Dieudonné
Reputation
1,911
Top tag
Next privilege 2,000 Rep.
Access moderator tools
 Jan 24 awarded Yearling Apr 18 answered Astronomical databases for machine learning? Apr 17 comment Variable and Multiple Stars in Hipparcos Orbital solutions are used to calculate the orbit of one star in the binary with respect to the other star. They can be found in the file 'hip_dm_o.dat'. Apr 17 comment Variable and Multiple Stars in Hipparcos Yes on variable stars, see my edit of the answer. Apr 17 revised Variable and Multiple Stars in Hipparcos added 1399 characters in body Apr 17 answered Variable and Multiple Stars in Hipparcos Apr 11 revised Equations for coordinates of solar system objects deleted 1 character in body Apr 11 answered Equations for coordinates of solar system objects Apr 7 answered Orbital Elements Mars near June 2014 Mar 26 answered Do we still use the term “astronomical unit” nowadays? Feb 23 comment Fourier Transform of Galaxy Images I'm not sure what you're question is? Why have you chosen to use fourier transform for segmentation? If I look at the bottom picture, the fourier transform does not seem to be a very good feature. Feb 10 reviewed Approve L4&L5 positions? Feb 9 answered Is it proper to refer to objects beyond Neptune as “Kuiper Belt Objects?” Feb 2 comment Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog The value for the temperature $4000\textrm{K}$ was read from a graph so it is very imprecise. There are some empirical relations that transform colour index to temperature but they depend on the type of star. They can be found in the scientific literature and they will often involve transforming the colour index used in Hipparcos to other passbands (such as those used by SDSS). Feb 2 comment Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog To calculate $L$ you will need the apparent (V) magnitude $m_V$ and the parallax to calculate first the distance (first equation above, and don't forget that Hipparcos gives the parallax in MILLIarcseconds) and then the absolute magnitude $M$ (second equation above). Together with the absolute magnitude of the Sun ($M_{\textrm{sun}}$ = 4.83) you can calculate the Luminosity with $L = 10^{M-M_{\textrm{sun}}}$. This is expressed in Solar Luminosities, i.e. the Sun will have a luminosity of $L=1.0 L_{\textrm{sun}}$. (The SI unit is Watt = Joule per second.) Feb 1 comment Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog First: Yes $-2.5\log(L/L_{\textrm{sun}})$ is $-2.5\times \log()$. As to converting this into computer code, I'm not sure what you do not understand. Maybe the conversion from $M-M_{\textrm{sun}} = \log{\frac{L}{L_{\textrm{sun}}}}$? This would become $\frac{L}{L_{\textrm{sun}}} = 10^{M-M_{\textrm{sun}}}$ so if you have a value for $M-M_{\textrm{sun}}$, say $2.056$, then you can calculate the luminosity (in solar luminosities) with 'pow(10,2.056)'. Feb 1 answered Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog Jan 28 comment Retrograde motion and Kuiper Belt Objects To add to my comment: astronomical twilight is defined when the Sun is 18 degree below the horizon. If the Sun is lower than that, observations should be possible from a site that is dark enough. When the KBO a starts or ends its prograde motion the angle is 90 degrees. That should be enough. Jan 28 comment Retrograde motion and Kuiper Belt Objects Good point, the prograde motion of an object infinitely far away will take 6 months. And it will occur when the Earth is on the half of its orbit that is opposite to the KBO. That is when the KBO will be most difficult to observe. The worst circumstances will be when the KBO is exactly on the ecliptic. But even then it will still be observable when it just starts the prograde motion or when it is about to end the prograde motion. At least from somewhere on the Earth, for instance, one of the poles. At that point the angle KBO-Earth-Sun will be about 90 degrees. Jan 28 answered Retrograde motion and Kuiper Belt Objects