Is communication with proxima b possible with todays technology? Let's say we have already a base on an exoplanet (like Proxima B). Would it be possible to send information from that base to the earth with todays technology? Not to talk about communication with such a distance ...

And how would it be done? Light or Radiowaves? Would it be possible to direct it so well, that some very small fraction of that wave could reach the earth? Or is it only a question of how much energy is needed?

This is a pretty complex question.  You might have better luck at space exploration than here.

Since you asked about current technology, how NASA does it now is they have very large microwave receivers, 3 stations, 70 meters each and the transmission is in microwave frequencies.    Microwaves are likely the best wavelengths to send long range space communications because it's a comparatively low background noise and because it travels through Earth's atmosphere, so receivers can be built on the ground, which is much easier than building a 70 meter (or larger) dish in space.   I'm using this as a source, it seems reliable, though it's apparently a hobbyist, not anything official.

This picture below is a page from NASA, showing the microwave window, which is used for communication.

And the microwave receivers, called the Deep Space Network (DSN), operated by NASA's Jet Propulsion Laboratory (JPL)

The precise band used is a little tricky (and I'm not an expert), but bands for satellite communications are divided up by countries.  Article on that here.   I would think they might try to use a band in the Water Hole, 1.42 to 1.666 GHz, which is in the L band, but most deep spacecraft communication and photos sent from distant space crafts is done in the S band.    I'm not sure why that is.

As distances double, the strength of the signal is reduced by four and the signal has to stand out against background radiation, which is constant, while the signal weakens by the square of the distance.   This isn't an easy problem over vast distances.   But when you say "communicate", there's different kinds of communication, Morse Code or Binary Codes are very simple, only a few bytes per letter and messages can be transmitted that way at far greater distances.   NASA is generally interested in photographs, and our spacecraft relay pictures back to NASA by ones and zeros via microwave messages, this can be compared to internet connections for speed.  New Horizons, which recently flew past Pluto, takes about 1 hour to send 1 picture.   Very good write-up on that here.

The Mars Reconnaissance Orbiter (per Quora answer), can communicate with as fast as 2 million bits per second, but it's much closer.    How far and/or how fast (the 2 questions are related) a space craft can communicate with Earth depends on how big a transmitter they give it, how much power it has and how long it's power source is required to last.

Now, I'll try to answer your question badly, using Pluto and New Horizons as a baseline.

New Horizons is a relatively recent launch, (at least, compared to the Voyagers), it was launched in 2006, so it's using 10 year old technology.   It's currently a bit over 35 astronomical units from earth and it's still sending pictures because it sends them very slowly.   This article is 5 months old, but it says that it took New Horizons about 8 months to send half of the pictures it took of Pluto and Charon (and Nix and Hydra and . . . the others), by the equivalent of a tediously slow internet connection, (from the article)

  New Horizons is transmitting data at a poky 2,000 bits per second (March 2016), and by summer that will drop to 1,500—a fraction of one percent of
  broadband data rates here on Earth. It takes about an hour for an
  average image file to arrive, from the first bit to the last.

It's also relatively low power.   I couldn't find specifics on the wattage of it's transmitter/receiver unit, but the entire New Horizons power usage is currently about 200 watts Source and there are several components on it that share that energy.   Source.  Obviously we have the means to build much more powerful transmitters than 30 or 50 watts, or whatever it uses, but that's all that was needed for this specific mission.   New Horizons is expected to operate for 20-25 years, till about 2030, at which point it will be over twice as far from Earth as it is now, and transmitting data to us at about 500 bits per second.   Source.

A final bit of information is the angle of the beam.   (from the first New Horizons link above):

  The system includes two broad-beam, low-gain antennas on opposite
  sides of the spacecraft, used mostly for near-Earth communications; as
  well as a 30-centimeter (12-inch) diameter medium-gain dish antenna
  and a large, 2.1-meter (83-inch) diameter high-gain dish antenna. The
  antenna assembly on the spacecraft’s top deck consists of the high,
  medium, and forward low-gain antennas; this stacked design provides a
  clear field of view for the low-gain antenna and structural support
  for the high and medium-gain dishes. Operators aim the antennas by
  turning the spacecraft toward Earth. The high-gain beam is only 0.3
  degrees wide, so it must point directly at Earth.

Now, onto (approximately) Proxima Centauri.   It's about 268,000 AU from Earth, which is about, 7,600 times further from Earth than New Horizons currently is.   That requires a beam, everything else being equal, some 57 million times stronger.   That's not going to work.

I couldn't find any good information on the technology.   If a beam could be tightened from 0.3 degrees wide to 0.1, or 0.01, the first would provide 9 times as strong a signal.  The 2nd, 900 times.   A .01 degree wide beam from Proxima Centauri would spread out to about 46 astronomical units by the time it reaches Earth, so I would think, the tighter they could make the beam the better, but at .01 degrees, the signal strength would still be about 1/60,000th the strength of New Horizons current transmission.   .001 degrees, and I have no idea if that's possible, but that tight a beam, you're getting into the "this could work" range, about 1/600th the strength of the New Horizon's signal, and with greater power behind the signal and, perhaps, a larger satalite dish built on Earth, then it's in the range of workable.  (but I have no idea if microwave beans can be sent over just .001 degrees)

Presumably with a much more energetic transmitter, say a few thousand watts and a very tight beam and they might be able to get a good enough transmission for Morse code, but, I would think, no pictures, only a letter at a time, like the old telegraph communication.

Do we have the technology to do it today?   Well, to me, it looks pretty touch and go.   I think, maybe, if there was a human colony already there, with nuclear power plants and equipment to build a large transmitter/receiver then, if they can get the beam tight enough, we probably could exchange dots and dashes with today's technology.

Can we build a spacecraft that could communicate to us from 4.2 light years away - no, I don't believe we could do that with today's technology, not at that distance.

All that said, communication is a far less difficult problem than actually getting there.

I hope my layman's answer isn't too far off.


Computing radial-velocities from Cross Correlation data How can I better fit a Gaussian curve to a CCF so that I get the most precise RV value? The image below shows the fitting where I compared the fitting by weighting by the uncertainties and not. There is not a big difference between them because the errors are nearly the same for all data points.

The RV is the $\mu$ and the uncertainty was gotten from the covariance matrix first time in the diagonal. Performing a similar procedure but for summing several orders (such as fig. below) gives me the RV time-series which has too much spread. I need to find a way to reduce the noise as much as possible. I am not showing the errors in RV time-series because I thought it to be coming from the covariance matrix but it looks unrealistic (too big).

Why not fit the orders you are interested in separately and then use the standard error of the mean (possibly weighted by the signal-to-noise in each CCF) as the precision in the final, averaged RV.
In terms of what to fit, I don't see why a Gaussian is so bad? You probably need to limit the fit to the inner $\sim \pm 1 \sigma$ to avoid noise outside the peak pulling the fit one way or another. Other options are to just use the numerical centroid (but this may be affected by asymmetry) or you could use a sinc function to the central region (sometimes a better model of the peak of a CCF).
Edit: You haven't followed my advice, which was to limit the Gaussian fit to $\pm 1$ sigma from the peak (do it iteratively). At the moment, the "wings" of the Gaussian are just adding noise.
As a rule of thumb though, you are not going to do much better than $2.2\sigma$  divided by the signal-to-noise ratio. It looks like your $\sigma \sim 2$ km/s and your signal-to-noise ratio is about 30, so I don't see how you can have gotten a scatter as small as you have from data like this? And the graph below contains no points with the value $-10.76$. There is still something missing from your question.


Identifying cataclysmic stellar phenomena Cataclysms such as novas, kilonovas, supernova, magnetars, quasars, gamma ray bursts are rarely visible from Earth.   Under what conditions (such as distance, day vs. night, naked eye vs. level of optics) would such phenomena be visible from the Earth's surface, and what would they look like in terms of shape, initial duration, brightness, and so forth?

For instance, I have seen questions like how long does a supernova last and I have read that they can be visible for some time afterwards as a nebula forms, but this doesn't tell me what the initial burst would look like.  For instance, would it be possible to see a supernova in a neighboring galaxy such as Andromeda?  If so, how long would it be visible with the naked eye?  Would it look like a flash, last a few seconds, several?  I assume with a supernova not in our own galaxy there would be no subsequent light from the expanding shell since it would be too far away, whereas in our own galaxy if it were close enough we would subsequently see it, but in that case would there be a short burst of light, following by darkness and then the re-emergence of light, or would it be a continuously visible phenomena?

Any of these phenomena would be visible to the naked eye if they occured in the Milky Way, or a neighbouring galaxy (such as the Magellanic clouds) and was not obscured by dust. At the distance of stars, all of these would appear star-like.

Supernovae have been seen, most recently in 1987, when there was a supernova in the Large Magellanic cloud. It had a peak magnitude of 2.9, (about as bright as gamma ursa minor, the star at the bottom of the Little bear constellation, furthest from the polestar). It was visible, but not spectacularly bright. Supernovae that have occurred in the Milky way have been very bright, outshining all the other stars. A supernova in 1006 may have been brighter than Venus. If it is close enough a supernova could be seen during the day.

All of these are highly energetic, and if you were close enough to see any detail (such as a nebula) with the naked eye, you would be toast. The exposive events (Supernovae, Kilonovae, Gamma ray bursts) would fade over time. Magnetars and Quasars are stable and would continue to shine. Remnant nebulae from Supernovae are visible in telescopes, the most well known is the Crab Nebula, M1.

A supernova in Andromeda would likely be naked eye visible, but not impressively bright, though it depends on the acutal power of the explosion, and other factors, like dust in Andromeda.

Supernovae are visible for months. Kilonovae fade faster. Gamma ray bursts may only last for a few seconds (though may be associated with a supernova that lasts longer). The visible element of a GRB might be visible, there was one that might have reached naked eye visibility, though there is no evidence that anyone actually saw it (it would only have been marginally visible in ideal conditions). Quasars are stable objects that don't fade over time, they are active supermassive black holes.

There are lots of other, more local things that can produce a "star" that brightens then fades: Planes have headlights, and as they turn the appear like a star that brightens then fades over a few seconds. If they are distant enough, the other lights might not be visible.


Is the Sun homogeneous? Does the Sun have the same composition and density everywhere within it? How does it vary in other stars?

No, the sun and all other stars do not have the same composition and density within them.

