| Orientation of astronomical coordinates |
In astronomy, coordinate systems are used for specifying positions of celestial objects (satellites, planets, stars, galaxies, etc.) relative to a given reference frame, based on physical reference points available to a situated observer (e.g. the true horizon and north to an observer on Earth's surface).[1] Coordinate systems in astronomy can specify an object's relative position in three-dimensional space or plot merely by its direction on a celestial sphere, if the object's distance is unknown or trivial.
Spherical coordinates, projected on the celestial sphere, are analogous to the geographic coordinate system used on the surface of Earth. These differ in their choice of fundamental plane, which divides the celestial sphere into two equal hemispheres along a great circle. Rectangular coordinates, in appropriate units, have the same fundamental plane and primary (-axis) direction, such as an axis of rotation. Each coordinate system is named after its choice of fundamental plane.
The following table lists the common coordinate systems in use by the astronomical community. The fundamental plane divides the celestial sphere into two equal hemispheres and defines the baseline for the latitudinal coordinates, similar to the equator in the geographic coordinate system. The poles are located at ±90° from the fundamental plane. The primary direction is the starting point of the longitudinal coordinates. The origin is the zero distance point, the "center of the celestial sphere", although the definition of celestial sphere is ambiguous about the definition of its center point.
| Coordinate system[2] | Center point (origin) | Fundamental plane (0° latitude) | Poles | Coordinates | Primary direction (0° longitude) | ||
|---|---|---|---|---|---|---|---|
| Latitude | Longitude | ||||||
| Horizontal (also called - or -) | Observer | Horizon | Zenith, nadir | Altitude or elevation | Azimuth | North or south point of horizon | |
| Equatorial | Center of the Earth(geocentric), or Sun(heliocentric) | Celestial equator | Celestial poles | Declination | Right ascension or hour angle | March equinox | |
| Ecliptic | Ecliptic | Ecliptic poles | Ecliptic latitude | Ecliptic longitude | |||
| Galactic | Center of the Sun | Galactic plane | Galactic poles | Galactic latitude | Galactic longitude | Galactic Center | |
| Supergalactic | Supergalactic plane | Supergalactic poles | Supergalactic latitude | Supergalactic longitude | Intersection of supergalactic plane and galactic plane | ||
See main article: Horizontal coordinate system. The horizontal, or altitude-azimuth, system is based on the position of the observer on Earth, which revolves around its own axis once per sidereal day (23 hours, 56 minutes and 4.091 seconds) in relation to the star background. The positioning of a celestial object by the horizontal system varies with time, but is a useful coordinate system for locating and tracking objects for observers on Earth. It is based on the position of stars relative to an observer's ideal horizon.
See main article: Equatorial coordinate system. The equatorial coordinate system is centered at Earth's center, but fixed relative to the celestial poles and the March equinox. The coordinates are based on the location of stars relative to Earth's equator if it were projected out to an infinite distance. The equatorial describes the sky as seen from the Solar System, and modern star maps almost exclusively use equatorial coordinates.
The equatorial system is the normal coordinate system for most professional and many amateur astronomers having an equatorial mount that follows the movement of the sky during the night. Celestial objects are found by adjusting the telescope's or other instrument's scales so that they match the equatorial coordinates of the selected object to observe.
Popular choices of pole and equator are the older B1950 and the modern J2000 systems, but a pole and equator "of date" can also be used, meaning one appropriate to the date under consideration, such as when a measurement of the position of a planet or spacecraft is made. There are also subdivisions into "mean of date" coordinates, which average out or ignore nutation, and "true of date," which include nutation.
See main article: Ecliptic coordinate system. The fundamental plane is the plane of the Earth's orbit, called the ecliptic plane. There are two principal variants of the ecliptic coordinate system: geocentric ecliptic coordinates centered on the Earth and heliocentric ecliptic coordinates centered on the center of mass of the Solar System.
The geocentric ecliptic system was the principal coordinate system for ancient astronomy and is still useful for computing the apparent motions of the Sun, Moon, and planets.[3] It was used to define the twelve astrological signs of the zodiac, for instance.
