Is there an orbit that could cause regular debris showers like in the movie Gravity? [some spoilers]

If you were in the ISS and orbiting the Earth roughly every 90 minutes, would it be possible for debris to be orbiting such that it passes you every 90 minutes?

While Neil DeGrass Tyson pointed out scientific flaws in the movie on twitter, he praised many things the movie got right in interviews, including "The 90-minute orbital time for objects at that altitude."  I think I recall an interview on a podcast that I can't remember where he specifically said that referring to the debris recurring every 90 minutes.

In reality wouldn't the debris have to be in a different orbit?  If the ISS orbits every 90 minutes (92.87 really but I'm saying 90) then after 90 minutes they would be in the same relative position.  The Earth would have rotated in the mean time so they wouldn't be over the same point on Earth, but they would be in the same position in orbit about the Earth.  Is there an orbit that would allow the same debris to cross the same spot in their orbit every 90 minutes?  Would it be nearly the same orbit, just inclined slightly?  Would that explain why the debris was moving so slowly relative to the ISS as well?

note:

There are problems with the orbits and location in the movie, I simplified my question to ask about the particular orbit of the ISS.

If I understand you correctly (and I haven't seen the movie ^_^), you want to know if it is possible for two different objects in orbit to come close to each other periodically, with roughly the same period as one of the object's orbital period.

Yes, that is possible, but not probable.

Obviously, if the objects are in the same orbit, they are always close to each other
If they are in slightly different orbits, one of the objects will seem to move about the other object in regular or irregular patterns, depending on orbit specifics
If they are in completely different orbits, the necessary conditions are

the orbital period of both is equal
they have at most 1 point in common (or at least very closely approach each other)
the phasing of both objects is such that the objects are in the close approach region at the same time.

In ordinary two-body Keplerian celestial mechanics, an orbit's semi-major axis is the only orbital element that determines the orbital period. Therefore, one orbit may be circular and the other elliptic and inclined in all sorts of ways, but as long as the semi major axis of both is the same, they will have the same period.

However, this whole scenario is completely unstable for any real-world celestial body. The (relatively small) asphericity of the Earth, as well as the presence of the Moon for example, causes all orbits around it to drift in some way or another. If you put two objects in orbits like you describe, they will not likely meet up very often before never seeing each other again for centuries.


What are the current accepted theories of the formation of the Uranian moon Miranda? Miranda, a moon of Uranus is unique in that it has a very fractured surface

Source: University of Oregon

The surface is said to be jagged and fractured, with comparatively large disjointed cliffs and great faultlines.

What are the current accepted theories for the formation of such a strange world?

Please include authoritive links in any reply

As Miranda is a rather small object (only 235km diameter) it may simply have not enough mass to make it matter settle into hydrostatic equilibrium under the influence of its gravity, thus it basically retains the irregular shape most asteroids have. This nice arxiv paper discusses the mass limits necessary for a stellar body to settle into a nice round shape:

http://arxiv.org/ftp/arxiv/papers/1004/1004.1091.pdf


How big is nebula dust? Whenever I see the term dust for the particles of a nebula, I ask myself whether this is actually based on some reliable measure of particle size. Of course we can all agree it looks like dust from this distance. And we all know how such appearances often lead to assumptions which are incorrect and whose incorrectness we are often blind to.

Is there any reliable evidence by which this "dust" is known not to be pebbles, rocks, boulders etc.? Which in reality would seem vastly more likely.

How might we measure size from such distances?

The size of cosmic dust grains is in general given not by some size, but by a size distribution. The only direct measurements of such a distribution are made on dust collected on plates of satellites, which is of course a very local measurement. When we think that distributions look similar — though not exactly alike — in other locations of the Universe, so far away that we will never be able to go there, it is because we know how a given distribution and composition affects light traveling through an ensemble of dust grains.

A particle has a given probability of absorbing a photon with a given wavelength, and in general this probability peaks around wavelengths of the order of the size of the particle (for small particles). Thus, if we know how the spectrum of some light source looks if there were no dust around (and we do for many sources, e.g. stars), then the difference between the known intrinsic spectrum and the observed spectrum can be modeled assuming some distribution and composition. Often the composition needs not be assumed, but can be constrained from the emission of the dust at infrared wavelengths.

Usually, the model that fits best the observations is a steep power law of the form $P(r)\propto r^{-a}$ with an index of roughly $a\sim3.5$; that is, the probability of finding a small grain is much larger than the probability of finding a large grain (more precisely, for $a=3.5$, grains of size $r=x$ is $10^{3.5}\simeq3\,000$ times more common than grains of size $r=10x$, and $10^7\!$ more common than grains of size $r=100x$).

For very large grains, the probability of absorbing a photon becomes independent of the wavelength of the photon. Whereas the small grains as described above have a "color preference", large grains are said to be "gray". This is the case for boulders, rocks, pebbles, and even sand-grain-sized particles. Thus, if a cloud consisted of such particles, the spectrum of a background source would simply be diminished by a constant factor at all wavelengths. This is very rarely observed — rather the sources are diminished much more at the short wavelengths than at the long wavelengths, as expected if there are more small grains than large grains.


How far should the source be, for the gravitation waves to be visible on Earth? If I understand it well enough, gravitational waves cause ripples in the space-time fabric. To our eyes, that would look like those rippling-effects on amateur videos.

Where does a source have to be, in order for the effects of gravitational waves to be visible, here, on Earth.

The waves pass by at the speed of light. So you you would'nt see ripples, they would pass too fast, and remember the waves would be passing through you too. The wavelength was (relativly) long about 3000km. The wave doesn't pass you, you are inside the wave.
The amplitude of the waves detected by LIGO was small, one part in $10^{21}$, Now while the intensity of the waves follows an inverse square law, because intensity is proportional to amplitude squared, the amplitude is inversely proportional to distance from source.
We are familiar with strains that are one part in 1000, that is a variation of one millimetre for every metre. To get that we would have to be a lot closer to the source, $10^{18}$ times closer. Since the black holes were about 1 billion light years distant, to get a strain of 1 part in 1000, the black hole merger would have to be about 10000km away
Now if you are 10000km from a merging pair of 30 solar mass black holes, the strain of gravitational radiation is the least of your problems. If it is any consolation, your problems won't lost long.


What would the sky have looked like when galaxies first started to form? If I stood on a theoretical earth at the time that galaxies were just starting to form, what would the sky have looked like?

Due to the smaller size of the universe would it have been densely packed with objects? If yes, what might I have seen?

In early times of galaxy formation you would have seen a sky similar to this Hubble Extreme Deep Field image. A sky filled with small, irregularly shaped, young galaxies and galaxy predecessor fragments (sub-galaxies). Galaxies probably formed bottom-up from smaller fragments.

Further back in time it would have been dark. Hydrogen clouds would begin to contract, as predecessors of first stars (population III) and clusters of stars.

Even further in the past (redshift between 300 and 400) the sky would have been glowing red by what we now call cosmic microwave background.


What should the mass of a planet be in-order for its escape velocity to be near the speed of light? What should the mass of a planet be in-order for its escape velocity to be the speed of light?  Is it even possible? What will it look like from an outside viewer? Will it even be visible in the human eye spectrum?

As @KeithMcClary says in his comment, escape velocity depends on both mass and radius. The smaller the radius for a given mass the higher the escape velocity. So if you could somehow compress the Earth until it was just a few centimeters across, its escape velocity would approach the speed of light.
At the other extreme, if you filled a space twice the diameter of Earth's orbit with copies of the Earth, without doing any compressing at all, it would already be a black hole.
If you just pile matter onto a planet and let its own gravity compress it, you need about 2 solar masses before it is close to being a black hole (at which point it is 10-20 km across).


Tests for code that computes two-point correlation function of galaxies I'm currently writing a code to compute the two-point correlation function of a set of galaxies. As I work with a loooot of galaxies, it was suggested to me not to do it in real space, but to compute the power spectrum in Fourier space and get the correlation from the power spectrum, as is explained e.g. in appendix B in this paper (arXiv:1507.01948).

What I need are some very simple test cases with known results to test my code at each step in order to ensure that I really get what I want. Can anybody point me to any such simple tests? Or at least to a already existing code (preferably in Python or Fortran) that does exactly this, so I can cross-compare the results?

A very simple example is to fill a 3d density with a regular plane wave, which should result in a single peak in the power spectrum. Here is a simple python scripts that does exactly that.

Computing the two point correlation function should be only one inverse fourier transform of the Power spectrum away. (Depending on which FFT library you use, you might need to renormalize the results. FFTW and numpy.fft for example have unnormalized fourier transforms: $F^{-1}[F[f(x)] = Nf(x) $, where $N$ is the number of samples.)

#!/usr/bin/python3

import numpy as np
import matplotlib.pyplot as plt

#==================================
def main():
#==================================

    nc = 128                # define how many cells your box has
    boxlen = 50.0           # define length of box
    Lambda = boxlen/4.0     # define an arbitrary wave length of a plane wave
    dx = boxlen/nc          # get size of a cell

    # create plane wave density field
    density_field = np.zeros((nc, nc, nc), dtype='float')
    for x in range(density_field.shape[0]):
        density_field[x,:,:] = np.cos(2*np.pi*x*dx/Lambda)

    # get overdensity field
    delta = density_field/np.mean(density_field) - 1

    # get P(k) field: explot fft of data that is only real, not complex
    delta_k = np.abs(np.fft.rfftn(delta).round())
    Pk_field =  delta_k**2

    # get 3d array of index integer distances to k = (0, 0, 0)
    dist = np.minimum(np.arange(nc), np.arange(nc,0,-1))
    dist_z = np.arange(nc//2+1)
    dist *= dist
    dist_z *= dist_z
    dist_3d = np.sqrt(dist[:, None, None] + dist[:, None] + dist_z)

    # get unique distances and index which any distance stored in dist_3d
    # will have in "distances" array
    distances, _ = np.unique(dist_3d, return_inverse=True)

    # average P(kx, ky, kz) to P(|k|)
    Pk = np.bincount(_, weights=Pk_field.ravel())/np.bincount(_)

    # compute "phyical" values of k
    dk = 2*np.pi/boxlen
    k = distances*dk

    # plot results
    fig = plt.figure(figsize=(9,6))
    ax1 = fig.add_subplot(111)
    ax1.plot(k, Pk, label=r'$P(\mathbf{k})$')

    # plot expected peak:
    # k_peak = 2*pi/lambda, where we chose lambda for our planar wave earlier
    ax1.plot([2*np.pi/Lambda]*2, [Pk.min()-1, Pk.max()+1], label='expected peak')
    ax1.legend()
    plt.show()

#==================================
if __name__ == "__main__":
#==================================

    main()


What's the difference between the apparent horizon and event horizon of a black hole? The Wikipedia page for apparent horizon is pretty sketchy and requires some GR knowledge. Is there any simple definition?