The composition varies with depth. Most stars overall composition reflects the interstellar medium from which they form. Convection, settling and nuclear processes then lead to layers of different elements at differing depths.

Fusion occurs at the core of our sun (where pressure and temperature are greatest) therefore there is a greater abundance of Helium at this depth (through conversion of hydrogen). Other stars that have entered the red giant phase will also have many more heavier elements. Very late stage stars are thought to fuse these elements progressively, leading to layers of decreasing atomic number elements stacked on top of each other.

Convection can also lead to mixing of elements between layers. The diagram below shows theoretically how these convective layers differ over a range of stellar masses

The density increases as you move closer to the center of the star. Note how the density is less than that of water for up to half the sun's radius!


Is there a name for this type of orbit of a natural satellite? Is there a name for the orbit of a natural satellite of a planet who's period is the same as the rotational period of the planet, but the orbital motion has the opposite sense than the planet's rotation?

If there isn't an existing term or name, how would it at least be described, technically?

How much does the tidal effect affect satellites in a retrograde orbit?

No.  Such an orbit would be co-incidental and dynamically uninteresting.

For a planet of given mass and rotation period, there is a class of orbits, whose period is equal to the rotation period of the planet. If the orbit is prograde this gives an interesting effect: the satellite would remain in the same position in the sky (or if it had an elliptical orbit, it would return to the same position each day)

If the orbit is retrograde, the orbit would not have any particularly interesting features. While it is possible for tidal locking to produce a synchronous orbit, there is no mechanism to stabilise a retrograde orbit like this, and tidal effects will slowly disrupt it.


Move asteroid away with white color? If we find asteroid flying towards us, can we change it's direction with white painting? Will the ray from sun do something?

@called2voyage's comment points you to an in depth discussion. I'll add to that a simple summary of some things.

If you imagine that the astroid is in a simple, circular orbit around the Sun that's defined only by gravity, let's call that the nominal orbit.

Now consider the effects of sunlight.

If the asteroid were white or reflective, then a lot of the incident sunlight will reflect back towards the Sun, pushing the asteroid away from the Sun. Since both gravitational attraction and the Sun's brightness vary as $1/r^2$, so to the asteroid it feels like gravity is just a little weaker or the Sun is a little lighter, but otherwise still feels like gravity. If the asteroid became reflective slowly, it would just move to a slightly larger radius and remain in circular orbit.

Remember that the incident photons, and the reflected photons both contribute to the push!
If the asteroid were black and absorbing, then the absorption of the photons would give a push, and then they would warm the asteroid. Warm things radiate heat, and once equilibrium were reached, the radiation would also be a source of pressure.

If the astroid didn't rotate fast, but instead turned slowly so that it had a warm side always facing the Sun and a cold side always facing away, then the radiated infrared photons would also result in a push away from the Sun! So a dark asteroid that turned so slow that it had a permanent hot and permanent cold side would also be at a larger radius than nominal. The difference between the two depends on the functional form of the diffusivity and roughness of the surfaces.
However if the dark asteroid were rotating fast, say every few hours or tens of hours, then the radiation would be a little stronger in the forward or backward direction with respect to the orbit. If it was rotating the same sense as its orbit, the hot side would turn backwards, radiate backwards, and push the asteroid forward. That would slowly, but continuously raise the orbit of the asteroid. The longer this happened, the larger and larger its radius would become.

This is called the Yarkovsky effect and over long periods of time (millions of years) can have a big impact. For an asteroid to just miss the Earth, maybe a few years would be enough.

It turns out that @ BobJacobsen's answer to Why paint only one-half of Bennu? is, in a highly condensed and stylized form; "because the Yarkovsky effect is strong enough that there's no point in wasting good 'space paint' on the other half."
If you are a tiny dust particle near the Sun, there is yet another thing that happens,  called the Poynting–Robertson effect.

Tiny bits of dust have a higher area to mass ratio, so even though they may be in thermal equilibrium and not have a hot and cold side, tiny forces make a big difference. And if you are close to the Sun, then the forces are less tiny i.e. larger. @honeste_vivere's excellent answer points out that the Poynting–Robertson effect can cause the radius of the dust particles' orbits to decrease over time. An effect similar to (identical to) astronomical aberration will cause more pressure at the front of the dust particle, pushing it backwards and causing it to spiral closer to the Sun over time.


Why can we see stars but not astronaut on the moon A question came up today in astronomy class and it got me wondering all night.

How is it that if stars are so far away, we can see them so clearly but we cannot see astronauts on the moon.

An astronaut on the moon could only be seen by reflecting the Sun's light towards Earth. Stars on the other hand emit their own light.

To first order, the amount of flux incident upon the Moon from the Sun is the same as that at the Earth - about 1.4 kW/m$^{2}$.

Let us assume that an astronaut is perfectly reflective and that the relevant reflective area that we can see from the Earth is 1 m$^2$. NB: If the astronaut is not lit up by the Sun, then there is obviously no way that they can be seen.

Treating the astronaut on the Moon as an isotropic point source emitter of reflected light, we have a light source of power 1.4 kW at a distance of 400,000 km. The flux at the Earth is therefore $7\times10^{-16}$ W m$^{-2}$.

How does that compare with starlight? Well, the total luminosity of the Sun is $3.8\times10^{26}$ W. It has an absolute magnitude of 4.8. This means that if we put the Sun at a distance of about 20 pc, it would be about as faint as the faintest naked eye star in the sky. The flux received at the Earth from such a star would be $8\times 10^{-11}$ W m$^{-2}$ and thus 100,000 times brighter than the astronaut.

No need to worry about the resolution of the eye, since both the star and the astronaut (at the distance of the moon) are unresolved points.

Also no need to go into the problems of contrast against the moon's bright surface (which you would need to consider if an astronaut's reflective area was 100,000 times bigger), the reflected light from the astronaut is just too faint to be seen at that distance.


How to plot celestial equator in galactic coordinates? Why does my plot appear "backwards"? I'm trying to plot celestial equator in galactic coordinates. Here is my code (based on this answer):
import matplotlib.pyplot as plt
import seaborn as sns; sns.set()
import numpy as np

from astropy.coordinates import SkyCoord
from astropy import units as u

ra_all = np.linspace(-180, 180, 100)
dec_0 = np.zeros(100)

# Transform equatorial coordinates to galactic
eq = SkyCoord(ra_all, dec_0, unit=u.deg)
gal = eq.galactic
l_plot, b_plot = gal.l.wrap_at('180d').radian, gal.b.radian

lon = np.linspace(0, 360, 100)
lat = np.zeros(100)

# Transform ecliptic coordinates to galactic
ecl = SkyCoord(lon, lat, unit=u.deg, frame='barycentricmeanecliptic')
ecl_gal = ecl.transform_to('galactic')
l_ecl_gal, b_ecl_gal = ecl_gal.l.wrap_at('180d').radian, ecl_gal.b.radian

plt.figure(figsize=(14,7))
plt.subplot(111, projection='aitoff')

plt.scatter(l_plot, b_plot, s=5, label='Celestial Equator')
plt.scatter(l_ecl_gal, b_ecl_gal, s=5, label='Eclptic')

plt.grid(True)
plt.legend(fontsize=16)

plt.show()

However, the map I drew seems to be flipped left to right compared to the one bellow (taken from here)

Question 1: Why does the result of my implementation appear to be the reverse of the other image? Do I have something wrong, or is this simply an issue of mapping the surface from the inside versus from the outside?
Question 2: What is the best way to change my code to make it draw galactic map in a conventional format?

Ok, community helped me to figure out that I faced the “mapping from the inside” issue (explained here). I'll use this answer to show my final code.
import numpy as np
from astropy.coordinates import SkyCoord
import astropy.units as u
import matplotlib.pyplot as plt
import seaborn as sns

sns.set()

def eq2gal(ra, dec):

    '''
    Transforms equatorial coordinates to galactic ones.
    Then prepares them for matplotlib aitoff projection.
    '''

    eq = SkyCoord(ra, dec, unit=u.deg)
    gal = eq.galactic

    # Minus appears because of “mapping from the inside” issue
    l_gal, b_gal = -gal.l.wrap_at('180d').radian, gal.b.radian

    return l_gal, b_gal

def ecl2gal(lon_ecl, lat_ecl):

    '''
    Transforms ecliptic coordinates to galactic ones.
    Then prepares them for matplotlib aitoff projection.
    '''

    ecl = SkyCoord(lon_ecl, lat_ecl, unit=u.deg, frame='barycentricmeanecliptic')
    gal = ecl.transform_to('galactic')

    # Minus appears because of “mapping from the inside” issue
    l_gal, b_gal = -gal.l.wrap_at('180d').radian, gal.b.radian

    return l_gal, b_gal

# Equatorial plane
ra_all = np.linspace(-180, 180, 100)
dec_0 = np.zeros(100)

l_eq_gal, b_eq_gal = eq2gal(ra_all, dec_0)

# Ecliptic plane
lon_ecl = np.linspace(0, 360, 100)
lat_ecl = np.zeros(100)

l_ecl_gal, b_ecl_gal = ecl2gal(lon_ecl, lat_ecl)

plt.figure(figsize=(14,7))
plt.subplot(111, projection='aitoff')

plt.scatter(l_eq_gal, b_eq_gal, s=4, marker='v', label='Celestial Equator')
plt.scatter(l_ecl_gal, b_ecl_gal, s=4, marker='^', label='Eclptic')

# Essential thing is to rename RA axis ticks to transform them to conventional format
plt.xticks(ticks=np.radians([-150, -120, -90, -60, -30, 0, \
                             30, 60, 90, 120, 150]),
           labels=['150°', '120°', '90°', '60°', '30°', '0°', \
                   '330°', '300°', '270°', '240°', '210°'])

plt.grid(True)
plt.legend(fontsize=16, loc='lower center')

plt.title('Galactic', fontsize='16')

plt.show()


How can we tell that the Milky Way is a spiral galaxy? From Earth, we can distinguish the type of another galaxy very easily simply by observing the shape, colour, and structure of the galaxy when we image it. But since we are located inside the Milky Way, we cannot get an "outside view". Thus, how can we tell if we are located in either a spiral or an elliptical galaxy?

