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396B Posssibility of Asteroid Hitting Earth (2)
256. Konstantin TsiolkovskyI h a v e o n e q u e s t i o n f o r y o u . W h y e v e r y t i m e w h e n I r e a d i n g a r t i c l e s a b o u t s p a c e e x p l o r a t i o n "A m e r i c a n a u t h o r s" p r o b a b l y c a n m e n t i o n S e r g e i K o r o l o v' s n a m e , b u t n e v e r n a m e o f K o n s t a n t i n T s i o l k o v s k y ?
P r o b a b l y y o u c a n e x p l a i n t o m e w h y w e g i v e o u r c h i l d r e n' s o n l y i n f o r m a t i o n a b o u t A m e r i c a n s c i e n t i s t s ?
ReplyI don't know what American authors led you to your conclusion, but I don't think my writing is like that. For a comprehensive account which seems to give fair credit where deserved, get "Blazing the Trail--The Early History of Spacecraft and Rocketry" by Mike Gruntman, published by the American Institute of Aeronautics and Astronautics. The author, by the way, grew up in Russia (part of his childhood, in Baikonur), worked at IKI and is now professor at the U. of Southern California. I also highly recommend the Cambridge Encyclopaedia of Space, which however is out of print.
I have certainly never tried to bias my account towards American contributions. I tell the story as I see it, concisely but fairly: Congreve, Goddard, Oberth, Von Braun, Karmán, Korolyov etc., not to slight De Laval on whose work modern rocketry is based--their roles are clearly stated, and most were not American.
However, I am not sure what the legacy of Tsiolkovsky is. He was not a scientist, nor a discoverer or engineering innovator. Indeed, it is difficult to evaluate him fully, because Soviet propaganda deliberately tried to make his work seem prominent. He was a visionary, sure, but if he did not exist, the history of rockets may have evolved more or less the same way.
If you feel my account misses anything substantial, please tell.
257. The Color IndigoI teach high school physics and chemistry, but am also conducting a critical review and error-checking of the proofs of a new 9th grade science text for the publisher.
How many named colors are there? The issue is the "color" indigo. British and American schoolchildren have learned a number of mnemonics with seven words or letters to remind them of the seven named colors, including indigo. American kids seem to learn ROY G. BIV. I have read that Isaac Newton first described seven colors, perhaps with a bit of a mystical or numerological preference for the number seven. All the high school and college-level physics books that I have in my library list seven colors. All of the Internet sites that I have checked (except for one, written by a peculiarly outspoken chap) list seven colors.
However, I recall reading some years ago that scientists had decided that indigo was not a separate color. I do not recall when this was, or the name of the publication in which I read it. Here is the question: Do you know whether any organization of scientists has indeed decided as a convention that there are only six named colors and that indigo is not one of them? If so, what was the organization and when was this decided? Since your website discusses seven colors, I would appreciate your point of view on this subject.
I have no particular bias here, but I would like to recommend a balanced approach to the publisher, who wishes to state that the "I" in ROY G. BIV is there simply so one can pronounce the name
ReplyThe numbering and naming of colors is a social and cultural convention, and I can think of no reputable scientists who would decree whether indigo is or is not to be counted.
And I also hope 9th grade students will learn more than the names of the main colors and their order. For instance, tell them that indigo is a dye derived from the indigo plant (today also made chemically) which was used (and still is) to color "blue jeans." After that piece of trivia, they will be sure to remember "indigo" even if they forget "blue."
But more important--I hope that 9th grade students will get to know a little better what color is, and that there exist two definitions of color. "Physical" color is a position in the spread produced by a prism (or by a finely ruled grating), the so-called "spectrum." It is a continuous spread, with no limit to the number of divisions into which you can break it up--as long as, of course, you have an instrument sensitive enough to separate neighboring divisions. The place in the spread is connected to the wave nature of light, associated with wavelength, and a lot more can be said about those waves. The analogy in sound would be pure tones (the ones of a flute come close).
On the other hand, "perceived" color is what the eye tells you, a sensation which our brain receives. The human eye has 3 kinds of cells, so any color you see is the sum of 3 responses only, by 3 types of cells, each of which responds in a certain way to a different spread of wavelengths. It is as if your ear only responded to chords of 3 specific tones (or tone combinations). That is why a color TV or color computer monitor can reproduce a wide array of colors, using only 3 types of "color pixels."
