Happy lunar new year! When will "sunset" happen for Chang'e-4 and Yutu-2? Firecrackers all night and all libraries being closed here are two good indicators to a nerd like me that Lunar New Year has come again, and that means that it's the beginning of a lunar month and new Moon as seen from Earth.

That also means that the Chinese lander Chang'e-4 and rover Yutu-2 should be in sunlight since they are on the far side of the Moon.

What phases of the Moon as seen from Earth correspond to times of sunrise and sunset for those two?

tl;dr: To an accuracy of about a half of a day:

Third quarter is sunrise for Chang'e-4 and Yutu-2
First quarter is sunset for Chang'e-4 and Yutu-2

Moon seen from Earth:
From Wikipedia's article Selenographic coordinates

Longitude
Longitude on the Moon is measured both east and west from its prime meridian. When no direction is specified, east is positive and west is negative.
Roughly speaking, the Moon's prime meridian lies near the center of the Moon's disc as seen from Earth. For precise applications, many coordinate systems have been defined for the Moon, each with a slightly different prime meridian. The IAU recommends the "mean Earth/polar axis" system,2 in which the prime meridian is the average direction (from the Moon's center) of the Earth's center.3
2 https://astrogeology.usgs.gov/Projects/WGCCRE/WGCCRE2003preprint.pdf
3 http://lunar.gsfc.nasa.gov/library/451-SCI-000958.pdf

Simplifying further, seen from Earth, 0° longitude wobbles near the center of the lunar disk. (see also 1, 2, 3)
Chang'e-4:
Chang'e-4 is in Von Kármán crater (also here) at about 44.8° South, 175.9° East.
So Chang'e-4 and Yutu-2 sit almost diametrically opposed to the Earth longitude-wise, and so sunset on the Moon's prime meridian is close to sunrise for the pair, and vice-verse.
Answer:

What phases of the Moon as seen from Earth correspond to times of sunrise and sunset for those two?

First quarter is sunset for Chang'e-4 and Yutu-2
Third quarter is sunrise for Chang'e-4 and Yutu-2

Location Confirmation:
We know where Chang'e-4 landed because it returned video of it's own landing.
YouTube Video: https://www.youtube.com/watch?v=jyjyf4ANGIo
From How did Chang'e-4 hover, rotate, and then descend so gracefully? (Video).
Additional verification that Chang'e-4 and Yutu-2 are in Von Kármán crater:
From The Lunar Reconnaissance Orbiter Camera recent post First Look: Chang'e 4:

Arrows indicate position of Chang'e 4 lander on the floor of Von Kármán crater. The sharp crater behind and to the left of the landing site is 3900 meters across (12,800 feet) and 600 meters (1970 feet) deep. Image was shrunk by more than a factor of ten (full resolution available inside the post). LROC NAC M1303521387LR [NASA/GSFC/Arizona State University].

Area around lander enlarged by a factor of two relative to the native pixel scale (see zoomify below), bright speck between two arrows is the lander. The large crater in the center (just right and below arrows) is about 440 meters (1440 feet) across. LROC NAC M1303521387LR [NASA/GSFC/Arizona State University].


Why was Uranus named what it was? Why was Uranus named what it was, and who came up with the name?

Herschel, the discoverer of the planet named it Georgium Sidus, "George's star" after his patron, King George III of Great Britain.

The name was not popular outside of Great Britain, and there were various other proposals. It was Johann Elert Bode who proposed Uranus, the Latin form of the Greek god of the sky.  It fits with the existing planets having the names of Roman Gods, and just as Saturn was the father of Zeus, so Uranus was the father of Saturn.


What is a typical value for core-to-star efficiency? I was reading Unfolding the Laws of Star Formation: The Density Distribution of Molecular Clouds by Kainulainen et al., which discusses star formation rates and efficiencies.

One variable used is $\varepsilon_{\text{core}}$, the core-to-star efficiency, describing how much gas above the critical value of $s$, the logarithmic mean-normalized density, forms a star.

I'm using this parameter in some calculations, but I'd like to use an average value, as opposed to a value for a given molecular cloud.

What is a typical value for $\varepsilon_{\text{core}}$?

From this book it seems you can have $\epsilon_{core}$ values raging from ~$30\%$ (a low value) up to $70\%$ (which is more "usual", i.e., requested by theories).

More information can be found here and here.


What is the specific term for time of the night when there is no moon? There are times when the sun goes down and it takes a while for the moon to rise, because the moon obviously doesn't rise with the sunset whole year. I guess this happens only when the moon is full. Also, in some days, moon sets in the midnight and there is no moon on the sky for the rest of the night. So, I was wondering what that time period of the night is called when the moon has gone down, or has yet to rise? Can we call it 'moondown'? I am not talking about the new moon when there is no moon the whole night. Please help, and thank you in advance!

You are right that the moon only rises as the sun sets when it is full. At other times there is some time each night when both the moon and the sun are both set.

The terms moonrise and moonset are well established, by analogy with sunrise and sunset. The time when the moon is at its highest point in the sky is known as lunar culmination. You can use "lunar day" for the 24.84 hour period between lunar culminations. I know of no specific term for the period when the moon is set, or for when it is risen, instead you can say "after moonrise" or "before moonset", as appropriate.


"Official" time for other planets I want to make clock that shows current time on different solar bodies. For that I need to know how to count time on those planets. Is there any official definition of times on other planets in our solar system? Brief google search did not help.

I am capable of implementing library for this myself but I wonder about current state of the art of time telling on other planets. So basically I want to know these things:

Is there any official time standard?
If so, where can I find it?
If not, how should I approach this problem? Should I make something like Unix epoch at some time in the past for every planet?

Thanks for any tips.

EDIT: Ok, so on Mars, seconds, minutes and hours were made longer to keep 24-hour day. Is this viable to other planets as well? Venus aside for now :)

I was also thinking about keeping the hours the same, having days as long as many whole hours fits in them plus smaller compensation hours which would have length of (total_minutes_in_solar_day % 60). This concept is from David Weber novels, if you are familiar with them.

EDIT2: Based on discussion in the comments, I'll try to make some basic model that would be if possible applicable to any planetary body (thinking ahead :)). This is obviously not going to have any serious real world application but I would still like it to have some sense.

Thanks for your input, I will leave the question open for some more time if anyone wants to add their opinion.

Time on Mars is measured in sols.

Venus could be a little trickier, depending on your application.

I'm not aware of any timekeeping system for any other planets. There hasn't been a need for one yet. So, again depending on your application, you could come up with various different possible schemes.

Speculation (based on the OP's input regarding application):

Mercury - because of the length of the day, I doubt would-be colonists would measure time in days. They would probably use Earth time (because of communications) and also keep track of Mercury years.
Jupiter, Saturn, Uranus, Neptune - no colonization, no timekeeping


What fraction of starlight, seen from Earth, is actually reflected light? Thanks to reflected starlight, many planets and comets in the Solar System have been visible to humans since long before the development of modern astronomy.

Some of the starlight from outside the Solar System should be reflected light as well. That is, light emitted by one star in the Milky Way may encounter another star, and be reflected toward Earth. Normally, stars are assumed to be perfect blackbodies, but in reality they must bounce off some radiation.

Also, starlight should be reflected off exoplanets, moons, asteroids, meteors, comets, brown dwarfs, etc. What fraction of starlight, seen by the naked eye from Earth, is actually reflected light?

A truly negligible amount. You only need compare the brightness of the planets as viewed from the Earth with the brightness of the sun. A very rough calculation (considering the relative magnitudes of the planets and the Sun) suggests that light emitted directly by the sun is 100 million times brighter than light reflected off planets.

You note that the sun is not a perfect black body, and so does reflect some light. However it only reflects light reflected on to it from the stars, which are a very weak light source. And most starlight that does fall on the sun is absorbed. I haven't tried to estimate how much of the sun's light is reflections of starlight, but it is a very very small proportion. I'd guess much less than a billionth.

There is one situation where reflected light is significant, and that is when very bright stars illuminate dust and gas, to form a reflection nebula. These are common and at least part of the orion nebula is reflected starlight. Other reflection nebulae are too dim to see.

While a I don't have an exact figure, it is clear that nearly all starlight is directly emitted from the star, and it takes special equipment to see any reflected light at all.


What is the exact time of sun stay in each zodiac every year on sidereal year duration? I tried to reverse engineer from astro-chart times at which sun stays in a zodiac.
Here is the list.

Sun_Stay_zodiac  timeofstay_sec   timeofstay_days
  1  Aries         2672700     30 day 22 hour 00 minute
  2  Taurus        2714700     31 day 10 hour 05 minute
  3  Gemini        2734080     31 day 15 hour 28 minute
  4  Cancer        2719440     31 day 11 hour 24 minute
  5  Leo           2680080     31 day 00 hour 28 minute
  6  Virgo         2586960     29 day 22 hour 00 minute
  7  Libra         2626020     30 day 09 hour 27 minute
  8  Scorpio       2548020     29 day 11 hour 47 minute
  9  Sagittarius   2533080     29 day 07 hour 38 minute
 10  Capricorn     2544240     29 day 10 hour 44 minute
 11  Aquarius      2576520     29 day 19 hour 42 minute
 12  Pisces        2622540     30 day 22 hour 25 minute

Since the orbit of earth is elliptical and therefore also variation is orbital speed the stay of sun in each zodiac is different i.e Maximum at Gemini and after 6 months minimum at Sagittarius. However when I add them up sums up to 31558380 seconds which is 230.495 seconds than time in seconds in sidereal year ie 31558149.504 Now what I am looking for is more exact time duration of sun stay in each zodiac that includes seconds. I hope this data that I have in my question will help others too.

A sidereal zodiac is not the same as the IAU constellations but corresponds more closely to them than the Western tropical zodiac does, e.g. the eastern edge of Virgo remains near Spica despite precession.

I assumed equal 30° blocks of ecliptic longitude with the J285.25 equinox as origin (Lahiri ayanamsa).
Your local tradition may differ slightly, but any difference should affect all signs equally and have little effect on Sun residence times.