There are several clues.
The Milky Way is a flat disk
The first one, and the simplest one, is that we live in a disk. As you can see on images like the ones from the 2MASS survey in the infrared range:

We clearly see that we are inside a disk, since what we see when we look around us is a disk seen edge-on. It does not look like an ellipsoid or any other shape; it looks like a disk.
The Milky Way has arms
The other strong clue we have is the velocity of stars of the Milky Way, we can measure. First, we see an overall rotational motions, and, when you look closely to the patterns (like in the following image, that is observations in the radio range of the Milky Way at Green Bank Observatory), you can deduce that stars are even ordered into spirals:


Why can't we see distant galaxies with the naked eye? If light keeps travelling in a straight line, why can't we see distant galaxies with the naked eye? Surely if you stared long enough, the light from them would eventually hit your eye? I apologize if this is a dumb question:)

Surely if you stared long enough, the light from them would eventually hit your eye?

Collecting light over a long span of time is how telescopes can see very dim objects. The human visual system doesn't work that way.

For one thing, even when you think you are staring at something, your eyes still dance around a bit. It's a built-in response called ocular microtremors. These microtremors appear to be an essential part of making the vision system functional.

For another, your eye does not and cannot collect light for arbitrarily long periods of time (the way a photographic telescope can). There's an immense amount of signal processing that happens in the eye and along the way to the brain. This signal processing depends on light being collected for short spans of time.

Our vision system evolved to see food, friends, and dangers under well-lit conditions. We are very good at seeing motion in broad daylight. We're not so good at seeing stationary objects, and we're not good at all at seeing barely visible sources under very dark skies.

Naked eye astronomy is limited by the nature of the human vision system. The most remote object we can see is the Triangulum Galaxy, and that's only under conditions of extremely dark and extremely clear skies.


How can a mirror see things that the naked eye can’t? How does a mirror pick up light from vast distances away? And, if that light can reach a mirror, why can’t we see that light?

Your eye has two functional parts: a lens (that is about ¼cm²) and a light-sensitive surface (the retina) that is covered in rod-cells that detect light. It takes about 10 photons arriving at about the same time (0.1sec) for a rod to react to the light and send a signal to the brain.

This means that, unless about 400 photons arrive from a star onto every square centimetre then there isn't enough light to see the star, even under ideal conditions.  In practice, many more photons are needed for your brain to be able to interpret the signal from your eye as a point of light.

In contrast the mirror of a telescope is much larger. A small amateur telescope could have a mirror that has an area 1000 times larger than the lens of your eye. This means that a star can be 1000 times dimmer, but still be visible when you look at it in the telescope. Large, professional, telescopes have mirrors that are millions of times greater than the area of an eye.

Moreover a telescope can be fitted with a camera, and film (or digital sensors) that are both more sensitive to light than your retina and can integrate the light arriving over a long period of time.  If you make a long exposure, of several minutes, or even hours, then even dimmer objects will become visible.  This can increase the sensitivity by a factor of many thousand again.

The combination of the large light gathering area of the mirror, with the sensitivity and long exposures make possible to "see" things using the mirror of a telescope that can't be seen with the naked eye.


Tools for simulation of theoretical astronomical problems? I tried looking around for open source tools that would allow me to run simulations and build on top of the code with my own need(mainly mathematical simulations, but 3D visuals would be a bonus.), unfortunately I haven't really found many open source projects, so I think maybe this tools are somehow available on websites like NASA or other Foundations for astronomical research. Where would one go to find the actual professional software for astronomy research and simulations ?

Give GADGET-2 a check if you haven't already.

GADGET is a freely available code for cosmological N-body/SPH simulations on massively parallel computers with distributed memory. GADGET uses an explicit communication model that is implemented with the standardized MPI communication interface. The code can be run on essentially all supercomputer systems presently in use, including clusters of workstations or individual PCs.
GADGET computes gravitational forces with a hierarchical tree algorithm (optionally in combination with a particle-mesh scheme for long-range gravitational forces) and represents fluids by means of smoothed particle hydrodynamics (SPH). The code can be used for studies of isolated systems, or for simulations that include the cosmological expansion of space, both with or without periodic boundary conditions. In all these types of simulations, GADGET follows the evolution of a self-gravitating collisionless N-body system, and allows gas dynamics to be optionally included. Both the force computation and the time stepping of GADGET are fully adaptive, with a dynamic range which is, in principle, unlimited.
GADGET can therefore be used to address a wide array of astrophysically interesting problems, ranging from colliding and merging galaxies, to the formation of large-scale structure in the Universe. With the inclusion of additional physical processes such as radiative cooling and heating, GADGET can also be used to study the dynamics of the gaseous intergalactic medium, or to address star formation and its regulation by feedback processes.


Why does time pass faster in moon than the earth? One of my friends said that,

time passes faster in moon than the earth

Why does time pass faster on the moon than on the earth?

Okay, this question is actually quite simple and requires some background on the relations between gravity and time.
The property that can be used to answer this question is :

Time is affected by gravity in the following way : Imagine you are on an exoplanet. The gravity there is twice that gravity here on Earth. Now, assuming that we take Earth Time as standard, the Exoplanet's time should be slower than Earth-Time.

Basically, the property states that the greater the gravity of the place, the greater will be the time dilation effect.
Here, for your question, we can apply the same principle. Since Moon-Gravity is 1/6th of Earth's, time should pass faster there.
@MarkAdler has given an excellent answer here, on the Stack Exchange for Space Exploration. To quote his answer :

Time passes about 0.66 parts per billion faster on the Moon than on Earth, due to not being in as strong a gravity field. The dilation, including all the effects of being further from the Earth's gravity field, being in the Moon's gravity field, and the velocity of the Moon, could easily be measured in sum by comparing good atomic clocks on the Earth and the Moon.

He has also posted some maths on the same, so I recommend checking out the link above :
Hope this answer helps,


Why do spectroscopic binaries have approximate circular orbits? So for an assignment I have to answer the question what I can conclude concerning the shape of the orbit, knowing that it is both an eclipsing and spectroscopic binary.

Now, I think the answer I supposed to be that, since we know it is a spectroscopic binary, the orbits will most likely be circular. And I've read on various places that this is true, because of the tidal forces that cause for circular orbits. These tidal forces are quite strong, because apparently spectroscopic binaries are quite close to each other.

So here is where I lose it, I don't understand why spectroscopic binaries should per se be close to each other. I have considered that this is because they are not visual binaries; if they were visual binaries we would be able to tell them apart. But we can't tell them apart, so they are not visual binaries. From this you could conclude that this means that they are close to each other, but I think it could just as well mean that they are very very far away, and that that is the reason we cannot tell them apart. So I can't find a proper explanation for this.

Many thanks if you can help me out!!

Spectroscopic binaries are not necessarily close and not necessarily in circular orbits. If the orbit is circular, then a time series of radial velocity measurements for one or both stars would be well modelled with a simple sine wave.

However, eclipsing binaries are almost certainly close, since eclipses in a system with widely separated components are improbable (but not impossible). If you have the eclipsing light curve, then the primary and secondary eclipses will be equi-spaced in time if the orbit is circular.

The key factor for circularisation is the orbital period. If it is less than about 7 days (for solar-type stars), then circularisation due to tidal forces is almost certain.


What is the accuracy of an artificial satellite at a certain magnitude? When we say that 'accuracy of gaia satellite is 10 $\mu$as at $V = 10$ magnitude in position and annual proper motion'. What does this mean? Does it change over time?

What you are referring to is the precision with which the astrometry (position) of stars will be measured with the Gaia satellite.

The sentences you refer to in Perryman et al. (2014)  are "Gaia will achieve accuracies of some 10μas (microarcsec) in positions and annual
proper motions for bright stars (V ∼ 10), degrading to around 25μas at V = 15, and to around 0.3 mas (300μas) at V = 20 (Lindegren et al. 2008)."

What this means is that after the 5-year Gaia mission is complete, the angular position (or coordinates) of 10th magnitude stars on the sky will be determined with a relative precision of 10 millionths of an arcsecond.  This will also be the accuracy with which the annual tangential motion (the so-called proper motion) of stars can be measured. i.e. 10th magnitude stars will have an uncertainty in their proper motion of $\pm 10 \mu$as/year.


Low mass cut-off for the Dharma Planet Survey's detection of habitable planets around 40 Eridani A? News of the discovery of a potentially habitable planet around the star 40 Eridani A (about 16 light-years away) is interesting for reasons including a 1991 letter described below in Sky & Telescope's recent article Super-Earth Discovered in (Fictional) Vulcan System, part of which is shown below.
Question: Is there any way to estimate a low mass cut-off for the Dharma survey for this particular star? Could there also be a 1 or 2 Earth mass planet there as well, within a habitable zone, as-yet undetected? Or does data there exclude this possibility?
The paper about the recent discovery: https://arxiv.org/abs/1807.07098

Almost three decades ago, Gene Roddenberry (producer of the Star Trek universe) wrote a letter to Sky & Telescope, along with Harvard-Smithsonian Center for Astrophysics astronomers Sallie Baliunas, Robert Donahue, and George Nassiopoulos. In their Letter to the Editor, they argued that 40 Eridani A — an orange-ish star 16 light-years away — would make the ideal home for Vulcan, the home planet of Science Officer Mr. Spock.
Now, a new discovery puts a little more science into that science fiction assertion.
A Long-ago Letter
In the July 1991 issue, the three astronomers and one movie-maker made the case for what star should be considered Vulcan's home:

The star around which Vulcan orbits was never identified in the original series or in any of the feature films based on it and so has never been officially established. But two candidates have been suggested in related literature.

Two Star Trek books named the star 40 Eridani A as Vulcan's sun, while another publication named Epsilon Eridani instead. However, Roddenberry and the astronomers made an argument for 40 Eridani A:

We prefer the identification of 40 Eridani as Vulcan's Sun because of what we have learned about both stars at Mount Wilson. . . . The HK [Project] observations suggest that 40 Eridani is 4 billion years old, about the same age as the Sun. In contrast Epsilon Eridani is barely 1 billion years old.

Based on the history of life on Earth, life on any planet around Epsilon Eridani would not have had time to evolve beyond the level of bacteria. On the other hand, an intelligent civilization could have evolved over the aeons on a planet circling 40 Eridani. So the latter is the more likely Vulcan sun. . . . Presumably Vulcan orbits the primary star, an orange main-sequence dwarf of spectral type K1. . . . Two companion stars — a 9th magnitude white dwarf and an 11th magnitude red dwarf — orbit each other about 400 astronomical units from the primary. They would gleam brilliantly in the Vulcan sky.

