convert equatorial coordinates into decimal degrees
ConvertEquatorialCoordinatesIntoDecimalDegrees
2025-02-26 05:56:32
2025-02-26 05:56:32
Example
equatorial coordinate system
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
A star is observed to have the following equatorial coordinates:
\begin{itemize}
\item Right Ascension (RA): 5 hours 34 minutes 12 seconds
\item Declination (Dec): \ang{22}{15}{48}
\end{itemize}
\begin{enumerate}
\item Convert the Right Ascension (RA) from hours, minutes, and seconds into decimal degrees.
\item Determine the star's position in the sky relative to the celestial equator (is it north or south of the equator?).
\item If the Local Sidereal Time (LST) at your location is 3 hours 45 minutes, calculate the hour angle of the star.
\end{enumerate}
\section*{Solution}
1. \textbf{Convert RA to decimal degrees:}
\begin{itemize}
\item RA = 5 hours 34 minutes 12 seconds
\item Convert hours, minutes, and seconds to decimal degrees:
\begin{align*}
1 \, \text{hour} &= 15^\circ \quad (\text{since } 24 \, \text{hours} = 360^\circ) \\
1 \, \text{minute} &= 0.25^\circ \quad (\text{since } 60 \, \text{minutes} = 15^\circ) \\
1 \, \text{second} &= 0.0041667^\circ \quad (\text{since } 60 \, \text{seconds} = 0.25^\circ)
\end{align*}
\item RA in decimal degrees:
\begin{align*}
5 \, \text{hours} &= 5 \times 15^\circ = 75^\circ \\
34 \, \text{minutes} &= 34 \times 0.25^\circ = 8.5^\circ \\
12 \, \text{seconds} &= 12 \times 0.0041667^\circ \approx 0.05^\circ \\
\text{Total RA} &= 75^\circ + 8.5^\circ + 0.05^\circ = \boxed{83.55^\circ}
\end{align*}
\end{itemize}
2. \textbf{Determine the star's position relative to the celestial equator:}
\begin{itemize}
\item Declination (Dec) = \ang{22}{15}{48}
\item Since the declination is positive, the star is \boxed{\text{north of the celestial equator}}.
\end{itemize}
3. \textbf{Calculate the hour angle (HA):}
\begin{itemize}
\item Local Sidereal Time (LST) = 3 hours 45 minutes
\item Convert LST to decimal degrees:
\begin{align*}
3 \, \text{hours} &= 3 \times 15^\circ = 45^\circ \\
45 \, \text{minutes} &= 45 \times 0.25^\circ = 11.25^\circ \\
\text{LST} &= 45^\circ + 11.25^\circ = \boxed{56.25^\circ}
\end{align*}
\item Hour Angle (HA) = LST - RA
\begin{align*}
\text{HA} &= 56.25^\circ - 83.55^\circ = \boxed{-27.3^\circ}
\end{align*}
\item A negative hour angle means the star is \boxed{\text{east of the meridian}} (not yet crossed the observer's meridian).
\end{itemize}
This example was generated by DeepSeek, an AI, on February 25, 2025.