Get a Straight Answer

    I am currently studying astronomy, and found your article on Lagrangian points thoughtful and very useful in helping me understand. I do have one question for you if you don't mind, however. You mention that were it not for other influences, the Lagrangian points would be stable. How can this be? If would appear to me that as an object starts to move away from one of the points, the change in the gravitational pull from the sun would cause its orbital velocity to change, which in turn would cause it to move farther away from the L point. A closely related question is this: how can an object orbit a L point without having some mass to which the object is attracted to?
    I am sure the answer is simple, but my brain is hurting trying to figure this out. Your answer will be most appreciated.
    Regards, Larry

Reply

  I wrote in "Stargazers" that if it were not for other attractions, L4 and L5 would be stable--but one should add that L1 and L2 are unstable. (Still, I am not sure about some "halo orbits" near them--see "The Art of the Orbit" by Gary Taubes, p. 620-622 Science, vol 283. 29 January 1999, section after the subhead "Three-body perfection.").

  If you are studying astronomy at the college level, you might find a relevant derivation in Symon's text "Mechanics." For objects that keep fixed positions in a ROTATING frame, the equilibrium can be studied in that frame by adding a centrifugal force, and then you can obtain a potential function and draw its contours. The problem then resembles that of a small ball rolling with no friction on a curved surface: if the L4 point is the center of a pit, small displacements would cause the ball to roll back, so the equilibrium is stable. Or else it could circle the pit, like a marble in a bowl: it needs no attraction from the middle.

  If instead it is on top of a dimple, a small displacement will cause the ball to roll even further away, never to return, which signifies an unstable equilibrium.

    If a meteor of significant mass hit the earth wouldn't this cause the earth in turn to move. Would its orbit be disrupted?

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    To give a short answer to your long question--not likely. Asteroids are far too small. An asteroid with a 10 km radius would have a volume less than one part in 200 million of the Earth, and if its mass were similarly scaled, the impact on the Earth would negligibly affect its orbit. Anything large enough to shift our orbit would have to be larger than any known asteroid, and the collision would be violent enough to wipe out all life.

    Personally, I would think so. The logic being that if I assumed that the mass of the upper crust were zero, the closer the object moves towards the core the greater the gravitational pull (till the object penetrates the core).

Ron

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    That is some neat question you have asked, and your qualitative argument is absolutely right. A short calculation (using some elementary calculus) makes it more precise.

    Suppose we are at a distance R from the center, the local density is D(R), and we move a test mass m downward by a small distance dR. If G is the constant of gravity and M the attracting mass, does gravitational attraction increase or decrease?

    In a spherically symmetric mass, any mass closer to the center than the attracted one acts as if they were concentrated at the center, while any which is more distant has no effect. That result is part of the theory of the potential, although Newton cleverly derived it from elementary considerations even before such a theory existed.

Therefore, as our test mass advances a distance dR towards the center, the mass that is attracting it diminishes by dM = 4 p R2 D(R) dR, and the attracting force decreases by

where K = 4p Gm. On the other hand, the closer approach to the center adds to the force

    Let us ignore signs and just recognize the contributions are in opposite directions (the fact R is positive upwards while the force of gravity points downwards can confuse). If the average density of the mass M below is , then

Substituting in the equation, canceling the cube power and introducing K gives

Thus if D(R) is smaller than (2/3), gravity increases, if larger it decreases, which includes the case of constant density, D(R) = . A nice problem!             David

    When I saw those movies, there was always something that confused me so much. What's the differences between LIGHTSPEED, HYPERSPACE, and WORMHOLE?

I can understand about lightspeed, but I don't know if a wormhole could be used in space travel. As far as I know, quantum theory was just used to prove other dimensions of our world (parallel worlds), so is there any connections here between this wormhole and space travelling?

Well, Mr. Stern, I think these are the questions to which I'd like to know the answers. Can you please help me?

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    The stories of science fiction movies come from professional writers, not from scientists. About 100 years ago Einstein found (something confirmed since then in many ways) that no material object can move faster than light, 300,000 kilometers per second. (If YOU moved that fast, time would pass at a different rate, so TO YOU the speed might seem greater--but not to someone in the outside world).

    Writers of fantasy stories, and later of fantasy movies, felt restricted by that fact, which suggested that back-and-forth travel or communication with civilizations on planets outside the solar system was impossible on the short time scale of travel and communications between countries on Earth. As seen now, a projected trip to another world (even using technology we do not have yet!) might take many thousands of years.

