Q. Could you tell me which telescope would be the best buy for under $300?

A. It's hard to give you advice on this because it depends completely on what kind of features that you would like to have. What I would suggest is that you buy a copy of Sky and Telescope magazine. Not only do they review different kinds of telescopes and telescope paraphrenalia, but there are lots of advertisers selling telescopes and they have a whole section of personal ads of people selling their own telescopes, so that you may be able to get a more full-featured telescope for less money, although it would be used. Meade and Orion are both very reputable companies, so definitely look into telescopes built by them.

Q. Since I am interested in astronomy (a very beginner), I decided to buy a telescope, but I couldn't find a good one. What kind of telescope would you recommend? Are there good condition second hand telescopes?

A. The best advice I can offer you on telescopes is to wait. First, learn your way around the sky with your own eyes. If you are not familiar with the sky and identifying objects of interest to you, you will have difficulty finding them in a telescope. Telescopes only show you a very tiny piece of the sky at any one time and upside down to boot. When you have mastered locating the constellations, you might want to invest in a quality pair of binoculars. Binoculars will help you see more objects and to see some familiar objects more clearly. They have a wider field of view than a telescope and allow you to see right-side up. For looking at the sky, you would want a pair with large front lens and good quality optics.

Finally, when you have thoroughly explored the sky with your binoculars, you may want to consider investing in a telescope. When selecting a telescope, look for a smoothly working mount and "diffraction-limited" optics. You also want to invest in a large aperture. You should also consider how convenient the telescope will be to transport and set-up; if the telescope is too unwieldy, you will not get enough use out of it. Avoid the cheap refractors available from most general retail stores that have objectives around 50 millimeters. Once you are comfortable with your telescope, you can consider all sorts of devices to attach at the focus, such as CCD's, instead of your eyes.

You might also consider constructing your own. While grinding lens is time-consuming, I understand it is very rewarding to use equipment you built yourself.

For detailed information on equipment, you might want to look at "SkyWatch" or "Sky & Telescope," which are published by Sky Publishing Company ( Astromart ( and Starry Messenger (ceased publication <--!> has classified advertisements of equipment for sale. The Internet Telescope Exchange ( lists both new and used equipment. <--!> I am certain there are many more sources of used equipment.

I personally have a Celestron C90 1000 mm f/11 telescope my grandfather gave me in elementary school. Although this was too much for a child, my father enjoyed it and taught me how to use it. I have continued to enjoy it through high school and college. However, transporting it safely to a site with decent viewing is sometimes difficult. I am not recommending for or against this particular instrument. Many Astronomers do not own personal instruments and one is certainly not necessary for enjoying Astronomy as either a hobby or a career.

Q. If I measure the angle of the sun one day and again in a few days later at the SAME exact time, would the angle be the same?

A. The Earth is tilted on its axis, which is responsible for the seasons. As the Earth orbits around the Sun, the angle between the Sun and the horizon measured from any point on the Earth at a given time of day changes every day. The noontime height of the Sun in the sky is highest on the summer solstice and lowest on the winter equinox.

If you took a picture of the Sun (being careful not to look at the Sun or magnify its light) every day at the same time, for a year, you would see a pattern called an analemma:

Q. I need to write code in Visual Basic that will return the position of the sun given observer's position and time. Do you have any algorithms or formulae I can use? I would like to work in ECEF coordinates, since I work with GPS, and return azimuth/elevation for the sun position. However I can convert coordinates as necessary. Any help would be appreciated.

A. The conversions from Equatorial Coordinates hour angle (H) and Declination (d) to Altitude-Azimuth coordinates for an observer at latitude L are:

sin(alt) = sin(L)sin(d) + cos(L)cos(d)cos(H)
cos(alt)cos(az) = cos(L)sin(d) - sin(L)cos(d)cos(H)
cos(alt)sin(az) = -cos(d)sin(H)

Hour angle (H) = LST (local sidereal time) - RA (right ascension) of Sun

Q. There is no preferential coordinate system for linear velocities. It does seem possible to establish a preferential system for rotation, however. If I am in a closed box with a spring attached between two masses, and if I can alter the rotation rate of the box with a system of rockets, I can find a rotation that causes the spring to go slack. If I do that, and then open the box I will find that I am not rotating relative to the stars. Does that not establish a preferential system for rotation? If so, isn't it curious that the local environment in and around the closed box "knows" about the very distant stars?

A. Einstein's strong equivalence principle states that all inertial reference frames are equivalent and no experiment can be devised to distinguish between them. Your above proposed experiment compares a non-inertial frame (the rotating box) to an inertial frame (a non-rotating box). Moreover, inertial frames are preferential to non-inertial frames. Thus, zero rotation is a preferred system, but only because it is an inertial system.

A rewording of your second question might be: How does the box "know" that it is rotating? This question deals with how local inertial reference frames are defined. There have been three primary schools of thought on this subject. Newton believed in absolute space and that your box knows it is rotating with respect to absolute space, even if all that was in the universe was just your box. Ernst Mach developed what is known as Mach's principle since he believed that absolute space made no sense at all. Mach said that inertial reference frames are established by the distribution of matter in the universe. So if all there was in the universe was just your box then rotation makes no sense. The box would not be rotating with respect to anything. Einstein's theory of general relativity is sort of the middle ground between these above theories. He decided that the geometry of spacetime determines local inertial frames and this geometry was set by the distribution of matter and energy in the universe.

For more information about these issues please check out the following URL: It is a website for a book written by Dr. John Hawley, one of our faculty members, and it contains answers to these and other cosmological questions.

Q. I have found many sites that list the equinox and solstice dates for the next several years. I am looking for algorithms to calculate these for any given year. I am specifically using Visual Basic, but any readable pseudocode will work for me.

A. The solstices and equinoxes occur, not on specific dates, but when the sun is at a certain point in the sky. Specifically:

The vernal equinox is when the sun is at (RA,DEC)=(0,0) The summer solstice is when the sun is at (RA,DEC)=(6,+23.5) The autumnal equinox is when the sun is at (RA,DEC)=(12,0) The winter solstice is when the sun is at (RA,DEC)=18,-23.5)

The general dates of these events are about March 21, June 21, September 23 and December 21 each year. As an aside, Christmas is located on December 25 so that it could block out Pagan rituals that celebrated the solstice.

The actual dates change slightly because a year is not exactly 365 days long. It is 365.24219. So the date and time of equinox moves ahead .24219 days each year, then moves back one day each time we have a leap year. Leap years occur every four years with the exception of years that are divisible by 100 (1900 was not a leap year). However, years that are divisible by 400 (2000) are leap years.

What I would recommend is: (1) Set a zero point by finding the date and time of the most recent round of equinoxes and solstices, setting the dates as decimal points (example, January 4, 1989=1989.011) (2) For whichever year you want, figure out how many years have passed. Add .24219 days to the date and time of your zero point year, subtract 1 day for each leap year.

This will give you reasonable accuracy - the accuracy depending on how accurate your first dates and times are and how accurate the measure of a year is.

Q. "Tidal forces cause orbits to go to a state of lowest energy while conserving angular momentum. This results in the circularization of originally elliptical orbits". I think this is what is happening to the moon, which rotates in a synchronous orbit around the Earth and is slowly moving away from it. I reckon that the enlargement of the Moon's orbital radius depends on the tidal bulges it produces on Earth, while the circularization of its orbit depends on the tidal bulges EARTH produces on it. How is the orbital energy lost?

A. You seem to have the basic concepts correct. Tidal forces in the Earth-Moon system have caused both the Earth and the Moon to develop bulges. The energy from these differential motions is dissipated into heat through friction or compressional effects. This heat is then radiated out into space. On the Earth, most of the energy is lost when the ocean tides hit the coasts of continents. It is this tidal friction that allows the Earth and Moon to synchronize. Conservation of angular momentum causes the Moon's orbital radius to increase when the rotational rate of the Earth decreases.

A more fascinating example of tidal forces is in the Jovian system. Jupiter's moon Io is actively volcanic and this is believed to be the result of Jupiter's tidal influences. Jupiter has three other large moons which are believed to perturb Io enough to not allow it to circularize its orbit. Thus, Io is constantly being stretched and compressed by Jupiter's tidal forces introducing a tremendous amount of heat (through friction) into its interior. Resulting in the most volcanically active body in the solar system with the youngest surface.

Q. What are the Tropic of Cancer and Tropic of Capricorn and where do their names come from?

A. The tropics of Cancer and Capricorn are lines of latitude on which the sun is directly overhead at noon on the summer and winter solstices, respectively. Specifically, they are 23.5 degrees north latitiude (Cancer) and 23.5 degrees south latitude (Capricorn). When the tropics were named, the sun was in the constellation of Cancer on the Summer Solstice and in the constellation Capricorn on the winter solstice, hence the names. Because the earth wobbles on its axis (it takes about 26,000 to make one complete wobble), the sun is no longer in these constellations on the solstices. The names, however, are still with us.

Q. This year's equinox occured on Sept. 23, 12:37 AM CDT. Why is it that the actual equinox (12 hours of day and 12 hours of night) occurs a couple of days later, maybe today or tomorrow?

A. I quote from Norton's 2000.0 Star Atlas by Ian Ridpath 18th edition:

"Firstly, sunrise and sunset are calculated for the upper limb of the Sun, not its centre; secondly, the effect of refraction in the Earth's atmosphere lifts the image of the Sun about half a degree at the horizon. These both have the effect of slightly lengthening the day."

Norton's is an excellent reference on this sort of question and I highly recommend it, even if it is a bit pricey ($50).

I would add that equinox is when the Sun reaches zero declination, not when we have equal day and equal night.

Q. Why didn't the planet Venus create an atmosphere for life like in the Earth?

A. Venus and the Earth are very similar, in fact they are often referred to as sister planets. The main difference between the two planets is their distance from the Sun, Venus orbits the Sun about 30% closer than the Earth does. For this reason, the amount of solar energy hitting the surface of Venus is larger than the amount hitting the surface of the Earth. This difference is what is believed to have caused the differences in the two planets' atmospheres. The larger surface temperature on Venus would cause any water to evaporate, increasing the levels of water vapor and carbon dioxide in its atmosphere. These two gases are known as "greenhouse gases" and trap heat in the planet's atmosphere. The trapped heat would evaporate even more water, increasing the water vapor and carbon dioxide content of Venus' atmosphere again. This process ran away, that is it kept going indefinitely, until the atmosphere on Venus reached its present state of containing approximately 99% carbon dioxide. Since the Earth is further from the Sun than Venus, the surface temperature stayed low enough to form oceans, trapping much of the carbon dioxide that would wind up in the atmosphere of a hotter planet, such as Venus.

Q. It's an interestingly difficult task to find information on what is probably a relatively simple problem:

There are exactly 24 hours in an earth day as measured by Cesium 133 clocks - (I have heard this referred to as Mean Solar Time). If you string together 365 of these exactly measured earth days, the earth is still about 1/4 earth day from completing a full orbit of the sun. Thus, to keep the seasons consistent, we need to add one earth day to our calender approximately every 4 years.

This would all seem relatively simple if not for the fact that a day is also measured by a full turn of the earth on its axis. In order for this equation to reconcile, one Mean Solar Day (allowing for variances in earths' orbital speed) would have to, over an orbital time of 365.24 days, average exactly 24 hours. Is this the case?

I believe that a solar day is about 4 minutes longer than 24 hours. However, this may be an average Apparent Solar Day. If not, we would have to add one day each year to our calenders in order that the sun rose and set at about the same daily time throughout the year. Of course, this would mess up our seasons, as we would now have about three too many days every four years.

As the world seems to have figured this all out centuries before my birth, I'm confident that I must be missing something basic. Please help!!

A. There are in fact 24 hours in a solar day. A solar day is measured from noon on one day (where noon is defined as the time when the Sun is transiting your local meridian) to noon the next day. In this time, the Earth actually rotates more than 360 degrees. The reason for this is simple, in one solar day, the Earth has also moved approximately 1 degree along its orbit around the Sun, and therefore the Earth has to rotate through more than 360 degrees for you to see the Sun in the same part of the sky as you did the previous day.

If you were to instead measure a day as the time between successive transits of a star (e.g. Vega), you would find that one day measured this way would in fact be approximately 23 hours and 56 minutes. A day measured this way is called a sidereal day. This is most likely where your confusion about a solar day containing an extra 4 minutes comes from; it is not that a solar day is 24 hours 4 minutes long, instead a solar day is in fact 24 hours long, however this is 4 minutes longer than a sidereal day.

A tropical year is the time for the Earth to have completed a complete revolution of the Sun, and is about 365.242 solar days long. By introducing the leap year correction to the calendar, we correct for the fact that our calendar is based on a year consisting of 365.0 solar days.

Q. We have been puzzled by a simple fact that we can't understand. We wonder why lunar eclipses can only occur on a full moon?

A. Lunar eclipse occurs when the Earth's shadow falls across the moon. For this to happen, the Earth must be between the Sun and the moon. Full moon always occurs at this point in time as it the only time that the entire illuminated part of the moon can be seen from Earth.

Q. Would like to know about the meteor shower that occurred on Oct. 8, 1998.

A. The Giacobinid or Draconid Meteor Shower reached its peak on October 8 with a peak zenithal hourly rate (a measure of the number of meteors) exceeding 500. The shower is a result of the Earth passing through the orbit of Commet Giacobini-Zimmer. This comet has a 6.6 year orbit and will reach perihelion (its point of closest approach to the sun) in November 1998. The Giacobinid Meteor Shower has only been detected in those years during which the comet reaches perihelion. In particular, the comet created spectacular outbursts in 1933 and 1941 but its return in 1985 was less impressive. You can find some historical information about the comet and its meteor shower and some reports of last week's observations on the web pages maintained by Sky & Telescope (dead link, Astronomy ( Magazines, and the Dutch Meteor Society (

Q. When was the last time (if ever) there were 2 new moons in the same calendar month? Is this event possible? If yes, when will this event occur again?

A. It is possible to have the same phase of the moon occur twice in the same month. The synodic period of the moon (the time it takes the moon to go through all phases, e.g. from full moon to full moon) is 29.5 days. Since this is shorter than most months, you will have a repetition of one of the phases of the moon during a month that is 30 or 31 days long. From published tables of the lunar phases, it appears that the last time we had two new moons in one month was in October of 1997, October 2nd was a new moon, and so was October 31st. The next time this will occur is in July of 2000, July 1st 2000 will be a new moon and so will July 31st 2000.

You may have been told that when two full moons occur in the same month, the second is called a blue moon, hence the phrase "once in a blue moon". However, I recently read that this definition of a blue moon is more modern than most of us think. For more information, please refer to this article in Sky & Telescope.

Q. I need you to settle an argument for me. We were wondering why, when, and by whom Daylight Savings Time was created.

A. I've always learned that Ben Franklin discussed the idea in the US. In fact, he proposed the idea while serving as US minister to France, as a way of conserving the cost of lighting for shops. The purpose has been to essentially utilize more of the daylight hours in Spring and Summer for daytime activities. Simply put, the sun is up longer in the sky (rises earlier and sets later) in summer. Clocks and common mechanical timepieces keep what's called Civil Time, which roughly corresonds to the passage of a fictious "average" sun through our sky. This is necessary since the Sun does not rise and set on a regular basis, but changes by as much as 15 minutes through out the course of the year.

That being said, if clocks are advanced by an hour, The sun, according to Civil Time, rises and sets later, providing more useable hours of sunlight for various activities.

It was generally adopted by countries around WWI, but the US repealed it's mandatory daylight savings law, signed by President Wilson on March 31, 1918. During WWII, it was reestablished by law on a year-round basis, and later seen as a fuel saving measure during the old crisis of 1973-74. In late 1974, standard time was brought back so we didn't live in the dark during winter, which has the fewest daylight hours.

Q. Where can I find an explaination for the force that causes the Earth and other celestial objects to rotate?

A. Any basic Physics textbook covering mechanics should contain an explaination of the conservation of angular momentum. For any isolated collection of particles, their total angular momentum will remain constant and is dependent on the collection's moment of inertia and angular velocity. The classic example of this law is an ice skater spinning slowly with his arms outstreached (large moment of inertia). As the skater reduces his moment of inertia by bringing in his arms, he spins faster (large angular velocity). For our Solar system, the Sun, Earth, and other bodies are believed to have formed from the gravitational collapse of a slowly rotating cloud of interstellar gas and dust. Similarly, other rotating bodies probably formed from the collapse of other clouds which had some initial angular momentum. The angular momentum found in galaxies may have its origins in tidal forces applied to the clouds from which they formed before the clouds began to collapse.

Q. I understand now that a rocky asteroid through the friction of comming through the atmosphere can be crushed and, in effect, explode. But how is the resultant radiation accounted for?

A. As a meteoroid falls through the atmosphere, it collides with particles in the atmosphere. During these collisions, the meteoroid's kinetic energy is converted to thermal energy, and it heats up. The heating of the meteoroid's surface causes atoms to boil off of it. In turn, these released atoms collide with atoms in the atmosphere causing ionization and collisional excitation. The meteoroid continues to fall through the atmosphere converting deeper and deeper layers of itself into glowing envelope around its core and a trail in its wake. Depending on the size and content of the meteoroid, it may disintergrate completely before reaching the Earth's surface or it may crash into the Earth. Several meteorites strike the Earth daily but massive ones are rare.

Q. I was told the Mayan Calendar ends in 2012. Is this true? If so, Why?

A. Yes, it probably does. The Mayan calendar is composed of multiple cycles of different lengths, which all meshed together for agricultural, religious and other purposes. Some of the more important cycles were 260 days, 365 days, 52 years, and (probably) 5126 years. This last one is known as the Long Count, and it started on August 11th 3114BCE, which would mean that the end of this Long Count would be December 21st, 2012. Some archaelogists believe that the Mayan predicted an age of enlignment at the end of the Long Count. Others believe the Mayans would have merely constructed yet a longer cycle as they approached the end of the Long Count. For more on this, see:

Q. On Saturday night (11/7/98) at about 11:00 pm, I happen to be gazing with my binocs basically straight up. I could have sworn that I saw what looked like a comet. It was a cloudy looking object, not like a bright pin-point of a star or planet. I was wondering if I may have come upon the Tempel-Tuttle comet. Is it observable to us like the Hale-Bopp was? It was just to the left of a star or planet. Any info on this would have been greatly appreciated.

A. Comet P55 Tempel-Tuttle passed by the earth on March 8, 1998. What is creating the Leonid meteor shower is the debris left behind by the comet as it went past; the comet itself is long gone. What you saw was most likely M31, the Andromeda Galaxy. It is the nearest normal spiral galaxy to our own, about 2 million light years distant (that's pretty close on astronomical scales). The "M" in its catalog name means that it's part of the Messier catalog, a list of 110 astronomical objects compiled by French astronomer Charles Messier from 1771-1784. Messier was looking for comets at the time, since the discovery of a new comet brought a large cash reward from the French crown in those days, but was largely unsuccessful in his search. He published the Messier catalog as a list of items which are NOT comets and were therefore deemed uninteresting. Ironically, these are some of the most widely-studied and interesting objects in modern astronomy.

Q. How far does light travel in a calendar year?

A. One lightyear is about 5880049720323.18 miles.

Q. What formula describes the velocity of an object while it is in orbit?

A. For a two-body system, the orbital velocity can be obtained using the vis viva equation:

v^2 = G(m1 + m2)[2/r - 1/a]

where m1 and m2 are the masses of the two objects, r is their separation, a is the semi-major axis of the elliptical orbit, and G is the gravitational constant. Note that for non-circular orbits the velocity varies with time. There are many other equivalent ways to express the above equation (see "Introductory Astronomy & Astrophysics" by Zeilik, Gregory, & Smith, 3rd Ed. pg 16 for more info).

Q. What is the distance between earth and mars??

A. The distance between Earth and Mars varies between about 10^8 km and 5.5 x 10^7 km.

Q. I would like to know if there is a simple formula to figure out exactly how many days since certain events have happened. I don't want to get too technical about it but I would like the leap year days figured in. Just things that have happened in the last 100 years.

A. Julian Date is a date that is determined solely in days, for example October 28, 1987 is 2447907. If you can convert two dates to Julian, you can then subtract them to get the numbers of days passed.

The formula for this is: jd = ( 1461 * ( y + 4800 + ( m - 14 ) / 12 ) ) / 4 + ( 367 * ( m - 2 - 12 * ( ( m - 14 ) / 12 ) ) ) / 12 - ( 3 * ( ( y + 4900 + ( m - 14 ) / 12 ) / 100 ) ) / 4 + d - 32075

Where y is the Gregorian year, m is the month and d is the day.

Q. What is the dividing line between the formation of white dwarves and black holes?

A. First of all, there is an object between those two called neutron stars. All of these objects are the remains of dead stars, although black holes may be created by other means at the centers of galaxies.

