Luminosity Schechter function for galaxies Just a question I am having trouble understanding. I have the Schechter luminosity function for galaxies, given as:

$$\Phi(L)dL=\Phi_{0}\left({\frac{L}{L_{\star}}}\right)^{\alpha}e^{-\frac{L}{L_{\star}}}\frac{dL}{L_{\star}}$$

I need to consider the case when $\alpha=-1$. And then show that the average luminosity of a galaxy is exactly $L_{\star}$. Could somebody explain how I could go about doing this and perhaps a hint or some part of a setup would be excellent. I really need to understand this.

Another part, which is related to the above question, is asking me to explain why the total luminosity is a finite number, whereas the total number of galaxies diverges. By this does it mean that the total number is infinite? Any extra comments on this would also be really appreciated.

Total Luminosity with $\alpha=-1$:
$$
\begin{aligned}
L_{tot}&=\int_0^{\infty}L \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\
&=\int_0^{\infty}\Phi_0 e^{-\frac{L}{L^*}}dL
\end{aligned}
$$
Now put $L'=L/L^*$ and you get:
$$
L_{tot}=\Phi_0L^* \int_0^{\infty}e^{-L'}dL'=\Phi_0L^*
$$
Total number:
$$
\begin{aligned}
N_{tot}&=\int_0^{\infty} \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\
&=\int_0^{\infty} \Phi_0 \frac{1}{L}e^{-\frac{L}{L^*}}dL
\end{aligned}
$$
which diverges at the lower limit.

Observe with $\alpha=-1$ we have no gamma functions.

Also IIRC with $\alpha \in (-1,0)$ when computing $L_{tot}/N_{tot}$ the gamma functions cancel leaving $(1+\alpha)L^*$


Fresnel lenses for a makeshift Galilei telescope? For a kindergarden (pirate) party, friends of me want to build together with the 4 year old kids a functional pirate's telescope for each of the kids. On the web I found full constuction sets, e.g. http://www.astromediashop.co.uk/Telescopes.html (I am not affiliated), but that is only half the fun than improvising and building with household things.
Does anybody has experience building a makeshift telescope using the cardboard tube from a paper kitchen roll, in particular which lenses to use and how to fix them?
The rolls are e.g. 25.5cm long and have an approximate diameter of 4cm. Ideally, we would need two fresnel lenses, I guess, but I am not sure which ones to choose, how to fix them nicely to the tube etc.
Looking forward to your ideas.
Edit: In the meantime, I also found View the Stars Through a DIY Telescope which just states:

[You neeed]: 2 convex lenses (you can get these from a pair of magnifying glasses or by ordering lenses online)

Finding suitable lenses (with right focal length, diameter, weight) is actually the hard part.
Thinking further how I recycle things from home, I got the rough idea to build water lenses but this is clearly out of scope of a kindergarden project, or is it?

My first telescope was made from the lens and eyepiece of an old discarded pair of binoculars. I used a paper towel roll cardboard tube with the objective taped on the front and a flat, circular piece of card stock (with radial cuts near the edge for folding) taped over the other end. It had a hole big enough for the eyepiece to fit, and pieces of tape slightly covering one edge of the hole for friction.
Sliding the eyepiece in and out slightly against the friction provided focus.
It's exciting to see the world big and upside down.
I'd go ahead and try the fresnel lenses for both objective and eyepiece if you can find a pair with different focal lengths, but I think you can also find short focal length lenses in novelty, "dollar", toy, or other random-object store as pocket tools or toys, or in hobby or tool stores as cheap imitation jewler, gemologist or stamp-collector's loupes.
For a short focal length lens you can even try a drop of water stuck on a glass slide, or a transparent material that hardens transparent.
Or a spherical transparent marble or bead: Ball lens calculator, Ball lens.
If you'd like to "go small" afterwards and explore the microcosm, see foldscope.com, teamfoldscope, Foldscope for more ideas how low-cost projects can be fun and educational.


If Planet Nine indeed exists, how large a impact event there should be to be seen from Earth? Recently I've found this news article, A meteorite crashed into the Moon during total lunar eclipse in January, about a collision into the moon that happened while a lunar eclipse was under way. According to a paper{1}, it produced a flash that peaked about 4.2 magnitude, what puts it in the range visible to the naked eye, assuming a threshold around magnitude 5:

Despite visible from Earth, it was caused by a pretty small object, about 45kg, with impact energy about 1.5 ton TNT equivalent:

That made me wonder, if we were lucky enough, could we pinpoint Planet Nine position just by looking in its general direction and waiting for a large enough impact with a KBO, able to produce a visible flash?
Or instead of a visible flash, perhaps we could at least hope to detect the infrared glow from the heat produced. For example, assuming a Planet Nine about twice Earth radius at 800 AU, I estimate the amount of energy received from the sun in a Earth day to be about $10^{17}J$:
Python 3.8.5 (default, Sep  4 2020, 07:30:14)
[GCC 7.3.0] :: Anaconda, Inc. on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> solar_constant_earth = 1362
>>> solar_constant_9th = solar_constant_earth * (1/800)**2
>>> solar_constant_9th
0.0021281250000000002
>>> 9th_cross_section = 3.1415 * (2 * 6350000)**2
  File "<stdin>", line 1
    9th_cross_section = 3.1415 * (2 * 6350000)**2
     ^
SyntaxError: invalid syntax
>>> cross_section_9th = 3.1415 * (2 * 6350000)**2
>>> daily_energy_input_9th = solar_constant_9th * cross_section_9th * 60 * 60 * 24
>>> daily_energy_input_9th
9.316555641045002e+16

This is about the estimated yield from the Tunguska event, so I think even a impactor as small as 100m across delivers the equivalent to a day worth of 9th total solar irradiation in a split second, so the infrared glow of the planet could be boosted a lot, briefly.
{1}. Madiedo, José M., et al. “Multiwavelength observations of a bright impact flash during the 2019 January total lunar eclipse”. Monthly Notices of the Royal Astronomical Society, vol. 486, no 3, julho de 2019, p. 3380–87. Silverchair, doi:10.1093/mnras/stz932. link

We can compare the brightness of a comet hitting planet 9 with the 1994 impact of Shoemaker-Levy 9 / SL9 on Jupiter. It was seen by Hubble, and Galileo which was orbiting Jupiter at the time.
View from Galileo of SL9-W, a 49 second event involving a 350m comet fragment at 215,000km/h:

info from NASA:

_Galileo RADIOMETER

brightness of L-impact > that of G-impact by 20 percent   * G impact about 15 percent of total light from Jupiter at
maximum

_CCD (late 1980's tech, 800x800) time resolution 2.5 sec

data returned for K, N, V and W impacts   * W-impact clearly seen as light point with 1 percent of total luminosity of jupiter, and afterwards the fireball  roughly constant during 49 secs at 15 percent of total Jupiter intensity.

_UV Spectrometer at 292 nm

G-impact seen at 07:33:32; 20 percent increase in total
Jovian brightness   * secondary impact at 07:34:36, 1/4 as intensive as G
itself

The fireball events from SL9 reached 3200km above the cloud deck the total energy released was 60,000,000 Megatons of TNT.
Jupiter is 5 Astronomical Units away and Planet 9 is estimated at ≈500AU away, so an SL9 equivalent impact would be 100 times further away and 100x100 times less bright than SL9 because of the inverse square law, corresponding to 13.6 apparent magnitude difference compared to SL9, which was 20% of Jupiter's brightness.
Perhaps i'm wrong, i'm suggesting a maximum visual magnitude of 13.4 / 11.2 for an SL9 event of comets ranging from 350m / 3.5km for planet 9.
Asteroids are probably very rare at 400AU, otherwise a processor could check historical records for matching flashes of light where Planet 9 would emit single or multiple flashes ranging from 5 to 20 apparent magnitude, conforming to a characteristic signal of a 5-10 second rise, a 30 to 120 second constant brightness and a fade.


Photometer vs. CCD-camera The Wikipedia article on photometry says that

  These have largely been replaced with CCD cameras that can simultaneously image multiple objects, although photoelectric photometers are still used in special situations, such as where fine time resolution is required.

How much the research community is using photometers today? I looked up the prices of some photometers from here. The device, filters, etc. can cost nearly $2500. One could get a good CCD camera with that price.

What are some applications where the good time resolution (apparently around 1 ms) would be needed? Are photometers equally useful with small telescopes (e.g. 8-inch) as CCDs?

You have correctly identified a niche for photometers. Another point in their favour used to be that they were much more sensitive in the U-band than CCDs, but I think that newer CCDs can almost match or surpass the U-band response of photometers.

CCDs take quite a time to readout. The faster you read them out, generally speaking, the higher the readout noise. By comparison, the output from photometers can be very fast indeed with little impact on the noise characteristics. One of the beauties of photometers is also that they give you an instant readout of how many photons have been detected, requiring little analysis to provide real-time monitoring of signals.

However, things move on. So now there are CCD instruments that can work on extremely fast timescales. The primary example of this is a UK-built and operated instrument called ULTRACAM, which allows data-taking at 100 Hz and which uses a special "drift" mode of readout to achieve this.

Back to your main question - what are the scientific objectives of such instruments? On the link I provided above you will see a series of press releases that describe some of the science. These range from measuring the transits of Trans-Neptunian-Objects in front of background stars; examining detailed time series of flares on magnetic stars; investigating the eclipses caused by white dwarfs passing in front of companion stars; looking at the flickering output from accreting black holes; attempting to get phase resolved light curves of the optical emission from rapidly rotating pulsars and much more.

A rule of thumb is, that if you are able to obtain a time resolution $\Delta t$, then you are probing size scales of $\leq c \Delta t$, where $c$ is the speed of light. Thus if you can get a resolution of 0.01s, this corresponds to size scales of $\leq 3000$ km.

As to whether a photometer would be good on an 8-inch telescope, I'm not sure. When I was a student I used a photoelectric photometer on a small telescope to (i) study the scintillation of stars as a probe of conditions in the upper atmosphere - e.g. Stecklum (1985) - which has timescales of order $0.01-1$ seconds. (ii) To monitor "flare" stars with a time resolution of about a second in an effort to get real-time information so that spectra could be obtained at various points during a flare.