I computed these statistics for the years 1970-2070 using Skyfield and the DE430 ephemeris:

     mean  stdev   mean
       seconds     days
 1  2667075 ±186  30.8689
 2  2702097 ±174  31.2743
 3  2717490 ±139  31.4524
 4  2708635 ±122  31.3499
 5  2678191 ±139  30.9976
 6  2635085 ±167  30.4987
 7  2591328 ±182  29.9922
 8  2558345 ±168  29.6105
 9  2544241 ±125  29.4472
10  2552351 ±101  29.5411
11  2580755 ±128  29.8698
12  2622569 ±167  30.3538

Gravitational interactions with other planets tend to make any given year slightly longer or shorter than the mean sidereal year of 31558150 seconds.
For the same 100 years as above, I computed a standard deviation of ±320 seconds in sidereal year length.


Are brown dwarfs which don't sustain any fusion considered stars? I was reading about the coolest stars, and was surprised to find about stars with a surface temperature lower than a candle, like 2MASS 0939-2448 A/B, of about 100 Celsius like CFBDSIR 1458+10B and even below 0 Celsius temperature where the water freezes like WISE J0855-0714. Then I thought, they might be stars where fusion only happens very deep in the core, and then I found out this:
Brown dwarf

In addition, many brown dwarfs undergo no fusion; those at the low end
of the mass range (under 13 MJ) are never hot enough to fuse even
deuterium, and even those at the high end of the mass range (over 60
MJ) cool quickly enough that after 10 million years they no longer
undergo fusion.

Are these brown dwarfs which don't sustain any type of fusion considered stars? If so, what's the reason?

Brown dwarfs are technically stars, but not in your typical sense, as they cannot start or sustain nuclear fusion. Think about it this way: stars are formed in clumps inside giant gas clouds called nebulae, along with planets, but what if one of these clumps is too big to become a planet and too small to become red dwarfs. These are brown dwarfs. Some consider them to be stars, others consider them to be planets, and some neither. So, opinions vary.


Population of excited H levels in a Strömgren Sphere In chapter 2.2 of Astrophysics of Gaseous Nebulae and AGN Ostriker and Ferland claim that, as far as ionization is concerned, one can assume all atoms to be in the ground state in a Strömgren Sphere setting (static, homogeneous $n\approx 10/\mathrm{cm}^3$, isothermal $T\approx 10^4\,\mathrm{K}$ pure hydrogen cloud around a single star). While I do not doubt the validity of this assumption, I have some trouble with how they justify it.

They estimate the inverse ionization rate for ground state hydrogen at a typical distance from the star, which is of course the lifetime against ionization from the ground state $\tau_\mathrm{ion}^{1^2S}$. Then they say that the estimate also holds for ionizations from excited levels and they find

$$\tau_\mathrm{ion}\gg\tau$$

where $\tau_\mathrm{ion}$ and $\tau$ are the typical lifetime against excitation and the typical lifetime of excited states respectively.
So far so good. In the next step, however, they end their chain of argumentation by saying that consequently any atom in an excited state will have enough time to decay to the ground state before it can be ionized, so all ionizations are from ground state. That is my understanding at least.

My problem with this is the following: suppose I have some mechanism that keeps a good fraction of my atoms in an excited state. In that case, even if $\tau_\mathrm{ion}\gg\tau$, I will have ionizations from this excited level simply because it is always populated. So what one should really look at in my opinion is the lifetime of an atom against excitation.

My first question is: Am I missing something or did I get something wrong?

And secondly: If my argumentation makes sense, one has to compare the lifetime against excitation with $\tau$, right? Where can I find the corresponding cross sections for radiative and collisional excitation? I was only able to find ionization cross sections.

There just aren't any other mechanisms to keep the population excited in a Stromgren sphere.  In principle collisions can also lead to excitations, but if you do an order of magnitude calculation you'll find that the time between collisions is much longer than the time to receive an ionizing photon.  In a laboratory setting you might be able to coerce some particles into remaining in an excited state using the Quantum Zeno effect, but that won't happen in interstellar space.

So without something to keep the atoms in an excited state, even if you start with everything in the excited state, in time $\tau$ you will find that everything has decayed to the ground state, with the exception of some very small fraction $\tau_{\textrm{ion}} / \tau$, which may have moved up to a higher energy state.  Assuming the lifetimes of those very excited states are comparable, after a few more times $\tau$, you will find that even those will have decayed to the ground state.


Would we have more than 8 minutes of light, if the sun "went out"? The common theory is, that if the Sun "shut down", we would see the light for eight more minutes (the time that it takes the photons to reach the Earth).

However recently I have read that photons need around 100 000 years to reach the Earth, since the reactions are happening at Sun's core, and gamma rays can't leave the Sun without interacting with other particles, unlike neutrinos for example.

Is that theory correct? If the Sun's core "shut down", would we still receive photons (light) for another 100 000 years, with only neutrinos disappearing immediately?

If nuclear fusion were to suddenly stop in the centre of the Sun, then the only clear signature we would have of this is the lack of detectable neutrinos received at Earth, starting about 8 minutes after the reactions ceased. The Sun however would continue to shine for tens of millions of years at roughly its current luminosity.

The power source is not "stored" photons. The Sun itself would simply resume the slow gravitational contraction that was halted about 4.5 billion years ago when nuclear reaction rates at the centre were able to increase sufficiently to supply the radiative losses from the surface of the Sun.

The characteristic timescale for the contraction is about
$$\tau_{\rm KH} = \frac{GM^2}{RL},$$
which is 30 million years. i.e. The Sun has enough gravitational potential energy to supply its current luminosity for tens of millions of years.

While this is happening, the Sun would approximately maintain its current luminosity, but decrease in radius, meaning that its surface temperature would increase.

Once the Sun had contracted to a few times the size of Jupiter (so about 30% of its current radius), the contraction would begin to slow, because the electrons in the core become degenerate and the pressure increases with density by more than expected for a perfect gas. The slowing contraction decreases the rate of potential energy release and hence the solar luminosity. The contraction continues at a slow rate until the Sun becomes a hot "hydrogen white dwarf" a few times the size of the Earth, which then cools to a glowing cinder, with no further contraction, over billions of years (see What would the Sun be like if nuclear reactions could not proceed via quantum tunneling? for some more details).

Even if you were to not allow the Sun to contract, it would take some time to radiate it's thermal energy. This timescale is approximately
$$\tau_{\rm therm} \simeq \frac{3k_B T M}{m_H L},$$
which assumes the Sun is a perfect gas of protons plus electrons, with an average temperature $T$. If we take $T =10^7$ K and the current solar luminosity, then $\tau_{\rm therm}=$ 40 million years.

On the other hand, if your scenario is just that light from the Sun stops being emitted, then of course it goes dark on Earth about 8 minutes later.


Is the discovery of Alien or Extra Terrestrial Life a function of the science of Astronomy? Is the discovery of Alien or Extra Terrestrial Life a function of the science of Astronomy? When i read astronomy articles on the search for new Planets or discussions on Mars a lot of the topic is on the planets occurring in the habitable zone and the possibility of life on them or stories on the possibility of previous life on Mars. Is this actually part of the science objectives for the Astronomy community, or is there another branch of science who is tasked with this? Is Astronomy only focused on the mechanics of celestial bodies and the discovery of Alien or Extra Terrestrial life is just and interesting sidebar?

The search for extraterrestrial life is usually covered by "exobiology". But astronomy can provide clues of whether a planet or an exoplanet provides some of the conditions, which allow life or indicate life.

The habitable zone - in its various versions - defines, in which distance from a star a planet needs to be to allow for e.g. liquid surface water, a condition regarded as necessary for most life-forms we know from Earth. With the discovery of extremophiles the habitable zone for that kind of life has been extended.

Astronomy can provide information about the central star and the distance of a planet from the central star. From these data a decision is possible, whether the planet fulfills the surface temperature condition for habitability.

Astronomy can provide information about the stability of a star, necessary for life to evolve on its planet.

Astronomy can - in principle - provide information about the chemical composition of a potential atmosphere of a planet. Oxygen and methane at the same time would be regarded as an indicator of possible life on the planet.

Astronomy can determine size and density information about the planet. That way it can be decided, whether it's a rocky planet or a gas planet, and infer the gravitational acceleration at the surface. The gravitational acceleration is relevant, e.g. for constraints of the atmospheric composition, or for possible properties of life, if extant.


What are optical counterparts? There surprisingly no proper definition of this term available on google. There are many research papers that are based on "finding the optical counterparts".

What exactly are optical counterparts and how to "find" them?

Observations can be made using different parts of the electromagnetic spectrum. Many of them not within the range of the spectrum that can be made with optical telescopes. Other messengers than photons (e.g gravitational waves, neutrinos,...) can also be observed.

Optical counterparts are observations made in the spectrum observable using optical telescopes that can be matched up with observations made using other observational methods, including radio telemetry, Gamma ray detectors, and gravity wave detectors.

Observers are interested in lining  up observations made up of any combination of observational methods as this provides the opportunities to identify correlations that are a rich source of new discoveries.


who invented the Hubble telescope I know that the Hubble telescope was named after Edwin Hubble, due to his astronomical discoveries of astronomy, but who designed it? Which people were most directly responsible for the inspiring and planning the HST project.

There was no single designer of the Hubble telescope. It was designed and build by NASA, with contributions from the European Space Agency (ESA).

The advantages of a telescope in space have been clear for over 100 years. These ideas were crystallized in Lyman Spitzer's 1946 paper. Spitzer continued to push for a telescope through the '50s and '60s.

When Congress eventually funded the telescope in '78, the task of design was given to the Marshal Space Flight Center. The head of the space telescope project was Dr. Bob O'Dell, chair of the Astronomy dept. at the University of Chicago. In a video from NASA, Dr O'Dell talks about the early years of HST development.

Over the years others have worked as Project Manager. There never was a single "inventor". Hundreds of scientists have been involved in its design and planning. Something as complex as the HST could never be designed by a solo inventor.


Potential Re-Capture of Rogue Planets Rogue planets can be designated as rogue planets due to one of two possible scenarios: 1) the "rogue planet" formed as a sub-brown dwarf, or 2) the planet was ejected from its home star system. I want to make it clear that here, I'm interested in objects of the second category - objects that are similar to the planets we find in our solar system.