The dividing line for all three is mass. A white dwarf is the electron degenerate core of a star. As it grows in mass, it shrinks. When it reaches a mass 1.4 times that of the Sun, electron degenerate pressure can no longer hold it up against the pull of its own gravity and it collapses to a neutron star. Neutron stars are held up by degenerate neutron pressure. As they grow in mass, they also shrink. The calculations become extremely complex, but it is thought that beyond 3 solar masses, a neutron star can no longer support its own weight and it collapses.

As an aside, degenerate pressure is the quantum mechanical pressure of particles. You can think of it as electrons or neutrons pushing against each other.

Q. What is our best guess for the size of the universe? I recall hearing this estimate being revised upwards by a factor of 50 in recent years related, I believe, to observations of the Hubble Telescope

A. The size of the Universe depends on its geometry. If it is open, as some maintain, then it is infinite. The "observable Universe" however, is only that for which the light has had time to reach us. The age of the Universe is a source of heavy debate, but is generally around 15 billion years - so 15 billion light years is the furthest we can see.

If the Universe is closed, then it is finite and its size depends on the rate of expansion, which is complicated by inflation theories and acceleration or deceleration. As a rough approximation, we could note that the current value of the Hubble Constant (call it 60 km/sec/Mpc) implies a universe about 5 billion light years in radius. Now, what you may be thinking of is that the Hubble Constant has been revised by HST.

You'll notice I've mentioned two points of debate - open vs. closed and the value of the Hubble Constant. HST has weighed in on these matters, but there are no easy conclusions at this point. The observational evidence has to constrain our imaginations. However, the main points are agreed upon. We know it is billions of light years. The last change to our cosmological paradigm occured about 80 years ago when Hubble showed that our galaxy did not contain the entire Universe, revising the size from hundreds of thousands of light years to billiions.

What you may be thinking of is the distance _scale_. We measure distances in astronomy by the use of standard candles. These are objects for which we know the intrinsic brightness well. By comparing the apparent brightness and the intrinsic brightness of an object, we can derive a distance.

Occasionally, standard candles are revised. The most famous incident occured 50 years ago when Baade divided the Cepheid variables into two classes - doubling the distance scale. However, there have been only minor revision since - most notably through the recent data from the HIPPARCOS satellite.

Q. How fast do meteorites travel?

A. Metoroids hit the Earths atmosphere with a velocity of anywhere from about 12 to 72 km/s. Since their orbits are not necessarily circular, their speed varies as they orbit the Sun, moving faster when they are closer to the Sun. Thechnically, these objects are not called meteorites until after they've struck the Earth's surface.

Q. Can you describe how GPS systems work? How accurate and reliable are the calculated positions? What is the hottest debated topic in astronomy today? Where can I read more about these topics?

A. GPS systems work using the concept of triangulation. If you have in your hand a GPS receiver, it receives signals transmitted by one of several GPS satellites, each in a different orbit. By measuring the precise travel time the signal from each satellite took to reach your receiver and multiplying by the speed of light, the GPS receiver determines very accurately your distance from each of the satellites. By using your distance from several satellites in different places, you can determine the angle between you and each individual satellite. These angles can then be converted into your angular position on the Earth (longitude and latitude). The quoted accuracy of the GPS for civilian use is 25 meters (82 feet).

I can't accurately answer your second question because there are several very hotly debated topics in astronomy today. I can list some of these for you (in no particular order):

What produces gamma ray bursters?

Is there a cosmological constant?

When did galaxies form?

Why do we measure a deficit of neutrinos from the Sun?

What makes up "dark matter"?

To read about these topics at a popular level, I would check the magazines Astronomy, Sky & Telescope, or Scientific American.

Q. What is the exact time for the line of nodes of the moon and earth orbit to be the same relative to the stars (the most precise answer I have found is the very imprecise 18.6 years). And what is the time of the anomalistic month of the moon (the time from perogee to perogee)?

A. A Saros cycle (the time for the line of nodes to return to its original position) is 6585.32 days. The anomalistic month has a mean value of 27.55455 days.

The Saros cycle is not actually the time between successive alignments of the line of nodes with respect to the stars, it is instead the time between repetitions of the solar and lunar eclipse cycles. The reason eclipses repeat after 6585.32 days is that for two eclipses to appear similar from the Earth, the moon must be at the same phase, at the same place in its orbit with respect to the node, and at the same relative distance from the Earth. To satisfy these three conditions you need an integral number of synodic months (the time between successive full moons), draconic months (the time between successive passages through the ascending node), and anomalistic months (the time between successive perigees). 6585.32 days is approximately 223 synodic months, 242 draconic months, and 238 anomalistic months, so after this amount of time (the Saros cycle), nearly identical eclipses will occur. The "nodal period" is the exact time for the line of nodes of the moon and earth orbit to be the same relative to the stars, and is 6798.38 days.

Q. What is the best way to clean the mirror from a Dobsonian reflector telescope? The mirror can be easily removed, but I am concerned about scratches or mineral deposits from improper cleaning techniques.

A. You can clean your mirror by simply washing it in warm water with a very mild detergent. You can also remove particles on the surface using the softest cotton cloth you can find. Check at an autoparts store, the cloths used to baby delicate car finishes should be soft enough for your mirror.

Q. What is the exact day and time that the winter solstice took place in 1965?

A. In 1965 the winter solstice occurred on December 22nd at approximately 8:37 AM.

Q. Why do the dates of earliest sunset and latest sunrise not occur on the winter solstice?

A. Referring again to Norton's, pg. 51:

"The reason is to be found in the equation of time [which is why the Sun is not exactly at the meridian at noon - see above], as a result of which sunrise and sunset are both getting later each day at the solstices, although by different amounts. The net result is that the latest sunrise actually occurs after the shortest day, and the earliest sunset before it, while the earliest sunrise occurs before the longest day, and the latest sunset after it. In each case, the exact date depends on the observer's latitude, but the offset from the solstices is greater at lower latitudes. The effect is more pronounced at the December solstice than the June solstice because the day-to-day increase in the equation of time is greater in December.

Q. I want to know about Sirius , the brightest star and where I can see this time of the year in the sky. Would be grateful for your help. I am in North Carolina.

A. Sirius is the brightest star in the sky, after the Sun of course. It is a part of the constellation Canis Major, which is a winter constellation. This time of year, you can see it just above the southern horizon.

Q. What are the delta time corrections in seconds (UT to TDT) for the years 1983 through 2000 for July 1?

Universal Time (UT), which is also called Greenwich Mean Time (GMT) or Zulu, is determined by the rotation of the Earth. Astronomical observations are used to determine it. However, several versions of UT exist that differ by less than 0.03 seconds. Terrestrial Dynamic Time (TDT) replaces the older Ephemeris Time, which attempts to account for slow variations in the Earth^s rotation. TDT includes relativistic effects. For more detailed information on these and related time systems, read the relevant section of the Time, Calendars, and Terrestial Phenomena FAQ of site ( This FAQ provides the details, including delta times, for converting between systems. My ephemerides only provide ecliptic longitude and declination (RA). What are the mathematical formulas for finding ecliptic latitude and right ascension? The Right Ascension/Declination System is an equatorial system centered on the center of the Earth. Its longitude-like component, right ascension (RA or alpha), is measured in hours from the vernal equinox. Because 24 hours is equal to 360 degrees, right ascension may be converted to degrees as follows

RAdeg = (RAhours X 360)/24

Its latitude-like component, declination (Dec or delta) is measured in degrees north or south of the celestial equator. The Ecliptic Coordinate System is based on the ecliptic, which is the plane of the Earth^s orbit. Celestial longitude (Clong or lambda) is measured in degrees from the vernal equinox in the same direction as right ascension. Celestial latitude (Clat or beta) is measured in degrees north or south of the ecliptic. To transform the Ecliptic Coordinate System into the Right Ascension/Declination System, the Ecliptic Coordinate System values must be rotated through an angle equivalent to the tilt of the Earth^s axis (Tilt or epsilon), which is approximately 23.5 degrees. This transformation results in the following trigonometric equations
tan (Clong) = cos (Tilt) X tan (RA) + sin (Tilt) X tan (Dec) X sec (RA)
sin (Clat) = - 1 X sin (Tilt) X cos (Dec) X sin (RA) + cos (Tilt) X sin (Dec)

Q. I still do not understand why the dates of the earliest sunset and latest sunrise do not occur on the same date.

A. In addition to the previous response to this question and the answer to why the day with equal hours of day and night does not occur on the equinox, I would also refer you to the response to why sun dials run slow. You may also wish to consult Norton's 2000.0 Star Atlas and Reference Handbook. Apparent solar time is the time measured by a sun dial showing the real motion of the sun across the sky. Mean solar time is the time measured by a clock. The irregularities in the solar time are smoothed out of mean solar time. The difference between apparent solar time and mean solar time is called the equation of time, which varies from several minutes ahead to several minutes behind. Around December 21, mean solar time is behind by about 2 minutes. In December, the times of sunrise and sunset are both getting later but at different rates. Consequently, the earliest sunrise will occur before the solstice and the latest sunset will occur after the solstice.

Q. Is my new Celestron F80 EQ WA telescope sufficient to see objects in space that will be gratifying, such as Saturn or Jupiter? What is the latitude of Boston, Mass? Do you have a recommendation of books for the novice?

A. You should be able to see Saturn, Jupiter, comets, and some deep-sky objects with your current telescope. It should be appropriate for a beginner. Before trading up at this time, review our archive questions regarding purchasing telescopes and check out the references provided.

According to that latitude, longitude search provided by, Boston, MA is located at 42.33602 N, 71.01789W.

I find recommending books difficult without knowing your personal interest or how technical you would like the material to be. My best friend shruggs off galaxies, globular clusters, and nebulae as boring blobs but can ooh and aaah over Saturn and Jupiter for hours. I suggest you check out the offerings at your local public library to find some appropriate materials for you. Both Sky & Telescope and Astronomy magazines have good articles and monthly observing charts (on-line versions of both magazines are available). You might also consider joining the Amature Telescope Makers of Boston ( to meet people with similar interests and to learn more about amateur astronomy.

Q. I heard that Jupiter is star that failed; had it been ~30 times more massive, it would have under gone thermonuclear reactions and it would have become a star. However, since Jupiter has a solid iron core, how could fusion ever take place? After all, isn't iron too heavy to induce fusion? Doesn't a star's death come about when the fusion process begins yielding iron in its core?

A. It's true that if Jupiter were about 80 times more massive it could burn hydrogen in its core, thus becoming a star. However, the process by which planets form is quite different from that which forms stars, so the amount of hydrogen in Jupiter's core is very small. Stars form when a cloud of gas collapses under its own gravity and eventually the pressures at the center get high enough to allow fusion to take place. Planets, on the other hand, are formed out of a ring of gas and dust that forms around a young star. This ring starts developing clumps of rock and gas that grow over time and eventually become planets. Since Jupiter started with a rocky core instead of the hydrogen core needed for fusion, it couldn't have become a star even if it were massive enough.

If a ball of gas the same mass of Jupiter had collapsed, it would have had the right composition to become a star, but the central pressure would not have been great enough to begin nuclear fusion. The minimum mass for a star is about 0.08 times the mass of the Sun; Jupiter is about 0.001 times the Sun's mass.

Yes, iron the ultimate biproduct of nuclear fusion at very high temperatures, but only very massive stars will ever get hot enough to produce significant amounts of iron through fusion. Whereas everything lighter than iron releases energy when fused with another atomic nucleus, iron actually absorbs energy in the process, so we do not find iron undergoing fusion in the cores of stars. The Sun will end up with a core made mostly of carbon and oxygen, the heaviest elements it can produce.

Q. How many days will this millenium have lasted? My own calculations from 1 January, 1000 to 31 December, 1999 (or from 1001 to 2000 as some people believe, but this is not the problem) say that this millenium lasts 365327 days (Julian Calendar till 1582, then Gregorian Calendar, leaving out the days between 4 October, 1582 and 15 October, 1582).

But somehow this result seems wrong, it should be 365325 days, because pope Gregor wanted to correct the error that came from the Julian Calendar.

So which one is the correct result (or are both wrong) ?

A. The current millenium began at midnight on 1 January, 1001 and will last until midnight of 1 January, 2001. The Julian Day at the beginning of the millenium was 2086673.70833 and will be 2451910.70833 at the end of the millenium. This means that there will have been 365237 days in the second millenium C.E. This accounts for the missing days in October 1582.

Q. I have a few question about stars and galaxies. It was mentioned in an astronomy class that there are approximately 20-30 billion stars in our galaxy (this was two years ago). It was also mentioned that the number of galaxies in the universe corresponds to the number of stars in our galaxy (20-30 billion). I think the estimated number of stars in our galaxy has recently increased to 200 billion.

So, my questions are:

1) how many stars are there in our galaxy (approximately)?

2) how many galaxies are there in the universe (approximately)?

3) has the number of galaxies also changed?

A. The most recent estimate for the number of stars in the Milky Way is indeed about 200 billion. The initial estimate from the Hubble Deep Field for the total number of galaxies in the universe was about 30-40 billion, but as more work has been done on the HDF, that number has been revised upward to about 50 billion, and may keep going up as the data in the HDF are better understood.

Q. A few years ago I was headed north towards Helena Montana at about 7:30Am. Looking to my left was a second sun. Not real bright but very detailed. It evedently was some sort of a reflection, but nothing I ever seen before. Others I told about it just thought I was seeing things. Can you help?

A. I can think of several possible explanations. It's possible that what you saw was a reflection either off of your car window or perhaps off of ice (if it was winter at the time). A more likely possibility is that what you saw was in fact the full moon. The full moon and the sun are opposite each other in the sky, so if the conditions are right, one can see them both above the horizon at the same time. In addition, the moon (and sun, if you could look directly at it) often acquires strange distortions near the horizon due to the fact that it's being viewed through more of the earth's atmosphere than when it's overhead.

Q. We know that radio waves travel at about 300,000 km/sec, but is that speed altered by the medium the wave is traveling through (such as the Earth's atmosphere as compared to space)? And about how far out is the first radio signal ever sent from Earth? And how far will it keep going?

A. In answer to your first question, the standard speed of light is the speed of light in a vacuum. Light slows down in air, glass or any other medium. This is what makes lenses and other optics possible. The velocity is equal to the speed of light in a vacuum divided by the index of refraction (1.0002926 for air, 1.33 for water, 1.5 for glass, etc.)

I'm not sure when the first radio transmission was made or if it was possible for it to penetrate the Earth's ionosphere. If you find out such date, then all you need do is figure out how many years it has been since it was broadcast and that gives you the distance in light years.

These signals will continue to move out into infinity. For example, the first TV transmission - Hitler opening the 1936 Olympic Games, is now 63 light years out and just passed Aldebaran in 1995.

However, the signal fades. That 1936 broadcast is now spread out over an area of 10^29 square kilometers!!! Even a modern 50,000 watt radio signal would be dimmed so that a radio dish the size of the Earth at the nearest star would only perceive it as the faintest of whispers - a millionth of a billionth of a watt. And considering that broadcast power was very dim back in 1936 or earlier, I doubt that anyone could ever pick it up. Modern signals, according to Clifford Pickover, have an effective range of about 1000 light years if they are specifically targeted at a star.

The strongest broadcast from Earth is, strangely enough, the Superbowl, which is broadcast all over the globe. This signal might be marginally detectable . . . if anyone's looking.

Q. Is there a simple formula to determine the energy output (bolometric magnitude, luminosity) of a brown dwarf, if given the mass? What are the upper and lower mass limits for a brown dwarf?

A. To estimate the energy output, you really need to know the mass, radius, and chemical composition. Assuming the brown dwarf is a self-gravitating mass of gas in hydrostatic equilibrium, you can estimate the temperature using

-(G * M) / R = (k T)/ m
where G is the Gravitational constant (6.67*10^-8 dyne cm^2/g^2)
      M is the mass of the brown dwarf
      R is the radius of the brown dwarf
      k is Boltzmann's constant (1.38*10^-16 ergs/K)
      m is the mean molecular mass of the particles making up the gas
           for pure Hydrogen gas this would be 1.672*10^-24 g
           for the Sun's gas this would be 0.6 * 1.672*10^-24 g 
      T is the central temperature of the brown dwarf
If you assume the brown dwarf is a perfect blackbody radiating at its central temperature, then its luminosity would be
Ld= 4 * Pi * (R^2) * s * T^4
where Ld is the luminosity of the brown dwarf
      s is the Stefan-Boltzmann constant (5.672*10^-5 ergs/cm^2 s K^4)
Then by comparing the luminosity of the brown dwarf to the luminosity of the Sun, the bolometric magnitude may be calculated using
Md - Ms = -2.5 * Log [Ld / Ls]
where Md is the bolometric magnitude of the brown dwarf
      Ms is the bolometric magnitude of the Sun (4.75)
      Ld is the luminosity of the brown dwarf
      Ls is the luminosity of the Sun (3.96*10^33 ergs/s)
A brown dwarf is expected to have a mass that is between 1% and 8% of the mass of the Sun (1.98*10^33g). Less massive objects are likely to be planets; more massive objects should be able to sustain nuclear burning of Hydrogen and be true stars.

Q. The Moon orbits the Earth so it must be going around the "sunny side" of the Earth for half the time of each orbit. So, why can we always see the Moon at night or are there times when we cannot see the Moon at night because it is on the opposite side of the Earth.

A. The Moon takes the same amount of time to rotate around its axis as it does to revolve around the Earth. Each side of the Moon experiences periods of sunlight and darkness with each period lasting about 2 weeks. The Moon takes 27.32 days to complete one revolution around the Earth compared to the background stars. However, in this time, the Earth moves about 27 degrees around the Sun. Therefore, the Moon requires 29.53 days to complete all of its phases. The different phases of the Moon are caused by its alignment relative to the Sun as seen by us on Earth. During the New Moon phase, the Moon is between us and the Sun so we do not see it. All of the sunlight the Moon receives is reflected away from us. During the Full Moon phase, we are between the the Moon and Sun so we see the largest amount of the Moon's surface during the cycle. All of the sunlight the Moon receives is reflected towards us. The other phases of the Moon are caused by the partial illumination of the face pointing towards us. A fraction of the sunlight the Moon receives, which depends on the specific phase, is reflected towards us and the remainder is reflected away from us.

Q. If I understand correctly, according to Carl Sagan's book "Cosmos", stars in our galaxy do not travel around the galaxy at the same rate as the arms. Therefore, old and new stars are mixed with each galactic revolution. How intense is this mixing? Were any of our neighboring stars born in the same cluster as our sun? Neighboring, as I use it here, means within 25 light years.

A. This is correct. Spiral arms are not solid objects - they are like traffic jams in the motions of stars and gas aroud the center of our galaxy. The compression of gas in these arms causes star formation to occur. The brightest stars will destroy themselves in supernovae within a few hundred million years and will not drift very far away while small stars (like our Sun) will last billions of years and drift far apart from each other.

Most of the stars in the Milky Way were probably once in clusters of hundred of stars (much like the Pleiades or Hyades now). But these clusters disperse themselves within a few billions years and over the course of 5 billion years (our Sun's age), the members could be spread out over thousands of light years. So it is possible that some nearby stars were formed with us, but not many.

Q. How do geocentric and heliocentric cosmologies explain the retrograde motion of the planets?

A. The answer to this will be in the first chapter of any introductory astronomy text. In brief, the geocentric cosmology posits a complex systems of epicycles (wheels within wheels) so that the planet orbits around a point that orbits the Earth. Heliocentric cosmology explains this as the Earth passing the other planet in its orbit, giving it a relative backward motion.

Q. When we changed calendars from the lunar to the solar one we use now, did the dates move any? If I were to ask when George Washington was born, it would not be the 22nd of February because of this theory.

A. The switch from the Julian calendar established by Julius Ceasor to the Gregorian one we use now did involve shifting dates. However, we were switching from one solar-based calendar to another more accurate one. The Gregorian calendar was adopted in different countries at different times starting in 1582. In the Gregorian calendar, New Year's Day falls on January 1 and an extra day is added after February 28th every leap year. In the Julian calendar, leap year occurred every four years. In the Gregorian calendar, leap year occurs in years that are divisible by 4 but not by 100 or are divisible by 400. Therefore, 1600 and 2000 are leap years but 1700, 1800, and 1900 are not. The Gregorian calendar was officially adopted by Britain and her colonies in 1752. Eleven days were skipped in September so that September 2, 1752, was followed by September 14, 1752. The reason for skipping the 11 days was to resynchronize the date of Easter, which is calculated using a lunar calendar, with the seasons. Prior to adopting the Gregorian calendar, New Year's Day had been been celebrated on March 25 in the colonies. Therefore, George Washington was born on February 11, 1731 according to the Julian calendar or February 22, 1732 according to the Gregorian calendar. When comparing historical dates, you must ask what system of dating is being used.