It looks like XEphem's format of databases as described in the manual's file format description. Does the filename end in .edb ? That is a common extension for these sort of files. Assuming it is, then the format breaks down as (taking the first line for example):

A name (Akamar)
A "type designation" for the sort of object, in this case f for fixed (as opposed to orbital elements or a bound satellite)
A subfield for the fixed type indicating it's a Star
The spectral type of the star (A3)
Right ascension (RA) and proper motion in RA (02:58:15.696 and -44.6 milliarcsec/year)
Declination and proper motion in Dec (-40:18:16.97 and 19.0 milliarcsec/year)
Magnitude (brightness) of the object (3.2)
Reference epoch for the proper motions (2000.0)


Easy interactive visualization of orbital elements To get a grip on orbital elements, I'd like to play with a ready interactive visualization like in Universe Sandbox. E.g. drag the inclination slider and see immediately what happens to the satellite orbit. Are there any free tools capable of doing this? I wasn't able to find such a function in Celestia or Stellarium so far.

If all you need is a quick and free tool to visualize orbits around earth, websites such as the following might be enough:
First, the very simple orbitalmechanics.info, where you can just ignore the "add launch" option:

And then, the more artsy Harmony-of-the-Spheres, which can also visualize some other systems.

Maybe that already helps.


How did Eratosthenes know that the sun is far away? The famous measurements and calculations done by Eratosthenes around 300 BC are very widely known. He concluded correctly that the circumference of the Earth is about $252\,000$ times the length of an athletic stadium.
But what Eratosthenes did would make no sense if the Sun were (for example) only $6000$ miles from the Earth. How did he know it was much farther away than that?

The sun and the moon go around the observer once a day, Eratosthenes knew that the apparent size of moon doesn't change.  This must mean that Alexandria is near the centre of the moon's orbit.  But the apparent size also doesn't change when viewed from anywhere.  So everywhere is close to the centre of the moon's orbit.  Thus the moon must be much further than the radius of the Earth. If the moon were 6000 miles from the Earth, then it would seem to grow and shrink in size as it passed by (such an effect can be seen on Mars, where the moon really does orbit close to the planet)
And the Sun is further still. At half moon, the sun seems to be at $90^\circ$ to the moon.  This is only possible if the sun is much further away than the moon.
In conclusion, the distance to the sun must be very very large in comparison to the radius of the Earth, and we can assume that that the rays of light from the sun are parallel.


Why does the Moon appear to be flat? There's no way around it: when I look at the full moon at night it looks like a mostly flat disc, with at most a bit convexity in the middle. Phases of moon look like a full moon seen through a cut-out.

Why is that the case? I recall, when I was really young, looking at it through what at the time seemed to be a very strong telescope, and it appeared not only a lot more textured but also a lot rounder. A similar thing can be seen in a video I saw a couple months ago, at 2:05: https://www.youtube.com/watch?v=XCrJ3NflOpE . So it's clearly a naked eye thing.

I presume it's related to the optics, which made me hesitant to post it on this stack exchange (thinking it might be more suited to the physics one), but chose against it presuming people here would have more experience with celestial bodies and "large" distances.

I have a hunch that the answer is very simple, but I really don't know what that might be, especially since it's done over what I view as large distances.

It is an optical illusion.

We perceive nearby objects in 3d because we have two eyes. As we see objects from two different viewpoints, our brain can put the images together to make a 3d image.

Objects that are more distant perceived as 3d if they are moving fast enough for us to see them change appearance. Again it is not in the eyes but in the brain that the perception of depth occurs

The moon is a long way away, far too far for any 3d from our two eyes, and doesn't change fast enough for to see any motion. It is too small for us to see shadows or any other clues that we could interpret as 3d structure, so it looks like a small  flat shape.

Telescopes can make shadows visible, and time-lapse can make the motion of the moon apparent, and so add to the perception of a solid.


Why does this graph for sunlight intensity on land has a steeper slope during sunrise as compared to sunset? I got this image while checking weather data for a city in North India using  Mathematica's Wolfram Alpha query

I noticed one feature in the graph which i could not explain .
Why does the encircled part 'A' which denotes  sunrise has slightly higher slope as compared to encircled part 'B' which denotes sunset !

I also checked the graph for summer month and the pattern was exactly opposite

Summer Month

Am i interpreting the data wrong or is it so that sun achieves its highest intensity during the day faster in winter months than in summers ? What could be the reason behind this ?

Too long for a comment, alas:

https://www.mesonet.org/index.php/site/about/other_measurements/#srad notes:

  Solar radiation measurements in the morning and evening are sensitive
  to obstruction (e.g., trees) on the east and west horizons. A delay of
  sunrise or early arrival of sunset at a particular station may be
  explained by examining panoramic site photos available at
  http://www.mesonet.org/index.php/site/sites/mesonet_sites.

Also dig around a bit starting at http://mesowest.utah.edu/cgi-bin/droman/meso_graph_climo_ndb.cgi?stn=SEAM5&unit=0&hours=24&day1=0&month1=&year1=2014&hour1=00&windred=&time=LOCAL&var=SOLR&lastyear=1&vnamev=Solar%20Radiation&stationname=SEAGULL&vlabel=%C2%A0W/m*m and you should find more data.


Theoretical limits for natural satellites having natural satellites Is there an inherent instability or limit that prevents natural satellites like the moons of Jupiter from having their own natural satellites?

I looked at other questions such as do moons have moons but they focus on discovery rather than theoretical limits analogous to the Roche limit for satellite formation.

Wikipedia says there may be a reason (emphasis mine):

  No "moons of moons" (natural satellites that orbit a natural satellite of another body) are currently known as of 2014. In most cases, the tidal effects of the primary would make such a system unstable.

This seems to be because the secondary would have to be very close to the primary - close enough that it doesn't become simply another satellite of the actual primary (i.e. the planet).

The page then goes on to talk about Rhea, a moon of Saturn that may or may not have a ring system. This gives some more information.


Has the Milky Way collided with another galaxy? Has the Milky Way ever collided with another galaxy? Or will the collision with Andromeda be our first?

According to our current understanding of physical cosmology (as contrasted with cosmologies of various cultural references) the Milky Way must necessarily have had collisions in a process called hierarchical clustering.
How we're certain of this is the following:

No matter what exactly happened at the big bang, we know the state of the universe at roughly the age of ~380.000 years, where the universe becomes transparent (more or less instantaneous, but let's omit that discussion). At this moment, matter emits light that teaches us the distribution of matter in the young universe.
This is the Cosmic Microwave Background Radiation.
In the CMB we see that matter is remarkably homogeneous, apart from very small fluctuations, that help us build the cosmological standard model. This means at the time of the CMB emission, the universe had nearly the same matter density everywhere. Which is the contrary of clustered matter blobs (galaxies) that we see today.

Fig. 1: CMB temperature fluctuations as seen by WMAP.
Those variations can be translated into density fluctuations, given a physical model. To our current understanding the form the seeds of galaxy growth.

Now we take those small fluctuations, add dark matter and evolve a model of the universe over time, in large-scale compuational facilities. Then we find that galaxies are a natural outcome of the small CMB fluctuations that become gravitationally unstable. Those unstable blobs of matter amplify the initial fluctuations, until whole pockets of matter collapse into dense, rotating objects.
Those objects would be dwarf galaxies, and those collide and merge and thus grow into larger, mature galaxies. This is the process of hierarchical clustering, I was referring to above. This process can be simplified to "small things collapse first", as opposed to "large things collapse first". In the latter case the Milky Way would not have had collisions before.

Summarizing
Yes, our galaxy had collisions before, and they were necessary to grow to it's current size. Scientists are searching for the relics of those collisions in a process called galactic archaeology (the link goes to a personal website, which I find neatly made).


Are rogue planets ever born in isolation? I'm using the term "rogue planet" here loosely to mean anything of planet-level mass that is not orbiting around a star. Popular discussions of these objects always describe them as forming around a star and then being ejected by gravitational interaction with other planets. But I would have thought that, in principle, they could form in interstellar space just by the collapse of a sufficiently small cloud of dust and gas.

I realize that a small cloud will have less self-gravity, and would therefore take more time to collapse [see the answers for a correction to this assumption] and have a greater chance of being disrupted by other forces. But has a theoretical analysis been done of the minimum mass object that could be expected to form in isolation over the time frame of the current lifetime of the universe? It would be a surprising coincidence of that minimum threshold were the same mass as the mass required for hydrogen fusion.

The minimum mass of a "planet" forming from a gas cloud (definitions of what a planet is are rather slippery, and some would say this is not a planet at all) is not determined by the time available. The collapse process is rapid - less than a million years. There is a minimum mass though, and what you are referring to is something known as the fragmentation limit.

A cloud becomes unstable and collapses if its mass exceeds the Jeans mass. The Jeans mass depends on temperature to the power of 3/2 and inversely on the square root of the cloud density.
$$ M_J \propto \frac{T^{3/2}}{\rho^{1/2}}$$
When a cloud of gas collapses, its density increases. If it is able to radiate away heat efficiently, then its temperature can remain more-or-less constant and so the Jeans mass decreases. This allows the cloud to fragment into smaller pieces.

The minimum mass that can collapse in isolation will therefore be set by the smallest value that the Jeans mass can attain as the collapse proceeds. This fragmentation limit is in turn set by the cloud becoming opaque to its own radiation, which occurs when the density becomes large enough. At this point, the cloud can no longer efficiently get rid of all the heat that is generated in its interior by the work done by gravity in squashing it. The temperature rises and the Jeans mass stops decreasing. Now, the cloud may still collapse, but it won't break into smaller chunks.

The fragmentation limit is difficult to calculate with any accuracy, because it depends on the 3D turbulent dynamics of a collapsing cloud and also on whether the cloud is spinning. It is generally thought to be in the range one-to-a-few times the mass of Jupiter (e.g. Whitworth & Stamatellos 2006). This is far below the minimum mass for hydrogen fusion of about 75 Jupiter masses or deuterium fusion of about 13 Jupiter masses.