    So writers picked up some scientific terms, suggesting some day in the future the limitation of light speed may be overcome, by using hyperspace or wormholes. However, these are just ways for literature and films to imagine things which physics says (at least right now) cannot be done. I am not sure about wormholes, which have to do with general relativity: the added dimensions proposed by some theories extend only a very short distance into our universe, and are not likely to help us navigate the three principal dimensions of our universe (or 4--though time is a different kind of dimension)

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My almost 8 year young son Adam and I have a question about the revolution of the sun. We know that the planets revolve around the sun, and all have rotational periods also. We see that the sun aside from having a rotational period, also has a revolution of some 250 million years. We are curious what it is that the sun is revolving around?

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    What do they rotate around? Good question. There is SOMETHING at the center of the galaxy, and radio astronomers have determined it is very compact--I read somewhere, smaller than the orbit of Saturn, or maybe Jupiter. It also seems massive, but does not shine brightly, and most astronomers favor a humongous black hole, created in the early years of the universe (yes, Adam, we are safe from it).

    Still, what holds galaxies together is a bit of a mystery. If it were just the gravity of something pulling it towards the middle, a galaxy would rotate like the solar system--fast motion near the middle, slower and slower as one gets away. Vera Rubin has examined the light of galaxies and has determined (by the Doppler effect) that many of them, apart perhaps for the outer edges, rotate together, like a spinning dish, which is SLOWEST near the middle.

Love your pages; they're very useful and educational.             Michael

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    The fact the orbital planes of all planets and of most of their moons are so close to each other (though not exactly the same) suggests that they all were created from the same swirling cloud of dust, gas and flying rocks of assorted sizes. Different theories exist about how it happened, but I believe astronomers have observed such clouds, which one day may become planetary systems.

    The fact the Earth, and you, and I, contain fairly heavy atoms (oxygen, chlorine, even iron) suggest that at least some of the material of that cloud was previously part of another star, which "burned up" its hydrogen fuel and then exploded. See

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    Space rockets are narrow and long to reduce air resistance. They are inherently supersonic--orbital velocity is 24-25 times the speed of sound. That means they do not use wings during ascent, wings only help at low speeds, and just create more air resistance later on (though the first stage of the Pegasus launcher does have short wings). Also, they have sharp noses, to create the weakest shocks in front--again, shocks create resistance.

    Out in space, more variety exists: spacecraft can spin or not, some are drum-shaped (those usually spin), some have solar panels that stick out. But all that does not involve aerodynamics.

    If the spacecraft is to reenter the atmosphere safely, a lot of energy must be dissipated. A blunt front creates a strong shock wave, and much of the energy goes to the heated air in the shock wave, it does not heat up the spacecraft. Still, the heating of the front of the spacecraft is strong enough to require protection, by ceramic tiles in the shuttle and by material that ablates (wears away) on the re-entry capsules of Apollo, Mercury and Gemini.

    First of all please accept my thanks and regards as you have clarified many astronomical puzzle for which I was searching the correct answer. I am writing to you after a long time.

    As I know from Internet web site that plenty of space debris is revolving around the Earth at various altitude and definitely at different speeds.

    Frequently NASA or ESA etc sends an artificial satellite or space shuttle around Earth's orbit at a distance of more than 200 KM to 40000 KM. Even in 1994 astronaut Mark Lee was found flying over earth's surface as satellite.

    How do they avoid collision and monitor the movement of such unwanted space debris as the danger appears due very high speed?

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    Space debris is gradually being recognized as a serious problem, and at least one collision has already been reported, involving a French satellite. The density of spacecraft is still low, so the risk is small, but it is not zero. The US Navy is monitoring such objects by radar and yes, the number is increasing.

    The solution is uncertain. Low altitude orbits and highly elliptical ones reenter the atmosphere after a while, but communication satellites in synchronous orbit, of which hundreds now exist, will stay around for millions of years unless picked up.

    As a 7th grade science teacher, I have been looking through many websites, to find activities to teach sun's fusion reaction "in a nutshell." That is how I came across yours sections S-7 and S-7A.

    Actually, I have been looking for a more kinesthetic "hands on" approach but hopefully I can take your material and "soften the edges" to make it more middle school friendly (although our population of students tends to be academically inclined and I hope that I won't have to take off too many edges). I hope to be able some way to come up with something like M&M's for them to experience fusion tastefully!

    Thanks for the info.

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    Dear Sharon

    You have my respect for teaching nuclear fusion in middle school! However, the only hands on demonstration I can think of is to use a bunch of those small cylindrical magnets used for pinning messages to a steel partition (or refrigerator). They will all stick together, but the forces are short range--once you pry a magnet a short distance off the bunch, you have no problem pulling it all the way.