Q. As a photographer, I've always been interested in knowing exactly where the Sun would appear in the sky on any given day; where it would rise and set, what bearing and altitude at 2 pm. I realize that it changes from Summer to Winter due to the tilt of the Earth and the orbital elipse. Is there a simple formula I can use to determine the Sun's rough position based on my position? Do I need a nautical almanac?

A. You could use a calculator with trigometric functions and the formulas with accompaning tables found in The Astronomical Almanac (published by the Naval Observatory) to calculate the Sun's position and times of rising or setting. Alternatively, you could use the calculators provided by the Naval Observatory web page (

Q. Where is the best place to look for a single polynomial equation for the geocentric right ascension and declination of themoon with respect to time? "The Astronomical Almanac" gives the equation for single days only. Is there a single polynomial equation that accurately describes these coordinates over a long period of time ... say one Saros cycle?

A. "The Astronomical Almanac" gives a set of equations for calculating the position of the Moon at a specific time because the Moon's position is changing constantly. However, these equations should be useful over a reasonable range of Julian dates. "The Astronomical Almanac" or its explainatory supplement give information about converting from various calendars to Julian dates and Universal Time. You should be able to modify them easily for your purposes. The Saros cycle is a period of about 6585.32 days at the end of which the centers of the Sun and Moon have returned to almost the same relative position. The Babylonians noticed that the pattern of eclipses will recur each Saros cycle with their longitudes shifted about 120 degrees west from one cycle to the next. The cycle therefore is simply a period of days over which the relative positions of the Sun and Moon shift with a recognizeable pattern.

Q. Suppose that the Earth was moved to a distance of 10 AU from the Sun. How much stronger or weaker would the Sun's gravitational pull be on Earth?

A. Gravity is determined by Newton's Law of Gravity:

F= G * M1 * M2 / R*R

Where G is the graviational constant of the Universe, M1 and M2 are the masses of the two objects and R is the distance. In this case, the masses are constant. By increasing the distance by a factor of ten, you decrease the force of gravity by 10*10=100.

Q. What is the angle of the earth to the ecliptic? Is it the same as the angle of latitude of the Tropic of Capricorn/ Cancer?

A. The Earth is tilted at 23.5 degress with respect to the ecliptic. This is equal to the latitude of the Tropics of Capricorn/Cancer (23.5S/23.5N). Note that one must be between these latitudes for the Sun to ever pass directly overhead.

Q. When are the Perseid Meteor Showers this year? Is there a web site where the time of the showers is listed each year??

A. Actually, I will answer your second question first. The Sky & Telescope web page ( is a great resource for answering questions such as this. I actually found the answer to your question on their meteor page.

Now to answer your first question, Sky & Telescope says that the morning of August 13, 1999 should be the best time to see the shower, but the meteor rate will be high for a day or so before and after the peak.

Q. At what time did the sun set on December 16, 1998 in Norfolk, Virginia? In which phase was the Moon on that date?

A. On December 16, 1998, the Sun set at 4:50 PM Eastern Starndard Time in Norfolk, Virginia. The Moon was in the waning crescent phase with 5% of the visible disk illuminated. The US Naval Observatory provides a web page ( for easily calculating this type of information.

Q. Could a coronal mass ejection do serious damage to the Earth if it hits the planet? The magnetic field of the Earth protects us but could a strong enough hit penetrate? Could the Earth's magnetic field be disabled leaving the planet exposed? What would happen if something penetrated the magnetic field?

A. A Coronal Mass Ejection (CME) is an eruption of plasma from the corona of the Sun. The corona is the highest and hotest layer of the atmosphere of the Sun. Although the exact mechanism driving CME's is unknown, scientists believe interactions in the magnetic field of the Sun cause CME's. Once a CME takes place, the ejected material moves through space at speeds up to 5 million miles per hour, which produces shocks. Depending on the activity level of the Sun, CME's can occur as frequently as several times a day. The Earth is frequently in the path of these ejections. The magnetic field of the Earth protects the planet from most damage by deflecting the on-coming ejecta. However, some particles do penetrate the magnetic field, especially near the poles. CME's can cause auroras, magnetic storms, and alter the magnetosphere of the Earth. CME's can affect satellites through increased atmospheric drag and damaging sparks. CME's can disrupt radio signals. CME's can produce power surges in electrical lines and transformers resulting in the loss of service. CME's can produce electrical currents in gas and oil pipelines increasing the rate of deterioration. NASA provides some general information about CME's at NASA has also featured CME's in the Astronomy Picture of the Day; the relevant entries may be found by searching at

Q. Can you prove that the Earth revolves around the Sun only using knowledge that an astronomer had approximately 200 years ago?

A. In approximately 350 BC, Aristotle argued that a stellar parallax would be observed if the Earth revolved around the Sun. Tycho Brahe, who died in 1601 AD, opposed the heliocentric model proposed by Copernicus because Brahe failed to observe any stellar parallaxes. In 1838, Friedrich Bessel published the parallax of 61 Cygni, which he measured to be one-third of an arcsecond. Brahe's naked-eye observations were acurate to one arcminute. The parallaxes of the nearest stars are sixty times smaller, on the order of one arcsecond. In 1609, Galileo began using a telescope that made objects appear thirty times closer than they were, which is not quite good enough to observe stellar parallaxes. However, Bessel's first parallax is better than the largest values of about one arcsecond. Therefore, an astronomer between Galileo and Bessel could have observed a stellar parallax if he had choosen the right star and had good enough optics in his telescope. Stellar parallax, or heliocentric parallax, is the apparent shift in the position of a star due to the Earth's motion around the Sun. Over the course of six months, the apparent angular separation between two stars grows from some small angle B to a larger angle A before shrinking back to the orginial angle B. However, the stars are so far away that the shift in angular separation is impossible to see with the naked-eye. A good, basic astronomy text (such as Zeilik & Gaustad's Astronomy: The Cosmic Perspective or Harmann's Astronomy: The Cosmic Journey) should include a set of diagrams explaining stellar parallax. Most such texts will also give a brief history of the development of cosmological models and the arguements associated with them.

Q. Given that parallax shifts prove that the Earth revolves, how can one prove the Earth revolves around the Sun using only knowledge that was available 200 years ago?

A. Johannes Kepler's heliocentric model combined with Isaac Newton's gravitational laws makes the most accurate predictions for the motions of the Sun, Moon,and planets of any of the earlier competing models of the Solar System. The Ptolemaic and Copernican models are roughly equal in predicting planetary positions and those errors are observable to the naked eye. The now commonly accepted Keplerian model accounts for both the retrograde motion of Mars and the phases of Venus in a simple fashion. The geocentric model of Tycho Brahe correctly predicts the phases of Venus but the Ptolemaic model does not. William of Occam stated the principle now known as Occam's Razor in 1340. In accordance with Occam's Razor, we prefer the model that fits the observations using the fewest assumptions. Therefore, rather than postulate another center of mass or add additional deferents, epicycles, or equants, we prefer Kepler's system. In addition, Newton's gravitational laws, which can be demonstrated, provide a physical basis for the Keplerian system. However, Kepler had suggested a magnetic force tot keep the planets in their positions.

Q. Can you explain why the Sun is one of the two foci of an elliptical orbit instead of the center? Is there something that serves as the other? Any idea what lead Newton to this fact?

A. Actually, a planet and the Sun both orbit their common center of mass. However, the center of mass lies inside the Sun so we say that planets orbit the Sun. An orbit may take the shape of any conic section with the center of mass at one focus. Circles, ellipses, parabolas, and hyperbolas are all conic sections. Circles may be considered a special type of ellipse; one in which both foci are located at the same point. In our solar system, planets and periodic comets travel in elliptical orbits. Comets that appear only once travel on parabolic or hyperbolic orbits. Circular orbits are observed in some systems of binary stars. For planetary orbits, the Sun is located at one focus. The other focus, which differs for each planet, is located in empty space. The other focus is simply a geometric concept. Johannes Kepler discovered that the planets travel along elliptical orbits. Issac Newton's gravitational laws supplied a demonstratable force that keeps the planets orbiting the Sun. Kepler spent approximately four years trying to devise a model that would reproduce Tycho Brahe's observations of the orbit of Mars. Kepler began using circular orbits in accordance with the physics of his day and the Copernican model. He attempted to reproduce the motions using geometric devices like eccentrics and equants. After learning about the magnetic force, Kepler thought it might hold the planets in their places. He eventually found that a magnetic force analysis would work if an elliptical orbit was used instead of a circular orbit. Because both the gravitational force and the magnetic force have similar effects with respect to distance, Kepler was able to build an accurate model in this way.

Q. What are the names of the stars in the Big Dipper?

A. Actually, the names of most stars are quite boring. They're simply indexed in order of brightness with greek letters (alpha being the brightest) followed by the constellation names: alpha ursa major, beta ursa major, gamma ursa major, ect. (ursa major means "big bear", it contains the big dipper). Some lucky stars get individual names. These stars include Mizar and Alcor, which are the stars at the crook of the handle in the big dipper (if you look closely, you'll see that there are in fact two stars there. If you look with a good telescope, you'll discover that each star is in turn a double star).

Q. Hi, i'm trying to fix my digital satellite dish to recieve some free chanells on hotbird, nilesat,and others. we have been told that the HOTBIRD is on direction of 30E and NILESAT is on 70E, so my question is how can i fix my dish on this direction? can you help me in that?

A.You need to understand a few things to answer this question. (1) What do these `30E' coordinates mean, (2) how do we find these coordinates for the particular satellite you want, and (3) how do you get compass directions and altitude from these coordinates.

(1) To understand these coordinates we need to know a little about communication satellites. Comm satellites are in geo-synchronous or geo-stationary orbits. This means they revolve around the Earth as the same rate the Earth turns giving the illusion to an observer on Earth that the satellite is stationary. You can get a good idea about what I mean by going to the web-sites

The 30E means that the satellite is located over the 30 East longitude line on the Earth. (Remember, it stays above this line permanently due to it's orbit). You may wonder where on this line of longitude the satellite is located. Most communication satellites are over the equator, or roughly zero degrees latitude. You also need to know the satellites distance above the surface of the Earth. This can be found by expressing the angular velocity, omega, of the satellite:

omega = sqrt(G*(mass_of_earth)) / (radius_of_orbit)^(3/2)

where G is the gravitational constant. Setting omega equal to 2*Pi radians per 24 hours, we can solve for the radius of the satellite orbit. The radius turns out to be roughly 42,000 meters. So, now we know the altitude of the satellite and the line of longitude and latitude it is directly over.

(2) But how do we know the coordinates of a particular satellite? check out the web-page

This page lets you see a view of Earth from any satellite. It also gives it's altitude above the Earths surface, and the lines of longitude and latitude directly below the satellite. After visiting this web-page I must call into question your coordinates for HOTBIRD and NILESAT. The coordinates for NILESAT are 1' south lat and 6 degrees 55' west long at an altitude of 35796 km. I found that there are 5 HOTBIRD satellites, HOTBIRD1 -> HOTBIRD5. There coordinates are all roughly the same however. Thus, I will assume you meant HOTBIRD1. Its coordinates are 3' north lat, 13 degrees 9 minutes east long at an altitude of 35794 km.

(3) Now that we know where the satellite is located, we can calculate where it is in the sky with respect to your location. Your email stated you were from Dubai. The latitude and longitude of Dubai are 25 degrees and 15 minutes north latitude, 55 degrees and 16 minutes east longitude. It takes a bit of math to use this info to find altitude in degrees and correct compass direction. To sum up: you must (a) find the vector from your location on Earth and the satellite, (b) transform this vector into a cartesian coordinate system at your longitude and latitude (see Davis and Snider, 'Introduction to vector Analysis' chapter 5.4), (c) finally, now that you know the satellites positional vector in your reference frame, you should be able to figure out it's altitude and compass direction by using familiar dot and cross product definitions. I have figured out the positions of HOTBIRD1 and NILESAT for you hear. To do so I wrote some c-code to make the calculate easy and fast. Here are my results.

NILESAT     |  16.6 degrees   |    257 degrees    |
HOTBIRD     |  33.6 degrees   |    244 degrees    |

The compass direction should be understood as follows,

0 degrees = North
90 degrees = East
180 degrees = South
270 degrees = West

With this key you should be able to find an appropriate compass direction for your dish. Have fun!

Q. Why do all galaxies rotate clockwise?

A. Clockwise and counterclockwise direction depends on the arbitrary selection of an axis about which the rotation will occur. If you look directly at the face of a clock, the hands appear to turn "clockwise." If you were to stand behind the same clock and look through the face, the hands would appear to turn "counterclockwise." The apparent clockwise rotation of a galaxy is a result of selecting the appropriate axis and not an indication of a universally "preferred" direction.

Q. I have been pondering the following three matters in relation to the Earth for some time. Perhaps you can provide some explanation for me.

1. The Earths magnetic field. Since Maganetism is due to the aligment of magnetic domains in materials like steel, or created by the presence of an electric current. Since also the core of the earth is theorecticaly molten and therefore cannot sustain a magnetic field in the absence of an electrical current. From whence do we get it the magnetic field?

2. If the Earth's Magnetic field is due to electrical currents, how are they generated?

3. Is there any evidence that an Earth Year was at one stage 360 days? If there is evidence of a change, was the change in the "days-in the-year" due the the slowing of the earth's orbit around the sun or an increase in the speed of earth's rotation. What caused the change??

A. To answer your first two questions, the magnetic field of the earth (and the sun, and other planets and stars) is not created by simple induction as happens with "ordinary" magnets. They are created instead by a mechanism known as a "dynamo." This theory states that differential rotation of the molten core of the planet allows a magnetic field to be sustained there even in the absence of an electrical current.

As for the third question, the earth's rate of rotation is slowing down due to tidal friction with the moon, at the rate of 0.002 seconds per century. The similar interaction between the earth and the sun is too small to be noticable. So, since the earth used to rotate faster than it does, the year used to have more days in it, but the time it takes for the earth to complete one orbit around the sun has stayed constant.

Q. High or low tide vary in size so greatly in two different locations on the same tide... i.e. 2' high in SF Bay -- same high tide in B.C., Canada is 15'.? Why are there different lengths of time between high and low tides in 24-hours?. What is the total time of hi/lo tides in a 24-hour period!!!???? Does this amount of total time vary from day to day? What affect has the sun on the tides?

A. The variation of the strength in tides from region to region is caused by differences in the local geography--shape of the coastline, depth of the water, etc. When the tides happen and how strong they are from one day to the next in a given place depend on the relative positions of the sun and the moon in the sky. The moon's gravitational pull on the earth creates high tides on the earth around the time when the moon is either transitting (passing through the imaginary line in the sky than goes from due south, through straight overhead, to due north) at that location on earth or when it is transitting on the exact other side of the globe. Since the moon orbits around the earth, it transits at a different time each day, always about 50 minutes earlier than it did the day before. This causes the time between successive high tides (and between successive low tides) to be around 11 hours 35 minutes.

The effect of the sun on the tides is about half as strong as that of the moon, so it doesn't much affect the time between tides, but it does affect their strength. When the sun and moon are in the same direction or in the opposite direction in the sky the tides are stronger. These happen at full moon and new moon and are called Spring tides. (They have nothing to do with the season.) When the sun and moon are 90 degrees away from each other, at first or third quarter of the lunar cycle, the tides are especially weak. These are called Neap tides.

Q. What is the position of each of the stars assigned a Flamsteed Number?

A.Rev. John Flamsteed of England was the first Astronomer Royal. He catalogued nearly 3000 stars at the new Royal Observatory at Greenwich. These measurements were of unrivalled precision but were only published posthumously in 1725 as Historia Coelstis Britannia (with an associated set of charts in Atlas Coelestis in 1729). The so-called Flamsteed numbers were not introduced until J.J. Lalande's French edition of Flamsteed's catalogue (1783). Lalande simply introduced a consecutive number for each star in a constellation.

For a complete set of current positions for stars with Flamsteed Numbers, try Richard Dibon-Smith's The Flamsteed Collection. This book contains over 4900 entries, including all of the stars in the Bayer and Flamsteed catalogs. It also includes all other known stars with apparent magnatudes greater than 5.50. The positions in the book are based on the Hipparcos satellite data.

Q. I would like to know if there was anything special about the year 1962 as far as the planets or something out there,is concerned. My mother once told me scientist wrote her to inform her of their desire to monitor my behavior because of the special time I was born. 2/6/62

A. Loretta,

Thanks for your question. To answer your question one must first define what it means for something to be "special" about the planets. For example, there are some who believe that it is a special occasion if and when certain planets are found near the same location in the sky. These people call themselves astrologers. I am not one of these people and have a very strong negative opinion towards astrology in general. I don't feel there is anything special about planets being in certain position at certain times. Indeed there is no scientific reason to feel that way. The only reason people feel this way is because humans love to see meaning in aestheticly pleasing patterns. For example, I used to work at a drug store in high school. Once in a while a sale would come to an even dollar amount such as $10.00 or $15.00. Customers would always look so surprised an exclaim "WOW! An even dollar amount! I'll bet that doesn't happen very often!" To which I would reply, "It happens about as often as every other number actually." I have coined this phenomena "The Whole Dollar Amount Effect". Kinda catchy, ain't it? Keep this mind when dealing with astrologers.

Going back to your question, while an astronomer would tell you there is nothing "special" about your birth date an astrologer would note that the Sun, Venus, Jupiter, Mercury, Saturn, and Mars were all located near the same area of the sky. Uranus, Pluto, and Neptune were in other parts of the sky. Again, for the same reason that customers get excited when their sale comes out to $13.00 even, people get excited over this "coincidence". It's not as coincidental as you may think, however. Venus and Mercury, being closer to the sun than Earth, are always located "near" the sun, so don't get too excited. These celestial objects were in the same area of the sky for about a month, after which they slowly drifted apart. There is a bit more to the story though. I believe there was an eclipse on the 4th on February that year as well. An astrologer, being a slave to the "Whole Dollar Amount Effect", will jump up and down and get all excited about this. A true scientist, however, will not. As for you comment about scientists telling your mother that they wanted to monitor you behavior, I truly do not believe that these people were scientists in the common sense of the word. They could have been astrologers or other holistic "new-age" types, but I'll bet my next paycheck it wasn't Richard Feynman.

Q. While talking to a co-worker, I was trying to grasp the idea of the physical boundries of the universe. If it is finite, then it must be contained within something. If it is expanding, then it must be expanding into something. His explanation was something to the effect of: if you started walking in a straight line from one end of the universe and continued walking (albeit a LONG time later!) you would eventually arive BACK at the spot at which you began.

A. Turns out your co-worker is right! According to Einstein, sapce is really a 4-dimensional object called "spacetime". Spacetime is curved in the 4th dimension, even though we cannot directly perceive this curvature (we precieve it indirectly in the form of gravity; mass bends space in the fourth dimension). A common analogy is to imagine you are a two-dimensional creature on a balloon. If you go out in any direction, you eventually come back around to where you started but, not understanding 3-dimensional space, you wouldn't understand how this could have happened (unless you were a two-dimensional Einstein and could desribe the phenomenon mathematically). This analogy not direct BECAUSE the balloon has an "outside", but the Universe is everything. In a sense, it contains itself and therefore does not expand "into" anything.

There are a few things to keep in mind: 1) there hasn't been enough time to make such a trip "around the Universe". The maximum speed is the speed of light, and the universe just isn't old enough to have been traveled through completely. 2) What I described above is for a closed Universe, one which has enough mass in it such that it will eventually recollapse to a pre-Big Bang state. If the universe is "open", it expands forever, and the topology is more complicated. Here it is event horizons which would limit our travels in principle.

Q. What are the positions of the now obsolete constellations: Cancer Minor, Musca Borealis, Sagitta Australis, and Triangulum Minor?

A. A useful source of information about obsolete constellations is The Cambridge Guide to the Constellations by Michael Bakich. Cancer Minor consisted of the stars seen between Cancer and Gemini. Musca Borealis, also known as Apes, consisted of Flamsteed 41, 33, 35, and 39 in Aries. Sagitta Australe consisted of stars to the north of Aquila. Triangulum Minor consisted of three stars to the south of Triangulum in the direction of alpha Aries. Plancius formed Cancer Minor, Apes, and Sagitta Australe around 1614. Hevelius included Triangulum Minor in his atlas published in 1687. The Cambridge Guide to the Constellations includes an illustration of Triangulum Major and Triangulum Minor from the Atlas Coelestis of 1742.

Q. Can you tell me the exact value of Mars' axial tilt?

A. The 2000 edition of the Astronomical Alamanac and the (US) National Space Science Data Center list Mars as having an axial tilt of 25.19 degrees.

Q. I saw in your web site that the earth is about 13 miles "thicker" at the equator than at the poles (difference between equatorial and polar radius). I assume that this is caused by the spinning forces of the earth "piling up" mass along the equator .

