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    Hoping to settle a bet I stumbled across your website

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

Please help!

    Many people suggest that during each equinox every point on the Earth receives equal amounts of sunlight and darkness. This is not quite right, even at the equator, due to the refraction of the sun's rays through the Earth's atmosphere. However at the poles I think that this must be very wrong. During each equinox and at each pole I think that the Sun is, at least partly, above the horizon for the entire day (due to the Sun's size and distance, but also the refraction effect).

    Is this correct? I've convinced myself of this with a bit of geometry but I'll spare you (as it didn't convince my friend!)

    thanks,

    p.s. I can't resist explaining -- skip this if you want: during each equinox the line between the centers of the Sun and Earth is perpendicular to the Earth's axis. Consider the line of longitude in the plane containing these two lines. Call it L.

    Now, the sun always illuminates more than one-half of every line of longitude (actually it always shines on more than one-half of every great circle. This is because the Sun is bigger than the Earth, and due to the refraction effect.) So I just have to show that the "lit" half of L includes both poles.

    Well, the Earth's axis cuts L into two equal arcs and is perpendicular to the line connecting the centers of the Earth and the Sun. Done.

    Is that right?

Reply

    From:

            http://itss.raytheon.com/cafe/qadir/q406.html

    I get

    In 1975 I met for the first time Sir Edward Bullard, distinguished British scientist with an exquisite sense of humor (he communicated by colorful postcards of museum exhibits, and always signed his name "Teddy"). It was at a conference in Colorado Springs, where he told of an experience at his university in England. One day there he entered the office of a colleague from India and found him sitting in front of a meal set out on his desk, next to a ticking alarm clock.

    The man explained that he was a Moslem, and this was the holy month of Ramadan when Moslems were not supposed to eat before sundown. So he looked up the time of sunset, set a meal on his desk and was waiting for the proper time.

    I have just finished studing 12th grade according to the Indian syllabus (CBSE-Central Board OF Secondary Education) and am keen in taking up astronomy as her career.My subjects in the 12th grade included Physics, Maths, Chemistry, Computer Science and English as a langauge. My ambition is either to become an astronaut or serve either ISRO Indian Space Research Organization ? or NASA.

    I would like to know what are the procedures or the degrees to taken up In order to get into the field of astronomy.

    I would appreciate your kind help at your earliest.

Reply

    To become a professional astronomer or space scientist, you essentially MUST have a science doctorate, which required finishing your regular university study and then being accepted as graduate student in a good program, which takes a few more years of study.

My advice to you has 3 parts:

1.     Study hard, whatever you do (being in 12th grade, most of this is behind you, so this advice does not say much). Your letter suggests that you may need a better command of English, and more reading and writing will help you get it.

2.     Read about physics, astronomy and the math they use. The recent book "Seeing in the Dark" by Timothy Ferris is a nice introduction. My own web collection "From Stargazers to Starships" should help you too. Read it all, and try solve as many as possible of its problems (there exist two sets).

3.     I do not know where you are writing from--presumably from India, but I do not know where in India. If you have a local university or space-related institute, see if you can get a summer job there--even without pay. Ask your physics teacher to help you find such a job and perhaps write an introductory letter. By working with scientists and talking to them, you will learn more about the field you choose. You might well decide on a technical field less glamorous than space flight or front-line astronomy. The world has a much greater need for good engineers and technical workers.

Hello,

    I have been looking at your website for information that would help me in my science project.

    Recently, I finished experimenting with different colored lights on solar panels. It seemed likely that blue and darker colored lights (with short and energetic wavelengths) would produce the most volts. However, it did not turn out this way. Consistently--we ran this test over 30 times--Red, which has the longest wavelength of the visible electromagnetic spectrum, gave the largest number of volts.

    Now, I need to explain why this happened. I am really having a hard time explaining this. I know a little.......but I would really appreciate if you could send me some ideas or information about why this is happening! Thanks...