By giving each of the 3 pixel types an appropriate brightness, the TV tube can (for instance) make the eye feel it is watching the pure green from a specific part of the rainbow spread. But the tube is just fooling the eye! A spectrometer, an instrument accurately resolving contributions from various parts of the rainbow spectrum, will clearly show the two colors are not the same. And of course, if all the three types of color detector in the eye are stimulated equally, we see white, while if no light is detected by any of the three we see black.
By the way, the July 2006 issue of "The Scientific American" has an article, telling how the eyes of birds have 4 types of cells, so the colors they see are 4-fold combinations (like 4-tone chords in music), while most mammals only have two. Presumably deer can't distinguish the bright red jackets of hunters. As for indigo--perhaps those scientists who would rather abandon "indigo" as a label for part of the rainbow are simply saying that the color of indigo dye is too complex to use as a label for a single pure color. That's their personal judgment, no more.
258. Electromagnetic Waves and Electromagnetic InductionYour page on electromagnetic waves was insightful, pointing out the connection with the discoveries of Hertz, Einstein etc.
However I do not understand how an EM wave exactly moves across empty space--as shown by the moving diagram on your web page.
When you use the term "oscillation," do you imply that the magnetism of a source induces an electric current, which induces a magnetic field and so on ? If my rambling sentence is correct, then what exactly does the magnetism induce into, to produce that current? This is particularly unclear if the EM wave spreads in empty space, where there is no material to induce currents into! I would greatly appreciate it if you could help clarify to me, how exactly an EM wave would move through empty space.
ReplyI guess the web site did not say enough about electromagnetic induction, an omission which needs to be remedied.
Electric currents do create magnetic forces, which you can observe (say) using a compass needle. This was found by Oersted and Ampére, and later Faraday (perhaps) and Maxwell (certainly) called the region of magnetic forces a "magnetic field."
Many people also tried the opposite: if an electric current flowing in a coil around a bar of iron turns it into an electromagnet (a similar effect, but less strong, also exists without the iron)--perhaps if we put a bar magnet inside a coil of wire, it will create an "induced" electric current in that wire.
It did not work that way, and today, knowing about energy, we can guess why: an electric current in a wire needs a constant supply of energy, and where would that come from?
However, Faraday found a somewhat similar result, if the magnetic field in the coil was undergoing change, as happens (for instance) as the bar magnet is being pushed into the coil, or being pulled out. During those changes, a momentary current did flow. Faraday called this electromagnetic induction, creating an "induced current." See for instance
That led to Faraday's law of induction, by which a changing magnetic field can generate a changing electric field during its change (there exist formulas, too). If a closed conducting wire occupies the space where the change occurs, the changing electric field may cause a changing electric current to flow. That of course is how we generate AC electricity, and also how transformers work--a high voltage current flows in a coil around an iron core, causing a changing magnetic field, and a "secondary" coil wrapped around the same core senses the changing magnetism and an induced electric current flows in it, perhaps with a different voltage.
In that case we have the chain
Changing magnetic field -- > changing electric current -- >
changing magnetic field caused by that current -- > and so on
That is also described in
Up to here, the cause-effect chain involves magnetic fields and electric currents. However, in empty space (as you noted) there exist no electric currents--only "electric fields," regions where a current might flow if (say) a closed copper wire passes the location, but otherwise nothing remarkable seems to happen there.
James Clerk Maxwell however guessed, that maybe this became a modified type of space--an "electric field." Under certain conditions, a varying electric field can also carry an electric current, through empty space. For instance, an electric current can flow through empty space between the plates of a capacitor. That is a device for storing electric charge, and in the simplest case, consists of two parallel metal plates, with empty space between them. Ignore for now the details and theory, let it just be said that if the voltage of one plate (that is, the level of its electric field) rises or falls rapidly, the voltage of the other plate is "dragged" up or down to match the changing values. It can be shown that a variable electric current then flows between the plates, through the empty space. The faster the change, the more efficient is the generation of such a current.