Could a rogue planet, having been ejected from its home star system, be gravitationally re-captured by another star system? If so, would it be likely to disrupt other bodies in the stellar system (I know this depends on size; let's say that the rogue planet is a "Hot Jupiter")?

I realize that we don't know a whole lot about these objects (let's face it, we only know of a handful), so I'm not looking for examples, but any logical answers would be pretty helpful.

The processes by which planets can be removed from their parent stars are discussed in some detail by Davies et al. (2014). These include direct ejection through interactions with other stars in dense birth environments; the ejection of planets due to planet-planet interactions, again usually taking place fairly early in a star's life; the later ejection of planets after an instability due to a passing star. There is also a possibility of exchange interactions, i.e. a planet will swap stars during an interaction. Again, this is most likely to happen in a dense birth environment rather than later in a star's life.

If the ejection of the planet happens while they are still in the dense environs of their birth, then recapture by another star is possible. A very recent paper by Wang et al. (2015) explores this very possibility using N-body simulations, but there are other papers to look at by Parker & Quanz (2012), or Perets & Kouwenhoven (2012).
These papers seem to agree that the process of recapture is not common (but not rare either) and will happen sooner rather than later, before the cluster disperses. The latter of these papers has some particularly eye-catching results: they find that if the number of free floating planets is comparable to the number of stars then about 5% of stars recapture a planet. The capture tends to put the planet in a wide, eccentric orbit (hundreds of au or more). Capture is more likely for massive stars, but they claim that even brown dwarf-planet or planet-planet pairs could be formed in small numbers.

Once the planet has left the cluster environment then, as is the case of star-star interactions in the Galactic field, the probability of encountering another star and being captured is very remote.


How can the orbit of a Kuiper Belt Object be differentiated from the transit of a rogue planet? Related to the question "Are any Pluto-sized objects remaining to be discovered in the Kuiper Belt?" and the fact that most of the Kuiper Belt objects have very elliptical orbits, the question begs, what observational techniques are used to confirm that the object being viewed is an orbiting Kuiper Belt Object, as opposed to the transit of a rogue or 'orphan' planet?

I'm not speaking from an informed position here, but two things come to mind.

First is that a rogue planet is likely to be traveling very fast, relative to our solar system. An object from within our solar system (a Kuiper/Oort object) is going to have an orbital velocity. Something that isn't a part of the system at all could be traveling much, much faster.

Secondly, if a rogue planet somehow slowly drifted into our system at almost no speed, and then got pulled into a sort of orbital speed by the sun, the direction of its orbit might give it away. For example, it's orbital inclination could be way off--even perpendicular to the normal orbital plane. It also might be orbiting in the wrong direction--that is--clockwise, instead of ante-clockwise.

That being said, there's no reason that some disturbed Kuiper belt object might not have an eccentric orbital inclination or even an anti-clockwise orbit. But the larger the object, the less likely this is, because it would need a larger and larger disturbance to throw it so out of whack.


What are the azimuths of the planets' orbits? I am creating a virtual solar system model and I want it to be as realistic as possible (e.g. orbits are ellipses, not circles, and orbits are oriented correctly, not all coplanar).  In order for me to do this, I need to know the eccentricity of the orbit, the perihelion, the semi-major-axis, inclination and azimuth.  I have all of this information, except for the azimuth of the orbit, which is the angle from the sun to the aphelion.  Are there any reliable sources with this information?

As someone commented, you are actually short a term for fully defining the orbit.  The "azimuth" as you describe is commonly defined as a "longitude of perihelion", and you're forgetting the orientation of the inclination, commonly defined as "longitude of the ascending node." (Both of these values are included in the first link in the previous posted answer)

For a more detailed look at the math which uses the defining values, JPL made a whitepaper which goes into the math, and also allows you to account for changes in the orbital terms.  (Most of the math you're probably looking for is at the end of the document, beginning about section 8.10, on page 25)

ftp://ssd.jpl.nasa.gov/pub/eph/planets/ioms/ExplSupplChap8.pdf


If Saturn's rings cannot coalesce into a moon because of tidal forces, then how are shepherd moons able to exist? From Wikipedia:

In celestial mechanics, the Roche limit, also called Roche radius, is the distance within which a celestial body, held together only by its own force of gravity, will disintegrate due to a second celestial body's tidal forces exceeding the first body's gravitational self-attraction.
Inside the Roche limit, orbiting material disperses and forms rings, whereas outside the limit material tends to coalesce.

However if tidal forces are strong enough to prevent the rings from coalescing, then how are shepherd moons such as Pan and Daphnis able to exist without being disintegrated by Saturn's tidal forces despite literally orbiting within the rings themselves?

There's two parts to this question, the first part is relatively easy.  The 2nd part, more tricky.

  How are shepherd moons such as Pan and Daphnis able to exist without
  being disintegrated by Saturn's tidal forces despite literally
  orbiting within the rings themselves?

I'm going to focus on Pan because it's larger and closer.  It's unusual, ravioli shape suggests, at least to me, an unconventional origin.  It's unusually symetrical for an object that size and probably formed while rotating, but that has nothing to do with your question.

Pan is small enough that it's able to withstand the tidal forces inside the Roche limit.   Phobos, around Mars is in a somewhat similar situation, though not as deep inside Mars Roche limit and outside the solid Roche limit, the principal is the same.  The integrity of the moon is strong enough to resist the tidal forces.  The relatively small size of the object is a big help in that regards.  Smaller objects experience smaller tides.  That's basically why oceans have visible tides and lakes have tiny tides, often not visible at all.

Pan doesn't look like a normal 30 km object in space

Source.

Source

This is what a 30 km object should look like:

An interesting property of being on a small moon inside the Roche limit is that you'd float off the surface on either the planet side or the far side of the moon.   Stand on the right spot on Pan and you'd float away.   Pan doesn't break apart because of the integrity of it's ice, but it, in effect, has negative gravity over parts of it's surface - the tidal bulge parts.  That's true with Phobos as well.

But the forces are small.  You'd drift off the moon very slowly, unless you were to find something wet and stick yourself to the moon - the wet would freeze, creating a grip that would hold onto you.

It's thought that Pan accrued it's ice-ring around the moon by holding onto ice particles it collected from Saturn's rings.   Ice is sticky and while the Moon does clear out a band in Saturn's rings, any wayward ice doesn't need much velocity to reach the moon and upon making contact, some of the ice sticks.    As a result, Pan kind of, almost magically, slowly accrues ice in a region where a larger moon would break apart.

The more tricky part of the question is how Pan got where it is, because it couldn't have formed inside the Roche limit.  The origin of Pan is a mystery.


Will Sirius B start accreting from A and become a supernova type Ia? Sirius B is a massive white dwarf of 1 Solar mass, orbiting at about 25 AU distance from the 2 Solar mass Sirius A. As it evolves and expands, will the A star start shedding matter to the white dwarf, and when will this start happening? Will the Sun be at a safe distance when/if it does happen, or is Sirius our doom?

Will Sirius B start accreting? Yes, it is doing so now. Sirius A will have a wind and some of that wind will be captured by the white dwarf.

The effectiveness of wind capture is a strong function of relative wind speed. An analytic approximation to the accretion rate, known as Bondi-Hoyle accretion, goes as the inverse cube of the relative speed. In its present evolutionary state, the mass loss from Sirius A will be relatively weak (like the Sun) and relatively fast (like the Sun). This disfavours any significant accretion by the white dwarf.

However, in the later stages of its life, Sirius A will swell to become an asymptotic giant star. The envelopes of such stars are gradually (on timescales of millions of years) blown away quite slowly by a dusty wind. If Sirius A is around 2 solar masses now, it will lose about 1.4 solar masses during this phase at speeds of only 10-20 km/s.

Only a fraction of this mass may be accreted by the white dwarf, because the separation between the stars is still large at 25 AU (and will become larger as mass is lost from the system) compared with the likely terminal size of Sirius A (probably of order 2 au). If you look at the likely Roche lobe size, then Roche lobe overflow would require A to reach about 40% of the separation, which isn't going to happen. Exactly what fraction is captured by the less efficient wind accretion process (the majority will likely disappear into space and widen the orbit) depends strongly on the wind speed, which is hard to predict.

Even if Sirius B could accrete the 0.35-0.4 solar masses (I think that is unlikely, but lack the wherewithal to do the hydrodynamic simulation) it needs to become unstable, it is not clear if that mass will "stick". A build up of hydrogen-rich material can ignite and explode in a nova (not supernova) on the surface of a white dwarf, causing mass loss!

Finally, when will this happen? Well Sirius is probably about 300 million years old now and has another perhaps 500 million years before it starts to evolve in the way I described. It will be nowhere near the Sun then.


How can we tell that a short-period binary is tidally locked? Observationally, how can this be established?

What is known theoretically about the process?

Are there any reference papers to read?

The phenomenon of tidal locking occurs primarily in short period binary systems with convective envelopes. The tides raise bulges in the stars and these interact with convective motions to dissipate energy and to synchronise the orbital and rotational periods of stars and circularise their orbits. The process is much less efficient in stars with radiative envelopes.

How can we tell observationally? Well, we have to show that the rotation period of a star is the same as its orbital period. The orbital period is easily found, either by measuring eclipses, in the case of binary systems with a high inclination, or by measuring the motions of the star(s) using the doppler shift in their spectra.

Measuring rotation periods is also not so difficult. Fast-rotating stars with convective envelopes have magnetic activity and starspots (analogous to sunspots, but covering a larger area). As the star rotates then the light from the star goes up and down depending on what starspots are on the visible surface. This modulation gives the rotation period. In eclipsing binaries it can be difficult to separate out spots from eclipses, but here one can use the known orbital inclination and known radii of the stars to see whether their projected rotation velocities (measured from the widths of spectral absorption lines) matches the predictions from a tidally locked orbit.