The black hole region of a spacetime is defined as a region where nothing can escape to infinity and an event horizon at a given time is the boundary of a connected region of space which is part of the black hole region. As you're after a simple answer I won't give a formal definition of a black hole or an event horizon, but they can be found in Wald.

The problem though is to know whether anything in a region of space can escape to infinity depends on precise knowledge of the future, also not all spacetimes have a suitable notion of infinity. Yet clearly even when we don't know the entire history of the spacetime or when there isn't a suitable notion of infinity, we can still identify objects that are functionally equivalent to black holes. The apparent horizon if you like is the spatial boundary of what we might consider to functionally be a black hole.

As nothing can escape a black hole, even light directed away from it is pulled back towards the singularity so we know that a black hole can cause even light directed outwards, relative to a point in space, to move inwards to that point. The apparent horizon is the boundary between where outwardly directed light moves outwards and where it moves inwards.

In Schwarzschild spacetime, or more generally Kerr-Newman spacetime (in standard coordinates) the apparent horizon and the event horizon coincide. However, more generally, the location of the apparent horizon depends on how you 'slice' spacetime (it is observer-dependent if you like - unlike the event horizon). Apparent horizons needn't be associated with (formally-defined) black holes, however as long as the spacetime has certain properties, they will indeed be associated with black holes and will lie at or inside the event horizon.


Will time that moon crosses meridian always be periodic? If I observe the moon pass my celestial meridian at a specific north polar clock 'time' will I observe every subsequent passage at ( 'current time' + 13.2* X 'days passed' mod 360* ) ? Does the moon pass celestial meridian at exactly the same time if observed from every point on this longitude?

Your formula is a crude approximation. It's ok if you just want a rough idea of the number of orbits the Moon makes over a long period of time, but it's not useful for calculating the daily motion of the Moon.

The Moon's orbit is fairly eccentric (about 0.0549), so its daily speed varies quite a bit. The Babylonians discovered that the mean daily motion is approximately 13° 10' 35", but that it can range from 11° 6' 35" to 15° 14' 35". The Moon's motion is rather complex, the Wikipedia article on lunar theory is a good introduction, with a rundown of the history, and quite a few relevant formulae.

The fact that the Moon's motion is so hard to compute accurately (compared to the motions of the Sun and the visible planets) is a mixed blessing. On the one hand, it would have been great for navigators if it had been easier to prepare Moon position tables. On the other hand, it made it rather obvious that early models of the celestial motions were incorrect. No matter what they tried, the early cosmologists weren't able to produce a model of the Moon's motions that was consistent with observation. Eg, in the best lunar theory of Ptolemy's model of the cosmos, which was good enough at estimating the Moon's position to use for eclipse prediction, the apparent size of the Moon ought to change by around 100% every month, which it obviously doesn't do. :) So if it weren't for the Moon's tricky orbit we might still be using Ptolemy's geocentric model of the cosmos today, and the whole scientific revolution may never have happened.


Calculation of hour angle I need to determine Right Ascension and Declination from Azimuth and Altitude, working in C#. The problem is that the formula for calculating hour angle, for some reason, doesn't work. Here's the code:

        az = az * DEG_TO_RAD;
        alt = alt * DEG_TO_RAD;

        lati = latitude * DEG_TO_RAD;

        // Julian day
        JD = CalculateJDN(year, month, day, h, m, s);

        // Greenwich mean sidereal time
        GMST = CalculateGMST(JD);

        LST = GMST + longitude / 15;

        dec = Math.Asin((Math.Sin(lati) * Math.Sin(alt)) + (Math.Cos(lati) * Math.Cos(alt) * Math.Cos(az)));

        ha = Math.Atan2(Math.Sin(az), (Math.Cos(az) * Math.Sin(lati) + Math.Tan(alt) * Math.Cos(lati)));

        ha = ha * RAD_TO_DEGREE / 15;
        dec = dec * RAD_TO_DEGREE;

        ra = LST - ha;

        // Input data for Mintaka (delta Ori):
        // az = 47.5, alt = -35.3 on 13:57 UTC, 1 Dec 2015
        // latitude = 43.897, longitude = 20.344
        // Required output: dec = -0.19, ra = 5.5
        // Given output:
        // dec = -0.19, ra = 17.5, ha = 2.5

Az and alt are given in degrees, so they are first converted into radians. Functions for calculating Julian day number and GMST are correct, since I've already tested them. Formula for declination is good, but for some reasons formula for hour angle (ha) doesn't work. I don't know where's the error.

I would suggest that this:

  LST = GMST + longitude;

line is wrong. 360 degrees of longitude correspond to 24 hours, so it should be something like:

  LST = GMST + longitude/15.0 ;

Assuming these are all in floating point and you are subsequently handling the potential overflow of time into the next day correctly...


When calculating absolute magnitudes of distant objects, do you use the light-travel distance or the comoving distance? I want to calculate the absolute magnitude of GN-z11. Its light-travel distance is 13.4 billion ly (4.1 Gpc), its comoving distance is 32 billion ly (9.8 Gpc), and its apparent magnitude is $25.8$. Calculating its absolute magnitude, we get $M=-76.86$ for light-travel, and $M=-94.09$ for comoving. Which one do I use? I am assuming it is the former. Is this correct, or is there a different formula for this?

Neither. You use the Luminosity Distance.
$$M = m -5\log D_L + 5$$
This assumes bolometric magnitudes. If you are trying to estimate it in some photometric band then you must also calculate and apply a K-correction that will depend upon the intrinsic spectrum of the source.
For $z=11.09$ this cosmology calculator gives a model-dependent luminosity-distance modulus of 50.39 (corresponding to $D_L=1.2\times 10^{11}$ pc).
NB: The distances you quote in the question do not give the absolute magnitudes quoted.


Satellite in orbit in front of and behind the Moon Would it be possible to put a satellite into the same orbit as the Moon, but far enough ahead or behind the Moon to remain in place?

Has this ever been done in practice?

Yes. Trojan orbits (60° before or after the Moon) are even stable in most cases. These points are called the L₄ and L₅ points, which are two of the five Lagrangian points.

These orbits are mostly impractical (nothing is there).

Today the most known is the Chinese Queqiao satellite, which is a communication relay to their probe on the far side of the Moon. Instead of using L₄ or L₅, Queqiao is in a halo orbit around another Lagrangian point; Earth-Moon L₂.


Hubble HST scheduling algorithm NASA developed SPIKE: Intelligent Scheduling of Hubble Space Telescope Observations (HST) and this.

My question is simple, is there a programming library, preferably in python or C#, to implement the algorithm? AFAIK  NASA never released their code.

There is apparently an option on the STSI (Space Telescope Science Institure) website to try their Spike implementation.

This doesn't clearly answer your question, as it's not open, and I don't know the language that has been used.

However, there is a lot of information in the documents you provided and on the linked website, so maybe you can start your own implementation of the algorithm? I'd be happy to help!


Distance between objective lens and eyepiece I am trying to make home telescope, I have a convex lens of focal length 1000mm and eyepiece (plano-convex) with 25mm focal length. What is the ideal distance between objective and eyepiece to see objects more than one kilometre away.

tl;dr: 1025 +1/-3 mm
If you were to focus at infinity, then the primary focus is 1000 mm from the primary lens and you would put the eyepiece lens one of its focal lengths away from that point to make a new parallel beam. That's 1025 mm. If your eyes can not focus at infinity without glasses, then wear them so that you can see the parallel beam and image it on your retina.
If you don't have your glasses or would like to try without, then your eye needs a diverging beam which means the eyepiece should be a little less than its 25 mm focal length from the primary focus, i.e. a tiny bit closer to the objective, perhaps 1 or 2 mm.
Now you mentioned that objects are over 1 kilometer away from your telescope. If they were exactly 1 km (1,000,000 mm) then the primary focus would be farther from the objective by a little bit.
The thin lens equation 1, 2 is
$$\frac{1}{d_{img}} = \frac{1}{f} - \frac{1}{f_{obj}}$$
so your image distance would then be 1001 mm, and so you'd have to move your eyepiece away from the objective by 1 mm to correct for that.


What inaccuracies are in this Stellarium depiction of the sky? I found this depiction on Stellarium of the sky as seen from near the Apollo 12 landing site on January 11, 2131. The earth is the black circle in the arm of the Milky Way.

Here is a similar view with labels.

Assuming a person had a naked eye view of this, and the most of the light coming directly from the sun could be attenuated, what inaccuracies are evident in this image? As I understand it the colors to the Milky Way could not be seen by the naked eye. I take it the size of the objects are illustrative of apparent magnitude and not the visible size. Are there any others?
If the "night-sky" tag is not appropriate because this is not the view from Earth please remove or let me know and I'll remove it.
Here is the full screen shot, at horizontal FOV of 90°.

In regards to the sizes of the sun and the earth, at a narrower FOV Stellarium shows the earth as being substantially bigger than the sun.

