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1

In this answer you can find a calculation for how bright "Nemesis" would be at near-infrared wavelengths. This calculation assumed we were looking for a 20 Jupiter mass object with a similar age to the Sun at a distance of 1.4 light years (to fit in with the Nemesis hypothesis). The calculated magnitudes were H=14 and W2=8 (in the WISE infrared satellite ...


2

It's pretty easy to put upper limits on how bright any nearer stars could be. As a rough guide, consider a survey like Hipparcos, which is complete to something like 9th magnitude. If there's another star closer than Alpha Centauri, the brightest possibility is if it's at the same distance. Wolfram|Alpha tells us that a magnitude 9 star at Alpha Centauri's ...


3

It's begining to look like there aren't as many brown dwarfs as was once thought: NASA's Wide-field Infrared Survey Explorer, or WISE, has been turning up a new crowd of stars close to home: ... Previous estimates had predicted as many brown dwarfs as typical stars, but the new initial tally from WISE shows just one brown dwarf for every ...


4

No, black holes are not the only cause of HVSs, although it is thought to be the most common mechanism. Hyper velocity stars are believed to be caused when binary stars come close enough to a supermassive black hole for one of the pair to be captured while the other star is ejected at high velocity. This appears to the main mechanism for HVSs. See for ...


12

Human color vision is based on three types of "cones" in the eye that respond differently to different wavelengths of light. Thus, not counting overall brightness, the human color space has two degrees of freedom. In contrast, the spectra of stars are very close to a black body, which depends only on effective temperature. As one varies the temperature, the ...


1

The European Southern Observatory has catalogues with image data available from http://www.eso.org/qi/, you will have to register before you are able to access them. I'd suggest you look at other observatory's websites for their data. You will have to look past the pages targeted at the general public and find links for data or science, or user portal or ...


1

Yes. If you have the Hipparcos data from ftp (http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I/239&-to=3), you will have several data files. In hip_main.dat you have a field MultFlag at position 347 that indicates whether the star is a double or multiple star: Note on MultFlag: indicates that further details are given in the Double and ...


1

Basically you need to convert between luminosities (which you can add) and magnitudes using $$M-M_\odot=-2.5\log_{10}(L/L_\odot)$$ Let's call the total luminosity $L_0$ and magnitude $M_0$ and the individual luminosities and magnitudes $L_1$, $L_2$ and $L_3$ and $M_1$, $M_2$ and $M_3$. Then, you have the total luminosity of the system, directly ...


2

In the case of multiple images of a background, distant object, the answer is relatively simple. You take a spectrum of the multiple images or parts of an extended lensed image and you see whether the spectrum looks the same, and in particular whether the redshift of the multiple images are the same. Gravitational lensing affects light of all wavelengths ...


1

Yes, and not only from space but from the Earth surface too. Stars emit in almost all wavelengths depending on their surface temperatures. The hotter the star is the shorter (higher energy) wavelengths it'll emit. You can try this simulator to check this: http://astro.unl.edu/naap/blackbody/animations/blackbody.html


5

The answer is: frequently. There are many amateur astronomers that make it their ambition to discover new supernovae or to observe and report on new variable stars. As an example, let me cite amateurs Robert Evans, who apparently holds the record for most supernovae found by visual observation, or Tom Boles, who holds the record for supernova discoveries by ...


2

I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star? The answer is, you can sort of estimate this, by using something called the Rossiter-McLaughlin effect (see also Rossiter 1924; McLaughlin 1924). You can find plenty of information on the web - I'll add a couple of links when I have a moment - ...


0

Is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star? It's unclear what precisely you're asking. The planet will be orbiting (to very good approximation) on a Keplerian orbit (closed ellipse co-focal with the parent star). Such an orbit has two relevant angles. One is the angle between the orbital plane and the ...


1

Stan has essentially answered this in his comment, which I will attempt to spell out a little more laboriously. The significant majority of our Sun's energy output comes from the proton-proton chain. This was advocated by Eddington back in the 1920's, but at that time your basic concern was a very real and major problem. Objects with like electrical ...


-4

Most of the answers I've read to this question seem to come from people that don't have all the facts. If you were to look at our Sun from Pluto, it would look like just another star in the sky. stars have a limit on how far away they can be seen. Most of the stars in our own galaxy are too far away to be seen with the naked eye. what you see in the sky at ...


1

Let's assume that what is falling onto the neutron star is "normal" material - i.e. a planet, an asteroid or something like that. As the material heads towards the neutron star it gains an enormous amount of kinetic energy. If we assume it starts from infinity, then the energy gained (and turned into kinetic energy) is approximately (ignoring GR) $$ ...


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The Gliese catalogue can be found at CDS here. Given an RA and Dec then you can use the fortran program found here to get the constellation.


3

This question is very broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do ...


1

There is no equation, you need a detailed model of the interior physics of very low mass stars. Very roughly, you can say that hydrogen fusion occurs when the central temperature exceeds about $10^{7}\ K$ (the density dependence is secondary) and that from the virial theorem, the central temperature is given approximately by $T \simeq 1.6\times 10^{7} M/R$, ...


1

A proposal for the minimum mass of a gas cloud before collapsing to become a star was made by James Jeans; because of this, it was termed the Jeans mass. It is calculated as $$M=\frac{4 \pi}{3} \rho R^3$$ where $\rho$ is density and $R$ is the radius of the cloud - one half of the Jeans length, which is dependent on the speed of sound and the density of the ...


2

I think it's an interesting question. The trick would be a sustained fission reaction, faster than half life, but slower than a chain reaction. A chain reaction could hardly be considered "star like" - it would just explode. Lets say you had a planet sized object, maybe 100 parts Iron, basically inert, to 1 part Uranium , which would generate heat ...


2

It is very unlikely with the normal fission process for most of the elements. They can be divided into 2 groups: slow reaction and fast reactions. The elements with slow reactions do not generate enough energy in a short enough time to be able to heat sufficiently to provide light. The elements with fast reactions would disappear before enough accumulated ...


3

The Sun has lots of features "like the red spot", but they are dissimilar too. Similarities: The Sun's photosphere - the bit we can see - is entirely gaseous; the photosphere rotates differentially with solar latitude; the gas is turbulent. There are features that can be seen quite easily - these are the dark magnetic sunspots, typically of size a few ...


1

Ahh got it. The coordinates are given as latitude, longitude. Fortunately, the longitude hardly matters (depending what precision you need), but the latitude does. A star at the zenith will have a declination equal to the latitude on Earth. As far as Right Ascension goes, well around the March equinox then stars with 12 hours in Right Ascension will reach ...


0

There's a couple of celestial coordinate systems you should learn about, which are the most commonly used: Horizontal coordinate system, observer-centered. Equatorial coordinate system, earth-centered. Depending on the system you use, the coordinates of your celestial object will be represented in one way or the other and should be read accordingly.



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