Maxwell guessed that this unusual electric current, flowing through "empty space," can also generate magnetic field, just like the flow of a current in a wire. He called it a "displacement current" and included it in his formulation of the fundamental equations of electricity.
He then modified the above chain of cause-and-effect, to
Changing magnetic field -- > changing electric field -- >
changing magnetic field -- > changing electric field -- > and so on
With the displacement current replacing the ordinary electrical current, theory predicted that electromagnetic waves could exist in empty space. Those waves spread in 3 dimensions like sound--except that where sound contains pressure variations along the direction of propagation, the electromagnetic wave contained variable electric and magnetic fields perpendicular to the direction of propagation, a bit like the jiggling of Jello.
Unlike sound, it was sensitive to electrical properties--for instance, a good conductor of electricity tended to reflect the wave rather than let it pass (hence mirrors). The wave spread with the velocity of light (which turned out to be related to measured properties of electricity and magnetism) and Maxwell guessed it was light. From that he deduced some laws of reflection, and also explained the polarization of light. Then came Hertz, and radio, and the rest as they say is history.
I hope this short description makes sense
259. Why do orbits curve?I am trying to figure out why the orbits of the Space Station and the Shuttle Orbiter are curved - from South to North then North to South in nature. There are satellites that circumnavigate the earth in parallel orbits - why not these two?
I realize that the earth tilts on its axis, which creates a wobble, but the orbits of these 2 man made space objects against a map of the earth should be a straight line from West to East. Since it is not, I just have to know ... why?
ReplyI think your problem is concerned not with the shape of the orbit but with the way it appears on a flat map.
Suppose the Earth was a perfect sphere, and a spacecraft kept a fixed altitude of, say, 300 miles. Then its orbit would always be a circle around the center of the Earth, and the plane of that orbit (like any flat plane that includes the center of the Earth) would cut the sphere of the Earth into two equal halves.
Two such planes, representing the orbital planes of two different satellites, will therefore always intersect, the way two planes through the center of the Earth must do. They cannot be parallel!
They may however appear parallel on a flat map. Many satellites which observe the Earth--e.g. observe clouds, weather, magnetic fields, civilian or military objectives-- follow orbits that go over both poles of the Earth, and sooner or later pass above any global location. In a Mercator map of the Earth, their trajectories aseem to be straight parallel lines in the north-south direction. They are often launched from NASA's Vandenberg center in California, not far from Santa Barbara, starting southwards over the ocean. If anything goes wrong, the debris falls into the water and does not hit anyone.
The shuttle however is usually launched eastwards from Cape Canaveral, taking advantage of the Earth's rotation
(see http://www.phy6.org/stargaze/StarFAQ13.htm#q204 and links given there).
The latitude of Cape Canaveral is 28.5° N and that is the farthest north the orbit reaches. It gradually bends towards the equator, crosses it, bends the other way until it reaches latitude 28.5 south, its furthest reach south, then towards the equator again. On a map using the Mercator projection, you get a wavy line.
The orbit of the space station also follows such a wavy line, but I think this one reaches to something like 52 degrees, because the station is resupplied by Russian rockets, and Russia does not have launching sites as close to the equator as Cape Canaveral. Those orbits can be reached from Cape Canaveral, too, by moderately slanting the orbit northwards, and when the shuttle resupplies the space station, that is how it goes. The launch trajectory is still above water, but the advantage of the Earth's rotation is only partially realized (though the difference is not great)..
The only east-west straight orbit on the map is one which stays over the equator. It is somewhat hard to achieve it from off-equator launch sites. NASA once created a special equatorial launch station near Mombassa, Kenya, to launch the small Italian "San Marco" scientific satellite into such an orbit.
And it's highly unlikely that any manned mission would choose a polar orbit, because if the Sun loosens a blast of high-energy ions (very rare, but can happen), and it arrives while the mission is near a magnetic pole, astronauts could be exposed to those ions and could suffer a high radiation dose.
260. The "Sundial Bridge" in Redding
I would not include the so-called "Sundial Bridge". The following is the description of the bridge on the NASS (North American Sundial Society) web site.
