moves with the Moon's relative velocity - at least, close enough.

The Earth rotates in the same direction that the Moon moves, peak velocity at the equator of a bit over 1,600 kph.   In effect, the rotation of the Earth tries to keep up with the movement of the Umbra during an eclipse, but the Umbra moves West to East about twice as fast as the Earth turns West to East.

So the solution for your 3 hour eclipse?   Make the Moon bigger and push it further away so it orbits the Earth more slowly and it's Umbra moves closer to the Earth's rotation speed, making a 3 hour eclipse, at mid day, close to the equator . . . possible.

If we replace the Moon with Venus (3.5 times the diameter of the Moon) and we push it some 3 times further away (which would be pushing the limits of the stability of the Earth's hill sphere), then you'd be a lot closer, as our new "Venus Moon" would move across the sky and across the sun still to fast, but it would move more slowly and closer to Earth's rotational velocity leading to eclipses that stayed in the sky much longer.  If you did that, you might get your 3 hours of total eclipse.  You wouldn't want to push Venus too far, cause you'd want it a big bigger than the Sun in the sky, but 3 times the Moons distance, it would be about 15% larger and close to the limit of orbital stability.

Other silly ideas.

Now if you were to Move the Planet Uranus into Earth's L1 Trojan point, which . . . well, the Trojan point would be further away due to the larger planet's mass dominating so . . . that wouldn't work and because L1 points aren't stable, it would also be highly dangerous.   Moving on.

If you were to move Uranus to where the Earth is and make the Earth a satellite of Uranus . . . we might get longer than 3 hour eclipses every time we were on the far side of Uranus making an orbit.   Seems a lot of work though just for an eclipse. . . . but that would work too.

OK, I'll stop now.


Are planet-destroying quakes possible or observable? The Gutenberg-Richter-Law is a very-well studied relationship between the magnitude and total number of (earth)quakes. In theory, there is no upper bound for the size of a quake other than the size of the planet. In other words: There is a non-zero probability (which is rather low) that a quake can rip appart the whole planet. There are numerical evidences with toy models such as the Olami–Feder–Christensen model suggest, such things can really happen.
Therefore my question: Are they any hints for (exo)planet-destroying quakes? And what about star-quakes which occur during stationary phases of star evolution?
References

cond-mat/0608324: Transient and stationary behavior of the Olami-Feder-Christensen earthquake model - not the publication which claims that planet-destroying quakes are possible, but a possible introduction to the kind of numerical methods used

A planet-ripping earthquake isn't possible on Earth.  For a magnitude 10.5 Richter scale earthquake, one would need twice the fault lines that exist on Earth according to the USGS.
The same website relates energy $E$ in joules to earthquake magnitude $M$ with Baths equation: $log E = 5.24 + 1.44M$.  A 10.5 magnitude earthquake would produce about 2.29e20 joules of energy.  The Earth's gravitational binding energy is about 2e32 joules. So if all possible earthquakes went off at the same time twice, it would only generate a trillionth of the energy necessary to rip the planet apart. Earthquakes occur just under the surface of the earth, and they release their energy over time. So, even if an earthquake could generate 2e32 joules of energy, it wouldn't be sufficiently directed to destroy the planet.
Earthquakes are the product of built up energy in tectonic plates by convecting magma.  This is a small effect compared to the power of gravity. The IAU requires a planet to have "sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape." That means gravity is a dominant force over the rigidity of the materials that compose the planet. But the seismic moment is governed by the modulus of rigidity of the materials. So localized earthquake forces and energies will be small compared to the gravitational forces from the planet itself.  So, we wouldn't expect planet ripping earthquakes on exo-planets either.
Notes:

If we consider nuclear fusion as a source of quakes, then supernovas might be called star-ripping quakes.

If we consider collisions as a source of quakes, then the asteroid 16 Psyche might be an example of a proto-planet destroyed by earthquakes.


What magnification is required to see detail on Mars / other local bodies? I have a 125mm Jones Bird newtonian telescope (very shoddy) with 6x lens (2× + 3×) essentially we were able to see a very small O (yeah that's to scale) with varied shades of red.

Is Mars far away from earth right now relative to the average distance or is it just my telescope not powerful enough?

If I wanted to seriously have a chance at picking up significant features on Mars or perhaps pickup the rings of saturn or the red spot of jupiter, whats the minimum setup that I am going to need?

This setup would run me around 3,000 dollars:

Its named...

Celestron CPC 1100 StarBright XLT GPS Schmidt-Cassegrain 2800mm Telescope with Tripo... https://www.amazon.com/dp/B000ARFND2/ref=cm_sw_r_sms_c_api_x7mSBb39EAS9H

Would this be overkill? For $3,000 is this one of the better choices for local planetary observation?

