# Tag Info

3

The helium core becomes significantly denser and hotter after a star has left the main sequence. Providing the overall mass of the star is greater than about 50% that of the Sun, the helium core will become hot enough to begin fusion whilst the star is a red giant. From there, the star will pass through phases of burning helium in its core, then hydrogen and ...

2

Another link worth sharing comes from ESO and is called The Large Magellanic Cloud before and after SN1987A. The link mainly shows the following

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Yes, it is clearly visible in photographs of the LMC from the 1980s. It is shown in this comparison from the Australian Astronomical Observatory We know it was this star in the most simple possible way: The supernova occurred exactly at the same location as this star, and when we look now, that star has gone (replaced by a little bipolar supernova remnant)

2

This is a highly uncertain prospect, since the physics of the system itself relies on processes that are themselves uncertain to varying degrees, e.g., stellar nuclear evolution of very high-mass stars, accretion of gas onto compact objects, etc... A relevant paper, cited by the wiki article for quasistars, is the work by Begelman $\it{et~al.}$ (2007). They ...

0

There is a rich variety of types of stellar evolution models. The modern model of a star is the spherically symmetric quasi-static gas assumed to be in local thermodynamic equilibrium (Chandrasekhar, 1967). There are four equations for how mass, pressure, temperature, and luminosity vary with radius as the star's thermal and radiative pressures support its ...

1

This depends on the initial mass of the star, but generally the gradient of the core is not sufficiently strong on the main sequence to justify a well defined core-envelope boundary, which is why if a star on the main sequence experiences Roche-lobe overflow the entire star is essentially destroyed, e.g., in a binary system. The core of a main sequence star ...

4

The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star. Therefore, the second relation for the two fluxes is about the apparent magnitudes (...

2

There is no simple answer to these questions - although I could be brief and say (i) No it doesn't and (ii) no they won't. If you make a simple two component atmosphere then the observed spectrum will be the flux-weighted combination of two spectra. $$S_{\rm obs} = \frac{A_1 T_1^4 S_1 + A_2 T_2^4 S_2}{A_1 T_1^4 + A_2 T_2^4}\ ,$$ where $A_1, T_1, S_1$ are ...

20

This is because your image is not in focus. So you’re seeing the shadow of the secondary mirror. You should have a knob near the eyepiece, that you can turn to adjust focus. You need to turn it, one way or another, until the image is as small as possible and as bright as possible. Good luck!

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This is a heavily de-focused image (possibly taken through cloud looking at the orangeish "glow" to the right). The dark center of the "donut" is the shadow of the secondary mirror centered within the image/pupil coming from the primary mirror. The focus needs to be adjusted until the "donut" shrinks down to sharper points of ...

0

The night sky is the hemisphere of the sky which is pointed away from the Sun. I note that the Eath orbits the Sun. It takes one year to orbit around the sun. So today one hemisphere of the sky will pointed toward the Sun, and it will be imposible to see stars in the daylight in that hemisphere, while the opposite hemisphere of the sky will be pointed away ...

1

One way of thinking about this is to think of what, for instance, a digital sensor would see (does see!) when looking at a galaxy. Even if there is no dust to scatter light (which often there is), if there is, in most cases, one or more (generally 'more') stars in the bit of the image covered by a pixel in the sensor, then that pixel will be at least partly ...

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The way the star chart is drawn, the points labeled 1 and 3 are on the celestial equator, the 0 degree declination reference. The tip-off is that a line between them passes through Orion, the top-right star in Orion's belt, which is very close to the celestial equator. (Orion is the constellation labeled B.) North is, therefore, the point labeled 2. The ...

0

First determine the North (South) direction by drawing a line from the center through Polaris (Sigma Octantis). The other principal directions follow from that. You need to obtain the coordinates of a well-recognized star in equatorial coordinates (right ascension and declination), not too near the celestial pole. Given your chart, it's no problem as it ...

4

You are mostly empty space. Every atom in your body is very tiny compared to the relatively vast spaces between it and its nearest neighbors. And the same goes for every "solid" object you ever saw. But when you look at yourself in a mirror, the number of atoms you see is so incredibly vast that you don't see the spaces between them but instead ...

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The reddening line is the locus that a star would move along in a colour-colour diagram as the interstellar reddening towards it increases. i.e. It is an equation on the colour-colour plot of the form (for this colour-colour plot) $$E(U-B) = f(E(B-V)) + c\ ,$$ where $E(U-B)$ is the colour excess in $U-B$ due to interstellar reddening, $E(B-V)$ is the colour ...

-1

Don't trust your eyes when they tell you something is solid. Please remember that the stars are suns. So, if our galaxy were really solid, you'd directly look into a star's surface in every spot of the Milky Way, and that would mean a band in the sky where every spot shines as brightly as the sun. Then the Milky Way would be 100 times brighter than our sun's ...

1

The actual transparency of something depends on whether a light ray will pass through the object or hit something, becoming absorbed or scattered. The total cross-sectional area of the stars of a galaxy is microscopic. This is why one can see the background galaxy through the foreground galaxy in NGC3314: However, dust can be actually opaque in this sense. ...

-1

First, you are correct that galaxies are mostly empty space, at least in terms of stars. The Milky Way may have as many as 400 billion stars while Andromeda has roughly two and a half times that number. For the Milky Way that comes out to roughly one star every four light years, on average. Viewed from a distance, however, those stars are extremely bright ...

2

No mass blob of stellar mass is transparent at any wavelength of interest. Opacities $\kappa_{\nu}$(inverse transparency) as function of wavelength becomes really high and broad band at pressures above > 0.1 bars, for all wavelengths. This leads to the optical depths $\tau_{\nu}$ being enormous and as transmission is $T=1-\exp(-\tau)$, you won't be able ...

3

Color is a difference, not an absolute value Being "blue" doesn't (necessarily) mean that a light source has a large flux in the $B$ band. A color is not an absolute value; it is the ratio between two fluxes or, equivalently, the difference between two magnitudes. Being blue means "More flux in some short wavelength band (e.g. $B$) than in a ...

2

Yes it would. It is that way because the effective temperature is defined to be $(L/4\pi \sigma R^2)^{0.25}$. The radius of a neutron star is about 10 km $(1.4\times 10^{-5}R_\odot)$. They are born with surface temperatures of around $10^8$ K. The coldest white dwarfs have effective temperatures of about 3000 K. The luminosity ratio is  \frac{L_{\rm NS}}{...

3

I don't think this is feasible, because of two competing effects. To have significant time dilation, the velocity needs to be large. The velocity in a Keplerian orbit of radius $r$ scales as $r^{-1/2}$, so the star would need to be close to the black hole. Tidal forces from the black hole (the difference in gravitational force on the near side vs. far side ...

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We want it so that $1 \text{ year at the speed of x% of c} = \dfrac{10^{10}}{80} \text{ stationary years}$. The Lorentz factor (in terms of c) is $\gamma=\dfrac{1}{\sqrt{1-v^2}}$. Solving for $v$ with $\gamma = \dfrac{10^{10}}{80},$ we get $v=0.999999999999999968c = 299792457.999... \text{m}\cdot\text{s}^{-1}$. This is definitely impossible, as the energy ...

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