This chapter was copied with permission from Nick Strobel’s Astronomy Notes. Go to his site at www.astronomynotes.com for the updated and corrected version. This chapter has been edited for content. Text that has been altered from the original is denoted in green font.
Astronomy Without a Telescope
I discuss the celestial sphere, motions of the Sun (solar and sidereal days, time zones, equation of time, and seasons), motions of the Moon (phases and eclipses), and planetary motions.
Now that you have some feeling for the scales of time and space that astronomy encompasses and some of the difficulties caused by being Earth-bound (well, okay: solar-system bound!), let’s take a look at what is up there in the sky beyond the clouds. In this chapter, you will learn where to find the key points on the night sky, how to use the coordinate system that astronomers use, how the Sun’s position among the stars changes and how that affects the temperature throughout the year, and about the phases of the Moon and eclipses. At the end of chapter, you will learn about the motions of the planets among the stars. All of the things in this chapter, you can observe without a telescope—naked eye astronomy (note to Jesse Helms and Sen. Exon: that means astronomy without the use of a telescope). You just need to observe the objects carefully and notice how things change over time. The vocabulary terms are in boldface.
3.2 Celestial Sphere Defined
Imagine the sky as a great, hollow, sphere surrounding the Earth. The stars are attached to this sphere—some bigger and brighter than others—which rotates around the stationary Earth roughly every 24 hours. Alternatively, you can imagine the stars as holes in the sphere and the light from the heavens beyond the sphere shines through those holes. This imaginary sphere is called the celestial sphere, and has a very large radius so that no part of the Earth is significantly closer to any given star than any other part. Therefore, the sky always looks like a great sphere centered on your position. The celestial sphere (and, therefore, the stars) appears to move westward—stars rise in the east and set in the west.
Even though it is now known that this ancient model of a stationary Earth is incorrect, you can still use this model because it is a convenient way to predict the motions of the stars and planets relative to a location on the Earth. A star’s apparent brightness is actually determined by its distance, as well as, its physical size and temperature. It is also now known that the stars apparent motion around us is due to the Earth rotating once every 24 hours on its axis. The stars are stationary and the Earth rotates from west to east. This rotational motion makes the stars appear to move from east to west around us. The celestial sphere model is used by planetaria to simulate the night sky. I hope you will be able to distinguish between the convenience of the celestial sphere model and the way things really are.
Why a sphere? The Earth is spherical! This was known much earlier than Columbus’ time. Sailors had long known that as a ship sailed away from the shore it not only diminished in apparent size, but it also appeared to sink into the water. The simplest explanation to use was that the Earth was curved (particularly, since those ships did come back without falling off some edge!). They also knew that if one traveled in a north-south direction, some stars disappeared from view while others appeared. The difference in the height of a star’s height above the north or south horizon is directly proportional to the difference in the north-south distance of observers looking at the star at the same time. The simplest explanation said that the Earth is round, not flat. Pythagoras noted that the shadow of the Earth falling on the Moon during a lunar eclipse was always curved and the amount of the curvature was always the same. The only object that always casts a circular shadow regardless of its orientation is a sphere. This Pythagorean argument is passed on to us through the writings of Aristotle.
- celestial sphere
Review Questions 1
- How does the sky appear to move around us?
- What motion of the Earth produces the apparent motion of the stars around us?
To measure distances on the imaginary celestial sphere, you use “angles on the sky” instead of meters or kilometers. There are 360 degrees in a full circle and 90 degrees in a right angle (two perpendicular lines intersecting each other make a right angle). Each degree is divided into 60 minutes of arc. A quarter viewed face-on from across the length of a football field is about 1 arc minute across. Each minute of arc is divided into 60 seconds of arc. The ball in the tip of a ballpoint pen viewed from across the length of a football field is about 1 arc second across.
The Sun and Moon are both about 0.5 degrees = 30 arc minutes in diameter. The pointer stars in the bowl of the Big Dipper are about 5 degrees apart and the bowl of the Big Dipper is about 30 degrees from Polaris, the North Star that is very close to the North Celestial Pole. Some angles using your hand held at arm’s length are described in the figure below. The arc from the north point on the horizon to the point directly overhead to the south point on the horizon is 180 degrees, so any object directly overhead is 90 degrees above the horizon and any object “half-way up” in the sky is about 45 degrees above the horizon.
Review Questions 2
- How many degrees is 30 arc minutes?
- How many degrees is 10 arc seconds?
- How many Moon diameters would it take to span the distance from a point on the eastern horizon to a point directly opposite on the western horizon?
3.4 Reference Markers
Now for some reference makers: The stars rotate around the North and South Celestial Poles. These are the points in the sky directly above the geographic north and south pole, respectively. The Earth’s axis of rotation intersects the celestial sphere at the celestial poles. The number of degrees the celestial pole is above the horizon is equal to the latitude of the observer. Fortunately, for those in the northern hemisphere, there is a fairly bright star real close to the North Celestial Pole (Polaris or the North star). Another important reference marker is the celestial equator: an imaginary circle around the sky directly above the Earth’s equator. It is always 90 degrees from the poles. All the stars rotate in a path that is parallel to the celestial equator. The celestial equator intercepts the horizon at the points directly east and west anywhere on the Earth.
If you joined Santa last Christmas at the north pole (90 degrees latitude), you would have seen Polaris straight overhead and the celestial equator on your horizon. The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian. The positions of the zenith and meridian with respect to the stars will change as the celestial sphere rotates and if the observer changes locations on the Earth, but those reference marks do not change with respect to the observer’s horizon. Any celestial object crossing the meridian is at its highest altitude (distance from the horizon) during that night (or day). The angle the star paths make with respect to the horizon as they rise up or set down = 90 degrees minus the observer’s latitude. At the north pole, the latitude = 90 degrees so the stars paths make an angle of 90 minus 90 degrees = 0 degrees with respect to the horizon—i.e., they move parallel to the horizon as shown in the figure above. For locations further south you will see in the figures below that the stars will rise up (and then set down) at steeper angles as you get closer to the equator.
During daylight, the meridian separates the morning and afternoon positions of the Sun. In the morning the Sun is “ante meridiem” (Latin for “before meridian”) or east of the meridian, abbreviated “a.m.” At local noon the Sun is right on the meridian (the reason why this may not correspond to 12:00 on your clock is discussed a little later in this chapter). At local noon the Sun is due south for northern hemisphere observers and due north for southern hemisphere observers (though observers near the Earth’s equator can see the local noon Sun due north or due south at different times of the year for reasons given in the next section). In the afternoon the Sun is “post meridiem” (Latin for “after meridian”) or west of the meridian, abbreviated “p.m.”
For each degree you move south with Santa in his sleigh, the North Celestial Pole (NCP from here on) moves 1 degree away from the zenith toward the north and the highest point of the celestial equator’s curved path in the sky moves up one degree from the southern horizon. This effect has nothing to do with the distance between a celestial object or marker and you at different points on the Earth (remember that the celestial sphere has a practically infinite radius). In fact, observers on a spherical world only ten miles across would see the same effect! The picture above shows the celestial sphere for the far northern city of Fairbanks in Alaska. Since it is 25° south of the north pole, the NCP is 25° away from (north of) the zenith for Fairbanks observers.
By the time you reach your hometown, the NCP has moved away from the zenith so it is now a number of degrees above the horizon equal to your latitude on the Earth. Remember that your position on the Earth is specified by a latitude and a longitude coordinate. The latitude is the number of degrees north or south of the Earth’s equator. On a map or globe, lines of latitude run horizontally, parallel to the equator. The longitude is the number of degrees east or west of the 0° longitude line (the “prime meridian” on the Earth) that runs through Greenwich England. On a map or globe, lines of longitude run vertically, perpendicular to the equator. The celestial sphere for observers in Seattle and any other observer at the same latitude (47° N) on the Earth is shown above.
