Finding angle between a given sky position and ecliptic pole

Question

Given a right ascension in degrees and declination in degrees (say, 88 degrees RA, -63 degrees DEC) I want to find the angle between that location and the ecliptic pole (90 degrees RA, -66 degrees DEC... are these the right values? What are the decimals following the 66? Most of my values that I'm working with are similar to the one I give, so which ecliptic pole, if any, does this correspond to?) How would I find this angle, I need a convenient formula (such as a dot product to find this angle). Thank you!

Answer 1

The angle between an ecliptic pole and another sky position is complementary to the ecliptic latitude of that position. To find ecliptic (λ, β) coordinates, one method is to convert the equatorial (α, δ) to Cartesian form: $$ \begin{align} x_\mathrm{equ} &= \cos \alpha \cos \delta \\ y_\mathrm{equ} &= \sin \alpha \cos \delta \\ z_\mathrm{equ} &= \sin \delta \end{align}$$ then rotate by ε = 23.44° on the x axis: $$ \begin{align} x_\mathrm{ecl} &= x_\mathrm{equ} \\ y_\mathrm{ecl} &= y_\mathrm{equ} \cos \varepsilon + z_\mathrm{equ} \sin \varepsilon \\ z_\mathrm{ecl} &= z_\mathrm{equ} \cos \varepsilon - y_\mathrm{equ} \sin \varepsilon \end{align} $$ and convert to spherical form: $$ \begin{align} \lambda &= \mathtt{atan2}(y_\mathrm{ecl}, x_\mathrm{ecl}) \\ \beta &= \arcsin z_\mathrm{ecl} \end{align} $$

If you are not interested in the ecliptic longitude λ, you can express the ecliptic latitude as $$ \beta = \arcsin (\sin \delta \cos \varepsilon - \sin \alpha \cos \delta \sin \varepsilon) $$ Then the angle off the north ecliptic pole (α=18h0m, δ=+66.56°) is 90° - β.

Working your example (α=5h52m, δ=-63°), I get (λ=75.63°, β=-86.34°), and the angle off the south ecliptic pole (α=6h0m, δ=-66.56°) is β + 90° = 3.66°.