Location: At Turtle Bay Exploration Park crossing the Sacramento River
Remarks: The 217 foot high suspension span called Sundial Bridge wants to be a sundial, and has come very close. The suspension pylon is aligned true north, but unfortunately performs as an inaccurate gnomon with an inclination of 49 deg (for bridge functionality) rather than for the 40.6 deg latitude of the site. The bridge is 700 feet long and weighs 1,600 tons.
Why would anyone promote a "wanna-be" sundial?
ReplyWhat you wrote about the "Sundial Bridge" is interesting and was not known to me. However, I wonder if the pylon may still act as a good sundial gnomon, even at its present angle. It depends on where the center of the dial circle is, the circle on which hours are marked.
Imagine a tight string at the correct angle of 40.6 degrees, from the top of the pylon to a point P somewhere on the ground. The shadow of that string would be a "good" sundial gnomon. If a circular band divided into hours and fractions is drawn with its center at P (rather than at the base of the pylon), the tip of the shadow will give the correct time. Of course, the band must be wide enough to accommodate the seasonal variation of shadow length, but as long as you observe only the tip, you have a working sundial.
At least, that is what it seems to me. Am I right?
I have not been to Redding and do not know how the people there designed the dial circle. But I would suspect they would not miss the above point.
ResponseSunny Day David,
What you say is true if the top of the pylon is used as a nodus. But then this technique can be used for any vertical object and may require a very large area if the nodus is up very high. The point is that they are promoting the pylon as being a gnomon. This is very misleading to the many people who are not familiar with sundials. There are so many sundials in the world that are worthy of promotion. Why pick something that is not?
I have visited a number of sites for the "Sundial Bridge" and have not been able to find a description of how it has been made to work. Your method is the only way. But from the written descriptions this does not appear to be what has been done.
261(a). Temperature in spaceWhat is the ambient outside air temperature at the Space Stations mean altitude?
(see also next question, #261(b))
ReplyA strange question-- and I am not sure the answer will help you with whatever you are really after!
The air temperature at an altitude of 250 miles or so may be 700 deg absolute--or more or less, depending on the solar cycle. At sunspot maximum, the Sun emits more soft X-rays, which heat up the upper atmosphere, and which also cause it to expand, increasing the air resistant to low-orbit spacecraft. The space station "Skylab" came down a bit early (1977?) because of increased sunspot activity, and a similar expansion brought down in 1967 the remnants of the artificial radiation belt produced by the "Starfish" nuclear blast in space, in 1962.
But that air is very, very rarefied--see
Where you are sitting, the atmosphere is dense, its molecules constantly collide and share energy, and the temperature is a reasonable indicator of the average energy. Up at 250 miles there hardly remain any collisions (between neutral molecules): they mostly rise up like tiny ballistic missiles and then fall back. By the way, they move appreciably more slowly than the shuttle in its orbit, so most of the velocity of any collision with the shuttle is from the shuttle itself. And the atmosphere there contains free ions, too--free negative electrons, and positive atoms or molecules which have lost an electron (also negative ones which have temporarily attached an extra one). The ions are much more rarefied, but they dominate electric and magnetic phenomena.
Thus, if you want to describe the behavior of the atmosphere at that level, you need much more than just the number giving the temperature.
261(b). Exposure to space environment: Freeze or burn?(see also preceding question, #261(a))
I have always wanted to know what the temperature in space was. What I ask is: if an astronaut were to step out of the space station wearing nothing but a t-shirt and jeans, what would happen to him? Would he freeze or burn, or would it be room temperature? I know there is no oxygen but I don't understand how there can be nothing. I hope that you can explain this to me.
ReplyThe temperature of outer space varies, but the value does not matter much because the amount of matter sustaining it (and therefore, of energy involved) is tiny. See preceding question here, or http://www.phy6.org/stargaze/StarFAQ5.htm#q98
The moment a massive body (like your own) touches the rarefied gas of outer space, it and not the gas will determine the heat flow around it.
"If a astronaut were to step out of the space station wearing nothing but a t-shirt and jeans," I suspect he would first choke (all the air will escape his/her lungs) and then after a while he/she would end up freeze-dried, as all water would also evaporate from the body. Museums sometimes freeze-dry the carcasses of small mammals by suspending them in a vacuum chamber.
Needless to say, I don't encourage the experiment.