For another more detailed example, let’s choose Los Angeles at latitude 34° N. The NCP is therefore 34 degrees above the north horizon. The diagram for latitude 34° N is shown above. Notice that finding the angle of the NCP above the horizon provides a very easy way of determining your latitude on the Earth (a fact used by navigators even today!). Because the Earth’s equator is 90° away from the north pole, the celestial equator as seen in Los Angeles will arc up to 90 – 34 = 56 degrees above the southern horizon at the point it crosses the meridian. It still intercepts the horizon at the exactly east and west points. The stars rise in the east part of the sky, move in arcs parallel to the celestial equator reaching maximum altitude when they cross your meridian, and set in the west part of the sky. The star paths make an angle of 90 – 34 = 56 degrees with respect to the horizon.
If you are in the northern hemisphere, celestial objects north of the celestial equator are above the horizon for more than 12 hours because you see more than half of their total 24-hour path around you. Celestial objects on the celestial equator are up 12 hours and those south of the celestial equator are above the horizon for less than 12 hours because you see less than half of their total 24-hour path around you. The opposite is true if you are in the southern hemisphere.
Notice that stars closer to the NCP are above the horizon longer than those farther away from the NCP. Those stars within an angular distance from the NCP equal to the observer’s latitude are above the horizon for 24 hours—they are circumpolar stars. Also, those stars close enough to the SCP (within a distance = observer’s latitude) will never rise above the horizon. They are also called circumpolar stars.
To warm Rudolph’s frozen nose, Santa heads down to the equator (0 degrees latitude). At the equator, you see the celestial equator arcing from exactly east to the zenith to exactly west. The NCP is on your northern horizon. At the equator you see one-half of every star’s total 24-hour path around you so all stars are up for 12 hours. All of the stars rise and set perpendicular to the horizon (at an angle = 90 – 0 = 90 degrees).
Continuing southward you see the NCP disappear below the horizon and the SCP rise above the southern horizon one degree for every one degree of latitude south of the equator you go. The arc of the celestial equator moves to the north, but the arc still intercepts the horizon at the exactly east/west points.
Here is a summary of the positions of the celestial reference marks (note that “altitude” means the number of degrees above the horizon):
- Meridian always goes through due North, zenith, and due South points.
- Altitude of zenith = 90° (straight overhead) always.
- Altitude of celestial pole = observer’s latitude. Observers in northern hemisphere see NCP; observers in southern hemisphere see SCP.
- Altitude of celestial equator on meridian = 90 minus the observer’s latitude.
- Celestial equator always intercepts horizon at exactly East and exactly West points.
- Angle celestial equator (and any star path) makes with respect to the horizon = 90 minus the observer’s latitude.
- Stars move parallel to the celestial equator.
- Circumpolar object’s distance from celestial pole = observer’s latitude.
(You’ll want your web browser to fill most of your screen!)
Check out the Rotating Sky module from the University of Nebraska-Lincoln’s Astronomy Education program to explore the connection between your position on the Earth and the path of the stars (link will appear in a new window—choose the “Rotating Sky” part).
- celestial equator
- North Celestial Pole
- South Celestial Pole
Review Questions 3
- How do the positions of the celestial equator, celestial poles, zenith, and meridian depend on the latitude of the observer?
- Would their position with respect to the horizon change if the Earth were only 200 miles in diameter? How about 80,000 miles in diameter? Why is that?
- During a night, how do the stars move? What angle does their nightly path make with respect to the horizon? How does it depend on latitude?
- What reference point is a celestial object on when it is at its highest position above the horizon?
- Why do observers in the northern hemisphere see celestial objects above the celestial equator for more than 12 hours?
- For northern hemisphere observers, which celestial object would be above the horizon for the greatest amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is 70° above the celestial equator, or one that is 40° below the celestial equator? Which one would be above the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.
3.5 Motion of Our Star the Sun
Now that you have your bearings, let’s take a look at the position and motion of the closest star to us, the Sun. Every day the Sun rises in an easterly direction, reaches maximum height when it crosses the meridian at local noon, and sets in a westerly direction and it takes the Sun on average 24 hours to go from noon position to noon position the next day. The “noon position” is when the Sun is on the meridian on a given day. Our clocks are based on this solar day. The exact position on the horizon of the rising and setting Sun varies throughout the year (remember though, the celestial equator always intercepts the horizon at exactly East and exactly West). Also, the time of the sunrise and sunset changes throughout the year, very dramatically so if you live near the poles, so the solar day is measured from “noon to noon.”
The Sun appears to drift eastward with respect to the stars (or lag behind the stars) over a year’s time. It makes one full circuit of 360 degrees in 365.24 days (very close to 1 degree or twice its diameter per day). This drift eastward is now known to be caused by the motion of the Earth around the Sun in its orbit.
The apparent yearly path of the Sun through the stars is called the ecliptic. This circular path is tilted 23.5 degrees with respect to the celestial equator because the Earth’s rotation axis is tilted by 23.5 degrees with respect to its orbital plane. Be sure to keep distinct in your mind the difference between the slow drift of the Sun along the ecliptic during the year and the fast motion of the rising and setting Sun during a day.
The ecliptic and celestial equator intersect at two points: the vernal (spring) equinox and autumnal (fall) equinox. The Sun crosses the celestial equator moving northward at the vernal equinox around March 21 and crosses the celestial equator moving southward at the autumnal equinox around September 22. When the Sun is on the celestial equator at the equinoxes, everybody on the Earth experiences 12 hours of daylight and 12 hours of night for those two days (hence, the name “equinox” for “equal night”). The day of the vernal equinox marks the beginning of the three-month season of spring on our calendar and the day of the autumn equinox marks the beginning of the season of autumn (fall) on our calendar. On those two days of the year, the Sun will rise in the exact east direction, follow an arc right along the celestial equator and set in the exact west direction.
When the Sun is above the celestial equator during the seasons of spring and summer, you will have more than 12 hours of daylight. The Sun will rise in the northeast, follow a long, high arc north of the celestial equator, and set in the northwest. Where exactly it rises or sets and how long the Sun is above the horizon depends on the day of the year and the latitude of the observer. When the Sun is below the celestial equator during the seasons of autumn and winter, you will have less than 12 hours of daylight. The Sun will rise in the southeast, follow a short, low arc south of the celestial equator, and set in the southwest. The exact path it follows depends on the date and the observer’s latitude.
Make sure you understand this. No matter where you are on the Earth, you will see 1/2 of the celestial equator’s arc. Since the sky appears to rotate around you in 24 hours, anything on the celestial equator takes 12 hours to go from exact east to exact west. Every celestial object’s diurnal (daily) motion is parallel to the celestial equator. So for northern observers, anything south of the celestial equator takes less than 12 hours between rise and set, because most of its rotation arc around you is hidden below the horizon. Anything north of the celestial equator takes more than 12 hours between rising and setting because most of its rotation arc is above the horizon. For observers in the southern hemisphere, the situation is reversed. However, remember, that everybody anywhere on the Earth sees 1/2 of the celestial equator so at the equinox, when the Sun is on the equator, you see 1/2 of its rotation arc around you, and therefore you have 12 hours of daylight and 12 hours of nightime everyplace on the Earth.
The geographic poles and equator are special cases. At the geographic poles the celestial equator is right along the horizon and the full circle of the celestial equator is visible. Since a celestial object’s diurnal path is parallel to the celestial equator, stars do not rise or set at the geographic poles. On the equinoxes the Sun moves along the horizon. At the North Pole the Sun “rises” on March 21st and “sets” on September 22. The situation is reversed for the South Pole. On the equator observers see one half of every object’s full 24-hour path around them, so the Sun and every other star is above the horizon for exactly 12 hours for every day of the year.