Evidence that such objects may exist can be found in answers to the related questions How are rogue planets discovered? and Is there any hard evidence that rogue planets exist?


How large will our Sun get during the red giant phase (ie which planets will it engulf) before the helium flash? I have read that our Sun will increase in size to a red giant.  But after it goes through the helium flash it will contract somewhat.  Then after that grow again to be even larger before finally losing its outer layer as a planetary nebula and becoming a white dwarf.  I've seen estimates that the red giant phase will consume Mercury, Venus, and probably Earth.  (Even with Earth's expanded orbit due to the Sun losing mass to the solar wind.)  My question is how large will that first stage be prior to the helium flash?  Which planets, if any, will it engulf?

Low mass stars like the Sun do become very large prior to He ignition in the core. The exact value depends a bit on models for mass loss from the extended atmosphere (e.g. Guo et al. 2016), but estimates of 250 times its current size are possible (Schroeder & Smith 2008; Spiegel & Madhusudhan 2012).
At this radius both Mercury and Venus are engulfed. The mass loss from the Sun means Earth would orbit a bit further out than now. Nevertheless, it too would be engulfed in Schroeder and Smith's model at that point, because tidal drag dissipates energy from the orbit and it "falls" into the Sun very quickly.
Interestingly, the Sun achieves its maximum size at the tip of the RGB, rather than in the AGB phase.


Where does the radiation in space come from and can we observe it? I have recently been reading that space travel is strongly influenced by "space radiation" and how it poses a threat to human space exploration.

Does this radiation originate from stars like our Sun, or is it an omnipresent — let's just call it — "force" in space (such as cosmic noise) that doesn't have any specific source?

Also, can an amateur astronomer visualize this radiation in some way so as to be able to observe it?

Cosmic rays consist of both electromagnetic radiation (i.e. photons) of different frequencies (radio waves, IR, light, UV light, x-rays, gamma rays), as well as charged particles (protons, electrons, maybe even ions of light elements), and other stuff like neutrinos.

The vast majority of the radiation we encounter around earth will be from the sun, because it is so very close and basically a large radiating blob. Usually with isotropic (equally in all directions) radiating sources, the radiation intensity falls of with the square of the distance. That means radiation diminishes very, very fast. Go twice as far from the sun, and you only get a fourth of the radiation.

The EM radiation from UV and up (X-rays and gamma rays) is probably the most harmful. The earth's magnetic field shields us from these rays, but interplanetary travel will not have this benefit. X-rays and gamma rays may also come from supernovae and other stellar objects, which are far away, but will probably be much too faint to have an effect on astronauts. However, it can be picked up by sensitive specialized telescopes and satellites.

The charged particles may be a problem to spacecrafts and electronics onboard them, but can probably be dampened by shielding in the spacecraft, as to protect the astronauts.

Neutrinos are I think of no concern, since they hardly interact with other matter.

As an amateur, you will have problems detecting UV and above. Mainly because we are mostly shielded from this kind of radiation by the magnetosphere and the atmosphere.

You could detect particle radiation, by taking photos of northern lights, though... :)


Space Travel and length of time So I read that if an astronaut travels at the speed of light and goes to a nearby star system , when he or she returns, he or she will have aged 10 years but 1000 years will have gone by on Earth. Travel at the speed of light is not achievable ,so I was wondering if the spacecraft was traveling at an achievable speed of 10% the speed of light , how much will the astronaut have aged and how much time has elapsed here on Earth upon the return of the astronaut ?

You're actually describing the twin paradox.
That's a little beyond Special Relativity, since the astronaut accelerates, and deaccelerates. Since Special Relativity treats only inertial frames of reference, accurate treatment would require some extension towards General Relativity.
But you may consider two astronauts, one travelling with constant velocity towards Earth, the other one with the same (by amount) velocity away from the Earth, and then apply the formula for time dilation of Special Relativity to both astronauts, as an approximation, and to avoid the twin paradox.
Then you get the almost ubiquitous factor $\sqrt{1-\beta^2}$ of Special Relativity, with $b=v/c$ the ratio of the relative velocity to the speed of light. For $\beta=0.1$, hence 10% the speed of light, we get a factor of

$\sqrt{1-\beta^2}=\sqrt{1-0.1^2}=\sqrt{1-0.01}=\sqrt{0.99}\approx 0.9949874371$.

Divide 10 years by 0.9949874371, and you get 10 years plus 18.4 days on Earth.

For $\beta=1$ this division would fail, implying a break-down of the formulas of Special Relativity. That's why travelling with the speed of light doesn't work for astronauts respecting Special Relativity.


Beginner looking for extra eyepieces for telescope I have a Bushnell Voyager Reflector telescope. It came with 2 eyepieces: k4mm and k20mm. And I am trying to find additional eyepieces to zoom in further on planets and stars, but the bushnell website doesnt offer any. I am unsure if other brands will work with the telescope. Or is my teleacope not powerful enough to look further out.

First of all, any eyepiece of the correct diameter will work with your telescope. Typical diameters are 1.25" and 2".

The eyepieces you have listed tell you the focal length of the lens. This relates to the magnification - the focal length of the telescope divided by the focal length of the lens will tell you the magnification. So the 20 mm eyepiece will give you lower magnification than the 4 mm eyepiece. Thus to work out the magnication you will need to know the size and focal length of the telescope for starters. A Google search showed there are at least a couple of models but your owner's manual will give you this information.

Now a caveat: Forget (at least a bit) about magnification. Many telescope brands will [try to] sell the magnification, because this is a big selling point. BUT higher magnification might just turn a small fuzzy blob into a bigger fuzzy blob.

The most important features of a telescope are the aperture (diameter) and lens/mirror quality. Bigger aperture means more light means easier to see faint stuff. And a quality lens means the difference between a fuzzy blob and a crystal-clear image.

You will probably find that the mirrors in your telescope are reasonable quality; the lenses are usually where the sellers save on cost. I am confident that if you buy a good quality lens in the 10-15 mm range and one in the 20-25 mm range you will see very nice, clear images of the Moon, the planets out to Saturn, nebulae and star clusters.


What did the night sky look like 100,000 years ago? Question in the title. Would any constellations be identifiable?

(Here I assume by "constellation" you mean "pattern of stars", rather than "region of the sky". To be a pedant one should use "asterism" for a pattern of stars.)

Some asterisms will be recognisable, but there can be some significant proper motion over that period.

Some stars will move more than others. Generally, closer stars will appear to have moved further than more distant stars. So the stars that make up the asterism of "the plough" (which are fairly close) will have moved more, but will still be in the same general region of the sky. On the other hand the bright stars of Orion are more distant. In 100000 years ago the general appearance of Orion would already have formed.

Videos showing
The past and future of the Plough https://www.youtube.com/watch?v=txJH8RlIoXQ

The future of Orion https://www.youtube.com/watch?v=5KFQD6Bpnm0


Is there a relation with the noun planet and to be in a plane? This morning I was in the shower and it struck me that the word planet might be related to the fact that the PLANEts orbit the sun in a PLANE. Did this naming somewhat happen by accident or was it consciously chosen? If it is the former then I'm simply mindblown by the fact that I didn't see this relation before.

I quickly looked up the etymology on the word planet, which comes from Greek and means wanderer/wandering. Nonetheless, could there still be a relation, both words might stem from the same origin...?
Thanks in advance!

There is no close connection. Planet is from Greek asteres planetai meaning wandering stars (via Latin and French). Plane is from Latin planus meaning "level or flat".

It is possible that both words are derived from the proto-indo-European root *pele-, (flat) However the evidence linking *pele to planetai is weak. Thus there is no more connection between planet and plane than between planet and field (field is also derived from *pele, via the sound change in German from p to f)

http://www.etymonline.com/index.php?allowed_in_frame=0&search=planet

http://www.etymonline.com/index.php?allowed_in_frame=0&search=plane

http://www.etymonline.com/index.php?term=*pele-%20(2)&allowed_in_frame=0


Why does an irregular moon point its longest axis towards its parent planet? I'm reading the book Ice Worlds of the Solar System. Page 62, it states that an irregular moon should point its longest axis towards its planet. Why is this the more stable configuration?

Rotation of a rigid body around the major axis (the principal axis with the highest moment of inertia, which will tend to be close to the longest axis of the satellite) minimises the kinetic energy for a given angular momentum. As a quick way to visualise this, write the equation for rotational kinetic energy $T$ in terms of angular momentum $L$ and moment of inertia $I$:
$$T = \frac{L^2}{2I}$$
At constant $L$, $T \propto I^{-1}$, therefore picking the axis with the highest moment of inertia will give the lowest kinetic energy.
As a satellite is not perfectly rigid nor perfectly elastic, deformations will tend to dissipate energy but angular momentum is conserved, thus leading it to evolve into this state.
Some lecture notes on the relevant dynamics can be found here, with the relevant result (the "major axis theorem") given in §4.11.

Any freely rotating body that is not perfectly rigid, will lose kinetic energy and whist its
angular momentum remains constant in space, it moves with respect to the body until the
body is rotating about its major axis.

This explains the rotation, likewise orienting the long axis towards the planet minimises the gravitational potential energy. As an example to demonstrate this, consider a rod of length $2r$ of linear density $\rho$ with its centre located at a distance $R$ from a mass $M$, with the rod at angle $\theta$ to the radius vector. Neglecting variations in the gravitational field perpendicular to the radius vector, the potential energy is
$$\begin{aligned}V &= \int_{-r}^{r} \frac{-GM\rho}{R+x \cos \theta} dx \\
&= -\frac{GM\rho}{\cos \theta} \log \left( \frac{R+r \cos \theta}{R-r \cos \theta} \right) \\
&= -2GM\rho\left( \frac{r}{R} + \frac{1}{3} \left( \frac{r}{R} \right)^3 \cos^2 \theta + \ldots \right)
\end{aligned}$$
where the last line uses the MacLaurin series for $\log (1+x)$.
All the terms in the sequence in brackets are positive and proportional to increasing even powers of $\cos \theta$, so the potential energy decreases as $\cos \theta$ increases. The minimum potential energy is therefore when $\cos \theta = 1$, i.e. when the rod is aligned with the radius vector.
The state of rotating around the major axis with the major axis aligned with the radius vector (long axis pointing towards the planet) is therefore the minimum energy state.