Let's get the ball rolling! Hopefully some of these are what you are looking for.
Point light approximation

I take it the size of the objects are illustrative of apparent magnitude and not the visible size

This is important, and it does not only appear to the planets. Everything in this image, except the Sun, the Earth, the Milky Way, and to some degree Venus are going to be point lights to the naked eye.
Computer monitors or paper are not very good at illustrating point lights. They have by far too few nits.
Instead, brightness is approximated by giving all the stars some visible area, proportional to their brightness. It's a useful approximation, but I would definitely count it as an inaccuracy compared to the naked eye.
Relative size of the Sun and the Earth
The Moon is famously the same relative size in the sky as the Sun. However, in this image, we are viewing the much larger Earth from the same distance, yet it still appears the same size as the Sun. The size of either or both has to be wrong
Sun glare
This image shows glare from the Sun. This will vary a lot depending on your eyes. And if the light form the Sun is blocked out like you propose, none of the solar glare in this image will be visible.
Light from the lunar landscape
Unless you also block out the surface of the Moon, the reflected light is going to severely impact the view.
Smooth edges of objects
Real life has no smooth anti-aliased edges. Cutoffs by planetary discs and stars are very hard, unless an atmosphere is present.
City lights on the dark side of the Earth?
Apollo astronauts could not see this, but admittedly, it was because of limiting observational conditions. Stellarium like does not simulate this
Image compression artefacts
These screenshots have two kinds of compression artefacts introduced by the JPEG format. Needless to say, these are not present in real life.

Banding, due to the limited number of dark colours available in the  YCbCr transform used by JPEG:

Macroblocks of 8x8 pixels, due to the discrete cosine transform

Colour space
All the way from the colour handling in Stellarium, to the YCbCr space of JPEG, to how your monitor is calibrated, decisions are made about colour. The end result may be far removed from how they really appear. Notably, Stellarium is likely considerably increasing the saturation of the colours of the Milky Way.
Black level
The black in this image is merely grey. Even the brightest parts of the Milky Way will appear darker than what can be found in the screenshot.
Field of view
"Personally I don't see in rectangles" as a comment says. The human field of vision is not rectangular, and it is almost twice as wide as the 90 degrees in this crop.
Brightness and colour sensitivity is also very different between the centre and edges of human vision.
mac status bar: this is not visible in real life :)


Is Jupiter a dwarf planet? I know the question seems silly, but as far as I've understood, is this the definition of a dwarf planet:

circling around a star, and not being a star itself
having a round shape (otherwise every comet or asteroid could be a (dwarf) planet)
not being able to clear its own orbit (otherwise it's a planet)

If we have a look at Jupiter, then we can easily confirm the two first points, but the third one is an issue: in Jupiter's orbit there are two sets of asteroids, called the Trojans, which are located, together with Jupiter, in some kind of regular triangle, turning around the sun. Following the laws of Kepler, Jupiter and the Trojans have the same speed, so Jupiter can't catch them and although Jupiter's gravity is very strong, it does not stretch to the other side of the sun, which means, following the definition, that Jupiter is a dwarf planet.

I know, Pluto is called a dwarf planet because of the many objects in its orbit, and Pluto's gravity is so small that it can't get rid of those objects. Nevertheless in a scientific you can't use vague wordt like "vicinity", "close", ..., so you can't say Pluto is a dwarf while Jupiter isn't because of the vicinity of the mentioned objects (being "too close").

Do I have a point here or am I missing one? :-)

I suggest taking a look to https://en.wikipedia.org/wiki/Clearing_the_neighbourhood .

Quoting it,

  In the end stages of planet formation, a planet (as so defined) will have "cleared the neighbourhood" of its own orbital zone, meaning it has become gravitationally dominant, and there are no other bodies of comparable size other than its satellites or those otherwise under its gravitational influence.

The Trojans are textbook example of objects under the influence of a planet (Jupiter).

For another example, Pluto is said to not have cleared its neighbourhood because Neptune is in it. Neptune has cleared its neighbourhood although Pluto is in it because Pluto is a lot smaller and Pluto orbit is governed by resonance with Neptune.


Defining a nautical mile Astronomy is new to me so my question might be stupid: I read that a nautical mile is defined as one minute of latitude along any line of longitude. What if it was the other way round, i.e. one minute of longitude along any line of latitude? Would it make any difference?

Yes, very much so.
The lines or circle of constant longitude always form a great arc which intersects with the poles; thus it always has the same length. A circle of constant latitude varies in circumference: the largest is at the equator while it has intermediate length at intermediate latitudes and no length anymore when you reach the pole.

See e.g. that image from wikipedia and follow a circle in longitude, thus from North to South. And compare the different sizes of different circles for different lattitudes (thus the circles which are parallel to the equator)
Mind though that the nautical mile is no longer defined via the fraction of the polar circumference of Earth. It is defined as exactly 1852m in length, and thus can in principle be determined as accurately as one can measure time (1852 / 299 792 458 seconds).
In earlier times it was a MUCH easier task to determine a difference in latitude as it directly relates to the (culmination) height of celestial objects above the horizon. It was VERY DIFFICULT to impossible to determine difference in longitude unless you had a very accurate clock and an almanach of the rise and set times of celestial objects.


Are there any electrons inside the sun / a star? Like the question says, are there electrons inside an active star, or has all the matter been so thoroughly ionized / plasmarized that the electrons are long gone? I ask because it would make sense when thinking about the magnetic field of a star; naked protons (neutrons are of no magnetic concern) would create a very one-way magnetic effect when circling inside the star, while protons + electrons would seem to counteract one another, magnetically.

Come to think of it, I guess the same question could be asked for the Earth's liquid core, considering its temperatures...?

Plasma consists of ionised particles. The electrons are not bound to their nuclei and are free to move around within the star.

However, they are not free to leave the star. Without the electrons to maintain a neutral charge, there would be massive electrical repulsion from all the positive protons and helium nuclei, the result would probably be something like the formation of a black hole larger than the observable universe... The "What if", blog considered something like this in Proton Earth, electron moon and concluded that it would be one of very few things that could actually destroy the universe.

In conclusion, the sun contains about $10^{57}$ electrons. One for every proton.


What kind of star would allow for life like our Sun/Earth and then go nova? In the Star Trek Prime Universe, Star Trek:  The Original Series, the home stars of the Fabrini, Platoans and Sarpeidons went nova, while in Star Trek:  The Next Generation one of the stars in the Bynar system (which might have been a multiple system) went supernova.

What type of star would allow sentient life to evolve to an advanced level, roughly to the same as twenty-first century Earth, and then go nova.

In the real universe, stars that go supernova are very large stars that are short lived, or, they could be white dwarfs that accrue matter and overcome the Chandrasekhar limit.

Very large stars, 8 solar masses or more - about the minimum size for a star that goes supernova, the life span of that star is too short for a planet to form, cool and develop life.   The life of an 8 solar mass star is about 8 to the 3rd power or 1/512th the life of our sun.   On the order of 20 maybe 30 million years.   You can't get a planet to form and have advanced life that quickly.

Now if there's a 2 star system, like lets say, Sirius A and B, the larger star reaches it's red giant stage first prior to becoming a white dwarf and the smaller star keeps burning and in time, clears out much of the red-giant solar nebula left by the first star, then you can have a threat of a supernova binary system where the remaining star feeds the white dwarf, and that star lives long enough for planets to develop life.   Binary systems may have some issues, like the planets need to be fairly distant and/or the stars quite close and you have the white dwarf stealing matter from the star, so the star loses mass over time and the white dwarf likely emits intense UV and perhaps gamma rays, but in this way, it's theoretically possible to have a super-nova on the way and have a planet with life on it in orbit around the binary system.

Sirius A & B are actually not a good example.  They're too far apart and Sirius A is in the direction of too large and short lived and Sirius B is not expected to go supernova (maybe after Sirius A goes red giant - just maybe - that would put on quite a show if we're still around to see it, but I digress).

More realistic, much more common and just as deadly would be a star going red giant.


Could an impact have resurfaced Venus 300 million years ago? Venus surface isn't older than about 300 million years. The only explanation I've come across is that some kind of global volcanism resurfaced the planet. But couldn't it have been an impact event? A huge comet hitting it by chance, maybe making Venus tip over and stop spinning too.

Or could Venus have had a moon like Phobos which spiraled inwards, disintegrated and rained down during a few million years? Phobos will do so in only 50 million years, maybe it is a common feature for terrestrial planets to form with such moons that hit them billions of years later. Maybe Earth is an exception because of the large Moon and how it formed.

Would an impact event leave visible traces like impact basins, or could the entire surface melt and reform as it is today, as I suppose Earth did when the Moon formed? Could Venus have been a very different planet up until 0.3 billion years ago? How could one find out, what kind of investigation would be needed?

Pluto's mass is about 1/50th of Mars (and Theia), so I suppose that it would be enough for a Pluto sized comet to have had about 7 times the impact speed compared to the pre-Earth/Theia collision. Comets could have up to 70+35 km/s speed relative to Venus. Someone maybe can make a rough estimate of what's needed to resurface Venus and how likely such an impact event would be?

And could reformation and redifferentiation after an impact have eliminated a previous magnetic field?

The diversity of the planets and moons, I suggest, depend more on their vast history of random events, than on where and how they once formed. I suspect that planetary scientists generally want to discount a unique planet reforming event only 0.3 billion years ago, just to make their job a bit easier ;-)

Would an impact event leave visible traces like impact basins, or
  could the entire surface melt and reform as it is today, as I suppose
  Earth did when the Moon formed? Could Venus have been a very different
  planet up until 0.3 billion years ago? How could one find out, what
  kind of investigation would be needed?

Certainly giant impacts were fairly common, though I don't think they were common any more as recent as 300-500 million years ago.   That would have been a kind of unusual event.   Mostly they occurred when the solar-system was young.

Mercury probably had one

Venus may have too - and I recommend this article as very loosely related.

The Earth had one - as you know.

and Mars too, but not quite as "giant", though still very big and it left a visible mark that spans nearly 1/2 of Mars' surface.  See Here and Here.

The Moon even had one and there is evidence in the thickness of the Moon's crust and location of it's volcanic magma pools/dark spots.   See Here and Here.

The point I'm trying to make with all these examples is that unless the Impact is very very large, it would leave a impact footprint of sorts.   The impact has to be large enough to not only liquefy the surface but to also allow the surface enough time to reset in mostly uniform layers so you don't get 1/2 of the planet with a thicker crust than the other half.   Impacts of that size are statistically improbable, within the last 300 million years.  Most of the big impacts happened long before then.