The heliocentric ecliptic system describes the planets' orbital movement around the Sun, and centers on the barycenter of the Solar System (i.e. very close to the center of the Sun). The system is primarily used for computing the positions of planets and other Solar System bodies, as well as defining their orbital elements.
See main article: Galactic coordinate system. The galactic coordinate system uses the approximate plane of the Milky Way Galaxy as its fundamental plane. The Solar System is still the center of the coordinate system, and the zero point is defined as the direction towards the Galactic Center. Galactic latitude resembles the elevation above the galactic plane and galactic longitude determines direction relative to the center of the galaxy.
See main article: Supergalactic coordinate system. The supergalactic coordinate system corresponds to a fundamental plane that contains a higher than average number of local galaxies in the sky as seen from Earth.
See also: Euler angles and Rotation matrix. Conversions between the various coordinate systems are given.[4] See the notes before using these equations.
\begin{align} h&=\thetaL-\alpha&&or&h&=\thetaG+λo-\alpha\\ \alpha&=\thetaL-h&&or&\alpha&=\thetaG+λo-h \end{align}
The classical equations, derived from spherical trigonometry, for the longitudinal coordinate are presented to the right of a bracket; dividing the first equation by the second gives the convenient tangent equation seen on the left.[5] The rotation matrix equivalent is given beneath each case.[6] This division is ambiguous because tan has a period of 180° whereas cos and sin have periods of 360° (2).
\begin{align} \tan\left(λ\right)&={\sin\left(\alpha\right)\cos\left(\varepsilon\right)+\tan\left(\delta\right)\sin\left(\varepsilon\right)\over\cos\left(\alpha\right)}; \begin{cases} \cos\left(\beta\right)\sin\left(λ\right)=\cos\left(\delta\right)\sin\left(\alpha\right)\cos\left(\varepsilon\right)+\sin\left(\delta\right)\sin\left(\varepsilon\right);\\ \cos\left(\beta\right)\cos\left(λ\right)=\cos\left(\delta\right)\cos\left(\alpha\right). \end{cases}\\ \sin\left(\beta\right)&=\sin\left(\delta\right)\cos\left(\varepsilon\right)-\cos\left(\delta\right)\sin\left(\varepsilon\right)\sin\left(\alpha\right)\\[3pt] \begin{bmatrix} \cos\left(\beta\right)\cos\left(λ\right)\\ \cos\left(\beta\right)\sin\left(λ\right)\\ \sin\left(\beta\right) \end{bmatrix}&=\begin{bmatrix} 1&0&0\\ 0&\cos\left(\varepsilon\right)&\sin\left(\varepsilon\right)\\ 0&-\sin\left(\varepsilon\right)&\cos\left(\varepsilon\right) \end{bmatrix}\begin{bmatrix} \cos\left(\delta\right)\cos\left(\alpha\right)\\ \cos\left(\delta\right)\sin\left(\alpha\right)\\ \sin\left(\delta\right) \end{bmatrix}\\[6pt] \tan\left(\alpha\right)&={\sin\left(λ\right)\cos\left(\varepsilon\right)-\tan\left(\beta\right)\sin\left(\varepsilon\right)\over\cos\left(λ\right)}; \begin{cases} \cos\left(\delta\right)\sin\left(\alpha\right)=\cos\left(\beta\right)\sin\left(λ\right)\cos\left(\varepsilon\right)-\sin\left(\beta\right)\sin\left(\varepsilon\right);\\ \cos\left(\delta\right)\cos\left(\alpha\right)=\cos\left(\beta\right)\cos\left(λ\right). \end{cases}\\[3pt] \sin\left(\delta\right)&=\sin\left(\beta\right)\cos\left(\varepsilon\right)+\cos\left(\beta\right)\sin\left(\varepsilon\right)\sin\left(λ\right).