(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.

Comments

[makovette.livejournal.com]

According to: http://www.earthsky.com/scienceqs/browsefaq.php?f=170

It's about 5.5*.

IIRC we (the solar system) are on the described as orbiting Sag. A* on "north" side of the galactic plane, but that orbit is varies over time due to gravitational encounters with other stars &ct, so we wobble somewhat randomly more than we glide as gravity dictates.

CYa!
Mako

[shigeru.livejournal.com]

Well, the first bit's true, it was pointed out to me one day when we were up at the telescope. About the second bit...hmmm...luckily for me, I have a coordinate system transformation program I wrote a year ago...Ok, galactic center (l=0, b=0) has an RA of 17 45 37, and a dec of -28 56 10

At a latitude of 30, say, stuff at a declination of 30 passes right overhead. And it looks like an RA of 18 will be overhead in Arizona in a couple hours...ie, around midnight whereever you are. So you can probably see the galactic center up to an latitude of around +30 (Tucson: +32, so my story checks out), maybe up to +40 or +45 if you have really clear horizons.

Second bit is trickier...there's a 23 degree angle between dec=0 and the ecliptic, and about 30 degrees between dec=0 and galactic plane. Now, for a bit of basic astronomy...during the summer, the earth's north axis points away from the sun, and around that time, the galactic center is about 3 hours off from midnight, or 45 degrees off the line between the sun and the earth, in the plane of the ecliptic... call that angle (phi), and that's a tricky problem, so I'm going to take the easy way out. If phi was 0, we'd just add up the two angles and get 53. If phi was 180, the two would subtract, and you'd get 7 degrees. The very wrong way to solve the problem, then, is to take the value at phi=0 and subtract 1/4 the difference between these two values, which is precisely what I"ll do.

53 - .25*(53-7) = 41.5 So about 40 degrees. I'll check a textbook or something tomorrow.

So that's significantly off from makovette's answer.

What do you want from me? I'm tired and a pretty lousy astronomer.

[shigeru.livejournal.com]

Heh, I'm stupid. The north axis points toward the sun during summer. So my guess is now 7+11.5 = 18.5 ... still wrong, but closer this time. That's an error of 13 degrees, or 13/180 = 7%...I'm satisfied, so I'm going to watch anime.

[level-head.livejournal.com]

That 5.5% is a different number, I think. The cyclical serpentine coil up and down through the galactic plane has been statisticalyl linked to times of larger asteroid bombardment -- it's about 26 million years.

The Sun is apparently somewhere between 5 and 40 parsecs above the galactic plane, and this paper tries to narrow it down.
http://arxiv.org/PS_cache/astro-ph/pdf/0101/0101310.pdf

===|==============/ Level Head

[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.

[kelloggs2066.livejournal.com]

Thanks for all the help, guys!