Reply

    Here is one possible explanation. It is true that blue photons are more energetic, and therefore ideally red light should not be as effective in triggering off the release of an electron than blue light.

    Actually, this is most clearly seen in ejecting electrons from freshly cut metal surfaces in a vacuum, as Robert Millikan showed in his famous experiment. That was the experiment that opened the way to Einstein's equation "E = h times frequency," for which he won the Nobel prize. Inside the semiconductor crystals of solar cells, this may not be so simple.

    In fact, I suspect that your solar cell is sensitive to ALL visible wavelengths. The blue filter on a light bulb lets through only the blue light and absorbs the red and yellow, while a red filter lets through the red and absorbs the blue.

    But the average light bulb is not hot enough to radiate a lot of blue, and even the Sun has more yellow and red than blue.. Its spectrum is skewed towards the red and, and if I remember right, the peak may be in the infra-red. Thus you get much more light energy through a red filter, which absorbs the weak blue, and less through a blue filter, which absorbs the strong red and yellow.

What is "radiation"? And--if you "nuke" food in a microwave oven, can it become radioactive?

Reply

    The word "radiation" in common use has (unfortunately) more than one meaning. And to answer the 2nd question--no, microwave radiation and nuclear radiation (involving radioactivity) are very, very different.

    "Radiation" is essentially something which spreads radially in all directions, along straight lines (in empty space). Examples are light and its close relatives, infra-red (IR) radiation (emitted by hot objects) and ultra-violet (UV) radiation.

    In the middle 1800s James Clerk Maxwell concluded that light was an electromagnetic wave. That suggested such waves could be created by purely electrical means and in 1886 Heinrich Hertz did so, producing radio waves. The microwave radiation in your oven is also produced electrically, it is just a short-wave radio wave, developed in World War II for use in radar. When physicists talk of "radiation" they usually mean "electromagnetic radiation," the large family covering light, IR, UV, radio and microwaves.

    Then came the year 1895. Two discoveries stirred up the world of physics and seemed to upset Maxwell's neat scheme.

    It was already known that in a glass tube containing very rarefied air and two separated electrical contacts, if a source of high voltage is connected between those contacts, an electrical current flowed between them and a faint glow was produced. [In 1897 J.J. Thomson showed this current was carried by tiny pieces of matter, electrically charged, which were named "electrons."] One of the discoveries of 1895, by the German Konrad Roentgen, was that the tube also emitted something that could darken photographic plates, even ones kept in tight dark boxes. Obviously some radiation was produced, and Roentgen named it "X-rays", "X" for "unknown."

    That same year the Frenchman Henri Becquerel found that uranium ores also emitted some "radiation" able to darken photographic materials. It thus seemed (for a while) that Nature had perhaps other new types of radiation--in fact, some claims were made (and later retracted) for more kinds, and science fiction writers had fun with stories about "death rays."

    Gradually a clear picture emerged. X-rays were part of the electromagnetic family, with very short wavelength and high penetration. Radiation emitted by radioactive materials consisted of 3 kinds: of these the gamma-rays were similar to X-rays, but usually with even shorter wavelength and greater penetrating ability. The other two were fast particles, atomic bullets--fast electrons (negative charge), and "alpha particles", fast positive nuclei of helium.

    Those particles really should not be called "radiation," but it is too late to educate the public, especially since many detector instruments (such as Geiger counters) responded both the particles and to X- or gamma rays. When James Van Allen's instrument on the "Explorer" satellites I and III discovered high energy particles magnetically trapped in near-Earth space, they were called the "radiation belt" and the name stuck. The best we can do is call the kind which comes from radioactivity "nuclear radiation."

    But you will never find it in microwave ovens.

Question: How high does the atmosphere go?

Reply

As high as the sky ... no, just kidding. We live at the bottom of an ocean of air, but while the watery ocean has a well defined top surface, the atmosphere just dwindles away.