262. Remotes sensing of ThunderstormsHow can you tell if an electrical storm is approaching or receding?
ReplyThe simplest way of course is to stand outside and see which way the clouds are drifting, then decide if the sound of thunder comes from upwind or downwind. On level ground thunder can usually be heard from strikes up to 5 miles away. Thunder is loud enough to be heard further than that, but its propagation tends to curve away from the ground.
A map of radar reflectivity is often available on the web and can show storms in your vicinity. If you have a fast web connection, you can even download a loop of reflectivity maps from the past few hours, and let it cycle through, telling you which way the weather pattern is moving. I use
Thunderstorms often form a "squall line", an approximately continuous row of storms. If radar shows that row heading towards you, you may perhaps expect a storm--if it moves away, probably not.
Other lightning-storm detectors use radio. You can of course hear the crackling of lighning signals on an AM radio channel, but you need a more complex instrument to tell how the source moves. A bolt of lightning hit Apollo 12 as it was starting to rise from its launch pad, and soon afterwards Cape Canaveral installed a radio network to track lightning storms in its part of Florida, to allow launches to be postponed if a storm is near. See
Apollo 12 escaped damage, landed on the Moon and its crew returned safely.
263. Why do Electrons stay away from the Atomic Nucleus?I am getting interested in electromagnetism at the nuclear level. My question is:
If a nucleus has a proton (positive charge) and an electron (negative charge), why is there such a large open space between the two? Why do they not hug like opposite poles of two magnets? Is it related to one of the nuclear interactions or gravity?
ReplyYour question "why don't the electrons fall into the nucleus" and neutralize its electric charge--that was one of the clues to quantum theory.
It was not asked quite in the way you did, because before 1911, physicists did not know an atomic nucleus existed. They were aware that atoms as a whole were neutral, that lightweight negative electrons could be extracted from them, and that there also existed a positive electric charge, associated with the bulk of the atom's mass. But how positive and negative charges existed together, no one knew.
One idea (popular around, say, 1900) was that the positive charge was spread out over the bulk of the atom, and the electrons were embedded in it "like raisins in a pudding." When they vibrated around their equilibrium positions, in well-defined frequencies (the way a tuning fork does) they emitted light. It was already known that a vibrating electron acted like a small broadcasting antenna, and its vibration seemed to explain why atoms emitted well-defined colors of light (the "atomic spectrum">. But why such an arrangement was stable and preferred, no one knew.
We now know that the laws of Newtonian mechanics are modified for phenomena at very small distances, into the laws of "quantum mechanics." By those laws the electron is actually represented not by a little electrically charged ball (as was earlier proposed) but by a spread-out wave. The size of the wave depended on the associated energy, which can be estimated from the frequency of light emitted by atoms. The fixed frequencies of emitted light had a rather complex origin, and in the end they provided a lot of information about atoms. But the thing you want to note is that the size of that wave turned out to be much bigger than the nucleus, and it could not be collapsed to nuclear size.
It is hard to explain more of this without a lot of detail (and math, if you want to do those calculations). The best I could suggest is going to my web collection on the Sun
and read there a brief introduction to quantum phenomena, free of math. You might start with sections S4 and S5 on the color and nature of light. Later you may perhaps also look into other sections (like the first 3 related to weather and climate). The section on nuclear energy also will tell you how electric forces inside the nucleus (between protons) are the source of energy driving nuclear fission.
264. The Stars on the Winter Solstice of 2012I am an artist and am planning a new painting. I have been trying to find out which constellation the sun rises in on December 21, 2012. Can you help me? Any other information on planetary positions or eclipses on that date will also be helpful, such as which constellations the planets will be in at that time?
ReplyMy field of expertise is the magnetic environment of Earth in space, a bit far from what you ask. Let me try to answer your question, but better check it with an astronomer friend.
The position of the Sun at Winter solstice in ANY year is in the constellation of Saggitarius, close to where it meets the constellation of Scorpio. See star map in
You want the location where the blue line passes its lowest point, crossing the vertical line marked 18h (right ascension of 18 hours).