Will the Sagittarius A* Black Hole eventually swallow the entire Galaxy? In one of his interviews, Sir Roger Penrose mentioned that when the Milky Way and the Andromeda galaxies eventually "collide" and merge into one, their super-massive black holes will also eventually merge. He further mentioned that the newly formed super-massive blackhole at the centre of the new galaxy will eventually "hoover out" all matter within the galaxy.
I recall reading that this is generally the case with super-massive black holes at the centers of galaxies: they will eventually "eat out" all matter within their galaxies.
Question: why don't the stars and other objects within the galaxies just keep rotating around these black-holes, like they do now? Why will they eventually be swallowed by the black hole's gravity pull?

Black holes do not suck in matter any more than stars and planets do: an object in orbit would remain in stable orbit if nothing perturbed it.
However, in the long run ($10^{19}$ years and more) interactions between stars will perturb their orbits, making many of them end up in the central black hole. An easy way of seeing this is to consider a random close encounter between two stars. They will change velocities, and potentially one will get a sufficiently high velocity to escape the galaxy - but the other one, by energy conservation, would be more tightly bound and orbit in a closer orbit. Eventually the galaxy "evaporates" through this kind of interaction, with heavier stars congregating near and eventually into the black hole and the lighter stars mostly expelled. (For all the mathematical details of this and related processes, see Binney & Tremaine's Galactic Dynamics).
Even if this did not happen, gravitational radiation makes a star orbiting something slowly lose angular momentum and spiral in, but this takes a far longer timescale.
Hence stars either fall in or are expelled over very long timescales.


How to calculate what is the best date and latitude to observe a certain object in space? What is the mathematical formula used to do that? Is there a website/app that does the calculation for me?

It's a very general question, as a rule of thumb the higher an object is on the sky the light is going to be less dispersed as the atmosphere is thinner. So, the best point and moment to observe an object is when the object is in zenith (directly over your head). Usually it is not possible to travel around the globe to make your observations, so we are going to see better the objects that happen to be near zenith, or they that are crossing the meridian line (with circumpolar objects, it's better wait until the object the meridian line nearest to the zenith)

I don't have here my books so I can't give you a proper formula, but if you want, I can try to give you that formulae.


What made the recent supermoon so super? I understand that a "regular supermoon" is a full moon when the moon is at perigee (closest to earth in its orbit).

They seem to happen often, here's a post from one last year. But I've seen news articles saying this is the 'superest' super moon since 1948 and we won't see another super moon this bright until 2034.

What makes the recent super moon so bright?

The thing you have to really understand is that not all supermoons are equal. The official definition of a supermoon, as you state, is that the full moon must coincide with the moon being at perigee.

However, this never actually happens! The moon being at perigee happens for only a single instance of time - less than a second. The moon being officially "full", again only happens for a single instance of time (i.e., when the Sun, Earth, and Moon all form a single line). The chances of these two instances of time happening simultaneously are almost non-existent.

So, when someone calls a full moon a supermoon, they really just mean that the exact full moon instance is "pretty close" to the exact moment of reaching perigee. The closer the better and the more super your supermoon will be.

If you take a look at the Supermoon which occured on Nov. 14, widely proclaimed to be the "superest" supermoon for a long time, you can see that perigee was reached precisely at 11:23 UTC, while the full moon occurred precisely at 13:52 UTC, two and a half hours later. For pretty much every other supermoon, these two events are close enough for the moon to be called a supermoon, but not two-and-a-half-hours close. The closeness in time between reaching perigee and being truly full made this moon especially big and more important, especially bright.

There is also another thing which made this supermoon the superest and that was the actual distance of the Moon's perigee. As with anything in our solar system, the Moon's orbit is not perfectly elliptical, but instead has small variations and perturbations over time. This means that the perigee of the Moon's orbit is slightly changing over time. The November supermoon occured during a particularly close perigee that won't be matched until 2034 supermoon which you reference.


Apart from Io, will any Galilean moons have oceans and land if their ice melts When our sun expands earthlings, if still alive, will have to leave. The Galilean moons offer a lot of water but we would also need land.  As far as I can tell there will be no land for earthlings to live on when the oceans of these moons melt. Is anyone aware of land  masses on these moons and if so how large?

We're not even sure the Galilean moons will still be orbiting Jupiter in 5 billion years.   They probably will, but it's hard to say with certainty.   It's difficult to model their orbital stability over that long a time-frame.

But in the spirit of your question, it takes much more than oceans and land to make a planet or moon livable.   You also need atmosphere, proper gravity (Jupiter's moons have very low gravity) and the right mix of elements.   If, somehow, one of the Galilean moons had continents, it's "ocean water" would be full of ammonia and other elements and we'd find it both stinky and likely toxic.

But the answer to your question is no, there are no continents.   The ice/water to silicate and metal-oxide ratio for the Galilean moons is much too high for it to ever form continents, and even if it could, a moon that light would have some difficulty retaining it's oceans.   Some would boil of into an atmosphere and with that low gravity, the atmosphere would be stripped from the moon over time.

From a practical standpoint, there aren't any good options for mankind to move to in 1-4 billion years.   Terra-forming mars is possible but difficult and only a temporary solution.  In time the Sun would get too hot for Mars as well.   Titan is an interesting option because it has about the right atmospheric pressure.   That's actually very useful, though living on Titan would be kind of dreary.  It's under a constant thick blanket of clouds.

The good news is, there's a lot of time to find a solution and living in space doesn't require living on the surface.   Living under the surface of an icy body (like Pluto) would be workable too.  In fact, there's several advantages to living inside an icy world rather than trying to construct one where we live on the surface.   Much less work to set it up.

If someone wants to move this to world-building, feel free, but I felt the question was asked in good faith.


What is the temperature about 10 or 15 feet below Mars' surface? I was wondering this for a school paper I'm writing.

This is one of key questions that NASA's Mars InSight probe is trying to answer but it has only recently arrived and starting taking data and it will be a while before the data is analyzed and published in peer reviewed journals. The "mole" which would burrow below the surface and make the temperature measurements encountered some initial issues and NASA and the instrument's builders were running further tests and looking at those data before proceeding according to this mission press release.


Can space-time expansion destroy light? I know that light shifts from high frequency(like gama rays) to low frequency (like radio waves) when traveling throw a gravitational field, this is called red shifting.
My question is: Is there a limit to this, what happens to the light wave when it can no longer stretch, does it get destroyed ?

Thanks, and sorry for the dumb question, just curios.

Edit:
From my knowledge and the internet :) Electromagnetic spectrum it states that light is an electromagnetic wavelength and the frequency which the wavelength travels up and down and the distance it's travel it's called spectrum and in that spectrum there is a frequency that is visible to us, but the spectrum is very large, my question is just for curiosity, I want to know if we know what happens when the light gets beyond the spectrum, can it? And I know that the expanding of space-time makes the light stretch from a high frequency spectrum to a low one, but what happens when you go to the limit of the spectrum ? can the wavelength be stretch to the point of "breaking" ?

The stretching has no limit. Light can be redshifted to infrared, then to microwaves, to radio waves. There is no limit. This is because there is no medium that carries the waves to be stretched. There is no limit of the spectrum. It runs from arbitrarily long wavelengths, to arbitrarily short.

Instead of thinking of waves being stretched you could use a model suggested by Feymann. Each photon is like a small clock with one hand. THe colour of the light is the time it takes for the hand to go round. The photons in visible light turn a thousand trillion times a second. Radio waves only a million times a second. There is no limit to how slowly the hand can turn. Red-shifting is the slowing of the hand. There is no need to think of something stretching, and nothing to "break".


Typical wavelength of solar flare I read that solar flares are customarily viewed in H-alpha light, as a temporary brightening of a small portion of chromosphere.

What all can be interpreted from this? Is it because, energy of the radiation contained by the flare lies around this wavelength? And why chromosphere ?

Solar flares are observed at wavelengths right across the electromagnetic spectrum, not just H alpha.

The basic model for a solar flare starts with the magnetic field in the corona. You can think of the topology of the magnetic field to consist of loops that poke up out of the photosphere and extend into the corona. However, the photosphere of the Sun is turbulent and constantly in motion due to convection and differential rotation. Whilst a loop may be formed in a minimum energy state, it can get twisted and stressed by these motions.

At some point an instability is reached and the magnetic field can undergo a "reconnection" event, to flip back to a lower energy configuration. During this event, charged particles are accelerated and travel down the magnetic field lines towards the photosphere.

Before they get there, they encounter the chromosphere, which is where the bulk of the particle kinetic energy is deposited. i.e. the density increases as you go down towards the photosphere and once a certain column density is reached, the accelerated electrons are stopped and deposit their kinetic energy. This results in heating and excess H alpha emission from material at about 10 thousand kelvin in the flare footpoints. Any hotter than this and all the hydrogen is ionised. The H alpha is in emission because the only material above it is optically thin to the H alpha radiation. There is hotter, ionised material produced too, and much of this is evaporated such that it fills the magnetic loops with X-ray emitting plasma at temperatures of more than a million kelvin. Some of the flare energy may also be used to accelerate material away from the Sun in "coronal mass ejections".


Convert Ecliptic to Galactic Coordinates How to convert Ecliptic co-ordinates to Galactic? The wikipedia article on Celestial Coordinate systems contains various conversions, but this one is currently absent.

For a more specific example: I would like to convert Proper motion in ecliptic latitude and longitude from the ATNF catalog into proper motion in galactic coordinates. Using the following notation from the catalog documentation:

PMElong:     Proper motion in the ecliptic longitude direction (mas/yr)
PMElat:      Proper motion in ecliptic latitude (mas/yr)
...
ZZ:          Distance from the Galactic plane, based on Dist
XX:          X-Distance in X-Y-Z Galactic coordinate system (kpc)
YY:          Y-Distance in X-Y-Z Galactic coordinate system (kpc)

How would I calculate the following (my own addition) from PMElong & PMElat?