Wikipedia suggests that there's other reasons for believing the global resurfacing though interior heat idea.  Venus Global resurfacing event - Wikipedia - Venus lack of a magnetic field, it's influx and loss of water and D to H ratio and it's Sulfur content.     Do these have had to happened through internal heat resurfacing?   Now, the article says there's no proof, only some pretty good evidence, so it it possible Venus resurfacing was due to an impact?    I see no reason why it wouldn't be possible.

Given Venus' Heat trapping atmosphere, higher temperature and greater gravity, which also means, greater impact velocity, at least escape velocity of about 10 km/s where as the Mars impact that covers a hemisphere was thought to be about 5 km/s and the impacting object into Mars, at least 1,600 km in radius which is nearly the size of the Moon.   You might not need that big to impact Venus and create a complete resurfacing, but I wouldn't think you could go too much smaller.

I love the Venus had a moon that crashed into it idea cause that seems to work on a few levels.  A giant impact that created Venus could have happened 4 and change billion years ago and created a Venusian Moon, which over time, could both slow Venus' rotation and work it's way towards Venus in the process, eventually crashing into it.   It seems at least, possible, though I'm not sure what evidence could be gathered to prove it.    I like your idea a lot and I think it's much more likely than this one:  Earth stole the Moon from Venus.

As to Venus being very different prior to it's resurfacing.    The sun was only about 3%-5% less luminous when Venus resurfaced and given Venus' proximity to the sun, I have a very hard time seeing Venus as not already in a run-away greenhouse effect even then.   I love the Venus once had oceans and maybe life hypothesis, but with a CO2 rich atmosphere, it's hard for me to see it as likely.   My guess is that Venus was hot and virtually without water 300 million years ago too.   Unless it somehow stored a huge amount of it's CO2 underground, I have a hard time not seeing it as too hot for oceans 300-500 million years ago.   Maybe, but I'm skeptical.    It would be very neat to have a picture of what Venus was like prior to it's resurfacing and over the hundreds of millions and billions of years before that.

Edit

In this article it mentions that it's improbable that Venus once had a Moon based on angular momentum.   A planet-Moon system maintains it's angular momentum and if the moon spirals in towards the planet, it should carry significant angular momentum with it.   Such an impact should give Venus a pretty rapid rotation which would be difficult to explain given that it rotates so slowly now.  Unless perhaps, the Moon was in a retrograde orbit, Moon going in one direction the planet in another and the combined angular momentum largely cancelled out, but adding more specific conditions tends to decrease the probability.   I still like the hypothesis, but the article above says it's unlikely.


Can you recommend a book about big bang nucleosynthesis and chemical abundances? I am interested in learning about big bang nucleosynthesis, nuclear fusion up to iron in stellar cores and beyond iron in supernovas, and into the lithium problem (galactic abundance anomoly for atomic numbers 3,4,5). As these topics are interlinked, I have been hoping to find good resources on the subject. Can anyone recommend a good source for self-study? In case it's relevant (regarding difficulty level), I'm a physics grad student lacking exposure to related topics.

I highly recommend Nucleosynthesis and Chemical Evolution of Galaxies by Bernard Pagel. It contains the basics of nuclear reactions andstellar evolution, chapters on big bang nucleosynthesis and light element production, as well as covering the broad swathe of stellar nucleoynthesis and how these link together into predicting the chemical evolution of galaxies.


How well would the Moon protect the Earth from an Asteroid? Would the Earth fare better if the Moon blocked the meteor, comet, rogue planet, or otherwise rather than a direct impact? At what point would the Moon's debris would be an extinction event?

The limit of the impactor is the size, angle of attack, or composition in which to moon would no longer be of any protection.

It would be much better for Earth if the impactor hit the moon...

In this Worldbuilding answer, I used a paper on ejecta kinematics to do calculations for ejecta velocity upon impact. Without going into too much detail here, much of the ejecta from a large impactor would not exceed the moon's escape velocity of 2.38 km/s. You can examine Figure 7 from the linked paper which shows the logarithmic relationship between ejecta velocity and the edge of the crater. Only material within a few tens of meters of the edge of the formed crater can attain the velocity needed to escape the Moon's gravity.

The mass of the asteroid increases with the cube of its radius; while the mass of material within $n$ meters of the edge of the impact crater increases with radius; so it is evident that the the larger the meteor itself gets, the lower the danger of the any potential ejecta relative to the original impactor.

Furthermore, material ejected from the Moon has the potential to land back on the Moon, enter a stable orbit of Earth, or be ejected from the Earth-Moon system; so only a (probably small) fraction of Moon ejecta would threaten Earth.

Overall, fewer, smaller rocks are much better than large rocks when it comes to getting hit by things (except maybe for your car's windshield).

...unless the impactor knocked the Moon into Earth.

Of course, there is always the possibility that a very, very large impactor would knock the moon into a different orbit, potentially one that eventually impacts Earth. That would, obviously, be the worst case scenario. Significant changes to the orbit of the moon, even if these changes do not cause a collision with Earth, have the potential to do significant damage in terms of tides; not just ocean tides but also the poorly understood effect of tidal forces on Earth's mantle.

In any case; it is probably much better for a large impactor to hit the Moon instead of the Earth, but possibly much, much worse.


Which moons have cold traps? (i.e. low ecliptic inclination in orbital and rotational axes) Both the Moon and Mercury have polar cold traps in polar craters, with signs of volatiles. This is of course not as interesting on a moon in the outer Solar System which consists of volatiles anyway. But suppose that there is some use for very cold real estate shielded from the Sun out there.

Which moons are candidates to have eternally shadowed (or eternally sunlit) regions, and wouldn't they be even colder than those on the Moon? Jupiter and Neptune have less than 2 degrees inclination so their moons maybe are good candidates.

Many of the moons of Jupiter and Saturn are tidally locked and probably move ice from their equatorial regions to their polar regions. Ganymede is the best example where you see bright polar ice caps and a dark equatorial zone.  A migration process occurs on all of these moons in which light from the sun gets absorbed by water ice molecules and sends them on random jumps.  After a long random walk, they eventually end up near the pole where they stay for a long time even if they are not in a perpetual shadow, and they stay there for nearly forever if they are.

But, if the moon has a very thick coating of ice everywhere, this process may not have dug down deep enough, so it is still covered with ice everywhere, in which case it is hard to see the effect happening.


How much heat is generated from waxing and waning of reflected radiation from the Sun? According to this article, the greatest cause of global warming would be from the waxing and waning of reflected radiation from the Sun. (Placing carbon emissions at 10-20%) So really, how much heat is generated from the process that stays on Earth? And would it really affect it that much?

The article doesn't go into specifics and appears, not to be written by Dr. Evans at all, but I'll pull some quotes.

  When it is completed his work will be published as two scientific
  papers. Both papers are undergoing peer review

and

  He has been summarising his results in a series of blog posts on his
  wife Jo Nova’s blog for climate sceptics.

  He is about half way through his series, with blog post 8, “Applying
  the Stefan-Boltzmann Law to Earth”, published on Friday.

(footnote, I'm guessing in Australia, summarizing and skeptics are spelled differently than in the US, cause that's a copy-paste)

I've only read one of his summaries, and in Mathematics, you have to look at the details, which he's not provided, so, he's basically saying little more than "this is true, here's generally why, I've done the math, please take my word for it, I'll publish the numbers later".   Now Michael Mann is also presenting non peer reviewed work to Paris, so, what's good for the goose is good for the gander I suppose.

Now, I'm just a guy who likes science, but my understanding is that ideas have been presented before they are finished or peer reviewed fairly often.   Einstein did this in fact regarding general relativity and another scientist actually beat him to publishing the theory (though, the other scientist was gracious about it and gave Einstein full credit), so, I don't think it's necessarily bad to present a summary prior to peer review.

I think it is, however, unusual to post summaries of an idea on a blog saying "I figured it out, the majority of research on this subject is wrong".   Dr. Evans is saying he has proof, but he's acting like a junk science blogger.

Dr. Evans "trust me I did the math" claim kind of requires that we look
at his track record, and his track record isn't very strong, though he will say, he's being attacked by the establishment.

According to this site, he hasn't published anything peer-reviewed since the 1980s and he's on a their Climate Denier List.  He's also on Skeptical Science's Climate Misinformers List.

And here's a list of debunked claims he's made (and if this list is accurate, he's not a scientist at all, just a guy on a fishing expedition.   No good scientist would agree with pretty much every counter argument against climate change cause that's not how the scientific method works.  You can disagree with something, that's fine, but to agree with every counter argument - that's silly.  Here's another more detailed explanation of what he's gotten wrong, from 2011.    Evans seems to be in this debate to disagree with it much more than he's in it to do scientific research.

It's pretty much impossible to make a true scientific argument against Evans "proof" without his specifics, and, that kind of proof/disproof can get a little long and complicated, but for now, disproving him is impossible.   But should we listen to him?

It says above he's published 8 blogs related to his recent research.   I'm not going to dig up all 8, but here's the most recent one. Applying the Stefan-Boltzmann Law to Earth.   Now, I'm just a layman, but even I can see problems with his argument here (and he should too given that he's an engineer with a PhD).   The Stefan-Boltzmann law is an approximation.   It's a physical model to calculate radiation into space.

The problem with his approach is, the best way to measure how much heat/energy leaves the earth by radiation is to measure it directly, by satellite.   The amount of energy that radiates from the Earth into space varies with temperature, snow cover, cloud cover, even humidity, and probably 1 or 2 other things I'm overlooking.   If you try to calculate this energy leaving the earth into space by playing with with the Stefan-Boltzmann Law instead of relying on direct measurements, you're allowing yourself a lot of fudge factors and inviting a far greater error than direct measurements would give you.

On the quoted article, let me pull out an example:

  Dr Evans has a theory: solar activity. What he calls “albedo
  modulation”, the waxing and waning of reflected radiation from the
  Sun, is the likely cause of global warming.

OK, so, which is it, solar activity or albedo modulation, cause they're not the same thing.  The first takes place on the sun, the 2nd, the earth.   This paragraph makes no sense to me.

  He predicts global temperatures, which have plateaued, will begin to
  cool significantly, beginning between 2017 and 2021. The cooling will
  be about 0.3C in the 2020s. Some scientists have even forecast a mini
  ice age in the 2030s.

