41

Naked eye stars are not distributed uniformly in the sky. That is because the median naked eye star is at a distance of 440 light years, and this is far enough away that some of the details of Galactic structure start to become apparent. Most importantly, the density of stars increase towards the Galactic midplane and has a scale height of a few hundred ...


21

Having now looked at the paper by Aiola et al. (2020), it emerges that for that map, they filtered the data to exclude low frequency multipoles with $|l|<150$, corresponding to about 1 degree. This filtering was done to all the maps in the paper and will be responsible for the dramatic "hole" in your Fourier transform. As for the high frequency ...


19

For that specific E-mode map we have applied a Wiener filter to highlight the high SN modes (those "rings"). I also further apply the following filter: $((1 + (kx/5)^{-4})^{-1}) * ((1 + (k/150)^{-4})^{-1})$. This second filter gives the "hole" and a "thin" vertical line in your 2D PS. The image above is just for PR purposes. In ...


13

Saying why gets tricky beyond "because of Jupiter", but to clarify on the quote, the statement "Earth's eccentricity follows a 100,000 year cycle" is loosely true but it's also an oversimplification. From Wikipedia. The Earth's orbit approximates an ellipse. Eccentricity measures the departure of this ellipse from circularity. The shape of the Earth'...


9

The formula is Kepler's equation, but to understand it you need to know three values: $M$ is the "Mean Anomaly". It increases linearly from 0 to 360 over the period of one orbit, measured from periapse to periapse. So if a planet has a period of 100 days, then the mean anomaly at day 0 is 0, at day 50 it 180degrees, at day 25 it is 90 degrees. This is only ...


8

Alright I finally finished this program so I could take a look at each tier individually and see for myself. First of all, the projection type does indeed matter, so I will explain it here. It needs to be an equal-area projection. The whole point of the question was about a uniform distribution of stars over the surface of a sphere. In other words, each ...


8

Ok, here's my take on calculating the color of a blackbody, or any spectrum in fact: Disclaimer: I'm not a color theorist, and there may be more accurate methods. But the result, shown in the bottom, looks about right. Spectrum First note that since color is a function of the relative intensity in various wavelength bands, it doesn't matter whether we ...


7

The probabilites are unknown at the moment (March 2014), since there is only one known planet (Earth) harboring life. This doesn't allow any meaningful probability estimates for the occurence of life, based on empirical data. The overall formation of life is too complex to allow simulations based on current technology. Although some intermediate steps can ...


6

Random points on the surface of a sphere can be generated by allowing the azimuthal angle $\phi$ to take a uniformly distributed random value between 0 and $2\pi$. To convert this to RA in degrees you multiply by $180/\pi$. To convert to hours, minutes and seconds you divide the $\phi$ in degrees by 15, which gives the hours, divide the remainder by 60 which ...


5

Astronomer is moving towards the big(ger) data era because of many sky survey technologies. The coming ones include, e.g., LSST, JWST, and WFIRST. By the meaning of survey, it normally means observing the whole sky over a few days, and keep repeating over and over. Also, since most of the surveys are imaging technologies, every pixel in an image is ...


5

I agree that the common explanation of the longitude of periapsis, as the question gave it, appears at first sight to make no sense -- because it constructs a sum of two angles in two different planes, and that seems at first sight to lack geometrical meaning. In fact the fault, if any, is in the explanation rather than in the geometry. It is not so ...


5

The thermal eccentricity distribution was first calculated by Jeans 1919. The probability density function is indeed $$f(e) = 2e.$$ See this blog post for a nice derivation of this beautiful result, which is independent of the "temperature". The derivation relies on a small 'swindle' by assuming a population of only binaries and not any single or trinary ...


5

I think you can understand that factor as follows: When photons lose energy, they're spread out over a larger wavelength range. Since there is a fixed number of photons, that means that the number of photons per observed wavelength bin decreases by a factor of (1+z), and hence the flux density, which is really what $f_\lambda$ is, decreases by this factor.