My question is: of the 13 miles of "extra stuff" at the equator, how much of it is water and how much of it is rock? I have assumed that much of it was water since it has a much lower viscosity that the earth's mantle, but then there is so much more mantle that the thin veil of water. Once we have tacked this; how is this bulge distributed north to south? In other words, is the bulge zero after you travel 500 miles north of the equator or is it 1000? If you have an equation to define the profile of this bulge, it would be appreciated, especially if the profile differs ocean from crustal bulge.

A. The Earth's surface is what is called an equipotential. It is where the centrifugal potential of the Earth, the graviational pull of the Earth and the pull of Sun and Moon all add up to the same value. The moon's gravity is the primary element pulling it out of shape. The shape of the potential is only slightly distorted. That's a 13 mile bulge of out 3800 miles. The stratification of the Earth's layers will not change significantly. The effect that causes heavy elements to "sink" to lower layers and light elements to "rise" to higher layers will operate almost the same. The Earth's mantle will have a slight bulge, as will the core.

The equation describing the shape of the Earth is complex, but a rough description would probably be taking the cosine of the latitude.

Q. What is the diffence between a light telecope and a radiotelescope?

A. Well, the basic idea is the same. Radio and optical telescopes both observe "light", it's just that radio telescopes observe light with very long wavelengths. However, due to the differences in wavelength and frequency, optical and radio telescopes use different technology to detect photons. Optical telescopes gather and focus light, and we can look at this light directly wiht our eyes or, as professional astronomers do, use a device called a CCD, which is a computer chip much like the one you'd find in many video cameras. Put simply, radio telescopes gather and focus radio light, and the oscillations of these focused EM (electromagnetic) waves induces a current in the radio telescope's antenna, which can then be amplified and recorded. An interesting aspect of radio astronomy is that since radio light has such long wavelengths it is easy to do what is called interferometry, where an array of radio telescopes look at the same object and their data is combined to simulate a single larger radio

Q. Dear Sir Days ago I saw a magazine talking about the dangers of cosmos. I'm a portuguese teacher and I don't understand nothing of astronomy, so... I'm in panic. I'm very nervous because I don't know what to think. We have future, haven't we? I'm very nervous. Humanity has still long time to live, hasn't it? May I relax and keep living my life? Sincerelly Patri'cia

A. Dear Patri'cia

Well, let me start out by saying you have nothing to worry about. Yes, humanity has a future and yes you can keep on living your life. I'm not exactly sure what magazine you read so I can't comment on the article. It is true that there are many dangerous "things" in the cosmos. For example, impacts from large comets and asteroids could threaten life on Earth although the possibility of such a large impact is minuscule. Also, our sun will eventually grow so large that the present orbit of Earth will be located inside the red giant! Before you run off and buy 100 shares of coppertone stock, you should know that this won't happen for another 5 billion years. So sit back, relax, and enjoy yourself - there are plenty of other things to worry about!

Q. I have been searching the Web trying to find out how to report a comet or meteor sighting. At least I think it was a comet or meteor. While camping in S.W. Virginia on Friday 9/10/99 I and a few others saw what looked like a comet. I say a comet because I have seen meteor showers before and this looked very different. It was very close ( viewed by naked eye) and looked almost gold in color. It traveled from north to the east and had a large, thick tail. I have been trying to find other people who have reported this but have not found anything so far. Any information you could provide would be greatly appreciated.

A. Comets move very slowly through the sky, so they in fact do not appear to be moving at all at any given time. One can only tell that they're moving by watching their changing position from night to night. This is because comets never get very close to earth, so even a few thousand mile change in their position is viewed as a very small distance on the sky (a thousand miles viewed at a distance of a million miles is very small indeed).

What you saw sounds typical of brighter meteors. A meteor a little larger than a grain of sand will make a spectacular display as it burns up in the upper atmosphere. So-called "fireballs" are extremely bright and leave behind a gold-orange tail that often persists for a few seconds after the meteor itself is gone. During most meteor showers these are fairly uncommon, and on ordinary nights they are a very rare sight. Consider yourself lucky to have seen this.

Q. Could you tell me how scientists measure the distance between earth and other neighbouring planets?

A. The relative distances between planets have been known since the time of Copernicus and Kepler. If one measures how long it takes a planet to complete one orbit around the Sun (which can be measured observationally), and if one uses Kepler's third law (the square of the orbital period is proportional to the semi-major axis cubed), the distances to the planets can be measured in units of the Earth-Sun distance. For example, you can say Jupiter is 5.2 times further from the Sun than the Earth.

However, to truly measure the absolute distances in a unit like meters, you have to measure Earth's distance to at least one other object in the solar system. In recent years, distances to bodies in the Solar System have been measured by reflecting radar off of their surfaces. Since radar moves at the speed of light, if you measure how long it takes for a return radar echo from Venus, for example, then the distance to Venus equals the speed of light times half of the radar beam travel time. The factor of one-half comes in because the radar beam makes the trip from Earth to Venus and then back from Venus to Earth.

We know that at its closest, Venus is about 1/3 of the distance from the Earth to the Sun, so we can use the radar measurements to set the value of the Earth Sun distance, which is called an Astronomical Unit (AU). The value of the AU is 1 AU = 149,597,892,000 meters. From Kepler's law we know how far each planet is from the Sun in AU, so we can convert these distances to distances in meters using the value of the AU measured from radar distances to Venus.

Q. Could you give me the exact time for the winter solstice 1999 in Germany (northwest) or an address to find it?

A. The solstices and equinoxes occur at a single moment in time which is the same for everyone on earth. The 1999 winter solstice is on December 22 at 7:44 am Universal Time. Universal Time is equivalent to Greenwich Mean Time.

Q. Do you have any information about buying a star or having a star named for a loved one?

A. The International Astronomical Union is the only official organization that names astronomical objects, including stars, and does not engage in commercial transactions of this nature. For additional information on such ficticious offerings, review the FAQ at For additional information on the legitimate naming of celestial objects, review the FAQ at

Q. Does the University of Virginia have an Astronomy Club? If so, how does one get involved?

A. UVa does have a student-run astronomy club. For information, please contact Steven Thomas (smt2k at virginia dot edu).

In addition, you might consider the

Charlottesville Astronomical Society
The Leander McCormick Observatory is open on the first and third Friday of each month. For more information about these public nights, try

Both the Astronomy Department and the National Radio Astronomy Observatory hold weekly colloquia, the schedule for which can be found at Other local events of Astronomical interest, such as the annual Jansky lecture, are also listed there.

Q. Can you tell me something about Orien (is that how you spell it?) and also about the North Star. I am having trouble finding information on the Web so far, though I haven't looked at many sites.

A. Orion is spelled O-r-i-o-n. It is one of the most recognizable constellations in the winter sky. Orion, or "The Hunter" has seven bright stars that make up his shoulders, knees and his waist. The short row of three stars at Orion's waist is called "Orion's Belt". The brightest stars in the constellation Orion are Rigel and Betelgeuse. Rigel is the bluish-white star located at Orion's left knee and Betelgeuse is the reddish star found at the warriors right shoulder. On a dark night you can make out Orion's bow and his sword. His sword is located just below his waist, between his legs. To find information on the myths involving Orion try the following links:

The North Star marks the direction "north" on a compass. If you were to extend the axis of the Earth you would notice that is "almost" intersects a bright star in the northern part of the sky. This star is named Polaris and is the North Star. Because it is almost located along an imaginary line extending from the Earth's axis, it appears to be stationary while all other stars trace out circles across the sky as the Earth rotates.

Q. I would like to buy a filter for my small refracting telescope in order to cut down glare when observing the moon. Would you recommend a polarizing filter or something else?

A. Either a polarizing filter or neutral-density filter will cut down on the glare while observing the moon. Some of these filters are designed specifically for lunar observing. When choosing a filter, select a filter of comparable quality with your telescope.

For detailed information on equipment, you might want to look at "SkyWatch" or "Sky & Telescope" ( or "Astronomy Magazine" ( Astromart ( has classified advertisements of equipment for sale.

Q. We were looking at the sunrise sunset table from the Naval Observatory and noticed that even though sunset starts to get a little later every day after the winter solstice, sunrise also keeps getting later until January. Then, it starts getting earlier so that the days are getting longer at both ends - but the change is assymetric. Why isn't the amount of change the same at both ends of the day?

A. I refer you to our archive at www.astro.Virginia.EDU/AQuA/archive.php and the questions about why the dates of earliest sunset and latest sunrise do not occur on the same date, about why the day with equal hours of day and night does not occur on the equinox, and why sun dials run slow. Norton's 2000.0 Star Atlas and Reference Handbook might also help you. Apparent solar time is the time measured by a sun dial showing the actual motion of the sun across the sky. Mean solar time is the time measured by a clock, which smoothes out the irregularaties of apparent solar time. The difference between the two measures is called the equation of time, which varies from several minutes ahead to several minutes behind. Between January 1 and January 20, mean solar time is ahead of apparent solar time by between 3 and 11 minutes. During January, sunrises are getting earlier and sunsets are getting later but at different rates.

Q. What formula do I need to correct a tilted telescope if I have two stars with their altitude and azimuth values and the manual corrections that are necessary to bring them into the center of my telescope?

A. The difference between the sellar positions read from your setting circles when the stars are centered in the telescope and positions from a star catalog corrected for precession is the value you need to adjust the telescope. You can improve this correction by measuring the position of several stars near the zenith and performing a least squares fit.

For a least squares fit, you assume the relationship between the value on your setting circle (Y) and the value from a catalog (X) can be expressed as a line (Y = m X + b). Taking your n measurements (Xi, Yi), you can calculate the best m and b values as follows

p = sum of all values Xi * Yi
q = sum of all values Xi
r = sum of all values Yi
s = sum of all values Xi squared or sum of all values Xi * Xi
t = (s * n) - (q * q)

m = ((p * n) - (q * r))/t
b = ((s * r) - (q * p))/t

Additional information on adjusting your telescope may be found in Volume 1 of the Compendium of Practical Astronomy edited by Gunter Dietmar Roth (translated into English and revised by Harry Augensen and Wulff Heintz), which is published by Springer-Verlag Company. It also includes information about performing least squares analyses.

Q. Would you please provide me the latitude and longitude of Charlottesville? I received a 5" Schmidt-Cassigrain telescope for a Chistmas present.

A. The McCormick Observatory on the University grounds is at 38 degrees, 2 minutes north lattitude and 78 degrees, 31 minutes, 24 seconds west longitude. The full extent of Albemarle county is about 20 minutes in both longitude and latitude, with the observatory roughly in the center of the county, so you will need to adjust the above numbers slightly depending on where in Charlottesville you live.

Q. Who actually discovered the planet Jupiter? I know it was named by the ancient Romans, and that Galileo discovered its four largest moons, but who actually first discovered it? Because it is so large and visible, is it even possible to determine who discovered it?

A. Becuase Jupiter is visible to the naked-eye, people must have seen it in prehistoric times. Usually, only the Sun, the Moon, and Venus are brighter. Therefore, we do not know who first discovered it. However, the Babylonians began compiling star catalogs and recording planetary motions in about 1600 BC.

Our name Jupiter comes from name of the supreme god in the Roman religion. The god Jupiter was a sky god and the protector of the Roman state. Other cultures have different names for this planet.

In 1610, Galileo Galilei discovered the four largest moons: Io, Europe, Ganymede, and Callisto. Since his time, 12 additional moons and three rings have been discovered.

Q. Who actually discovered the planet Jupiter? I know it was named by the ancient Romans, and that Galileo discovered its four largest moons, but who actually first discovered it? Because it is so large and visible, is it even possible to determine who discovered it?

A. Becuase Jupiter is visible to the naked-eye, people must have seen it in prehistoric times. Usually, only the Sun, the Moon, and Venus are brighter. Therefore, we do not know who first discovered it. However, the Babylonians began compiling star catalogs and recording planetary motions in about 1600 BC.

Our name Jupiter comes from name of the supreme god in the Roman religion. The god Jupiter was a sky god and the protector of the Roman state. Other cultures have different names for this planet.

In 1610, Galileo Galilei discovered the four largest moons: Io, Europe, Ganymede, and Callisto. Since his time, 12 additional moons and three rings have been discovered.

Q. I want to know how to calculate the sun's declination with good precision for any given date.

A. First, I would refer you to our archives for the question about obtaining the Sun's position in altitude-azimuth coordinates. The Astronomical Almanac published by the United States Naval Observatory contains a set of equations for calculating the Sun's declination. The Naval Observatory, also, provides a calculator on their webpage ( for obtaining the Sun's position in altitude-azimuth coordinates, which may then be converted to right ascension-declination.

Q. I am looking for a large list of stars of the Milky Way including their location, distance and type. I have found small lists which concentrate on the constellations and the named stars. I am interested in numbered stars.

A. You can find many catalogues of stars that have been selected for different reasons.

Astronomers at the United States Naval Observatory have compiled very large catalogues of stars with positions and magnitudes (brightness) listed.

For example, there is the AC2000: (

and the B1.0: (

The Astronomical Almanac, which is published yearly, contains a table of the brightest stars that includes positions, magnitudes, and spectral types.

Larger catalogues include the "The Hipparcos and Tycho catalogues : astrometric and photometric star catalogues derived from the ESA Hipparcos Space Astrometry Mission", which is a 16 volume set listing the positions, distances, velocities, magnitudes, and colors of an enormous sample of stars. An older catalogue that includes spectral type information on its stars is the "Bright Star Catalogue" by Dorrit Hoffleit. Both of these references can probably be found in the Astronomy/Physics library of most major universities that have active research programs in Astronomy.

Q. What is the term for a star whose exact distance is a known quantity and is used to guage the distance of other stars. I was under the impression that such a star was known as a "sephiant" star or something to that effect but I could not find such a word in any dictionary I consulted.

A. Standard candle is the term used to describe a class of objects for which we know the actual luminosity well. By comparing the appearent brightness of a specific member of the class with the expected brightness for an object in that class, we can calculate the distance to the specific object. Comparisons of other known quantities may also be used. Several different classes of objects are be used as standard candles, including Cepheid variables, RR Lyrae variables, the brightest stars in a galaxcy, and HII regions (comparison of angular size instead of brightness).

The lumimosity of a Cepheid variable star is directly related to the period of time over which the luminosity of the star varies. Distances to Cepheids may be obtained by measuring the period and the average apparent brightness of the star. Using the period-luminosity relationship, the luminosity of the star (M) may be obtained. This value is compared to the apparent brightness (m) to determine the distance using the formula

m - M = 5 log d - 5
where m and M are given in magnatudes, d is given in parsec, and log is with respect to base 10. However, two classes of Cepheids exist: Type I or Classical Cepheids and Type II Cepheids. Each class has a different period-luminosity relationship. When this was discovered, the distance scale for the entire universe doubled.

Q. I have a designation and a Planet # of Asteroid, (774) Armor = 1913 TW but, I'm still having a difficult time in finding more details. Could you please provide any information about Asteroid Armor?

A. C. le Morvan discovered the asteroid or minor planet now known as (774) Armor on December 19, 1913 at Paris. "1913 TW" was the provisional designation given to the discovery. In the old-style of provisional designations, asteroids were identified by the year of their discovery and a two-letter sequence indicating the order of discovery. More information about designations is available from the Minor Planet Center at (especially the link to "Provisional Designation").

(774) Armor is an S-type asteroid. S-type asteroids compose about 17% of astroids and have albedos between 0.10 and 0.22. These asteroids are believed to be composed of metalic-silicates.

(774) Armor is located in the main asteroid belt. The semi-major axis of its orbit is 3.0 AU long. The eccentricity of its orbit is 0.17. Orbital ephemerides for this asteroid is available from the Minor Planet Center at

Q. I am currently reading the book In search of the edge of time: black holes, white holes, wormholes by John Gribbin. In chapter two, Mr. Gribbin explains in supposedly layman's terms Einsteins how acceleration is exactly equivalent to gravity by telling to story of the high-speed elevator and how a light's beam, in the elevator, would bend due to gravity. I haven't exactly gotten the point, and i was wondering if you could enlighten me on the subject. It would be greatly appreciated.

A. This is an example of what is known in General Relativity as the strong equivalence principle. It says that acceleration due to gravity (or lack thereof) and acceleration due to any other mechanism are indistinguishable. In the elevator thought experiment, we imagine that we are in an elevator somewhere in deep space, where there are no nearby masses creating an acceleration on the elevator. If we shine a flashlight across the elevator, the beam of light goes straight across. Now imagine the same elevator in free fall in a gravitational field. The strong equivalence principle says that the two situations are indistinguishable to someone inside the elevator, so if we again shine a flashlight, the light beam should go straight across the elevator. But the fact that the elevator is in a gravitational field means that it is falling, so in order to go straight across the elevator, the light must be falling with the elevator. To an observer outside the elevator, the light would appear to bend downward.

This brings us to the conclusion that light must fall in a gravitational field. When viewed by an observer who is stationary within that gravitational field, i.e. not falling, the path the light takes appears to bend. On everyday scales, this is not visible, but near very dense objects like neutron stars and black holes, the gravitational bending of light is easily observed.

Q. I am a designer for a major restaurant company and I need your help! If I have a 23' wall, north facing, here in Vancouver, BC, Canada, what length of shadow, from the sun, will it produce? (Max. & Min. values) I believe our coordinates are: 49,16 Degrees North 123,7 Degrees West Thank you so much for your time and for your help!

A. I work in the shadow division at the astronomy department here at UVa. The answer to your question is as follows:

As you may know, the length of your shadow changes during the day depending on where the sun is. When the sun is rising and setting your shadow is the longest because the sun is on the horizon, obviously. It may look 50 feet long or more. However, during the day the sun rises until it is at it's highest peak (typically around 1:00 PM). It is at this time that your shadow is as short as it will get during the day. Afterwards, the sun begins to set and your shadow increases.

Now, during the summer the sun's highest point is much higher than the sun's highest point in the winter (that's why it's warmer in the summer). As a result, your shadow gets much shorter in the summer. The shortest your shadow gets is during the summer solstice (June 21st). At this date the shadow of your wall will be as short as 10 feet (at around 1:00ish; it will be longer the rest of the day). During the winter, the shortest your shadow will be (at 1:00ish) will be 73.7 feet (the shadow will be longer at other times during the day however).

It is important to keep in mind though that, for the entire year, the LONGEST your shadow ever gets is very long, during sunrise and sunset, everyday. The shortest it gets is around 1:00 on June 21st.

P.S. My mom lives in Seattle, can she drive up to your restaurant and get a free meal sometime?

Q. Are stellar winds the same as we have here on earth? iI not, could you give me a simple explaination of the difference?

A. Wind on the earth is just a flow of air molecules; similarly, a stellar wind is a flow of subatomic particles released by a star. A stellar wind is a steady stream of electrons and protons that flows off the surface of a star into space. The impact of these particles on objects in space puts pressure on them in much the same way as air molecules put pressure on things they strike here on earth. This pressure is what causes wind to blow things around. A stellar wind also "blows" on things like dust grains, and even the earth's magnetic field.

Q. 1.Why are scientists interested in finding out accurate measurements of star positions? What do they use information for? 2. What is a star trail? Could you give me some addresses that might show star trails?

A. 1. There are several uses for accurate star positions. One simple example is that often these positions are used as reference positions by other astronomical instruments. What I mean by this is that for example, the Hubble Space Telescope uses a catalog of stars and their positions (the Hubble Guide Star Catalog) to help it determine where it is pointed at a given time.

A more complex example of what stellar positions are used for is the measurement of trigonometric parallax (distance) and proper motion (velocity). By accurately measuring the *change* in position of stars as a function of time, you can accurately determine two quantities: the distance to the star and the velocity of the star in the plane of the sky. If you measure the distance and velocity for a large sample of stars, you can determine the three-dimensional shape of the stars that make up the Milky Way Galaxy and also how those stars are moving within the Galaxy.

2. A star trail is a picture of the sky that has been taken over a long period of time so that the image of each star appears to be a streak of light instead of a point of light. To take a star trail photograph, you need to point your camera at a portion of the sky and let the shutter remain open for a long enough period of time (usually a few hours) so that the rotation of the Earth causes the stars to appear to move along circular paths from East to West. The Department of Astronomy had a star trail image as its "picture of the month" for March of 2000. To view this picture, go to The Picture of the Month page and click on the link for the March 2000 image. <-->

Q. Is there such a theory called, "Star-Mix Theory"??

A. Perhaps you are thinking about "mixing length theory." The average distance that a convective bubble of gas travels in the interior of a star is called the characteristic length or mixing length. The mixing length is proportional to the pressure scale height. These values are important in the development of models for main sequence stars. Materials in a convective layer will become well mixed in a short period. Materials in a radiative layer adjacent to a convective layer will be mixed more slowly as convective bubbles of gas overshoot the boundaries of the convective layer. Observations of the relative abundances of elements in stars can provide information about the size and temperature of the convective zones.