Now, this paragraph is particularly devious.   El Nino's tend to warm the earth, La Nina's cool it.  The effect is temporary and not huge, but enough to cause yearly variation.   A strong El Nino drove the big spike in global temperature for 1998 and we're in an El Nino now (edited my answer, since 2014 they've been talking about entering an El Nino, I gather it's officially started now).

We had more La Nina years than El Nino 2006-2013 with the only small El Nino coinciding with 2010, which set records for temperature.  A lot of the hiatus in warming that is often talked about is related to there being only 1 small El Nino over 7 years.

Predicting 2017 as the time when the cooling will "begin" is devious because that could be around the time the El Nino has ended and the oceans could switch back to a La Nina (which usually follows El Nino).   This will create a temporary cooling for a year or two, which he, no doubt, will take credit for if it happens.   Now, he also predicts 2021 which could go either way and he gives an amount, but that doesn't change the fact that he's making a prediction and hoping the El Nino of 2015 will end and make it prediction look good.

Real global warming or cooling can't be measured in 1 year anyway, unless, maybe, if it's ocean current and the occasional mega-volcano adjusted - then, maybe you can get some measure of warming/cooling based on one year, but it's still only one year.  That's a really really short period to make any predictions on and not something I'd trust very far at all.

and on the "scientists have predicted a mini ice age in 2030", that's not actually true.   There was a study on sun-spots and they predicted that we could see a sun-spot low period around 2030, perhaps similar to the Maunder Minimum that may have caused the mini ice age, but the scientists who predicted that were very clear that they were not predicting a new mini ice age, they said the effect would be smaller than the effect of CO2.

Here's the mini ice age prediction, which a few people made (but not the scientists who did the research).

Here's an article that explains why it isn't true.

So, there's a lot of bad and a handful of false statements in that article you quoted, which, granted, wasn't written by Dr. Evans himself, but still, it's hard for me to take it seriously.

Until he publishes his results, he can't be proved or disproved but based on what I've read, I find it hard to take him seriously.   My hunch is, he's not trying to reach scientists at all, but he's trying to reach his target audience.  Those who question climate change and he gives them a name and an alternate argument that they can stand on.   An argument doesn't have to be correct, it only needs to sound correct and with that, you can usually convince a percentage of people to agree with you.

Not sure how much that helps, but that's my take and I went through and tried to clean up my long answer a bit.   If you'll forgive me, it reminds me of the the old joke.   How can you tell Dr. David Evans is lying?   He's talking or writing.  :-)


Necessary steps to calculate photon's path by using null geodesic equation Can anyone please give me an explanation on how to calculate photon's path by using the null geodesic equation? N.B. I know all of the non-zero values of Christoffel symbols.

The geodesic equation below is a second-order partial differential equation for the spacetime coordinates $x^\mu$, $\mu = 0 \ldots n$, $${d^{2}x^{\mu } \over ds^{2}}+\Gamma ^{\mu }{}_{\alpha \beta }{dx^{\alpha } \over ds}{dx^{\beta } \over ds}=0 \;,$$
where $\Gamma ^{\mu }{}_{\alpha \beta }$ are the Christoffel symbols, and the proper time $s$ is the parameter of the curve (i.e. how ``far along'' you are on the path). You need to solve these $n+1$ coupled equations as any such system, e.g. numerically.

If the parameter $s$ is confusing, you can reparametrise the curve as a function of coordinate time $t = x^0$, as explained e.g. on Wikipedia's page on geodesics in general relativity, $${d^{2}x^{i} \over dt^{2}}=-\Gamma ^{i}{}_{{\alpha \beta }}{dx^{\alpha } \over dt}{dx^{\beta } \over dt} + \Gamma ^{0}{}_{{\alpha \beta }}{dx^{\alpha } \over dt}{dx^{\beta } \over dt}{dx^{i} \over dt} \;,$$
which gives you $n$ partial differential equations for the spatial coordinates $i = 1 \ldots n$. (Note that there are remaining greek indices, for which $dx^0/dt = dt/dt = 1$.)


Can we calculate the orbit of exoplanets? I'm not an astronomer, but the question came up and I'm interested if we are able (or how accurate we are able) to calculate the orbital parameters of an exoplanet. Since the transit method must yield some properties of its orbit, we must know some of it. I also found some information here regarding the measurement of the orbital plane (Can we know the orbital planes of extraterrestrial planetary bodies?).

But do we have a clear picture of its orbit? Like we have of the planets in our system?

Could we, for example, tell if the planet was born in this system or has been caught by it at some point (giving it a different orbit I suppose)?

The characteristics of the orbit of an exoplanet can be determined very accurately using measurements of the reflex motion (the "Doppler wobble") of its parent star. These measurements yield the orbital period and eccentricity of the orbit. If we can estimate the mass of the parent star, then the orbital semi-major axis is known through Kepler's 3rd law. Orbital periods, and hence semi-major axes, can also be measured very precisely from transit data, and eccentricities can also be estimated provided one has a good idea of the mass and size of the parent star (e.g. van Eylen & Albrecht 2015).

Additional details can be gleaned from transiting exoplanets, such as the inclination of the orbit to the line of sight and any difference between the orbital axis and the spin axis of the star it orbits. These measurements exploit something called the "Rossiter-McLaughlin effect", as explained in an answer to the question you linked to.

The answer to your broader question is yes, of course people are looking at the distribution of orbital parameters: the behaviour of eccentricities versus planet mass and semi-major axis; the prevalence of planets of particular mass at certain distances around different types of stars; the frequency of occurence of multiple planets and orbital resonances, the misalignment between the spin of a star and the orbits of its planets, and so-on, in order to try and understand how planetary systems form and evolve.

Some resources:

The HEASARC Exoplanet orbital parameter database

exoplanets.org


Opposite Constellation of Sagittarius A* Suppose an alien race fired a laser beam from the centre of the Milky Way, it went through the Sun and then off further out of the galaxy.  What Constellation would the light beam head towards, closest star also if possible?  Am not interested about how much devastation the light beam may or may not cause, just the journey and destination.

Would that direction be the quickest/shortest way out of the galaxy if we had a spaceship that could travel that fast?

Per https://en.wikipedia.org/wiki/Galactic_Center, the galactic center
is at RA 17h 45m 40.04s, Dec -29° 00' 28.1" in the J2000 epoch.

The point opposite this would be RA 5h 45m 40.04s, Dec +29° 00' 28.1"
(J2000).

The nearest star I could find to this position is a relatively
uninteresting 16th magnitude star:

The nearest "interesting" star is HIP 27088, which is an 8th magnitude
double star:

My calculations show this point is in Auriga, not Gemini, albeit just barely:

A slightly more zoomed out view:

As @zephyr notes, the fastest way out of our galaxy would be "straight up" (perpendicular to the plane of the galaxy), not away from the center, since the Milky Way's "height" is much smaller than its length/width.

If someone wants to replicate what I did on Stellarium and use RA/DEC grids, be sure to set the date to 1999 Dec 31 at noon GMT: the grids Stellarium displays are based on the current date, not J2000.

NOTE: http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad lists even more stars than Stellarium (Stellarium uses only part of this catalog), so you might be able to find an even closer star to this point, but I was too lazy to do so.

EDIT: Newer versions of Stellarium let you overlay galactic coordinates. I'm too lazy to do this myself, but if someone else wants to, it would be a better answer.


Appearance of sun at geologic past After origin of Earth ( 4.54 billion years ago ) to today; what are the changes  through-which the sun undergone with time? , ( if seen from Earth. )

I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; should I draw it just like present day? Or the look/ appearance (colour, size, flames, position, halo etc) varied in observable amount through the geologic time-scale(precambrian-cambrian-ordovician-silurian-devonian-carboniferous-permian-triassic-jurassic-cretaceous-palaeogene-neogene etc). (the timescale terms provided just to hint the answerer the level of details may NOT to go )

Update: Here I attach a  small version of geologic time table to indicate time ranges.

from here .

To see the entire chart; see international commission of stratigraphy charts

I'm doing some palaeobiology-landscape rendering; so I'm thinking if I want to add the sun in such pictures; should I draw it just like present day ?

Although the Sun was less luminous 4.5 billion years ago (about 70% of it's current luminosity, it is thought), and while this would affect it's color, making it perhaps a little more orange than yellow, the effect is small.

The Sun is still going to be a blazing near-white object in the sky of roughly the same size.

Although there was probably a color shift to a slightly more orange tint, the composition of the Earth's atmosphere was radically different (see for example this page on Paleoclimatology) and this would have had a much more significant effect on the appearance of the Sun (and the lighting of the landscape) than any minor change in the Sun's appearance.

There are some issues relating to the development of life and this change in the Sun's luminosity (like the Faint Young Sun Problem) but these won't make much difference to your rendering, I think, compared to the atmospheric and climate issues.

However the climate changes are not fully understood and it's rather speculative to try and make specific claims about what the overall effect would have been on the illumination levels and color of light reaching the surface in these periods.


Basis of color index As an amateur astronomer for 60 years and always as interested in the science as in the observing, I thought I had a pretty good understanding of the fundamentals. But I'm distance-auditing a third-year course in the fundamentals of applied astronomy, and I found that I had a misconception about the color index.

I had assumed the CI was derived by measuring the star's own magnitude in each of, say, the B and V passbands, then subtracting to get the CI (e.g., by taking the ratio of flux intensities, expressed as a magnitude).

Not so! It's derived by measuring the star's magnitude relative to standard zero points (the magnitude of Vega at each of those wavelengths) that are defined to be magnitude 0 for those wavelength, then subtracting to get the CI.

My question is: Why measure two fluxes relative to a standard and then compare those two values, rather than simply measuring them relative to each other? Why not just measure the flux at each wavelength, take the ratio, express this as a magnitude, and use that as the CI? The relationships are the same either way, just with a different zero point (think 0K and -273°C). It seems like finding how far apart Cleveland and Chicago are by measuring the distance of each from NYC and subtracting, rather than just measuring the distance between them. I don't see that the convention accomplishes anything other than assigning a CI of 0 to "neutral" white stars. Or am I missing something?