Since the ecliptic is tilted 23.5 degrees with respect to the celestial equator, the Sun’s maximum angular distance from the celestial equator is 23.5 degrees. This happens at the solstices. For observers in the northern hemisphere, the farthest northern point above the celestial equator is the summer solstice, and the farthest southern point is the winter solstice. The word “solstice” means “sun standing still” because the Sun stops moving northward or southward at those points on the ecliptic. The Sun reaches winter solstice around December 21 and you see the least part of its diurnal path all year—this is the day of the least amount of daylight and marks the beginning of the season of winter for the northern hemisphere. On that day the Sun rises at its furthest south position in the southeast, follows its lowest arc south of the celestial equator, and sets at its furthest south position in the southwest. The Sun reaches the summer solstice around June 21 and you see the greatest part of its diurnal path above the horizon all year—this is the day of the most amount of daylight and marks the beginning of the season of summer for the northern hemisphere. On that day the Sun rises at its furthest north position in the northeast, follows its highest arc north of the celestial equator, and sets at its furthest north position in the northwest.
The seasons are opposite for the southern hemisphere (e.g., it is summer in the southern hemisphere when it is winter in the northern hemisphere). The Sun does not get high up above the horizon on the winter solstice. The Sun’s rays hit the ground at a shallow angle at mid-day so the shadows are long. On the summer solstice the mid-day shadows are much shorter because the Sun is much higher above the horizon.
To check your understanding of the concepts in this section (and improve it!), go through the Motions of the Sun module of the University of Nebraska-Lincoln’s Astronomy Education program (link will appear in a new window). One section of the module will also cover solar and sidereal time that Astronomy Notes covers in a later section.
- autumnal (fall) equinox
- solar day
- summer solstice
- vernal (spring) equinox
- winter solstice
Review Questions 4
- How does the Sun move with respect to the stars during the year?
- Why does everyone have 12 hours of daylight on the equinoxes?
- Why is the length of daylight in the northern hemisphere so short on December 21?
- When will the Sun be at its highest altitude in the year in Los Angeles or Seattle? How about Singapore (on the Equator)? Why?
- On what date is the Sun above the horizon the shortest amount of time for the Southern Hemisphere? Why?
Early astronomy concentrated on finding accurate positions of the stars and planets. This was due in part to the influence of astrology, but later, accurate positions came to be important for determining the physical characteristics of the stars and planets. Accurate positions for the stars was also crucial for commercial and military navigation (navigation by the stars has only recently been replaced by the use of satellite systems such as the Global Positioning System). But probably of more importance to you is where to point your telescope or binoculars to find that cool object talked about in the newspaper or astronomy magazine.
There are a couple of popular ways of specifying the location of a celestial object. The first is what you would probably use to point out a star to your friend: the altitude-azimuth system. The altitude of a star is how many degrees above the horizon it is (anywhere from 0 to 90 degrees). The azimuth of a star is how many degrees along the horizon it is and corresponds to the compass direction.
Azimuth starts from exactly North = 0 degrees azimuth and increases clockwise: exactly East = 90 degrees, exactly South = 180 degrees, exactly West = 270 degrees, and exactly North = 360 degrees = 0 degrees. For example, a star in the southwest could have an azimuth between 180 degrees and 270 degrees. Since stars change their position with respect to your horizon throughout the night, their altitude-azimuth position changes. Also, observers at different locations looking at the same star at the same time will see it at a different altitude-azimuth position. A concise summary of this coordinate system and the numbers involved is given at the end of this section.
The second way of specifying star positions is the equatorial coordinate system. This system is very similar to the longitude-latitude system used to specify positions on the Earth’s surface. This system is fixed with respect to the stars so, unlike the altitude-azimuth system, a star’s position does not depend on the observer’s location or time. Because of this, astronomers prefer using this system. You will find this system used in astronomy magazines and in most sky simulation computer software.
Selecting the image link will bring up a short animation of a spinning celestial sphere. The lines on a map of the Earth that run north-south are lines of longitude and when projected onto the sky, they become lines of right ascension. Because the stars were used to measure time, right ascension (RA) is measured in terms of hours, minutes, and seconds instead of degrees and increases in an easterly direction. For two stars one hour of RA apart, you will see one star cross your meridian one hour of time before the other. If the stars are not circumpolar, you will see one star rise one hour before the other. If they were 30 minutes of RA apart, you would see one rise half an hour before the other and cross your meridian half an hour before the other. Zero RA is where the Sun crosses the celestial equator at the vernal equinox. The full 360 degrees of the Earth’s rotation is broken up into 24 hours, so one hour of RA = 15 degrees of rotation. The lines of RA all converge at the celestial poles so two stars one hour of RA apart will not necessarily be 15 degrees in angular separation on the sky (only if they are on the celestial equator will they be 15 degrees apart).
The lines on a map of the Earth that run east-west parallel to the equator are lines of latitude and when projected onto the sky, they become lines of declination. Like the latitude lines on Earth, declination (dec) is measured in degrees away from the celestial equator, positive degrees for objects north of the celestial equator and negative degrees for objects south of the celestial equator. Objects on the celestial equator are at 0 degrees dec, objects half-way to the NCP are +45 degrees, objects at the NCP are +90 degrees, and objects at the SCP are -90 degrees. Polaris’s position is at RA 2hr 31min and dec 89 degrees 15 arc minutes. A concise summary of this coordinate system and the numbers involved is given at the end of this section.
The Basic Coordinates module of the University of Nebraska-Lincoln’s Astronomy Education program provides a great way to make the connection between terrestrial coordinates (longitude and latitude) and the equatorial coordinate system (link will appear in a new window). The first part of the module has you drag a cursor around on a flat world map or globe and read off its terrestrial coordinate position. The second part of the module has you do the same sort of thing using a flat map of the sky or a globe of the celestial sphere and read off the right ascension and declination. Both parts also illustrate the distortion that happens when you project a curved spherical surface onto a flat two-dimensional map.
The UNL Astronomy Education’s Rotating Sky module has you explore the connection between the two coordinate systems. You can change your location on the Earth and adjust the position of multiple stars and see where the stars would appear and how they would move on the celestial sphere and around your position on the Earth as the Earth rotates beneath the stars.
An effect called precession causes the Sun’s vernal equinox point to slowly shift westward over time, so a star’s RA and dec will slowly change by about 1.4 degrees every century (a fact ignored by astrologers), or about 1 minute increase in a star’s RA every twenty years. This is caused by the gravitational pulls of the Sun and Moon on the Earth’s equatorial bulge (from the Earth’s rapid rotation) in an effort to reduce the tilt of the Earth’s axis with respect to the ecliptic and the plane of the Moon’s orbit around the Earth (that is itself slightly tipped with respect to the ecliptic). Like the slow wobble of a rapidly-spinning top, the Earth responds to the gravitational tugs of the Sun and Moon by slowly wobbling its rotation axis with a period of 26,000 years.
This motion was first recorded by Hipparchus in 100 B.C.E. who noticed differences between ancient Babylonian observations and his own. When the Babylonians were the world power in 2000 B.C.E., the vernal equinox was in the constellation Aries and the star Thuban (in Draco) was the closest bright star to the NCP. At the time of Jesus Christ the vernal equinox had shifted to the constellation Pisces and the star Kochab (in the bowl of the Little Dipper) was the closest bright star to the NCP. Now the star Polaris is close to the NCP and the vernal equinox is close to the border between Pisces and Aquarius (in 2600 C.E. it will officially be in Aquarius) which is what a popular song of years ago refers to with the line “this is the dawning of the Age of Aquarius.” In the year 10,000 C.E., the bright star in the tail of Cygnus, Deneb, will be the pole star and Vega (in Lyra) will get its turn by the year 14,000 C.E. Horoscopes today are still based on the 4,000-year old Babylonian system so even though the Sun is in Aries on my birthday, the zodiac sign used for my horoscope is Taurus. I guess it’s hard to keep up with all of the changes in the modern world!
Star Chart sites
(These will appear in another window)
- National Geographic Society’s Star Chart with Hubble image enhancements. Select a particular section of the sky on the image map. Locations of objects imaged by Hubble are hyper-linked on the section you selected. The grid lines are lines of right ascension and declination.
- Interactive Star Chart. Choose a constellation as your starting point for this viewer of the skies. Your browser must be java-enabled.
- Your Sky is a free interactive planetarium simulator on the web. Use it to create a sky map of the entire sky or a horizon view as seen from your location on the Earth.
- Monthly evening sky maps are available for free from Skymap.com. The star charts show the entire sky with a white sky background and black stars and include a calendar of astronomical events for the month.