    As most of us realize, at sea level the atmosphere exerts a certain pressure--about 1 kilogram per square cm, or about 15 pounds per square inch. It comes from the weight of the air piled up on top of us. Atop a mountain 5 kilometers high, half the atmosphere is below you, so the weight on top is only half as much, and the pressure and density are also only half of what they are as at sea level. (The value given as 5 km is approximate and also depends on temperature.) Another 5 kilometers, half as much again--that is 1/4 the pressure and density of sea level. And so forth.

    At this rate, 20 such halvings bring us to 100 kilometers and a millionth of the density, (actually it's about twice that). Air still contains 12 million million molecules in each cubic centimeter, but up to this height, its composition has remained about the same--78% nitrogen, 21% oxygen, nearly 1% of argon, some 0.1% carbon dioxide and other gases. From here on, however, collisions between molecules are less frequent, and different gases tend to settle at different rates, the ones with heavier molecules at the bottom, with a smaller "halving distance." Both oxygen and nitrogen form molecules of two atoms, but oxygen molecules tend to get split up into separate atoms, and these (being lighter than the molecules) extend further up. Nitrogen dominates to 200 kilometers, but then up to 600 kilometers it's atomic oxygen, and after that--would you believe helium? Down at the ground its concentration is insignificant, but being lightest, it outlasts the others. Still higher, the main component is hydrogen, even lighter than helium.

    By then, the density is so low that atoms and molecules rarely collide, they just rise above the atmosphere like tossed stones, then fall back. But even at 600 kilometers some atmosphere remains--ultra-violet photographs from the Moon have seen a hydrogen "geocorona" glow extending to 3-4 Earth radii, gradually fading away. And satellite orbits at 400 kilometers are still degraded by air resistance. Of course, all this concerns the "neutral" atmosphere, held by gravity. Free electrons and ions (atoms missing electrons) are held by the Earth's magnetic field to even greater distances, though their density (at least near Earth) is smaller.

    So, how far does the atmosphere go? Depends on how you measure it!

   

    I was looking for information on the De Laval nozzle as they are applied in rocket engines and came across your wonderful web site. It's the only place I was able to find a picture of his turbine!

    You write "Luckily for Goddard, this problem had been solved by Gustav De Laval ..." Does that imply that Goddard got the idea for that shape for his nozzles from reading about De Laval's work?

    Do you know if the Germans also used De Laval's work? Or did they come up with the V2 nozzle shape by trial and error?

    I would greatly appreciate it if you could recommend a good source on the technical aspects of Goddard's engines? The books I have read on the German effort seem to dwell on the organization and give almost no technical details about the A4 engine.

    I am trying to trace the technological advances leading to modern liquid rocket engines.

Reply

    I am not sure I can help you! My source is the first volume (of 3) of "The Papers of Robert H Goddard", edited by Esther C. Goddard and G. Edward Pendray, published in 1970. It is not a commonly available book and even in 1970 it cost $150--I got my copy by reviewing it for a journal. Being in Los Angeles, you might try the library of the University of Southern California, and could ask Prof. Michael Gruntman there for advice. He has a very useful web site on space technology, by the way,

        http://ame-www.usc.edu/bio/mikeg/astromike/index.html

    You may also look up Milton Lehman 's biography of Goddard, "This High Man", but I do not recall how much it discusses the nozzle. If you are interested in the design of later rocket, the "Papers" collection is your best bet.

    In Germany, the De-Laval nozzle was used by Oberth in 1929 in his liquid-fuel rocket, as "Kegeldüse" (mind the umlaut on the "u"), meaning "cone-nozzle." I am not sure when the nozzle was first tested--I think it was in the early 1920s and I am fairly sure it was inspired by Goddard's work. You can find many references if you type that word into Google, but as far as I can see, all are in German.

    Good luck

Further Question

Dear David,

    Thanks for your helpful response. I will give those references a try. It's a shame no one has written a general book on Goddard's rockets comparable to the many works that have been written detailing the technology used in the Wright Flyer. Maybe for the 100th anniversary in 2026.