\\[6pt] \begin{bmatrix} \cos\left(\delta\right)\cos\left(\alpha\right)\\ \cos\left(\delta\right)\sin\left(\alpha\right)\\ \sin\left(\delta\right) \end{bmatrix}&=\begin{bmatrix} 1&0&0\\ 0&\cos\left(\varepsilon\right)&-\sin\left(\varepsilon\right)\\ 0&\sin\left(\varepsilon\right)&\cos\left(\varepsilon\right) \end{bmatrix}\begin{bmatrix} \cos\left(\beta\right)\cos\left(λ\right)\\ \cos\left(\beta\right)\sin\left(λ\right)\\ \sin\left(\beta\right) \end{bmatrix}. \end{align}
Azimuth is measured from the south point, turning positive to the west.[7] Zenith distance, the angular distance along the great circle from the zenith to a celestial object, is simply the complementary angle of the altitude: .[8]
\begin{align} \tan\left(A\right)&={\sin\left(h\right)\over\cos\left(h\right)\sin\left(\phio\right)-\tan\left(\delta\right)\cos\left(\phio\right)}; \begin{cases} \cos\left(a\right)\sin\left(A\right)=\cos\left(\delta\right)\sin\left(h\right);\\ \cos\left(a\right)\cos\left(A\right)=\cos\left(\delta\right)\cos\left(h\right)\sin\left(\phio\right)-\sin\left(\delta\right)\cos\left(\phio\right) \end{cases}\\[3pt] \sin\left(a\right)&=\sin\left(\phio\right)\sin\left(\delta\right)+\cos\left(\phio\right)\cos\left(\delta\right)\cos\left(h\right); \end{align}
In solving the equation for, in order to avoid the ambiguity of the arctangent, use of the two-argument arctangent, denoted, is recommended. The two-argument arctangent computes the arctangent of, and accounts for the quadrant in which it is being computed. Thus, consistent with the convention of azimuth being measured from the south and opening positive to the west,
A=-\arctan(x,y)
where
\begin{align} x&=-\sin\left(\phio\right)\cos\left(\delta\right)\cos\left(h\right)+\cos\left(\phio\right)\sin\left(\delta\right)\\ y&=\cos\left(\delta\right)\sin\left(h\right) \end{align}
If the above formula produces a negative value for, it can be rendered positive by simply adding 360°.
\begin{align} \begin{bmatrix} \cos\left(a\right)\cos\left(A\right)\\ \cos\left(a\right)\sin\left(A\right)\\ \sin\left(a\right) \end{bmatrix}&=\begin{bmatrix} \sin\left(\phio\right)&0&-\cos\left(\phio\right)\\ 0&1&0\\ \cos\left(\phio\right)&0&\sin\left(\phio\right) \end{bmatrix}\begin{bmatrix} \cos\left(\delta\right)\cos\left(h\right)\\ \cos\left(\delta\right)\sin\left(h\right)\\ \sin\left(\delta\right) \end{bmatrix}\\ &=\begin{bmatrix} \sin\left(\phio\right)&0&-\cos\left(\phio\right)\\ 0&1&0\\ \cos\left(\phio\right)&0&\sin\left(\phio\right) \end{bmatrix}\begin{bmatrix} \cos\left(\thetaL\right)&\sin\left(\thetaL\right)&0\\ \sin\left(\thetaL\right)&-\cos\left(\thetaL\right)&0\\ 0&0&1 \end{bmatrix}\begin{bmatrix} \cos\left(\delta\right)\cos\left(\alpha\right)\\ \cos\left(\delta\right)\sin\left(\alpha\right)\\ \sin\left(\delta\right) \end{bmatrix};\\[6pt] \tan\left(h\right)&={\sin\left(A\right)\over\cos\left(A\right)\sin\left(\phio\right)+\tan\left(a\right)\cos\left(\phio\right)}; \begin{cases} \cos\left(\delta\right)\sin\left(h\right)=\cos\left(a\right)\sin\left(A\right);\\ \cos\left(\delta\right)\cos\left(h\right)=\sin\left(a\right)\cos\left(\phio\right)+\cos\left(a\right)\cos\left(A\right)\sin\left(\phio\right) \end{cases}\\[3pt] \sin\left(\delta\right)&=\sin\left(\phio\right)\sin\left(a\right)-\cos\left(\phio\right)\cos\left(a\right)\cos\left(A\right); \end{align}
