galaxy - Do all the stars visible to the naked eye belong to Milky Way?

For decades the variable star S Doradus, in the Large Magellanic Cloud, was considered to be the most intrinsically luminous star known. But even though the Large Magellanic Cloud is the second or third closest external galaxy, S Doradus is still too far away from Earth to be seen with the naked eye from Earth. O fall the stars that always or sometimes are intrinsically brighter than S Doradus, only the variable Eta Carinae, in our own Milky Way Galaxy, is close enough to Earth that it is sometimes visible with the naked eye.

Omega Centauri, or NGC 5139, was cataloged as a star by Ptolemy in his Almagest about 150 AD and by Bayer in the Uranometria of 1603 - Bayer named it Omega Centauri. It is at a distance of about 15,800 light years from Earth. There is a theory that Omega Centauri is a remnant of a dwarf galaxy that as captured by our galaxy.

Thus Omega Centauri is an object that was listed and named as a star and may be the remnant of a former galaxy that has been captured by our own galaxy. Thus so far it seems to ALMOST fit the definition of an extra-galactic star visible from Earth with the naked (Human) eye.

Of course Edmund Halley noticed that Omega Centauri was not a star as early as 1677. Today it is classified as a globular Star cluster in our own galaxy (and possibly the remnant of the core of a dwarf galaxy). The light that makes Omega Centauri visible to the naked eye on Earth comes from the light emitted by hundreds of thousands or millions of stars, not one single star.

Even though a globular star cluster has the light of tens of thousands to millions of stars, and our Milky Way Galaxy has over a hundred globular clusters, only a few of them are visible from Earth with the naked eye (47 Tucanae mentioned by RichS is another, and it was also mistaken for a single star at first).

So a single star that was as far away as Omega Centauri 15,800 light years away, or 47 Tucanae 17,000 light year away, would have to shine as bright as tens or hundreds of thousands, and maybe even millions, of ordinary stars to appear just barely visible to the naked eye like those two clusters.

The Canis Major Dwarf Galaxy is believed to be 25,000 light years from earth and the nearest external galaxy, if it really is a galaxy. The Saggitarius Dwarf Elliptical Galaxy is about 70,000 light years from Earth and the Large Magellanic Cloud is about 163,000 light years from Earth. A star in the Large Magellanic Cloud, about ten times as far away as Omega Centauri or 47 Tucanae, would have to be as bright as millions to hundreds of millions of ordinary stars to be seen by the naked eye from Earth.

I suppose it is theoretically possible that one of the only about 6,000 stars visible with the naked eye from Earth might actually be in one of the 2 or 3 closest galaxies or floating alone outside of the disc of our galaxy. But it would have to be a supergiant or hypergiant star, and astronomers would have to somehow not notice oddities in its spectrum that would point to it being so rarely luminous.

I would estimate that the odds against that would be "astronomical".

observation - How to calculate the LST of an astronomical object at a given height above the horizon [in degrees]?

My situation:

I want to observe M52 at RA = 23h24m48s, DEC = +61deg35arcmin36arcsec from, let's say, Calar Alto at 37.23deg N and 2.546deg W.

How can I calculate the Local Sidereal Time (LST) at which M52 is at 40degrees height above the horizon?

Thank you for your help!

ascension - How to find maximum and minimum right acsension and declination based on the telescope's location?

In the Southern hemisphere, the maximum declination you can see is
90-L where L is your latitude. The minimum is -90, since you can see
the south celestial pole.

As the earth rotates and revolves, you can see any right ascension
within those declinations.

The only exception is that you ordinarily can't see stars that are too
close to the Sun. However, the Sun moves enough between October and
March that all right ascensions will be visible at night sometime
between October and March.

genetics - What's the aim of genetically modifying of foods/organisms?

GMO foods have a huge potential to make food cheaper to produce and more nutritious.

The most common GMO foods have at least one gene added to them - an enzyme that makes the plant resistant to RoundUp - an herbicide made by the same company (Monsanto). this makes the farmers able to grow their crops with much less intensive labor to keep the plants healthy. It does cost some money and people wonder whether using so much roundUp is good. I won't come down on the benefits of this, but you can see how it might be more economical way to grow crops. roundUp is biodegradable and does break down in about 3 weeks, just FYI.

Other GMO foods can make the crops more resistant to drought, disease or insects. This might enable crops to be grown in areas and or with longer growing seasons - a big advantage for thirsy crops like tomatos or rice. Other GMO project may allow us to make the crops more nutritious. A famous example of this is golden rice, which has been enhanced to produce pro-vitaminA, which will help malnutrition in millions of children who can die annually because of a lack of vitamins in their diet.

There is also an effort in nutriceuticals, where vaccines and common drugs maybe produced by edible plants for easily processing or even direct distribution of pharmaceuticals.

For people who can afford organic food and free range beef, I think its great that its available, but at for a hungry world, GMO foods can help solve some vital problems.

radio astronomy - Why (actually) aren't ground-based observatories using adaptive optics for visible wavelengths?

Adaptive Optics (AO) techniques allow ground based observatories to dramatically improve resolution by actively compensating for the effects of Astronomical Seeing.

The atmospheric effects are quite variable in both time and location. A parameter called Isoplanatic Angle (IPA) is used to express the angular extent over-which a given wavefront correction optimized for one point (usually a guide star, artificial or natural) will be effective. As an example, Table 9.1 in this Giant Magellan Telescope resource shows values for IPA scaling almost linearly (actually: $simlambda^{6/5}$) from 176 arcseconds at a wavelength of 20 microns to only 4.2 arcseconds at 0.9 microns.

This suggests an IPA of 2 to 3 arcseconds for visible wavelengths, which taken by itself is not a killer limitation.

However, it seems almost all currently active AO work is done exclusively in various infrared wavelengths, apparently down to 0.9 microns but no further. (AO is also implemented computationally to array data in radioastronomy.)

Is this because the observed wavelength needs to be longer than the guide star monitoring wavelength? Because it is simply much harder and there is always Hubble above the atmosphere for visible work so it's not worth the extra effort, or is there another more fundamental reason?

I'm not looking for speculation or opinion, I'd like a quantitative explanation (if that applies) - hopefully with a link for further reading - thanks!

probe - How will Yuri Milner and Stephen Hawking's nanobots decelerate and transmit data upon arrival at Alpha Centauri?

Solar sails often suggest a couple of modes for slowing down.

1) Have a launch laser at the target location to slow it down the same way. Assumes much infrastructure.

2) Use the solar energy of the target sun to slow the sail down. I.e. Turn it around and spend a longer time decelerating as you get closer and closer.

3) There was a model, where the main sail had a second larger sail that would fly with it, during acceleration, but upon cruise would separate, and advance in front of the main sail. Then the launch laser would target the larger sail, to reflect the energy back on the smaller sail to slow it down. (Lots of tricky bits in that approach).

black hole - Large hadron collider

Even if the LHC did create a black hole, its mass would be no greater than the sum of the particles that formed it, i.e. very tiny. It would evaporate almost instantly due to Hawking radiation. Secondly, subatomic particles have been colliding with atoms high in earth's atmosphere for eons, and we're still here.