You have several questions here:

Mars is relatively close to Earth right now (October 2018) - about 90 million km.
Is your telescope powerful enough for you to see detail on Mars?  From what you have said, it's difficult to tell.  You say you have magnification of 6x.  This is almost certainly wrong.  If you state the focal lengths of your telescope and its eyepieces we can work out the correct magnification available.  To see much detail on Mars you would need in excess of 100x magnification (same for Saturn, less for Jupiter), and ideally a lot more.  Your scope should be able to manage 100x.  However the Jones Bird design is known to be difficult to align (collimate) properly, so even with that magnification, you may just see a blur.  How much detail you can see will also be heavily affected by the state of the atmosphere - ideally you need a night when the air is calm. Sometimes a slight mist can indicate a steady atmosphere.
Would the CPC 1100 be overkill etc?  This is much more opinion based, but in short, yes - for a beginner anyway.  For an experienced observer, the answer is much more nuanced, and I'm not going to go into any of that.  The scope is generally regarded as pretty good for what it is.


Terminology for empty space at the edge of the observable universe? Is there a name for a point at the edge of the observable universe at which there is no visible matter, that appears completely dark?

I'm thinking something like "empty space", but that could equally apply to an empty void between galaxies, which isn't quite what I'm looking for.

So it would refer to a theoretical line in space from the observer to the edge of the universe which contains no visible matter. Is there a term for this?

There is no name for this.

As we look further in space, we also look further back in time. The "edge of the observable universe" is in the early universe. At an early time in the evolution of the universe, the whole of space was so hot and dense that it was opaque.  In whichever direction we look, we can't look back further than this. The is called the cosmic microwave background. There is no direction in which we can look in which we don't see the cosmic microwave background. There is no direction that contains no visible matter.

There are directions that contain no galaxies or other more nearby objects, but there is no name for these directions.


What causes the raise of temperature in the outer layer of a red giant when the core gets contracted? A star emits energy through the fusion of hydrogen to helium in the core through the proton proton cycle or the carbon nitrogen cycle. Over time, hydrogen in the core is increasingly burned, which causes the fusion process to decrease over time. As a result, the star cools down and decreases the radiation pressure in the core. As a result, the core draws a little more together under the influence of its own weight, which increases the temperature. This temperature rise leads to a hydrogen fusion in a shell outside the core.

But how is the temperature increased as the radiation pressure is decreased? Is this increase caused by conduction or convection or....?

The temperature increase of the core and shell is governed by the virial theorem. As contraction occurs, gravitational potential energy is released. According to the virial theorem, half of this is radiated (or conducted or convected) away and half goes into the internal energy of the contracting material, thus raising its temperature. You can find a simple proof here.

Radiation pressure is unimportant, but continues to increase in any case. Only the surface of the star becomes cooler, not the interior.


Initially non-flat space-time makes dark matter obsolete$\dots$ After the Big Bang, dark matter clusters and build the seeds for visible matter, that then forms stars, planet, you, me, everybody...

But do we really need dark matter to build cluster first. What if spacetime isn't flat per se on cosmlogical sclaes but rather crumbled or heterogenous, like a 4-dimensional version of folded cloth. We don't need dark matter, if we accept that space time didn't came along at the big bang in a clean way, as mathematical text books on general relavity suggest.

Is it possible to support this idea with astronomical data of galaxy mergers?

I think there a few problems with this idea.

First off, as Stan Liou indicates, the early universe expanded extremely rapidly, potentially through hyper-inflation during the inflationary epoch. While we're still looking for observational evidence that this period occurred (e.g., the failed BICEP2 results a few years ago), the existence of such an epoch early in the universe neatly explains away several problems with the standard Big Bang theory (see the horizon problem, monopole problem, and the flatness problem). Assuming such a period existed, then any region of space-time that was "crumpled" would have been quickly and cleanly smoothed out during the hyper-inflation. Such crumpled-ness would not have been around long enough to affect cluster growth and structure. What's more, we very clearly measure that the Universe is flat over large, cosmic scales so it seems unlikely that there are constant and pervasive wrinkles as you suggest there could be.

A second issue I potentially see with your proposal is that it would imply a top-down approach to galaxy formation. That is, first you get huge clusters of mass (conglomerated by your "crumpled" space-time), then you get fragmentation and individual galaxies can form, then galaxies fragment into individual stars, etc. In other words, big structures form first, then smaller structures derive from the bigger structures. The opposite formation model would be "bottom-up" formation where first small things clump, then those clump into larger bodies and soon you have galaxies clumping into clusters. I don't think there is definitive evidence that galaxy formation is either top-down or bottom-up, but simulations seem to me to suggest bottom-up is more likely (of course those simulations use dark matter). I'll leave you to think about that one.