PMZZ:        Proper motion perpendicular to Galactic plane
PMXX:        X-direction proper motion in X-Y-Z Galactic coordinate system (kpc/yr)
PMYY:        Y-direction proper motion in X-Y-Z Galactic coordinate system (kpc/yr)

Probably it's easiest to use some package for this, for example, if you use Python, you can use astropy:
http://docs.astropy.org/en/stable/generated/examples/coordinates/plot_galactocentric-frame.html.
Or if you are interested, here is a description on how to do it by hand, including the propagation the uncertainties:
http://adsabs.harvard.edu/abs/1987AJ.....93..864J

Of course, you also need radial (line-of-sight) velocities, but I don't know if that is in the ATNF catalogue...

Edit:
About the radial velocities: we only see the pulsars move across the sky and know the distance, so we get two velocity components perpendicular to our line of sight, a $v_\alpha$ and a $v_\delta$, from the proper motion in Right Ascension ($v_\alpha=d\cdot \mu_\alpha$) and the propermotion in Declination ($v_\delta=d\cdot\mu_\delta$).

Probably, it is useful for you to convert these proper motions from $\mu_\alpha, \mu_\delta$ to galactic longitude ($l$) and latitude ($b$) $\mu_l,\mu_b$, so you can get the velocities in cartesian coordinates: $v_y=d\cdot\mu_l$ and $v_z=d\cdot\mu_b$. This way, you have at least the two velocity components perpendicular to our LOS. Oh, and you maybe also want to subtract the rotation of the sun around the MW center.

There are ways to estimate the radial velocity sometimes: if the pulsar lives in a binary system, a spectrum of the companion maybe. Or a very crude way would be maybe something like http://adsabs.harvard.edu/full/1977ApJ...216..842H... But I don't know much about it.


Working with high-magnification eye-pieces My telescope (Newtonian reflector) has a focal length of 900 mm. So, using a 25 mm eye-piece, I get 900 mm / 25 mm = 36X magnification. Using a 10 mm eye-piece, the magnification would be 90X.

My problem is that I cannot see a thing with the 10 mm eye-piece. If I would insert the 25 mm eye-piece, find a bright star, and adjust the eye-piece distance from the secondary mirror, I would get the things ready after some trial and error. With the higher magnification eye-piece, it's just all black. What should I do to get around this?

The aperture is 130 mm if it matters.

Try the moon first.

If you see nothing but black, assuming that you don't have a lens cap on or something, then most likely, you are zoomed in on, well, relative blackness. The star you were viewing is probably off to the side now. Or, you may just be looking through the eye-piece at the wrong angle or something. The moon is too big a target to miss, and you have nice rough features to use to tune in your focus. And if you see black when you are looking at the moon, it's much easier to troubleshoot. You should definitely see something!

Now, secondarily the reason you expect to see something, but instead see nothing may be a viewfinder problem. If you have a small, low magnification viewfinder scope attached to your telescope, you usually can't just trust that what you see in the center of that view is exactly what your telescope will be aimed at. You may have to align it. Search "align telescope viewfinder" for help with this. From your description, it sounds like you are not using a viewfinder, but this seems to be a common problem, so I thought I should mention it.


How can apparent magnitude be negative? What is the reason for that scale? Is it because otherwise defining an maximum would be too hard(?). Why do objects that are (apparently) brighter get assigned a smaller number (and down to negative)?

Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us.  The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6).  Each successively lower magnitude was twice as bright as the one before, meaning the scale was logarithmic.  We still use magnitudes for historical reasons, though the scale was later standardized to use the formula

$m_x - m_{x,0} = -2.5\log_{10}(\frac{F_x}{F_{x,0}})$

where $m_x$ and $F_x$ are the magnitude and flux of the object of interest and $m_{x,0}$ and $F_{x,0}$ are the magnitude of flux of a reference object (where usually Vega is used to define the 0 point in magnitude).  This means any object that appears brighter than Vega has a negative magnitude.  There is no limit to how bright an object can appear, so there is no lower limit to magnitudes.  The sun, for example, being the brightest object in our sky, has a magnitude of roughly -27.


Halley's Comet current location in sky and visibility? I am an amateur in astronomy, but recently I wonder if it is possible to see (in a photo, maybe, if not in a telescope) Halley's Comet when its not close to earth. If there is such photo, how I can find it. Thanks in advance.

In case you happen to have a very large telescope, here's where to look for Halley's Comet in 2019. I got these from JPL's Horizons ephemeris.

date (UTC)    h  m  s        d  m  s        T-mag    N-mag
2019-Jan-01   08 26 12.95   +01 38 11.0     25.48    28.99
2019-Feb-01   08 22 37.55   +01 48 26.3     25.48    28.97
2019-Mar-01   08 19 35.31   +02 04 26.5     25.49    28.99
2019-Apr-01   08 17 21.46   +02 24 37.7     25.52    29.04
2019-May-01   08 16 49.80   +02 41 21.1     25.55    29.07
2019-Jun-01   08 17 59.21   +02 51 29.8     25.58    29.10
2019-Jul-01   08 20 21.84   +02 52 27.8     25.60    29.10
2019-Aug-01   08 23 27.41   +02 44 30.4     25.61    29.10
2019-Sep-01   08 26 24.57   +02 29 35.3     25.60    29.10
2019-Oct-01   08 28 23.88   +02 12 04.1     25.58    29.10
2019-Nov-01   08 28 56.98   +01 55 29.1     25.56    29.08
2019-Dec-01   08 27 45.99   +01 45 32.8     25.53    29.04
2020-Jan-01   08 24 59.69   +01 45 05.9     25.51    29.00

T-mag N-mag =
   Comet's approximate apparent visual total magnitude ("T-mag") and nuclear
   magnitude ("N-mag") by following standard IAU definitions:
    T-mag =  M1 + 5*log10(delta) + k1*log10(r)
    N-mag =  M2 + 5*log10(delta) + k2*log10(r) + phcof*beta
   Units: MAGNITUDES

From Wikipedia's Halley's Comet; History:

  After 1986

  Grainy, white-on-black image showing Halley as a barely distinguishable black dot
  Halley's Comet observed in 2003 at 28 AU from the Sun

  On 12 February 1991, at a distance of 14.4 AU (2.15×109 km) from the Sun, Halley displayed an outburst that lasted for several months, releasing a cloud of dust 300,000 km across.49 The outburst likely started in December 1990, and then the comet brightened from magnitude 24.3 to magnitude 18.9.112 Halley was most recently observed in 2003 by three of the Very Large Telescopes at Paranal, Chile, when Halley's magnitude was 28.2. The telescopes observed Halley, at the faintest and farthest any comet has ever been imaged, in order to verify a method for finding very faint trans-Neptunian objects.9 Astronomers are now able to observe the comet at any point in its orbit.9

  9New Image of Comet Halley in the Cold". European Southern Observatory. 1 September 2003. Retrieved 26 February 2018.

  49J. C. Brandt. "McGraw−Hill AccessScience: Halley's Comet" (PDF). McGraw-Hill. Retrieved 27 November 2009.)

  112D. Prialnik; A. Bar-Nun (1992). "Crystallization of amorphous ice as the cause of Comet P/Halley's outburst at 14 AU". Astronomy and Astrophysics. 258 (2): L9–L12. Bibcode:1992A&A...258L...9P.

You can view Halley's Comet's orbit in 3D here: https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=1P;orb=1

  caption: Seventeen years after the last passage of Comet Halley, the ESO Very Large Telescope at Paranal (Chile) has captured a unique image of this famous object as it cruises through the outer solar system. It is completely inactive in this cold environment. No other comet has ever been observed this far - 4200 million km from the Sun - or that faint - nearly 1000 million times fainter than what can be perceived with the unaided eye. This observation is a byproduct of a dedicated search 1 for small Trans-Neptunian Objects, a population of icy bodies of which more than 600 have been found during the past decade.

Source

  A record observation
  At 28.06 AU heliocentric distance (1 AU = 149,600,000 km - the mean distance between the Earth and the Sun), this is by far the most distant observation ever made of a comet 2. It is also the faintest comet ever detected (by a factor of about 5); the previous record, magnitude 26.5, was co-held by comet Halley at 18.8 AU (with the ESO New Technology Telescope in 1994) and Comet Sanguin at 8.5 AU (with the Keck II telescope in 1997).

  Interestingly, when Comet Halley reaches its largest distance from the Sun in December 2023, about 35 AU, it will only be 2.5 times fainter than it is now. The comet would still have been detected within the present exposure time. This means that with the VLT, for the first time in the long history of this comet, the astronomers now possess the means to observe it at any point in its 76-year orbit!


Should a station in orbit around the Sun travel in the same or the opposite direction to the planets? If we put a station in orbit around the Sun, and wanted to send transports as regularly as possible without the travel costs being prohibiting; should we have it rotate around the Sun clockwise, or anti-clockwise like the other planets?

(I'll be checking the page frequently so feel free to post comments and questions should you have them. I'll respond as quickly as possible.)

It must orbit the sun in the same direction as the Earth and other planets.

The main difficulty in sending a transport to the station will be the the velocity change needed to put the transport into the station's orbit. If the orbit is retrograde with respect to the Earth, then the transport needs to make a massive change of velocity to reach the station, and this needs a massive amount of fuel.

If, on the other hand, the station is prograde with respect to the earth, then the transfer can be made with an economical Hohmann transfer. This is much more important than any apparent benefit of a retrograde orbit passing close to the Earth more often.


How do I modify redshifts to gain corrected line of sight velocities? I'm currently trying to collect the data to run an N-body simulation for 11 of the galaxies in the Local Group where proper motions are known, however I don't understand how to get the required line of sight velocities for the simulation.

I have redshift data from NED for the required galaxies, however by judging what sources tell me, this won't be the required information for the simulation due to our motion around the Milky way's galactic centre.