There are many known tidally locked bnaries. Any low-mass binary with an orbital period of less than a few days is almost certainly tidally locked, but the situation at orbital periods greater than about about 6-8 days becomes more complicated (e.g. Meibom et al. 2006). The plot below is taken from a review by Mazeh (2008), showing the rotation vs orbital periods from some small, but well-defined samples.

The absolutely classic, must-read, theoretical work on this is by Zahn (1989). The timescale for synchronisation depends on structural properties of the stars, their moments of inertia, the binary mass ratio and the ratio of their separation to their radii to the power of 6. It is this latter property that means tidal locking is nearly bimodal; binaries with periods shorter than some threshold (about 6 days) are usually tidally locked, but those with slightly longer periods almost never are.


Conditions on Hot Jupiters Hot Jupiters are close to their parent star - after all, that's why they're so hot. Their atmospheres are thought to be slowly blown away by the parent star, and they will eventually become chthonian planets.

Are there any other effects on Hot Jupiters because of their proximity to their parent star? Examples I'm looking for include issues with solar flares or coronal mas ejections, or even difficulties relating to the solar wind. I would also be tangentially interested in any effects the parent star would have on moon formation (or lack thereof).

I think we can broadly distinguish two classes of effects here:

Flares directly affecting the planet
The by-products of the flare affecting the planet

I can loosely tell you what I know (although my knowledge here is far from being complete)

Direct effects include:

Magnetic field geometries: Far-away from their host star (>0.05AU) planets usually have isolated dipoles. The solar wind then must penetrate the magnetosphere of a planet to erode its atmosphere. This is however much easier for the wind if due to small distance both star and planet-magnetospheres connect. What this will lead to, is quasi-free flow of a significant portion of the stars total ejected mass towards the planet and huge energy input into the planet's upper atmosphere.

Indirect physics can be:

An active young star can have frequent outbreaks of CMEs, those are accompanied by strong UV / X-ray emission and this will affect the planet:

Increased photochemistry: This can generate a very active photochemistry in the upper atmospheres. Doubly ionized molecules like $O_2^{++}$ or $N_2^{++}$ can then be produced more efficiently and eject their constituents into space upon dissociation due to high dissociation energies
UV-X-heating: Additional heat through thermalized high-energy photons can add to the already happening 'natural' atmospheric erosion.

We don't know however, to which extent those effects play a role. I'd add tomorrow some references if you're interested.

There could be signs however of such activity in young stellar systems, as verious combinations or extensions of the above have been proposed to explain the radius anomaly in super-close extrasolar giant planets.


A smudge near Jupiter I was looking at the Jupiter for the first time today and I have noticed a smudge near it. Could someone explain what it is, please? I was looking online but did not find anything. So I thought that some kind of defect in my setup but I couldn't get rid of it. I mean the thing below its moons.

Edit1:
My telescope is a Celestron AstroMaster 76 ESQ. The smudge was there no matter what eyepiece I have used. I was able to see it even without a camera.

Edit2:
I was watching it for about 10 minutes, and the smudge was moving along with the rest of the objects.

I took just this one image for the StackExchange purposes. I have put the camera in front of an eyepiece to take the picture.

Thank you

The faint green smudge looks to me like a flare (that is, an optical defect called a flare).

However I'd expect a flare to be symmetric with the bright object it relates to about the center of the field of view.  This does imply either the image is off center (e.g. the camera is off center or the image is a crop off center) or there's a misalignment of your optics.

This question on Physics SE is about correcting flare in a telescope and may be relevant.

The similar question Does this smartphone photo show Mars just below the Sun? and its answers contain some other examples.


Explanation for the mathematics behind Venus' retrograde motion I've always been interested by the pattern within Venus' orbit around the sun. I found this image and formula of its retrograde motion in the upper right hand corner which confused me. What does it mean? can someone break this down in layman's terms?

Thank you

The really simple explanation is that Venus' orbital period is 225 days. Earth's is 365 days.

That ratio is 13:8 (well, more accurately 12.9777777777..:8) so you can see that your complete cycle is 13 orbits of Venus to 8 of Earth, but within that cycle our view of Venus travelling round the sun gives the pretty pattern as shown in your image.

This YouTube video is quite a nice visualisation - should make it easy to understand.


Is it possible to witness a star's death? Given that the stars' distances to Earth are measured in light-years (for example, Sirius is 8.6 light-years away from Earth), what we are seeing as Sirius now is actually its state 8.6 years ago, right?

So it is possible that a star (maybe not Sirius, I don't know, it's just an example) somehow explodes and creates a supernova, and if this is the case, we will see this event 8.6 years later (I assume everything is right up to this point).

So my question is, is it possible for me while looking at the sky on a lucky day, suddenly see the explosion of a star that happened x years ago and be the first eye witness of this event? In other words, is there a technology on Earth (emphasis on "on Earth" here, the satellites or space shuttles do not count since they might be slightly closer to the star than the Earth is) that can see this before me?

My logic is that even the greatest telescope "sees" whatever light it receives. So since a telescope cannot increase the speed of light it receives, it shouldn't be more fast than me. And since light is the fastest way of transferring information, I assume that I am as possible as NASA to see such an event. Is there any way this assumption is wrong?

Naked eye nova are fairly common, several per year. Here's one.
Naked eye supernova are far rarer. SN1987a in the large Magellanic cloud was naked eye visible (vid). From this list, it appears the supernova in 1987 was the most recent naked eye supernova.
There was a naked eye gamma ray burst in 2008, but I don't think anyone actually got outside in time to see it.

If you have 50 years to look at the stars, you might see a supernova.
If you have a small telescope, you can pick them up pretty regularly in nearby galaxies.


What does a solar eclipse at 99% totality look like? People smarter than me say the sun will be 99% covered during the 2017 solar eclipse where I live (Portland, OR metro). I am curious to know what it might look like (photo or drawing). Does this photo approximate 99% coverage? Or will there be slightly more sun? Will I still see some of the corona?

At 99% the sun will be reduced to a small sliver. There are simulations on the internet. It will become very notably darker. However, the sun is 400000 times brighter than the full moon, so 1% of the Sun is still 4000 times brighter than the Moon, and so no corona will be visible. You can still damage you eye looking at that sliver.

Since the centre of totality is less than an hour's drive from Portland, if you have a car, and can get out early, you can see totality and the corona (weather permitting)


Can you still see Polaris even if you are in the south pole? I haven't been to south pole but can the Polaris still be viewed if the viewer is in the south pole? Or this question makes no sense at all?

Currently Polaris is at a declination of a bit over 89 degrees, which means that no one south of 1 degree south latitude can see Polaris. That's almost all of the Southern hemisphere, let alone the South Pole.

Polaris won't be the North Star forever, thanks to axial precession. In about 13000 years or so, Polaris will have a declination of about 46 degrees or so (twice the 23 degree axial tilt). Polaris will thus be visible in 13000 years or so as a wintertime star to all of Africa, all of Australia, and most of South America, but none of Antarctica.

After millions of years, proper motion may make Polaris visible over Antarctica. But then again, being a yellow supergiant, its unlikely that Polaris will be visible anywhere (without a telescope). It will instead be dead.


Infrared Astronomy for non transiting objects? I understand how infrared spectroscopy works for determining molecules in the atmosphere of a transiting object (like Venus from Earth). But how does this work for objects that don't pass between us and the sun? If the IR spectra is an absorption spectra, is the light emitted by a planet, like Uranus, enough for us to determine its atmoshpere's composition via IR spectroscopy?

Yes it is. As long as light is produced by a source and then passes through a medium, we can perform spectroscopy on it. When considering a planet like Uranus, most of the light that it emits is actually reflected light or re-emitted light from the Sun. Uranus produces very little light by its own merit. For light that is directly reflected (by that I mean the light from the Sun goes into the atmosphere, hits molecules, and gets scattered out of the atmosphere again), this is directly equivalent to the standard transiting planets you mention in your question. For light that is re-emitted, you still see absorption spectra, but the black body background has changed. In this scenario, the light is absorbed lower in the atmosphere and re-emitted at a longer wavelength, but that emitted light must still try to leave the atmosphere and will experience some absorption along the way, creating absorption lines.


Calculating Air Mass at Zenith for observing sites above sea level? Air mass relates to how much of the Earth's atmosphere the light from an observed object must pass through before being observed.

At Zenith, and sea level, the value of "air mass" is set to 1.

Light from objects lower to the horizon at an observing site must travel through more atmosphere, and hence the brightness of the object is somewhat diminished. In this situation the value of air mass increases above 1.000, presenting greater dimming for the observed object from increased atmospheric light scattering.

As example, when making precise observations of the magnitude of an object in astronomy, there is a need to calculate the air mass for an object at the time of observation, determined in large part by its altitude angle from zenith, or from the geometric horizon.

There are various formulas for calculating air mass. Precise calculations consider several factors and can be quite complicated.

For more details see - Air Mass (Astronomy). Wikipedia.

My question:  Is there a simple formula to calculate the air mass at Zenith (only), for observing sites about sea level.

Such a calculation provides a comparison of the merit of an observing site at higher elevation over one at lower elevation in regard to the factor of air mass.

I imagine for most elevations the factor is quite small. Naturally there are plenty of other factors to consider when it comes the the quality of an observing site - humidity, frequency of clear skies, human light sources and light pollution, etc.

To my understanding, the atmosphere in most calculations is considered to be 9 km in thickness.

A very simple calculation could be the site elevation subtracted from 9 km, divided by 9 km, providing a ratio value somewhat less than 1. However this calculation does not account for the changing density of the atmosphere with increasing elevation.

If there is a simple formula, what is it?

I would use the wikipedia variable density atmosphere model

https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere

Which gives airmass as a function of zenith distance for an exponentially decaying atmosphere of scale-height $H$. This formula is valid for all heights above sea level but the airmass given will be in terms of airmass at zenith from that height. To make your adjustment to an equivalent sea-level airmass then you would multiply by the ratio of the density at your observing altitude to the density at sea level.


Hydrogen Line observation - Interference from trees I am setting up a 2.4 metre parabola pointing straight up to observe the 21cm hydrogen line at 1420MHz with meridian drift scans.

I'm planning to place the dish on a steel pole embedded in the ground and protruding 2 metres from the ground surface.

