# Tag Info

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 $... 16 It is true that a surprisingly large number of stars are smaller (and thus less massive) than the Sun. However, the stars that are bigger than the Sun are often much bigger. Look at this chart: Image courtesy of Wikipedia user Jcpag2012 under the Creative Commons Attribution-Share Alike 3.0 Unported license. Notice how small the Sun is compared to some of ... 9 The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. This Physics SE answer, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this "by eye" to see that the CMB contribution is the largest followed closely by an optical/IR bump. A ... 8 I am not an expert, but it seems that the fundamental plane is a relation among characteristic quantities of the galaxy, showing a correlation that is the analogous of the Tully-Fisher relation for spiral galaxies. You can start from the virial theorem:$\frac{M}{R}\sim v^2$($M$mass contained in the radius$R$,$v$velocity dispersion of the stars, also ... 7 The original Hertzsprung-Russell diagrams constructed by Henry Russell and Eijnar Hertzsprung consisted of absolute magnitude on the y-axis and a spectral type or an indicator of spectral type on the x-axis. Below you can see an original HR diagram produced by Russell in 1913. When the diagrams were constructed, it was not at all clear what the sequence of ... 7 Originally, what was plotted was luminosity against colour, and by colour I mean the wavelength of the peak intensity. Short wavelengths on the left, and long on the right, as you would expect. Now since stars emit (nearly) black body radiation, there is a close relationship between colour and temperature. I suspect that the reason that the x axis isn't ... 6 Here are some values found by taking the hue from images, and adjusting the brightness to fit the albedo: Mercury #1a1a1a Yes it is really that dark Venus #e6e6e6 or perhaps a bit darker Earth tricky as it is a mix of colors, and changes over the year seems to average out as about #2f6a69 Mars #993d00 Jupiter #b07f35 Saturn #b08f36 Uranus #5580aa ... 6 I think the argument is the following. The central temperature can be estimated from a form of the virial theorem. At least dimensionally speaking, total thermal energy$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 ... 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 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 ... 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

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 ...

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 $\frac{\mathrm{HJD}-T_0}{P}-\mathrm{INT}\left(\frac{\mathrm{HJD}-T_0}{... 3 The ugriz values are just measures of the stellar brightness taken in 5 bands from the near ultraviolet u through to the near infrared z. You need the distance to the star to convert the apparent SDSS magnitudes to absolute magnitudes. Then you need to find a relationship (or relationships) between bolometric correction in a particular band and the colour ... 3 If the normalised filter response function is$R_{\nu}$then the measured flux is $$F = \int f_{\nu} R_{\nu}\ d\nu$$ The integration is done over the frequency range of the filter. If you measure a flux through a filter then the process cannot be inverted exactly. However the average flux density can be found by dividing the total flux by the effective ... 3 You can easily convert flux to luminosity and vice versa using the spherical nature of the propagation of light. The equation you need is: $$L=4\pi D^{2}f$$ Where$f$will be your flux, and$D$is the distance to the object, Tau Ceti in this case. From Pijpers (2003) Tau Ceti's luminosity is$0.52\pm0.03L_{\odot}$. You should have everything you need ... 3 Caveat this with the fact that I am not familiar with this code. The input asks you for the flux through the filter, but a filter has a bandpass, so if you have a 10 nm bandpass filter and you measure$1 W/m^2$, then you measured$0.1 W/m^2/nm\$ (sorry, I'm a physicist so I never got used to measuring wavelengths in Hz). As for your second question, the ...

3

This graph is not the light curve of a cepheid. It is used to find the average brightness of the star, and from that, estimate its distance. The rate at which cepheids pulse is related to their average luminosity (and so their average absolute magnitude). This graph relates the period to their average absolute magnitude. To use the graph, observe a cepheid ...

3

I might venture into some math which is slightly above what you need, but I wanted to show you the reason behind the correct equation to use. To start with, the pressure due to radiation, in the most general sense, is given by $$P_{\mathrm{absorption}} = \frac{\langle S\rangle}{c} \cos(\theta)$$ $$P_{\mathrm{reflection}} = 2\frac{\langle S\rangle}{c} \cos^2(... 3 According to CalSky (https://www.calsky.com), there were two Iridium flares visible low in the south-east from Minneapolis at 21:18 (Iridium 60) and 22:28 (Iridium 55) on that evening. Both would have been about 15 degrees elevation and both in the right ballpark for brightness. Iridium flares occur when the rising / setting Sun's rays bounce off the ... 3 It's not hard to calculated brightness to distance, but that doesn't take into account cloud cover, which is important for Titan. Just looking at distance first, Titan (based on Saturn's distance) averages 9.6 AU from the sun, or about 1/91st as bright as the sun appears from Earth, assuming equal atmospheres. 1/91st isn't a bright sunny day, but it's ... 3 The luminosity L isn't estimated, it's observed. You make models of stellar populations with many different parameters and those give you a set of SEDs. For each set of parameters, you will have an SED covering a range of wavelengths (a continuous line). From your observations, you will have an SED of a few points depending on how many wavelengths you ... 2 The equilibrium temperature of the Earth, T_E, scales roughly as L^{1/4}, which is proportional to R^{1/2} T, where L, R and T are the solar luminosity, radius and temperature. The actual approximate relationship is derived by equating the power received by the Earth, which is proportional to the solar luminosity L, with the power radiated by ... 2 The easiest method of photometry is by comparing the measured instrumental intensity of the unknown target with one or more other known targets in the same field and then converting the difference into magnitudes. This is called differential photometry, and is easier because you don't have to worry about things like atmospheric extinction. The more ... 2 Total Luminosity with \alpha=-1:$$ \begin{aligned} L_{tot}&=\int_0^{\infty}L \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\ &=\int_0^{\infty}\Phi_0 e^{-\frac{L}{L^*}}dL \end{aligned} $$Now put L'=L/L^* and you get:$$ L_{tot}=\Phi_0L^* \int_0^{\infty}e^{-L'}dL'=\Phi_0L^* $$Total number:$$ \begin{aligned} N_{tot}&=\int_0^{\infty} \...

2

Sounds like homework (correct me if that assumption is wrong, as you may want a somewhat different answer then.), so here is a hint: Clean up the equation first, to something that looks more like an equation for mass: $$\frac{L}{L_{sun}} = \left(\frac{M}{M_{sun}}\right)^{3.5}$$ $$\left(\frac{L}{L_{sun}}\right)^{\frac{1}{3.5}} = \frac{M}{M_{sun}}$$

Only top voted, non community-wiki answers of a minimum length are eligible