In general, finding the low mass cutoff of a particular RV survey is normally quite complicated as there are a lot of selection effects on many different variables (such as planet period, planet eccentricity, host star type, signal-to-noise, stellar activity level etc) that determine whether you can recover a planet signal. In the case of this paper, this would be particularly true as they are combining datasets from several instruments, each with different coverage, sensitivity and systematic noise properties. The way this is normally handled is through a "bootstrap" technique where you inject planet signals of known parameters and see what your detection code finds and study the recovery percentage as a function of the varied parameters.

Using the online Habitable Zone calculator for a Teff=5072 K, Stellar luminosity=0.47 * L_Sun (luminosity of the Sun) matching the host star, produces 0.678 AU as the inner edge of a conservative habitable zone. If we use this, the star's mass from the paper (0.78 M_Sun) and the 1 M_Earth mass you would like to find, then making use of the formulas given on the Exoplanet Archive website, gives us a period of 230.9 days. Plugging this into an equation for the radial velocity semi-amplitude (also on the Exoplanet Archive), produces an amplitude of about 12 cm/sec. This is assuming an edge-on orbit, so the sin i term from the inclination is 1; given the lack of detected transits in the paper, the inclination is likely lower than 90 degrees, which would lower the RV amplitude further.

This RV amplitude is far too small to be detected with current instruments. The best instrument we have right now is HARPS on the ESO 3.6m telescope, whose data is included in the paper. If I am understanding their modeling correctly, they get error bars of 181 cm/s on the HARPS data for this star in their analysis, 15x the expected signal. Also note that the calculated RV signal is somewhat best case (unless it is more massive); decreasing the inclination from edge-on (likely) and moving the planet further out in the habitable zone from the inner edge I calculated (also likely) will both decrease the RV amplitude further. Even for future instruments such as ESO's ESPRESSO which will go the 8.4m VLT and reach 10 cm/s precision, this would be a challenging measurement.


Is there a significant temperature difference on the moon from apogee to perigee or vice-versa? Do those few thousand kilometers of difference going around the earth significantly affect the amount of thermal energy the moon receives from the sun to warm up its surface?

Also, are there more significant differences between these variables when the earth goes from perihelion to aphelion and vice-versa?

Roughly speaking, the moon's orbit has a semi-major axis of about 400,000 km. The semi-major axis of the Earth's orbit is 149.6 million km. The intensity of sunlight received by the Moon depends on the inverse of the square of the distance, so the difference amounts to $((1.496e8 + 4e5)/(1.496e8 - 4e5))^2$ = 1.011, or about 1%. So, not very much.

The variation in Earth's distance from the Sun is about five million km, so that's going to have more of an effect than the lunar orbit itself (as Chappo pointed out in their comment). That works out to a variation of about 6.9% in insolation from perihelion to aphelion. The maximum possible variation -- full moon at aphelion versus new moon at perihelion -- would be $((1.521e8 + 4e5)/(1.471e8 - 4e5))^2$, which works out to about 8.1%.

The difference in insolation from new moon to full moon at perihelion is about 1.09%, versus a difference of 1.06% at aphelion.


Could a moon be passed from one planet's orbit to another? Like if you had two planets that had a close orbital path could the moon be pulled away from the first planet?

Given the nature of 3 body orbital mechanics, it's possible, just unlikely and I hope you'll forgive me if I skip the math.

In our solar-system, it's pretty much not possible because our planets are too far from each other.   Speaking in solar-system distances, Moons have to be close to planets, within the stable regions of their planet's hill spheres in order to have long-lasting orbits.

I'm going to assume that I can ignore temporary moons, which in my opinion, aren't real moons anyway.

The difficulty with one planet giving another planet it's moon is that they'd have to pass quite close to each other, and one of the definitions of planets is that they have their own orbital region around their star, but orbits can move over time, so it's possible for two planets to pass close to each other but there would be other effects as well because, as they got close, the planets would interfere with each other as well.

In a just right scenario a moon could theoretically be passed from one planet to another during a close pass.   This is similar to the theory about how an asteroid in a binary-asteroid pair could get captured if it passes close to a planet.

A planet could also lose a moon and maybe in a few million years that moon could find it's way to being captured by another planet, but I don't think that's what you're asking.

You might think that around a red-dwarf where planets are quite close to each other and much closer to their star, that a transfer could happen due to proximity, but the difficulty there is that planets that close to their star have a hard time having moons at all because they have very short Hill radii.   It's probably more likely, but still unlikely, to happen between two distant planets.

Take the planet Neptune, as an example.  Neptune has a Hill radius of a bit over 100 million km, about the distance between Venus and the Sun, and it's true region of stability, between 33% and 50% of that, so Neptune could in theory have a moon in a stable orbit as distant as perhaps up to 50 million km.  At that distance, it's moon would orbit it very slowly, about .37 km/s with an orbital period of about 25 years.

(A fun mathematical quirk, which has nothing to do with your question, but if you calculate the maximum orbital period a moon can have inside the stable orbital region around it's planet, it works out to about 1/6th the period the planet has around it's star, it doesn't matter the mass of the planet or the distance to the star that 1/6th rule roughly applies to every scenario).

So in theory, if Uranus passed close to Neptune, say within 100 million km, the theoretical distant moon, especially if it had a somewhat elliptical orbit, could get transferred from Neptune to Uranus and as is passed.  In the process, it would become a wrong-way moon around it's new planet.  (insert wacky theory about Triton having once orbited Uranus here).

Uranus and Neptune never get that close, though perhaps they did when our solar-system was young and Jupiter was migrating.

It's more likely, if two planets got that close that they'd alter each other's orbits rather than trade a moon.  The moon could also be ejected from one planet and not captured by the other.  That's probably a more likely scenario.   If one of the planets was considerably smaller than the other, the entire planet could be tossed around significantly, perhaps tossed out of the solar-system.

What probably does happen, is binary systems like Pluto-Charon, when they pass close enough to a planet, one of them can be captured.  This article suggests that Triton, Neptune's wrong way moon, was captured by that method.  A capture involving 3 gravitational bodies isn't that strange, but the criteria that two of those bodies are planets makes it more rare.

3 body orbits allow for a variety of scenarios, and trading a moon is probably one of the least likely options but theoretically possible.  This just might happen when solar systems are young and chaotic and full of planetesimals.

On a related note, I've read an article a while back that stars in the center of our galaxy can trade planets, but I looked, I couldn't find that article.


Any possibility for life on moons of planets in our Solar System We all focused on our moon, to find water or life, or any thing related to life. But my question is, is there any chance for life in moons of other planets in our solar system, others have too many moons....

If you check 2001 A Space Odyssey saga, Europa is actually a satellite with Life.

Coming back to real space, the "comfort zone" for life is approximately from Venus' orbit to Mars' orbit. There, only Moon, Phobos and Deimos are satellites, and all of them are too dry to have any kind of life.

Beyond the comfort zone you need some other way for water not to frozen (a commonly referred prerequisite of Life), so you'll need a volcanic or tidal-heatened satellite. There are a few of these. Europa and Io come to my mind just now. Life there could be possible.


Sunrise - from earth - sun position or when sun rays reaches earth? Time for the sun rays to reach earth is approximately 8 mins.

If time of the sun rise is 7:30 - which time is this? Time when sun rays reaches earth or time when earth and sun comes into correct position where sun is visible again after night?

If it is time when sun rays reaching earth - that means sun is already 8 mins away from that first position.

If it is second option - it means it's still 8 mins to wait until we actually see 'sun rising' - sun rays reaching earth.

So which time sun rise really is?

It's when we see the Sun's rays and includes refraction. It also is the moment we see the first glimpse of the sun. Think of it this way: we humans need to know sunrise and sunset for our human activities; the fact that the sun is about eight light-minutes away doesn't concern us, doesn't affect the things we do. BTW, this is also true of solstices and equinoxes.


What would happen if an ice cube is left in space? Recently I boarded a flight and noticed outside air temperature as -53°C at an altitude of 36860ft (11.23km). I don't know what causes such a freezing temperature in that altitude but was wondering higher altitudes (space) may have even freezing temperatures. Here I got a doubt i.e what happens if an ice cube is left in space? Would it be melting or stay as it is?

It depends on where in outer space you are.

If you simply stick it in orbit around the Earth, it'll sublimate: the mean surface temperature of something at Earth's distance from the Sun is about 220K, which is solidly in the vapor phase for water in a vacuum, and the solid-vapor transition at that temperature doesn't pass through the liquid phase.  On the other hand, if you stick your ice cube out in the Oort Cloud, it'll grow: the mean surface temperature is 40K or below, well into the solid phase, so it'll pick up (or be picked up by) gas and other objects in space.

A comet is a rough approximation to an ice cube.  If you think of what happens to a comet at various places, that's about what would happen to your ice cube.


How long does it take for a white dwarf to cool to a black dwarf? I was reading on white dwarfs, and I came across this sentence—

  Without energy sources, the white dwarf cools to a black dwarf in a few billion years.[1]

However, when I looked into the Wikipedia page on White dwarf, it says

  Because the length of time it takes for a white dwarf to reach this state is calculated to be longer than the current age of the universe (approximately 13.8 billion years), it is thought that no black dwarfs yet exist.

So which is true?

And what is the proper definition of a black dwarf?

References:

[1] Introductory Astronomy and Astrophysics. Zellik, M. Gregory, S. 4th edition. Brooks/Cole. 1998

I don't think there is an accepted definition of a "black dwarf" - it is not a term used in the scientific literature.

A popular definition that appears to circulate on the internet is that it is a white dwarf that has cooled down to the extent that it no longer emits any radiation in the visible part of the spectrum. But this is an unworkable theoretical definition. Even the coolest objects emit some radiation in the optical part of the spectrum; whether or not it is detectable will depend on the size of the object and how far away it is from us.

If a black dwarf is simply to be given the practical definition of being unobserved/unobservable at visible wavelengths then such objects exist already. Kaplan et al. (2014) have found evidence for a $\sim 1M_{\odot}$ white dwarf companion to a pulsar that they are unable to detect at visible wavelengths even with the Keck telescope down to an absolute (red) magnitude of $M_R>19.1$.

They argue that this means the white dwarf has cooled below 3000 K. This is possible within the age of the universe because more massive white dwarfs cool more rapidly because their higher interior densities lead to more rapid crystallisation and a steep decrease in their heat capacity when they enter the Debye cooling regime. A more typical old white dwarf of mass $\sim 0.6M_{\odot}$ would still be at a temperature of $\sim 4000$K and emit a reasonable amount of light at visible wavelengths.