For example, one source has given the NED redshift as 229 km/s for Carina Dwarf, but the corrected line of sight velocity is 14.3 km/s.

Does anybody know how to convert these redshifts into the desired corrected versions?

Thanks!!

You need a model for the motion of the Sun with respect to the Milky Way centre. You then have to subtract the component of this that is resolved towards the galaxy in question.

The solar motion around the Galaxy is somewhat uncertain, but is roughly (11, 240, 7) km/s when expressed as a vector aligned with Galactic coordinates (i.e. towards the Galactic centre, tangential to the Galactic centre and perpendicular to the Galactic disk).


Practicality of Dobsonian telescopes I've been using binocular for more than a decade for visual observation. I'm thinking of upgrading to a telescope. The Dob caught my attention due to its simplicity in setup and also cost effectiveness. I was thinking of going for 8" or 10" Dob.
What I have also observed is - If I'm going to use dob, I need to keep moving the OTA frequently to keep the focus. If I'm going to use the dob alone, it is not an issue. But my concern is when I want to show the celestial objects to others, I need to frequently move the OTA to keep the objects in focus. It may become tiring if there are many people waiting to watch.
Has anyone encountered similar situation with dob? What are your thoughts? Did you ever feel that you should have gone for go-to mount instead of Dob? Any suggestions please?

An 8 or 10 inch Newtonian telescope is principally a fine thing. Much telescope for little money.
A Dobson is a Newtonian telescope mounted on a rockerbox (see Alt/Az mount for a discussion). Consequences are: cheap, lightweight, trivial setup, your're either on your knees or on a chair, too seldom in between, it needs constant adjustment because the sky doesn't move that way.
If you want to track stars you need an equatorial mount, a cross of 2 axes of which one points to the celestial north (or south) pole. The same tube from the Dobson can also be mounted there. They can be motorized on one or both axes or/and have a computer built in to automagically find an object out of a database, then they are decorated with the "goto" attribute.
They are heavy, expensive (multiple times the price of a newton tube) and need a setup ritual each time they are moved. But once oriented, a motorized mount, connected to some software and computer, can be left alone for hours and will track the stars it points to.

So, the question is: do I want it light and cheap and don't I care about gymnastics, then the Dob. Do I want to go further and track things, do photography, than the other way. Perhaps with a sexy apochromatic refractor. But that's easily more than 20 times the price of a dob without camera stuff.
I personally don't do goto. Finding things is half of the fun for me. I go with equatorial mounts, a small portable one and one for photography setup I can drag outside. How about, you start with the Dob. If you bite, you're going to invest more anyway :-)
Edit: there is a third way (and potentially others) that might be worth considering, that would be a telescope like the one in this thread, a small reflector mounted on a computerized alt/az mount.


Best locations for large telescopes at high northern latitudes Large optical telescopes want to be located at high altitude (get above as much of the atmosphere as possible). I get the impression that there is also a preference for low latitude (see as much of the sky as possible). And perhaps also for a location with good weather, so that there will be as few cloudy nights in the year as possible? So for example, many large optical telescopes are on the summit of Mauna Kea, Hawaii; there is also McDonald Observatory in Texas.

Suppose you wanted to put large optical telescopes as far north as possible, but still meet the other requirements, what would be a good location? Looking for high mountains at northern latitudes is an obvious starting point, but how well would that meet the requirement of clear skies? For example, Mount Washington in New Hampshire is the highest peak in the north-eastern United States, but has notoriously terrible weather; this would be an important downside?

You can (probably) do much better than Mount Washington, both in terms of latitude and air quality. For one, you're right about the weather. According to Wikipedia, it is a convergence point for several storm tracks and next to a breeding ground for low-pressure areas. This not only means a lot of clouds, but very turbulent air, which also lowers the quality of the sky. Also, at just a little shy of $2~km$ in height, you're going to find yourself in the middle of all these clouds, which won't be pleasant. Oh, and you should also consider the difficulties of actually building an observatory in such weather.

That being said, building a telescope at very high latitudes is certainly not a bad idea. On the other side of the globe, there is Ridge A on Antarctica, which is agreed to be "the best suited location on the surface of Earth for astronomical research".

Yes, being located on low altitudes allows you to see more of the sky, but you can never see all of the sky at once. Observatories near the equator are great for studying a vast number of different objects over the course of the year. If you're doing a more specific study though (for example, if you want to observe a certain object every night during an extended period of time), you'll have to find yourself a more northern (or southern) observatory. On these higher latitudes, there is a bigger portion of the night sky that is visible throughout the year (constellations in these areas are called 'circumpolar'). This may be desirable for some specific projects.

Looking for good northern spots to build a telescope, we can indeed start by looking at mountains in the northern part of the world. If you take any topographic map of the northern hemisphere, you should immediately notice Greenland sticking out. The big issue with Greenland is that it's covered in a giant layer of ice. You can't build on ice; you'll get in trouble when it melts. We'll disregard that for now though, and we're interested in its highest peak: Gunnbjorn Fjeld. We're in luck here; Gunnbjorn Fjeld is not covered in ice, but is a rocky peak protruding through the ice. The major difficulty would be actually building there, but I'm going to take a wild guess here and claim that modern building techniques and machines could do it. So this is probably what you're looking for. I'm not an expert on the weather in Greenland though, so I can't say anything about that (I do know that it's cold, which is good, because that means the air is stable).

Alternative locations at northern latitudes would be Iceland (but that's probably a bad idea given its geologic activity) or Siberia. The highest point of Siberia is Klyuchevskaya Sopka (which is also a bad idea given that it's an active volcano). Apart from that, googling mountains in Siberia yields lots of peaks which may or may not be well-suited for an observatory. You could probably find a mountain in the Urals as well, but they are not as high as one would ideally want.


Which is the closest exoplanet to Earth? The closest star to Earth (after the Sun) is well-known: Alpha Centauri, 4.3 light years away. But what about the closest exoplanet to Earth, outside of the Solar System?

Earlier this year (2016), scientists used the radial velocity method to discover a planet orbiting Proxima Centauri: Proxima Centauri b. It was announced in Anglada-Escudé et al. (2016). Here are some of its basic properties, as reported by the authors of the paper and known as of August 2016:

Mass: At least 1.3 Earth masses
Semi-major axis: 0.05 AU
Maximum eccentricity: 0.35
Orbital period: 11.2 days
Surface temperature: 234 K (see Table 1)

Now, as the authors note, the planet may not be habitable:

  The habitability of planets like Proxima [Centauri] b—in the sense of sustaining an atmosphere and liquid water on its surface—is a matter of intense debate. The most common arguments against habitability are tidal locking, strong stellar magnetic fields, strong flares and high ultraviolet and X-ray fluxes; but none of these have been proved definitive.

However, further observations might help us learn more about the planet and determine if it could be habitable for any type of life.

It should be noted that the controversy about planets around Alpha Centauri continued after this question was initially answered by user8. Alpha Centauri Bb was shown to be nonexistent by Rajpaul et al. (2015). However, in another twist, observations by Demory et al. (2015) seemed to show the existence of Alpha Centauri Bc (at the time, Alpha Centauri Bb was still a possibility, so now "Alpha Centauri Bc" would actually be "Alpha Centauri Bb" if it existed), an Earth-like planet. However, there has been no more evidence to confirm the existence of this planet - and at any rate, Proxima Centauri is closer to the Solar System than Alpha Centauri, so Proxima Centauri b wins no matter what.


Hamiltonian of general three body problem i try to extend my analysis from planar restricted three body problem to the general one.

Is there anyone that can help me to find out the Hamiltonian for this prolem in cartesian coordinates?

I need the hamiltonian of 3D case somethink like eq.(1) in this article or in this one paper again eq.1 . Because my symplectic integrator can be used only if I have the fixed cartesian coordinates.

Thanks

The condition $$\sum_{i=1}^{3}m_iq_i'=0$$
of this paper generalizes to
$$\sum_{i=1}^{n}m_iq_i'=0,$$
meaning the barycenter should be at the origin of the frame.

Equation (1) generalizes to
$$H=\frac{1}{2}\sum\frac{|p_i'|^2}{m_i}-\sum_{i=1}^n\sum_{j=1, j\neq i}^n\frac{m_im_j}{|q_i'-q_j'|}.$$
The first part is the total kinetic energy of the system relative to the barycenter, written in terms of momenta: $$E=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{p}{m}\right)^2=\frac{1}{2}\frac{p^2}{m}.$$
The second part is the sum of all self-potential energy between each pair of (spherical) objects, with Newton's gravitational constant $G$ set to 1.

See also n-body problem on Wikipedia.

The 3-dimensional 3-body (or n-body) problem is like the planar version. Just take 3-dimensional vectors for $p_i'$ and $q_i'$.


GW190521 black hole merger total mass calculation and missing mass, how does this happen? I have just read an article about that black hole merge event (it's in Italian):

Sette miliardi di anni fa, due mostri si unirono

What made me curious is that the article tell that a 66 solar mass black hole merge with a 85 solar mass black hole to form a 142 solar mass black hole.
If the datas are correct, that suggests to me that during a merge event, matter inside a black hole is ejected (66+85=151).
Is that (mass ejection) possible? I have always thought that escape form the inside of a black hole is impossible.
I suppose it is somewhat possible if we think about Hawking radiation, but due my amateur knowledge level I have not been able to dig into this very much.

I remember the first time I'd read about a gravitational wave detection and the resulting mass I also stopped when I realized the numbers didn't add up. At first it's shocking. While we might be aware that very tiny changes in mass are associated with huge amounts of energy release, it's absolutely amazing to think of a mass the size of our Sun radiated as energy!
But that's exactly what happens in the mergers of black holes. The fraction of mass converted to energy is much smaller when two normal stars merge, gravitational waves are produced but are small in comparison. But when the objects are extremely dense so that they can orbit very close before the merger, as in the case of neutron stars and black holes, those gravitational waves are powerful and a significant fraction of that mass will radiate away as ripples in space itself.
Answer(s) to Fraction of initial mass lost (radiated) by neutron star mergers compared to black hole mergers? indicate that this fraction can be up to several percent in these cases.
We can convert an absurdly tiny amount of those distortion waves back into other forms of energy as a thought experiment, though not practical. For more on that, see answers to

Does a gravitational wave loses energy over distance?
Transfer of energy from gravity back to other “more familiar” forms of energy?
Does shortening the path length of an excited etalon do work? What about LIGO?