    Sir

    Your website stargaze/Sun4spec shows and explains the visible spectrum.

    At present we are looking into the 1,400 Watts-sec/square metre constant and we would very much like to know the make-up of the light received on planet earth. [over full spectrum not only visible.]

    Ideally we would like to know what percentage in Watts of the 1,400 Watts is attributable to which of the elements in the periodic table 1] Helium 2] Hydrogen 3] Carbon 4] Oxygen etc.

    Only a very rough answer say within 10 % is necessary. If this information is already on an alternative website please point us in the right direction.

    Many thanks

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    Dear Clive

    Visible sunlight, coming from the photosphere, may have started as a specific emission of single atoms, probably of hydrogen or helium, in a deeper layer of the Sun. However as this light works its way to the surface it undergoes absorption and re-emission many times. The final spectrum reflects not the original emission and its energy levels, but the way energy is shared among many interacting atoms. A similar situation exists in a hot glowing solid, and in both cases the spectrum is smooth and depends only on temperature. I seem to recall the Sun's color distribution fits about 5800 degrees and that about 1% is in the ultra-violet range.

    Could you tell me how the Jewish calender originated ?

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    Dear Robert

    As I wrote in the "Stargazers" unit, the Jewish calendar is very similar to the Babylonian one--in using the Metonic cycle, in its names of months and the ambiguity of the new year (all of which I found in my 1967 edition of Encyclopaedia Britannica). The similarity is understandable, because according to the biblical scriptures (which are quite consistent on this point), Jews lived in exile in Babylonia for 70 years in the 6th century BC.

    They came to Babylonia speaking Hebrew and returned speaking (except in religious usage) the language of Babylonia, Aramaic, which is somewhat similar to Hebrew and which prevailed over 1000 years, up to the Arab conquest. They came to Babylonia with their own alphabet, angular like the Greek one, and returned using Babylonian letters (although again, Maccabean kings for instance continued using the old script on their coins). The script known today as "Hebrew" and used in Israel is, in fact, Babylonian. And most probably the calendar, too--the bible still mentions some old names of months (Ziv, Bul, Eytanim, perhaps Aviv) which did not persist. The bible itself uses primarily numbers ("second month") but the names we have today are very close to the Babylonian ones.

    Still, there are signs that it took a long time before the new system was completely accepted. For many centuries a new month was supposed to begin, not on a pre-calculated date as in the Metonic cycle, but only after reliable witnesses had seen a "new moon" (supposedly--it is unclear what was done during prolonged cloudy weather!). According to Jewish tradition, the final form of the calendar was introduced in 358/9 AD by the patriarch Hillel the 2nd (Encyclopaedia Judaica).

    The reckoning by which this is year 5762 to the creation of the world is even more recent, derived by bridging between the historical record and biblical chronology. A similar calculation was performed by the Christian Bishop Ussher.

    I hope this satisfactorily answers your question!

    I will argue that it is unnecessary for an object to achieve 8km/hr to leave the Earth's gravity, as long as it has a continued thrust which is greater than the pull of gravity. With such thrust a rocket could literally crawl from the Earth at one mile per hour. Obviously each rocket has this thrust or it would not leave the surface of the planet, where gravity is at its strongest. Escape velocity pertains only to objects without any additional thrust available.

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    Hello, J.D.

    Your argument is correct, but the conclusion you draw is not. Suppose you have a rocket of mass M accelerating from the pad with an acceleration a=g, which we will round off to 10 meter/second squared. That means its rocket must provide a thrust of 2Mg--Mg to support the weight of the rocket and Mg to accelerate it. To reach orbital velocity of 8000m/sec will take 800 seconds (8000/a = 8000/g). During that time the launch vehicle has to use half its thrust just to keep itself from falling--only half the thrust goes to accelerating.

    Actually the mass M of a rocket decreases as fuel is burned off, so the acceleration increases, making the time shorter (the space shuttle achieves orbit in about 6 minutes, less than half the above time). One reason stages are dropped in manned missions is to limit the acceleration to about 2-3g ; more than that is hard on the astronauts. See example of the V-2 rocket in "Newton's 2nd law", section 18 of "From Stargazers to Starships" at

    I'm an interested science teacher at St. Catherine's British Embassy School in Athens Greece. Is there any reliable information relating to when and by whom it was first proposed that the earth is a sphere?

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    Funny that YOU should ask--living in Athens, Greece, you may well have the best experts on the subject within walking distance. From what I can quickly find on the web (I am at home, away from any library), the first solid arguments were made by Aristotle:

    I've enjoyed your page on the precession of the equinoxes at

    Hello Mr. Stern!                 (Received 21 December 2001)

    Excellent website. Here's my question:

    What is the name of the figure-eight traced upon the earth by the combination of axis tilt and orbit in a year; where the figure represents the shortest distance between the earth and sun?