Thus, if you are happy to accept the definition above, then black dwarfs can result from the cooling of massive white dwarfs in about 10 billion years. The plot below, taken from the paper of Kaplan et al. (2014), shows some cooling models for massive white dwarfs with hydrogen atmospheres. They easily cool below 3000 K within 10 billion years.

To address Mark Foskey's point: Once in the Debye cooling regime, the cooling rate hardly slows because the heat capacity is $\propto T^3$, whilst the luminosity is roughly $\propto T^{3.5}$ (here, $T$ is the isothermal interior temperature). Luminosity is also $\propto T_s^4$, where $T_s$ is the surface temperature.

The relevant differential equation is that the rate of change of thermal energy equals (minus) the luminosity
$$ \frac{d}{dt}\ \int T^3\ dT \propto -T^{7/2}$$

Putting this together, we find that $dT_s/dt \propto -T_s^{3/7}$ and so the cooling rate does not change much once Debye cooling starts. As a consequence the temperature plummets as $T_s \sim -t^{7/4}$ and the time taken to cool well below 1000K from 3000K would only be a few times longer (maybe a billion years) than the time to cool from 4000K to 3000K (as suggested by a simple by-eye extrapolation of the models shown in the plot).

The same treatment in the "Mestel" cooling regime, where the heat capacity is constant yields a much shallower $T_s \sim t^{-7/20}$, and accounts for the sharp "knee" at the onset of Debye cooling.


extrasolar comets - basic question How can an extrasolar comet be detected - this is based on an article regarding the influence on a extra solar gas giant on these comets causing changes in their orbit and leading to them being swallowed by the star ..
Still stuck on understanding how we detect them .. they certainly have no measurable effect on the movement of the planet or the star .. and the chemical make-up of the comet seems irrelevant given the size of the star and observations we can make ? Just missing something Simple I think ?

The chemical make-up of the comet is indeed relevant. From Wikipedia (Exocomet):

  The exocomets can be detected by spectroscopy as they transit their host stars. The transits of exocomets, like the transits of exoplanets, produce variations in the light received from the star. Changes are observed in the absorption lines of the stellar spectrum: the occultation of the star by the gas cloud coming from the exocomet produces additional absorption features beyond those normally seen in that star, like those observed in the ionized calcium lines. As the comet comes close enough to the star, cometary gas is evolved from the evaporation of volatile ices and dust with it.

This is not to say that you can determine anything else (besides their existence and chemical composition) about the comets.


A strange star-like object So it was around 6 AM and i was looking at the sky and there were not many visible stars because the sun was about to rise. There were two stars that were brighter than the others and i was looking at them for about 5 mins. They weren't moving and the space between them seemed very little. Suddenly, one of those stars started moving down in direction towards the horizon and the other remained still. The star continued to go down. In 1 or 2 minutes it went behind the horizon and stopped being visible, while the other was still at the same place like earlier. At first I thought that the moving object was a satellite, but then i realised that a satellite would orbit the earth with constant speed, and the object i saw remained still for atleast 5 mins. So what else can it be???

If it disappeared towards east or so, as I  think because the other "star" should be Venus, that it can be a satellite - or even more an airplane - as seen from "its rear". Basically at first it went far along the line of your line of sight looking still, until finally you could see the curvature of its path. From crude geometric feeling it was most likely an airplane, a lower altitude make it easy the seeing "from the rear" (think about what you see observing the rear light of a car moving on an horizontal road towards a down hill. It looks more or less two lights in the dark then it disappears. Or a ship approaching horizon). About absence of blinking: if it indeed was an airplane mirroring the sunlight, not way to see the less intense "navigation" lights.


Gravity inside a star? Recently, I answered this question and need some clarification regarding the gravity when within the radius of a star. I thought that, when something is inside a star (ignoring drag), it will experience less gravity because as it goes deeper inside, more and more mass is outside (in the up direction) the distance left before it reaches the center. Is this true, and is there a formula that needs certain parameters to calculate it?

If you consider a radially symmetric mass distribution, the gravity experienced at a distance $R$ from the center is caused by the mass inside the sphere with radius $R$, thus the mass $$M(R) = \int_0^R \varrho(r)  4 \pi r^2 dr$$ In a radial-symmetric case one can show that the contribution of all masses outside $R$ cancel each other.
Thus the experienced gravitational acceleration at radius $R$ is
$$g(R) = \frac{G}{R^2}\int_0^R \varrho(r) 4\pi r^2 dr$$
The interesting part here is to know the radial density variation $\varrho(r)$ inside your body, thus the Sun in this case. This depends on the equation of state of the matter - which in turn is one of the least well known equations. Deriving that requires intricate modelling of the Sun. There are at least two nice answers which give the radial density variation. There's a link in the first answer to data for the standard solar model for radius, density, enclosed mass etc.
EDIT to add a way to address the density profile:
One may try to get a handle on the density by assuming a polytropic model for the star, such that the relation between pressure $p$ and density $\varrho$ reads like
$$ p = K \varrho^\frac{n+1}{n}$$
With a constant $k$ and the polytrop index $n$. This is a result of the solutions of the Lane-Emden equation
$$\frac{1}{\xi^2} \frac{\mathrm{d}}{\mathrm{d}\xi} \left(\xi^2\frac{\mathrm{d}\theta}{\mathrm{d}\xi}\right) + \theta^n = 0$$which is a dimensionless form of the Poisson equation for a radially-symmetric self-gravitating polytropic fluid, thus where density follows a function of the form $\varrho = \varrho_c \theta^n$ with a central density $\varrho_c$. This equation can be solved exactly for polytrop index 0 (isobaric polytrope), 1 (isothermal polytrope) and 5 (limited use as it results in infinite stellar radius) and can be addressed analytically via convergent series and can also be solved numerically for a wide range of cases.
Yet a main sequence star is believed to be around $n=3$ for the radiation zone. Fully convective stars follow more an adiabatic gradient with n = 5/3 being the ratio of the heat capacity at constant pressure and heat capacity at constant volume for an ideal mono-atomic gas. You will have to integrate the euqation numerically for these cases, though.
(Reinhard Meinel gives excellent lectures on relativistic astrophysics - and I still use my notes from back then as reference - but I'm sadly not aware of any published form of the notes on this particular topic... maybe this)


Sun's orbital elements I have the orbital elements of an asteroid's heliocentric motion (the orbit has non-zero inclination and is elliptical). But I want to simulate the motion of the sun around the asteroid. So without having the sun's ephemeris, and given only the asteroid's orbital elements, how can I determine an equivalent orbital elements for the sun's motion around the asteroid?

Edit:
I realize that the question wasn't clear enough based on the comment. So, what I am trying to do is to numerically solve the two body problem equation in the inertial frame. In the classic two-body problem 2nd order ODE, in my case, the position vector is defined from the centre of the asteroid to the sun, which is absolutely valid and upon integration gives me the position an velocity for the sun with respect to an inertial frame centred at the asteroid. My question is concerned with the initial conditions to begin the numerical integration. As of now I determine the initial state vector for the Sun from some random orbital elements. But to perform a more realistic simulation, Is it possible to determine the real orbital elements for the asteroid-centric motion of the Sun from a given set of orbital elements of a heliocentric-motion of the asteroid? I hope I was clear and not made this more confusing.

Strictly speaking the Sun's motion is going to be dominated by the gas giants and the planets and one asteroid is going to have negligible effect on it compared to those.

However if we assume there are no other items we can get an approximate idea of the effect.

You can use the similarity of the orbits.  The Sun ( in this simplified model ) follows a complimentary orbit around the center of mass of the two bodies.

The distances you have for the size of the obit scale like this :

$$r_1 = \frac \mu m_1 r$$

$$r_2 = \frac \mu m_2 r$$

where $\mu$ is the reduced mass :

$$\mu = \frac { m_1 m_2 } { m_1 + m_2 }$$

And as you can see $r = r_1 + r_2$ is the distance between the two bodies.

Now as you can see from those equations, the most massive body moves the least.

The mathematics comes from the way the standard two body problem is solved.

It might be instructive for you to compare the effect of your asteroid to that of Jupiter or Earth.


Frames per second capabilities cameras currently installed in Modern Telescopes? What are the present real-time capabilities of various modern telescopes as video cameras?  How do these compare with the capabilities various of space telescopes?

For example, if it was known that there would soon be a Saturn-comet impact (akin to the Shoe-Maker levy impact into Jupiter), what would the "frames per second" of the captured imagery be, regardless of imaging resolution?

Video of the Shoemaker-Levy impact: https://youtu.be/p7RP2SW_gSw?t=92.  Is this video presented in real time?

Animation of the Shoemaker-Levy impact:

https://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9#/media/File:Max_Planck_Institute_Shoemaker%E2%80%93Levy_9.gif

Live Streaming Jupiter (from Earth):

https://www.youtube.com/watch?v=ILh4lWHi_ag

Shoemaker-Levy image:

https://en.wikipedia.org/wiki/Comet_Shoemaker%E2%80%93Levy_9

ISS Live Stream [for comparison]:

https://www.youtube.com/watch?v=EEIk7gwjgIM

The types of sensors that are used for "Lucky Imaging" can work at frame rates easily capable of sampling changes in the speckle pattern due to turbulence in the atmosphere; 50 Hz is certainly possible.

The detectors used are electron-multiplying CCDs and are available on a number of telescopes around the world.

A video camera can of course be put at the focus of any ground-based telescope.

Such devices are not usually put on space telescopes because they have no need to deal with a turbulent atmosphere. Space telescopes are usually concerned with taking pictures of very faint objects, where minimising the readout noise of the device (to a few electrons) is a priority, and necessitates longer readout times - often tens of seconds.


Have I nearly found the event horizon of a black hole? The calculator on the website https://planetcalc.com/1758/ cites Wikipedia's Gravitational acceleration and implements:
$$g = G \frac{M}{(R + h)^2}$$
For a black hole with mass $M$ equal to 5 solar masses with a diameter $R$ of 1 meter at a distance $h$ of 1500 km, the acceleration given is 296000 million m/s^2 which is nearly close to trap light.
So one might expect the event horizon to be at least close to that. Is this correct?

