29

Tl;dr, From personal experience, it’s not worth it. It’s a lot easier to throw away bad data than to try to calculate when stop. There’s not a lot of ambiguity as to whether it messes up the data. As of right now, it’s not that big a problem either. Sure, some are particularly unlucky, as seen with a Magellanic cloud observation in early 2020. The concern is ...


19

It is not possible to know. The speed of light is the speed of information. The information "the star has exploded" cannot travel faster than the speed of light, so there is no way to know that a star has gone supernova before that information reaches us. Usually the first particles to reach us from a supernova are actually neutrinos, which can ...


7

The E-ELT is being built at Cerro Armazones, about 20 km from Paranal where the ESO VLT is sited. Unsurprisingly, the observatory that is already there is known as the the Cerro Armazones Observatory. This observatory is a collaboration between the German University of Bochum and the Chilean Universidad Catolica del Norte. It was inaugurated on December 2, ...


6

The magnitudes don't sum and subtract that way: consider two 0.00 magnitude stars closely together. Is their combined magnitude 0.00? Actually, it is -0.75. This paper might help you with the derivation of the formula for addition of magnitudes, but you are interested in subtraction. You just need to rearrange the formula in the paper to $$m_a=-2.5\log(10^{-...


6

The brackets refer to the average, so $\left< x^2 \right>^{1/2}$ is the root-mean-square (RMS) of $x$. That is the square root of the mean (or average) of the square of multiple $x$s. The RMS average is useful when a quantity can be either negative or positive. For instance, a sine or cosine wave has an average of zero over one cycle, but its RMS ...


5

Optical observatories -- or commissions that allocate time on a national or international basis to one or more telescopes -- generally accept proposals twice a year, during a one-month period with a deadline in the middle or end of March or in the middle or end of August (in both cases, these are for observations that will take place starting about six ...


4

How, precisely, do radio astronomers detect (and record) the phases of waves for interferometry? The popular press always talks about directly 'interfering' two waves as they come in, but can they tell the exact phase of a single wave? tl;dr: It's a good question. There is no such thing as "the exact phase of a single wave", it's only the phase ...


4

The Wind, STEREO, Parker Solar Probe, and Solar Orbiter spacecraft all carry radio instruments that observe radio frequency emissions from a few kHz up to ~10-20 MHz. There's an entire section in Wilson et al. [2021] devoted to discussing the novel discoveries that resulted from the launch of Wind in radio astronomy, specifically solar radio observations. ...


4

It's been 3 years and I wondered if you found further answers on this. It's sad the answers here don't relate to your question. I too wanted to find this out. On the non overlapping, the 12 cycles are deducted from saros 120, so 83-12 = 71 eclipses or 70 cycles of 1260 years. thus these same 12 eclipses will not be deducted within saros 80 since it's "...


3

In addition to @User123 answer, and assuming you are subtracting luminosities and not magnitudes, there is usually one more obstacle: the catastrophic cancelation. In short, if the difference between the total and the source B luminosity is near the accuracy of either of them, you will get no meaningful estimate for the source A. In practical terms, I wonder ...


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

Radio waves (approximately $\lambda=1\rm\,m$[1]) are very weak, about $E=\frac{hc}{\lambda}=\frac{6.63\cdot10^{-34}\cdot 3\cdot 10^8}{1}\rm\,J\approx2\cdot 10^{-25}\rm\,J$ every photon. An $1\rm\,mg=1\cdot10^{-6}\rm\,kg$[2] snowflake falling at $0.7\rm\,\frac{m}{s}$[3] has $$K=\frac{mv^2}{2}=\frac{1\cdot10^{-6}\cdot 0.7^2}{2}\rm\,J\approx2.5\cdot10^{-7}\rm\,...


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