- Sky and Telescope‘s Interactive Sky Chart (login required) is another free star chart service. Create a chart of the entire sky or a horizon view from your location on the Earth. You can also create a PDF of the chart for printing that has a white sky and black stars.
- right ascension
- Altitude varies from 0 to 90°. Vertical position of object.
- Azimuth varies from 0° to 360°. Exact N = 0°, exact E = 90°, exact S = 180°, exact W = 270°. Horizontal position of object.
- Right ascension varies from 0 to 24 hours, so every hour corresponds to a rotation angle of 15°. Horizontal position of object measured in time units.
- Declination varies from -90° (at SCP) to +90° (at NCP). Celestial equator declination = 0°. Vertical position of object.
- Meridian altitude of any object = 90 – (observer’s latitude) + declination degrees. If declination is negative, then addition of declination becomes a subtraction.
Formulae for Sun’s position
- Ecliptic tilted by 23.5° with respect to the celestial equator.
- Sun’s declination ranges between -23.5° and +23.5°.
- Vernal equinox right ascension = 0 hours; declination = 0°; Sun rises at 90° azimuth and sets at 270° azimuth.
- June solstice right ascension = 6 hours; declination = +23.5°; Sun rises at less than 90° azimuth and sets at greater than 270° azimuth.
- Autumnal equinox right ascension = 12 hours; declination = 0°; Sun rises at 90° azimuth and sets at 270° azimuth.
- December solstice right ascension = 18 hours; declination = -23.5° Sun rises at greater than 90° azimuth and sets at less than 270° azimuth.
Review Questions 5
- At what two azimuths does the celestial equator intercept the horizon?
- If a star’s position at 10 pm is 110° azimuth and 40° altitude, will its azimuth be greater or less at 11 pm? If the star is still east of the meridian at 11 pm, will its altitude be greater or less than it was at 10 pm? First assume you are in the northern hemisphere. Explain your answer. Then assume you are in the southern hemisphere and explain your answer.
- Why do astronomers prefer using right ascension and declination?
- What is the azimuth of any object when it crosses the meridian at any time of year in the southern sky?
- If a star has a RA of 5 hours and crosses the meridian at 10:45 pm, what is the RA of a star that crosses the meridian at 1:00 am? Explain your answer.
- What is the Sun’s altitude when it crosses the meridian in your home town and its declination is +23.5°?
- What is the altitude of the NCP at Fairbanks, Alaska (lat. = 65° N)?
- How do the positions of the equinoxes and solstices with respect to the horizon depend on the latitude?
- What is the maximum altitude of the Sun on the vernal equinox for people on the equator? What is the Sun’s right ascension at that time?
- What will the Sun’s declination be on the following dates: June 21, March 21, September 22, and December 21?
- If the Sun sets 10° away from due West on October 20, what is the sunset azimuth?
- If the Sun rises 12° away from due East on April 19, what is the sunrise azimuth?
- What causes precession?
- How does precession affect the positions of the stars?
- If a star on the celestial equator has a RA of 5 hours 33 minutes, what would you estimate its RA to be in 20 years and in 200 years? Explain your answer. (Remember that the Earth spins about 15°/hour.)
- Which star is the current pole star? Which star was the pole star 2,000 years ago? Which star will be the pole star 8,000 years from now?
- Are modern horoscopes based on the current motion of the Sun and planets with respect to stars?
The fact that our clocks are based on the solar day and the Sun appears to drift eastward with respect to the stars (or lag behind the stars) by about 1 degree per day means that if you look closely at the positions of the stars over a period of several days, you will notice that according to our clocks, the stars rise and set 4 minutes earlier each day. Our clocks say that the day is 24 hours long, so the stars move around the Earth in 23 hours 56 minutes. This time period is called the sidereal day because it is measured with respect to the stars. This is the true rotation rate of the Earth and stays the same no matter where the Earth is in its orbit—the sidereal day = 23 hours 56 minutes on every day of the year. One month later (30 days) a given star will rise 2 hours earlier than it did before (30 days × 4 minutes/day = 120 minutes). A year later that star will rise at the same time as it did today.
Another way to look at it is that the Sun has made one full circuit of 360 degrees along the ecliptic in a year of 365.24 days (very close to 1 degree per day). The result is that between two consecutive meridian crossings of the Sun, the Earth has to turn nearly 361 degrees, not 360 degrees, in 24 hours. This makes the length of time for one solar day to be a little more than the true rotation rate of 23 hours 56 minutes with respect to the background stars.
3.7.2 Solar and Sidereal Time as Viewed from Space
Let’s jump to a more modern view and take a position off the Earth and see the Earth revolving around the Sun in 365.24 days and rotating on its axis every 23 hours 56 minutes. The Earth’s rotation plane is tilted by 23.5 degrees from its orbital plane which is projected against the background stars to form the ecliptic.
Note that the Earth’s rotation axis is always pointed toward the Celestial Poles. Currently the North Celestial Pole is very close to the star Polaris. The figure above shows this view of the Earth’s nearly circular orbit from slightly above the orbital plane (hence, the very elliptical appearance of the orbit).
Imagine that at noon there is a huge arrow that is pointing at the Sun and a star directly in line behind the Sun. The observer on the Earth sees the Sun at its highest point above the horizon: on the arc going through the north-zenith-south points, which is called the meridian. The observer is also experiencing local noon. If the Sun were not there, the observer would also see the star on the meridian.
Now as time goes on, the Earth moves in its orbit and it rotates from west to east (both motions are counterclockwise if viewed from above the north pole). One sidereal period later (23 hours 56 minutes) or one true rotation period later, the arrow is again pointing toward the star. The observer on the Earth sees the star on the meridian. But the arrow is not pointing at the Sun! In fact the Earth needs to rotate a little more to get the arrow lined up with the Sun. The observer on the Earth sees the Sun a little bit east of the meridian. Four minutes later or one degree of further rotation aligns the arrow and Sun and you have one solar day (24 hours) since the last time the Sun was on the meridian. The geometry of the situation also shows that the Earth moves about 1 degree in its orbit during one sidereal day. That night the Earth observer will see certain stars visible like those in Taurus, for example. (Notice that the Earth’s rotation axis is still pointed toward Polaris.) A half of a year later Taurus will not be visible but those stars in Scorpius will be visible. (Again, notice that the Earth’s rotation axis is still pointed toward Polaris.) The extra angle any planet must rotate on its axis to get the Sun back to the meridian equals the angle amount the planet moved in its orbit in one sidereal day.
The amount of time it takes to spin the extra angle = (extra angle amount)/(spin rate). For the Earth, the spin rate = 360°/23.9333 hours = 15°/hour or 1°/4 minutes. Notice that I converted 23 hours 56 minutes to a decimal fraction of hours before I did the division. The amount of time between the solar day and sidereal day = (1 degree)/(1 degree/4 minutes) = 4 minutes.
The Earth’s sidereal day is always 23 hours 56 minutes long because the number of degrees the Earth spins through in a given amount of time stays constant. If you are a careful observer, you will notice that the solar day is sometimes slightly longer than 24 hours and sometimes slightly shorter than 24 hours during the year. The reason for this is that the Earth’s orbit around the Sun is elliptical and that the Sun’s motion is not parallel to the celestial equator. The effects of this are explained fully in the Equation of Time section below. The value of 24 hours for the solar day is an average for the year and is what our time-keeping system is based on.
The precession of the Earth’s rotation axis introduces another difference between sidereal time and solar time. This is seen in how the year is measured. A year is defined as the orbital period of the Earth. However, if you use the Sun’s position as a guide, you come up with a time interval about 20 minutes shorter than if you use the stars as a guide. The time required for the constellations to complete one 360° cycle around the sky and to return to their original point on our sky is called a sidereal year. This is the time it takes the Earth to complete exactly one orbit around the Sun and equals 365.2564 solar days.