What you may be missing is that no two telescope, filter and detector combinations are the same. A colour index defined by your method would be measured differently by different observers.

The solution is to define standards that  have defined magnitudes and colours that all observers can use to calibrate their observations.

In actual fact, the basic measurement is
$$ B-V = \alpha (b-v) + z + \beta X,$$
where $b$ and $v$ are native measured magnitudes (-2.5 log the detected adu with a correction for exposure time) and $X$ is the airmass. The constants $\alpha$, zeropoint $z$ and extinction coefficient $\beta$ are determined by observing a network of secondary standards with a range of colours at a range of airmasses.


Why is the Earth called terrestrial? Can we call the Earth an ocean planet? Why is Earth called a terrestrial planet even though it is 70% covered by water?

There really isn't a lot of water on the Earth. The image below, from the USGS science school, graphically illustrates the volume of the water on (and in) the crust of the Earth, relative to the volume of everything else.

The bigger blue ball is all the salt water, from all the oceans & seas, the small one is all the fresh water, from the rivers, lakes and aquifers. You may need to zoom in to see the fresh water ball...

There is more water dissolved in the mantle, but we can't get at it (with current technology), apart from what's released in volcanic eruptions.


Need advice on mirror-making for a home-made amateur reflector For example, what would be some good references to read first? What sort of glass slab should I use? What size; diameter and thickness. Im a first-time builder trying to build one

I second the comment about Texereau -- originally written in French; it's the best, bar none. You won't need anything else.

Remember the old adage that is quicker to make a 4" mirror and then an 8" mirror than it is to make an 8" mirror first. Don't bother with borosilicate. You should be able to get mirror kits and all the abrasives that you need online. Join a club if you can find one.

How to Make a Telescope, 2nd edition by Jean Texereau:

http://www.willbell.com/tm/tm3.htm
https://www.amazon.com/Telescope-second-Texereau-published-Atlantic/dp/B00EKYQCQS


Is carbon dioxide gas giants possible? Is it possible that a planet originally full of water, and the solar winds decompose the water and hydrogen escapes out, then the oxygen combines with carbon at crust, just like Venus but the whole planets are mostly carbon dioxide and hence become a gas giant?

While not absolutely impossible this is not a likely situation. The reason is that for a planet to be a "giant", then it will have enough gravity to hold on to its hydrogen and helium. Conversely, if you remove the H an He from a planet, it won't be "giant". There are only two gasses that are common enough in the universe to form a massive gas mantle, and those are Hydrogen and Helium.


What is an "arc" spectrum ? I sometimes hear about astronomers using an arc spectrum to calibrate observations. For example a "He-Ar arc spectrum". What is an "arc" in this context? I assume it's nothing got to do with angles (arcmin, arcsec, etc.).

An arc spectrum is one produced by a discharge lamp where the discharge is through ionised gas, in the case of He-Ar a mixture of Helium and Argon, which produces a predictable line emission spectrum.

They are often used to provide a calibration spectrum for spectrometers.


Why is a giant planet around a tiny star unusual when binary stars are normal? Inspired by this BBC article and the corresponding journal article about the M dwarf star GJ 3512.

  The Jupiter-like world is unusually large compared with its host star, contradicting a widely held idea about the way planets form.

On the other hand, binary stars are very common, comprising (according to Wikipedia) about one third of star systems.

Naively, the same processes that create binary star systems could be expected to also create star systems comprised of a single star and a Jupiter-like "failed star" - the only difference being that in the latter case the would-be second star wasn't large enough to ignite.

What is the difference between these scenarios?  What makes the small-star-and-large-planet, according to the standard models of planet formation, unlikely, when binary stars are common?

(In non-technical terms if possible, please.)

It's important to realize that binary stars form much differently than planets do. Assuming that both stars form in situ (i.e. excluding scenarios where one is captured from outside the system), there are several main ways for a binary star system to form from a molecular cloud. The most widely-accepted model at the moment is the fragmentation hypothesis, where the protostellar cloud splits during its collapse, forming two clouds which will each collapse into separate protostars.

Planets, on the other hand, form not from collapsing clouds but in protoplanetary disks around stars. The dynamics of these systems are substantially different, as there are interactions between the gas, dust, and protoplanets - perhaps a more complicated system than a collapsing gas cloud. We therefore see planets arise through different processes, such as gravitational instabilities within the disk or pebble accretion.

The paper being referenced found that the GJ 3512 system appears incompatible with the pebble accretion models the team tested - that is, rocky planetary cores being built up through collisions. In a low-mass star system, the cores should move inwards too quickly to accrete enough mass, and tweaking other initial conditions didn't help. That's what the BBC article focuses on.

What the BBC ignores is that the authors were able to reproduce the system by assuming that the planet(s) arise through gravitational instabilities within the protoplanetary disk, and this model is borne out by the simulations. It may not have been widely considered, but it still works. That's one of the key results to pay attention to.


What does "additive fringing from the (multiplicative) twilights" mean? The question How can I make the fringing pattern visible in my flat (twilight) images for filter i? includes:

  I have a bunch of twilight flats that have been preprocessed (dark, bias, overscan etc.) and need to remove the additive fringing from the (multiplicative) twilights. I normalized each one by dividing by the mean and then combined them using ccdproc in astropy. I then divided the normalized flats by this "new" flat and viewed them in DS9 but I still cannot see the fringing.

and the OP went on to post an answer where fringes are visible.

But I haven't any idea what causes "additive fringing" (which I assume means it's a fringing pattern that is additive to an otherwise corrected image) nor what "multiplicative twilights" means.

Could someone explain the basics behind both?

The fringing pattern is caused by thin-film interference within the CCD. The signal received in a pixel will be proportional to the light falling on it, multiplied by a sensitivity, but then some extra signal is added or subtracted which depends on how much of the incoming light is at particular wavelengths that are affected by the interference (i.e. the whole broadband signal does not contribute, only the flux at certain wavelengths).

The fringing pattern is therefore an additive modulation of the sky background, which in general, for broadband filters (e.g. the i filter is 100 nm wide), has both a broadband component (scattered sunlight, moonlight) and a component at discrete wavelengths (airglow lines). Only the latter make a significant contribution to the fringing.

The fringing does not affect stellar photometry measurements, other than by introducing structure in the background that can be difficult to deal with when performing background subtraction. However, it does affect the images cosmetically and disrupts photometric measurements of extended objects.

It is not possible to remove perfectly because the amplitude (and also the shape) of the fringing pattern depend exactly on the spectral make-up of light illuminating the pixels and this will vary depending on (e.g. time of night, whether the Moon is up).

A first-order correction is to make a "fringe frame" by median stacking all the moonless night time exposures, after scaling them by the median sky signal (or the exposure time). A scaled version of this can then be subtracted from every processed image frame to remove the background component that is affected by fringing. It is hence referred to as an additive correction.

Twilight flat fields are used for an entirely different purpose. At twilight the sky background is close-to-uniform and has a broadband continuum spectrum that is dominated by scattered sunlight. It is therefore very difficult to see any fringing in it with broadband filters.

That is good, because the purpose of the twilight flats is to calibrate out the broad band sensitivity of each pixel by dividing by the flat field (hence referred to as a multiplicative correction, and you would normally multiply by a "balance frame" constructed by dividing a smoothed version of the median twilight flat by the unsmoothed version). The twilight flats can also be used to calibrate any large spatial scale sensitivity variations caused for example by vignetting. Twilight flats that do show evidence of strong fringing in broadband filters should not be used to remove pixel to pixel sensitivity variations. In those cases one usually resorts to "dome flats".


Difference in gravity on both sides of the moon How much (if any) difference in gravity would you feel if you are on the surface of the moon and:

standing right 'under' earth (exaclty between the moon and the earth) and
on the opposite side of the moon (So both, the moon and earth are exactly under you)

Very, very little

It's important to keep in mind that the Moon is constantly accelerating toward the Earth. The component of gravitation that one feels on the surface of the Moon is the difference between the gravitational acceleration toward the Earth at the point of interest and the Moon's gravitational acceleration toward thbe Earth as a whole.

At the sub-Earth point (the point on the Moon closest to the Earth), the gravitational acceleration toward the Earth is slight greater than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's sub-Earth point.

At the antipode to the sub-Earth point (the point on the Moon furthest from the Earth; I'll call this the anti-sub-Earth point), the gravitational acceleration toward the Earth is slight less than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth once again very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's anti-sub-Earth point. The effect is nearly (but not exactly) identical to that at the sub-Earth point.

There are places on the Moon where the gravitational acceleration toward the Earth does slight increase the observed gravitational acceleration, which is the set of points where the Earth is more or less on the horizon. This effect here is about half that (in magnitude) of the effects at the sub-Earth and anti-sub-Earth points.

A qualitative discussion

Ignoring that the Moon is rather lumpy in a gravitational sense, the gravitational acceleration on the surface of the Moon due to the Moon itself is $$\vec a_M = -\frac{\mu_M}{{r_M}^2}\hat r_M$$
where
$\mu_M=G M_M$ is the Moon's gravitational parameter,
$r_M$ is the Moon's radius,
$\hat r_M$ is the vector from the center of the Moon to some point on the surface of the Moon, and
$\boldsymbol a_M$ is the gravitational acceleration due to the Moon at that point.

At that point, the gravitational acceleration toward the Earth is $$\vec a_E = \frac{\mu_E}{||R_E \hat r_E - r_M \hat r_M||^3}(R_E \hat r_E - r_M \hat r_M)$$
where
$\mu_E=G M_E$ is the Earth's gravitational parameter,
$R_E$ is the distance between the centers of mass of the Moon and the Earth,
$\hat r_E$ is the vector from the center of the Moon to the center of the Earth, and
$\boldsymbol a_E$ is the Newtonian gravitational acceleration toward the Earth at that point.