Kepler's Law, focal points - Heliocentric or Barycentric? It is known that all the mass in the Solar system moves around the Barycenter.
For the two focal points in Kepler's law; is the first focal point $F_1$ a Heliocenter? Or is it actually a Barycenter?
If the $F_1$ is a Heliocenter, then, can the same exact ellipse and eccentricity of orbit be maintained by adjusting the second focal point $F_2$ and taking Barycenter as $F_1$ instead of Heliocenter?

Planetary orbits are normally described as heliocentric, but it is possible to describe them from a barycentric point of view. JPL Horizons (https://ssd.jpl.nasa.gov/horizons.cgi) provides for both possibilities.


What is the definition of a "pole" of a celestial body? What is the definition of a "pole" of a celestial body?
Earth's pole is defined as it's rotational pole.  The North and South Poles are the two points on Earth where its axis of rotation intersects its surface.
Apparently, (according to everywhere I've read), the poles of astronomical bodies are determined based on their axis of rotation in relation to the celestial poles of the celestial sphere.  But this is an inadequate definition.
What about Pluto?  Pluto's axis of rotation does not intersect its surface.  Pluto is tidally locked with Charon, and Pluto orbits a barycenter that is located outside of Pluto.  And yet Pluto has a North and a South Pole.  Why are the North and South Poles of Pluto where they are, as opposed to any other location on the surface of Pluto?
How are these poles defined for Pluto?
What is the definition of a "pole" of a celestial body?

You are mixing the rotation of the body around a barycenter with its moons, and the rotation of the body around its own center.
For a bound rotation like Pluto and Charon both have to have the same rotation period - yet both have to be present. Thus Pluto's axis of rotation of course intersects its surface - it rotates around its own axis at exactly the same speed as it orbits around the common barycentre with its moon Charon. Were it different, it would not be a bound rotation.
The North pole usually is defined as the place where the body's rotation axis (about it's own center of mass) with the right-hand-rule intersects the surface (thus the vector of the rotational momentum in positive direction intersects the surface when placed in the centre of rotation).
There is virtually no non-rotating body.


Sources of astronomy-related dataset? I'm doing a project for a statistics class, and I thought it may be interesting work with a dataset associated with astronomy. I know absolutely nothing about astronomy, unfortunately, and I wouldn't even know where to look for any large sets of data. Preferably I would want some data that could conform to a regression model, ie explanatory variables plus a response variable. Does anyone know of a place to look for what I am describing?

I know that the question is kind of general - I guess the important part is that the data is somewhat 'interesting' (something a layman understands), and that there are multiple variables that might possibly have a correlation.

You have plenty of choices!

I am specialized in X-ray astronomy, then I can suggest you many facilities: they usually give at least some of their data publicly available. Tipically, you have data for light curves (photon count rates versus time), and/or spectra (that is flux versus energy).

I suggest you to work on the light curves: they are easily available, already reduced (you don't have to play with the data to make them usable, you can directly use them), and they fit your request of conforming to a regression model. Also spectra are enjoying, still I know some theoreticians who are not familiar with spectra, plus they usually need some more knowledge, because of response files, spectral fitting packages, etc. It is up to you (and your available time)!

Here a list of some facilities with available data:

Rossi X-ray Time Explorer

Swift/BAT

INTEGRAL

Chandra

MAXI

Also here you can find a list of many facilities with NASA participation, so the list is much larger.

Of course you can also use other kind of data (optical, infrared, radio) by the MAST project, but I have no experience on those. I guess, in the optical band, the Hubble Space Telescope will have a huge amount of data!

If you need a hand, beyond the SE vote up/vote down neurotic fashion, let me know. Good luck!


How can I check if we can observe a certain star (given its equatorial coordinates) from a certain location on Earth (given its DMS coordinates)? I am given the equatorial coordinates of a star (e.g. Sirius has RA: 6h45m8.9s and Dec = -16°42'52.1"). Now my question is: can we observe this star from a specific location on Earth (e.g. from Brussels 50°51'1.62" N 4°20'55.61" E)? If yes, how can I justify this using the coordinates of the star and location on Earth?

You can check whether the star is anti-circumpolar i.e. if it never rises above the horizon (if its upper culmination altitude is less than zero). If it is, you will never see it from that location. If it is not, at some point during the year you will be able to see it in conditions with minimal light pollution if its upper culmination altitude is around zero (and, of course, the conditions do not have to be as good if the upper culmination altitude is greater, so you may want to increase the upper culmination bound by a few degrees to make sure you can actually see the star).
A way to justify being able to observe a star whose upper culmination altitude is big enough is to state that the Sun goes along the ecliptic across a fixed celestial sphere during the year, and at some point must be on the other side of the celestial sphere compared to the star. That day, when the Sun sets, the star will be visible.
Sirius, in particular, can be seen from Brussels tonight at around 2 AM:


Planet orbit circumbinary planet I was wondering about a planet who orbits two stars. Could it be that it moves as in the image?

While such orbits can be proven to exist mathematically, they are unstable, and so no real planet has such an orbit. Such an orbit is analogous to balancing a pencil on its point. Mathematically there is a point of balance, but it is still impossible to do.


When did the Scutum constellation receive its current name? While cleaning my desk, I found an out of date star map (puts the cleaning into perspective, huh?) where the Scutum constellation is still referred to as Scutum Sobiescianum).  This is the original name Johannes Hevelius gave it in 1684

My map is (unfortunately) modern and comes from the 2nd half of the 20th century. It's a Polish one so it's not odd for it to still use the traditional name. The main illustration of the Scutum article in Polish Wikipedia happens to do so as well.

I know the name of the constellation has been changed but I can't find any information on who did it and when. I suppose it must have been the IAU (International Astronomical Union) but I have no idea regarding the time of the change.

When was the English name of the constellation officially changed and who did it?

IAU's official website mentions that the names of the 88 current constellations (Scutum included) were standardised during the organizations inaugural meeting in 1922 and apparrently haven't been changed ever since.

  Celestial nomenclature has long been a controversial topic. At its inaugural meeting in 1922 in Rome, the IAU standardized the constellation names and abbreviations. More recently IAU Committees or Working Groups have certified the names of astronomical objects and features. In the following links you can find further information on how different objects and features are named.


Can IR sensitive cameras read signals from Venus's surface emissions? This question got me thinking about this.

I know that we measure the Earth's surface temperature by satellite (perhaps somewhat inaccurately, but it's done all the same).   Using Venus as an example, can we see see evidence of Venus' 800 degree surface from space or is Venus' temperature deduced more by modeling and the Russian ship that landed there.

All things with heat give off a thermal signature and that signature isn't a single wavelength but a range of wavelength.   I assume, by measuring the overall color emitted and/or the peak wavelength, it's a fairly straight forward calculation to work out the temperature.

But when you have an atmosphere, and this is the main part of of my question.   I assume, an atmosphere is to some degree transparent, so you'd be seeing in a sense, heat through a medium and inside as well as on the upper surface.  But the heat you see diminishes depending on how opaque/transparent that atmosphere is to the specific wavelength.

Are there any good ballpark estimates to how thick (or how dense or pressurized), we can see temperature signatures through an atmosphere?     In the case of Venus, does any of it's 800 degree surface temperature generated IR make it through it's atmosphere, where it would be observable by spacecraft, giving a good measurement of it's surface temperature, or was it's surface temperature deduced by different types of models and that mistreated Soviet spacecraft that didn't last long once it had landed.

I imagine some atmospheric gases are more transparent than others.

Just to clarify - a specific "yes/no we can detect IR from Venus' surface" and a more general answer as to how far we can see heat through an atmosphere, are both fine.

Yes. A thermal image from the surface can be detected, even by amateurs.

The ground is heated to about 700K, and thermal emission at a wavelength of 1 micron is just about observable by skilled amateurs, with the right equipment. Mountains, which are cooler by about 30K, can be seen at this wavelength

From orbit, more details can be seen. Akatsuki used the 1 micron band to observe the surface in 2015

At other wavelengths, cloud patterns are revealed. Galileo took images at 2.3 microns during its flyby. They show sulphuric acid clouds at 50km above the surface (10km below the visible cloud layer) and at about 270K, with gaps exposing the lower atmosphere, glowing at 500K.


Why do the planets in the Solar system stay in the same orbital plane? An earlier question addressed why all planets formed in the same orbital plane, but how is this angle maintained? What prevents the planets from taking on a different orbital plane?

Answer to the NEW question:

the Angular Momentum Conservation Law states that, for any moving body, its angular momentum does not change unless you exercise an external force different from the central force.

For an orbiting body like a planet, this means that Sun's gravity, being the central force, does not modify Angular Momentum, but any other external force will do.

Examples of external forces are collisions or the forces made by Jupiter on another planet, or by Neptune on Pluto.

After the Solar System was formed, these external forces are quite small, and thus does not change greatly the Angular Momentum of any major body. But you can see how passing near a body can alter a comet's orbit.

Moreover, the external forces made by bodies that are in the same plane as an orbiting body does modify the value of its Angular Momentum, but not the direction. This causes that  the orbiting body changes its orbit but can not not change planes.

So if you add small forces from objects in the same plane, you end up with no changes to planes.


Do constellations change? Many stars are part of constellations. Due to dark matter the stars have velocities which are nearly the same unregardless their radius of orbit around the centre of the Milky Way. Also, it takes about 200 million years to orbit this centre. But nevertheless the angle from which we look at those stars could change a bit during this orbit.
Regarding all these effects, could you say that constellations don't change at all or does it just take a long while?  And of course stars could run out of energy and explode and vanish, so that we don't see them anymore. But is this the only way the constellations could change?