The slow shift of the star coordinates from precession means that the Sun will not be at exactly the same position with respect to the celestial equator after one sidereal year. The tropical year is the time interval between two successive vernal equinoxes. It equals 365.2422 solar days and is the year our calendars are based on. After several thousand years the 20 minute difference between sidereal and tropical years would have made our summers occur several months earlier if we used a calendar based on the sidereal year.
- local noon
- sidereal day
- sidereal year
- tropical year
Review Questions 6
- Which is used for our clocks—sidereal day or solar day? Why?
- Why is there a difference between the sidereal day and solar day?
- If the Earth rotated twice as quickly as it does now, what would be the difference in minutes between the solar and sidereal days? Explain how you got your answer.
- If the Earth’s year was twice as long as it is now, what would be the difference in minutes between the solar and sidereal days? Explain!
- Mars rotates once every 24.623 hours (its sidereal day) and it orbits the Sun once every 686.98 solar earth days. Show how to find out how long a solar day is on Mars.
3.7.3 Time Zones
People east of you will see the Sun on their meridian before you see it on yours. Those in Denver, Colorado will see the Sun on their meridian about 52 minutes before people in Los Angeles will see the Sun on their meridian. Denver residents experience local noon about 52 minutes before those in Los Angeles. That is because Denver is at longitude 105° West longitude while Los Angeles is at 118° West longitude (or 13° difference). For each one degree difference in longitude a person is from you, the time interval between his local noon and yours will increase by 4 minutes.
It used to be that every town’s clocks were set according to their local noon and this got very confusing for the railroad system so they got the nation to adopt a more sensible clock scheme called time zones. Each person within a time zone has the same clock time. Each time zone is 15 degrees wide, corresponding to 15 degrees × 4 minutes/degree = 60 minutes = 1 hour worth of time. Those in the next time zone east of you have clocks that are 1 hour ahead of yours. The Pacific timezone is centered on 120° W longitude, the Mountain timezone is centered on 105° W longitude, etc.
3.7.4 Equation of Time
There is a further complication in that the actual Sun’s drift against the stars is not uniform. Part of the non-uniformity is due to the fact that on top of the general eastward drift among the stars, the Sun is moving along the ecliptic northward or southward with respect to the celestial equator. Thus, during some periods the Sun appears to move eastward faster than during others. Apparent solar time is based on the component of the Sun’s motion parallel to the celestial equator. This effect alone would account for as much as 9 minutes difference between the actual Sun and a fictional mean Sun moving uniformly along the celestial equator.
|Another effect to consider is that the Earth’s orbit is elliptical so when the Earth is at its closest point to the Sun (at perihelion), it moves quickest. When at its farthest point from the Sun (at aphelion), the Earth moves slowest. Remember that a solar day is the time between meridian passages of the Sun. At perihelion the Earth is moving rapidly so the Sun appears to move quicker eastward than at other times of the year. The Earth has to rotate through a greater angle to get the Sun back to local noon. This effect alone accounts for up to 10 minutes difference between the actual Sun and the mean Sun.|
However, the maximum and minimum of these two effects do not coincide so the combination of the two (called the Equation of Time) is a complicated relation shown in the figure below. The Equation of Time explains why, according to your clock, the earliest sunset and latest sunrise is not at the winter solstice. Yet, the shortest day is at the winter solstice. Rather than resetting our clocks every day to this variable Sun, our clocks are based on a uniformly moving Sun (the mean Sun) that moves along the celestial equator at a rate of 360 degrees/365.2564 per day. Aren’t you glad that your watch keeps track of time for you?
The seasonal temperature depends on the amount of heat received from the Sun in a given time. To hold the temperature constant, there must be a balance between the amount of heat gained and the amount radiated to space. If more heat is received than is lost, your location gets warmer; if more heat is lost than is gained, your location gets cooler. What causes the amount of energy reaching a given location during the day to change throughout the year?
Two popular theories are often stated to explain the temperature differences of the seasons: 1) the different distances the Earth is from the Sun in its elliptical orbit (at perihelion the Earth is 147.1 million kilometers from the Sun and at aphelion the Earth is 152.1 million kilometers from the Sun); and 2) the tilt of the Earth’s axis with respect to its orbital plane. If the first theory were true, then both the north and south hemispheres should experience the same seasons at the same time. They do not. Using the scientific method discussed in chapters 1 and 2, you can reject the distance theory.
A popular variation of the distance theory says that the part of the Earth tilted toward the Sun should be hotter than the part tilted away from the Sun because of the differences in distances. If you continue along with this line of reasoning, then you conclude that the night side of the Earth is colder than the daylight side because the night side is farther away from the Sun. This ignores the more straightforward reason that the night side is directed opposite the Sun, so the Sun’s energy does not directly reach it. But let’s examine the tilt-distance model a little more. The 23.5° tilt of the Earth means that the north pole is about 5080 kilometers closer than the south pole toward the end of June. This is much, much smaller than the 152 million kilometer distance between the Sun and the Earth’s center at that time. The amount of energy received decreases with the square of the distance.
If you calculate (152,000,000 + 5080)2/(152,000,000 – 5080)2, you will find that the north pole would get slightly over 1/100th of one percent more energy than the south pole. This is much too small a difference to explain the large temperature differences! Even if you compare one side of the Earth with the opposite side, so you use the Earth’s diameter in place of the 5080 kilometers in the calculation above, you get 3/100th of one percent difference in energy received. Clearly, distance is not the reason for the large temperature differences. Notice that I used the aphelion value for the distance between the Earth and Sun. That is because the Earth is near aphelion during the northern hemisphere’s summer! This is known by measuring the apparent size of the Sun. You can safely assume that the Sun’s actual size does not vary with a period that depends on the orbital period of a planet thousands of times smaller than it, or that it would choose the Earth’s orbital period as its pulsation cycle.
Earth reaches perihelion in the first week of January (during the north hemisphere’s winter!) and aphelion in the first week of July (during the north hemisphere’s summer!). The distance theory predicts the opposite seasons from what’s observed in the north hemisphere. Precise dates and times for the perihelion and aphelion events can be found at the US Naval Observatory’s Applications Department’s Earth’s Seasons page (link will appear in a new window; make the appropriate time adjustment for your time zone.)
Even though the distance model (in any variation) is incorrect, it is still a “good” scientific theory in that it makes testable predictions of how the temperature should change throughout the year and by how much. However, what annoys scientists, particularly astronomy professors, is ignoring those predictions and the big conflicts between predictions and what is observed. Let’s take a look at a model that correctly predicts what is observed.
The tilt theory correctly explains the seasons but the reason is a little more subtle than the distance theory’s explanation. Because the Earth’s rotation axis is tilted, the north hemisphere will be pointed toward the Sun and will experience summer while the south hemisphere will be pointed away from the Sun and will experience winter. During the summer the sunlight strikes the ground more directly (closer to perpendicular), concentrating the Sun’s energy. This concentrated energy is able to heat the surface more quickly than during the winter time when the Sun’s rays hit the ground at more glancing angles, spreading out the energy.
Also, during the summer the Sun is above the horizon for a longer time so its energy has more time to heat things up than during the winter.
The Seasons module of the University of Nebraska-Lincoln’s Astronomy Education program enables you to understand these concepts by manipulating such things as the position of the Earth in its orbit and your position on the Earth (link will appear in a new window—choose the third part of the module). You can switch between an earth-centered view showing the Earth at the center of the celestial sphere with the Sun traveling along the ecliptic and a Sun-centered view showing the Earth moving around the Sun. Both views show how the amount of daylight and the angle of sunlight upon the ground change with the passing days and the location on the Earth.
The rotational axes of most of the other planets of the solar system are also tilted with respect to their orbital planes so they undergo seasonal changes in their temperatures too. The planets Mercury, Jupiter, and Venus have very small tilts (3° or less) so the varying distance they are from the Sun may play more of a role in any seasonal temperature variations. However, of these three, only Mercury has significant differences between perihelion and aphelion. Its extremely thin atmosphere is not able to retain any of the Sun’s energy. Jupiter’s and Venus’ orbits are very nearly circular and their atmospheres are very thick, so their temperature variations are near zero.
