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Inclination of Equatorial North Pole relative to Galactic Equator 
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#1
Jun1614, 08:06 PM

P: 27

If the axis of the Earth is tilted at 23.4 degrees away from the North Ecliptic Pole, and the Ecliptic Plane is tilted at 60.2 degrees away from the North Galactic Pole, then why is the Earth's rotational axis inclined at 27.13 degrees relative to the Galactic Equator instead of (90(60.2+23.4)) = 6.4 degrees? Is there a formula that will allow me to calculate this, and if so, can anyone show me how to do it? I know it involves spherical trigonometry.
I'm somewhat familiar with transformations of galactic, ecliptic and equatorial coordinate systems and spherical trigonometry. But I can't find the formula that will allow me to transform Equatorial coordinates to Galactic coordinates. Here is a diagram which may or may not help. It at least illustrates my dilemma. Thanks! 


#2
Jun1614, 11:49 PM

Sci Advisor
P: 2,847

Can you double check your statement "The Ecliptic plane is tilted at 60.2 degrees away from the North Galactic Pole"? In your picture the Ecliptic plane is tilted 60.2 degrees away from the Galactic plane, not the North Galactic pole.
It seems though that the Wikipedia page on the Milky Way suggests that what your drawing, and not your statement is correct about that point though. So it might be that you just misstated that fact. The only other thing I can think of right now is that the Earth's 23 degree tilt might be in the other direction relative to the galactic plane (i.e. it might be 23 degrees counterclockwise from the perpendicular line to the ecliptic, rather than 23 degrees clockwise from this line), but applied to your drawing this would get a declination of the galactic north pole of 53 degrees rather than 27 degrees... 


#3
Jun1714, 12:09 AM

Sci Advisor
Thanks
P: 3,755




#4
Jun1714, 12:12 AM

Sci Advisor
P: 2,847

Inclination of Equatorial North Pole relative to Galactic Equator
Yes, Nurgatory is right! I was looking at this in a 2D point of view as well. The tilt might not only be above and below the line you drew, but might be into the page or out of the page as well.
So in this case, the relative angle between the Galactic north pole and the celestial north pole might be anywhere between 6 degrees and 53 degrees. It happens to be 27 degrees. 


#5
Jun1714, 08:03 AM

Mentor
P: 15,170

On January 1, 2000, the galactic north pole had a right ascension ##\alpha## of 12^{h} 51^{m} 26.00^{s} and a declination ##\delta## of 27° 7' 42.0". Note that right ascension and declination are expressed in celestial (earth equatorial) coordinates. The unit vector along the north galactic pole has celestial coordinates of ##\hat z_g = \bigl[ \cos\delta\cos\alpha, \cos\delta \sin\alpha, \sin\delta\bigr]##. The celestial coordinates of the celestial north pole is just ##\hat z_c = \bigl[0, 0, 1\bigr]##. The inner product between ##\hat z_g## and ##\hat z_c## is ##\sin\delta##. Since these are both unit vectors, this is the cosine of the angle between these two vectors. Since ##\arccos(\sin x) = 90^{\circ}x##, the angle between the north celestial pole and north galactic pole is 90°  (27° 7' 42.0") = 62° 52' 18" (or 62.8717 degrees). The ecliptic plane is formed by rotating the celestial plane about the xaxis by the obliquity of the ecliptic ##\epsilon## of 23° 26' 21.406", making the celestial coordinates of the ecliptic north pole ##\hat z _e = [0, \sin\epsilon, \cos\epsilon]##. The inner product between ##\hat z_e## and ##\hat z_c## is ##\cos\epsilon##. Thus the obliquity of the ecliptic is the angle between the celestial and ecliptic north poles. What about the angle between the galactic and ecliptic north poles? Now we want to find the inner product between ##\hat z_g## and ##\hat z_e## and take the inverse cosine of that inner product. The inner product is ##\hat z_g \cdot \hat z_e = \sin\delta \sin\epsilon  \sin\alpha \cos\delta \cos\epsilon##. Letting WolframAlpha do the math for me (http://www.wolframalpha.com/input/?i...6%2F60%29%2F60), the angle is 60.1889 degrees. 


#6
Jun1714, 10:03 AM

P: 27




#7
Jun1714, 12:57 PM

P: 27

I plugged your formula into an Excel spreadsheet (which uses radians for angles, which then have to be converted degrees), and this is what I got: acosd(sind(d)*cosd(eps)sind(a)*cosd(d)*sind(eps)) where a = 192.858 degrees (3.366007089 radians) d = 27.1283 degrees (0.473478155 radians) eps = 23.4393 degrees (0.409092959 radians ) = 1.085552547 radians = 62.1976 degrees instead of 60.1889 degrees, which is odd. Maybe I got something wrong, or maybe an error crept in through all the conversions. Another calculation I've seen for the inclination between the ecliptic and the galactic plane is this: = arccos (sin (delta1) * sin (delta2) + cos (delta1) * cos (delta2) * cos (alpha1alpha2)) where alpha1 = 270 degrees delta1 = 66.5607 degrees alpha2 = 192.8 degrees delta2 = 27.1 degrees Plug these values into the equation and you get 60.1891 degrees (very close to your estimate). I found this calculation at http://answers.yahoo.com/question/in...3110530AAMkEYV This whole quest began when I first wondered how we on Earth are oriented relative to the Galactic Equator. A simple estimate is the declination of Polaris above the GE (27.4 degrees), but living in the city, the Milky Way is not visible, so it's not possible to just look up and see it. Instead, one has to resort to visualizations and calculations. I've been chewing on this question for a long time, and I'm just beginning to wrap my thoughts around it. 


#8
Jun1714, 01:30 PM

Mentor
P: 15,170

http://www.wolframalpha.com/input/?i...09092959%29%29 Note that I did not input 192.858 degrees and such into wolfram alpha. My input was



#9
Jun1714, 07:06 PM

P: 27

This is the formula I used : =DEGREES(ACOS(SIN(d)*COS(eps)SIN(a)*COS(d)*SIN(eps))) [noting that Excel will only let you use radians], and where a = 192.8583333 degrees = 3.3660129065754 radians d = 27.12833333 degrees = 0.473478737245195 radians eps = 23.43927944 degrees = 0.409092600600583 radians The result is 60.18894305 degrees (1.05049523 radians). Thanks for your help, this is great. I'm finally getting somewhere after all these years. Now if I could only figure out how to backcalculate these numbers so the result would be "d"... is it possible to do this? 


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