17

It is true that a surprisingly large number of stars are smaller (and thus less massive) than the Sun. However, the stars that are bigger than the Sun are often much bigger. Look at this chart: Image courtesy of Wikipedia user Jcpag2012 under the Creative Commons Attribution-Share Alike 3.0 Unported license. Notice how small the Sun is compared to some of ...


17

The easiest way to determine the magnitude of a given star is probably to use the Pogson relation. The idea is to determine the magnitude of a star knowing the magnitude of a reference star; it is thus quite easy, using a well-known reference as Vega or Sirius. The Pogson relation is given by: $$m_1-m_2=-2.5\ log\ \left({\frac{E_1}{E_2}}\right)$$ where $...


13

Nearly all the helium in the photosphere of the sun comes from the helium in the interstellar gas that collapsed to form the sun. That helium was produced shortly after the Big Bang (in about the first 20 minutes) while the universe was hot and dense enough for hydrogen to fuse to helium. That produces a universe in which ordinary matter is about 25% helium ...


11

You are right that most of the dots of light that you see in the at image of Andromeda are stars in our galaxy that happen to be on the same alignment. However we can resolve stars in the Andromeda galaxy. (Here I am understanding "resolve" means "to see as individuals and not just a haze") https://www.spacetelescope.org/images/heic1502a/zoomable/ Even ...


10

There is some confusion here about the word "resolve". In astronomy, to resolve an object means either to establish details of its structure and physical extent, rather than seeing it as a point source; or it means to separate a single entity into its component parts. The former depends on the size of the object in question and how far away it is. The ...


9

The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. This Physics SE answer, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this "by eye" to see that the CMB contribution is the largest followed closely by an optical/IR bump. A ...


8

I am not an expert, but it seems that the fundamental plane is a relation among characteristic quantities of the galaxy, showing a correlation that is the analogous of the Tully-Fisher relation for spiral galaxies. You can start from the virial theorem: $\frac{M}{R}\sim v^2$ ($M$ mass contained in the radius $R$, $v$ velocity dispersion of the stars, also ...


8

Yes, there are monotonic relationships between mass and luminosity and radius on the "deuterium burning main sequence". Deuterium "burning" begins when the core temperature exceeds just over $10^6$ K. This happens very early in the life of a contracting protostar and, because it is fully convective and thoroughly mixed, all the D is "burned" in less than a ...


7

The original Hertzsprung-Russell diagrams constructed by Henry Russell and Eijnar Hertzsprung consisted of absolute magnitude on the y-axis and a spectral type or an indicator of spectral type on the x-axis. Below you can see an original HR diagram produced by Russell in 1913. When the diagrams were constructed, it was not at all clear what the sequence of ...


7

Originally, what was plotted was luminosity against colour, and by colour I mean the wavelength of the peak intensity. Short wavelengths on the left, and long on the right, as you would expect. Now since stars emit (nearly) black body radiation, there is a close relationship between colour and temperature. I suspect that the reason that the x axis isn't ...


6

Here are some values found by taking the hue from images, and adjusting the brightness to fit the albedo: Mercury #1a1a1a Yes it is really that dark Venus #e6e6e6 or perhaps a bit darker Earth tricky as it is a mix of colors, and changes over the year seems to average out as about #2f6a69 Mars #993d00 Jupiter #b07f35 Saturn #b08f36 Uranus #5580aa ...


6

I think the argument is the following. The central temperature can be estimated from a form of the virial theorem. At least dimensionally speaking, total thermal energy $MkT/\mu$ (where $\mu$ is the average mass per particle) is proportional to gravitational potential energy $GM^2/R$. So $T \propto GM/R$. But for a given mass, standard polytropic theory ...


6

The question comes near the centennial of a famous debate about the nature of so-called spiral nebulae. In 1920, Shapley argued that that they were clouds within our own galaxy, and Curtis argued that they were distant galaxies in their own right. Edwin Hubble's observations a few years later settled the issue. Using the 2.5 m reflector at Mt. Wilson to ...


6

I have to imagine that the glow from the galaxy comes from the scattering and reflecting of all the collective starlight inside the galaxy off the dust and gas and whatnot contained therein, and that the actual individual stars would be kind of like individual droplets of water in a cloud. This is not correct. The glow from from the galaxy is overwhelmingly ...