There is a tree some metres away that has a few branches well overhead of the antenna.  It's a Eucalyptus tree, so not heavy with green leaves.

Question:  Should I be concerned about interference from a few overhanging branches (say about 15 metres above the dish)?  My understanding is that shorter lines in the cm range should be less affected by interference (e.g weather), but I'm not sure about trees.

The 21 cm line is from atomic hydrogen which is going to be free, neutral atoms in vacuum, where the electron is bound only to the proton.

In water, or any other hydrogen containing molecule, the electron's orbit will be dramatically modified (or even "stolen") by the atom to which it is attached, and so the transition will no longer exist as a narrow line.

Certainly there might be all kinds of RF interference from objects encroaching on your antenna's reception pattern, which may have side-lobes as well as the main beam pointing upwards, so like @Kozaky's comment suggests, you should to some experimentation if possible to understand contributions to background.

Your sensitivity will depend on the bandwidth you choose for your receiver, as well as the thermal and Doppler broadening of of the source that passes by, so make sure you have an understanding of what the bandwidth should be to see a peak, and also if there are significant offsets in frequency due to a Doppler shift.

above: From here

  Ground state hyperfine levels of hydrogen (parallel and antiparallel) with the spin-flip transition, emitting radiation at 1420 MHz. The corresponding wavelength is 21 cm. (21-cm line, hydrogen line)


Potential formation mechanisms of the interstellar asteroid 'Oumuamua NASA recently discovered 'Oumuamua, an interstellar asteroid that appears to have a very elongated shape, unseen in solar system asteroids.

What physical mechanisms, in the progenitor stellar system or from interstellar transit, could lead to the formation of an asteroid with such a shape?

If you don't mind speculation, it seems to me that a massive impact - like the one that is thought to have formed our moon, but even larger, could eject shards like that.    That such a shard could escape it's solar-system and fly through space between solar-systems isn't unreasonable.   Asteroids (to my understanding) are somewhat loosely formed and an elongated shape like that is unlikely.

A small moon or larger asteroid near a super-nova might get turned into an elongated shape like that by the blast as another speculative answer.   A 3rd possibility might be a ejection from a high velocity and somewhat violent debris disk, perhaps from a planet that fell inside the Roche limit of a white dwarf star.

It's still an object that's strange for two completely separate reasons.  It being the first ever object observed to be passing through our solar-system and it's thought to be very unusual shape (footnote - better images would be nice).   Still, the intersection of two unrelated firsts scores high in statistical improbability, even if the shape can be explained, it still leaves open the question of why two never before seen events coincided with the first ever observed exo-asteroid.


Where do the heavier elements come from? Partly inspired by this question, the end result of stellar fusion is Iron-56 (eventually, after some decay). Then it explodes. How/where then to the heavier elements come from. The answer here mentioned two colliding stars form gold. What process is taking place to form this gold (and I'd assume the rest of the heavier elements)?

The processes at work are the r-process and the s-process.

The r-process occurs in supernovae. Under the extreme conditions leading up to it, heavy elements are initially blocked from undergoing beta decay to other elements. If a heavy atom captures a nearby neutron, it can go to a heavier isotope or element. This happens multiple times to an atom, producing heavy elements.

The s-process is slower, and occurs under less dramatic conditions in asymptotic giant branch (AGB) stars - a certain type of red giant. It also uses neutron capture and the subsequent decay of that neutron into a proton. Over thousands of years, atoms climb to become heavier elements. The s-process, though, requires that there already be some heavy elements in existence, unlike the r-process.

For more information, Hyperphysics, this lecture and this German site (in English) are very useful.


What uncertainty does an error bar signify in astronomy? When an astronomer talks about her/his topic and shows an X/Y-plot with error bars. What should one assume that those error bars represent?
1 standard deviation? Or 2? Or some specific significance level like 95% or 99%?
Is there a generally understood convention for this, or does it vary depending on topic? I've noticed that it is very rarely explicitly stated what the meaning is of the range which the error bars cover.

And how are the cases treated, where exact observations reveal a variation in the actual population? Not "errors", but natural true certain variation. Are there other bars to illustrate that range of true values, as opposed to the variation due to observational uncertainty?

In a published scientific paper, the significance of any error bars should be explained. If this has not been clearly stated then peer review is not doing it's job. However, in some fields there may be unstated conventions which slip through without explanation.

In my own field (star formation) plots which shows averages over a large data set would show means with error bars representing one standard deviation, while plots which show single measured values might have error bars representing calculated uncertainties on the measurement, for example due to detection limits or instrumental uncertainties.


Can my Sky Watcher Infinity 76P "make the sky bright" & do I want it to? I've seen quite a few time-laps films where the Milky Way is seen as bright and filling the sky, as some fantastic glowing cloud full of glittering stars.

I wonder if it is possible to see this (or something a little similar) through a telescope using my eyes (not a camera)?

It seems to me that you would want as little magnification as possible to see the sky like this, and it would be all about gathering as much light as possible? This answer describes this scenario as "Low Brightness" (and low resolution) objects.

I have a Sky Watcher Infinity 76P telescope. At the moment, it has a 30x erecting lens that tends to wobble in its helical mount, and doesn't give a good image. I've read claims that with a non erecting lens and a bit of care, this telescope can give good images.

Are non magnifying lenses available for the Sky Watcher? Or would I be better off taking a different approach completely, such as buying binoculars?
I'd like to be able to share this with my children, and they find the existing Sky Watcher lens hard to use – as well as its wobble, I think this is because it has a narrow "exit pupil" (?), so that you need to get your eye in just the right place. Can I get a low magnification lens that is more forgiving for the user?
Can I expect to be able to see something from my backyard in a small UK town? Some of the shots in that film I link to seem to show clear bright sky pictures when there are other sources of light on the ground, or even the setting sun.

Many thanks.

To make the images in the time-lapse (and similar Milky Way photos), the photographer uses time exposures with wide angle lenses, may stitch together multiple images from longer lenses, and uses extensive post-processing to balance contrast and include foreground or bright objects in the scene.

The closest you can get to the same thing visually is to go someplace with very dark skies (middle of the ocean, or a wilderness with no nearby towns) and use the naked eye or wide-angle binoculars.  In dark skies, the Milky Way overhead is obvious and impressive, but it still won't match the color or vividness of the photos.

Your telescope isn't really useful for looking at the Milky Way as a whole, but it can probably give a satisfying view of the many star clusters that dot the Milky Way--the Pleiades and the Beehive are naked eye objects and so shouldn't be too hard to find with your telescope.  When these clusters are "resolved" into their individual stars, they are impacted less by the light-pollution you probably have in your backyard location.  I can't tell if the 76p eyepiece is one of the standard sizes that can be replaced (1.25", 0.965" OD).  If it is, an inexpensive eyepiece of about 25mm focal length might improve your views of astronomical objects.


At what distance would it be safe for humans to stare at the sun? Let's say we could travel anywhere in the solar system, at what distance would it be safe for a human to stare at the sun (without eye protection) and not get eye/retina damage?

(Not sure where this question is good for, here, or perhaps Biology.SE, so let me know.)

If the sun is visible as a disc, then it is certainly can cause damage. As you move further from the sun the disc appears smaller, but the surface brightness remains the same. The eye can resolve a disc of about 1 arcminute, and the sun is about 30 arcminutes when viewed from Earth. So you would need to be at least 30 times further from the Sun than the Earth is, which puts you almost exactly at Neptune.

From that distance, the sun would be a magnitude -19 star. If you stared at it, it could probably still damage your eyes, however, your eyes are constantly jumping (saccades) which might move the image of the sun around the retina enough to prevent damage. There is, of course, no atmosphere to protect you from UV, so let's hope that's not a problem.

Beyond Neptune the sun gets dimmer, as it can't appear to get any smaller, and there is probably little risk by the time you reach the main part of the Kuiper belt.


If suddenly "knocked" or perturbed from its orbit, would gravity eventually return the Earth to its original orbit? If suddenly "knocked" or perturbed from its orbit, would gravity eventually return the Earth to its original orbit?
I am curious as to whether this is even possible.
It seems to me that since the orbit is based on gravity, the same gravity will pull it back to correct orbit eventually. Is this reasoning right?

There's a few parts to this question so there's more than one answer.
Earth gets knocked a little bit out of its orbit all the time by gravitational influence of other planets in our solar system. Jupiter and Venus are the primary two, but all the planets have some effect.  These are called orbital perturbations and they tend to alternate, not add up.  They're the causes of Earth's Milankovich cycles and the effect isn't negligible.  These variations in Earth's orbit, which take thousands of years to move back and forth can cause the formation and recession of glaciation, sometimes called ice ages.
These orbital perturbations, in addition to being cyclical, as a rule, don't effect Earth's semi-major axis which is very consistent, and may only change slightly as the Sun loses mass.
Orbital perturbations can to lead to much greater variations of eccentricity and axial tilt.   Mars, for example, undergoes much bigger variations than Earth, but it still undergoes a cyclical back and forth that leads to general stability within a range.   All eight known planets in our solar system are thought to be relatively long term stable.
The exception to this cyclical back and forth is if there's a resonance where the effect can grow over time.   The closest example of this in our current solar system is Jupiter and Mercury, where they're not in, but they're close to resonance and it's possible that Mercury will be tossed out of its orbit in a few billion years.  It's the most unstable planet in our solar system.
This article isn't published, but I still think it's a good summary of both the stability of orbits in our solar system and the possible (but perhaps unlikely) destabilization of Mercury in a few billion years.
Planetary migration is another means for tossing a smaller planet out of its orbit.  This is thought to be fairly common, based on observations of other solar systems and some uncertainty that gas giant planets could form close to their suns.  (there may be room for debate on that), but migrating planets is thought to be fairly common, if somewhat slow. It's never been observed but it can be modeled.  It's been suggested that when our solar system was young, Jupiter moved inwards, perhaps tossing Uranus and Neptune outwards and reducing the material available that would eventually become Mars and when migrating back outwards, leading to the late heavy bombardment which brought water to Earth. This is called the grand tack hypothesis.
An inwardly migrating Jupiter could certainly toss Earth about easily into a completely different orbit, but there's no evidence that suggests Jupiter is likely to migrate that much in the future.
A massive enough object from beyond the solar system could toss Earth into a different orbit and in doing so, not only change Earth's eccentricity and orbital plane, but change Earth's semi-major axis as well and that change would effectively be permanent.
A gravity assist of that kind could give Earth both a new eccentricity, a new orbital plane and a new semi major axis and perhaps a new set of Milankovich cycles, though I suspect over time, the orbit could re-circularize and the Milankovich cycles, largely reset to where they were, so there could be some correction not by the sun, but by the other planets orbiting the sun, but not everything would go back to where it was.   The Semi-major axis change would likely be permanent and there would also be the question of whether the new orbit was in near resonance with other planets which would lead to further changes over time.
A higher eccentricity could create stronger seasonal variations and perhaps trigger a new ice age, or maybe fix man made global warming.  :-)
The effects of a change in semi-major axis would change the length of a year, so, we'd need new calendars, and could warm or cool if it was big enough.
There would also be tidal concerns if a massive object passed close enough to significantly change Earth's orbit and perhaps even, pulling the Moon out of its stable orbit.
The good news is, space is big and quite empty and an object massive enough to change Earth's orbit passing close enough to do so is extremely unlikely.