This is the Newtonian model of gravity. It is a very good model, it is used for accurate calculating the motion of objects in the solar system to a very high degree of accuracy.
However, for very strong gravitational fields you need to use Einstein's model, which accounts for things like the constant speed of light for all observers. I'm not going to go into mathematical details (not least because they are far to hard for me!)
Now you can calculate the gravity of a sphere and you find that at a radius of $r=2GM/c^2$ something strange seems to happen, nothing can travel from a radius of less than this to a radius of more.  It is the event horizon.  This can only happen in Einstein's theory of gravity.  Newtonian theory has no "speed limit" so there can be no event horizon.
Now it so happens that this is the same radius that Newtonian gravity would predict to have an escape velocity $v_e = \sqrt{\frac{2GM}{r}}$ equal to the speed of light.  That is mostly a coincidence(there was a recent question about this, that I haven't been able to find), but a neat one.
You seem to be comparing the Newtonian acceleration due to gravity with the speed of light.  The units of acceleration and speed are different, so the comparison is not valid.


How far is the nearest black hole from our solar system? Is there an official list? Which is the nearest known or suspected black hole from our solar system and how far is it from us? Is there an easily-viewed catalog that astronomers use to keep track of these?

According to table 2 of Corral Santana (2015) (The BlackCAT catalogue of stellar mass black holes in X-ray transients - which other than gravitational waves, is how they are found), the nearest known black hole is the primary of the V616 Mon system (otherwise known as A0620-003) at a distance of $1.06 \pm 0.10$ kpc ($3460 \pm 390$ ly).

Black holes can only be found at present when they are in close binary systems with other stars (or black holes). This is the tiny minority of an estimated Galactic population of 100 million black holes.

The nearest black hole is likely invisible to us at present. A statistical estimate can be made (see my answer here) that the nearest will be within 20 pc or so, depending on the parent population and detailed physics of supernovae.

The online BlackCAT catalogue.


How space scientist keeps the track of the stars? In billions of stars, how space scientist keeps the track of the stars ?

There is a star called KIC 8462852, I saw that picture, star is so small from the earth. It has higher chance to miss that star in the universe.

Scientists make lists of stars in a catalogue. In the past this would be in a book nowadays it is more likely to be a computer database.

The stars don't move (much) so it is possible to make a list of them by position Modern catalogues are online (see SINBAD older ones were on paper (see the 1918 Henry Draper catalogue)

The Henry Draper catalogue is typical and easier to follow. It is just a table, the first column is the catalogue number (used instead of a "name")  the second column is a cross reference to another database. Then there is information on the location of the star in the sky using the RA and Dec system. Next comes information on the brightness of the star (estimated in two ways). Some letters come next to describe the spectrum (or "colour") of the star and finally some short "remarks", and a "plate number (the number of the photograph that contains an image of the star"

Scientists need to be systematic and careful to avoid missing anything.


Where will the planned Deep Space Atomic Clock orbit "after" Earth? I've seen multiple sources repeat the information that the Deep Space Atomic Clock will 'at first' orbit Earth, before moving on out of Earth orbit.

  The Deep Space Atomic Clock (DSAC) is a miniaturized, ultra-precise
  mercury-ion atomic clock for precise radio navigation in deep space.
  It is orders of magnitude more stable than existing navigation clocks,
  and has been refined to limit drift of no more than 1 nanosecond in 10
  days.

Wikipedia: "Its development will include a test flight in low-Earth orbit"

JPL: "the instrument will be tested in Earth orbit for one year, with the goal of being ready for future missions to other worlds"

What would it then orbit? The Sun? Or some L point? Other planets? Or will multiple ones orbit all of the above?

Edit: for those wanting to close based on "not astronomy", have you considered astronomy does not have to be done from earth and indeed much of the data astronomers use often comes from probes and satellites in space.  Therefore 'how/where will this space beacon be used?' has strong impact on future of astronomy.

"The instrument will be tested in Earth orbit for one year, ..."
What would it then orbit?

It will continue to orbit the Earth, at least for a while. It will eventually end up not orbiting the Earth because it will be placed in low Earth orbit, where orbits decay due to atmospheric drag.
The project is a technology demonstrator, which means a limited amount of funding.


What are reliable resources for weather and all other atmospheric obstructions? Everyone knows you need clear skies to see what the stars or the Sun is doing. Weather is not too particularly hard to determine. There are many resources out there that will tell us the weather in any city. However, they very often lack other atmospheric obstructions. Naturally, light pollution is one of the biggest concerns. But another major concern is erratic atmospheric changes due to things like forest fires.

I'm looking for good resources that will tell me not only weather, but also light pollution and other obstructions such as smoke from forest fires.

(There's a solar eclipse coming up and I'm driving 500 miles to see it.)

You might try https://clearoutside.com/forecast/50.7/-3.52  It gives weather and visibility among other things.

They have an app for iPhone and android. Advertises site was developed by astronomers.


Are black holes solely responsible for hyper velocity stars? I read that the stars orbiting around Sagittarius A* (a.k.a the supermassive black hole located at the heart of the Milky Way) move very fast (many times faster than our Sun moves across the galaxy), and it is believed they have enough speed to wander off. My question is, is any exception to this rule, where somehow some stars may break a stellar speed limit not due to a black hole in their vicinity?

No, black holes are not the only cause of HVSs, although it is thought to be the most common mechanism.

Hyper velocity stars are believed to be caused

when binary stars come close enough to a supermassive black hole for one of the pair to be captured while the other star is ejected at high velocity. This appears to the main mechanism for HVSs. See for example this Harvard Smithsonian Center for Astrophysics page, and this University of Utah page.
when one star of a binary goes super nova, the second star may be ejected at high velocity. See US 708: Hypervelocity Star Ejected by Supernova Breaks Galactic Speed Record and How do Hypervelocity Stars End up Breaking The Speed Limit?.

There may be other causes, since not all HVSs seem to be explained by these mechanisms. Perhaps some of them are not from this galaxy at all - they may be just passing through.


Are there pictures from a solar eclipse of the earth taken from a moon landing unit or orbiter? Did any moon landing vehicle or orbiter took pictures of a solar eclipse happening on the earth, i.e. the moon's shadow on the earth? (Alternatively from other spacecraft showing the whole globe). A link to those pictures would be nice.

Such images are most readily available from full-time Earth-observing satellites.

The DSCOVR climate observatory sees the Earth from Lagrange point L1, 3.9 lunar distances toward the Sun.
Its EPIC camera captured sequences of images half an hour apart for the solar eclipses of 2016-03-09 and 2017-08-21.

Geostationary satellites orbit 0.11 lunar distance from the center of the Earth.
The CIMSS group at Wisconsin has a collection of various weather satellites' images of several eclipses.

A lunar point of view would fall between these two examples, qualitatively more like the former than the latter.


Will the UK be able to witness the tetrad in April 2014? I live in the UK and am aged 15. I have become extremely interested in astronomy, astrophysics, the lot. Basically I have never seen a tetrad and believe this will be my first time - at least understanding what it is. I'm hoping it will be possible and if so where will it be seen in the sky? As in North, South etc. Thanks!

It is great that you have an interest in astronomy.

Unfortunately, the UK will be on the wrong side of the Earth to see the lunar eclipse on April 15. If you you want to experiment with various places on earth that will see it, and what they will see, get a planetarium application like Stellarium or SkySafari or TheSkyX.

A lunar eclipse is always easy to spot - it is where the full moon is. Also, because it is the result of the Earth getting in the way of the Sun shining on the Moon, it is always after the sun sets and after the full moon rises in the East. That is, if you're going to be in a place to see the whole sequence, not just a part of it.

Just so you know, the tetrad is not one but a series of lunar eclipses over 2 years, this is just the first lunar eclipse in the series. The last tetrad was 2003-2004, so you were around, but probably not paying as much attention at that time :-)

Unfortunately, Cambridge UK will not see any of the first 3 eclipses in this tetrad. However the 28 Sept 2015 eclipse will be visible from that location.

Incidentally, one of the better places to see this lunar eclipse will be Easter Island (Rapa Nui) which will see the whole eclipse, and with the totality occuring high in the sky, near the meridian. Also, not much light pollution.


Excluding a rectangular region from an image in FITS I am a complete beginner in astronomy. While learning DS9, I came up with the following question.

Say I want to exclude a boxed region from an image FITS. The corresponding parameters according to Chandra is

Box = (xcenter,ycenter,width,height)
Box = (xcenter,ycenter,width,height,angle)

Although the name is self-explanatory yet I am confused about their respective units. What unit is used to label these parameters?

Usually pixel coordinates.
What I would do is to let DS9 make a region file, as your documentation site says it supports them.

To do so:

Open file in ds9
From region->shape select the kind of region you want.
Click on edit->region
Draw the shapes you want (just click on the img)
Click region->save. In the coordinates box choose physical pixel coordinates.

Probably that should work... Good luck!


What is the reason the Chinese want to measure the <30Mhz radiation behind the moon? On the Earth it is difficult to detect radiation below 30MHz from space due to our atmosphere. But why do they want to measure this radiation behind the moon? It has something to do with the big bang, but isn't the frequency of CMB not around 160GHz?

The Chinese are planning to launch a satellite with a 30 MHz antenna on it in 2018.

Aren't there already enough decent measurements of the CMB? So what would they like to find concerning the Big Bang?

Frequencies below 30 MHz have a hard time penetrating the so-called ionosphere - the series of layers of ionized atmosphere that exist at high altitudes.

This is good news for folks like me, amateur radio operators, because it allows us to speak into a transmitter at home, and let the ionosphere bounce the radio waves back to Earth, then again up into the ionosphere, and so on, eventually reaching another radio operator on the opposite side of Earth.

But it's bad news if you're a radio astronomer trying to receive such frequencies from distant objects - the ionosphere simply bounces those waves back into space.

So, apparently, you could raise a receiver above the ionosphere, in a satellite, and have it receive signals from outer space. But there's a catch. At 30 MHz and below, the wavelengths are huge. 27 MHz - the wavelength is 10 meters. 14 MHz - wavelength is 20 meters. 3.5 MHz - 80 meters. And so on, the lower the frequency, the bigger the wavelength.

The size of the antenna needs to be comparable to the wavelength if you want to receive signals from one direction only, so as to be able to tell where the signal is coming from. This also helps with efficiency. A very good antenna needs to be much bigger than the wavelength for it to be directional and efficient.