Mars, Saturn, and Neptune have tilts that are similar to the Earth’s, but Saturn and Neptune have near zero temperature variation because of their very thick atmospheres and nearly circular orbits. Mars has large temperature changes because of its very thin atmosphere and its more eccentric orbit places its southern hemisphere closest to the Sun during its summer and farthest from the Sun during its winter. Mars’ northern hemisphere has milder seasonal variation than its southern hemisphere because of this arrangement. Since planets move slowest in their orbits when they are furthest from the Sun, Mars’ southern hemisphere has short, hot summers and long, cold winters.
Uranus’ seasons should be the most unusual because it orbits the Sun on its side—its axis is tilted by 98 degrees! For half of the Uranian year, one hemisphere is in sunlight and the other is in the dark. For the other half of the Uranian year, the situation is reversed. The thick atmosphere of Uranus distributes the solar energy from one hemisphere to the other effectively, so the seasonal temperature changes are near zero. Pluto’s axis is also tilted by a large amount (122.5 degrees), its orbit is the most elliptical of the planets, and it has an extremely thin atmosphere. But it is always so far from the Sun that it is perpetually in deep freeze (only 50 degrees above absolute zero!).
- Equation of Time
- mean Sun
- time zone
Review Questions 7
- How many minutes difference is there between local noon in Seattle, WA and somebody’s local noon 3 degrees longitude east of Seattle? Will their local noon occur before or after Seattle’s?
- The Eastern Standard Time zone is 3 hours ahead of the Pacific Standard Time zone. The Pacific timezone meridian is at 120° W longitude. What is the longitude of the Eastern Time zone meridian?
- Is the Sun’s drift eastward greater at the solstices or the equinoxes? Why is that?
- Is the solar day longer at perihelion or aphelion? Why is that?
- What causes the temperature differences between the seasons? How so?
- If you shine a flashlight on a flat tabletop, which gives you a smaller concentrated beam: one directed perpendicular to the tabletop or one directed parallel to the tabletop? Which one produces the longer shadow of a pencil on the tabletop?
- How would the fact that the Sun’s angular size is largest around January 4 contradict the popular theory that the Earth’s distance from the Sun in its elliptical orbit causes the seasons?
3.8 Motions of the Moon
The Moon moves rapidly with respect to the background stars. It moves about 13 degrees (26 times its apparent diameter) in 24 hours—slightly greater than its own diameter in one hour! The image below illustrates this rapid motion as well as the fact that you can see the Moon during the day. (Select the image to bring up an enlarged version.) Closer inspection of the image shows that the first image is a waning gibbous phase one day past full phase and the sequence progresses toward less of the visible part of the Moon being visible as the Moon moves closer to the Sun as seen from the Earth. Among other things, this section will explain why these images had to taken in the early morning and why I had to be facing toward the west-southwestern direction with the Sun more or less behind me.
Its rapid motion has given it a unique role in the history of astronomy. For thousands of years it has been used as the basis of calendars. Isaac Newton got crucial information from the Moon’s motion around the Earth for his law of gravity.
If you watch the Moon throughout the year, you will see the same face of the Moon all of the time. It is the “man in the moon”, “woman in the moon”, “rabbit in the moon” etc. One thing this shows you is that the Moon turns exactly once on its axis each time that it goes around the Earth. Later on you will find out how tidal forces have caused this face-to-face dance of the Earth and Moon. The Moon drifts eastward with respect to the background stars (or it lags behind the stars). It returns to the same position with respect to the background stars every 27.323 days. This is its sidereal period.
- sidereal period
Review Questions 8
- How does the Moon move with respect to the stars?
- How does the fact that we always see one side of the Moon prove that the Moon rotates once every orbital period?
- In a particular year the Moon is in the constellation Aries on June 1st. What date will it be in Aries the next time?
- Why did the Apollo missions to the Moon always have landings on the same side of the Moon? (Requires some thought and/or research.)
3.8.1 Phases and Eclipses
One of the most familiar things about the Moon is that it goes through phases from new (all shadow) to first quarter (1/2 appears to be in shadow) to full (all lit up) to third quarter (opposite to the first quarter) and back to new. This cycle takes about 29.53 days. This time period is known as the Moon’s synodic period. Because the Moon moves through its phases in about four weeks, the phases of new moon, first quarter, full moon, third quarter occur nearly one week apart from each other.
Select this link to find the phase for any date and time between 1800–2199 (will display in another window). A picture of the Moon will be shown.
The phases are due to how the Sun illuminates the Moon and the relative positioning of the Earth, Moon, and Sun. The figure below shows that as the Moon orbits the Earth, the fraction of its illuminated side that you can see from the Earth changes. From high above the Earth and Moon orbit, you can see that the Moon is always half lit by the Sun and the lit half (the illuminated side, or day side) always faces the light source—the Sun. The other half (night side) faces away from the Sun. The figure below combines two view points. The half-lit moon on the inner circle around the Earth is the Moon as viewed from high above the Earth and Moon orbit. The outer ring of Moon pictures in various phases is the view of the Moon as we would see it from the Earth. Of course, this drawing is not to scale.
Select here for a nice simulation of the Moon phases (will display in another window). Be sure to choose “both” in the point of view pop-up list.
You will observe only a small fraction of the Moon’s illuminated side when it is close to the Sun. In fact, the smaller the angular distance between the Moon and the Sun, the less of its illuminated side you see. When the angular distance is less than 90° separation, you on the Earth will see less than half of the Moon’s illuminated (day) side and it will look like a a curved sliver of light—the crescent phase. You will see mostly the night side of the Moon. Because the Moon is spherical, the boundary between light and shadow (night) is curved. Note that the figure of the phase angles above shows just one of the angles possible for the crescent phase (at 45°). When the angle between Sun and Moon is within about 6 degrees you on the Earth see it in a new phase and it is the beginning of the phase cycle. Sometimes that angle = 0 degrees and you have a solar eclipse—the moon is in new phase and it is covering up the Sun. The Earth’s shadow always points directly away from the Sun (and it tapers down to a point). At new phase the Moon is in the same direction as the Sun as seen from the Earth, so the Earth’s shadow cannot be why you just see the night side of the Moon.
The sequence of pictures below shows a technique you can use to figure out how a Moon phase will look from the Earth when you start from an “orrery” or space view of the Moon, i.e., how to translate from the orrery (space) view of the Moon in its orbit around the Earth to the ground-based view. To show the technique, let’s use the Waxing Crescent example.
At 90° angular separation from the Sun, you on the Earth see half of the Moon’s illuminated (day) side and half of its night side. The phase is called a quarter phase because you can see a quarter of the Moon’s entire surface (and 90 degrees is one-quarter of 360 degrees). The quarter phase a week after the new phase is called first quarter.
The greater the angular distance is between the Moon and the Sun, the more of the Moon’s illuminated side you can see from the Earth. When the angular distance between Sun and Moon is more than 90° separation, you on the Earth will see more than half of the Moon’s illuminated (day) side—the gibbous phase. You will see a small amount of the Moon’s night side. “Gibbous” means a shape that is convex (bulges outward) at both sides. Again note that the figure shows just one of the angles possible for the gibbous phase (at 135°). Also, note again that the Moon is always half lit up by the Sun. How much of the lit side (day side) we see from the Earth depends on where the Moon is in its orbit around the Earth.
Around 180° angular separation from the Sun, you on the Earth see the entire illuminated (day) side of the Moon—the full phase. Sometimes (about twice a year) the Sun-Moon angle is exactly 180 degrees and you see the Earth’s shadow covering the Moon—a lunar eclipse. Sometimes a descriptive term is added to the crescent and gibbous phases. If the amount of illuminated side you can see increases with time, it is waxing as in waxing crescent or waxing gibbous. The daylit side of the Moon will be facing toward the west (toward the right for observers in the northern hemisphere and toward the left for southern hemisphere observers). If the illuminated fraction decreases with time, it is waning as in waning crescent or waning gibbous. The daylit side of the Moon will be facing toward the east (toward the left for observers in the northern hemisphere). Readers in the southern hemisphere need to reverse “left” and “right” in the figure below.