6

A confirmation of the explanation is given slightly further on where it talks about how the uncertainty is derived: In equation (4) we use the 0.6 dex scatter of $\log L_{\rm UV}$ around the $L_{\rm UV}$ versus $L_{\rm [O\ III]}$ relation, as an estimate for the uncertainty in deriving $L_{\rm bol}$ from $L_{\rm [O\ III]}$. The term "dex" ...


5

Here is how you do it "properly" for the Hipparcos data. As Warrick correctly points out, what you have done in your question is massively biased towards giant and supergiant stars, which actually form the very small minority of stars. You must form a volume limited sample. To do this, sort the stars by distance (1/parallax) and choose a cut off point. Your ...


5

The Eddington limit represents the maximum luminosity that can be achieved by a body (such as the star) when there is hydrostatic equilibrium (http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium). For luminosity greater than Eddington limit, the radiative force of the luminosity on matter exceeds the gravitational force on the matter. If the ...


5

The solution to your question is surprisingly simple, I think: A quasar that puts out energy around Eddington luminosity or higher, must accrete at a certain rate, corresponding to the energy output. When consulting a lecture and a random paper from the archive on this topic, it is evident that those accretion rates correspond to 1-10 solar masses per ...


5

Small quibble to the (rightfully) accepted answer by James K that was too long for a comment: To be fair, $x=24^{+1}_{-3}$ doesn't mean that $21 \le x \le 25$, but that with a particular amount of certainty (usually 68%), $21 \le x \le 25$. Correspondingly, $x=24^{\times 2}_{\div3}$ would mean that, with some certainty, $8 \le x \le 48$. Symmetric vs. ...


4

All of these effects are related to the 11 year solar cycle. And while we know there are times when the sun is "active" versus "inactive" we aren't necessarily predicting exactly where sunspots or solar flares will occur, but how many we see in total.


4

The creator seems to be referring to flare stars. Flares may be magnetic in origin, like various manifestations of the Sun's magnetic field. These flares can be quit luminous across the electromagnetic spectrum, including in the x-ray wavelengths. Some red dwarfs are flare stars, though flare activity may be largely restricted to a small part of a red dwarf'...


4

There are other techniques to construct what you're interested in than making a volume limited sample. What you're trying to construct is called a "luminosity function," it's the distribution in luminosity normalized so that the area under the curve integrates to the volume density of stars. Constructing a volume limited sample is, perhaps the simplest ...


4

Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area). Simply setting the flux to 0 outside $\lambda_1$ and $...


4

It's not hard to calculated brightness to distance, but that doesn't take into account cloud cover, which is important for Titan. Just looking at distance first, Titan (based on Saturn's distance) averages 9.6 AU from the sun, or about 1/91st as bright as the sun appears from Earth, assuming equal atmospheres. 1/91st isn't a bright sunny day, but it's ...


4

Surely this is exactly what the V-band bolometric correction assesses. The numerically larger the bolometric correction, the more of a star's flux (as a fraction) is in the V-band. The Sun's bolometric correction is about -0.06 to -0.11 mag, depending on which sources you look at. Cool stars and hot stars have numerically smaller BCs and emit a bigger ...


4

To calculate a "luminosity light curve" from a time series of V-band photoetry, you need two things. You need to know the distance. The distance to Betelgeuse is uncertain and that means the absolute value of the luminosity you get will also be shifted systematically up or down by whatever distance you adopt. The V-band only contains a small ...


4

A notation like $x=24^{+1}_{-3}$ is quite common, it means $24-3<x<24+1$ with a best estimate of 24, and is a way of indicating uncertainty. The example you give is less common, by analogy $x=24^{\times 2}_{\div3}$ means $ 24\div3 < x < 24\times 2$ ie $x$ is between 8 and 48, with a best estimate of 24


4

I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet. I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the ...


3

INT refers to the floor function. So $\mathrm{INT}(x)$ is the largest integer not exceeding $x$. In this case, the function is used to get the fractional part of $\frac{\mathrm{HJD}-T_0}{P}$, i.e. the part after the decimal point. So if for example $\frac{\mathrm{HJD}-T_0}{P}=3.4$, then $\frac{\mathrm{HJD}-T_0}{P}-\mathrm{INT}\left(\frac{\mathrm{HJD}-T_0}{...


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