The last point of contention I have is that dark matter plays more of a role than just providing extra gravity. It contributes to the total mass-energy of the universe. Current theoretical models perform extremely well when we include dark matter. I believe a strong piece of evidence for this fact is our ability to theoretically match the CMB power spectrum using a Lambda-CDM model. I don't see that crumpled space-time could reproduce this result nearly as well.

I think you'd also have a very hard time describing a mechanism that could allow "crumpling" to exist in the space-time of our universe which exactly matches the observed gravitational effects in galaxies. We can easily see that they appear to have a halo of matter which follows something akin to an NFW halo or else an Einasto profile. If you can find away that crumpled space-time would naturally form to match these profiles and somehow not be smoothed out by hyper-inflation, I'm all ears.


How to tell between annular and total eclipse (directly)? An annular eclipse is similar to total eclipse except our Moon "apparently" appears smaller than the Sun, my question is since in both cases there will be a glowing ring forming... so how to tell them apart?

There is a massive difference in the brightness of the "ring". During a total eclipse, the photosphere of the sun is completely hidden and the corona comes into sight. The corona is much much less bright than the photosphere of the sun, so during a total eclipse, the sky becomes dark.

During an annular eclipse the photosphere is never completely obscured. The bright 'surface' of the sun is still visible, so the sky never becomes completely dark, the sun remains dangerously bright, and the corona is not visible. An annular eclipse is like the partial phase of a total eclipse, because part of the sun is still visible.


IRAF Like software that work on Window My professor asks to me install IRAF for the photometry project but We have a hard time installing it and also I'm more comfortable with the Windows environment (because there is other software that not in Linux that currently installed on my pc). So I'm looking for software that substitutes IRAF and works on windows. It would be a great help.
note: While comments have proposed adding a Linux environment and running there, I'm asking for an IRAF alternative that runs directly in Windows.

I followed the suggestion given by @planetmaker but still some of the things that I do not understand as I'm new to the Linux environment. Hope someone will help me: If you go to IRAF, then I have followed all the step: The last step I have executed is ./install this leads me to

Welcome to the IRAF installation script.   This script will first
prompt you for several needed path names.   Once  the  installation is
complete, you will be allowed to do some minimal system configuration.

For each prompt: hit  to accept the default value, 'q' to quit,
or 'help' or '?' to print an explanation of the prompt.

The next steps from the installation guide, I do not understand it. Can any help me though this?

It looks like @planetmaker's suggestion of using WSL will work.
On my Windows machine I've had Ubuntu 18.04 installed for a while, so have tested using that; but this should work with the latest version of Ubuntu as well.
Here goes:

Start Ubuntu from the Windows Start Menu
In the Ubuntu terminal window which appears, type sudo apt install iraf
The sudo command will ask for your password, so type it in and press enter.
Now you'll see a whole load of text, like this:

