Altitude of North Celestial Pole equals Latitude of Observer
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Theorem
Let $O$ be an observer of the celestial sphere.
Let $P$ be the position of the north celestial pole with respect to $O$.
Let $a$ denote the altitude of $P$.
Let $\phi$ denote the (terrestrial) latitude of $O$.
Then:
- $a = \phi$
Proof
Let $z$ denote the zenith distance of $P$.
Let $\psi$ denote the (terrestrial) colatitude of $O$.
By definition we have:
- $a = 90 \degrees - z$
- $\phi - 90 \degrees - \psi$
Then:
| \(\ds z\) | \(=\) | \(\ds \psi\) | Zenith Distance of North Celestial Pole equals Colatitude of Observer | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 90 \degrees - z\) | \(=\) | \(\ds 90 \degrees - \psi\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds \phi\) |
Hence the result.
$\blacksquare$
Sources
- 1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text {II}$. The Celestial Sphere: $18$. Altitude and azimuth.