Reply

    Dear Neil You are probably thinking about a figure known as the analemma. You can read all about it at

    Hello David,

    I was hoping perhaps you might be able to point me to a reference [if it exists]. I am doing research with a professor and looking for a site or reference that would state the coordinates of the polar axis location against year (ie 100 BC, 1000BC, 10,000 BC, etc.) (not the magnetic axis) Thank you so much!

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    Dear Eve

    Your question is not completely clear: what do you mean by "location"? The DIRECTION of the spin axis, determined by the angular momentum of the Earth, is almost constant. The Moon's pull (and maybe the Sun's too) on equatorial bulge of the Earth, created by the Earth's rotation, causes a 26,000 motion of the axis around a cone, expressed in the precession of the equinoxes. See

    Dear Sir:

    Would you please explain how the Van Allen Belt effected the first manned space flights. How were they protected?

    Thank you,

Reply

    Dear Belinda

    All manned flights (except those of Apollo) have stayed below the radiation belt: the Space Shuttle, for instance, orbits at about 215 miles. The atmosphere is very rarefied there, and radiation belt particles descending to that level may well come back without encountering anything. However, such particles have thousands of Earthward excursions each day, so the only ones which are likely to survive long are those that are always confined to higher levels.

    A more subtle effect is also at work. The equations governing the motion of trapped particle indicate that each has a characteristic value of magnetic intensity, below which is cannot penetrate. Suppose a particle is reflected by the intensity existing at 215 miles. As it happens, the Earth's magnetic field--its region of magnetic forces--has some irregularities, so in some regions that intensity is only reached at 100 miles. Now and then the particle's orbit will happen to descend in that region, where it penetrates to much deeper (and denser) layers of the atmosphere, and may be quickly lost, even if elsewhere it stays at safe heights. One such notorious region exists above the southern Atlantic Ocean.

    So the radiation belt does not reach the levels where Mercury, Gemini, Soyuz and Mir used to orbit and where the Shuttle and Space Station do so now. The early Russian Sputniks failed to discover the radiation belt because they too stayed in such low orbits and Explorers 1 and 3 only detected it because they were rather poorly controlled and rose above 1500 miles.

    Dear expert,                 (received 21 December 2001)

    Please answer this question which has been set at our school in East Sussex,UK. What is the nearest star outside our galaxy? I am a year 5, age 10. Thanks for your help and time

Reply

    Dear Oscar

    Nice of you to call me "expert." Actually, I am a space physicist, not astronomer, but will try to answer you anyway. All stars observed from Earth are in galaxies, so the nearest one outside our galaxy should be in the galaxy closest to us. I would guess that would be the Large Magellanic Cloud, so called because it was observed by Ferdinand Magellan after he crossed the equator (opening to his view stars never seen from Europe) as one of two fuzzy glows in the night sky.

    The LMC became famous in 1987, when a supernova exploded in it, allowing interesting phenomena to be observed. Of course, technically the explosion happened 164,000 years ago, because the LMC is 164,000 light years from us.

    The trouble is that stars in the LMC do not have names. Astronomers presumably have their designations (possibly, numbers in a catalog of stars), but they are not publicized in star atlases etc., because you only see such stars with powerful telescopes and perhaps time exposure photographs.

    To see some of these stars, go on the web to "Astronomy Picture of the Day", appropriately titled "Pick a Star." The address (one of several) is

1. I believe earth-orbit satellites are launched in either polar or basically west-to-east orbits. Why do we not launch in a westerly direction?

2. What is meant by "sun-synchronous" orbits?

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    Israel has launched two satellites so far (maybe more). Lacking the choice, it must launch westward over the Mediterranean, and those 360 (or so) meters/sec hinder rather than help its rockets, reducing the available payload. Yes, it can be done, but when a choice exists, eastward is preferable.

    To answer your second question, let me paste from the glossary of "Exploration of the Earth's Magnetosphere":

    Sun-synchronous orbit--a near-Earth orbit resembling that of a polar satellite, but inclined to it by a small angle. With the proper inclination angle, the equatorial bulge causes the orbit to rotate during the year once around the polar axis. Such a satellite then maintains a fixed position relative to the Sun and can, for instance, avoid entering the Earth's shadow.

(b)     Why are satellites launched from near the equator?

    When you calculate the centripetal force at the equator it is 0.033 m/s² (m v²/R) This means that there is approximately 0.3 % variation in the force that attracts a body towards the center of the earth between the equator and the north pole (where without precession no rotation would be felt). This would mean that it would be attractive to send rockets from the equator because it would save 0.3% in fuel. Is this true?