Reading package lists... Done
Building dependency tree
Reading state information... Done
The following additional packages will be installed:
  blt cpp-7 fontconfig-config fonts-dejavu-core g++-7 gcc-7 gcc-7-base iraf-noao javascript-common libasan4 libblas3
  libcfitsio5 libcilkrts5 libcminpack1 libdrm-amdgpu1 libdrm-common libdrm-intel1 libdrm-nouveau2 libdrm-radeon1
  libdrm2 liberfa1 libfontconfig1 libfontenc1 libfuntools1 libgcc-7-dev libgfortran4 libgl1 libgl1-mesa-dri
  libgl1-mesa-glx libglapi-mesa libglvnd0 libglx-mesa0 libglx0 libice6 libjbig0 libjpeg-turbo8 libjpeg8 libjs-jquery
  libjs-jquery-datatables liblapack3 libllvm10 libpciaccess0 libsensors4 libsm6 libstarlink-ast-err0 libstarlink-ast0
  libstarlink-pal0 libstdc++-7-dev libtcl8.6 libtiff5 libtk-img libtk8.6 libubsan0 libwcs5 libwcstools1 libx11-xcb1
  libxaw7 libxcb-dri2-0 libxcb-dri3-0 libxcb-glx0 libxcb-present0 libxcb-shape0 libxcb-sync1 libxcomposite1
  libxdamage1 libxfixes3 libxft2 libxi6 libxinerama1 libxmu6 libxpa1 libxpm4 libxrandr2 libxrender1 libxshmfence1
  libxss1 libxt6 libxtst6 libxv1 libxxf86dga1 libxxf86vm1 python3-astropy python3-numpy python3-pluggy python3-ply
  python3-psutil python3-py python3-pyraf python3-pytest python3-pytest-arraydiff python3-pytest-astropy
  python3-pytest-doctestplus python3-pytest-openfiles python3-pytest-remotedata python3-stsci.tools python3-tk saods9
  saods9-doc saods9-tclpackages tcl tcl-signal tcl-xpa tcl8.6 tcllib tclxml tk tk-html1 tk-mpeg tk-table tk8.6
  tk8.6-blt2.5 tkblt tkcon tksao x11-common x11-utils xbitmaps xterm
Suggested packages:
  blt-demo gcc-7-locales g++-7-multilib gcc-7-doc libstdc++6-7-dbg gcc-7-multilib libgcc1-dbg libgomp1-dbg libitm1-dbg
  libatomic1-dbg libasan4-dbg liblsan0-dbg libtsan0-dbg libubsan0-dbg libcilkrts5-dbg libmpx2-dbg libquadmath0-dbg
  iraf-dev apache2 | lighttpd | httpd lm-sensors libstdc++-7-doc libtk-img-doc libxml2-utils python-astropy-doc
  python3-astropy-affiliated python3-beautifulsoup python3-h5py python3-matplotlib python3-pandas python3-scipy
  python3-tz gfortran python-numpy-doc python3-nose python3-numpy-dbg python-ply-doc python-psutil-doc subversion
  ipython3 python3-urwid tix python3-tk-dbg python3-pyds9 xpa-tools tcl-tclreadline tcllib-critcl mesa-utils
  xfonts-cyrillic
The following NEW packages will be installed:
  blt fontconfig-config fonts-dejavu-core iraf iraf-noao javascript-common libblas3 libcfitsio5 libcminpack1
  libdrm-amdgpu1 libdrm-intel1 libdrm-nouveau2 libdrm-radeon1 liberfa1 libfontconfig1 libfontenc1 libfuntools1
  libgfortran4 libgl1 libgl1-mesa-dri libgl1-mesa-glx libglapi-mesa libglvnd0 libglx-mesa0 libglx0 libice6 libjbig0
  libjpeg-turbo8 libjpeg8 libjs-jquery libjs-jquery-datatables liblapack3 libllvm10 libpciaccess0 libsensors4 libsm6
  libstarlink-ast-err0 libstarlink-ast0 libstarlink-pal0 libtcl8.6 libtiff5 libtk-img libtk8.6 libwcs5 libwcstools1
  libx11-xcb1 libxaw7 libxcb-dri2-0 libxcb-dri3-0 libxcb-glx0 libxcb-present0 libxcb-shape0 libxcb-sync1
  libxcomposite1 libxdamage1 libxfixes3 libxft2 libxi6 libxinerama1 libxmu6 libxpa1 libxpm4 libxrandr2 libxrender1
  libxshmfence1 libxss1 libxt6 libxtst6 libxv1 libxxf86dga1 libxxf86vm1 python3-astropy python3-numpy python3-pluggy
  python3-ply python3-psutil python3-py python3-pyraf python3-pytest python3-pytest-arraydiff python3-pytest-astropy
  python3-pytest-doctestplus python3-pytest-openfiles python3-pytest-remotedata python3-stsci.tools python3-tk saods9
  saods9-doc saods9-tclpackages tcl tcl-signal tcl-xpa tcl8.6 tcllib tclxml tk tk-html1 tk-mpeg tk-table tk8.6
  tk8.6-blt2.5 tkblt tkcon tksao x11-common x11-utils xbitmaps xterm
The following packages will be upgraded:
  cpp-7 g++-7 gcc-7 gcc-7-base libasan4 libcilkrts5 libdrm-common libdrm2 libgcc-7-dev libstdc++-7-dev libubsan0
11 upgraded, 108 newly installed, 0 to remove and 277 not upgraded.
Need to get 103 MB of archives.
After this operation, 526 MB of additional disk space will be used.
Do you want to continue? [Y/n]

Press enter to accept the default (capitalised) answer of "Y"
Watch as it downloads and installs over one hundred packages...
When it's done then you can run the new program by typing irafcl and pressing enter:

This package contains tasks for viewing and manipulating 3d images.
It is a pre-release version, and does not reflect the ultimate
partitioning of n-dimensional image tasks within IRAF

  NOAO/IRAF Debian 2.16.1+2018.03.10-2 Wed, 14 Mar 2018 09:02:35 +0000

  This product includes results achieved by the IRAF64 project in 2006-
  2009 directed by Chisato Yamauchi (C-SODA/ISAS/JAXA).