The problem is, even with the ionosphere, there's a lot of signal "leaking" out of the Earth, through ionosphere, into space; some of it is bounced back down, but some of it penetrates the ionosphere and escapes into space. A satellite in low orbit would be swamped by transmitters from Earth, completely erasing any signals from deep space. You'd need a humongous antenna to ignore the signals from the nearby Earth. This is not feasible.

It would be a lot better if you could just put the Moon in between Earth and your antenna. Even these low frequency transmissions cannot penetrate the Moon, and can't go around the Moon either. So the Moon acts like a screen, casting a shadow on the antenna and eliminating parasite transmissions from Earth.

Now your satellite can receive low frequency transmissions from deep space.

TLDR: Too much noise near Earth at these frequencies, need to hide the antenna behind the Moon.


If Earth had no atmosphere, how many people would be killed by meteors every year? Suppose those people engage in similar activity, walking on surface etc.

Three millions.

For this answer to be any fun, let's first assume that before removing the atmosphere, we have equipped all humans with oxygen supplies. In return, they are not allowed to go indoors or otherwise protect themselves.

The size of meteors — as well as most other things in the Universe — has a distribution that is well described by a power-law. The larger the meteor, the rarer they are.
Small meteors can be lethal only by a direct hit, while large meteors in the real world can be lethal due to the effect they have on the environment; dust can cover the sky for years and prevent the Sun from keeping us warm. However, since we have removed the atmosphere, the debris will fall down immediately, and the only way to kill us is by the explosion. But, as will be shown below, the dangerous ones are the small ones because the rate at which large ones falls off so quickly.

The number $N$ of meteors entering Earth's atmosphere every year with a diameter $D$ (measured in meters) or larger is given by (Brown et al. 2002)
$$
\frac{N(>D)}{\mathrm{year}} \simeq 37 \left( \frac{D}{\mathrm{m}} \right)^{-2.7}.
$$

Minimum size needed to be lethal

What makes a meteor lethal is really outside astronomy, and someone may be able to come up with a better answer, but for the sake of the argument, let's assume that the energy needed to kill a human by a direct hit is equal to the energy of a bullet fired from the world's smallest firearm, the 2mm Kolibri, which has a muzzle energy of $E_\mathrm{lethal} \simeq 4\,\mathrm{J}$.

Earth orbits the Sun at 30 km/s. Depending on the direction from where the meteor arrives, it will have different speeds, but it will at least enter with the escape velocity of Earth, i.e. 11 km/s, and a typical speed is $v_\mathrm{impact} = 17\,\mathrm{km/s}$ (Marcus et al. 2010). Thus, the minimum mass to kill by a direct hit is as small as
$$
m_\mathrm{lethal} = 2 \, E_\mathrm{lethal} \, / \, v_\mathrm{impact}^2 \simeq 30 \,\mu\mathrm{g}.
$$
At an average density of 4 g/cm$^3$, this means that a small grain of a diameter of 0.2 mm will kill you. According to the formula above, the number of such grains (or larger) that hit Earth every year is $2\times10^{11}$.
Since the surface area of Earth is 510 million $\mathrm{km}^2$, that's $n_\mathrm{hits}\sim5\times10^{-4}\,\mathrm{hits}\,\mathrm{m}^{-2}\,\mathrm{year}^{-1}$.

Some 7 billion people live on Earth. They each occupy roughly $1\,\mathrm{m}^2$ (and weren't allowed to protect themselves, remember), so in total they occupy an area of $A_\mathrm{folks,tot} = 7 \times 10^9\, \mathrm{m}^2$.

Thus, the number of people being killed by direct hit meteors every year is
$$
N_\mathrm{kill} = n_\mathrm{hits} \, \times \, A_\mathrm{folks,tot}
 = \mathbf{3\,million\,year^{-1}},
$$
or one person every 10 seconds.

Death by explosion

How about non-direct hits? Let's assume that the minimum size to kill you would have an energy equal to that of a hand grenade holding 100 g explosives, and that you're killed if it hits within a few meters of you, so that you "occupy" 10 m$^2$. Similar to above, you can work out that this corresponds to a meteor of ~1 cm being lethal, and that the rate at which such meteors is only $2\times10^4$ per year, so only 2-3 people would be killed in this way. Even larger explosions are even more rare. Thus, the probability of killed by a explosion is negligible compared to direct hits, so the final answer would be the above, i.e. 3 million per year.

Conclusion: Let's not remove the atmosphere.


Can we tell how fast bodies are moving away by measuring their frequency? It's clear the universe is expanding because light from distant bodies, like galaxies, are shifted towards the red side of the spectrum.  In other words, the light measured is more red than it should be when compared to laboratory results.  This is great for measuring the distance to other bodies, but can't we also measure their velocities of departure by measuring how fast the light from these bodies is becoming more and more red over time?  That is, v is directly proportional to the change of frequency divided by the time elapsed?

Your understanding is correct. The doppler shift observed from a galaxy is the sum of its peculiar velocity with respect to the "Hubble flow" and the redshift due to the Hubble flow, which is caused by the expansion of the universe.

There is no direct way from a spectrum to separate these two components - they have the same qualitative result.

In principle, the expansion of the universe (or a change in the peculiar velocity) could be directly measured by looking for a change in redshift with time, which would depend on the cosmological parameters.

This is an extremely small effect and is confused by the peculiar motions of individual galaxies. Nevertheless, measuring this redshift drift is one of the prime goals of the Codex Instrument on the E-ELT (see Pasquini et al. 2010, http://esoads.eso.org/abs/2010Msngr.140...20P )
using Lyman alpha absorption systems towards distant quasars. This experiment is also planned for the Square Kilometre Array, using the 21cm line (Kloeckner et al. 2015 http://arxiv.org/abs/1501.03822 ).

In both cases, to overcome the experimental uncertainties (eg at 21 cm, it amounts to line drifts of 0.1 Hz over a decade), then observations of millions of galaxies must be combined.

There is no prospect of measuring this effect in an individual galaxy, furthermore I fear your understanding of cosmological redshift is flawed. The dependence on distance is a statistical average, not an absolute dependence. Individual galaxies are moving in individual gravitational potentials from objects around them. This gives them their peculiar velocities with respect to the flow. This velocity could increase or decrease as a galaxy got further away, but is never expected to be large enough to be detectable on human timescales for any individual galaxy. In addition, any change in peculiar velocity should average to zero when looking at millions of galaxies, leaving the redshift drift due to the expansion.


Spinning black hole vs non spinning black hole Does a spinning black hole have a stronger gravity than a non spinning one.After all, a spinning black hole is stretching space-time more, so does that make a difference is gravity?

It makes a difference, but not to the strength of the gravity.

Around a spinning black hole (or any other spinning mass) space time is dragged. This effect has been measured around the (spinning) Earth by Gravity Probe B. It would cause a plumb line not to point directly at the centre of the Earth, but slightly forward.

For a rotating black hole, there is a region close to the event horizon, but outside it, in which the gravity (a combination of downwards and sideways accelerations) is so strong that it is not possible to remain stationary, relative to a distant observer, for to do so one would have to travel faster than the speed of light (relative to the local spacetime)

A similar situation exists in a non-rotating black hole, but only behind the event horizon.

So a rotating black hole does not have stronger gravity, but it is different: It is sideways.


Formation of elements in the Sun other than helium I just learned that the Fraunhofer lines of the Sun's spectrum indicate that the Sun contains various elements other than just hydrogen and helium (for example, Na and Fe) but don't the Sun's p-p chain reactions only involve hydrogen and helium? So where do these other elements came from?

The Sun is currently turning hydrogen into helium. There are no other nuclear reactions taking place at any significant rate in the Sun. The Sun will not start to make heavier elements until it reaches the tip of the red giant branch in about 7 billion years time.
The elements heavier than helium that are present in the Sun were almost all made inside other stars. These stars lived and died before the Sun was born 4.5 billion years ago. The nucleosynthesis products from these previous stars were recycled in stellar winds, supernovae, kilonovae and novae, and then mixed into the turbulent interstellar medium.


Regions and size of an AGN In this diagram of the regions and approximate scale of an AGN:

which is from this paper, what do the x and y scales log(z/pc) and log(r/pc) mean? I'm guessing z is the z-direction ('upwards') and r is radius, but what is the meaning of the pc bit?

pc is the abbreviation for parsec, a common unit of distance objects outside the Solar system. In order to compress the scales, they have also taken the logarithm (to base 10) of the distances in parsecs. So the scale in the figure on the x-axis in the radial (r) direction from the central black hole spans from 1e-5 to 100 parsecs. The z-axis, which is perpendicular or normal to the plane of the galaxy and the accretion disk, spans from 1e-5 to 1000 parsecs.


How loud would a black hole be? If we could hear black holes, how loud would they be? I know that this is basically the same question as the question "How loud would the Sun be?", except this time there is a black hole. But since the black hole sucks in stuff like a passenger jet engine sucks in air, about how many decibels would it be? Also, would it lethal to humans if we were 1 AU away and what would it do to our planet?

Nothing comes out from the BH (well, with a single known, very small exception).

The accretion disk of the Black Holes most likely has various streams, pressure differences and so in it. Calculating it or measuring it is surely a very hard task, if you have luck, a professional explains the results in a better answer.

The accretion disc of the typical, star-sized black holes is much smaller as 1AU. And sound doesn't propagates in vacuum. Thus, from 1AU distance from the Black Hole we most likely wouldn't hear anything.

Other radiations (mainly gamma from the disc) may be deadly even from 1AU, but not its sound.


How small can a planet be and still have Earth-like gravity? A planet made from denser material than Earth might have equal gravity to Earth but a smaller radius. How small can a planet be and still have Earth gravity?

Ideally it should be habitable by humans, so not made of anything dangerous, radioactive or unstable.

The surface gravity of a planet is very close to
$$g=\frac{4\pi G}{3}\rho r.$$
With $g$ to be kept constant, and $\frac{4\pi G}{3}$ a constant, we need
$\rho_Pr_P=\rho_Er_E$, or
$$r_P=\frac{\rho_E}{\rho_P}r_E,$$
with $\rho_E=5.515 \mbox{ g}/\mbox{cm}^3$ the mean density of Earth, $r_E=6371.0 \mbox{ km}$ the mean radius of Earth, $\rho_P=22.59\mbox{ g}/\mbox{cm}^3$ the density of densest natural element osmium, and $r_P$ the radius of the fictive osmium planet.

