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Brown dwarfs have historically been difficult to detect (directly) simply because of how faint they are. Typical luminosities may range from $10^{-3}L_{\odot}$ to $10^{-5}L_{\odot}$ depending on spectral type. Any object that dim will be tough to find, regardless of spectral type or what sort of telescope you're using - you can have the largest, highest-...

12

There is no true upper bound. You could argue that a galaxy is a vast multiple star system, with hundreds of billions of stars. Natural galaxies are not perfectly stable (random encounters eject stars over very long timescales), but one can construct stable systems by taking pairs of similar-mass binaries and placing them into a remote circular orbit around ...

7

Names are common enough for solar system bodies, but generally stars don't have names, they have identifiers from various catalogues. In the case of Pulsars. they are named by their location in the sky. Just like Earth has longitude and latitude, every point in the sky has a "right ascension" and "declination". Due to the slow movement ...

6

The rules from the the IAU for official designations are posted neatly here. More casual names, like the Black Widow Nebula, are not standardized, as far as I know. For astronomical objects outside of our own Solar System: the designation of astronomical objects beyond the Solar System should consist of at least two parts — a leading acronym and a sequence ...

6

It just means that in a star that is an unresolved binary system (i.e. the light you receive comes from both stars), the secondary contributes a fraction of that light. This fraction could be expressed as a fraction of the total luminosity or a fraction of the flux at a particular wavelength or in a particular waveband. How do you work it out theoretically? ...

5

The ratio $T/\Omega$ tells you about the acceleration of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity. If the system collapses because it has $T/\Omega<0.5$, then when it reaches $T/\Omega=0.5$ it stops accelerating. That doesn't mean it stops collapsing. It overshoots in the ...

5

As jmh has answered, stars naturally form at large distances from each other. To add to the answer, what is the reason for this particular distance scale? If we imagine a very large, homogeneous cloud of gas it will be unstable to gravitational collapse over the Jeans length scale $\lambda_J=c_s/\sqrt{G\rho}$ where $c_s$ is the speed of sound in the gas, $G$ ...

5

The initial star formation regions were regions that have a high enough mass density to form a star. The density of the early universe was not constant at different locations. Some regions had high enough density to form a star, and some didn't. When a star forms it draws in matter from a large distance away. This forms an accretion disk and leaves a ...

4

The manual says the original eyepieces had 1¼ inch barrels and focal lengths of 25 mm, 20 mm, 12.5 mm, 8 mm (all reasonable) and 4 mm (not so much). If I had to pick two replacements for a beginner, I'd get a 25 mm and a 10 mm. You can probably get an entry-level Plössl by Meade, Celestron, or Orion for US \$20-25 from the Cloudy Nights classifieds or \$30-...

3

TL;DR: No, not always, but most of the time. Well, most planets are in either an equatorial orbit (~0° in relation to the star's rotation vector / equator). Some are in polar orbit (~90° and almost perpendicular to the star's equator/rotation vector), and even fewer are in retrograde or strange (highly eccentric or inclined) orbits. For example, all the ...

3

If another star with the brightness or lumminosity of the Sun orbited only 5 to 10 million miles from the Sun, and Earth orbited around both of those stars at the same distance as it orbits, Earth's goose would be cooked, as the saying goes, by the additional heat. But the question doesn't specify where the other star would orbit or how luminous it would be. ...

3

To start, if I'm reading your question correctly you've got the general galaxy evolution model backwards, at least as far as morphology goes. At a very high level the picture goes like this: most if not all (large) galaxies form as spirals, then at varying points in their lives they merge with other large galaxies (either ellipticals or other spirals) and ...

2

For the moon, you would normally not need a very high power eyepiece. 20mm would seem sufficient, on a telescope with a focal ratio of f/9.8. As this is a refracting telescope, if you get too much magnification you end up just magnifying the chromatic aberration. Smaller numbers mean more magnification. Planets would require more magnification, on the other ...

2

All these stars orbit the galaxy center of mass. The galaxy outweights any possible star by many orders of magnitude and dominates the orbital motion, unless some stars come pretty close together. The orbits of the stars are more or less stable in the same sense that orbits of planets are stable in the solar system. They are not absolutely stable, but stable ...

2

$\eta$ Cassiopeiae A has an estimated mass of 0.972 M$_\odot$, an estimated temperature of 5973 K, and a B-V color index of ~0.58[1]. In addition to the spectrum of the star, we look at these and other properties when attempting to classify main sequence stars. You can see a table [here] which shows the bulk properties of each spectral type that we can use ...

2

I think there are a few misconceptions floating around here. The Hubble Sequence is not a sequence in time. Hubble did not mean to imply that galaxies flow from one side to the other in the sequence (He may have thought it was a possibility though). It is just meant for classification. As it turns out a small fraction of galaxies have changed class, ...

2

I was wrong, hot systems are where members have velocities which have magnitudes in all directions in the same order (causing them to have roundish shape), while cold dynamical systems are where random motions do not have the same average magnitude in all directions (ie disc galaxies).

1

Short answer: Many stars and other objects beyond the Solar System have proper names, but the vast majority of listed objects have only designations in one or more catalogs, and the vast majory of all the billions or trillions of objects observable with modern instruments have not been listed in any catalog and have no designation. Long Answer: So how many ...

1

The spectral type of a star is determined by looking at its spectrum. Sometimes authors will use other, approximate, relationships between spectral type and colour or mass, or they will look at the spectrum compared with standard templates in different wavelength regions. These are all possible reasons why different sources might suggest slightly different ...

1

Short Answer: The most stars in a known system are seven, and eight seems like the theoretical maximum. But it is possible that in extreme situations stable systems with more than seven stars might be theoretically possible. Long Asnwer: Part One of Four: Stable Star Systems are Hierarchical. In stable star systems the orbits of individual stars and pairs of ...

1

Either three, or unbounded, depending on your definition. A three-star system would be one of the stable solutions to the n-body problem: a large star dominating the system, a smaller star orbiting it, and a very small star (eg. a red dwarf) in the smaller star's L4 or L5 point. In the unbounded case, you have two stars orbiting a common center of mass (a ...

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