  Welcome to IRAF.  To list the available commands, type ? or ??.  To get
  detailed information about a command, type `help <command>'.  To run  a
  command  or  load  a  package,  type  its name.   Type  `bye' to exit a
  package, or `logout' to get out  of the CL.

  Visit http://iraf.net if you have questions.

  The following commands or packages are currently defined:

      dataio.     images.     lists.      obsolete.   proto.      utilities.
      dbms.       language.   noao.       plot.       system.

ecl>

Now, although it's installed and runs its command-line interface, I think that you'll encounter various issues when trying to run it: among the packages installed were some which require X Windows, and those won't work when running under Windows (you'll see error messages talking about being unable to access devices, or "Can't open display. DISPLAY is not set").
I advise you to test anything 'graphical' - that is, anything that tries to draw graphics to a display - before you invest too much time.
I think you might end up having to install a virtual machine instead.


Daylight Time Expansion I live in northern hemisphere around 44th latitude. Starting from Winter Solstice, December 21, 2018 the duration of daylight is naturally increasing. What I observe, however, is that the duration increase is happening more in the evening rather than in the morning. In other words as days pass, the sunset seems to move forward much faster to a later hour than the first morning light moving backward to an earlier hour.

Is what I am observing a true phenomenon or is it merely a wrong observation? If so why is this happening in an uneven fashion?

EDIT

I found the sunrise/sunset times for my area. It in fact confirms the situation. In the first half of January sunrise moved back only 3 minutes while sunset moved forward 15 minutes with respect to civil time.

This phenomenon is the equation of time, the annually varying difference between apparent solar time (what a sundial indicates) and mean solar time (related to civil time).
It has two components:

The eccentricity of the Earth's orbit around the Sun causes a cycle with 7.5-minute amplitude and 12-month period, with zeroes at perihelion (Earth moving fastest and closest to the Sun) and aphelion (Earth moving slowest and farthest from the Sun).
The oblique angle between the Earth's equator (rotation plane) and ecliptic (orbital plane) causes a cycle with 10-minute amplitude and 6-month period, with zeroes at equinoxes (Sun crossing Earth's equator) and solstices (Sun farthest north or south of the equator). This answer explains how that works.

These combine so that apparent solar time is 17 minutes ahead of mean solar time in early November and 14 minutes behind in mid February.
Between those times of year, the apparent Sun falls several seconds behind the mean Sun each day, affecting both sunset and sunrise times as you observed.

For further explanation, see the US Naval Observatory or Wikipedia.


Does anybody know how to get Sun-Jupiter Lagrangian points L4 and L5 RA and Dec? I wanted to point at the L4 and L5 of Sun-Jupiter to find Trojan asteroids but I don't know how to calculate the RA and Dec of these.
Do you know of any software that does it for you or how to calculate them?

The L4 and L5 points are unstable orbits. The Trojans move in perturbed (ie slightly non-keplerian) orbits that move them around the L4 and L5 points. At any time you will find Trojans in a pair of lobes between about 30 and 90 degrees ahead of Jupiter.

You can find individual asteroids using NASA Horizons ephemeris
For example 624 Hektor, the largest Trojan (and placed in the Greek Camp around L4) will be found at RA 20 47 54.08, Dec -28 04 20.1 on Feb 10 2018. And 884 Piramus (in the Trojan Camp, L5) will be at RA 10 43 51.39 Dec +01 04 11.3.

(Jupiter is between them at RA 15 18 55.32 Dec -17 07 47.1)

Neither are easy to find, with a Magnitude of 15 or 16, due to their greater distance than main belt asteroids, and very dark colour. The image below shows the distribution of Trojans around Jupiter.


Why does each and every planet of our solar system have an unique axial tilt angle? Why do planets have an axial tilt? From the above image we can see that each planet's axial tilt angle varies and differs from the others.

What was the cause of this, was this from the beginning of our solar systems birth? Or did it happen later?

In the early stages of the formation of the solar system, planetesimals start condensing and everything rotates with angular momentum inherited from the collapsing cloud of gas and dust, so the planetesimals all have their orbit and spin axes closely aligned with that of the proto-Sun. And while a planetesimal continues to grow by attracting nearby material its spin orientation isn't affected very much.

But eventually the planetesimals get large enough that they start attracting one another, and collisions occur. These collisions can disturb the spin axis of a planetesimal, especially if the collision is off-centre, and if the two bodies have roughly similar momentum.