Again, in solving the equation for, use of the two-argument arctangent that accounts for the quadrant is recommended. Thus, again consistent with the convention of azimuth being measured from the south and opening positive to the west,
h=\arctan(x,y)
where
\begin{align} x&=\sin\left(\phio\right)\cos\left(a\right)\cos\left(A\right)+\cos\left(\phio\right)\sin\left(a\right)\\ y&=\cos\left(a\right)\sin\left(A\right)\\[3pt] \begin{bmatrix} \cos\left(\delta\right)\cos\left(h\right)\\ \cos\left(\delta\right)\sin\left(h\right)\\ \sin\left(\delta\right) \end{bmatrix}&=\begin{bmatrix} \sin\left(\phio\right)&0&\cos\left(\phio\right)\\ 0&1&0\\ -\cos\left(\phio\right)&0&\sin\left(\phio\right) \end{bmatrix}\begin{bmatrix} \cos\left(a\right)\cos\left(A\right)\\ \cos\left(a\right)\sin\left(A\right)\\ \sin\left(a\right) \end{bmatrix}\\ \begin{bmatrix} \cos\left(\delta\right)\cos\left(\alpha\right)\\ \cos\left(\delta\right)\sin\left(\alpha\right)\\ \sin\left(\delta\right) \end{bmatrix}&=\begin{bmatrix} \cos\left(\thetaL\right)&\sin\left(\thetaL\right)&0\\ \sin\left(\thetaL\right)&-\cos\left(\thetaL\right)&0\\ 0&0&1 \end{bmatrix}\begin{bmatrix} \sin\left(\phio\right)&0&\cos\left(\phio\right)\\ 0&1&0\\ -\cos\left(\phio\right)&0&\sin\left(\phio\right) \end{bmatrix}\begin{bmatrix} \cos\left(a\right)\cos\left(A\right)\\ \cos\left(a\right)\sin\left(A\right)\\ \sin\left(a\right) \end{bmatrix}. \end{align}
These equations[9] are for converting equatorial coordinates to Galactic coordinates.
\begin{align} \cos\left(lNCP-l\right)\cos(b)&=\sin\left(\delta\right)\cos\left(\deltaG\right)-\cos\left(\delta\right)\sin\left(\deltaG\right)\cos\left(\alpha-\alphaG\right)\\ \sin\left(lNCP-l\right)\cos(b)&=\cos(\delta)\sin\left(\alpha-\alphaG\right)\\ \sin\left(b\right)&=\sin\left(\delta\right)\sin\left(\deltaG\right)+\cos\left(\delta\right)\cos\left(\deltaG\right)\cos\left(\alpha-\alphaG\right) \end{align}
\alphaG,\deltaG
lNCP
\alphaG=192.85948\circ \deltaG=27.12825\circ
| \circ | |
| l | |
| NCP=122.93192 |
If the equatorial coordinates are referred to another equinox, they must be precessed to their place at J2000.0 before applying these formulae.
These equations convert to equatorial coordinates referred to B2000.0.
\begin{align} \sin\left(\alpha-\alphaG\right)\cos\left(\delta\right)&=\cos\left(b\right)\sin\left(lNCP-l\right)\\ \cos\left(\alpha-\alphaG\right)\cos\left(\delta\right)&=\sin\left(b\right)\cos\left(\deltaG\right)-\cos\left(b\right)\sin\left(\deltaG\right)\cos\left(lNCP-l\right)\\ \sin\left(\delta\right)&=\sin\left(b\right)\sin\left(\deltaG\right)+\cos\left(b\right)\cos\left(\deltaG\right)\cos\left(lNCP-l\right) \end{align}