The Moon itself accelerates Earthward due to Earth gravity by $$\vec a_{M,E} = \frac{\mu_E}{||R_E \hat r_E||^2}\hat r_E$$
The specific weight (force measured by a spring scale divided by mass) at the point of interest is $$\begin{align}
\vec g &= \vec a_M + \vec a_E - \vec a_{M,E} \\
&= -\frac {\mu_M}{{r_M}^2}\left(\hat r_M-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{\hat r_E - \frac{r_M}{R_E}\hat r_m}{||\hat r_E - \frac{r_M}{R_E}\hat r_m||^3}-\hat r_E\right)\right)
\end{align}\tag{1}$$
In the case where $\hat r_E = \hat r_M$ (the point on the Moon closest to the Earth) equation (1) simplifies to $$\begin{align}
\vec g
&= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{1}{||1 - \frac{r_M}{R_E}||^2}-1\right)\right)\hat r_m \\
&\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1+\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m
\end{align}$$
In the case where $\hat r_E = -\hat r_M$ (the point on the Moon furthest from the Earth) equation (1) simplifies to $$\begin{align}
\vec g
&= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(1-\frac{1}{||1 + \frac{r_M}{R_E}||^2}\right)\right)\hat r_m \\
&\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1-\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m
\end{align}$$
The dominant perturbing factor, $2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3$ is the same for both points and is rather small, about $1.5\times10^{-5}$. People at these two points would weigh a tiny bit smaller than they would if the Earth was not present. The difference between the two points is tinier still, a factor of about $2\times10^{-7}$.


How to calculate center (not focus) of planetary orbit? I'm developing a website showing the movement of Solar System planets according to the passage of time. I'm using this reference to calculate all the necessary orbital parameters.
These are two images some days apart.

(Only planets up to Mars are shown here. The moon orbit has been highly exaggerated to become visible. Celestial bodies radii are not to scale either, except among themselves—except the Sun.)
The images show the high eccentricity of Mercury's orbit, causing it to be very far from the Sun in the first image, and very close to it in the second.
I want to draw the correct ellipse, instead of circles that constantly change radius. I have all the “orbital elements” given in the link above (N, i, w, a, e, M, r, v, x, y, z, lon, lat), but don't know how to extract an ellipse from them. I'm drawing the planets using x, y heliocentric coordinates. Even if I had the ellipse foci, JavaScript needs the geometric center of the ellipse to draw it, and I don't know either how to convert the former into the latter.
How to calculate the ellipse center from the parameters given above?

I assume that the Sun is at the origin: (0,0).  Then we can look at this image of an ellipse:
The foci are at +-c, where c = ea, by the definition of eccentricity. To put the perigee on the negative y-axis, we can then say the center of the ellipse is at (0,ea).  Then, to account for the argument of periapsis, w, we need to rotate the ellipse (and ellipse center) counter-clockwise around the origin to get the coordinates for the ellipse center, (x,y) =(-easin(w),eacos(w)).  A better approximation would include the inclination, i, for a 2-D projection, but the inclination is small for Mercury so it may not matter for the purposes of your animation.  This diagram for the orbital elements should also help:


Planetary orbital resonances The Kirkwood gaps in the asteroid belt are associated with orbital resonances with Jupiter yet planets seem to prefer resonant locations. Why don't asteroids accumulate near the "Kirkwood locations" rather than avoid them??

This is actually a very subtle question, much more so than the answers to the similar questions provided in the comments give it credit for.  When I was in graduate school at Ohio State I routinely asked this question to visiting dynamicists and invariably got different answers.

The very basic answer is that if you have two sufficiently strong resonances sufficiently close together, then the resonance will be unstable.  Otherwise, the resonance will be stable.  But what determines "sufficiently strong" and "sufficiently close" is where things get very complicated quickly.  A basic criterion is the Chirikov criterion.  (The Scholarpedia article is somewhat more detailed.)  However, the Chirikov criterion is not universally valid.

If you have overlapping resonances, then an object gets bounced back and forth between these two resonances chaotically.  These different resonances perturb the orbit in different ways, and eventually they will perturb the orbit into an unstable orbit, thus leading to depletion of the resonance.  If a resonance is "distant" from other resonances, then the resonance tends to keep objects locked in place, leading to an excess of objects in the resonance.

Most of the resonances in the asteroid belt are fairly close together, which leads to them being unstable.  The Kirkwood Gaps are the most prominent manifestation of these instabilities.  For example, the Alinda family of asteroids are in a 1:3 resonance with Jupiter, and are very close to a 4:1 resonance with the Earth.  This leads to instability, and hence very few asteroids in this family.  However, in the outer Solar System, the resonances are generally far apart, and so are mostly stable.  The plutinos are one example of such a stable resonance, being in a 3:2 resonance with Neptune.


A good place to start? I am a high school student (going into 11th grade next year), who will be taking an introduction to astronomy and astrophysics as part of my core curriculum. As such, I would like to get a head start now, but I am not sure where to start.  Are there any specific learning resources that are recommended? Books, videos, documentaries?

Find a local astronomy club; they will help you get started, learn the sky, and show you different types of equipment. Also, read "Turn Left At Orion", H.A. Rey's "A New Way to See Them" and "Find The Constellations" to learn the sky.

And get a pair of 10x50 binoculars. These are the best "telescope" (actually, two telescopes!) to start with: inexpensive, wide-field of view to find objects (anti-soda straw effect), and 100x the light gathering power of the naked eye.


Evolution of galaxies with time I am doing an Introductory course in astrophysics UG level. I have studied that mass of the gases ($M_{gas}$) decreases with time as new stars are formed but mass of stars ($M_{star}$) doesn't increase because stars die out too. What about the dark matter halo ($M_{halo}$)? Does its mass increase or decrease over time?. Also how do $M_{gas}/M_{halo}$ and $M_{star}/M_{halo}$ change as the galaxy ages?

There is recycling of mass going on. As stars form they deplete the gas mass, but as they age they replenish the gas mass with a fraction through mass loss and supernovas.
Current thinking is that most halo formation has already occured, so the mass stays constant. There is perhaps a trickle of loss due to capture in massive objects (if there is some cross section between the DM particles and dense matter) including black holes, but this is vanishingly small over the stelliferous era (the next $\sim 10^{14}$ years or so). Over the degenerate era beyond this may matter more: if dark matter consists of particles and antiparticles they would slowly annihilate in the halo, and more rapidly inside the dense objects. Even if it doesn't annihilate, it will eventually very slowly fall into the central black hole in the black hole era due to gravitational radiation.
In the large, $M_{star}/M_{halo}$ begins low, as stars form it increases towards the current value and will keep on increasing across the stelliferous era (but still, it is $\ll 1$). Then around $10^{20}$ years in the future it will start declining as stars are randomly ejected from the galaxy or absorbed by the central black hole due to rare encounters, eventually becoming zero (or halving, if you still count the ejected stars). $M_{gas}$ will keep declining across the stelliferous era as it is used up, but infalling interstellar gas may slow the decline and make more stars for some time; in the degenerate era the small amount of remaining gas will keep on condensing on bodies or get ejected with them.
(See the excellent Adams & Laughlin paper for an overview of the long-term issues.)


How many arc seconds is Sirius "spiralling motion" around Sirius B? I know that Sirius A and B orbit each other. I'd like to ask how large is their relative apparent motion in arc seconds, seen from Earth?

From Sirius A to Sirius B on the sky is around 3 arcsec on average.

The separation between Sirius A and Sirius B actually varies from 3 to 11 arcseconds b/c of their orbital motion over a long period of time (Here).

I believe you're referring to that. There's a lot of information (and even some graphs and diagrams) here: Here. The information on that website is sort of vague, but I think it'll be helpful in visualization.

Hope that helps!


How far in arc seconds is a planet from its star? The star is 12 parsecs away from earth. The planet is 1 AU from its star I've been reading the book Welcome to The Universe by Niel De Grasse Tyson. In the tenth chapter he has mentioned "Forty light-years is 12 parsecs. A planet at 1 AU from its star, 40 light-years away, will be 1/12 of a second of arc away from the star in the sky."

Can anyone explain how he got the figure 1/12 arc seconds?

Because a separation of 1 au subtends an angle of 1 arcsecond at 1 parsec. That is the definition of a parsec. As you move further away, the angle gets smaller by a factor of the increase in distance.

Just draw a right-angled triangle where the base length is the distance and the height is the star-planet separation. Now increase the base length and ask yourself what happens to the angle subtended at the Earth? NB All angles are small so $\theta \simeq \tan \theta$.


Could there really be a preferential direction to the speed of light? This question is inspired by a recent video by Veritasium Why no one has measured the speed of light linked below.
To summarize the video, Derek points out that it is impossible to measure the one-way speed of light, and shows examples of what the consequences of a preferential direction for the speed of light might be.
While the examples given of the measurement problem and consequences all appear valid on local scales, they seem to me to fall apart when you look at the scale of the observable universe. Wouldn't we be able to observe the effects of a preferential direction to the speed of light at cosmological scales? Take the extreme case: the speed of light in one direction is 1/2c and infinite in the opposite direction. If this was true, we should be able to observe the entire universe in the direction where light is approaching us at infinite speed, and see no redshift due to the expansion of the universe (because the photons would reach us instantaneously, there would be no time for the expansion of the universe to stretch them out). Even in less extreme cases, we should still see "more" universe and less redshift in one direction than its opposite.
Since we don't (to my knowledge) observe any difference in the "quantity" of universe (for lack of a better term, feel free to edit if there is a better term for this) in any direction, or any difference in the amount of redshift in any direction, then if there exists a preferential direction to the speed of light, it must be small enough that its effects lie within the error bars for our ability to measure the universe on cosmic scales.
Is my reasoning correct here, or is there some effect I did not take into account that would adjust things so we wouldn't see any difference? (or perhaps my understanding of cosmology is entirely flawed?)