For the planets, I found a web page dedicated to December 21, 2012, a day when "the calendar odometer turns around" on the Maya calendar:
The positions of the planets on 21 December 2012 are not remarkable. Here are the geocentric ecliptical longitude E' and the elongation E of the Sun and all planets on 21 December 2012, measured in degrees:
Perhaps more meaningful is a star map for that day, shown on that site, with positions of Sun, Moon and planets marked. As you can see, these objects are all close to a continuous curve--the blue line on the map of the preceding web site--and the elongation E is the angle between them and the Sun. Mercury is west of the Sun and therefore above the horizon, as are Venus and Saturn; the other planets are below the horizon and not visible.
On the star map of the previous page, 1 hour is 15 degrees. The hours there are measured along the celestial equator (straight across the page, while the blue line is curved) but if you neglect the difference, Mercury is around 17 hours, Venus around 16:26 and Saturn around 14:36 . To get more accurate numbers, consult your friendly neighborhood astronomer.
265. Mars SunsetGood day Dr David,
Would like to know currently, where does the sun sets in Mars? Thank you in advance for your answer.
ReplyIf you expect the answer to be either "east" or "west," it is west. Mars rotates in the same sense as the Earth does, namely, counterclockwise if viewed from far north of the planet (or clockwise when viewed from far south of it). So do the Sun and other big planets, and the orbits of all those planets go around the Sun in the same direction too, suggesting that they all originated from matter which rotated that way.
Of course, on Earth the Sun does not always set exactly in the west--depending on your location and on the season of the year, it can set further north or further south. The cause is the tilt of the Earth's axis--the axis is not exactly perpendicular to the plane of our orbit around the Sun, but about 23.5 degrees away from perpendicular. The tilt angle of Mars happens to be very close to that of Earth, 25.1 degrees, and the rotation period is just over 24 hours (also close to ours), so the dependence on latitude and season is similar. For instance, Mars too has season-long nights and (half a year later) long days at its poles. The difference is that a "year" on Mars is nearly twice as long--1.88 Earth years, because the planet is more distant from the Sun and also moves a little more slowly.
266. Gravity at the Center of EarthHi David,
As it just so happened, I was thinking about the laws of gravity.
Since mass exerts gravity, would this not mean that somewhere close to the absolute center of the earth (or any spheroid), there would be an area or low or zero gravity, and as you got closer to this point, gravity would decrease? I thought about this since at that point, you would be actually surrounded by approximately equal mass on all sides. Wouldn't this also mean that as you got closer to the center of the spheroid, the relative pressure may increase, but also at a certain point, diminish as the mass above you begins to cancel out the gravity pulling you into the center.
If this is the case, then perhaps, the center of our planet is solid because the gravity is low enough, and the pressure decreased enough that there is not enough friction to keep the iron molten?
I'm not sure if these are normal thoughts on this subject. I have not been able to find any material that would help confirm them.
ReplyAs Groucho Marx used to say, "close, but no cigar" (cigars were the prizes he gave out). There indeed exists no gravity at the center.
As Newton showed, a uniform hollow ball of constant thickness creates no gravity anywhere in its interior (anywhere--not just at the center). Therefore, if the Earth has uniform density and you go inside it to some distance r from the center, you can divide matter into two spheres: a hollow sphere of all the matter more distant from the center than r, and a smaller sphere of all the material closer than r. Only the latter exerts gravity, so as r gets smaller and smaller, the pull to the center gets smaller too.
But this does not take away the pressure felt from the weight of the layers above you. It just means that, mass for mass, matter at distance r contributes less to the weight felt at the center than matter near the surface. But it still contributes, so the pressure keeps increasing with depth, and is greatest at the center.
Because of that pressure, matter gets denser as you go deeper--also, the composition changes, the core seems to consists mostly of iron and you know iron is heavier than rock, even if not further compressed by an enormous weight heaped on top of it. Because of such density changes, the calculation of gravity inside the Earth is more complicated--as you go to distance r, the smaller ball inside it is also denser. Some student sent me a problem based on calculating such gravity, asking if a depth exist where gravity is strongest (possible).
The inner core of the Earth indeed seems solid, as we know from the propagation of earthquake waves, but is has nothing to do with friction. The melting point increases with pressure (just as the boiling point of water increases in a pressure cooker), while the temperature of the core does not rise much, because most of it is due to the decay of radioactive materials, most of which are (apparently) in the crust of the Earth.