Let’s do one last example of translating from the orrery (space) view of the Moon’s position in its orbit to the Earth (ground) view with the Moon in a Waning Gibbous position. Here are the sequence of steps in the figure below.
You can use the illustration of the lunar phases at the top to find out the time of day when the Moon will be visible. The Sun is at the right of the figure so a person at position (A) on the Earth (e.g., Los Angeles, CA) sees the Sun on the meridian. When an object is on the meridian, your part of the Earth is pointing directly toward that object. The Earth rotates in the counterclockwise direction (A to B to C to D). A person at position (B) (e.g., Sao Mateus in the Azores) sees the Sun setting since he is one-quarter turn (6 hours) ahead of the person at position (A). The person at position (C) (e.g., Zahedan, Iran) is at the midnight position (half a turn, 12 hours, ahead of position (A)) and the person at position (D) (e.g., Sydney, Australia) is experiencing sunrise (three-quarters of a turn, 18 hours, ahead of position (A)). If the Moon was at its new phase position, person (D) would see the new moon rising, person (A) would see the new moon on the meridian, and person (B) would see the new moon setting within a few minutes of sunset.
If the Moon was at its first quarter position, person (A) would see the Moon beginning to rise, person (B) would see the Moon on his meridian at sunset, and person (C) would the first quarter moon setting because it is already midnight at her position. The figure below illustrates these views.
Using the same method, you can see that the full moon is rising for person (B) at sunset, is on the meridian at midnight for person (C) opposite the Sun, and is setting for the person (D) at sunrise. Now try to figure out when the third quarter moon will rise, cross the meridian, and set using this method. Remember that each of the persons A, B, C, D are each six hours apart from each other. Also, remember person (D) is facing down in the orrery view on the left side of the figure above and if the person is facing south, the Sun will be rising on his left.
If you are having a hard time visualizing this, try using a white ball (e.g., a styrofoam ball) for the Moon, a bright light bulb for the Sun, and your head for the Earth in a room shut off from other lights. When your eyes are facing the bulb, that would be noon. While facing the bulb, move the ball to your left ear so half of it is lit up. That is the first quarter phase. If you move your head counterclockwise 90° so you are facing the half-lit ball, you will see the bulb out of the corner of your right eye (in the “west” direction). That would be sunset. Move the ball around so it is opposite the bulb but out of the shadow of your head. You should see all of it lit up—a full phase. If you face the same direction that you faced the half-lit ball, the full phase ball would be visible out of the corner of your left eye (in the “east” direction). As you turn your head counterclockwise, you will see the ball “rise” and the bulb “set”. When you face the full-lit ball, that would be midnight. How would you simulate a third quarter phase?
The table gives a summary of approximately when the Moon is visible and where to look (the crescent and gibbous phases are in between the table values). You may be surprised to find out that the Moon is sometimes visible in broad daylight!
|Phase||Time the Moon is
ahead/behind the Sun
|Moon Crosses Meridian
|New||within few minutes||Sunrise||Noon||Sunset|
|First Quarter||6 hrs behind||Noon||Sunset||Midnight|
|Full||12 hrs behind||Sunset||Midnight||Sunrise|
|Third Quarter||6 hrs ahead||Midnight||Sunrise||Noon|
The phase diagram seems to show that a solar and lunar eclipse should happen every month but eclipses actually happen only twice a year. You can see why if you look at the Moon’s orbit from close to edge-on. The Moon’s orbit is tilted by 5 degrees with respect to the Earth’s orbital plane (the ecliptic). In order for an eclipse to occur, the Moon must be in the ecliptic plane AND exactly at the new or full phase. Usually, the Moon crosses the ecliptic plane at another phase instead of exactly at new or full phase during its approximately month-long orbit around the Earth.
During a year the Moon’s orbit is oriented in very nearly the same direction in space. The position of the Earth and Moon with respect to the Sun changes while the Moon’s orbit direction is approximately fixed. So in one month the Moon will be below the ecliptic at full phase and above the ecliptic at full phase about six months later. Though the Moon crosses the ecliptic twice a month, an eclipse will happen only when it is exactly at full or new phase when it crosses the ecliptic. The tilt of the Moon’s orbit explains why eclipses happen only twice a year.
The direction of the Moon’s orbit slowly shifts (precesses) over time. Because the Moon’s orbit precesses, eclipses will occur on different dates in successive years. However, even if there was no precession, eclipses would still happen only twice a year. The figure above shows another complication—the elliptical orbit of the Moon around the Earth means that the new moon can occur at different distances from the Earth and the Moon’s shadow may not reach the Earth if it is too far away.
Why are the synodic and sidereal periods not equal to each other? For a reason similar to the reason why the solar day and sidereal day are not the same. Remember that a solar day was slightly longer than a sidereal day because of the Sun’s apparent motion around the Earth (which is really due to the Earth’s motion around the Sun). The Sun’s eastward drift against the stars also means that the Moon’s synodic period is longer than its sidereal period.
At new moon, the Sun and Moon are seen from the Earth against the same background stars. One sidereal period later, the Moon has returned to the same place in its orbit and to the same place among the stars, but in the meantime, the Sun has been moving eastward, so the Moon has not yet caught up to the Sun. The Moon must travel a little over two more days to reach the Sun and establish the new moon geometry again.
The modern model has the Moon going around the Earth with the Sun far away. At different positions in its orbit you see different phases all depending on the relative positions of the Earth-Moon-Sun. Another possible model was presented by highly-esteemed Harvard seniors at their graduation. They seriously proposed that the dark part of the Moon is the result of portions of the Moon lying in the shadow of the Earth. Many other people have also explained the phases with this Earth shadow model, but I will call this the “Harvard model” below.
Since the Moon would need to be opposite the Sun for it to be in the Earth’s shadow, the “Harvard model” predicts Sun-Moon angles that are very different from the observed angles. In addition, the model predicts that the Moon would need to be one-half a rotation (or 12 hours) away from the Sun. The Moon should rise 12 hours after sunrise (i.e., at sunset), cross the meridian 12 hours after the Sun, and set 12 hours after sunset (i.e., at sunrise) for all of the phases except full. How is this different from what is observed?
Test and improve your understanding of the lunar phases with the UNL Astronomy Education program’s Lunar Phase Simulator. A space view from high above the Earth showing the Moon’s orbit and the on Earth view are used to show how the Sun-Moon geometry gives rise to the phases (and how the Earth’s shadow cannot!).
- lunar eclipse
- solar eclipse
- synodic period
Review Questions 9
- Why does the Moon have phases?
- Why are New Moon phases longer than a sidereal period (27.3 days) apart from each other?
- If the Moon was full 7 nights ago, what time of day (night) should you look for the Moon to be up high in the sky in the south today? Explain your answer.
- What are the positions of the Earth-Moon-Sun during an eclipse?
- What would the Sun-Moon angular separation be for the New Moon if the Earth’s shadow caused the lunar phases? How about Gibbous phase?
- What are the real angular separations for New and Gibbous phase?
- About how much difference in time is there between moonset and sunset at first quarter phase? Does the Moon set before or after the Sun at that phase?
- About how much difference in time is there between moonset and sunset at new phase?
- If the Earth’s shadow caused the lunar phases, what would be the difference in time between moonrise and sunrise at new and first quarter phases?
- About when will the Waxing Crescent Moon be on the meridian? Explain your answer.
- The Moon is low in the western sky at sunrise, what is its phase? Explain!
- Why do we not have eclipses every month?
3.8.2 Eclipse Details: Lunar Eclipse
Let’s explore a little more about lunar and solar eclipses. Remember that an eclipse happens when an object passes through another object’s shadow. Any shadow consists of two parts: an umbra, which is the region of total shadow, and the penumbra, which is the outer region of partial shadow. If the Moon were to pass through the Earth’s umbra, an observer on the Moon would not be able to see the Sun at all—she would observe a solar eclipse! An observer on the Earth looking at the Moon would see a total lunar eclipse. The Earth’s shadow is pretty big compared to the Moon so a total lunar eclipse can last up to about 1 hour 45 minutes.