Why is moon light not the same color as sunlight? The light from the moon is light being reflected from the sun. The sun, in space, is white. But on Earth, when the light is filtered through an atmosphere, the light appears yellow. So then, why is moonlight white through the atmosphere?

The light from the moon is light being reflected from the sun.

This is at least one reason you should not expect the Moon to have the same color.  Sunlight hitting an e.g. blue object would appear predominantly blue and similar.  So the color we might expect to see from the Moon is going to be adjusted by the color properties of the Moon's surface.

As it happens lunar regolith is essentially gray in color (i.e. neutral), so it more or less reflects the spectrum it receives.

  The sun, in space, is white

No it is not.  There are a couple of reasons.

The Sun is an exceptionally bright object (for us) and will saturate our color vision so that it seems like it's white when we look at it directly.

The Sun is actually a yellow star.  It's spectrum peaks in the (visual) blue-green range.  There are stars that have a red or blue color.

White is a tricky one.  White is a perceived color.  There is no "white" wavelength in the EM spectrum.  White is a balance between the three different color sensors your human vision uses.  The Sun itself is not white because the color combination "white" we're balanced for is based not on the light from the Sun in the vacuum of space, but on the light that reaches the surface of the Earth and also what is reflected from the Earth (which is part of the ambient lighting we normally experience).

  why is moonlight white through the atmosphere?

It seemed a nice clear sky last night around my neck of the woods and the Moon looked yellow to me.  This will vary with your location and atmospheric conditions, but also with your personal perception of color.


Why are perihelion and the shortest day in the northern hemisphere different? The orbit of Earth is elliptical and the nearest distance to sun is perihelion position. Should it not be the shortest day in the northern hemisphere?

We have "days" because the Earth is rotating. The shape of the orbit has little effect on the daily cycle of light and dark.

The length of the day is determined by the tilt of Earth to the orbit of the Earth around the sun.  When the Northern hemisphere is pointed away from the sun, then the sun is low in the sky and is below the horizon for longer. So the shortest day happens when the Earth is at the point in its orbit when the angle that the Earth's axis makes with the line from the Earth to the Sun is greatest. This is on about the 21 December.

The longest and shortest days of the year occur because of the tilt of the planet, which also controls when the Equinoxes and Solstices occur. The tilt of the planet has absolutely nothing to do with its location in orbit, and it's only pure coincidence that they're so close together. Also, perihelion moves over time (apsidal precession) and in 56,000 years it would occur on the longest day in the Northern Hemisphere. Except for the fact that the Earth's axis will have also precessed (with a period of around 26,000 years) so the shortest day will occur somewhere else on in the Earth's orbit

For the sake of clarity: The equinox does not occur on the minor axis. The solstice does not occur on the major axis. The tilt of the Earth to then orbital plane does not change over the course of the year, and the Earth's rotational axis is no aligned with the major axis of the Earth's orbit.


What is the tightest orbit binary planets can orbit each other? If there were two Earth-like planets in a tight orbit around each other, how close could they be to each other without colliding?
How quickly would they have to orbit to be stable?
Would they be habitable?
Could something similar to a Roche lobe form from their atmospheres?

One can certainly imagine double planet "Rocheworlds" (like the one in Robert Forward's eponymous novel) where two terrestrial worlds fill their Roche lobes until they touch.
On its own this is perfectly fine: there are indeed a fair number of asteroids and trans-Neptunian objects forming tight pairs or a merged dumbbell shaped contact binary. The planets would be tidally locked to each other, not dissipating energy through tides.
The problem is that when the rest of the solar system is taken into account tidal dissipation begins to matter. The tides will cause the planets to move a bit relative to each other, and strong tidal dissipation will dampen that motion - dragging them inwards. Hence, relatively quickly since the tidal dissipation would be strong in big and soft terrestrials, the double planet would merge.
The orbital speed calculation looks a bit messy. Basically, the mass of the touching planets (that we need for the speed) is their density times the volume of their Roche lobe, which lacks analytical expression. However, one can approximate them fairly well as spheres of radius (see also Eggleton) $r=0.38 R$ where $R$ is their orbital distance. So we get $$M\approx 0.2298 \rho R^3.$$ Kepler's law gives us the period $$T^2 = \left(\frac{4\pi^2}{2GM}\right)R^3$$ (note the $2GM$ term rather than $GM$ - this is for a double pair), or $$T = \sqrt{\frac{2\pi^2}{ 0.2298 G \rho}}.$$ For Earth-density worlds I get $T=4.2512$ hours. Note that this is independent of the size of the system!
Can systems like this form? Probably. Can they remain long enough that life emerges on the planets and they become Earth-like? Hard to tell, but I suspect the answer is something like "yes, if they start sufficiently far away from each other then they will spiral in slowly enough that there is a biosphere when things get really tight." I suspect such coincidences in time and formation are rare. But the universe is large.


Measuring star distance by parallax using a small telescope I was wondering if it's possible to measure star distance using parallax at home with a small telescope (Something like a 3-inch reflector)? Or is this not really practical?

Short answer - not really, parallax for the closest stars is right on the limits of resolution for a good ground-based amateur equipment.

The nearest star would show a parallax angle of under an arcsecond.  (The Parsec even takes its definition based on one arcsecond of parallax.)

Ground based amateur observations are probably limited to an arcsecond at most, so it's probably not possible to measure them.

A special (and totally fictional) case - if a nearby star like Alpha Centauri were exactly between us and a much more distant star, and it were possible to see them both without Centauri overpowering that distant star, then maybe it would be possible to observe them as a close double.  That's not true in practice however.

Update:

Prompted by Rob's much better answer, the earliest parallax determination I've been able to find was using

  the 6.2-inch (157.5 mm) aperture Fraunhofer heliometer at Königsberg

(Source: Heliometer article on Wikipedia.)

In purely aperture terms, that's an amateur-sized instrument (with some admittedly very precise measuring gear built in.) So my suggestion now is to ignore my initially doubtful response and try measurements, perhaps using a camera at fairly high magnification as Rob suggests...


How is the mass and semimajor axis of a directly imaged exoplanet found? I can't find information on this for the life of me.

How do astronomers approximate the mass and semimajor axis of a directly imaged exoplanet? Would several photos need to be taken to find the period and find with Kepler's third law or can it be inferred through the temperature of the planet (or another method)? Also if you have any sources of this that would be great.

Thanks

A single direct image of an exoplanets can give an estimate of the mass and orbital period of an exoplanet.

For the former, you need a model for how the luminosity of a giant exoplanet (which all the directly imaged ones are) evolves with time, as a function of mass, and an estimate of the age. One simply looks up what mass of planet would have cooled to its current luminosity at the age of the star that hosts the exoplanet. The luminosity measurement needs an accurate distance to the system and benefits from observations at multiple wavelengths. Realistically, these masses are uncertain by a factor of two at least, because of cooling model uncertainties.

Here is an example of this applied to various directly imaged exoplanets. Evolutionary tracks for various masses are shown in the luminosity versus age plane. Note the size of the age error bars and that is before you factor in uncertainty in the models (two flavours, "hot start" and "cold start" are shown as solid and dashed lines respectively - from Bonnefoy et al. 2013).

The semi-major axis can only be estimated (a lower limit), from a single image, and only then if you have a good distance to the star/planet system. The angular separation is converted to a projected linear separation.

A perfect example of the above is the estimate of the mass and orbit for the exoplanet GJ 504b (Kuzuhara et al. 2013).

With more images spread over an appreciable fraction of the planet's period one can be more accurate. The projection of the elliptical orbit can be mapped, yielding the orbital inclination and thus a better semi-major axis measurement. The orbital period can then also be roughly estimated and checked for consistency with Kepler's third law and an estimate of the stellar mass. Note though that currently imaged planets have orbital periods of at least decades.

Wang et al. (2016) give an example of this approach for beta Pic b. A dozen precise images constrain the semi-major axis to a few percent and give the total system (star plus planet, but dominated by the star) mass.


Where to get rate of change for calculating ephemeris from JPL Horizons I have a working implementation of Schlyter's method for calculating the Sun's position.   I would like to know how Schlyter got the 3.24587E-5_deg in the following calculation for Mercury.

N =  48.3313_deg + 3.24587E-5_deg   * d    (Long of asc. node)

I could not find this number in JPL Horizons.   I'm guessing it's the difference between the N at two epochs divided by number of days between the two epochs, but I'm not sure.

My email to Paul Schlyter and his reply, posted with permission:

  On http://www.stjarnhimlen.se/comp/tutorial.html you note that
  Mercury's ascending node is given as:

  N =  48.3313_deg + 3.24587E-5_deg   * d

  According to https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt Mercury's
  ascending node precesses in the negative direction.

  As Where to get rate of change for calculating ephemeris from JPL Horizons notes, NASA's
  value matches the negative of your value almost exactly.