Q. Are the darker spots on the moon in the same place at each of the phases of the moon? Can one use his/her observations to show that the moon keeps the same face towards the earth?

A. The physical features of the Moon that are visible to an Earthbound will be in the same location every night. The area of the Lunar surface illuminated and associated shadows depends on the phase. Regular observations of the surface features should demonstrate that the Moon keeps the same face towards us. Using a good map or chart helps identify the distinguishing features. Until the advent of Lunar orbiters, we had no knowledge of the features on the opposite side.

Q. Does the sun rise or set (occur) at the same time and place each day? If not, where is it moving?

A. The sun does not rise and set at the same time each day. The relative amount of daylight changes throughout the year. For more information on this topic, I refer you the question in our archive regarding why sundials run fast and slow and the questions about sunrise and sunset times for the solstices.

Q. How do Saturn and/or Jupiter move against the background stars?

A. Saturn and Jupiter rise in the east and set in west on a daily basis. Most of the time, they move eastward against the zodiac. However, both planets demonstrate retrograde motion at some times during which they move westward with respect to the zodiac. Sky charts or planetarium programs can help you locate the planets at a specific time or track the motions over time.

Q. In what direction does the Winter Triangle seem to be moving?

A. The Winter Triangle is an asterism created from Sirius in Canis Major, Betelgeuse in Orion, and Procyon in Canis Minor. On a daily basis, the stars rise in the east and set in the west. Asterisms, like constellations, move with respect to the Sun over the course of a year. During the winter, the Winter Triangle will move westward towards the Sun until it is no longer visible in the night sky. During the summer, the Winter Triangle is below the horizon at night.

Q. I would like to know if anyone there knows of anything that may of happened in September of 1993 concerning Astronomy or Astrophotography. Were there any breakthroughs, was anything new invented in reguards to high-powered telescopes, Do the words "Be camer" or perhaps "Be camera" ring any bells? If so then where can I go on the web to get this info.? Did any comets swing by, maybe an eclipse? Were there any new galaxies discovered. . . . . Anything you can think of or even if you could direct me down any productive avenues would be great!

A. In September 1993, a number of astronomical events occurred of varying degrees of consequence. As a complete listing would be prohibitive, I have selected a few of these that might interest you. Mars and Jupiter were in conjunction on September 6 and appeared approximately 0.9 degrees apart. A nearly full Venus was visible within 18 arcminutes of Regulus on September 21. The autumnal equinox occurred at 8:22 pm EDT on September 22. Uranus and Neptune were visible within 1.1 degrees of each other on September 28. In the far north, an occultation of the asteroid 89 Julia with the star SAO 39807 was visible.

A 9th magnitude nova, Nova Sagittarius 1993 was discovered on September 14 independently by Matsua Sugano in Japan and William Liller in Chile when it was already several days past its peak. Four new icy planetesimals were found beyond Neptune by Jan Luu of Stanford University and David Jewitt of the University of Hawaii. The Orbital Retrievable Far and Extreme Ultraviolet Spectrometer (ORFEUS) flew a six-day mission from the bay of the space shuttle.

NASA selected four possible Small Explorer Missions: Joint Ultraviolet Night Sky Observer (JUNO), Positron Electron Magnetic Spectrometer (POEMS), Transitional Region and Coronal Explorer (TRACE), and Wide-Field Infrared Explorer (WIRE). TRACE launched successfully on April 2, 1998, and continues to provide information about our Sun. WIRE was launched on March 4, 1999. However, design errors prevented the spacecraft from accomplishing its primary mission. It has been rededicated to astroseismology and planet-finding.

Although additional information on NASA missions may be found on the NASA web pages, additional information on events in 1993 may require you to review back issues of astronomy publications, such as Sky and Telescope or Astronomy.

Amature and professional astronomers continue to use cameras to record images. However, camera technology continues to change from the photographic plates used early this century to charged-coupled devices (CCD) used now. According to Webster's Revised Unabridged Dictionary, a bellows camera is one that can be expanded or contracted like an accordion. However, I am unaware of a specific astronomical application for such a device.

Q. I am curious about when astronomers say they "See back in Time" - what does that mean? With the Hubble being so powerful they can "See back to when the universe started." Can you explain how one can see back in time?

A. Light from objects that are really far away take a lot of time to reach the Earth. For example, if you have an object you know is 10 light-years away, you know the light you are seeing now had to leave the object 10 years ago. Thus, you are "looking back in time" because you are looking at light from 10 years ago. The fainter an object is, the farther away it is (usually - sometimes you can just have a really faint object that is actually close) so since the Hubble telescope can see objects that are so faint, they must be really far away. Thus, the light Hubble sees from these objects left these objects a really long time ago so we are seeing light from a long time ago and essentially "looking back in time."

Q. I'd appreciate information about "lunar standstill." I've heard about it in conjunction with Chimney Rock, an archaeological site in Colorado. I'd like to know what it is, how often it happens and when it will happen again. I've looked for these answers in lots of places and hope you have them.

A. The moon's orbit is tilted about 5 degrees from the ecliptic. You know how the sun appears to be in a different constellation each month? Well, the ecliptic is just the line that connects all the points the sun appears to be over the course of a year. Because of the tilt of its orbit, the place we see the moon with respect to the background stars and the ecliptic changes. The moon wanders from being 5 degrees above the ecliptic to 5 degrees below the ecliptic. A major standstill is when the moon is 5 degrees above the ecliptic and a minor standstill occurs when the moon is 5 degrees below the ecliptic. There are about 18 years from major standstill to major standstill, so there are about nine years between a major standtill and a minor standstill. The last minor standstill was in 1996-1997 and the last major standstill was in 1988, so the next major standstill should be in 2005-2006.

This is often referred to along with Chimney rock because the moon appears between the pinnacles during a major standstill. It also is significant because the 11th century construction of the Chimney Rock Pueblo was during a major standstill.

Q. Planetary nebulae eject what two heavy elements into the galaxy?

A. They eject many heavy elements. Their principle ejecta are hydrogen, carbon, oxygen, neon magnesium and nitrogen.

Q. What Satellite Galaxy is moving fastest toward the Galactic center? What Satellite Galaxy is moving fastest away from the Galactic center?

A. Difficult to say. We can easily measure the radial velocity of a galaxy (the motion toward or away from us) but the transverse motion is very difficult. The motions are tiny and require very careful and accurate measurements. At the present time, only four galaxies have reliable measured motions. Sagittarius is probably the fastest moving as it is at is pericenter, but that motion is transverse, not toward or away.

Q. What are the two most distant galaxies in the Local Group of Galaxies?

A. At present, the two most distant galaxies are the dwarf galaxies NGC 55 and EGB0427+63.

Q. What planet's atmosphere is virtually all helium?

A. None. Planetary atmospheres have a similar chemical composition to the Sun - i.e., mostly hydrogen and hydrogen compounds. Earth, Mars and Venus are dominated by Nitrogen and carbon dioxide.

Q. What are "IRS 7" and "IRS 16?"

A. IRS is an infrared source. In these two cases, IRS 7 is a supergiant stars toward the Galactic Center, IRS 16 is a cluster of bright blue stars in the same area. Both produce powerful winds that affect the morphology of the Galactic Center and are affected by the extreme environment in that area.

Q. Who coined the astronomical term "subdwarfs?"

A. Subdwarfs are stars that are faint for their measured color - usually because of a chemical composition that has not been enriched with iron and other heavy elements. The term has been in use since at least 1940. The first publication I see it in was by Kuiper, but it was probably in use long before then.

Q. I read recently in an astronomy book that the Milky Way's companion galaxy, Andromeda, is approaching the Milky Way galaxy (due to the gravitational force) at some appreciable speed (1000s of kilometers per second?). Since Andromeda is approaching the Milky Way, does that mean its spectrum is blue-shifted because of the doppler effect? I have never read anywhere about any galaxy's spectrum being blue-shifted. In fact, wouldn't all the galaxies in the "local group" be blue-shifted since they are all affected (i.e., attracted) by gravity.

A. As I recall Andromeda is blueshifted, but I don't remember the speed without looking it up (1000's of km/sec sounds believable). This shift is in fact due to the Doppler Effect. All the galaxies in the local group orbit one another (or, more precisely, the center of mass of the group), so at any point in time some will be moving towards us while others are moving away from us, just as in our solar system other planets will sometimes appear to be moving towards us at at other times away from us depending upon where they (and we) are in their orbits. These so called "peculiar velocites" can therefore produce a blue or red doppler shift.

The reason you most often hear about galaxies being redshifted is because the universe is expanding, so like raisins in baking bread all galaxies are moving away from one another (EXCEPT those which are close enough together for the gravitational attraction to overcome the cosmological expansion). As you get further and further from the Milky Way, the expansion velocity continues to grow, while the peculiar velocity of any given galay remains about the same, so the expansion velocity will eventually overcome the peculiar velocity thereby causing all distant galaxies to appear redshifted.

Q. If humans were to travel at the speed of light to a planet, say, 50 light years away, that would mean that from an earthbound perspective the travelers had taken 50 years to reach the destination, right? If so, what would be the subjective time-lapse for the travelers? Is there a formula for calculating subjective time at the speed of light?

A. An object moving near the speed of light will experience a phenomenon known as time dilation. A clock that is moving at a fixed velocity with respect to a stationary observer will run slower than it would if it were at rest.

The time measured by the moving clock can be calculated using the time dilation formula:

t_r = t_m / sqrt(1 - (v^2/c^2))

here, t_r means time in the rest frame, t_m means time in the moving frame, sqrt means take the square root, v is the velocity of the clock, and c is the speed of light.

Q. A. Can you tell me how wide (and narrow) each of the constellations of the Zodiac are and give a fine reference too?

B. What is the most exact number that can be given for the "wobble cycle" (of equinox retrogression) of 25,800 - 26,000 years?

A. To answer your first question, the International Astronomical Union (IAU) adopted boundaries for each constellation in 1930. The accepted boundaries are published in the following work: E. Delporte, in, Delimitation scientifique des constellations (tables et cartes), Cambridge University Press, 1930. The boundaries are irregular polygons, so the width varies depending on where you choose your starting and ending points. However, we can get a rough estimate by taking the angular size of a great circle on the sky (360 degrees) and dividing this by the number of zodiacal constellations (12) to find that each is roughly 30 degrees wide.

To answer your second question, I also referred to an IAU publication. The IAU system of constants sets the general precession rate to be 5029.0966 arcseconds per century. At this rate, a full cycle (360 degrees) is completed in 257.7003591 centuries, or 25770.03591 years.

Q. What year was the first planet *outside* the solar system found?

A. Although no one has yet directly imaged a planet outside our solar system, the presence of planets around other stars has been inferred due to their gravitational pull on their parent star. In 1992, Alex Wolszczan and Dale Frail proposed that planets were orbiting around a pulsar, named PSR B1257+12, based on irregularities they observed in the timing of the pulses emitted by the neutron star.

Q. Could you possibly tell me what the most current scientific opinion is on what the rate of change or movement of the O/E is?

A. The Astronomical Almanac (Library of Congress number QB8.A75 2000) gives an equation to determine the obliquity of the ecliptic as a function of date. From the 2000 edition, page B18,

"The obliquity of the ecliptic of date (with respect to the mean equator of date) is given by

23.439291 - 0.0130042 T - 0.00000016 T2 + 0.000000504 T3

where T = (JD - 2451545) / 36525, and JD is the current Julian Date."

More details about the precessional motion of the ecliptic can be found on the same page in the Almanac.

Q. How long would it take to travel to Venus?

A. The amount of time required to reach the surface of Venus depends on how fast the vessel goes, the relative positions of Venus and the Earth, and the path chosen. The average time for the Soviet Venara missions (Venera 3, 4, 7, 9, 10, 11, 12, 13, and 14) to reach the surface of Venus was 118 days. The shortest time was 98 days for Venera 12. The longest time was 134 days for Venera 9. The US Pioneer Venus 2 mission took 123 days. The National Space Science Data Center provides a space exploration chronology at

Q. Sir, plaese tell me the orbital radius of the binary pulsar PSR 1913+16. I am trying to find the ORBITAL angular momentum of the system. Can you tell me what the orbital angular momentum is?

A. Because the orientation of the orbital plane of this binary system is not know, the semimajor axis of the orbit cannot be determined exactly. The semimajor axis (the orbital radius at the largest separation between the two neutron stars) is 2.34186 sin i, where "i" is the angle that the orbital plane makes with the plane of the sky. The stars are also not in circular orbits around their center of mass, but are in very elliptical orbits. They have an orbital eccentricity of 0.61714. In comparison, a circular orbit has eccentricity 0, and a radial orbit (head-on collision) has an eccentricity of 1.

Angular momentum, usually denoted by "J", is given by the square of the semimajor axis times the change is orbital angle with time. The latter can be calculated using Kepler's third law, which can be found in any introductory astronomy text.

Q. Could you possibly tell the the strength of the gravitational pull of the sun, and if there are different strengths on the different planets of our solar system.

A. The gravitational pull between any two objects in the Universe is described by Newton's equation:

F = G * M1 * M2 / R^2

Where G is the gravitational constant of the Universe, M1 and M2 are the objects and R is the distance between them. In cgs units, G is 6.67 x 10^-8 gm^-1 cm^3 s^-2.

An example calculation relevant to this question. The Sun has a mass of 1.99 x 10^33 gm, the Earth has a mass of 5.98 x 10^27 gm, they are separated by 1.5 x 10^13 cm. So the gravitational force is 3.53 x 10^27 Newtons.or 1.6 x 10^28 pounds of force.

Q. For a project that I am doing, I need to know what Mercury's weather is like, but I can't seem to find the answer. Can you please help me?

A. Mercury doesn't have weather, per se. It has no atmosphere, therefore no clouds or precipitation or anything else. What Mercury does have is temperature extremes. In direct sunlight, the surface of Mercury is 800 degrees F. In shadow, the temperature is -280 F.

Q. Do you think there will be a supernova soon?

A. The rate of supernovae in our own galaxy is approximately 1 every 50 years but not all supernovae in our own galaxy will be visible to us. However, supernovae from other galaxies may be visible to us. For 1999, the International Astronomical Union (IAU) lists reports of 204 supernovae. For 2000 the count was 183. However, many of these supernovae are only visible using large telescopes. You can find some lists at Therefore, another supernova will occur shortly but it probably will not be visible with the naked eye.

Q. There have been amazing advances in astronomical observations in the past thirty years. What specific advances do you think will occur in the next thirty 30 years?

A. In the October 2000 edition of On-line APS News (available at, Virginia Trimble, a professor of physics/astronomy at University of California, Irvine, and visiting professor of Astronomy at University of Maryland, College Park predicts the top ten astronomical discoveries discoveries of the 21 Century, as follows

Top Ten Predicted Hits of the 21st Century

  1. Neutrino astronomy will find a third source and become routine.
  2. Extremely high energy cosmic rays will reveal new kinds of physics or new particles.
  3. Gravitational radiation astronomy; backgrounds and sources.
  4. What came before the Big Bang? How did large scale structure form? What is the dark matter? It is also predicted that these will turn out to have the same answer.
  5. Theory of star formation. (Initial mass function, binary populations, etc).
  6. Chemical and dynamical evolution of galaxies will be freed from the Curse of the Variable Parameter.
  7. Spectroscopy of extra-solar, earth-mass planets reveal non-equilibrium chemistry (or not).
  8. Additional insights into processes under extreme conditions, including astrophysical masers, two-photon processes, induced Raman scattering, Landau levels, and astronomical dynamos.
  9. Something I haven't thought of.
  10. Something even you haven't thought of.

Note that Dr. Trimble's 9th and 10th predictions are of things beyond our current imagining. Part of the excitement of astronomy, and other sciences, is that we do not always know where we are going until we get there. With the development of different and more sensitive detectors and with increased access to wavelengths longer and shorter than visual light, we expect to observe objects we cannot now know exist. Each waveband shows different aspects of the Universe.

Q. I really love astronomy ever since I took it my freshman year. It is an interest of mine but I like Psychology better. Do you know of a good astronomy book or magazine? I am always in stores and I cannot find any sort of Astronomy book? Thank you!

A. I find recommending books or magazines difficult without knowing your personal interest or how technical you would like the material to be. The classic astronomy magazines written on a popular level are Sky & Telescope and Astronomy. Other general science magazines, such as Discover and Scientific American, provide coverage of astronomical topics. You should be able to find a wealth of astronomy-related books at your local "general interest" bookstore. For instance, a search for "astronomy" at produced a list of 6,281 titles and at produced a list of 6,470 titles. In selecting a book, you know best where your own interests lie. One book I enjoyed reading recently was The Darkness at Night by Edward Harrison. He, also, wrote Masks of the Universe, which I did not enjoy as much.

Q. I lived in Hawaii in 1999, one day while walking the shore line I found what I believe is a metorite. It is black like the lava rocks so at first glance that's what I thought it was, but when I picked it up it was heavier than any lava rock. It has what looks like silver glitter on it, and you can cut into it and see what appears to be a nickel inside. I don't know if you are familiar with meteorite's, but maybe you can tell me who to contact.

A. You can always take a suspected meteorite to a local museum and or college to have an expert look at it. David Kring of the Department of Planetary Sciences at the University of Arizona has an extensive web document, "Meteorites and their Properties," available at, which includes tests for suspected meteorites. A rock with a black surface that is heavier than expected and that is metallic silver on the inside and is different from nearby rocks may be a possible candidate. However, you did not mention whether it is whether it is solid without pores or whether it is magnetic, which are also characteristics for iron meteorites.

Q. Why when a person is on the moon they look at the Sun and it is white, but when on Earth it is a yellow colour? What relevance does space have on colour??

A. I'm not sure the appearance would change as dramatically as you indicate. However, the Earth's atmosphere does change the overall color of the Sun by blocking out most of its light outside of the visual spectrum (infrared, ultraviolet, X-rays, etc.). This is also why observations in most parts of the infrared, X-ray, far-ultaviolet and gamma ray regions of the spectrum have to be carried out above the Earth's atmosphere.

Space has no effect on color. However, interstellar dust can attenuate blue light, causing stars to appear redder than they should - much like the Earth's atmosphere making the Sun appear red at the horizon.

Q. What are the rotation (mean) volocities of our nine planets? I have orbital volocities (ex Mercury 47.8724 km/s) but I cannot find rotation volocities. I would like the values to to the ten thousandth of a km/s, similar to the orbital volocities.

A. Orbital speeds are the average speed with which a planet travels around the Sun. The speed with which a planet moves varies along the orbital path, increasing as the planet approaches the Sun and decreasing as it moves away. The rotational velocity of a planet is the average speed at the equator with which it rotates about its axis. These values below were obtained using the mean equatorial radius and sidereal period of rotation listed in The Astronomical Almanac. The relationship among these values is

Vr = (R 2 Pi)/P

where Vr is the rotational velocity, R is the mean equatorial radius in kilomenters, Pi is the physical quantity pi (3.14159...), and P is the rotation period. The Astronomical Almanac quotes rotation periods in days, each of which includes 86,400 seconds.

Mercury		0.0030 km/s
Venus		-0.0018 km/s
Earth		0.4651 km/s
Mars		0.2408 km/s
Jupiter		12.5720 km/s
Saturn		9.8710 km/s
Uranus		-2.5875 km/s
Neptune		2.6829 km/s
Pluto		-1.3606 km/s

The negative signs associated with Venus, Uranus, and Pluto indicate that these planets have retrograde rotations. In other words, they rotate in the opposite direction from the Sun and other planets. Because The Astronomical Almanac only reports five significant figures for the radius of Jupiter, the rotational velocity of Jupiter would be better reported as 12.572 km/s.

Q. Sir, is there a reason the Russian didn't send Mir to the sun to be destroyed?

A. Mir was in a low earh orbit where the gravitational pull of the earth on the station was quite strong. It would have required many many times more fuel than the station had, not to mention much more powerful rocket engines, in order to break away from the earth's gravity and send Mir to the sun. In comparison, it only took a few short bursts from the manuevering rockets to accelerate Mir's descent into the atmosphere. The atmosphere then put enough drag on the station to slow it down and pull it out of orbit.

Q. A science teacher at my high school says that light has weight, very very little but he says light has weight. Is that true? If so, There is a theory that man could never travel the speed of light because when you go the speed of light you gain infinate mass. So if light has weight, it has to have some sort of mass, right? So if this is all true. How can light go the speed of light?

A. It sounds like your science teacher has a few concepts confused. The first is the difference between mass and weight. Mass is an intrinsic property of an object and doesn't change under any circumstances. "Weight" is the nontechnical term to describe the force exerted on a mass in a gravitational field. The gravitational field at the surface of the earth is greater than than on the surface of the moon, so someone with a mass of, say 80 kg (176 lb) would weigh more on the earth than on the moon. That is to say that they'd feel heavier.

