(no subject)

[den]

shigeru, Can you answer this question? If you can spare some time from looking through telescopes at planets around other stars, that is.

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[shigeru.livejournal.com]

I did a silly thing, and just tried to find the angle to any old point in the galactic plane, which is why I got such a wrong answer.

Luckily for me, I have my officemate's textbooks now, so according to "Introductory Astronomy & Astrophysics" by Zeilik, Gregory, and Smith (Sadly, none of my graduate-level texts had this little piece of information), the galactic equator is tilted by 63.5 degrees to the celestial equation, which is tilted by 23.5 degrees to the ecliptic, and they claim "almost perpendicular" for the galactic plane and the ecliptic.

I'm not satisfied, so on I go. They give the coordinates of the galactic north pole as 12 49 00 +27.4, and I know that on June 21 at midnight, LST is about 18 hours, so the north pole of the ecliptic is 18 00 00 +66.5. These define two vectors in celestial coordinates, which I can put into Cartesian coordinates, defining the z-axis along the north celestial pole, x-y plane along the celestial equator, with x=0 at RA=0. Now, the one trick is I have to switch angles, so that theta (zenith angle) = 90 - dec, and phi (azimuthal angle) = ra * 15

Then, a quick transformation ( x=sin(theta) cos(phi) , y=sin(theta) sin(phi) , z=cos(theta) ) gives us the two vectors (0,-0.399,0.917) and (-0.868,-0.188,0.460). And, the dot product of any two vectors is just the angle between them, so (0)*(-0.868) + (-0.399)*(-0.188) + (0.917)*(0.460) = cos(epsilon) = 0.497, or epsilon = 60 degrees, which seems about right, given the figure.

Information I found elsewhere suggests that if you actually look at the direction of rotation of the solar system and the ecliptic, the angle is around 120 degrees, again consistent with an answer of 60.

There, 60 degrees inclination. I'm going back to reducing spectra, now.