5

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

If you plot the log of the period against the log of the semi-major axis then it is obvious that $P^2 \propto a^3$. Any other power law relationship simply wouldn't fit. The following passage (from https://www.mathpages.com/rr/s8-01/8-01.htm ) seems relevant: Is it just coincidental that John Napier's "Mirifici Logarithmorum Canonis Descripto" (...


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

I want to say that's a Mollweide projection. I know that they're pretty common in astronomy; many images of the CMB use them. I was actually working with one recently. Given latitude $\varphi$ and longitude $\lambda$, the $x$ and $y$ coordinates of an object are $$x=R\frac{2\sqrt{2}}{\pi}(\lambda-\lambda_0)\cos\theta,\quad y=R\sqrt{2}\sin\theta$$ for ...


4

The article references a paper by Huo et al. (2004; that's the reference number "33" in your quoted text). Huo et al. merely list a redshift of "0.0103" for Arp 299 in their Table 1 (without any apparent sources); combined with the fact that the authors of the Science paper say what Hubble constant they assumed (73 km/s/Mpc), this tells us that the distance ...


4

Heres more python than you can shake a telescope at. I just used @RobJeffries' algorithm. This is just a python script, the real answer to the question is @RobJeffries' answer and I've just scripted it. The mathematics behind generating statistically uniform distributions is explained very nicely there as well. Python is below the plots. You can see on an X-...


4

If you like Python then see Skyfield; Observing from a Moon location which I believe will do exactly what you need! Also see for example this answer to Ephemeris for lunar body-centered body-fixed coordinates? written by Skyfield's author. For example, you posted your question at 2020-04-20 15:21:36 UTC. Let's say your rover has just driven up to Surveyor ...


4

A full-text search using the astronomical-literature collection on the Astrophysics Data System (more complete and more astronomy-specific than Google Scholar) yields 10,845 hits for "Lagrange point" and 2,302 for "libration point" (not sure how long that URL will be valid, but it's easy enough to redo the search). So it's clear the former is more common ...


4

If I understand correctly, you are trying to create a realistic looking view of the stars rendered (in the game) on a dome surrounding the game play area. So you don't need to fly through the stars or interact with them in a 3D environment, is that right? If so, it isn't the star size you want to reproduce - it is the brightness. We can't see the width of ...


4

Great project! and welcome to Stack Exchange. I'll post a short answer but I think someone can add a more detailed, thorough and insightful answer. I think that website is not well suited, so I'll answer based on you switching to Horizons. If you like Python then it's more fun to use Skyfield. If you want apply an equation based on a Kepler orbit model, you'...


3

This is for satellites with unknown size and orientation but known standard magnitude (Standard magnitude can be found on the satellite info page of heavens above, the number is called intrinsic magnitude) The proper formula is double distanceToSatellite = 485; //This is in KM double phaseAngleDegrees = 113.1; //Angle from sun->...


3

I'm an astronomer and I get lots of job offers to retrain as a data scientist, but it might be more tricky to go the other way. Astronomy is definitely a field in which 'big data' is important, and the analysis and visualisation techniques we use every day are probably decades behind what is taught to computer scientists. However, most astronomy software is ...


3

It's the good old physical distance — also called proper distance — i.e. what you'd measure if you stopped the expansion of the Universe and patiently laid out meter sticks from us to Arp 299. It's also the comoving distance — i.e. the distance measured in coordinates that expand with the Universe and hence don't change with time — since that distance is ...


3

To the precision that you need, you can calculate the azimuth A and altitude h of the Sun from the following equations: $$ \tan(A) = \frac{\sin(H)}{\cos(H)\sin(\phi)-\tan(\delta)\cos(\phi)}\\ \sin(h)=\sin(\phi)\sin(\delta)+\cos(\phi)\cos(\delta)\cos(H) $$ where A is the azimuth measured from due south (positive towards the west) H is the hour angle which is ...


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