Big planetesimals grow faster than small ones, and are less likely to have their spin axis disturbed. So proto-Jupiter manages to capture most of the matter that doesn't end up in the proto-Sun, and suffers little disruption to its spin axis. The smaller planets aren't so lucky.

That's the rough picture, and it misses a lot of details. Some of the details are still unclear, and more work is needed to clarify them.

The biggest mystery is Uranus. We don't know why Uranus is rolling around on its side, but we presume that was caused by some interaction fairly late in its history because its major moons orbit in its equatorial plane, pretty much. One theory is that it suffered collateral damage in whatever event upset Saturn's spin axis. Note that Saturn's rings and large moons orbit in Saturn's equatorial plane.

Mercury has a short orbital period and hence a high orbital speed. It's tidally locked to the Sun, so that it spins 3 times on its axis in 2 revolutions around the Sun; that tidal locking also locks it's spin axis fairly closely to its revolution axis.

Venus is also a bit of a mystery. It was formerly believed to be tidally locked, with its year equal to its day, but actually its rotational period is about 243 (Earth) days, about 20 days longer than its year, and it's spinning backwards! We don't know why, but suspect it may be due to a strong coupling between it's surface and very dense atmosphere. Also note that Venus experienced planet-wide severe tectonic activity not so long ago (on an astronomical timescale), so its surface is relatively new, and a lot of the planet's angular momentum may have been transferred to its atmosphere.


Historical Day Lengths on Earth I recently read something about a major earthquake in Japan having possibly shortened Earth's day due to an increase in rotation. Here's an example:

https://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html

When I read some more about Earth's day length, I learned that the length of Earth's day is believed to have been shorter in the past and getting longer over time. Here's an example from this site:

What was the length of the solar day 73 million years ago?

As some may know from an introductory psychology course, humans tend to settle into a day that is longer than 24 hours when placed in an environment with no external time or light cues. For most people, their circadian system is able to adjust their day to the 24-hour Earth day. Some blind people lose this ability, as seen from commercials about medication to help them gain a 24-hour rhythm. Here is some background about that:

https://en.wikipedia.org/wiki/Non-24-hour_sleep%E2%80%93wake_disorder

At the risk of asking an unacceptable "opinion" question, how convincing is the evidence of historical Earth day lengths? Is the evidence at a point where shorter Earth days in the past are an accepted certainty? Or are things still at a point of being a best guess without really knowing? It would seem rather odd for our nervous systems to have evolved with a preference for days longer than 24 hours. It is certainly possible, but I would appreciate some input here before spending time contemplating why evolution might have worked that way. I'm only a hobbyist, so there would be no need to go into too much depth for me specifically. Thanks in advance.

Edit: The current answers and comments are interesting and appreciated, but to clarify, the intended focus is the extent and quality of the astronomical evidence that could be used in a discussion (elsewhere) about circadian issues.

From a physics point of view, there's no obvious mechanism that could provide significant, consistent increases in the earth's rotation speed.

The moon's gravitation appears to be the main driver of changes to the earth's rotation.  And since it revolves more slowly than the earth's rotation, it tends to pull energy from the rotation, lengthening the day.  This is completely consistent with the evidence that points to ancient years on earth having more (and therefore shorter) days than is true in the modern era.   This is also consistent with the observed gradual recess of the moon away from the earth.

It is also possible to directly measure the length of a day and see that days on earth are significantly longer than they were in 1900.  Again, consistent with a slowing rotation.

The implications of this for the evolution of sleep cycles would indeed be better in another location (perhaps Biology SE).  But I'd suggest that this is a very slow process.

Remember that this is something that increases the length of the average day.  But there are lots of other variations between days during a year.  An organism might for instance synch to something like sunrise.  But in spring, sunrises come more than a minute earlier than 24 hours apart each day, and in fall they come more than a minute later than average each day.

Meanwhile, it was probably more than a million years ago that the average day was shorter by a full minute.  So the increase in day length over time is definitely happening, but at a rate that is almost irrelevant to biology.


What could I resolve with my Telescope I recently won a Common 50mm refracting telescope with Solar Filter too.. I'm trying to Understand things I can skygaze with it. It's pretty basic and I should have known. Unfortunately, I don't possess this knowledge now. I had searched on various good websites which detailed and other requisite! Given, I stay in Navi-Mumbai, India, and less cloudy and less dusty area, what would my telescope's eye see?

And how to maintain and keep the Telescope in a good state.

What things would make me a better amateur-observational-astronomer?

