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[parent] celestial sphere and zenith example problem (Example)

Celestial Sphere and Zenith Example Problem

Let's examine a problem [1] involving the celestial sphere, an imaginary dome surrounding Earth where celestial objects are projected, and the zenith, the point directly overhead for an observer.

Suppose you're stargazing at latitude $35^\circ$ North on March 1, 2025, at 10:00 PM local time. You observe a star exactly at your zenith. The task is to determine its equatorial coordinates right ascension (RA) and declination (Dec) and assess if it's circumpolar (always visible) from your location.

Step 1: Understanding the Zenith and Celestial Sphere

The Celestial Sphere rotates around the north and South Celestial Poles, aligned with Earth's axis. At latitude $35^\circ$ N:

A star at the zenith aligns with this overhead point. In equatorial coordinates:

  • Declination (Dec): Measures north or south of the Celestial Equator ($0^\circ$ to $+90^\circ$ at the NCP).
  • Right Ascension (RA): Measures eastward from the vernal equinox along the celestial equator (0h to 24h).

Step 2: Declination of the Star

Since the star is at the zenith, its altitude is $90^\circ$. The declination of a star at the zenith equals the observer's latitude because:

  • The celestial equator is $90^\circ - 35^\circ = 55^\circ$ south of the zenith.
  • A star at $90^\circ$ altitude has a declination matching the latitude.

Thus:

Dec$\displaystyle = +35^\circ$    

Step 3: Right Ascension of the Star

The RA depends on the star's position along the celestial equator at that time. A star at the zenith is on the meridian, so its RA equals the local sidereal time (LST) at 10:00 PM on March 1, 2025. Estimating LST:

  • Sidereal time runs faster than solar time (1 sidereal day $\approx$ 23h 56m).
  • Around March 1, RA = 0h is near the meridian at midnight. At 10:00 PM, LST is approximately 2 hours earlier, so LST$\approx 22^$h.

Thus:

RA$\displaystyle \approx 22^$h$\displaystyle 00^$m$\displaystyle 00^$s    
(Exact LST requires longitude and precise calculations, but this is an approximation.)

Step 4: Is the Star Circumpolar?

A star is circumpolar if its declination exceeds $90^\circ -$   latitude:

$\displaystyle 90^\circ - 35^\circ = 55^\circ$    
With Dec$= +35^\circ < +55^\circ$, the star rises and sets (between $-55^\circ$ and $+55^\circ$).

Solution

The star's approximate coordinates are:

  • RA$= 22^$h$00^$m$00^$s
  • Dec$= +35^\circ$

It's visible part of the night but not circumpolar.

Bonus Twist: Altitude 6 Hours Later

Six hours later (4:00 AM), the celestial sphere rotates $6 \times 15^\circ = 90^\circ$ westward. The star, originally at $90^\circ$ altitude, is now near the western horizon, with altitude $\approx 0^\circ$ (adjusted for refraction).

This problem demonstrates how the zenith connects an observer's position to the celestial sphere, aiding in sky mapping.

[1] This example was generated by Grok, an AI developed by xAI, on February 24, 2025.



"celestial sphere and zenith example problem" is owned by bloftin.

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Cross-references: position, Celestial Equator, horizon, North Celestial Pole, South Celestial Poles, Celestial Sphere, zenith, objects

This is version 1 of celestial sphere and zenith example problem, born on 2025-02-28.
Object id is 958, canonical name is CelestialSphereAndZenithExampleProblem.
Accessed 21 times total.

Classification:
Physics Classification95.10.-a (Fundamental astronomy)

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