  Are you using a different sign convention here, or is this an error?

Reply:

  It's neither - instead I use a somewhat different reference system.

  I use the "epoch of the day" instead of the fixed epoch of, say,
  J2000.0. So add the rate of precession to the negative rate of change
  you found at NASA's site, and you should get my rate of change.

  Since I use the "epoch of the day", the positions I get are suitable
  to compute e.g. the rise, transit, and set times for the planets. Of
  course they are also good for computing the positions of several
  planets relative to one another. But if you need the positions to plot
  on a star map drawn for some fixed epoch, you need to apply a
  correction for precession to these positions.

  Yes, it can be confusing to distinguish those different astronomical
  coordinate systems from one another. But that's a consequence of
  living on the planet Earth which orbits the Sun, rotates, and wobbles.


Are we moving away or towards each other? There are numerous theories about the shape of the Universe. I've read somewhere most probably in 'A Brief History of Time' that there is a possibility that the Universe is spherical. So is it possible that all galaxies are moving towards each other? Imagine the Universe as a sphere. Now, imagine a diameter AB. The Big Bang took place at A and the galaxies started drifting away and as they moved away further they stared attracting each other on the other side i.e.Gravitational attraction. So we feel that the farther galaxies are the faster they are moving away(Hubble's Law) but actually they are moving towards each other and will eventually meet at point B. Also, space-time is spherical and so light travels along the sphere.

So, my question is: Is it possible for the universe to be spherical with galaxies apparently moving away but actually moving closer?

At the present time this doesn't hold for the observable universe, since the observable universe is only a tiny fraction of the surface of the 3-sphere (assumed it's a sphere at all). The observable universe is flat within a 0.4% error.

Based on this, we are moving away from each other. And the universe is expanding. So even if it would be curved, we would move away from each other. Only in a future Big Crunch scenario, we would eventually move towards each other.

Finite flat cyclic spaces can be constructed, but there is no significant hint, that the universe is of such a shape, or any other non-trivial topology.


Pictures of galaxies, nebulae: which way is up? We are used to viewing pictures of the Earth and the planets with their North Pole pointing towards the upper portion of the photo. But when a space telescope snaps photos of cosmic entities, do they rotate the photo around to get an attractive composition? Is it a subjective rotation, or are there rules? Which way is up?

Images are often presented with (Earth's) North at the top but there is no correct orientation. For example, in this image from the VLT north is 0.1 degrees from the vertical, the sidebar on the image page has details on the position, field of view and orientation.

However, this image from the same telescope has been rotated left 9 degrees, which is likely an artistic decision.

Amateurs also regularly align the cameras on their scopes so that north is up as it can make it easier to diagnose tracking problems. However, it's fairly common to change the camera alignment on the telescope to frame particular targets or groups of targets.

Many classes as astronomical objects, such as elliptical galaxies, irregular galaxies or nebulae don't have a well defined or known axis of rotation, which is why Earth's poles are used as a reference.


Laser Collimator producing oval not circle on the Primary I own an orion laser collimator which I had not used in a very long time. I decided to collimate an old Celestron NextStar 114 and I realized that the laser did not produce a circle but an oval.
Apparently the telescope is roughly collimated using a collimation cap.
There seems to be a corrector less in the focuser which I don't dare to disassemble. Could this oval be produced by this corrector or perhaps the laser collimator needs collimation too or maybe there is a problem in the secondary mirror?

First of all, what do stars look like through the scope, particular just inside and outside focus?  If they are oval, then read on.

I'm not familiar with this telescope, but I suspect the most likely sources of the problem are
1. The corrector lens you mention (which is a relatively unusual thing to have) is distorted.  If you can't remove this, then you are stuck with it.
2. The adjustment screws on the secondary mirror may be too tight, thereby distorting the shape of the .  Try loosening each screw by a small amount, say 1/8th of a turn, and see if this improves things.  Make sure the telescope tube is horizontal when you do this in case the secondary slips off its mount.


How feasible is it that we could see the Central Black Hole Sgr A* occlude one of it's closely orbiting stars? I'm thinking about the recent famous photo of a black hole, some months ago. It got me wondering if there isn't a way to more easily make a direct observation of these.

We detect exoplanets in part by phenomenae that occur when they transit their star.

We also have excellent time lapse photography of numerous stars orbiting the Milky Way's central Black Hole (Sgr A*), with timescales quite amenable to human timescales.

Could we observe Sgr A* directly by observing one of these stars passing behind, or almost behind it?

The diameter of the BH is estimated at 60m km, more than 40x the diameter of our sun (1.4m km). At that distance, parallax related effects and relative sizes wouldn't be an issue, as the star and BH are virtually at the same distance both very far from earth. A star also wouldn't have to be occluded, just pass close enough behind, for gravitational lensing to be detected on that fraction of its orbit (but not other places).

Additionally, a star could also be occluded by an accretion disk (if any), and its light affected in a way that would presumably show something of the nature of what it was passing behind. Admittedly Sgr A* may have little or no disk, but we might detect a disk of low enough density/size/activity not to be active, if a residual disk did in fact exist.

Are we likely to be able to see in time lapse photography, stars vanishing and reappearing behind an apparently empty patch of space, in this way? Has it been done?

I'm sure someone will write a longer/better answer, so I'll keep this short and just mention that when a black hole passes between the observer an a star, we usually say that it becomes gravitationally lensed rather than simply occulted. This can potentially provide much richer data than an occultation but requires more complex analysis.

Optical/Infrared

It also takes heroic efforts to try to image near Sgr A* optically. You need to use infrared wavelengths because of dust and an array of telescopes using adaptive optics, combined interferometrically. See How did they make a video of the center of the galaxy, and what is it exactly that's flashing there?

If there was a dedicated, second Very Large Telescope array pointed at SgrA* all night every night, there is certainly the possibility that a gravitational microlensing event could occur, and an otherwise invisible star could suddenly brighten substantially as it passed directly behind Sgr A*.

Are there candidates?

As far as I know there are no currently known stars closely orbiting Sgr A* that are likely to pass directly behind it. However, I suppose that there could be one so dim that it has not yet been observed, and a lensing event would be the first time it gets noticed, so this is certainly a possibility.

But lensing can brighten any background star, it would not have to be in a close orbit around Sgr A* at all.

Radio

Gravitational microlensing works essentially the same at all electromagnetic wavelengths, so it can be observed in radio as well as it can optically.

However, individual stars tend to radiate most of their energy optically and are not so bright at radio wavelengths. It would be hard to pick out the radio radiation of a background star above the strong emission of heated plasma in SgrA*'s accretion disk.

To this end I have just asked How far have individual stars been seen by radio telescopes?

I think (but I'm not sure) that you'd have to pick a time of extreme quiet (little/no infall) to even try.

above: homemade GIF from ESA video ESOcast 173: First Successful Test of Einstein’s General Relativity Near Supermassive Black Hole


How to compute satellite coordinates (lat,long) given antenna's coordinates, angles and satellite height Given an Earth station (antenna) coordinates (in lat,long) ant its Elevation (El) and Azimuth (Az), how to compute the satellite coordinates (lat,long), known its height?

For simplification purposes, the Earth can be considered as an sphere.

So its given:

Earth Radius = 6371000 meters
Satellite Height, in meters (perpendicular to the Earth's)

Station lat, long (as parameters)
Station height (0 = Zero, for simplification purposes)
Station Elevation (angle, radius or degrees)
Station Azimuth (angle, radius or degrees)

In summary: I'm trying to calculate satellite's coordinates projected on celestial sphere. I.e. it's current ground track position from the horizontal coordinate system values and the position of the observation.

The problem can be shown in this two figures:

A way to do this is to first find the xyz in ECEF of the point on the ground (a transform for a spherical Earth is just the spherical to xyz coordinates transform from most physics text books where lat = 90-theta and lon = phi and r = radius of Earth).  Then, find the satellite slant range (which depends only on the elevation angle and the altitude of the satellite).  Then compute the unit vector toward the satellite in local ENU (east, north, up) coordinates (use Gram Schmidt to find local E and local N given U is the local point and the north pole is (0,0,1). Now, reuse your spherical transform to change az and el into local ENU.  That part is a little weird, but if you think about it, the local horizon elevation is just like the theta of a spherical system, and az is phi (U corresponds to elevation = 90deg).  Finally, compute the same unit vector in xyz (multiply each component by your ENU unit vectors and add the terms), and then "walk along that vector" by the slant path range.  You will now have the xyz of the satellite.  Finally, from the xyz of the satellite compute the lat and lon (and as a check, you should get back your same altitude).


What if the two Black Holes spiraling around each other are evaporating via their Hawking radiation? A non-technical explanation
In a Gravitational Wave event, the point is that in this event when Black Holes are spiraling around each other, they will cause the very fabric of Space-time to ripple and this event is an energetic event and those ripples travel at the speed of light and  we can describe mathematically that how this could happen and how these Waves of Gravity travel. Now, my question stops here and my question is that as we all know Stephen Hawking came up with a Beautiful explanation that Black Holes aren't so much Black as we thought of, in time, they can evaporate and that would bear the name, ''Hawking Radiation''. Now, this implies to Black Holes, and still we haven't detected Hawking Radiation yet and Hawking didn't had a Quantum Theory of Gravity.
So, if we consider for example, that Black Holes can evaporate, so they must evaporate when they are also spiraling and what will happen in this case? If the spiraling Black Holes can evaporate via Hawking Radiation, how will we detect it and what will the mathematical equations describing this Event be?

