6

Without information about stellar radii, I think it's reasonable to assume $R_A \approx R_B$. Then your equation becomes $$ m_p - m_s = -2.5 \log \frac{F_A}{F_B} $$ and you can compute $k$ and the non-eclipsed total magnitude.


5

I assume that the diagram indicates what the observer sees (if they had a big enough telescope!). i.e. The viewpoint is nearly in the orbital plane but not quite. Why then are the eclipses asymmetric, with the secondary eclipse being shallower than the primary? Well probably because the surface brightness of the two stars is different - i.e. they have ...


5

It depends. For many categories of stars, it is possible to determine the approximate mass of the star from its non-redshifted spectrum. Stars on the main sequence: this table of spectral classifications lists the approximate mass for the various classifications. A more precise mass estimation should be possible with a more precise classification. Sol, for ...


4

Stars are far from perfect blackbodies due to scattering/reflection. This is especially true for hotter stars, because of all the free electrons, but even cooler stars can reflect a significant amount. For example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see that they use a reflection albedo of 0.30 for the K star and 1.00 for the F star, ...


4

It seems to me that you're mostly asking about star naming conventions, which is unfortunately a difficult thing to master because there are many many conventions. What makes this process difficult is that the conventions you seem to be referencing don't come from Hipparcos at all. Likely you're seeing the names of these stars as they're referred to by some ...


4

Gravitational wave detectors have a frequency range that they are sensitive to. In the case of LIGO it is about 10Hz to 1kHz. The lower limit is imposed by seismic noise, the upper limit by "shot noise" (basically not having enough photons to sample the interferometer path difference at high frequencies). LISA is in space and doesn't have the problem with ...


3

Most distance methods are luminosity-based: you measure an object's flux, assume it has a particular luminosity, and then determine the distance from that using the inverse-square law. So the question becomes: how do you determine the luminosity? The eclipsing-binary method uses the idea that a star's luminosity can be defined as $L = 4 \pi f_{s} R^{2}$, ...


3

The AAVSO eclipsing binary section is a good place to start reading. Their how-to articles address eclipsing binary specific issues such as predicting times of minimum. To get a list of observable Algol-type stars, try their observation planner tool with variable type EA. When you've selected a target, you can use their variable star plotter to make finder ...


3

www.skymaponline.net might be the one you want. Main goal when taking images in terms of timing and exposition: 1. Choose time with good sky condition (e.g., clear sky, no wind) 2. Expose long enough for good S/N ratio, but not too long that CCD is saturated. 3. If you want to measure the length of the eclipse, you can follow since it is coming in until ...


3

For this, you should create a lightcurve, a graph of brightness over time, to view the data. For Kepler data, the bjd(date) column is the time in BJD. The dtr_flux stands for detrended flux, meaning that it should contain "cleaned" data. First, I would plot the dtr_flux column over bjd(time) to obtain a lightcurve. Depending on the type of eclipsing binary, ...


2

For all of these direct imaging results, the critical parameter is the contrast as a function of separation. This lets you know how much fainter an object you can see around the much brighter primary object whose light has been suppressed by the coronograph (the black circle in the center of the star). From the change in eclipse timings (Figure 1 in their ...


2

The period (at least in physics) is defined the time an oscilating system needs to get back to its starting point (for a sinus curve its 2*pi). Now when they say an eclipsing binary has a period of T years, it means that its brightness varies as seen from earth when they are eclipsing (thats how we detect eclipsing stars and planets by the way), and that ...


2

Moreno et al. in their paper Eccentric binaries: Tidal flows and periastron events [2011] define the orbital phase as from -0.5 to 0.5, where periastron is at phase 0, and apastron is at -0.5 and 0.5. Orbital phase here is $\phi = \dfrac{t}{P}$, where time $t=0$ at periastron and $P$ is the orbital period. This definition allows them to plot the behavior ...


2

Usually, how is phase=0 defined? It is relative to the line of apses perhaps with phase=0 at periastron? The true anomaly, $\theta$, is the angle between the current location of the orbiting particle and its location in the orbit at which it is closest to the central body (called the periapsis/periastron). The word "phase" is used variously in ...


2

Unless I've done my maths wrong, the period of total eclipse is about 18 seconds. The CHIMERA camera at Mt Palomar, the instrument which followed up the discovery of this system, can take exposures at up to 8 full 1k$\times$1k frames/second and considerably higher if windowed on an object. There is no need for photon-counting equipment for such a slowly ...


2

The key here is mass. In nonbinary systems (or other external factors) it can be a bit tough to determine mass of a star. You just see a point source of light; we can get spectral features which can in some instances be used to give us some idea, but it’s rather tough to otherwise get mass. If you do have mass (which you can get because binary stars are ...


2

The condition that a binary eclipses is (approximately) just a piece of trigonometry, assuming the stars are spherical and in circular orbits (this is quite likely for close binaries). The condition that an eclipse will occur is that the inclination angle $i$ is given by $$ \cos i < \frac{r_1 + r_2}{a}\ , $$ where $r_1$ and $r_2$ are the stellar radii ...


2

No, it is correct. The inclination of an eclipsing binary can be estimated from the light curve (and is likely to be close to 90 degrees in order to produce an eclipse). The radial velocity curves can then give $M_{1,2}\sin(i)$ and hence the component masses.


1

The reason the KEBC doesn't have semi-major axes is because it's based entirely on photometric data with no EB modeling done. Determining the SMA requires spectroscopic data and EB modeling. Furthermore, Kepler mostly looked at solar-type stars, so there are few (if any) massive binaries in the KEBC. Unfortunately, I'm not aware of any catalogs of the ...


1

The answer comes from the number of stars examined by each method. Kepler in the first part of its mission examined 150,000 stars. After the extended mission it has examined 503,506 https://en.wikipedia.org/wiki/Kepler_space_telescope. Kepler stared at one patch of sky at a time measuring the brightness of many 10's of thousands of stars at a time. The new ...


1

This is a long-standing (and complicated!) problem in modeling eclipsing binary light curves. Here’s a full review from 1985 (with later citations here), which suggests (from a very quick skim) that an albedo of about 0.5 has been found for some systems, but the details depend also on temperature and wavelength. For the current state of the art in modeling ...


1

Complementing @KenG's answer, Here's an actual datapoint. The new paper in Nature Polarized reflected light from the Spica binary system (downloadable here) is notable in that the measurement of the polarized component of the reflected light speaks to it being an actual reflection, rather than one star heating the other producing a more brightly radiating ...


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