Although we know that the objects we see in the sky are of different sizes and at different distances from us, it is convenient to visualize all the objects as being attached to an imaginary sphere surrounding the Earth. From our vantage point, the sky certainly looks like a dome (the half of the celestial sphere above our local horizon). The celestial sphere is mapped in Right Ascension (RA) and Declination (Dec). Declination is the celestial equivalent of latitude, and is simply the Earth's latitude lines projected onto the celestial sphere. A star that can be directly overhead as seen from the Earth's Equator (0° latitude) is said to be on the Celestial Equator, and has a declination of 0°. The North Star, Polaris, is very nearly overhead as seen from the North Pole (90° North latitude). The point directly over the North Pole on the celestial sphere is called the North Celestial Pole, and has a declination of +90°. Northern declinations are given positive signs, and southern declinations are given negative signs. So, the South Celestial Pole has a declination of -90°.

Right Ascension is the equivalent of longitude, but since the Earth rotates with respect to the celestial sphere we cannot simply use the Greenwich Meridian as 0° RA. Instead, we set the zero point as the place on the celestial sphere where the Sun crosses the Celestial Equator (0° Dec) at the vernal (spring) equinox. The arc of the celestial sphere from the North Celestial Pole through this point to the South Celestial Pole is designated as Zero hours RA. Right Ascension increases eastward, and the sky is divided up into 24 hours. This designation is convenient because it represents the sidereal day, the time it takes for the Earth to make one rotation relative to the celestial sphere. If you pointed a telescope (with no motor drive) at the coordinates (RA=0h, Dec=0°), and came back one hour later, the telescope would then be pointing at (RA=1h, Dec=0°). Because the Earth's revolution around the Sun also contributes to the apparent motion of the stars, the day we keep time by (the solar day) is about four minutes longer than the sidereal day. So, if you pointed a telescope at (RA=0h, Dec=0°) and came back 24 hours later, the telescope would now be pointing at (RA=0h 4m, Dec=0°). A consequence is that the fixed stars appear to rise about four minutes earlier each day.


Both RA and Dec can be divided into minutes and seconds. One minute of RA is equal to 1/60 of an hour, and 1 second of RA is equal to 1/60 of a minute. At the Celestial Equator, Right Ascension can be measured in degrees equal to those of Declination (since the Celestial Equator is a great circle. With 360° in a circle, each of the 24 hours of RA is equal to an angle of 15°. This angle decreases until it becomes zero at the poles (just as the meridians of longitude meet at the poles on Earth).

The angular distance between any two points on the celestial sphere can be expressed in degrees, minutes, and seconds of arc by placing the two points on an imaginary great circle of the celestial sphere. The Full Moon's diameter, for example, is about 30 arcminutes, or 30', or 0.5°. A close double star might have a separation of 5 arcseconds (5").


For rough estimation with the naked eye, it is possible to "eyeball" angles in the sky. For example, one index finger tip at arm's length subtends an angle of a little more than 1° in the sky. A fist (with the thumb flat against the end) at arm's length subtends about 10°. This can be useful if you know an object (like a planet in bright twilight) should be 20° above the western horizon or 5° south of a brighter object.

In a telescope, we often want to know the field of view with a given eyepiece. So, we point the 'scope at a star near 0° Dec and put the star on the eastern edge of the field. Because of the Earth's rotation, the star will appear to drift westward. The time in minutes it takes the star to cross the entire field is equal to the telescope-eyepiece combination's field of view in minutes of RA (at the celestial equator only). Because each hour of RA at the equator is equal to 15 degrees of arc, and each minute of RA is 1/60 of an hour, one minute of RA at the equator is equal to (15/60 = 1/4°). So, if the field of view is 2 minutes of RA, the field of view in degrees is 2/4=0.5°. Distances between stars or objects can then be measured in fields of view and converted to degrees of arc. If a new eyepiece is used, the field of view for that eyepiece will need to be determined.


If we know the actual distance to an object, and measure its angular diameter in the sky, we can determine its actual diameter by solving a right triangle, as shown in Figure 1:

Always remember to convert the angular size (theta) to degrees of arc before taking the tangent. In other words, if theta is given in minutes it must be divided by 60. If theta is given in seconds, it must be divided by 3600.

In practice, we usually have only a rough estimate of D, so the estimates of actual size are rough as well. Table 1 shows some rough diameters (long axes for elongate objects) of a few prominent deep-sky objects.


Points on the celestial sphere remain fixed with respect to the stars, except for small changes due to precession of the Earth and independent proper motions of the stars. To an Earth-bound observer, there are a couple of dynamic points and circles that are useful for describing the appearance of the sky. The point directly overhead is known as the zenith. The declination at the zenith is equal to the site's latitude; therefore, the zenith for an observer at 45°N will be +45°. The Right Ascension at the zenith is not fixed, and varies as the Earth rotates and revolves. If we draw an arc from the North Celestial Pole (in the Northern Hemisphere) through the zenith down to the southern horizon, we describe what is known as the meridian. Objects rise in the East, reach their greatest altitude at the meridian, and set in the West. The Right Ascension of the meridian is known as the sidereal time. A clock adjusted to gain four minutes each day will keep rough sidereal time. It is important to know the time an object crosses the meridian, or culminates, since the object will be highest in the sky and most easily visible at that time. The edge of the visible portion of the celestial sphere is, of course, the horizon. An object will never be visible from a given latitude if it is below the horizon when it is on the meridian. The declination of the horizon at the meridian is equal to the site's latitude minus 90° (Southern Hemisphere observers will have to treat their latitude as negative and add 90° to get their meridional horizon). Certain objects near the visible celestial pole will not rise or set. These are known as circumpolar objects. Any star with a positive (negative for the Southern Hemisphere) declination greater than a site's latitude will be circumpolar from that site. An object's altitude when on the meridian (Am) can be determined by the formula Am= 90° - |L - Dec|, where L is the site's latitude and Dec is the object's declination. Note the absolute value notation. From 45° North latitude, an object at -30° Dec would attain an altitude of 15°. An object at +30° Dec would reach 75°, and one at +75° Dec would reach 60° altitude. An object at +45° Dec would, of course, reach the zenith at 90° altitude.

The next time you look at the night sky, try to visualize the horizon and meridian. Use the fist=10° method to gauge altitude. Note the altitude of Polaris if you are in the Northern Hemisphere. Is it equal to your latitude? Find the zenith, and trace the meridian. At what altitude does the celestial equator meet the meridian? Watch for a couple of hours as stars rise and set. Which stars are circumpolar from your location? If you have a telescope or binoculars mounted on a tripod, try to measure the field of view. Remember to pick stars near the celestial equator; Orion's belt stars work nicely. Using this information, about how large is the Orion Nebula in angular size? If the Nebula is 1300 light years away, what are the actual dimensions of the part you can see?

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