The constellations change slowly, over thousands of years, due to the proper motion of the stars with respect to the Sun.  The main cause of these motions is the orbit of the stars around the centre of our Galaxy (with or without dark matter).  This video shows these changes for the constellation of Ursa Major over 400,000 years.

Another way the constellations can change is indeed through stellar evolution $-$ either the birth or death (or brightening/dimming) of stars.  However, the timescale for stellar evolution for naked-eye stars is typically hundreds of millions to billions of years (the Sun's life expectancy is around 12 billion years), so that this effect is much slower (and less well predictable).


View from inside the black hole If an observer is falling toward a black hole with its face away from singularity then what will he observe after crossing the event horizon? The reason that why I am asking this question because as far as I know for an outside observer, the falling observer appear to freeze at event horizon i.e. time appear to stop for falling observer. So if the falling observer is able to look outward after crossing event horizon then he will be able to see an infinite amount of time which is impossible. So what will the observer see after crossing the event horizon?

Your premise is incorrect:

The external observer's view does not imply anything about an observer within the event horizon.

The event horizon itself is just an indicator of the edge of the region where light can no longer escape. Light can head inwards as you would expect.

So an observer inside the event horizon will be able to see light reaching them as normal (well, not exactly as normal, as there will be lensing effects) - there will not be an infinite amount of time visible.


Searching dataset for Constellation detection I need to develop an Machine Learning (ML) model for predicting constellation seen in a picture taken from a mobile. This I need to develop from scratch. Firstly, I need help finding a database which can provide me images of stars with constellations labelled in it. Those can be from a ground based telescope or from a camera. This is going to be a multi-labelled problem. Even after spending hours of searching I can't find a good dataset. I want a dataset with several images (preferably a few 100), each labelled with the constellation seen in them. Then I am planning to refine them and use TensorFlow for prediction. I am a grade 12 student and still new to ML, so I really need your help.

If actual photographs are not required, Stellarium could render a suitable set of reference images.
Use the Perspective (gnomonic) projection for the closest match to a photograph, and display only the features you want.
Its scripting capability may help you avoid repeating some manual operations.
Here are a sample pair of Stellarium images.
The first has only synthetic stars, which the online plate-solving service Astrometry.net successfully recognizes:

The second is the same part of the sky with only constellation labels and boundaries:

The unlabeled region between Ophiuchus and Scutum is Serpens Cauda,
part of the only discontinuous one of the 88 standard constellations.


Collision of two Earths Assume that two Earth-like planets orbited the same star, with elliptic orbits. Eventually, something occurs that caused the two planets to collide.

What would be the outcome of such a catastrophe? I figure that both planets (and any life if it exists) would be destroyed, but exactly how?

Would both planets explode, combine together into a larger planet, or be able to rebound and either reestablish an orbit or be pulled into the sun?

To start with, such two orbits are extremely unlikely. First of all, any object orbiting a star and deviating strongly from a circular orbit has a high risk of crashing into something. As you notice, there are very few planets in our solar system, and all of them except Pluto have a very regular orbit. Planets which have intercrossing orbits would have crashed into each other and fallen apart on the planetesimal stage, much earlier than becoming a planet.

Then, if you suppose that such an event does occur, the consequences will depend on multiple factors. For one, the composition of the planets - if the two planets are terrestrial (rocky), one would expect them to fragmentate and lose a large amount of their mass. Their cores would maybe fuse into one planet, and the debris would form one or several satellites. The exact outcome would be dependent on their speed, their density, and the angle of their collisions. A similar (but not the same) scenario happened during the late stage of Earth formation, when an object close to Earth's size crashed into it to create the Moon. If the sizes are the same and the formation is complete, chances are that most of the matter would fly away and the planet that would be left would be tiny.

It would be a different story if the two planets are gaseous. They have almost no solid matter in them, so chances are that the planets will fuse whilst losing (proportionally) significantly less matter than terrestrial planets. Still, there would be losses.

Lastly, if a terrestrial and a gaseous planet meet, the outcome is quite easy to predict. Gas planets are much larger than rocky planets, so the rocky planet would simply be engulfed. It will maybe cause a disturbance on the surface of the gas giant.

To wrap it up let me just note that such a collision would almost definitely change the orbit of whatever is left, so that it would either get pushed into the sun or away from it, causing it to float eternally in outer space. Have a nice day.


What is the speed of gravitational waves ? I read that two huge black holes collided or merged and so it created gravitational waves billion years ago , So now we are able to detect it & detected it. So what is the speed of gravitational wave ? Is it faster than light ? and  are there many more to be detected ?

Gravitational waves travel at the speed of light.

There are many possible sources of gravitational waves. The two confirmed detections so far (14 Sep 2016) are merging black hole binary systems, but sources of (detectable) gravitational waves may include merging neutron stars, short period binaries containing neutron stars or white dwarfs, gamma ray bursters or supernovae.

For a brief introduction see Gravitational Wave Sources at the ALIGO site.


Vertical speed vector at which the sun travels to and from the galactic equator What is the speed of progression of the sun toward and away from the galactic equator during its orbit?

The Sun moves at about 15-20 km/s with respect to a local standard of rest defined by the general motion of stars in our vicinity around the Galaxy. In three-dimensions, this "peculiar velocity" is $U=10.00 \pm 0.36$ km/s (radially inwards), $V=5.25 \pm 0.62$ km/s (in the direction of Galactic rotation) and $W=7.17 \pm 0.38$ km/s (up and out of the plane).  (Dehnen & Binney 1998). Different authors arrive at velocities that differ by $\sim 1-2$ km/s from these values and so this would probably be a more conservative estimate of the uncertainties.

The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. The Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100 pc; it will be roughly 30 million years before we cross the plane again.


Correlation of planet sizes with star sizes? I did my own amateur Kepler data analysis in May of this year of data from Kepler.nasa.gov/Mission/ discoveries. I found a strong correlation between star size and planet size (percentages). 37 of the 44 planets confirmed (84.1%) orbiting M dwarfs were less than 2.0 R(Earth) with only one hot Jupiter. 129 of 241 (53.6%) orbiting K dwarfs were less than 2.0 R(Earth) with 9 hot Jupiters (3.7%). 196 of 517 (37.9%) orbiting G dwarfs were less than 2.0 R(Earth) with 49 hot Jupiters (9.5%). 92 of 213 (43.2%) orbiting F dwarfs were less than 2.0 R(Earth) with 23 hot Jupiters (10.8%). I would like to know how much of this correlation is probably real and how much is biased by the limits of the study? Biases I am aware of are the fact that the Kepler study is primarily of systems with "close in" planets, (only 90 of 1015 confirmed as of 4/1/15 are over 0.3 AU) and that only 4.3% of the "Kepler stars" are M dwarfs. I also see biases that might arise from the methods of confirming the planets (Kepler stars from 110 to 405 had planets confirmed by the "multiple candidate" method). It also appears that there may be a greater difference between K and G star systems than between G and F. Where can I find a more professional analysis of this type?

I think there are a number of studies that look at these statistics; I'll try to dig some out. In the meantime I can give you some further things to ponder.

First, let's assume a null hypothesis that the statistics and mass distribution of planets were independent of stellar mass - what would we observe?

Well one selection bias you don't mention is that the likelihood of transit detection depends on the ratio of planet to star size. This means you are less likely to see small planets around F-stars than M-stars. Or to put it another way, you expect the fraction of large planets to increase with stellar mass - as you have found.

It is quite likely that the material available to form planets is correlated with the stellar mass - in other words there is a correlation between stellar mass and protoplanetary disc mass. In which case it could be quite difficult for smaller stars to have the requisite protoplanetary disc density to form giant planets before the disc disperses.


How do we know the current state of the universe by observing it from Earth? Since light travels at the constant speed and takes more than a billion years to travel from the distant parts of the Universe, how do we know the state of the distant regions? Do those still exist or not? Is our picture of the Universe outdated?

Since light travels at a constant and finite speed, looking into the Universe means looking back in time.

No matter in which direction we look, on average we see the same: Locally, we see evolved galaxies, far away we see young galaxies, and even farther we see the so-called cosmic microwave background, which is light from so far back in time that galaxies hadn't yet formed.

That is, the Universe seems to be isotropic. Unless we occupy a special place in the Universe, this implies that the Universe is also homogeneous, i.e. it would look more or less the same if you were located in another part of it.

This is turn means that the current state of the Universe can be inferred by observering the local region. The most distant observed galaxies can then be concluded be in the same state (statistically speaking; of course they don't look exactly the same, but any given property, e.g. the distribution of their sizes, will be the same as in our neighborhood).


Is it possible to use the stars to determine the passage of time? I'm writing a science fiction short story which involves a group of people being in suspended animation for a very long period of time, on the order of thousands of years.

My question is, would an astronomer on waking be able to determine that a significant period of time has passed purely by making observations in the night sky? How accurate would they be?

For the purposes of the story, I'm more interested in what an astronomer could observe unaided or with a primitive telescope of the sort Galileo might have had early in his career (around 3X magnification).

Thanks!

In the night sky in 10,000 years, two things will have changed in relation to the stars. The first, the rotational axis of the Earth will have changed, shifting the celestial sphere. The second, the stars themselves will have moved a bit relative to each other due to proper motion. So, the night sky will be quite different in 10,000 years, but still recognizable, particularly the constellations which will have changed somewhat but will still be identifiable. I would posit that as long as the person or persons were at least casual star gazers or astronomy enthusiasts (not even necessarily professionals) they could estimate how much time has passed in the course of their slumber, I'd say with a margin of error of about +/-2000 years. Should they be trained astronomers that margin of error should fall.


Dataset containing list of known globular clusters Does anybody know if there is an existing dataset that consists of all known star clusters, their name, and their apparent dimensions or similar? Is it possible to make a query to, or download all SIMBAD objects that are of type "globular cluster"?

Thanks

So far we have about 150ish known globular clusters in the milky way. A list of all the known ones can be found HERE. This includes their locations (may be slightly shifted by now as they were correct as of 2011, the time of writing) and where applicable, the Messier/NGC/IC numbers are included for ease.