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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 ...

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'...

8

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 ...

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

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

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.

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 ...

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

The time it takes the sun to cross the horizon is given at: Expression for length of sunrise/sunset as function of latitude and day-of-year How long does a sunrise or sunset take? Re the time it takes to cross the transit line, this doesn't really answer your question, but, to a good approximation, the time varies between 128 seconds at the equinoxes and ...

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 ...

3

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 ...

3

I looked for a source without luck, so I'll just post cause I've read this answer before. The diameter is the solid surface, not the atmosphere for rocky planets. For gas giants it's the atmosphere up to 1 bar. To determine Venus' angular diameter, you'd need to add something for it's atmosphere. It's difficult to say how high it's atmosphere stops ...

3

The power radiated from the planet, assuming it is at the same temperature across its entire surface: $$P_{\mathrm{rad}}=\epsilon\sigma T_{\mathrm{eq}}^4 \cdot 4\pi R_{\mathrm{pl}}^2$$ Where $\epsilon$ is the emissivity (to match the formula in the question, set this to 1), $\sigma$ is the Stefan-Boltzmann constant, $T_{\mathrm{eq}}$ is the equilibrium ...

3

If you know the length of a sideral day and the length of a year, you can determine the length of a solar day. From Wikipedia: After one year, there will have been one more solar day than sideral day. Therefore : solar day = sideral day / (1 - (sideral day/orbital period))

2

I think all this means is that up to some critical rotation rate, the equilibrium shape of a rotating, self-gravitating fluid is an oblate spheroid. That is, this shape defines a unique global minimum of energy. Above this threshold there are two possible equilibrium solutions. Apparently one is an oblate spheroid, whereas the other is a triaxial ellipsoid. ...

2

By latitude and longitude of the sun, I think you mean latitude and longitude of a point on the earth's surface where the rays of the sun fall perpendicular to the surface. In astronavigation, this point is called the Geographical Position (GP) of the sun and terminology to describe this point is slightly different:- Latitude of the GP is called ...

2

The key formula for signal-to-noise calculations in photometry can be written as something like $${\rm SNR} = \frac{S_{\rm star}}{\sqrt{S_{\rm star} + S_{\rm sky} + \sigma_R^{2}}},$$ where $S_{\rm star}$ is the photons counted from your target star, $S_{\rm sky}$ is the amount of unrelated photons in your photometry aperture (due to sky or other stars) and \$\...

2

If you are dealing with planets, or bodies with a pretty low inclination, it might make sense. If the inclination of all the planets was exactly zero the longitude of perihelion would be an actual physical angle in space. You would be able to compare the relative position of perihelion of all the planets. Now actually the planets are slightly inclined but ...

2

I found a table that was based on data from 2003 for Stellar Classification, and I can't get to the given data source ... The website that was the source of the data was last active in 2006. I've replaced the broken URLs with working URLs. I wouldn't recommend relying on data that old, there's likely a newer source. Original (404) Text and URLs: Credit:...

2

Congratulations to @NickBrown for his solution! Based on that equation and some additional references I'll just add a little more. https://apps.dtic.mil/dtic/tr/fulltext/u2/785380.pdf https://www.researchgate.net/publication/268194552_Large_phase_angle_observations_of_GEO_satellites Calculating visual magnitude takes three input parameters how good of a ...

2

Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as 0 (zero)? No, there isn't. Astronomy overlaps with cosmology and there can be some pretty big and small numbers. However most physical parameters also have units and the units chosen affect the numerical values. A great example is the Jansky: The jansky (symbol Jy,...

2

Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is (Douglas Adams) The wikipedia page notes that 'Oumuamua "has circulated the Milky Way several times" so it was unlikely to have formed around any nearby star. While it would take 'Oumuamua 600,000 years to travel here from a local star, the time before it can be ...

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