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    Your argument is well known, but it is usually phrased differently. The centrifugal force acts only in a rotating system. Once an object is detached from that system, putting the centrifugal force into a calculation may lead to incorrect result. Example: your bicycle wheel picks up a piece of mud from the road. As long as the mud is attached, it feels an outward force, a centrifugal force (in the frame of reference of the rotating wheel). Once it works loose, however, it does not fly in the direction of that force, but tangentially to the wheel!

    However... launching from the equator is still advantageous. A spacecraft located on the equator is carried by the Earth at about 400 meters/second, or about 5% of the orbital velocity. Even at Cape Canaveral, one still has a large fraction of that velocity, about 4/5. That is a significant advantage, and the reason all rocket launches from Florida are eastward. Polar satellites are usually launched from Vandenberg in California, and obviously cannot use that advantage.

    Launch sites, of course, are subject to political limitations. Cape Canaveral is as close as one can get to the equator from the continental US. The European Space Agency has its launch base at Kourou in French Guiana, about 5 degrees north of the equator. Israel, on the other hand, had to launch its satellites westwards, over the Mediterranean, requiring extra velocity to overcome the rotation.

    The only launch I am aware of that was conducted on the equator itself was of an Italian "San Marco" satellite, launched from an off-shore platform near Somalia. Today, with the "Sea-Launch" ship available for launches (a commercial partnership of Boeing and Russia), such launches are again practical.

    My name is Jason and I am currently in the 11th grade. My brother and I were wondering whether it was possible for people to grow infinitely tall. My brother believes there is a limit to height because muscle strength would be insufficient to support a person that tall.

    I contend that there should be no limit to height if they live on an extremely small planet with huge mass.

Please let us know who is right!

Reply

    Presumably, you meant "arbitrarily tall." Even with that change, I do not know what the proper answer would be, because the rules of your disagreement with your brother were not spelled out. On Earth, the answer is no: even trees do not grow arbitrarily tall, because of the difficulty of pumping sap very, very high. Pumping blood to comparable height would be very difficult (giraffes seem to have attained the limit) and even before that, the weight of an individual would make it hard to move or even stand.

    On another planet... if gravitation is very weak, the limits change. Of course a planet with such weak gravitation would probably not hold an atmosphere. A small planet with huge mass won't do, and in any case, we know no such planets.

    But... I have homework for you both. First of all, read "About Being the Right Size" by J.B.S. Haldane, written in 1928, on the web at
    http://ubmail.ubalt.edu/~pfitz/tell/gb/science/gb08_s.htm

    And secondly, you might enjoy "Food of the Gods" by G.H. Wells, a science-fiction novel about people growing to gigantic size.

    I'm a high school sophomore on Hawaii and I am doing a History Day project on the Gunpowder Revolution.

Some questions about the article you had written at this site:

        http://www-istp.gsfc.nasa.gov/stargaze/Sgoddard.htm

    Your work stated that rockets were a spin-off from the invention of gunpowder. Can I ask you to elaborate on that a little, as I am intrigued about this. For instance, what is the connection between rockets and gunpowder that would make rockets a 'spin-off'? On a much wider scale, how did this affect battles and how they were fought?

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    The Chinese made rockets by stuffing gunpowder into a hollow tube with one end plugged, aiming it skyward (with the open end towards the ground), lighting the gunpowder with a fuse or long match--and then standing safely aside and watching it rise (or explode, or do other unpredictable things). They used rockets for fireworks, and Europe and the rest of the world learned from them.

    Interestingly, the Chinese never invented guns, as the Europeans (prodded by an era of widespread warfare) did. One reason was that the Chinese did not at first get a very good gunpowder. You read that gunpowder consists of charcoal and sulfur, which provide the fuel, and salpeter, which provides the oxygen. True: but the salpeter needs to be refined, it starts out as a rather dirty mix. The Chinese only gradually learned to refine it, and one may guess that their earlier gunpowders were good for rockets and fireworks but not good enough to explode.

    Is it true that, as a result of the precession of the equinoxes, and because the earth's spinning axis remains roughly constant at 23.5 degrees, in about 13,000 years the northern hemisphere will experience the summer solstice in December?

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    One should not expect summer solstice to be in December in 13,000 years, because of the way the length of the year is defined, as the time between one solstice and the next, or one spring equinox and the next. In 13,000 the stars behind the Sun's position at the spring equinox will be quite different, but assuming the same calendar will still be current, the date will still be reckoned as 21 March, or near it.