Those materials form a tiny fraction of the Earth's material, but think about it! Suppose you stand on a square meter of surface (or square yard, if you wish), and imagine a cone-like volume extending from that area all the way to the center of the Earth. If heat generated inside the Earth flows out equally in all directions, the heat generated inside that cone (about 6400 km or 4000 miles long) has no escape except through the area on which you are standing! Thus even a very small rate of heat generation can raise the temperature quite appreciably--and even if most of it is just due to the topmost 100-200 kilometers, it's still a lot.
For more see
267. The Big BangI am not a scientist, but I have an interest in cosmology. I have a question that no one has been able to answer in a way that I can understand it.
Astronomers say that they can peer back in time toward the region of the Big Bang and see to within a billion years or so of the place where the universe started.
What I don't understand is this: Why hasn't all the light that was emitted at that time and that was headed in our same direction, and that is therefore available for us to see, long since zoomed past us?
After all, we, that is, the material of which we and our galaxy were made, started moving away from the spot of the big bang at the same time as the light that the astronomers are looking at but, presumably, the light was going faster. Why is there any still coming at us from that direction? Why hasn't it all zoomed past us?
Explanations involving objects being like raisins in a baking and expanding cake somehow don't seem to push the metaphor far enough for me to understand.
I'm sure that there is some basic misunderstanding at the root of my question, but I don't know what it is.
ReplyYour conclusion would be correct if the Big Bang was a sudden explosion in space, if at some moment, an enormous mass appeared in a very small region of space (ordinary, three dimensional space), and proceeded to expand with high speed.
Actually, the big bang is viewed as a sudden explosion of space. Space is not something which had always existed, stretching limitless in all directions, an expanse which by itself is just empty and devoid of physical properties. That was the view in Newton's time.
Today space, even without matter, is seen as having properties--for instance, electromagnetic fields. It's an arcane subject in which I myself am not an expert, but one of its not-so-simple properties is distance, more accurately the mathematical prescription by which distance is derived, known as the metric tensor.
Although distance seems infinite, it is not: its definition gives it an upper limit, which has been expanding since the Big Bang. An ordinary three-dimensional space does not have such a limit, but one which is curved into a 4th dimension can do so. I don't know if such a dimension actually exists (I am not counting time, an unusual kind of dimension, though it is involved in the definition), but space behaves as if it were curved into it.
The analogy usually given is between a flat two dimensional plane stretching in all directions, where distance has no limit, and a two dimensional surface of a sphere, where distance does have a limit. Imagine that generalized to three dimensions, if you can! Imagine also that the size of that sphere has been expanding ever since the big bang, when it was very, very small. Why the big bang should have happened no one knows: we only have the astronomical evidence for it.
Thus in any direction we look, we see traces of earlier times. A galaxy judged to be a billion light years away from us, emitted the light we see when the universe was a billion years younger, and so on, to earlier and earlier times. We can't see the big bang itself, because the early universe was very hot and opaque, and different in other ways. I vaguely recall that even the microwave radiation ascribed to the big bang actually started about 300,000 years later.
If you have points on an expanding sphere, every single one is on the front of the expansion. Similarly, so are we. Light which has gone past Earth may be captured by some alien telescope a billion light years away, but by then the universe will be a billion years older, and the aliens themselves will also be on its front of expansion (as will be Earth, still). It may sound weird, but the universe is not simple.
You may find more about this in earlier questions
268. "How often are Stars born?"My wife recently posed this question, and like many husbands at one time or another, I found myself completely stumped. After considerable time searching (and researching) the web, I am still answerless and hoping that perhaps you might be able to help.
'How often are stars born?"
My guess is constantly, but she is looking for a more definitive answer.
ReplyI guess stars are born all the time. If there are an estimated 100 billion stars in our galaxy, and the universe is a little over 10 billion years old, that averages to about 10 a year. And there may exist 100 billion galaxies (another number I vaguely recollect, though really no one is sure of it).
The more fundamental question is "How are stars born?" coupled with the inevitable "how do we know?" Stars apparently arise when a cloud of gas and dust is drawn together by its own gravity. Astronomers can see many such clouds in the galaxy around us, in various stages of contraction, illuminated by stars around them. As the gas is drawn together, it heats up (the stuff is falling towards its center, and its gravitational energy has to go somewhere), and if it gets hot and dense enough, nuclear energy is released. Hydrogen is then converted to helium, as it is in our own Sun, and this energy release heats the Sun or star, causing it to shine.