As for your question about speed-of-light travel, the "Speed of light" is defined to be the maximum possible speed that any particle or object can reach. Einstein's theory of Special Relativity tells us that this maximum attainable speed is only possible for particles with zero mass. The particles of light, called photons, are the most common truly massless particles in the universe. It is precisely their lack of mass that allows them to travel at this maximum speed, which consequently is referred to as the "speed of light," or "c" in physical equations. The reason why objects with finite mass can't reach this speed is not because their mass changes as they speed up (it doesn't, their mass stays constant) but because the amount of energy required to accelerate them to this speed would be infinite. There is an energy associated with movement, called "kinetic" energy, which is described by the famous equation E=gamma m c2. The gamma, or "boost factor" is equal to

----------------- ,
sqrt(1 - v2/c2)
where v is the speed of the particle in question. Gamma is zero for an object which is not moving and infinite for one moving at the speed of light. So for a particle accelerated from rest to the speed of light, its kinetic energy would have to go from zero to infinity. This energy would have to come from somewhere, for example a rocket engine. Since it's impossible to get an infinite amount of energy out of any propulsion system we know of, no object with mass can ever be accelerated to the speed of light.

Q. I've read that, because of the influence of the Sun and Jupiter, the Moon's orbit is quite complex. I'd like to know where could I find information about how was the Moon's orbit in the past. Is it possible that several million years ago the Moon turned in the opposite direction?

A. The moon's orbit could be perturbed by the effect of the Sun and Jupiter's gravity. In dynamics, this is known as the three body problem. There is no simple analytical way to analyze the behaviour of a system of three or more gravitational bodies. We have nine in our solar system. The only way to even approximate is through computer simulations and approximations. I'm not aware of any dramatic results of such research, but searching for lunar perturbation turns up over 1000 web sites

We do know several things about how the moon's orbit evolves with time. As the moon loses energy, its orbit becomes more circular. Also, the moon is orbitting the Earth more ane more slowly and therefore drifting further and further away. Both of these effects are very small.

At no point would it have been possible for the moon to orbit in the opposite direction.

Q. If you take an aquarium into space, will the fish in it float up or down (assuming there's a lid on it so the water stays in the aquarium in the first place)?

A. On Earth, fish move around in water in two ways: they propel themselves through the water using their fins, and they also change their depth by varying the gas pressure in their "swim bladders". In bodies of water on Earth, the water pressure varies as a function of depth, so by varying the gas pressure in a swim bladder, a fish can change from being neutrally buoyant to negatively or positively buoyant, which causes the fish to sink to a lower depth or rise to a shallower depth.

However, in a sealed aquarium in space (presumably on the space shuttle or ISS), the water pressure in the aquarium should be constant. This means that the fish can use their fins to propel themselves through the water, but since there is no change in pressure within the aquarium, the swim bladder will not allow the fish to change depth.

Another way of looking at this is that in space there will be no "up" or "down" in the aquarium, since a fish defines up as the direction where water pressure is lower than at the current location.

Q. How long would it take for a radio signal to reach the Earth from the Moon?

A. A radio signal is a type of light or electromagnetic radiation. So, radio signals move at the speed of light, roughly 300,000 km/sec.

The Moon has a slightly eccentric orbit, so its distance from the Earth varies from about 363,000 km to 406,000 km.

In either case, a radio signal beamed from the Moon to the Earth would take just over a second to reach the Earth (1.21 at perigee and 1.35 at apogee). However, most times, radio signals are beamed from the Earth to the Moon and we measure the reflected signal back on the Earth. In this case, the round trip would take twice as long, or 2.4 - 2.7 seconds.

Q. Can you tell me where I might be able to obtain a current site map of telescope and observatory locations in the United States, including radio telescope arrays?

A. I'm not aware of any maps per se. has a comprehensive list of observatories and planetariums in the United States. In addition, the Astronomical Almanac published jointly by the United States Naval Observatory and Royal Greenwich Observatory, has an appendix with the latitude and longitude of every observatory in the world.

Q. I recently have become interested in radio telescopes and listening to the stars for sounds of intelligent life. I am interested in building an amature radio telescope that would allow me to listen to the great out there. Do you have any suggestions on where I might start to gather information?

A. You may wish to join the Search for Extra-Terrestrial Intelligence (SETI) League, which exists to promote amateur SETI efforts primarily in the radio region of the spectrum. Their on-line technical manual ( provides links to equipment manufacturers' sites and suggestions for modifying commercial hardware. Another on-line source of information is the Small Radio Telescope project of Haystack Observatory ( You may also want to read such texts as John Kraus's "Radio Astronomy" or John Shields' "The Amateur Radio Astronomer's Handbook". Once you decide on a configuration, you have several options for making your own hardware or purchasing it from vendors, such as Radio Astronomy Supplies or Olde Antenna Lab.

Q. Do you know who would be interested in buying a meteorite?

A. There are two companies that I know of that buy and sell meteorites. They are:

New England Meteoritical Services
     (508) 478-4020
Bethany Sciences
     (203) 393-3395
I have no experience with either company, so I cannot recommend either one.

Q. What is of specific value of the X-ray part of the spectrum that makes it the most popular type for space-base telescopes?

A. Very little radiation in the x-ray portion of the electromagnetic spectrum reaches the surface of the Earth due to atmospheric absorption. Therefore, any astronomical observations must be performed in space. The earliest astronomical x-ray observatories were born aloft by high-altitude balloons and sounding rockets. However, high-altitude balloons do not completely escape the atmosphere and sounding rockets provide only a few minutes of data collection each. A number of interesting objects produce significant amounts of x-ray radiation. Rotating, magnetized neutron stars may produce x-ray pulses. Matter being accreted by a black hole may be heated to extreme temperatures and produce x-rays. The diffuse, hot gas found in clusters of galaxies produces x-rays. Active galactic nuclei are intense x-ray sources of intense, which may also be variable x-ray radiation. Supernova remnants, such as the Crab Nebula, can produce x-rays. In fact, x-rays have been detected from star forming regions, normal galaxies, and most stellar types. Although the x-ray sky is a relatively new observational area, it will be a fertile one for many years.

Q. My research shows that the SUMMER SOLSTICE is this Thursday morning 06-21-2001 at 3:24 a.m. EDT (7:24 hundred hours UT). 1. Is this correct? 2. So, is Wednesday or Thursday the longest day of the year?

A. According to the US Naval Observatory, the solstice will occur at 7:38 Universal Time (UT) on June 21, 2001, which corresponds to 3:38 AM Eastern Daylight Time. The length of day, as defined by the amount of time between sunrise and sunset, will vary only slightly over the week around the solstice. Actual sunset and sunrise times also depend on local conditions. The longest day possible for a particular location would theoretically occur if the summer solstice coincided with local noon, as defined by a sundial. Because the solstice occurs closer to local noon on June 21 than June 20, the longest day should be June 21. However, the difference may be a matter of seconds.

Q. I am in search of a definition for "in orbit." When does orbit begin?

A. Orbital motion is simply defined as the motion of two bodies around each other as a result of a mutual force between them. Let us restrict ourselves to gravitational orbital motion where one object is vastly more massive compared to the other, as in the case of a satellite or rocket around the Earth, as I imagine this is what you have in mind. A rocket will not achieve a particular orbit unless it has gained a certain amount of kinetic energy (read velocity) determined by the height of that orbit. Stable orbits for the Earth exist only at heights for which the density of the atmosphere is low enough so that air friction does not sap away the satellites orbital kinetic energy. The term "in orbit" is rather loosely defined. A comet could be in orbit around the Sun, even though it's actual orbit is not closed. For Earth orbiting satellites, to be in orbit usually means the satellite posseses enough kinetic energy that it maintains it's closed orbit over it's lifetime of operation. So a rocket or spaceship enters orbit when it manages to boost itself to at least the velocity required to remain in it's desired orbit. This minimum velocity actually falls the furthur away the desired orbit is. One may ask "If less velocity is needed to remain in a higher orbit, why do rockets need more power to get to one?" This is because a rocket loses it kinetic energy to the gravitational field at a drastic rate as it goes up, needing it to have that much more to actually be left over with enough to go into orbit. When the two bodies are of comparable masses, the situation gets more complicated, as both objects then orbit about a common center of mass. However, as before, to achieve a stable orbit the total kinetic energy of the objects must at least be that required for that orbit.

Q. I am writing to you to ask a question about a meteorite I discovered in my backyard. It was buried about 3 1/2 feet into the ground and I can supply photos, weight and size measurements. It is the size of a large softball, are you interested in more information regarding this?

A. Thanks for your message about your potential meteorite. Unfortunately, the Astronomy Department at the University of Virginia is not equipped to determine the authenticity of potential meteorites. The Mineral Sciences Division of the Museum of Natural History at the Smithsonian Institution in Washington, D.C. previously tested specimens at no charge, but given security concerns they no longer carry out such tests. You can find more information on the page:

If you live in the Central Virginia area, we would be happy to take a look at the meteorite. We are familiar with the basics of classifying meteorites and could tell you, based on a visual inspection, whether we think you have an actual meteorite.

Q. I recently moved to a place where I have a wide view of the Western horizon. I can watch sunset and moonset all year. I have begun noticing, in the last two months, that the new moon has appeared to set near the place where the sun sets, but the full moon has set much farther south, perhaps as much as 60 degrees. I had always thought, or assumed, that there was not this much "wobble" in the Moon's orbit. Could you tell me more about the "wobble" in the orbit, or what is causing this difference in the location of moonset?

A. The Moon orbits the Earth more or less in the same plane in which the Earth orbits the Sun. The axis of the Earth's rotation, however, is tilted by 23.5 degrees relative to this plane. This tilt causes the effect you observe, along with causing the seasons. Here's how it works: A crescent moon it quite near to the sun in the sky, so it sets in approximately the same place as the sun. A full moon, on the other hand, is opposite the sun in the sky. This means that the new moon and full moon can set as much as 47 degrees (two times 23.5 degrees) apart from each other. Near the solstices, when the sun is as high (summer) or low (winter) in the sky as it gets, the points at which the full moon and new moon set will be close to this maximum 47 degrees. At the equinoxes, when the sun sets due west, both the full moon and new moon will also set close to due west. You noticed this effect during the summer when it's at it greatest, but if you continue to watch during the fall, you'll notice the full moon and crescent moon setting closer and closer to the same point on the horizon. As fall moves on into winter, the effect will grow large again until it reaches its maximum around the winter solstice.

Q. I have been wondering how scientists are able to figure out the density of planets and stars.

A. To determine the density of any object, the mass and the volume of the object must be known. In most cases, mass is measured through gravitational interaction. Obtaining volumes can be more difficult. The density of an object is found by dividing its mass by its volume. To obtain a good measurement of the mass of a planet or a star, we must observe its gravitational effects on another body. The masses of stars may be measured in binary systems or through observations of orbiting planets. Similarly, the masses of planets may be measured by observations of their effects on the main star, on other planets in the same system, or any moons they may have. For solitary stars, masses may be estimated through theoretical means, such as mass-luminosity relationships and stellar structure models. Volumes are generally obtained from measurements or estimates of an object's diameter assuming that a planet or star is reasonably spherical. The diameters of planets within our Solar System and of the Sun may be measured directly. For other stars, the direct measurement approach is difficult because most stars appear to be points of light. However, the available technology is getting better. For some stars, their diameters may be measured by observing the passage of another object between us and the star. Such an occulting object might be the Moon, a planet in our Solar System, or an asteroid. Similarly, diameters for binary stars that pass in front of each other as observed from our perspective can be measured. This eclipsing-binary technique may be extended to measure the diameters of extra-solar planets and the stars they orbit.

Q. I have heard that there have been studies about a Planet X. Can you tell me about Planet X and any disasters it might cause?

A. Planet X is a name given to a hypothetical tenth major planet once believed to exist in an orbit beyond Neptune. The 'X' stands for unknown.

The motion of planets, comets, and asteroids in the Solar System are governed by the law of gravity. Since the Sun is so massive, it is the prime factor in determining an object's orbit; however, other objects of large mass (i.e. other planets) can slightly change an orbit. This requires accurate knowledge of the Sun's/planet's masses and positions. By comparing predicted orbits to these changes, or perturbations, scientists can see if they have accounted for all the major objects of large mass. Twice in the past, a new planet has been discovered when these predictions did not match the observed orbits (Neptune and Pluto).

In the late 19th century, Percival Lowell and others predicted the existence of a large unknown planet when they could not account for the perturbations of Uranus's orbit. Although Pluto was discovered close to where Lowell predicted this planet to be, it lacks the mass to account for the perturbations to Uranus's orbit. The need for a tenth planet to explain Uranus's orbit continued until the satellite Voyager 2 measured a more accurate mass of Neptune. Observational errors are now larger than any remaining discrepancies of Uranus's orbit. Because there is currently no need for the hypothesis of a nearby Planet X and there is no observational evidence supporting the idea of an unknown planet in the nearby solar system, the hypothesis of Planet X has been largely abandoned.

I have seen some scientific claims that there is is a collection of long-period comets that seem to come from the same direction. From this, two groups have inferred that that there is a massive object in the Oort Cloud (a cometary cloud whose mean orbit is more than 1,000 times farther from us than Pluto). The proposed object in the far away Oort cloud may be a planet, a brown dwarf (a star without enough mass to start shining brightly), or a statistical anomaly. If this massive object does exist, its only effect on the local solar system is to deflects some comets towards the local solar system. An extreme pessimist might argue that this planet increases the odds that a comet would strike the Earth, as Comet Shoemaker-Levy struck Jupiter. However, the total number of comets it introduces into the local solar system is much less than the number of comets and asteroids already in the local system. I don't think we have to worry about any disaster from this object, if it exists.

Although a local Planet X does not seem to exist, the search for it has led to many advances of knowledge about our solar system. Among these, a whole class of asteroids beyond Neptune (called Trans-Neptunian Objects) has been found.

Although a local Planet X does not seem to exist, the local solar system does end with Pluto. The Kuiper Belt, whose inner edge is defined by Neptune and outer edge is at least three times as far away as Pluto, is a collection remnants from the early solar system. It is believed that short-period comets come from the Kuiper Belt. It is estimated that the Kuiper Belt has more than 70,000 objects with diameters greater than 60 miles and that the total mass of these objects is 5 times the Earth's mass. Even without a Planet X, the solar system beyond Neptune is far from empty space.

For more detailed information on Planet X, follow this link,

Q. How do you explain that the the farthest objects in the sky are distributed in any direction when at the moment these photons were emitted, these same objects must be grouped into an enormously lower volume?

A. At the time the photons were emitted, the entire Universe was compressed into a much smaller volume. As the Universe expanded, these objects were increasingly separated. Correspondingly, the light emitted from them was increasingly stretched and redshifted. Assuming that the expansion of the Universe is uniform and constant in all directions, we wime the photons were emitted, the entire Universe was compressed into a much smaller volume. As the Universe expanded, these objects were increasingly separated. Correspondingly, the light emitted from them was increasingly stretched and redshifted. Assuming that the expansion of the Universe is uniform and constant in all directions, we would expect an initial uniform distribution of objects to remain uniformly distributed but more widely separated.

Q. Can you tell when the constellations will be the same as they were in 10500 B.C. and list the years?

A. The change in the position of constellations in the sky is the result of two basic processes. The first order change is from Lunisolar Precession. This is the slow wobble of the Earth's axis of rotation because of the gravitation of the Moon and Sun. The Pole describes a full circle in the sky over a period of 25770 years. So to answer your question, the next time the sky will be the in same "position" as it was in 10500 BC, at least to first approximation, will be in AD 15271, more than 13000 years from now. One can work out the progression beyond that. Note that this does not actually change the shape of the constellations, only their positions relative to a place on Earth.

The second order change is due to the stars actual physical motions. These are more or less random, but because the stars are so far away, these changes are very slow. So, in all truth, the sky we see today is unique and it has and probably never will be exactly the same.

Q. Why is it that you can see the moon all day long, but when nightfall comes around and you should see it, it is gone?

A. The moon goes through a full cycle of phases every 29.5 days, or once a month. As the Moon revolves around the Earth, the changing alignment of the Sun, Moon, and Earth causes the phases we observe. At new moon, the Sun and Moon appear close together on the sky. In this phase, the moon rises about sunrise and sets about sunset. Because the side of the Moon facing away from us is illuminated by sunlight, we do not see the Moon during this phase. Fourteen to fifteen days later, we observe a bright full moon all night long. During this phase, the Moon rises about sunset and sets about sunrise. Sunlight reflects off the side of the Moon facing us providing quite a bit of moonlight. Between new moon and full moon, the Moon progresses through the waxing crescent, first quarter, and waxing gibbous phases. We see a waxing crescent moon as less than half illuminated. At first quarter moon, we observe half of the Moon's face. When the Moon is in the waxing gibbous phase, we see more than half of its face as illuminated. As the month progresses, the Moon appears to rise later and later each day. As it appears to rise later, we see more of the face as illuminated. A later rising time corresponds to a later setting time. For instance, a first quarter moon rises about noon and sets about midnight. During the first quarter phase, you could see the moon in the afternoon and evening but not in the early morning hours. After full moon, the Moon goes through the waning gibbous phase, fourth quarter (also called last quarter), and waning crescent phase. The Moon continues to appear to rise later in the evening each day and set later in the morning. During these phases, we see less and less of the Moon illuminated until the new moon phase reoccurs. For example, the third quarter moon rises about midnight and sets about noon. During this phase, you would not see the Moon in the evening but could see it in the early morning and after sunrise. Additional information about the orbit of the moon including its phases is available in our archive.

Q. I would like to estimate the tidal forces between the binary stars in the YY Geminorum system. The stars are reported to be essentially identical, thus the mass of each is 1.192*10^30 kg, the radius of each is 4.3*10^8 meters and separation of the two is 2.7*10^9 meters. How do I go about calcualting this?

A. A tidal force can be characterized by the difference of gravitational force per unit mass between very small test masses on either extreme of a body in a non-uniform gravitational field. For the specific case of a spherical body of radius "r" at a center-to-center distance of "R" from a point mass "M", the formula is :

Tidal Force = 4*G*M*R*r/(R^2 - r^2)^2 where G is the Gravitational constant = 6.67*10^-11 N m^2 kg^-2. This force is along the line joining the centers of the two bodies. For the binary system YY Gem, the force comes out to be 7.3 Newtons. On Earth, this is a weight of 0.74 kilograms or about 1.6 pounds. The force by which one star in this close binary pulls the other apart is a little more than the weight of a basketball on Earth!

Of course, the mutual tidal distortions of the two stars will complicate the calculation a bit. Thus, the simple formula given above is not totally valid. The complete calculation is sufficiently difficult that a simple equation cannot be written down for it. Nevertheless, the actual force will not be very far from the value derived above.

Q. Are volcanoes and/or earthquakes still consistent problems on Mars?

A. Although Mars does not have the same type of plate tectonics as Earth, there are still ways to form faults. A result presented at a meeting in March of 2002 suggests that marsquakes do occur today. This paper claims that in a year there should be 4-40 marsquakes with an equivalent Richter scale magnitude of 4 and one marsquake with an equivalent Richter scale greater than about 6. This estimate, which comes from looking at the number of faults and their ages, places seismic activity at a level ten times that of previous estimates. Since this result is very new and has not gone through peer reviewal, this issue should still be considered open.

Scientists are still debating whether Mars has active volcanoes today. The main argument against active volcanoes is that there has been no direct observation of a volcanic eruption. There is ample evidence for past volcanic activity. With marginal reshaping by weathering, aqueous erosion and sedimentation, evidence of past volcanic activity, such as formerly active volcanoes has been preserved. Mars has many volcanoes, including the largest volcano in the solar system. Since Mars has little atmosphere, its surface is continuously bombarded by meteors, scarring the surface over time. There are many areas of Mars where those scars have been smoothed over by lava flows in Mars past. By counting the number of impact craters, some scientists have placed the last known volcanic activity within the last 10-30 million years (If all of Mars's history was compressed to one day, the last volcanic activity occurred at most about nine minutes ago). Since this is only an upper limit on the time since volcanic activity, many scientists are not willing to say the Mars has settled down into a time where its volcanoes are not active.

To learn more about Mars, you might want to go to your library and look at

or using the web, start at

Q. Perhaps you can help me.

I'm searching for an equation that accurately calculates the longitude from Greenwich of the projection of the north ecliptic pole on the earth's equatorial plane at any given instant (UT).

Any help would be greatly appreciated.


A. You want the longitude of the projection of the North Galactic Pole (NGP) on the Earth's equatorial plane at some time (Universal Time). This is same as finding the longitude of the place where NGP will be at the meridian at some given UT.

First we need to find out the Local Mean Sidereal Time (LMST) for a given UT. LMST is given by,

(1) LMST = GMST + L / 15.0,

Where L is the longitude of a place in degrees (negative if west), and GMST is the Greenwich Mean Sidereal Time.