Prior to Einstein's 1905 paper, the Lorentz transformation had already been worked out by Lorentz and others. Only their interpretation of it was lacking. They still clung to the idea that there was a Newtonian absolute time, and the times in the Lorentz transform were only apparent times. Einstein was the first to realize that there doesn't need to be a Newtonian time; the Lorentz transform stands perfectly well on its own.
The guy in this video is thinking the same way as Einstein's predecessors; he's an aetherist, though he doesn't realize it. He's stuck to the idea that there's a real time with respect to which the true speed of light is defined, but various "effects" prevent any experiment from actually determining what it is. This is most obvious starting from 11:32, where he says he wants to show how differently the universe works if light is anisotropic, but then shows that it works exactly the same in every experimentally measurable way.
The reality is that only what is operationally measurable matters. What we mean when we say that the speed of light is constant is that there exist coordinates with respect to which it's constant. In a Newtonian corpuscular world, no such coordinates would exist, so the fact that they do exist in the real world is physically meaningful. It's not necessary to use these isotropic coordinates, but it's frequently convenient. That's the only reason we use them. To put it another way, the Einstein synchronization convention really is a convention; it isn't an assumption.
There also exist coordinates with respect to which the speed of light isn't constant. This is not physically meaningful, because no theory could ever avoid them; you can always do a formal substitution of variables, as long as it's invertible and you're consistent about it. The result of any experiment in these coordinates is always the transformation of the result of the experiment in inertial coordinates, because they both describe the same reality.
If $(x,t)$ are standard inertial coordinates, then with respect to coordinates $(x,t')$ where $t'=t-x$, the speed of light $|dx/dt'|$ ranges from $c/2$ to $\infty$ depending on direction. Why don't we see this as an anisotropy in the sky? Because the universe in different directions has aged by different amounts, and their ages differ by just the right amount to compensate for the different light travel times. This is similar to the way that length contraction, the relativity of simultaneity, and so forth always conspire to make things consistent in different inertial frames.


Will there be public access to the New Horizons data from Ultima Thule? I imagine New Horizons data from its encounter with Ultima Thule (2014 MU69) has started arriving.

Does anyone know if this data is public?

And, if so, how you can access it?

The raw images1 from LORRI (high resolution greyscale) are available from JHU APL but the rest of the data such MVIC (wide angle in greyscale plus four color bands: near IR, methane, red, and blue), LEISA (infrared spectroscopic imagery), Alice (ultraviolet spectroscopic imagery), as well as other instruments seems not to be available yet.

Some processed imagery is also available elsewhere on APL's site.

Eventually all the data will be released via the NASA PDS2 but this may take some time since it first has to be transmitted to Earth (this will take 20 months) then processed and calibrated (several more months) or (for some images) converted to a public-friendly JPEG.  Once this is done the data will be released since, per this document (p. 15), the New Horizons data is not subject to embargo by the research teams:

  There are no proprietary data rights for the New Horizons mission. Selected uncalibrated (CODMAC Level 2) data, particularly image data, will be publicly released by the New Horizons project over the Internet in close to real time. ... Fully reduced, calibrated and corrected data products .. will be [released as it becomes available].

Quoting further (from Appendix D):

  Selected uncalibrated data, including image data, will be publicly released over the INTERNET in close to real time.
  Fully reduced, calibrated and corrected data products will be published and forwarded to the Planetary Data System as soon as they
  have been generated and validated.

I've asked a question on Space.SE regarding whether there are currently any archives of raw(ish) New Horizons/Ultima Thule imagery besides the one linked above.

1 Converted to JPEG files.

2 Thanks to @astrosnapper for that link.


With gravitational lensing, is it possible to mathematically compute the correct image of the galaxy that is being distorted? There's been another report of gravitational lensing in the news.  Wikipedia has a good image of an example of gravitational lensing taken by the Hubble Space Telescope.  This brings me to wonder that, since they understand the math behind the lensing, shouldn't they be able to reconstitute the original image of the galaxy behind the galaxy that is acting as a lens?  That is, take the image and run it through a computer program that reconstructs the image using known parameters of the mass and the position of the lensing galaxy?

That would be possible theoretically. However, in practice, the lensed images are highly non-linearly distorted and due to the limits in resolution, they are often limited to a thickness of a few pixels at best. Due to the non-linear transformations, this would cause the reconstructed original image to be even blurrier.

Having said that, I don't see any real need of reconstructing the original image. You can get most parameters (like a rough estimate of the size and type of the background galaxy) from the distorted image anyway.


Could planets explode in space? Could a planet like the Earth or any other explode or be destroyed and did we see a planet explosion with Hubble?

We never saw a planet explosion with Hubble.

To make something explode it has to produce a lot of energy in a very short time. Gravitational collapse and thermonuclear fusion in degenerate matter are two possible ways of producing enough energy to produce a supernova.

These mechanisms can not cause a planet to explode. Planets are solid and supported by their own matter, so can't collapse, and they don't contain large amounts of hot degenerate matter, so don't have any kind of fusion.

Planets don't produce much of their own energy, and there is no mechanism known that could cause a planet to explode.

The closest you may get is planetary collisions, such as that which likely formed the moon. But these have never been directly observed.


Is dust extinction/reddening caused by Rayleigh scattering or some other physics phenomenon (or both)? I know from Rayleigh scattering that in the atmosphere redder photons are scattered less than bluer photons by the particles that both make up and are floating around in the atmosphere.  This is why the sky is blue and sunsets are red.  My question is, is the physics behind Rayleigh scattering (which, according to wikipedia, is due to the electric polarizability of the molecules) the same physics behind dust extinction? Additionally, is it the only physics in play or is there additional physics that causes the phenomenon?

In an astronomical context, Rayleigh scattering does not take a huge place.
The extinction is mostly due to absorption and scattering, from gas and dust. Its effect is to diminish the flux from the source (that is, a change in magnitude). Please, take care this depends on the observed wavelength!

For simplicity, we can assume the extinction factor $Q_{ext}$ as:

$Q_{ext} = Q_{scattering} + Q_{absorption}$

Where $Q_{scattering}$ is in turn composed of many elements.

As the OP above said, blue light generally suffers of higher extinction than other frequencies, so causing reddened observed light. This is due to the cross-section dependence on the frequency: if the light has a wavelength that is comparable to the size of the grain dusts (high frequency), it is more effectively absorbed, while in the opposite case it can pass through undisturbed.
This is the reason why obscured fields are observed in radio (in our galaxy, especially) or in infrared (extragalactic).

So, the answer to you question, in space is: Rayleigh scattering is not the most responsible of the extinction/reddening, you have to invoke absorption/scattering by dust and gas.
For comparison, here some explanatory numbers about the extinction efficiency in different contests:

$Extinction \sim\lambda^{-1}$

$Thomson \sim\lambda^0$

$Rayleigh \sim\lambda^{-4}$

You basically would need impressive amount of gas to account for the extinction just by Rayleigh scattering.
Instead, Mie scattering is more important in this context. Here a figure to quickly grab the concept (from here):

And here a very popular plot about extinction versus frequency from here:

Of course, if you observe from ground, you should take care of Rayleigh scattering as well.

Other references:
Lecture 1
Lecture 2


When will the next transit of Earth be visible from Mars? Was the last one really on May 11, 1984? @PM2Ring's comment mentions the story Transit of Earth written by the famous British science fiction writer, science writer and futurist,3 inventor, undersea explorer, television series host, Fellow of the Royal Astronomical Society and Commander of the Order of the British Empire and celebrated polymath Arthur C. Clarke in which the character's task is to record a transit of the Earth across the Sun's disk as seen from Mars.
According to the story, this happens on May 11, 1984.
The story also mentions that the previous occurrence would have been in 1905, and the following one would be 100 years following this one, presumably in 2084.

The radio has just printed a message from Earth, reminding me the transit starts in two hours, as if I’m likely to forget, when four men have already died so that I can be the first human being to see it, and the only one, for exactly a hundred years.
It isn’t often that Sun, Earth and Mars line up neatly like this. The last time was in 1905, when poor old Lowell was still writing his beautiful nonsense about the canals and the great dying civilization that built them.

Question: Were there in fact transits of the Earth as seen from Mars on these dates (in 1905, on 11-May-1984, and in 2084)? If they do occur at roughly 100 year intervals as suggested, there may in fact be people on Mars to see the next one presuming that Sir Clarke did his maths right (did he ever not?). On what day would that be?

You can listen to Sir Clarke read the story himself in the YouTube video 'Transit Of Earth' by Arthur C. Clarke read by himself. You can also listen here: http://recordbrother.typepad.com/imagesilike/files/transit_of_earth.mp3 and it seems download HMIcomposite_2Kx2K.mov from NASA here (though I haven't verified it yet): https://svs.gsfc.nasa.gov/vis/a000000/a003900/a003941/

Yes. The Clarke story is quite correct. Earth (and Moon) would transit the sun from the viewpoint of Mars on 8 May 1905, 11 May 1984 and again on 10 Nov 2084.

full list over a period of about 1000 years, from Wikipedia Transit of Earth from Mars

November 10, 1595   May 5, 1621     May 8, 1700     November  9, 1800
November 12, 1879   May 8, 1905     May 11, 1984    November 10, 2084
November 15, 2163   May 10, 2189    May 13, 2268    November 13, 2368   May 10, 2394
November 17, 2447   May 13, 2473    May 16, 2552    November 15, 2652   May 13, 2678

They follow a pattern of occurring after 26, 79 then 100 years.


Did Tycho Brahe improve measurements twentyfold? I am reading a book on the Coopernican Revolution where the author states that Tycho Brahe was able to make twenty times more accurate measurements than any of his predecessors.

On the other hand, I remember having heard a lecture where the speaker said that Tycho did not significantly improve measurements since Ptolemy because no new equipment (telescope) had been invented.

So, which statement is correct?

Accuracy relative to other authors is hard to estimate.

However, here's a scholarly article that shows a modern determination of the actual accuracy of Brahe's measurements in absolute terms:

http://adsbit.harvard.edu/full/1978JHA.....9...42W

The conclusion is that it depends on the instrument and it depends on the actual measurement, but as a general rule he was able to obtain an accuracy better than 1 arc minute, often down to 0.5 arcmin.

For comparison, the resolving power of the healthy, young, average human eye is around 1 arcmin. So Brahe was able to somehow make measurements with an accuracy better than the smallest details resolved by the human eye, even though there were no magnifying instruments available in his time. I believe this was achieved by averaging multiple measurements.

In other words, he was very near the limit of what was theoretically achievable before the invention of the telescope. He was certainly the best of his era by a large margin, and far better than distant predecessors such as Ptolemy. What mattered here was the methodology, and the unprecedented scale of the instruments (the armillary sphere, etc).