The Moon is the easiest target to start with.
After that, try whichever bright planets are currently in the sky: Jupiter, Saturn, Mars, Venus.
Then get a star map and try to find some of the brighter star clusters, nebulas, and galaxies.

I recommend the monthly sky maps and object lists at Skymaps.com.
Latitude 19°N falls midway between their northern and equatorial editions, so try the equatorial one and remember that your southern horizon hides the southernmost ~10% of the map.
Most of the "Easily Seen with Binoculars" list are also good targets for a 50mm telescope.
Objects labeled e.g. "M42" are in the Messier catalogue.
Some objects are easier to find with a more detailed star atlas.

A telescope calculator indicates that in ideal conditions, 50mm aperture can theoretically show stars as faint as magnitude 11 or resolve double stars 2.5 arcseconds apart.
In practice it's reasonable to try objects as faint as magnitude 9 or as small as 4 arcseconds.

If the telescope has a finder or other aiming device, check its alignment periodically.
No other maintenance is required if you store the scope in a clean, dry place and use the dust caps to protect the lenses from dust and fingerprints.
A few specks or spots will not significantly degrade performance.
Only clean it if it gets very dirty, and then take care not to scratch or strip any lens coatings.

As you learn your way around the constellations and acquire more challenging targets, you can sharpen your observing skill by keeping a journal in which you describe or even sketch what you see.
A local astronomy club can give you ideas for observing projects and opportunities to try more advanced equipment.

Solar safety note: filters installed near the eyepiece can overheat and break; filters which securely cover the whole objective and make everything else invisible are usually OK.
With a 50mm aperture, the eyepiece projection method is a good alternative.


Should we expect clues from a gamma ray burst causing massive extinctions on Earth? I remember reading an article on the subject highlighting the periodicity of this event throughout history supposing it could be caused by the rotation of the sun around the galaxy.

My question is: Do we have any clues that can confirm this hypothesis? What should we expect to see? Shouldn't it leave any remains?

A gamma ray burst could cause an extinction event. There is no evidence that this has ever happened to the Earth.

Gamma ray bursts are rare events, we don't know exactly how rare, but to cause a mass extinction a GRB would need to be near (6000 light years or less is near in this context) and pointed right at Earth. GRBs are rare enough for it to be possible that this has never happened. It is possible that one or two of the extinctions in the last billion years were due to GRBs

The mass extinction that is most likely to have been caused by a GRB is the Orodvician-Silurian Event, 444 million years ago. There is no positive evidence of a GRB, but it can't be excluded. The best that can be said is that the pattern of extinction is consistent with a GRB.

A gamma ray burst would leave few signs that could be used to prove it occurred. The radiation would strip the ozone layer, leaving the surface vulnerable to UV radiation. But it wouldn't leave a crater, or a geological layer.

Gamma ray bursts can occur when a hypernova occurs. There is no reason to think that the Earth is more vulnerable to a hypernova when it is in the galatic plane. The evidence of a 26 million year periodicity to mass extinctions is weak, and there is no evidence that closely ties this to the sun's orbit around the milky way. There is good reason to tie other extinctions to particular impacts, for example there is no need to hypothesise that a GRB occurred 65 million years ago, at the K-T mass extinction

Gamma ray bursts are one of the less likely causes of a mass extinction. Researcher David Thompson likens them to the danger of finding a polar bear in his closet. It could happen; it would be dangerous, but the chance of it happening is so low that it is not a risk worth worrying about.


How do we know dark matter/dark energy exists? I've never quite understood the theory behind why dark matter and dark energy exist. I know it has something to do with gravitational pull being stronger than what we calculate it SHOULD be, could someone explain this in more detail?

Some additions to the answer of MBR:

In fact, we do not know that dark matter and dark energy do exist, but we have indirect clues. You will often see claims that dark matter and dark energy are two of the major problems of cosmology today, including by professional astronomers, but this is an epistemological misconception: you cannot call a hypothesis a problem.

Currently we have several observational anomalies (as defined per T. Kuhn): namely the rotation curve of galaxies, the luminosity of supernovae, the power spectrum of the CMB... (and some additional things like the difficulty to measure G or to detect directly gravitational waves).

According to the history of science, we have 3 ways to solve these anomalies:

ontological approaches (dark matter and dark energy)
legislatives approaches (extensions of General Relativity)
paradigmatic approaches (inhomogeneous Universes)

Dark matter and dark energy are two ontological approaches that seems to work pretty well: by adding two simple components we can fit a wide range of observations. This is why astronomers like them: their modelization are relatively easy and allow to fit everything well. But for the moment we do not know their nature.