To calculate GMST at 0h UT on a particular day, we use the following equation (from Explanatory Supplement to the Astronomical Almanac):

(2) GMST = 2.4110s.54841 + 60184s.812866Tu + 0s.093104Tu2 - 6.2 x 10-6 Tu3,


(3) Tu = du / 36525,

and du is the number of days of UT since Julian Day (JD) 2451545.0 (2000 January 1, 12h UT). (GMST calculated here is in seconds).

Thus for any given UT, we find the GMST for that time by first calculating GMST at 0h UT from equation 2, and then adding 1.0027379093*UT to it (UT in seconds).

Therefore, using equation 1, the LMST is GMST+L (L in degrees).

If an object has Right Ascension (RA) of R, then its Hour Angle (HA) is given by:

(4) HA = LMST - R.

For an object to be on meridian, HA = 0, which gives:

(5) LMST = R.

For NGP, we know R, and therefore we know the LMST from equation (5).

Since, we know all the terms now in equation 1 except L, we can solve for it.


  1. From NED Coordinate & Galactic Extinction Calculator, the RA of NGP is 12h51m26s.27915
  2. The number of UT days since JD 2451545.0 can be found using: (see Calculating mean and apparent sidereal time for example).<-->

    367y-int(7(y+int((m+9)/12))/4)+ int(275m/9)+day-730531.5 + (h + min/60 + sec/3600)/24,

    where the year is y, month is m, date in the month is day and h, min and sec are the hour, minute and seconds of UT.

  3. Strictly speaking, the UT used above is UT1, but actual UT and UT1 are guaranteed to be not more than 0.9 seconds apart (see the Explanatory Supplement to the Astronomical Almanac, p50 for details).
  4. As an example, on April 1, 2000 0h UT, I calculated the GMST to be 218204.227826. Subtracting 2*86400 (complete days) from this gives 45404.227826 seconds, which is 12h36m44s.2278.

    Therefore, (1) gives: 12h36m44s.2278 + L*60*60/15.0 = 12h51m26s.27915. Which gives me L as 3 degree 40 minutes 30.7702 seconds. Testing this out with a planetarium program such as Starry Night shows that this value is correct to about one minute of arc (or about 4 seconds in time).


  1. Explanatory Supplement to the Astronomical Almanac, Seidelmann, 1992.
  2. Calculating mean and apparent sidereal time
  3. NED Coordinate & Galactic Extinction Calculator
  4. Starry Night planetarium program

Q. We recently read that there are currently about 10,000 objects orbiting the Earth, about 2,800 of which are satellites and probes and the rest debris. This has raised a number of questions: 1) Is the amount of matter on Earth finite? 2) If so, what effects could displacement of matter into space have on the Earth and Moon? 3) If this displaced matter were not somehow compensated for, could it alter any of the physical aspects of the Earth (i.e., rotation, gravitation)?

A. First, the amount of matter on Earth is indeed finite. The typical measurement scientists used for describing the amount of matter is mass. The Earth's mass is about 6*1024 kg (6 million million million million kg) and the Moon's mass is about a 100 times less than the Earth's mass. As alluded to in your question, the mass of a planet is not constant. Some examples of why the Earth's mass changes are our launching of satellites into space, the planet's natural loss of some gas to space, and material falling onto the Earth.

Celestial orbits are determined by the force of gravity. The force between two objects is proportional to the masses of each object and inversely proportional to the square of the distance between the objects' centers of mass (Force goes like m1 * m2 / r2). In our solar system, the Sun, as the most massive object, is the dominant force in determining the orbits of planets. The effect of planets and their celestial satellites (ie. The effect of the Moon on the Earth) is merely a secondary effect since their mass is so much smaller than the Sun's. In a technical sense, the displacement of any matter causes the gravitational force, and thus the orbit, to change; however, this effect is trivial for even the largest of man-made satellites.

The spin of a planet is determined by conservation of angular momentum. Angular momentum is proportional to the mass of an object times the square of the object's angular velocity. Since the mass of a satellite is small compared to the Earth and its angular velocity relative to the Earth is also small, a satellite's effect on the spin of a planet is also trivial.

As an example, consider launching every human on Earth into geocentric orbit. The total mass of the human race is under 1011 kg. Although this seems to be a large number, it is very small compared to the Earth's mass. It's similar to comparing one blood cell to the number of cells in 1000 people. The effect on the Earth's orbit would still be less than 10-8 (10 parts in a billion) times the effect of the Moon. Even a wholesale exodus of the Earth's population would barely effect the orbits of any celestial objects.

Q. If all of the stars of our own galaxy were to disappear, what would we see in the sky? Obviously the Milky Way would be gone, but what would remain. It isn't obvious what is extra-galactic in popular guides.

A. I assume that when you say "all of the stars of our galaxy were to disappear", you mean that all the matter in our galaxy (excepting the solar system, if you will) vanishes. Because there is more to the Galaxy than the stars (dust, gas).

In any case, the naked-eye sky will be pretty boring, with most people being able to see only three objects--the Large and the Small Magellanic Clouds and the Andromeda Galaxy (M31).

With a telescope, however, people who look out for galaxies would not notice much difference---except that it will make their life much easier since they wouldn't have to care for the foreground starts to contaminate their field. Of course, if all the dust in the galaxy were to vanish too, the Zone of Avoidance would be gone too, and we would be able to see extragalactic objects there too.

If you have a desktop planetarium program, you can zoom out so that you see >= 90 degrees of sky, and then turn the stars, nebulae, clusters (globular as well as open), Milky Way etc. off and turn Galaxy display on. It will give you a rough idea as to how the extragalactic objects are distributed.


I am interested to know about the extreme future of the solar system and the Earth-Moon system.

  1. Isn't it true that the Moon will eventually be dragged back into the Earth from the effects of gravitational radiation? I know the effect is small for bodies as small as the Earth and Moon, but that would just mean that it will take a VERY long time. If I am not all wet on this, then how long would this take?

  2. I read "Distant Wanderers: The Search for Planets Beyond the Solar System" by Bruce Dorminey. The book indicates if the Earth is not engulfed by the Sun when it goes red giant, then it will be shoved out into a much larger orbit. Is this true and how does that happen if it is true?


  1. The prediction of Gravitational radiation comes from General Relativity. As such, the exact answer is very difficult to calculate. The key, as you pointed out, is that this will take an extremely long term for the Earth/Moon system. In fact, the time until this occurs is so large that our Sun will be long gone (My simple estimate of it is so large that it is difficult to comprehend the time, even compared to the current age of the universe) . In the mean time, the future of the Moon is more interesting due some other less exotic physics. The tides caused by the Moon dissipates energy (tidal friction). Since the angular momentum of the Moon must be conserved, this energy loss means the Moon must slowly drift away from the Earth and that the length of a day on Earth is increasing at about 0.002 seconds per century. In the future, the time it takes the Earth to rotate around itself (a day) will eventually equal the time the Moon takes to circle the Earth (a month) at about 50 present days. This is an example of synchronous rotation.
  2. I have not read this book and our library does not have a copy so I can not comment directly on what the author says; however, I think I might know what he tried to explain. Red giants have a tenuous hold on their outer atmospheres and can undergo a large amount of mass loss during this stage of its life. For instance, I have seen calculation that our Sun will only weight 3/5 = 60% of its current mass when it reaches the end of the red giant stage (becomes a white dwarf). Once again, the basic law of gravity comes into play. As the Sun loses mass heavily in the red giant stage, the Earth's orbit will increase such that M (solar units) / R (Astronomical Units = Earth-Sun distance) is constant. Thus, if the Sun did not engulf the Earth during the Red Giant phase, the Earth's orbit at the end of it would be 5/3 times the current orbit.


My 9 year old daughter is preparing a travel brochure to Pluto for her 4th grade science class. She wants to know the distance from earth to Pluto? I've looked several places on the web (including NASA) and in the dictionary. Can you help me?


There is detailed information about the planets at

If you click on each planet, there is a section called Observational Parameters, which has a minimum and maximum distance from the Earth. For instance, Pluto is roughly between 2.7 and 4.7 billion miles from Earth. Since planets orbit around the sun with different periods, the distance between them is not constant. Pluto has an orbit which is much more elliptical than the Earth's, causing a large difference between the minimum and maximum distance. Since light travels at 186,000 miles per second, light would take roughly between 4 and 7 hours to reach Pluto from Earth.

If you want to know the exact distance at any time, there is a web tool that NASA has made at Since the tool does not seem designed for someone with little background in astronomy, I have generated the distance for all the planets today. Here is a list with their rough values.

I hope this has helped.

Q. Given a true north - south line and a pole, how do you determine your latitude?

A. You will want to make several measurements of the shadow cast by the Sun around local noon. To avoid the complications caused by zone time, you can use the Naval Observatory web page at to get the times for Sunrise, Sun transit, and Sunset at your location. Local noon occurs at the time the Sun transits, or crosses nearly directly overhead. At local noon, the Sun will cast the shortest shadow for the day.

After determining the shortest shadow for the day, measure the length of the shortest shadow (A) and the length of your stick (O). Divide the length of the stick by the length of the shadow (O/A). The resulting number is the tangent of the altitude of the Sun. Use a calculator to convert from the tangent you obtained to an angle (ATan or Tan-1 function key). The resulting angle is the altitude of the Sun.

You will also need to know the declination of the Sun for the day on which you make the measurements. The declination of the Sun is the angle between the Sun and the celestial equator, which is a projection of the Earth's equator onto the celestial sphere. You can look the declination up in The Nautical Almanac published by the Naval Observatory or other suitable table. Alternatively you can interpolate, the declination of the Sun is 0 on the vernal and autumnal equinox, 23.5 S on the winter solstice, and 23.5 N on the summer solstice (dates available from the US Naval Observatory web page "Earth's Seasons - Equinoxes, Solstices, Perihelion, and Aphelion" at ).

The total angle from southern horizon to northern horizon is 180 degrees. The angle between the celestial equator and the north celestial pole is 90 degrees. The north celestial pole is a projection of the Earth's north pole onto the celestial horizon. The angle between the northern horizon and the north celestial pole is equal to your latitude. The angle between the southern horizon and the Sun (the altitude of the Sun) is either greater than or less than the angle between the southern horizon and the celestial equator. The difference is equal to the declination. Unfortunately, this description really requires a picture to understand; try looking in an introductory textbook where it describes sky motions and patterns. If the declination is negative or marked south, which will occur during the autumn and winter, subtract the absolute value of the declination from 90 degree. Then, subtract the altitude of the Sun from the result. This second and final value will be your latitude.

If the declination is positive or marked north, which will occur during the spring and summer, subtract the altitude of the Sun from 90. Then, add the declination to the result. This second and final value will be your latitude.

Q. I was in Big Sky, Montana for Christmas last month. Every morning, just before sunrise, a very bright object would appear in the sky very low on the horizon outside my window. I am dying to know what it was. I think I was facing south/southeast or maybe southeast/east. It rose over a ridge between Flat Iron Mountain and Andesite Mountain. Thanks.

A. That must be Venus. These day, Venus is very bright. Like our Moon, Venus also changes its phase, which means some day's Venus would be brighter than other day's. For more information about phases of Venus, please see this website; .

Q. Why is it that the moon does not revolve around the sun, even though the gravitation force of sun on moon is larger than that of Earth? Also, does the fact that all planets revolve around the sun in a counter-clockwise direction, which is same as direction of rotation of sun, suggest any relation between rotation of Sun and revolution of planets?

A. When you consider the forces on the moon from the sun and earth, you must count in one more effect, the centrifugal force of the moon due to the revolution of the earth-moon system around the sun. This component almost cancel out the gravitational force of the sun, so the resulting force on the moon is dominated by the earth. From the fact that the directions of solar rotation and planetary revolutions are the same, we can imagine the origin of the formation of the solar system. Astronomers believe that the solar system was formed by a single rotating cloud of gas and dust, which resulted in the same direction of solar rotation and planetary revolutions. Here is an example of the formation model of solar system--

Q. I received a T-adapter and camera mount for my Newtonian telescope. I have been unable to get it into focus as of yet. It seems as if the focal length is too long. If only I could extend the focuser about 2 inches, I think I would be fine. Do I need a focal reducer?

A. A focal reducer or Shapley lens would reduce the effective focal length of your set-up. Before purchasing a new piece of equipment, you may wish to consult the seller if you bought the pieces locally or your local astronomy club to ensure you have installed all of the pieces correctly and eliminate any other possible problems.

Q. What is the current common name spelling for Alpha Eridani? Is it "Achenar" or "AcheRnar".

A. The common name for Alpha Eridani is "Achernar". "Achenar" seems to be used at many places. A Google search for "Achenar" does result in many pages (many references to video/computer games).

"Achernar" is derived from "al-, the + nahr, river" (arabic), and literally means "The end of the river" (signifying that is is the last star in the constellation Eridanus which represents a river). See page for example.

Q. With the information that we have gained over the years, exactly how far do you think we have to go till we step foot on the surface of mars?

A. I think that a manned mission to mars would be at least 10 years away. First, we have the technology problem. The estimated time of travel to mars is about 150-180 days in the least. Moreover, after reaching mars, the astronauts will have to wait maybe 8 months to one year for an optimum time to begin the journey back. This means more food and supply. We don't have a spacecraft right now which can carry all of these loads for two or thress astronauts. Second, political problem! The manned mission to the moon was driven by a space race. But now, we don't have a space race like that and space missions are not as much craze as 30 years ago. Additionally, we have an economical problem. Economy is not blooming these day; the funding for such a record large mission is very difficult to be approved. Also, there is a debate on the reward of spending the money it needs. Is it neccesary? Is it appropriate? Are there not other more important issues, which need urgent money? etc. Therefore, I believe we have to wait more than 10 years to see this kind of space mission. Thanks.


Why do rockets go straight up into space instead of going up gradually like an airplane does?


When you want to plan the best route to take, whether for an airplane or a rocket, one of the primary concerns is minimizing fuel consumption. If you do not minimize how much fuel you will use, you have to carry more fuel to get where you want to go. This means you can carry less mass in the form of the things you are trying to transport, whether they are people or a satellite.

For launching rockets into space, the goal is often to get the rocket into an orbit around Earth. To do this, the rocket must reach the altitude of the desired orbit. At first one might think that this explains why rockets launch straight up; however, rocket launches are much more complex. This is because fuel consumption needs to be minimized. If the rocket travels straight up, it must spend a lot of its acceleration from its fuel just to beat Earth's gravity. This means that it does not reach high speeds easily. In turn, this means the trip takes a long time and a lot of fuels is spent. If the rocket was to travel parallel to the ground, the acceleration from the fuel greatly increases the rocket's speed, but it never reaches the desired altitude. Also, there is less air resistance at higher altitudes, which means that it is easier for the rocket to reach higher speeds at higher altitudes. So, a compromise must be reached. We launch rockets nearly straight up so that they spend the least amount of fuel to get to a higher altitude. Well before the rocket reaches the desired orbit, its trajectory flattens so it can reach its final destination quicker and thus conserve the most fuel.

Obviously an airplane is slightly different in that its goal is to safely land on another place on the surface of the Earth. You might think that means airplanes should not worry about getting too far above the ground. However, you must remember that it is more efficient to fly at high altitudes than low altitudes. This explains why airplanes ascend to their cruising altitude. If you think about the ascent to cruising altitude, you realize that, from a fuel consumption outlook, its not too different from a rocket launch. This gets to the crux of your question, why airplanes gradually ascend. This has more to do with passengers. In order to ascend to cruising altitude quickly and to then level off the airplane's trajectory would require large accelerations. Unlike in the space shuttle or military craft where the passengers are trained to handle large accelerations, regular airplanes need to worry about what type of accelerations you or I are capable of. The gradual ascent ensures that regular airplane passengers can handle the accelerations that occur during a flight.


If the Earth is suddenly displaced from its orbit, will the Moon's orbit change? If the Earth was quickly displaced, would the Moon still orbit the Earth?


The orbits of celestial bodies are governed by the force of gravity, which is inversely proportional to the distance between two objects. In other words, the effect of gravity decreases as two objects get further apart.

The Moon's current orbit of the Earth is in a roughly constant state. (The AQuA archive discusses some of the small departures from the constant state.) Any displacement to the Earth will affect the current equilibrium and the Moon's orbit would change. The detailed changes would depend on the encounter and, especially for the case of a large sudden displacement, would probably want to account for the gravity forces of the other celestial bodies; however, I will not go into those details because the basic physics can be better understood by treating the Earth and the Moon as the only two objects we care about. This is a good approximation because the gravitational force between the Earth and Moon is much larger than the Moon and any other body in our Solar System. Also, we will neglect the small difference between the center of mass of the system and the Earth's position, it is less than the radius of the Earth.

We'll use a clock to describe the positions of the Earth and Moon. Assume all rotations are clockwise. Imagine the Moon is at 3 o'clock and the Earth is at the center of the clock at one instance. If we were to measure the velocity of the Moon, it would be pointing down. If we were to suddenly give a very large displacement to the Earth, the force of gravity between the two objects would go down a lot, eventually becoming negligible. If we reach the point of negligible gravitational forces, there would be no force on the moon at all, and the moon would continue traveling down forever. Thus, we see that a large enough displacement will separate the Earth-Moon system.


Is our Galaxy rotating?


Yes, our Galaxy is in fact rotating. Our Sun travels at around 225 kilometers per second around the center of our Galaxy. Conservation of angular momentum makes it very difficult to have objects that are not rotating at some level (See other AQuA questions in the archive for more details). In fact, the measurements of the rotation of galaxies, commonly called rotation curves, are important tools in astronomy. They are important because they give us insight into the amount of mass at different points in a galaxy. For example, astronomers have used rotation curves to determine the amount of mass at the center of galaxies besides our own; one such measurement, that of NGC 4258, found the mass of a putative black hole to be almost that of 35 million Suns. Astronomers have also used rotation curves to show that there is some large amount of matter in galaxies that we can not see. The search for this "dark matter" is one of the more central questions in astronomy.

Q. I'm curious say we have a Brown dwarf (BD) with a Luminosity = 2.32431 * 10^-7 Solar and a Radius = 70940 km i.e. 7.094 * 10^9cm. If Solar Luminosity = 3.84 * 10^33 erg/sec and Stefan-Boltzmann constant is 5.67 * 10^-5 ergs cm^-2 s^-1 K^-4, radius of BD = 70940 km and taking Pi to be 3.1415926, does this mean for a BD with a Luminosity = 2.32431 * 10^-7 Solar, its luminosity can also be read as 8.9253504 * 10^26 ergs/sec (i.e. 2.32431 * 10^-7 * (3.84 * 10^33))?

A. What a very long detailed calculation you have chosen to share... You appear to have calculated the brown dwarf luminosity and effective temperature correctly. Your blackbody flux formula F=(sigma)T^4 gives you the energy per unit surface area. Yes, each square centimeter of the Sun produces more than 4000 x the energy of a similar area on the brown dwarf you describe.

To calculate apparent magnitude, you need mag(Sun) - mag(BD) = 2.5 Log (LuminosityBD/LuminositySun) and you need to know the apparent magnitude of the Sun, which is -26.7. Using this formula, I get an apprarent magnitude for the brown dwarf of -10.1, which is about 2.5 magnitudes dimmer than the full moon. The brown dwarf would still be a very bright object but would not wash out all other stars in the sky the way our Sun does.

To calculate the apparent magnitude of the brown dwarf at a distance greater than 1 AU, you can use the same equation but will have to divide the luminosity of the brown dwarf by the new distance squared. mag(Sun) - mag(BD) = 2.5 Log [(luminosityBD/4 PI distance^2)/(luminositySun/4 PI EarthSundistance^2)] Using this formula, I calculate an apparent magnitude for the brown dwarf of 10.4, which would be too dim to be seen with your eyes alone.

Q. I am wondering how astronomers determined the comparative apparent luminosity of stars. For instance, I found a site on the www that described Fomalhaut as the 17th brightest star in the sky. Was this discovered by the naked eye? Or is there somewhere a list of all the comparative luminosities of stars in descending or ascending order?

A. First a little bit of nomenclature. If you find this confusing, you can safely skip over it.

Greek Astronomer Hipparchus in about 150 B.C. ranked the stars he could see on a scale of 1 to 6, with 1 representing the brightest stars and 6 representing the faintest stars (yes, the scale goes upside down). Modern astronomers revised and extended this scale to define magnitude, which determines how bright a star (or any other object in the sky) is. Magnitudes are defined in a way that brightness increases with decreasing magnitudes, and a magnitude difference of 5 implies a brightness ratio of 100.

The definition of apparent luminosity of a star has to take into account several facts: First, we need to define the wavelength range over which we sum the light from the stars to measure the overall brightness. This is called the wavelength (or frequency) band, and indeed astronomers define magnitudes in different bands. For your question, we can either sum over a long range of wavelengths, or over visible part of the electromagnetic radiation. These two definitions, as it turns out, give different brightness values (magnitudes) to any star in question. Second, many stars are variable stars, that is, they change in intensity periodically (and some stars show random fluctuations too). Generally we take the average of the luminosity in that case to assign a mean magnitude to a star.