    I was visiting your site (http://www-istp.gsfc.nasa.gov) and I must admit, it is a great site. My question is: When designing solar sails and computing their top speed, is the resistance provided by the interstellar gas particles taken into consideration? Also, at such high speeds, how likely is it that the sail might rupture because of the impacts?

    Thanks for your time

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    I do not know your answer--you can calculate it. My guess is no, the flux is too small, and the extra push of the solar wind more than outweighs it.

    Take a solar sail moving at 10 km/sec. It intercepts about 1 interstellar atom per cc (depending on direction, since the solar system itself moves at about 20 km/s) or a million per second per cm-squared. From the other side, it is overtaken by solar wind ions at 400 km/s, density near the Earth orbit about 6 per cc, or about 240 million per sec cm^2. So the solar wind pressure is larger, and that of sunlight larger still.

    Fast particles of course can cross the sail, but I don't think such "radiation damage" is serious. I would worry more about tearing due to the degradation of the material from short-wave sunlight and the space environment.

    David

    My son is doing a science project on the Big Dipper. One of the questions is "How far is the Big Dipper from the Sun". We have checked every site and cannot find this answer. Please help us.

Chick

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    Dear Chick

    I do not know what answer the teacher expects to get to "How far is the Big Dipper from the Sun." The Big Dipper (or to astronomers, "Ursa Major," the big bear") is a group of stars which, when viewed from Earth, forms a striking pattern, but that by no means assures that they are all close to each other and have the same distance. It is quite possible that some are close to us, others distant, and only by chance are they found together in the same part of the sky. In fact, some of the stars do seem close to each other (in the real sense of the word), but others are just accidentally grouped together. They move differently, and if you could wait a few hundred thousands of years, those stars would be much further apart.

    Anyway: the numbers. Main stars are named in order of brightness in their constellation, according to the Greek alphabet, and those forming the "Big Dipper" are alpha, beta, gamma, delta, epsilon, eta and zeta of Ursa Major ("Alpha Ursa Majoris" etc.). A labeled map can be found at
    http://www.astro.wisc.edu/~dolan/constellations/constellations/Ursa_Major.html

    Their approximate distances from the Sun in light years, according to

    http://www.dibonsmith.com/uma_con.htm
are:
alpha--86 ; beta, gamma, delta--100 ; epsilon--64 ; eta--95 zeta--78

    Unlike you, I do not mind signing my name. ....

67.(b)     Big Dipper star names

    I'd be grateful if you could tell me the name of the brightest star in the Big Dipper, and the name of the brightest star in the Little Dipper. If you cannot, then please pass this on to someone who might know.

Reply

    According to my copy of "Naked Eye Astronomy" by Patrick Moore (W.W. Norton, 1965), starting to count stars from the front of the Big Dipper and ending at the tip of the handle, they are named Dubhe, Marak, Phad (aka Phekda), Megrez, Alioth, Mizar and Alkaid (aka Benetnash). The names are Arabic, and if "Alkaid" reminds you of "Al Qaeda" that is no accident--"Alkaid" means the commander and "Al Qaeda" means the command.

    Dubhe is the"alpha star," but it's practically the same brightness as Alioth, and just a tad less bright than Alkaid. Except for Megrez, the others are not far behind. Mizar is a double star and can be resolved with binoculars.

    The "alpha star" in the Little Dipper (Ursa Minor) is none other but Polaris, the pole star. See
http://www.phy6.org/stargaze/Spolaris.htm

    The other stars are fairly dim, except for the two "guardians of the pole" at the front of the dipper. The one closer to Dubhe is Kocab, again very close in magnitude to Polaris.

(b)     Was Moon landing a hoax?

    "The only time in history that an astronaut, Soviet or American, is said to have left the relative safety of Earth orbit and ventured through the Van Allen Radiation Belts, a twenty-five thousand mile thick band of intense radiation which surrounds the Earth beginning at an altitude of about one thousand miles, is going to the moon. The Soviets, with a five-to-one advantage in the early part of the space race, never once sent a human through the radiation belts to even orbit the moon.

    In 1998, the Space Shuttle flew to one of its highest altitudes ever, three hundred fifty miles, hundreds of miles below merely the beginning of the Van Allen Radiation Belts. Inside of their shielding, superior to that which the Apollo astronauts possessed, the shuttle astronauts reported being able to "see" the radiation with their eyes closed penetrating their shielding as well as the retinas of their closed eyes. For a dental x-ray on Earth which lasts 1/100th of a second we wear a 1/4 inch lead vest. Imagine what it would be like to endure an hour and a half of radiation that you can see with your eyes closed from hundreds of miles away with 1/8 of an inch of aluminum shielding!"