If the Moon only passed through the outer part of the shadow (the penumbra), then the observer on the Moon would see the Sun only partially covered up—a partial solar eclipse. The observer on the Earth would see the Moon only partially dimmed—a partial lunar eclipse.
Below is a sequence of images from the August 28, 2007 lunar eclipse as observed from Honolulu, Hawaii. The images were taken every five minutes through a telescope. The dark red-orange color is the color you would see if you observed the eclipse.
During a total lunar eclipse you see another interesting effect—the Moon turns a coppery (or bloody) red. The reason why some sunlight reaches the Moon despite the fact that the Moon is in the Earth’s umbra is that the sunlight refracts or bends as it passes through the Earth’s atmosphere. Dust particles in the Earth’s atmosphere remove much of the bluer colors in the sunlight so only the redder colors make it to the Moon. The amount of dust determines the deepness of the red colors. The dust in the air is also why the Sun appears redder at sunset on Earth. The observer on the Moon would see a reddish ring around the Earth even at mid-eclipse! Some nice visualizations of the lunar eclipse essentials are available at Goddard Space Flight Center’s Lunar Eclipse Essentials page.
The Moon’s shadow also has an umbra and penumbra, but its shadow is much smaller than the Earth’s. Only if the Moon is in the ecliptic plane when it is exactly New Moon will the Moon’s shadow hit the Earth. Where the umbra hits the Earth, you will see a total solar eclipse. Where the penumbra hits the Earth, you will see a partial solar eclipse.
Select the image to get information about this image of the July 11, 1991 solar eclipse taken by Fred Espenak (will display in another window).
In a total solar eclipse the bright disk of the Sun is completely covered up by the Moon and you can see the other parts of the Sun like the corona, chromosphere, and prominences. Unfortunately, only the tip of the Moon’s umbra reaches the Earth (the tip hitting the Earth is at most 270 kilometers [168 miles] in diameter) and it zips along the Earth’s surface at over 1600 kph (1000 mph) as the Moon moves around the rotating Earth. This means that a total solar eclipse can last a maximum of only 7.5 minutes. Usually total solar eclipses last only 2-3 minutes. Because of the orbital motion of the Moon and the rotation of the Earth, the umbra makes a long, narrow path of totality.
Sometimes the umbra does not reach the Earth at all (only the penumbra) even though the Moon is on the ecliptic and it is exactly in New Moon phase. A bright ring will be visible around the Moon when it is lined up with the Sun—an annular eclipse (because of the annulus or ring of light around the Moon). What do you think this implies about the shape of the Moon’s orbit?
The sequence below is of the May 20, 2012 annular solar eclipse as seen from Red Bluff, CA. Select the image to go to a full-size image of the eclipse sequence and an animation.
A video of the central 5 minutes of the total solar eclipse of November 13, 2012 as seen from Amaroo, outside of Cairns, Australia, is posted on YouTube. The video also shows that even astronomers have to contend with the weather.
The sequence below is of the May 20, 2012 annular solar eclipse as seen from Red Bluff, CA. Select the image to go to a full-size image of the eclipse sequence and an animation.
Eclipse Web Sites
The following sites are probably the best and most complete ones on the web. You’ll find eclipse predictions, photographs, observing instructions, and fairly complete indexes of all high-quality eclipse sites on the web. Between these three you should find whatever you may be interested in with regard to eclipses. They will display in another window.
- NASA’s Eclipse Bulletins.
- Fred Espenak’s eclipse site. A less technical version of the NASA site. Fred Espenak is also the main contributer to the NASA site.
- Sky & Telescope‘s eclipse site.
- annular eclipse
Review Questions 10
- What are the two parts of a shadow and in which part is the Sun partially visible?
- Why does the Moon turn orange-red during a total lunar eclipse?
- Would a person on the Moon ever experience an annular solar eclipse? Explain your answer.
- Why do we not have just annular solar eclipses or just total solar eclipses when the Moon and Sun are exactly lined up?
3.9 Planetary Motions
There are other celestial objects that drift eastward with respect to the stars. They are the planets (Greek for “wanderers”). There is much to be learned from observing the planetary motions with just the naked eye (i.e., no telescope). There are 5 planets visible without a telescope, Mercury, Venus, Mars, Jupiter, and Saturn (6 if you include Uranus for those with sharp eyes!). All of them move within 7 degrees of the ecliptic. This tells you something about of the orientation of the planet orbit planes with respect to the ecliptic—the figure below shows how flat the solar system is when viewed along the ecliptic plane. The planet positions, of course, do change as they orbit the Sun, but the orbit orientations remain the same.
The arrow pointing to Polaris in the solar system picture is tilted by 23.5 degrees because the Earth’s rotation axis is tilted by 23.5 degrees with respect to the ecliptic. As viewed from the Earth, two of the planets (Mercury and Venus) are never far from the Sun. Venus can get about 48 degrees from the Sun, while Mercury can only manage a 27.5 degrees separation from the Sun. This tells you something about the size of their orbits in relation to the Earth’s orbit size—their orbits are smaller and inside the Earth’s orbit. When Venus and/or Mercury are east of the Sun, they will set after sunset so they are called an “evening star” even though they are not stars at all. When either of them is west of the Sun they will rise before sunrise and they are called a “morning star”.
Planets produce no visible light of their own; you see them by reflected sunlight. True stars produce their own visible light. The planets inside the Earth’s orbit are called the “inferior” planets because their distance from the Sun is less than (or inferior to) the Earth’s distance from the Sun. Their closeness to the Sun enables us to see them go through a complete set of phases. Because they can get between us and the Sun, Venus and Mercury can be seen in a crescent or new phase. This also explains why the planets outside the Earth’s orbit, called the “superior planets”, are never seen in a crescent or new phase. When Venus is in crescent phase, it is the brightest object in the sky besides the Moon and the Sun. Even though you see a small fraction of its sunlit side, it is so close to us that you see it appear quite bright. At these times, Venus is bright enough to create a shadow! The fact that you can see Venus and Mercury also in gibbous and nearly full phase proved to be a critical observation in deciding between a Earth-centered model and a Sun-centered model for the solar system.
Because Mercury and Venus are closer to the Sun than we are (i.e., their orbits are inside the Earth’s orbit), they are never visible at around midnight (or opposite the Sun). The superior planets can be visible at midnight. At midnight you are pointed directly away from the Sun so you see solar system objects above the horizon that are further out from the Sun than we are. If you want to see where the planets are in their orbits today or any other date, then go to the Solar System Live site (will display in another window).
Ordinarily the planets “wander” eastward among the stars (though staying close to the ecliptic). But sometimes a strange thing happens—a planet will slow down its eastward drift among the stars, halt, and then back up and head westward for a few weeks or months, then halt and move eastward again. The planet executes a loop against the stars! When a planet is moving backward it is said to be executing retrograde motion. Perhaps it seemed to the ancients that the planets wanted to take another look at the stars they had just passed by.
The figure below shows Mars’ retrograde loop happening at the beginning of 1997. Mars’ position is plotted every 7 days from October 22, 1996 (the position on November 12, 1996 is noted) and the positions at the beginning and end of the retrograde loop (February 4 and April 29, 1997) are noted. An animation of this is available here. What causes retrograde motion? The answer to that question involved a long process of cultural evolution, political strife, and paradigm shifts. You will investigate the question when you look at geocentric (Earth-centered) models of the universe and heliocentric (Sun-centered) models of the universe in the next chapter.
- retrograde motion
Review Questions 11
- How do the planets move with respect to the stars?
- What does the fact that all of the planets visible without a telescope move within 7° of the ecliptic imply about the alignment of their orbital planes? What would an edge-on view of our solar system look like?
- Why are Venus, and Mercury never seen at midnight while the other planets can be visible then?
- What phase would Venus be in when it is almost directly between us and the Sun? Where would it be in its orbit if we see in a gibbous phase?
- Are the planet motions random all over the sky or are they restricted in some way?