43

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 ...


20

There are several factors determining the inner limit to moons. Perhaps the simplest is that it needs to stay inside the Hill sphere, the region around the planet where the planet's gravity dominates over the sun's. If the planet's orbit has semi-major axis $a$ and eccentricity $e$ the farthest the moon can orbit is $$r_H \approx a(1-e)\sqrt[3]{\frac{m}{3M}}$...


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 ...


16

Both ellipticity $f$ (also called flattening) and eccentricity $e$ are measures of how elongated an ellipse is, based on the semi-major axis $a$ and the semi-minor axis $b$ (figure from wikipedia). They are defined respectively as $$f=\frac{a-b}{a}$$ and $$e=\sqrt{1-\frac{b^2}{a^2}}$$ For a circle, $a=b$, which implies that $f=e=0$. In modern orbital ...


14

The relationship between $a$, $b$ and the eccentricity of an orbit $e$ is $$ a = b\left(\frac{1+e}{1-e}\right)\ .$$ It is clear therefore from a mathematical point of view that you cannot get a similar constant relationship by replacing $a^3$ by $b^3$ in Kepler's third law. You would have to use an expression involving both $b$ and $e$. Another way of saying ...


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'...


13

I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum. This is a relatively small increase in both quantities and it would send the Earth into a slightly more eccentric orbit than it has now. It would not come close to Mars. Using the vis-viva ...


12

Scale invariance and self-similarity Power laws basically mean that there is no preferred scale, i.e. that a physical property is scale invariant. Any deviation from a power law means that the Universe somehow thinks that the scale where it breaks down has some special significance. In other words, a power law describes self-similarity$^\dagger$. You can see ...


10

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 ...


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 ...


9

Consider this gif from wikipedia. All the orbits in the animation have the same orbital period $T$ and the same semi-major axis $a$, but different semi-minor axes $b_1,b_2...b_5$. This shows that the orbital period is independent of the semi-minor axis. As an analogy, imagine twin figure skaters spinning with the same angular momentum. The red figure ...


9

Ellipses have a "long radius" called the "semi-major-axis" which is the length from the centre to the ellipse measured along the long axis. And a "semi-minor-axis" which is measured along the short axis. Call the semi-major-axis "a" and the semi-minor-axis "b". Ellipses also have foci: which is where the ...


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 ...


8

I have to admit that power-laws (in general) used to be my shtick so I am happy to shed some light on their general importance in physics which obviously also hold for astronomy. The main idea of a power law is nicely written in Wikipedia, but the essential part is the I highlighted in the following quote: [A Powerlaw is] a functional relationship between ...


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 ...


7

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 ...


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.


6

Note that the Sun here is point-like and there is no refraction. What have I done wrong? Your supposition that the altitude of the Sun is directly related to the length of the day is wrong. Proof with counterexample: observe the altitude of the Sun on the equator and in the Moscow on the equinox; but still, the lengths are the same. And what is the correct ...


6

Using calculations from here: https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html New semi-major axis $$ a = \frac{150000000000 \cdot 0.0000000000667 \cdot 2\cdot 10^{30}}{2 \cdot 0.0000000000667 \cdot 2 \cdot 10^{30} - 150000000000 \cdot 30000 \cdot 30000)}\\ = 151.821\,\mathrm{million\,km} $$ New orbital period $$ p = \frac{\...


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

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-...


5

I can show you an ESA Series exercise I did a few months ago for an astronomy class. Given the light curve of 12 cepheid variable stars in the galaxy M100 (which are very nice standard candles to measure large distances): We can find the distance between us and the M100, a spiral galaxy in the Virgo cluster, using the apparent magnitude m and the absolute ...


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

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. ...


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