Hence $$r_P=\frac{5.515}{22.59}r_E=0.2441~r_E=1555\mbox{ km}.$$

Some compression of the core of an osmium planet due to pressure is neglected.


Saturn's tropical year? What is the value of Saturn's "tropical year" (as opposed to its "sidereal year").

I am seeing it listed as 10 days shorter (10,747 days) than the sidereal year (10,759.22 days).

How could this be if the precession of the axis (equinox) is so slow (~1.86M years)? Is this an "transient/osculating" value? Frame of reference (e.g. J2000)? Am I completely missing something?

I think I can calculate the tropical year by (close to, this works for Earth):

$P_{tropical} = P_{sidereal} \cdot (1-\frac{P_{sidereal} }{P_{precession}})$

Using NASA planetary facts data on Saturn
(these figures are re-used on many websites):

$10759.22\cdot(1-\frac{10759.22}{1.86\times10^6 \space\times\space 365.256})=10759.05$
Using JPL Horizons data:

$10755.698\cdot(1-\frac{10755.698}{1.86\times10^6 \space\times\space 365.256})=10755.528$

about 0.170 days less.

Also, interestingly, see the instantaneous sidereal year of Saturn from the graph of JPL Horizon's data, using the sun's centre (blue) and solar system barycentre (red). I think the instantaneous tropical year (at epoch) would always be 0.170 days less.

I think this is where the confusion is: texts have used particular values, and further the same text has picked a sidereal from one epoch (or average), and a tropical from another...

Saturn's instantaneous length of sidereal year in days, at epoch.

A quick skim of a few papers on Earth's tropical year, e.g. M.J.White, suggests to me that existing formulas are best-fit to data, not analytic.  How Saturn's ellipticity of orbit & axial precession(s) play into this is beyond my learning, but I'm getting the feeling that "it just is," and that any number of cofactors such as planetary layers, densities, and so on might affect what's being observed.


Why all the photos from 67p are black and white? I was wondering why every picture we saw on internet about Rosetta's landing on 67p just black and white ? Is there something i should know ?

True fact is that most of the images seen so far are taken from the navigation camera (NAVCAM) of the orbiter, Rosetta, and that takes black&white pictures.

Here is a very interesting article about how these pictures are taken (ironically it is on a NASA blog). They state that they show pretty much what you would see with a human eye from there, even thought the comet is actually all very dark. The white parts are due to the contrast with the very dark surroundings.

I'm not sure about the images received by the lander. They are captured by the CIVA cameras, where CIVA stands for Comet Infrared and Visible Analyser. So I guess they might show some infrared light not visible to the human eye, then processed to be in b&w. Unfortunately, I could not find a precise explanation on ESA's website so far. (here their page for Philae instrumentation).

Anyway, as a general rule you should not expect that all images from spacecraft are in true colors.


Parking a telescope at a Lagrange point: is this a good idea from a debris point of view? The James Webb space telescope is supposed to be located at the Earth-Sun L2 Lagrange point.

Do we expect the region around that point to have a higher concentration of space debris, asteroids, dust, etc...? Would this be a matter of concern for bringing the telescope safely in position (i.e. require extra dust shield)? Do we expect however the exact Lagrange point to be free of any matter since it would take an acceleartion/deceleration for matter to actually reach that exact location? How large would be this 'debris-free' zone?

Hmmm no, it wouldn't be cluttered with debris, and yes, it's a good idea to park the JWST (James Webb Space Telescope) at the Sun-Earth L2 point.

The five Lagrange points are unstable, for one because of the gravitational anomalies of the two massive bodies of the Lagrange system, eccentric orbits, and there are many other factors to their instability. At the same time, they are least gravitationally attractive points around two massive bodies.

Think of L-points as parking your car on a flat space at the top of the hill. You'll have to approach with some control and then try to balance at your parking spot, if you don't plan using your handbrake and also remain there stationary:

    Visualisation of the relationship between two massive bodies and their five Lagrange points (Source: Wikipedia)

There wouldn't be any debris clutter there, or any other matter like smaller particles, at least not any more likely than elsewhere around them, only transient in nature and possibly even less likely than elsewhere since all the other mass particles would gravitate towards the more massive bodies of two Lagrange system centers in their vicinity.

No body with rest mass would stay there on its own accord, not unless it has active attitude control to position itself there and constantly adjust for changes in gravitational attraction vector as the two mass bodies rotate around their axes, change distance while orbiting each other, or L points being influenced by other mass bodies of the same planetary system.

At the same time, JWST will be shielded by the Earth from any Solar activities and also Sun's interference with JWST's sensitive equipment as it starts observing the Universe in the infrared spectrum. Most of the orbital debris from our own space exploration missions is cluttered in the LEO (Low Earth Orbit) belt at roughly an altitude of 500-1500 km above the Earth's surface:

             Source: Active Debris Removal: EDDE, the ElectroDynamic Debris Eliminator, Jerome Pearson et al. (PDF)

Now, the JWST will be deployed to a very large 800,000 kilometres (500,000 mi) radius halo orbit around the Sun-Earth L2 point, that is 1,500,000 kilometers (930,000 mi) from the Earth, around 4 times farther than the distance between the Earth to the Moon. So not exactly in Earth's shadow and it will still use large deployable sunshield, but that's a long way away from where we're littering our LEO or even GEO/GSO (Geostationary/Geosynchronous orbits) with GEO being the farthest of these frequently used orbits, at 35,786 kilometres (22,236 mi) above the Earth's equator (and its graveyard orbit a bit farther than that). JWST will be placed in halo orbit nearly 40 times as distant from the Earth as you'd expect any space debris cluttering our planet's orbits at.

Yes, it is a very good idea to place JWST around the Sun-Earth L2 point.


What are the benefits of a large refracting telescope? Currently, refractor telescopes are limited in aperture because gravity will deform the main lens over time. According to Wikipedia the largest practical refracting telescope is 1m.

If we were to overcome this limitation, what would be the theoretical benefit of such a telescope that other types of telescopes currently can't give us?

This is a really interesting question!
The Dragonfly telescope described below takes advantage of the fact that on Earth, the optical surfaces of some lenses can generate less diffuse scattering than the optical surfaces of reflecting telescopes. If you were worrying about scattering from secondary mirrors you could make an off-axis primary, but at a nanometer scale a metalized mirror will have more roughness than a glass lens with antireflection coatings (or so the links say). I suppose there are broadband dielectric reflective coatings for telescope mirrors, but that's probably a topic for a new question.
note that in many cases surface brightness limitations are often the result of  skyglow (artificial) and airglow (natural) rather than telescope optics, so applications for large lenses might be found above the atmosphere.
If we ignore the fact that in the solar system space is filled with micrometeorites, then a lens in space might maintain it's lower diffuse scattering for quite a long time! But we can't, so it may not. It may get beaten up fairly quickly in fact. A long tube in front of the lens will decrease the solid angle of primary impacts, but impacts inside the tube may still cause trouble, and if you don't shade the tube from sunlight, diffuse scattering from the close-to-but-not-completely-black coating might also cause trouble.
So far there are no answers to How fast do optical surfaces get dirty or damaged in space? that I can point you to.

Below's discussion of and links to the Dragonfly Telescope is taken from the question What (actually) is the “deprojected half-light radius” of this almost-all-dark-matter Galaxy?

The recent news of the Ultra Diffuse Galaxy (UDG) Dragonfly 44 is an excellent example of what could be termed 'observe different' thinking. The dragonfly telescope is noted not for the size of its collective aperture, but for the absence of the diffracting effects of secondary mirrors and surface roughness that limit the contrast of dim objects in conventional telescopes when brighter sources are nearby. See here and here and here.

above: image of a Dragonfly refractive array telescope from here. Image: P. Van Dokkum; R. Abraham; J. Brodie

above: The Dragonfly 44 ultradiffuse galaxy from here. "Dragonfly 44 is very faint for its mass and consists almost entirely of dark matter. (Pieter van Dokkum, Roberto Abraham, Gemini Observatory/AURA)"


Extract anomalies from two-line element set (TLE) on ISS (International Space Station) I am faced to a problem of interpretation illustrated on figure below :

From wikipedia : Two-line element definition, I can extract the
following meaning anomaly  : 308.7017 ° (in degrees)

But in the 2 questions, I am asked to find :

1) at which anomaly is the ISS at the two lines epcoh

2) at wchich anomaly is the ISS on December 1st 2018.

I think the answer to question 1) is maybe what I have extracted above (308.7017 °)

But how to compute the anomaly of ISS on December 1st 2018 ?

I don't undertand very well the difference between these two anomalies requested.

If someone could help me, this would be nice.

Regards

UPDATE 1: Sorry, I must precise that I talk about mean anomaly.

First you should get an idea of what "Anomaly" means.  An anomaly is an angle which can describe the position of the satellite in its orbit. The true anomaly is the actual angle made between the position of the satellite and the perigee. The true anomaly changes quickly when the satellite is close to the Earth and slowly when it is far from the Earth.

The mean anomaly is much simpler: it just increases linearly with from 0 to 360. It is zero when the satellite is at perigee. So if a satellite takes 100 minutes to orbit and it is perigee at time t=0, then it will be at 90degrees at t=25, 180degrees at t=50.  The mean anomaly is just proportional to time.  "Mean" here means "average", not "meaning".

The orbital elements will tell you the mean anomaly at a particular time (given as "days in the year" In your data, at day 320.81254693 of 2018 (the "epoch") the mean anomaly was 308.7017.

You also know that the mean motion is 15.53998777. That is the number of orbits the satellite makes each day. One orbit is 360 degrees, so in 'n' days the mean anomaly will increase by 15.53998777 × 360 × n

You can now find the mean anomaly at any future time. For example 2.5 days after the epoch, at day 323.31254693, the anomaly will be 308.7017 + (15.53998777 × 360 × 2.5). You may need to reduce this angle to the range 0 to 360.

To add accuracy you can also use the fact that the orbital period is slowly changing (It is getting shorter as the ISS falls towards Earth due to drag). But this change is not quick and can be neglected over short periods of time.

Now you can work out how many days pass between the epoch and Dec 1st and you can work out the mean anomaly.

The question is worded strangely. It is an odd use of "which" and is unclear about the time. It should ask "What is the mean anomaly of the  ISS at the epoch" and "What is the mean anomaly of the ISS at 00:00 on Dec 18th 2018"