Gravitational waves are efficiently emitted by massive black holes orbiting each other - the power emitted increases with mass. Hawking radiation on the other hand is a process that increases with decreasing mass. As a result only very tiny black hole binaries would emit more power in Hawking radiation than they do in gravitational waves; at least towards the end of the inspiralling phase.
Details:
The characteristic timescale on which a black hole binary system spirals to merger is
$$\tau_{\rm GW} \simeq \frac{20c^5}{256 G^3}\left(\frac{a_0^4}{M^3}\right)\ ,$$
where $a_0$ is the separation, $M$ is the total system mass and I've assumed the binary components are of equal mass.
The characteristic evaporation timescale by Hawking radiation (again assuming each black hole is of mass $M/2$, is
$$ \tau_{\rm Evap} \simeq 640\pi \frac{G^2 M^3}{\hbar c^4}$$
For $\tau_{\rm Evap}$ to be shorter or even comparable with $\tau_{\rm GW}$, then
$$ 640\pi \frac{G^2 M^3}{\hbar c^4} \leq \frac{20c^5}{256 G^3}\left(\frac{a_0^4}{ M^3}\right),$$
which means
$$ \frac{M^6}{a_0^{4}} \leq 3.9\times 10^{-5} \frac{\hbar c^9}{G^5}$$
For black hole binaries, a reasonable value for $a_0$ would be a few times the Schwarzschild radius of the final merged black hole since most of hte power in gravitational waves is radiated in the final few orbits before merger, i.e. $a_0 \sim 10 GM/c^2$, so we can say
$$ M \leq 0.62 \sqrt{\frac{\hbar c}{G}} \sim 2\times 10^{-8}\ {\rm kg}\ ,$$
which is the Planck mass.
So the only merging black holes for which energy loss by Hawking radiation will play a significant role, close to the end of their inspiral, would have a mass of a few $\sim 10^{-8}$ kg or less. But the dynamics of such tiny black holes would not be controlled by gravity at all and they would evaporate in a fraction of a second.


How did the Mars satellites "take cover" from the recent comet flyby? After comet Sliding Spring's Oct. 19, 2014 close flyby of Mars, NASA reports that

  All three NASA orbiters around Mars confirmed their healthy status Sunday after each took shelter behind Mars during a period of risk from dust [from the comet's tail].

"Took shelter behind Mars"?  These spacecraft are in orbit around Mars, constantly in motion, so they can't very well just pack up and move to the other side of Mars to avoid getting hit with the dust.

So my questions are: What exactly qualifies as taking shelter behind Mars?  Does it simply mean being on the other side of the planet from the tail of the comet as dust from the tail passed Mars?

How did NASA get the (constantly in motion) spacecraft to hide behind Mars?  I assume that the orbits had to be modified in advance of the comet's arrival. I picture some orbital modifications that changed the phase of the orbits of the satellites so that they would be on the other side of Mars as the dust flew by, perhaps also making the orbits more elliptical with the apsis behind Mars, allowing the satellites to spend additional time in Mars' protection.

Will these orbital modifications affect the ability of the spacecraft to fulfil their primary missions as they relate to Mars?  I am not asking whether dust from the comet will affect the quality of the instruments on the spacecraft (indeed, this whole question focuses on steps NASA is taking to prevent that from happening) but whether the altered orbits the spacecraft have after their orbital modifications will affect their ability to pursue their primary missions (such as flying over different parts of Mars than was originally intended, which may or may not be beneficial to answering the questions the spacecraft were originally launched to answer.  Although scientists are nothing if not resourceful and I'm sure any data that come from the spacecraft, whether intended or not, will lead to important discoveries).

I've written an answer or two dealing with this within the last week, and I can't believe I never thought about this. Great question.

  These spacecraft are in orbit around Mars, constantly in motion, so they can't very well just pack up and move to the other side of Mars to avoid getting hit with the dust.

Believe it or not, they can just pack up and move. All they have to do is alter their orbital speed - i.e. either slow down, speed up, or change their orbit so they will automatically go a different speed, and so be in a different place. They execute a series of orbital maneuvers to ensure that they are in the part of their orbit that temporarily carries them behind Mars. Orbital maneuvers aren't hard; practically any spacecraft can modify its orbit slightly, if need be. According to this,

  The orbiters will be repositioned to a “safe zone,” where Mars will actually help shield the satellites from the cloud of dust particles.

In other words, the spacecraft will be moved so they are on the opposite side of Mars as the comet. They can't be there for long - after all, they orbit Mars extremely quickly. But, fortunately, they don't have to be there for long. As this says,

  The satellites will remain in this region for a period of up to 40 minutes, until the threat has subsided.

Why do short a period? Well, the comet will only be in the area for a short time: NASA estimates the worst part will last roughly 20 minutes. After that, the danger will have subsided. Any particles that could damage the spacecraft will be long gone.

  Did these orbital modifications affect the quality of research these satellites will be able to perform in relation to Mars?

I can't find a source for the answer to this, but I would expect there wouldn't be any problems. You should be able to undo any maneuvers that would change the orbit of the craft.


What does the filter name I+z' mean On the TRAPPIST telescope, the filter wheels are described to have a filter with the following description:

  NIR Luminance I+z' (>700nm)

After a bit of googling, I wasn't able to find what the specifics of this filter are (its FWHM and mean wavelength).

How would I go about finding these qualities of this filter?

They must be referring to two different filters:

the (Bessel) I filter which has a central wavelength of $\lambda \simeq 800\,\mathrm{nm}$  and a width of $\Delta\lambda\simeq 150\,\mathrm{nm}$, and
the (Sloan) z' filter, which has $\lambda,\Delta\lambda \simeq 970,255\,\mathrm{nm}$.

Hence their transmission is in the near infrared ("NIR"), and they're both above 700 Å.

The exact transmission curves vary a little depending on the producer, so if you want to calculate exact magnitudes, for instance, you need to find the transmission curves for the telescope in question. From the TRAPPIST equipment website, it seems they use filters from Astrodon. Astrodon don't seem to provide filter curves for these exact filter, but if you're okay with a good approximation, you can find the curves here:
Bessel I and
Sloan z'.


Need help with telescope mount; stars have tails If I take a shot longer than 20 seconds, the stars get a tail. I have a Celestron Advanced GT mount with starshoot autoguide scope. I use a cannon can on a 9" Celestron scope. I can't get a picture without having tails on my stars. It only looks good if I keep the pictures under 20 seconds.

EDIT :

I did the polar alignment. The mount has a polar alignment scope threw the center of it. It's also balanced, the mount has a brids eye level. I use the new PhD software with the starshoot autoguide.. There is no slop in the mount, the gears have almost no backlash.

If you can't get any longer than 20 seconds, chances are it's your polar alignment. The mount you have is a motorised GoTo EQ mount (I believe), but if these aren't aligned properly, then they can still have trouble.

Also, depending on your software, try a 2 or 3 star align to get a better accuracy.

Also check the scope is nicely balanced with the camera on the end, and that your mount can take the weight.

EDIT :

If you try all this and you are still getting trailing, it may be worth going to a professional to get the mount looked at. It may be faulty.


Conversion between Astronomical Frames, ex. IRCF, FK5, FK4, etc… I'm a little bit confused about reference frames, and I was wondering if someone could help clarify a few things?

So let's say I have equatorial coordinates that refer to the mean equator and equinox of an non standard arbitrary epoch. Now, if those coordinates are referred to the mean equator and equinox that epoch, then is the mean equator/equinox of the specific epoch considered the reference frame for those coordinates?

…and then how would you go about converting those coordinates to another reference frame like say the IRCF or FK5? I assume that I would need to precess them to J2000 accounting for any applicable proper motion (and if they referred to the true equator nutation). What else do you need to put them in either the ICRF or FK5 frames?

Also, something else I'm confused about is the epoch of a reference frame. So, let's say I have the IRCF astrometric coordinates of a planet at JD 2456963.5 TDB (or any other arbitrary date). Since they're IRCF coordinates, then the epoch of the reference frame is still J2000, but the coordinates refer to the instant of 2456963.5 TDB correct?

Any clarification would be greatly appreciated, and thanks in advance!

I'm not sure this answers your question, but ICRF is a non-precessing
reference frame.

On J2000 (2000-01-01 12:00:00 UTC), the z axis of the ICRF frame was
exactly parallel to a line connecting the Earth's center to the
Earth's north pole.

Since then, the Earth's north pole has precessed, so the ICRF z axis
and the geocenter-north pole line are no longer parallel.

To see this, you can visit http://ssd.jpl.nasa.gov/horizons.cgi and
use these settings:

(this is actually looking from the north pole "down" to the Earth's
center, so we would expect a negative z value).

The interesting part of the results:

$$SOE
2445701.000000000, A.D. 1984-Jan-01 12:00:00.0000,  1.007696415090285E+01, -3.712508526111991E-02, -6.356744326939433E+03, -2.994380889331686E-07, -2.570088696418019E-07, -4.731801778858530E-10,  2.120384334099952E-02,  6.356752314245179E+03,  5.329972950848752E-25,
2451545.000000000, A.D. 2000-Jan-01 12:00:00.0000,  1.601081935638483E-01,  1.746805655485580E-01, -6.356752309828780E+03,  2.230286001476848E-07, -8.078442785945724E-07, -1.658173608778433E-11,  2.120384334099953E-02,  6.356752314245181E+03, -5.549834335071262E-23,
2457389.000000000, A.D. 2016-Jan-01 12:00:00.0000, -9.877267476902714E+00,  3.153970319756332E-01, -6.356744632653835E+03,  1.794887917287106E-07, -5.409368308792473E-07, -3.057333255094828E-10,  2.120384334099952E-02,  6.356752314245179E+03, -5.736383388350969E-23,
$$EOE

It's a little hard to notice because of the scientific notation (and
the embarrassing fact that x and y aren't exactly zero at J2000), but
the x and y values at J2000 (the middle entry) are nearly zero.

The y value is nearly zero for all three entries, but the x value
changes. In 1984, the x value is about +10m, and, in 2016, the x value
is almost -10m, showing the effects of precession.

One benefit of a non-precessing frame like ICRF is that star positions
are relatively constant (excluding proper motion and the solar system's own
motion). For example, the z axis will always point (very roughly) to
Polaris, even after precession has changed the north star to, for
example, Deneb.

By default, NASA now uses ICRF coordinates to indicate where planets
are, so, if you're looking to map planets on a J2000 starchart, you
should use the ICRF coordinates.

The only problem you'll have using ICRF/J2000 coordinates (instead of
J2016 coordinates) is that any calculations you make for
stellar/planetary azimuth and altitude will be off slightly, not
enough to really notice unless you are doing precision work.