    Do you remember where did you find the drawing of De Laval's turbine?

    It was interesting to hear your opinion that some of Oberth's early work was inspired by Goddard. I have been trying to figure out how much of Goddard's work was picked up by the Germans. It's really hard of sort out. Did Goddard publish is work on rocket engines just prior to Oberth's experiments?

    Imagine being in Annapolis in 1945 when a captured V-2 arrived for Goddard to inspect. I suppose he was too sick at that time to record his response to it.

    All the best,

Reply

Yes, while Goddard's name is often invoked now by organizations and speakers, his actual work and personality get rather little attention. Some years back when I was still employed by NASA I proposed marking on 19 October 1999 the 100th anniversary of Goddard's inspiration, as a teenager perched in a cherry tree (see my web site). It was handed to a low-level education HQ official, who passed it to a local official at Goddard, who dropped the whole thing. It still seems an inspiring anniversary for kids in school to observe, rather than Dr. Goddard's birthday which has NASA's blessing.

    The drawing of De Laval's turbine I have seen in many places and believe it is from an old book. You can find it on the web at

        http://www.history.rochester.edu/steam/parsons/part1.html

an essay on turbines (about 100 years old, my guess) by Charles Parson, who designed the more practical turbine now used aboard ships and in power stations, and who also demonstrated it with his speedboat "Turbinia" (that remarkable story is somewhere on the web, too). As he notes, the drawback of DeLaval's turbine was that it ran too fast. One may design a nozzle efficiently converting the energy of heat to that of a supersonic jet, but it is hard to design a turbine wheel spinning fast enough to extract that energy, and many energy-robbing gears are needed to bring the rotation down to the useful range.

    By the way have you read "October Sky" (anagram of its original title "Rocket Boys") by Homer Hickam? It tells (among other things) how the discovery of the DeLaval nozzle made a huge difference to the rockets built by a bunch of high-school kids in West Virginia (a film version also exists).

    A highly recommended book of a different kind is the Cambridge Encyclopaedia of Space (1990), edited by Michael Rycroft, a large illustrated sourcebook covering almost anything related to spaceflight (up to 1986 or so). Nothing comparable exists, except maybe the French version which preceded it. The life and work of Oberth are on p. 28, which also contains a picture of the Kegeldüse. In 1919 Goddard published an article on rockets in "The Physical Review" (August) and "A Method of Reaching Extreme Altitudes" as a Smithsonian document; Oberth in his doctoral thesis of 1922 must have been aware of that work.

    You wrote "Imagine being in Annapolis in 1945 when a captured V-2 arrived for Goddard to inspect. I suppose he was too sick at that time to record his response to it." This actually happened (or were you aware of it when you wrote the above sentence?). In Goddard's biography "This High Man" this is described on p. 357. The end of that story quotes Henry Sachs (an assistant to Goddard) saying "It looks like one of ours" to which Goddard replied "Yes, it seems so."

    I was glad to help your interest, and only wish more people shared it.

Why don't space rockets use wings like airplanes, but overcome gravity by rocket thrust alone, as they rise vertically? Wouldn't wings provide extra lifting power?

Reply

"Wouldn't wings provide extra lifting power?" Yes they would, but mainly at low speeds. Once the speed of sound is passed, lifting power ("lift") drops and air resistance ("drag") rises steeply, and it soon outweighs any advantage a wing provides. To reach low Earth orbit takes a velocity 24 times that of sound, so overall, a wing raises more problems than it solves.

    One way of taking advantage of wings is to use an airplane to raise the rocket above the densest atmosphere before it is fired, avoiding wasteful air resistance. The "Pegasus" rocket is launched that way, from a specially adapted L1011 airliner (a B-52 bomber was also used). The first stage of that rocket also uses short triangular wings.