Thus, depending upon the brightness definition and method of calculation used, stars can have different brightness values, which may change their relative ordering in terms of brightness.

In order to measure the apparent luminosity of stars, astronomers have defined "standard stars", which have well-published magnitudes in different bands. These stars have been measured accurately using a wide variety of instruments.

Now, whenever one wants to measure the apparent luminosity of a star using one's own telescope, one can measure the object in question (by taking a timed exposure on a CCD [Charge-Coupled Device] camera for example), and do the same measurement on one or more of the standard stars (preferably close to the star being measured, since this would tend to cancel out the effects of Earth's atmosphere). Then one can scale the results to determine the apparent luminosity of the star.

For an example of lists of stars in decreasing order of brightness, you can see

As you can see, the order of stars in both the list is not exactly the same. This is due to the two effects mentioned above.

Q. I was wondering who are the researchers that have designed and constructed the solar millimeter wave radio telescope?

A. I have not been able to find any instrument specifically called the "solar millimeter wave radio telescope". However, I did find a similar instrument and I have described it below. I hope this is what you were looking for. The Solar Sub-Millimeter Wave Telescope (SST) is a new instrument that is being built for the Complejo Astronomico el Leoncito (CASLEO) in Argentina. It is a 1.5-m diameter radio telescope designed to observe the Sun at 210 and 405 Gigahertz (GHz). The Principal Scientist is Pierre Kauffmann. His contact address in on the SST webpage. Also, if you can, try to find the following reference : The Solar Submillimeter Telescope (SST): a New Instrument for the Investigation of the Atmosphere of the Sun and the Earth", Kampfer et al, 1997, 6th Symp. on Recent Adv. Microwave Technology, Beijing, China.

Q. Regarding moon first visibility, I found the following selection in the U.S. Naval Observatory web site "The date and time of each New Moon can be computed exactly but the time that the Moon first becomes visible after the New Moon depends on many factors and cannot be predicted with certainty." I ask you kindly to inform me about this issue and whether it is agreed upon by all scientists or not, and I would be more grateful if you provide me with more information about this matter.

A. The US Naval Observatory is a respected astronomical institution specializing in computing observable characteristics of the Sun, Moon, and other celestial bodies. Many professional astronomers rely on publications produced by the Naval Observatory for planning their own observations. Having read the FAQ entry to which you refer, I think it adequately explains the problems of observing the early appearances of a waxing crescent moon. The new moon phase has a precise astronomical definition and the times when the Earth-Moon-Sun system is properly aligned may be determined. The amount of the lunar disc that is illuminated increases slowly after new moon but this too can be calculated. Whether a particular observer can actually see the crescent moon within a day of new moon depends on a number of conditions, some calculable and some related to the observer and his location. You may wish to consult A note on the Prediction of the Dates of First Visibility of the New Crescent Moon published by Her Majesty's Nautical Almanac Office, see for more information.

Q. When was the star Thuban the North Star?

A. Thuban (Alpha Draconis) is a magnitude 3.5 star in the tail of the constellation Draco, the Dragon. Due to the phenomemon known as Lunisolar precession, the slow regular wobble of the Earths rotational axis, Thuban was once the star nearest the North Celestial Pole, getting its closest around 4700 years ago. At that time, its seperation from the Pole was even closer than the current North Star, Polaris, which actually lies about a degree (thats two Full Moon widths) from the true Pole.

Q. Are there any stars or constellations visible to the naked eye, from both the southern and northern hemispheres? In particular, from both England and Australia?

A. Yes, you can see some of the sky from both England and Australia. For example, in late September at about 21:00 local time in both England and Australia, it is possible to see constellations such as Aquila (the Eagle) and the Great Square of Pegasus. The brightest star in Aquila is named Altair. The zodiacal constellations Pisces and Aquarius are also visible. From London, look south to see these constellations. From Australia, look north. The full moon will interfere with good seeing, but in about a week it will move out of the way.

Q. What is the escape velocity from the Milky Way Galaxy for objects that are the same distance from the centre of the galaxy as our Sun is? I guess this depends on how much of the mass of the Milky Way is closer to the centre than we are and how much is farther away. Do we have a reasonable estimate of how the mass is distributed?

A. You are right that the escape velocity depends on how exactly the mass is distributed, over the entire galaxy. Astronomers are not decided yet on how the mass is distributed, but all of the models have some features in common. There is a bulge near the center of the Milky Way, and a disk surrounding it. The Sun is about two thirds of the way out from the center of this disk. The light from the galaxy comes from the bulge and the disk, but almost all of the mass is actually in a third part -- a spherical halo of "dark matter" that extends much further out than the disk. In order to escape the galaxy, an object has to have enough velocity to escape the gravitational pull of all of that dark matter. The actual number depends on the model you use, and there is still debate over which model is best. However, a good estimate for the escape velocity for an object at the radius of the Sun from the center of the galaxy is about 1000 km/s.

Q. One theory says that galaxy collisions can produce ultra-high energy cosmic rays. However can a galaxy gobbling up a satellite produce them too, where a satellite is a few percent the mass of the primary?

A. The primary formation mechanism of the highest energy cosmic rays (i.e those with energies of 10^15 electron volts or higher) is still a mystery. However, it is reasonably well established that high energy cosmic rays are produced primarily in supernovae. The collision of gas rich spiral galaxies of comparable mass can and will generate a massive starburst event in these objects; a good example of this is NGC 4038/39, called the "Antennae". Massive bursts of star formation are tied to huge numbers of supernovae (which come from massive, hot and short-lived stars) and other high energy processes like the fueling of nuclear black holes, all of which can produce cosmic rays. Collisions with small satellites can trigger bursts of star formation but they are rarely violent enough to produce massive star bursts. For example, our own Galaxy is in the process of devouring one or two small satellites, but the star formation rate is comparable to a normal galaxy of this size. However, in about a billion years, our collision with the Andromeda galaxy, M31, will result in a major starburst event and, very likely, produce many high energy cosmic rays. A final point about ultra-high energy cosmic rays. It is believed that such particles cannot survive for distances more than a 150 million light years or more, since they interact with the cool background radiation field that permeates the Universe and lose their energy. In this light, it is very likely that ultra-high energy cosmic rays are produced by some process in our own Galaxy and are not from faraway galaxy collisions and mergers. But not enough is known about the sources of the intergalactic cosmic ray field to verify this idea.

Q. What time would it be if a lunar eclipse was seen directly on your meridian?

A. Lunar eclipses occur when the shadow of the earth (due to the sun) falls on the moon. Thus, the geometry of the situation means that the sun, the earth and the moon are in a straight line.

Now, let's assume that the sun crosses the meridian (transits) at noon every day (which is not strictly true - see The Dark Days of Winter for example). Then, during a lunar eclipse, the moon would transit 12 hours later, i.e., at midnight. This is because of the geometry of a lunar eclipse as mentioned above.

Q. Is it true that Mars will appear larger this August than it has for 100,000's of years?

A. Hi,

Mars was very close to Earth in August 2003. According to NASA's Mars website, it was closest to the earth in almost 60,000 years. For this year (2005), Mars is still pretty close to Earth, but this is nothing extraordinary.

From Sky and Telescope's Sky at a glance page:

Mars (magnitude 0.3, in Pisces) rises around midnight and glows fiery orange very high in the southeast by dawn. Mars will rise earlier each week, will enter the evening sky by the end of summer, and will blaze brilliantly during a close pass by Earth in October and November. In a telescope, Mars is 11 arcseconds wide and showing its gibbous phase. In late October it will reach a maximum apparent diameter of 20 arcseconds.

So, once again, this August is not special as far as the apparent size of Mars is concerned, but the next few months are still very good for observing it.


NASA's Mars website:
Sky and Telescope:
This Week's Sky at a Glance:

Q. The sun was supposedly created following the explosion of a red giant about 7 BYA. Is there any evidence of this explosion today in remnant background radiation and if so might it be detected? Are there any known nearby stars that resulted from the same explosion and if so which ones?

A. The explosion of a massive star, such as a Red Supergiant, is called a Supernova. The influence of supernovae (a plural form of 'supernova') on the origin of the solar system and the Sun can come is two ways: a.) the initial gas cloud (nebula) from which the Sun formed almost surely contained gas from very ancient supernovae and b.) a supernova near this nebula might have caused the cloud to collapse and eventually form the Sun and its sisters (if there were any). The former cannot be conclusively dated; it was the sum effect of billions of years of gradual accumulation of gas from thousands of supernovae across the Galaxy. The latter scenario, if it did happen (and as yet, it is purely speculation, since many other processes can cause a nebula to collapse), must have occurred about 4.5-5 billion years ago, about the time when the solar system formed. Coming back to your questions: the typical lifetime for a supernova remanant (SNR), which is a vast expanding cloud of hot, shining gas that is driven by the power of a supernova and fed by the dead star, is less than a few hundreds of thousands of years or so. Any supernova that exploded 4.5 BYA would have long since faded into the thin gas between the stars. Also, since the Sun moves around the center of the Galaxy every 230 million years, it would leave any such SNR behind as it followed its orbital path. Which takes us to the second question. The Sun might very well have formed along with a cluster of stars from that very same nebula, but over the course of its lifetime, this cluster will have slowly dispersed due to the effect of nearby stars and mixed in with the dust of stars in the neighborhood of our Sun. I believe that, based on their chemical composition and motion in space, a few stars near the Sun have been postulated to have formed with our own star, but this work is not absolutely conclusive. A lot of work needs to done in improving our knowledge of star formation on the smallest of scales in order for us to make confident statements about the environment in which our Sun was born.

Q. I know that travel faster then the speed of light is not possible, but I have a question.:

I have heard that with very strong telescopes we have found stars that are moving away from us with a speed almost the same as the speed of light.

Now I wonder what the situation would be if a rocket were launched in the opposite direction from that of such a the star. In this case the rocket would only have to speed up a little to reach the speed of light sent from the moving star. Isn't the rocket now traveling faster then the speed of light transmitted from the star moving away from us?

A. This is actually a very loaded question and it is difficult to give a simple answer to it, but I will try.There are two key issues here: Does our measurement of the speed of light from an object change when we move with respect to that object? And, can we see objects that move away from us faster than the speed of light (if there are any such objects)?

The first answer is simple. According to the principle of Relativity, the speed of light is a constant in all frames of reference (which basically means, an observer and his measuring tools) and there are no preferred frames. Therefore, even if something is moving away from you, you would not see the speed of light from that object to be any different. Put simply, no matter how fast something is moving away (or towards) you, even if it is moving at 99% of the speed of light, the light from it would still move at 300,000 km/s! This is very counterintuitive, but it is a well tested part of the how Nature works.

Before we start with the second answer, I want to mention that the objects that we see moving at near the speed of light are usually very far away and are not stars, but usually galaxies and quasars (which are very massive black holes shining as they gobble gas from the surorunding galaxy). Stars would be way to faint to see at the distances under consideration.

When an object moves very fast with respect to us, what does change is the frequency at which we see its light. Therefore, all these far-away, fast-moving objects have their light shifted strongly to low frequencies (or redder colors, hence this is called a red-shift). One must be careful here. The red-shift is not because the object is moving but because the space that the light has travelled since it left the object has expanded along with the rest of the Universe. This is called a Cosmological Redshift. Now to the answer of your question. If we moved in a fast rocket in the direction opposite to a distant object in the sky, the velocities would be combined (not just added in the simple fashion that we are used to in everyday life, but in a slightly more complex way dictated by Relativity). The speed of light from that object would not change but the light itself would shift to the red. The faster we went, the redder this object would look till finally, all its light would be too red for us to see and it would become invisible. Therefore, we cannot see anything that is moving away from us faster than the speed of light. Having said this, Relativity guarantees that no matter how fast the object is moving with respect to us, the light from that object always moves at the speed of light.

A final note: Can there be galaxies moving from us at faster than the speed of light? Yes. We can't see them (yet), but remember that the objects themselves are not moving, but the space between them and us is expanding as the Unverse expands and ages. This expansion can happen at rates greater than the speed of light (and probably has happened in the very, very young Universe) and, in all appearances to us, it looks as if these galaxies themselves are moving faster than light.

I advise you to peruse a good text in modern cosmology for a more detailed description of these ideas: "Foundations of Modern Cosmology" by Hawley or "Introduction to Cosmology" by Narlikar, for example.

Q. After the new moon is born, is there any specific degree when the cresent is surely visible, and which theory is sure, Yallop's or Eliyas or Monzer or nome of them since I read the following in this link for the US Naval Observatory ( "The date and time of each New Moon can be computed exactly but the time that the new Moon first becomes visible after the New Moon depends on many factors and cannot be predited with certainty". So is the time when the first moon becomes visible is something absolute with no doubt at all or what?

A. The time at which the Moon first becomes visible is not absolute but depends on a number of factors as the US Naval Observatory has noted. Doctors Yallop and Ilyas (Eliyas?) have tried to establish guidelines indicating the best time to attempt to such a viewing. Monzur Ahmed's Mooncalc (Monzer?) program uses Yallop's criteria as a default but allows the user to select Ilyas'. Yallop (1998) and Ilyas (1994) indicate a need for additional observations and investigation. Ilyas recommends optimizing calculations for tropical latitudes because of calendar concerns. Improvements in these predictions will not lead to an absolute determination of the first visibility because they will not completely address local site conditions. In addition, local weather plays a part, as does the actual observer.

Q. Is there a website into which one can enter their latitude / longitude, and compute the azimuth and time of sunrise or sunset for a given day?

A. Try the US Navy's website: Click on "Data Services", and then click the link for "Altitude and Azimuth of the Sun or Moon During One Day". After you enter your location information, you'll get a table of the Sun's alt/az positions throughout the day. To get the exact time of sunrise, backtrack a step and click the "Complete Sun and Moon Data for One Day" link. Good luck!

Q. It seems to me that red-shift data has resulted in lots of complications. There's got to be lots of dark matter and dark energy, whatever they are?

A. At a very basic level, dark matter is stuff that does not give off significant light. Using the motions of stars and galaxies, astronomers can estimate the amount of matter is a galaxy or cluster of galaxies, respectively. These estimates are larger than the matter estimates based on the observed stars, gas, and dust in the galaxy or cluster of galaxies. Dark matter is what makes up the difference. You and I and brown dwarfs and supermassive black holes are baryonic, or normal, dark matter. Studies are currently attempting to detect more of this dark matter through its effects on luminous matter, such as gravitational lensing. Particle physicists are exploring more exotic forms for dark matter, so called Weakly Interacting Massive Particles (WIMPS), or non-baryonic dark matter. Measurements of fluxations in the cosmic background radiation made by the Wilkinson Microwave Anisotropy Probe indicate that the universe is 4% normal matter, 23% cold non-baryonic dark matter, and 73% dark energy. While these measurements place some constraints on the dark energy hypotheses, they do not clearly identify it.

Q. Why when we see videos of the Earth from space can't we see the Earth spinning?

A. If you were to hover above the equator, the Earth would be turning beneath you at a rate of about 1,042 miles per hour, or 17 miles per minute, which is small compared to its 25,000 mile circumference. You do not perceive its gradual change on a second by second basis. However, if you were check the view below every few hours or so, you would notice the progressive changes until the Earth completed a full day and the same location was beneath you again. The Sun and stars appear to rise and set as a consequence of the rotation of the Earth. You are not conscience of the changes in the position of the Sun during a few minutes (do not look at the Sun directly) but are aware that it shifts position noticeably between when you get out of school and when you have dinner. You can check this with the stars on a winter evening. Notice the position of a bright star with respect to a tree or building along the horizon early in the evening. Then, check it every hour or so until you go to bed for the night. You should see its position change throughout the evening.

Q. I'm looking for a diagram the clearly and simply shows how our Solar System is oriented relative to our Galaxy. I have seen many diagrams that show an approximation of our location in our our Galaxy, a map of our Galaxy so to speak, but I want to see how the plane of the Solar System relates to the Galactic plane.

A. You can get a good idea of the relative orientation of the Earth's rotation axis, the plane of the Solar System (the ecliptic), and the Milky Way by observing the night sky from a dark site. The Milky Way arcs across the sky marking the plane of our Galaxy. The motions of the Moon and planets stay close to the ecliptic and can mark out this plane; for a science project using these observations, review the materials available at The plane of our Galaxy is tilted about 63 degrees relative to Earth's equator; its projection on the sky is called the Celestial Equator. The North Galactic Pole has coordinates RA 12 h 51.4 m and DEC 27 deg 27 m 7.7 s and lies in the constellation Coma Berenices. The North Celestial Pole, which is a projection of the North Pole onto the Sky, has DEC 90 deg and lies close to Polaris (North Star) in the constellation Ursa Minor. The angle between these two poles is also about 63 degrees. The plane of our Galaxy is titled about 66 degrees relative to the plane of the Earth's orbit around the Sun, which is called the ecliptic. A picture of the entire sky taken by the Infrared Astronomical Satellite (IRAS) is available at; it shows the bright plane of the Milky Way along the center with an a S-shaped feature at about a 60 degree angle, which is dust in the ecliptic plane. If you look at this picture of the sky taken by Most planets orbit the Sun within a few degrees of the ecliptic. The North Ecliptic Pole has coordinates RA 18 h 0.0 m and DEC 66 deg 33 m 39 s and lies in the constellation Draco. The angle between this pole and the North Galactic Pole is also about 66 deg.

Q. 1. Is one light-year equivalent to approx. 5.8 trillion miles? And would that make Andromeda 11.6 trillion miles away? 2. What does this mean: "Phase of the Moon on 20 October: waxing gibbous with 63% of the Moon's visible disk illuminated." Does this mean that there will be a fair amount of moonlight on this evening? 3. I also have sunset on this day (Oct. 20, 2007) beginning at 6:23 pm and ending at 6:50 pm. Are these correct times?

A. 1. You are correct: one lightyear is equal to about 6 trillion miles. A lightyear is the distance that a photon (a particle of light) travels in space in one year. The Andromeda Galaxy is the larger companion of our home, the Milky Way Galaxy. Andromeda is considerably further than 2 lightyears away. The nearest known star to the Sun, Proxima Centauri, is more than 4 lightyears away. The Andromeda Galaxy is 2 million lightyears distant. Multiplying 2 million by the number of miles in a lightyear gives us the distance to Andromeda in miles: 12 quintillion miles. That's a 12 followed by eighteen zeroes, or 500 trillion trips around the Earth. When you look at Andromeda, your eye is being hit by photons that left Andromeda when the ancestors of modern humans were just learning to use stone tools. Meanwhile, the light from Proxima Centauri left the star only one presidential election ago! 2. There is a whole vocabulary for describing the phase of the Moon. You may already have discovered that the Naval Observatory's website is a great place for finding information about sunset times, the Moon phase, and other useful astronomical information: Here is how to break down this information: "Waxing" or "waning" describes how the Moon's appearance changes from one night to the next. If it appears to be getting bigger (as more of its illuminated surface becomes visible from Earth), it's waxing. That means that you can expect the moonlight to be brighter on the following evening. Waning means that it appears to be getting smaller, as more of the Moon's sunlit surface turns away from the Earth. "Gibbous" or "crescent" describes whether the Moon is more or less than half full. So a gibbous moon (more than half full) means that the Moon will look fat in the sky and there will be quite a lot of moonlight. The percentage of the visible disk that's illuminated (63% in your case) gives you a more specific number for how bright the moonlight will be -- 63% as bright as it would be for a full Moon. Then there are events for the phase of the Moon, like when it is full. Technically, the Moon is only full for an instant before its revolution around the Earth starts to turn its illuminated disk away from our eyes. The revolution is slow, though, so the Moon looks full for about three nights in a row. The cycle of the Moon's phases, about 29 days and 12 hours long, starts when it is a new Moon (meaning that it can't be seen at all), and the cycle is divided up into quarters. After that, the Moon is waxing crescent. First quarter is when you can see half of the Moon's disk. Then the Moon is waxing gibbous. Full Moon is halfway through the cycle. The Moon becomes waning gibbous. Third quarter is when you can see the other half of the Moon's disk, and then the Moon is a waning crescent until it's new again. On October 20th there will be a fair amount of moonlight, which means, unfortunately, that you may have a hard time seeing the Andromeda Galaxy during your bonfire gathering. However, the Moon itself makes a great target for small telescopes and binoculars! It's best to observe the Moon when it's not close to full or new, because then you can see dramatic shadows along its terminator (the line on the Moon where the sunlight ends). If you focus on the terminator, you can see mountains, valleys and craters in rich detail. 3. Sunset occurs at one time -- probably close to 6:30 pm on October 20th. You will probably need flashlights by 7 pm.