Some say that with WMAP (Wilkinson Microwave Anisotropy Probe) and Planck space observatory we have entered in the era of precision cosmology: but this is only true if our paradigm of dark matter/dark energy approaches is correct. If this is not the case, we currently do not interpret correctly what we measure, but ultimately more accurate measures can lead to a better view of current cosmological anomalies.

So dark matter and dark energy are two elegant and minimalist approaches to several fundamental problems. This is a very hot topic in cosmology, but we may realize, one day, that we only had problems because our current paradigm was erroneous.


When exactly star Sirius shines into the queen's chamber of Great Pyramid? I read that Great Pyramid was built to resonate with Sirius by aligning an airshaft that connected to queen's chamber. That is, the light of Sirius going through the shaft would be able to shine into the queen's chamber.

When exactly star Sirius shines into the queen's chamber of Great Pyramid?

The premise of this question is wrong: the "Sirius" shaft was not designed to point towards Sirius. Most of the shafts bend (sometimes several times), making it impossible for the light of a star to shine through. Furthermore, the angle of the shaft is not constant, making it possible for the shaft to be pointed towards tens of different stars, depending on which portion of the shaft you want to look at.

The Sirius shaft that leads to the queen's chamber faces south. It is at a 39.6° angle and the width of the window that it provides is 0.18°.

Due to precession, the maximal elevation of stars changes over time. Based on this document, Sirius would have been visible through the shaft between 2355 BC and 2295 BC. This is after the construction period of the pyramid (around 2550 BC), putting the idea that the shaft was "built to resonate with Sirius" even more in question.


Is there any phenomenon where a supernova doesn't anything leave behind? I have been researching this on google, and I think there is a process. What is this process called? I have forgotten.

You are probably thinking of a pair-instability supernova. Such a supernova releases more energy than the star's gravitational binding energy, completely disrupting the star and dispersing the contents.  Meaning that with nothing else in the universe, over infinite time every particle from the star will get infinitely far away from any/every other particle.  The mass, energy, etc. of the star doesn't vanish from the universe, though; it just gets spread throughout it.  But there will be no dense remnant left behind: no dwarf, no neutron star, no black hole; over long time scales there won't even be a discernible nebula remaining (though note the star may have shed off significant amounts of material prior to this event, which would interact with the ejecta and may result in some sort of permanent structure suggesting a star used to be there).


Could a dwarf galaxy host a star at the center instead of a SMBH? I don't think this would be possible, but I'm curious, and as what ever query I googled I only found posts about black holes indeed existing, what black holes are, that even the Milky Way has a black hole in its center or that We can't know if dwarf galaxies have non-SMBH at their center.

So sorry if it is a stupid question but I didn't want to go on with assuming something without knowing: So is it theoretically possible for a Galaxy to have a star at its gravitational center of mass?

A galaxy's center of gravity is not determined by the most massive object, but by all objects in the galaxy. Even supermassive black holes (SMBHs) do not dominate the gravitational field except very, very close to the center. By far, most of the stars in a galaxy couldn't care less about the SMBH.

The region within which a BH dominates over that of the stars (the "sphere of influence"$^\dagger$) is given by (e.g. Peebles 1972)
$$
r = \frac{G M_\mathrm{BH}}{\sigma^2},
$$
where $G$ is the gravitational constant, $M_\mathrm{BH}$ is the mass of the black hole, and $\sigma$ is the velocity dispersion.

In the Milky Way (MW), there's a SMBH (Sagittarius A*) of roughly $M_\mathrm{BH} \simeq 4\times10^6\,M_\odot$. In that region, the stellar velocity dispersion is roughly $50$–$100\,\mathrm{km}\,\mathrm{s}^{-1}$ (e.g. Genzel et al. 2010). Plugging in those values, you'll find that Sgr A* dominates the kinematic out to roughly 3 pc, or 10 lightyears, which is nothing compared to MW's radius of $\sim10^5$ lightyears.

If you take the most massive conceivable stars (a hypothesized Pop III star of $M\sim10^3\,M_\odot$) in Willman 1, the smallest known dwarf galaxy — which has stellar velocity dispersion of the order $5$–$10\,\mathrm{km}\,\mathrm{s}^{-1}$ — you'll find that such a star will dominate the gravitational potential out to a distance of only $\lesssim0.1\,\mathrm{pc}$, again completely negligible compared to the galaxy's radius of $\sim25\,\mathrm{pc}$.

In other words, although it's possible for a dwarf galaxy to host a massive star, it will just be a star like all the others, and will in no way define the galaxy's gravitational center.

$^\dagger$Not to be confused with the event horizon which is even smaller.