    If you wish to see the website, it's http://www.moonmovie.com.

Reply (to both questions)

    http://www.phy6.org /stargaze/StarFAQ2.htm#q30
and especially the 2nd question there is relevant.

    I have not seen the film you referred to, but looked at the web site you mentioned. Its producer sounds a rather shrill note, a single tone repeated again and again. Too much evidence points to the Moon landing being real. Besides the testimony of participants (including New Mexico's senator Harrison Schmidt), we have the famous video of feather and hammer dropped in vacuum, the reflected laser beam experiment (a dim memory tells me it was by Prof. Alley of the University of Maryland), scientists' analysis of moon rocks and much more. Of course, having worked as a scientist for NASA for 40 years (but nowhere close to project Apollo or manned space flight), I also know a thing or two about the competency of the agency, and can confidently say that a hoax like that is way beyond its capabilities.

    The Soviets indeed never sent a manned space flight beyond the radiation belt, but it was not for want of trying. Their rockets kept blowing up, and their space program was badly overextended before it shrank back.

    As for flashes of light seen by astronauts, I have never heard about that before. Flashes HAVE been seen on Earth, due to cosmic ray particles (muons) whose speed is close to that of light in vacuum. As I understand it (a dim memory of an article in "Nature", I believe) such particles pass through the eyeball fluid, where their speed is greater than the local velocity of light, which decreases in transparent media. That produces a flash of light, the "Cherenkov effect," which has been likened to the shock produced by a supersonic airplane or missile. I believe the subject of the experiment sat in a dark room and particle counters verified the path of the particle.

    It is highly unlikely shuttle astronauts would see such flashes. Anyway, inner radiation belt protons (the main radiation hazard) are too slow to produce Cherenkov light. I am not sure about fluorescent scintillations, though--you might have to ask someone. Cancer of the eye may perhaps be treated by radiation, and if so, there should exist experience from there.

    We are sitting around a bunch of us at work (EMS) and we got on the topic about the rotation of the earth on its axis and its rotation around the sun. So...are we correct??? We think that the earth rotates counter clockwise on its axis and clockwise around the sun. Seems like a question everyone would know the answer to, but we've checked so many sites and can't find the answer anywhere.

    Thanks,

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    As for the rotation of the Earth... you should first of all realize that "clockwise" and "counter clockwise" are not absolute properties, but depend on your point of view. Imagine a clock with a transparent face, with you watching it from the rear. The number 12 is still on top and 6 still on the bottom, but now 3 is on the left and 9 on the right. So when the clock hand moves from 12 to 3, it moves ... counterclockwise!

    To define rotations with no ambiguity, we can stipulate they are always observed from the NORTH, from some point far above the north pole of the Earth.

    I live on the Eastern seaboard of the US, while my son lives in California. I would not think of phoning him in the early morning--it may be day here, but the Sun has not yet risen where he lives, before that happens the Earth has to rotate an extra distance towards the Sun. To do that, the Earth must rotate COUNTERCLOCKWISE, as you have proposed.

    The motion of the Earth around the Sun is harder. Let us start with the APPARENT motion of the Sun around the celestial sphere. You go out in the early evening in winter and see Orion in the sky. By early springtime Orion has moved west and now it is the constellation of Leo the lion that is in full view in the evening sky. Look on a star chart: Leo is east of Orion.

    So the evening constellations open up eastward: you could not see Leo in the winter, because the Sun occupied that part of the sky, but now the Sun has moved eastward and opened up for viewing a new patch of the heavens. To move across the celestial sphere from west to east, the Sun has to go around the pole (as viewed from north) COUNTERCLOCKWISE.

    However... that is the apparent motion of the Sun. If the Sun SEEMS to move around the Earth counterclockwise, but actually the Earth is the one moving around the Sun, is the Earth's motion clockwise or counterclockwise? You need a sheet of paper to solve this problem, which is also problem #1 among the ones listed in "Stargazers." But I won't keep you in suspense: it is ALSO a counterclockwise motion.

    So your second guess was incorrect. In fact all orbits in the solar system are counterclockwise, even of the larger satellites (except for one of Neptune, suggesting a captured asteroid). The common explanation is that all planets and their satellites condensed from a swirling cloud of gas and dust, and their rotations reproduce that of the cloud ("conservation of angular momentum").

I'm a student. Could you tell me what kind of isotopes are disintegrating in center of the earth?

