Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Meta

Talkback

Downloads

Information
[parent] equatorial coordinate system example problem (Definition)

Equatorial Coordinate System Example Problem

The equatorial coordinate system locates celestial objects with two angular coordinates: right ascension (RA), analogous to longitude and usually measured in hours, minutes, and seconds along the Celestial Equator, and declination (Dec), analogous to latitude and measured in degrees, arcminutes, and arcseconds north or south of the celestial equator.

Consider an astronomer locating a star with these coordinates:

  • Right ascension:
            h  m   s
RA  =  5 32  15 .
  • Declination:
               ∘  ′  ′′
Dec =  +22  45 10 .

The goal is to determine the star’s position relative to the vernal equinox and celestial equator, and to discuss its visibility from an observatory at latitude 40 North on February 24, 2025.

Step 1: Understanding the Coordinates

Right ascension. The right ascension

5h32m 15s

means that the star lies eastward from the vernal equinox by the corresponding angular distance measured along the celestial equator. Since 24h = 360, one hour of right ascension equals 15.

Thus,

5h = 5(15) = 75,
32m = 32(    )
  15∘-
  60 = 8,
15s = 15(   ∘ )
  15---
  3600 = 0.0625.

Therefore,

RA  = 75∘ + 8∘ + 0.0625 ∘ = 83.0625∘.

Declination. The declination

+22 ∘45′10′′

means that the star is north of the celestial equator. Converting to decimal degrees gives

45 = 45
---
60 = 0.75,
10′′ = --10-
3600 0.00278.

Hence,

Dec =  22∘ + 0.75∘ + 0.00278∘ ≈ 22.75278 ∘.

Thus, the star is at approximately 83.0625 east of the vernal equinox and 22.75278 north of the celestial equator.

Step 2: Visibility from 40 North

At geographic latitude

      ∘
ϕ = 40 ,

the circumpolar declination boundary is

      ∘         ∘
δ ≥ 90 −  ϕ = 50 ,

while stars with

         ∘            ∘
δ ≤ − (90  − ϕ) = − 50

never rise.

Because

− 50 ∘ < 22.75278 ∘ < 50∘,

the star rises and sets. Its visibility at a particular clock time on February 24, 2025 depends on the local sidereal time.

Problem Twist: Altitude at Meridian Transit

At upper meridian transit, the altitude is

          ∘
hmax =  90 −  |ϕ − δ |.

For this star,

hmax = 90  ∘            ∘
|40  − 22.75278 |
= 9017.24722
= 72.75278.

The star therefore reaches an altitude of approximately 72.75 above the southern horizon at upper culmination.

This example illustrates how equatorial coordinates identify celestial positions and how declination and observer latitude determine basic observability.

Source note: This example was originally generated by Grok, an AI developed by xAI, on February 24, 2025.


"equatorial coordinate system example problem" is owned by bloftin.
(view preamble)
View style:

This object's parent.

Cross-references: horizon, meridian, local sidereal time, boundary, position, latitude, Celestial Equator, longitude, equatorial coordinate system

This is version 3 of equatorial coordinate system example problem, born on 2025-02-25, modified 2026-09-07.
Object id is 955, canonical name is EquatorialCoordinateSystemExampleProblem.
Accessed 1420 times total.

Classification:
Physics Classification95.10.-a (Fundamental astronomy)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Interact
rate | post | correct | update request | add derivation | add example | add (any)