    Ordinary rockets rise vertically, gradually tilting over into the direction of the orbit: the vertical ascent gets them out of the dense atmosphere in the shortest time, reducing losses due to air resistance. One possible way for a future space launcher to get more boost per gallon of fuel is to "breathe" atmospheric oxygen, in place of the liquid oxygen carried aboard nowadays. "Hypersonic ramjets" are still in the experimental stage, but engineers hope to create a design that saves fuel at up to 6-7 times the speed of sound, a reasonable speed for a first-stage spaceship. Like the Pegasus, it might be released by an airplane, continue to fly in the high atmosphere at the appropriate air density, and it may use short wings, or more likely, replace them with a flat "lifting body" providing enough lift at such high speeds.

    A forerunner of such a system was recently unveiled by Burt Rutan, the engineer who also designed the "Voyager" airplane which circled the globe nonstop and now hangs in the Smithsonian in Washington. Rutan seems incapable of designing an airplane which is conventional, nor one which isn't strikingly beautiful, and his twin-jet "White Knight" is no exception--see

        http://www.airventure.org/2003/news/rutan_forum.html
        http://space.tbo.com/space/2003/0419a.html

    Its elevated cabin straddles "SpaceShipOne", a 3-passenger spaceship with stubby wings of which parts hinge upwards to act as air brakes on reentry. It is meant to be released at 50,000 feet and to fly by rocket power to 100 km, then glide back to Earth. A ride might ultimately cost as little a $100,000 per passenger, rewarding riders with a view from near space, 3.5 minutes of weightlessness, and reentry stresses peaking briefly at 5 gravities. For Rutan it may earn a $10,000,000 prize offered for the first craft capable of such performance.

Hello, I've been trying to figure out the distance to the horizon on Mars.

    I imagine things would fall below it at closer distances, though it seems to me that some science fiction stories place too great an emphasis on this. Just how might I calculate distance to the horizon on Mars?
   

Reply

The horizon on Mars is closer than it is on Earth, but not by much, since (other things being equal) its distance is proportional to the square root of the planetary radius. On a planet with half the radius, the distance to the horizon drops to 70.7%.

    Rather than start the calculation from scratch, I copy below question 16 of http://www.phy6.org/stargaze/Sproblm2.htm together with its answer, and let you figure out the rest.

---

Q. An astronaut stands on the Moon, at altitude h. Assuming the Moon is spherical with radius R=3476 km, what is the distance D to the horizon? How far is the horizon for an astronaut on flat ground, with eyes 1.5 meters above the surface?

    A. The equation derived for Earth in section (8a)


D² = 2Rh
may still be used, provided R is the radius of the Moon, not of Earth. Any units of distance may be used, provide the same units are used on both sides of the equation. If we work in kilometers and denote square roots by the (1/2) power

D² = 6952 h
D = 83.38 h^1/2

    Suppose h = 1.5 meters = 0.0015 km. Then h^1/2= 0.03873 and D = 3.229 kilometers.

I hope I am not bothering you in any way, but there is a question that has been bothering me for awhile now. I have asked my teacher but he does not know the answer. I have been researching it online and in some text books for about a month now and have come up with nothing. I do not want this to seem like my math teacher knows nothing, he is a very intelligent man.

    I have taken Calculus 1 and 2 in high school and am leaving for college to study engineering in the fall. The question I have which I would appreciate any little bit of information or direction on is: Where does the 4 come from in the parabola equation:


(x – h) ² = 4 p (y–k)   ?

    This has been bothering me simply because I wish to know the why and how of things. I am sorry if I am taking up you time and I appreciate you for reading my question. Thank you.

Reply

You realize of course that the textbook could easily get rid of the factor 4, by simply introducing a new constant q = 4p, which leads to the same formula but without the puzzling factor. The fact that the "4" is included suggests that the constant "p" has some special significance, which would be lost if the equation were written otherwise.

    This indeed is the case.