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    That leaves mainly three: uranium, thorium, and potassium 40 (K40, with atomic weight 40). I do not know the numbers, but they all make important contributions to the heat inside the Earth. Uranium and thorium disintegrate in steps, creating intermediate elements such as radium, and these also contribute heat, but the ultimate source are these two elements. The end product is lead--as well as helium, ejected from unstable nuclei as fast alpha-particles. The helium we use to fill balloons comes almost entirely from this source: natural helium (e.g. of the Sun) has a certain percentage of helium with atomic weight 3, but the gas we get from underground sources is practically all helium with atomic weight 4. Potassium decays to argon, a gas which forms close to 1% of our atmosphere.

Hi,

    I am curious science teacher and I wonder "What is the coronal pressure? I mean for example 10 000 km of the photosphere."
        Best regards.

Reply

    Questions are easy, answers are difficult. I do not know the correct presure "off the top of my head" although I know the number is small. The encyclopaedia on the web says the density of the sun's photosphere is about 1/1000 that of the atmosphere on the ground, and even allowing for a 20 times larger absolute temperature, this is still quite rarefied. At 10,000 kilometers you are in the lower corona, and the answer is harder still.

    Since you are a teacher, I will go here into some more detail, maybe your class will be interested. The Sun is not my field, but I should know enough to explain.

    Why is the Earth's atmosphere near the ground under pressure? Because it is supporting the weight of all the air above it! You go up 10 kilometers and the pressure is down to 25%, because only 25% of the air is above that level, 75% is below it. In the atmosphere near the ground, pressure goes down by a factor 2 about each 5 kilometers (or by a factor e=2.71828... every 8 kilometers--that is called the "scale height"). The number depends on temperature, so it may go up and down a bit, but you can see it decreases very fast. Above about 100 kilometers the air is so rarefied that molecules and atoms rise up and fall like thrown stones, rather than colliding constantly, and then different rules hold and the decrease is slower.

    If the air were 20 times hotter (as in the photosphere of the Sun, 6000 degrees absolute against 300) the pressure would still be the same, because the weight above remains the same. The difference would be that the atmosphere would expand 20 times--"the halving distance" would rise to 100 km--while the density of the air would be only 1/20th. So with density dropping 20 times and temperature rising, the pressure SHOULD be the same.

    The same rules hold in the photosphere of the Sun. See:
        http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html

The gas is indeed has about 20 times hotter, trying to make the "scale height" 20 times larger than ours. However, gravity near the surface of the Sun (or what to the eye looks like one, it's really all gas) is about 28 times the gravity at the surface of the Earth, and that more than counterbalances the higher temperature. Most important, perhaps, is that the photosphere is mostly atomic hydrogen, about 1/30 times lighter than atmospheric molecules, and the scale height is larger by a corresponding factor. If you put all this together (and ignore temperature changes in the photosphere) you get a scale height of the order of 150 km and a "halving distance" of about 100 km.

    That is larger than in our atmosphere--but the Sun is much bigger too, and you realize that by the time you reach 10,000 km, something HAS had to change. As a matter of fact, the photosphere is only about 400 km thick. For the next 5000 kilometers you are in the chromosphere--hotter and very uneven, but still decreasing fast in density, and at 10,000 kilometers you are in the lower corona, temperature of about 1.3 million degrees, and who knows what pressure and density!

    I have looked for some references and found on the web an article
        http://www.aas.org/publications/baas/v31n3/aas194/481.htm
which claims to observe at 1.03 RS (solar radii; that is, at about 20,000 km above the surface) a density of 180 million electrons per cc (and if that is the density of atoms, that is more rarefied than any laboratory vacuum!) and a "nonthermal" velocity of 33 km/sec. Let me try and check it. Earth is about 200 RS from the Sun and is immersed in the solar wind, density about 10 per cc and velocity about 400 km/sec.

    If the flow is the same in all directions, in each second, the flow fills a spherical shell of radius 200 RS and thickness 40 million cm, containing

    I do not know too much about Tesla, but he had eccentric ideas, and one of them, I vaguely recall, was to tap energy from the atmospheric electric field. There exist relatively large vertical voltage differences across the atmosphere, the residual effect of atmospheric electric processes in distant storms, the same as are responsible for lightning.

    The reason they can persist is that the atmosphere is a very, very good insulator. Electric power companies string their cables in the air and never worry about any leaking away! To get any useful power from that voltage (as Tesla may have wanted) you need a closed circuit, part of which runs in the atmosphere, and the air would not allow it to flow there.

    I never heard about an electric car being run from this source. And although, when you stand, the voltage of the air around your head may be (say) 300 volts above the one at the ground, you never feel anything, because the tiny electric charge involved is immediately drained away by your conducting body, essentially short-circuiting the voltage to zero.