    When you define a parabola by its equation


(x – h)² = 4 p (y–k)

you are using coordinate geometry, invented by Descartes in the 1600s. Today we prefer such a definition, because it allows geometrical properties to be handled by means of numbers, and our age is very, very good in handling numbers.

    The ancient Greeks, the first to study parabolas, were not so good with numbers (maybe because they lacked the decimal system), but they were very good at pure geometry. They defined a parabola as the collection of points, each of which had the same distance -- call it "q"--from some point F, its focus, as it had from a straight line, which they called the "directrix" of the parabola.

    Let us try to translate this condition into the language of coordinate geometry. Get some paper, draw on it (x,y) axes, and add a freehand sketch of the parabola y = x² , passing through the origin and rising above it symmetrically, on both sides of the y-axis. You need not number the axes. After that please follow one by one the steps outlined below, on paper. Do not try to skip anything, and proceed past any point only after everything preceding is clear. Then when you are done, explain it all to your math teacher.

    On the graph you drew above, mark a point D= (0, –q) a distance q below the origin. Place D about as far from the origin as 1/4 of the height of the y axis you have drawn rising above it. Draw through D a straight line parallel to the x-axis (its equation is y=–q). That will be the directrix.

    By symmetry, you expect the focus F to be on the y-axis (only then does the Greek condition produce a curve symmetric with respect to that axis). The origin is one of the points of the parabola, at a distance q from the directrix, so by the Greek condition, the focus F is a distance q on the other side, at the point (0, q).

    Next, mark some point P = (x,y) on your parabola; to best illustrate the argument, choose it with y of about 2q or 3q. Also draw a perpendicular line from P to the directrix, a line parallel to the y-axis. Say it meets the directrix at point Q=(x, –q).

    You have drawn PQ, now add one last line PF. By the Greek definition, they are equal in length


PF = PQ Also
PQ = y + q Take squares
(PF)² = (y+q)²

    Expressing (PF) ² by the theorem of Pythagoras


(y–q) ² + x² = (y+q) ²

    Multiply out the squares! You find you can now subtract from both sides y² and also q², leaving just


–2yq + x² = 2yq giving
x² = 4qy

    But suppose we are working, not in (x,y) coordinates where the lowest point of the parabola happens to strike the origin, but in (X,Y) coordinates, at which the lowest point is at some arbitrary point (X,Y) = (h,k) . The connection between the two systems is a simple shift in x and y :


x = (X–h)         y = (Y–k)

    The equation of the parabola now becomes


(X–h)² = 4q(Y–k)

    That is of course your formula. You can see that when the factor 4 is included, then q (or p in your version) has a geometrical meaning, it is the equal distance which all points on the parabola maintain, from a point and from a straight line.

I was wondering, if our little planet had started orbiting in an ecosphere of an O class or B class star, instead of our G class, would the light emanating from the star affect the color of things (plants, human skin, etc.) on this planet? Or would it just produce a colored hue on everything? If we looked up into the sky would we see a blue or indigo sun?

Reply

    Hi, Chris

    No one could answer your question for sure--one can only speculate. Presumably, our planet would retain its water for it's hard to imagine life without it, and presumably, the distance from the star would be just right for liquid water, because it is hard to imagine life on a planet where oceans are eternally frozen or have boiled away.

    What then? An O star or B star is much hotter than the Sun, and would perhaps create so much UV that life would have to stay in the ocean forever. The star would not be blue, but dazzling white, and the sky would be much brighter and perhaps more violet, because blue and violet light are the kinds which scatter most easily in the atmosphere. I don't know what solar wind one could expect--whether it would be strong enough to affect the atmosphere in a profound way.

    The color of plants reflects the chemistry of chlorophyll, which helps plant absorb and convert the energy of sunlight. If more blue light were available, an alternate chemistry might have developed, based on higher energy levels of atoms and molecules, but I do not know whether that is feasible. It is an interesting idea.