As the cloud is drawn into smaller space, any rotation which is may have gets amplified by the conservation of angular momentum, a basic principle. When finally all this mass condenses into a star and planets (or a binary star system, a different possible outcome), that star rotates, and so do the planets. The fact that the Sun spins in the same sense as the planets orbit it and as the planets themselves tend to spin, all suggest they started from the same cloud.
For this scenario to start, a sufficient density of gas and dust is needed. It will never happen in the space between galaxies, which contains (apparently) very little matter. Our own galaxy is a disk (we see it sideways as the Milky Way) but ordinary matter in that disk is not evenly distributed either: radio observations have shown that the galaxy has a spiral structure (like its neighbor in space, the great spiral galaxy in the constellation of Andromeda). Matter is denser in spiral arms, so that is where stars are born, much more than in between. The center of our galaxy is also a fertile are for star creation.
There is more to astronomy, much more, and I highly recommend to you the book "Seeing in the Dark" by Timothy Ferris, which is eminently readable.
There is still more. Stars contain atoms much heavier that hydrogen and helium, yet apparently these heavier elements (like oxygen or iron) were not part of the universe shortly after the "big bang." They were "cooked" later in the burning of stars (more accurately, the nuclear conversion of hydrogen). The fact that they are found in the dust from which stars formed suggests that "in an earlier lifetime" these atoms were part of big stars, which grew unstable (as stars past a certain size do) and exploded as supernovas.
Sorry it takes such a long answer to such a short question, but you and I live in a very interesting universe.
269. The Path of LightningI would like to thank you for your web site. I have been enjoying it immensely. A friend of mine and his horse were struck and killed by lightning in 1997 and I have been reading about lightning since. Your web site is one of the most interesting and informative I have found.
My question about lightning is: does it travel from the negative ground to the positive upper part of the clouds? This makes sense to me because of the Edison effect. Do you know of anyone who has confirmed or done research on which way it travels? Thank you for your help.
ReplyThe Edison effect occurs in near-vacuum, while lightning is in the atmosphere, a completely different effect. I am not an expert, though I have tried to describe what little I know in
I advise you to read a more detailed book, like "Lightning" by Martin Uman or other books by that author (available at Barnes and Noble, I see). What I write below is from memory.
The top of a thunderstorm cloud usually has a positive charge, and the bottom a negative one: they are separated by updrafts inside the cloud. The positive charge of the ground may perhaps come from the return flow of air outside the cloud.
The lightning discharge has two parts: one is the breaking down of the electric insulation of the air, by advancing a "stepped leader" stroke, usually going down to the ground from the negative cloud base, in steps. Near the ground it is met by a discharge for below, and sometimes (e.g. with high buildings or peaks) the process is initiated from below. See:
The leader heats the air to where it conducts electricity ("plasma"). Once an electrical path is created, the main discharge occurs. That is where big currents flow and damage can result. I do not know much about cloud-to-cloud lightning, but suspect it is similar--first a leader, then the main discharge. The books probably tell more.
270. Launching a Rocket from an Airplane.Hi!
In a quotation from "NASA facts," the B52 is described as an "air launch carrier aircraft" What does this mean?
ReplyI am retired and not responsible for anything published by NASA, but both statements happen to be true. Let me explain.
(1) Launching a spacecraft is among those things which are "possible, but just barely," like clearing a high bar in a pole vault. Any slight advantage deserves to be exploited, such as launching eastward from the equator, taking advantage from the rotation of the Earth, which is close to 5% of the needed velocity.
Launching from a flying jet airplane adds a few percent more, and more important, it avoids the resistance of the lowed atmosphere. That is, for instance, why Burt Rutan in his project for the X-prize launched "Spaceship One" from an airplane at 50,000'.
The problem is, space rockets for big missions tend to be heavy, and no airplane exists able to carry them fully fuelled. But a relatively small rocket has been developed, the "Pegasus", which a B-52 can just lift (though nowadays a Lockheed L1011 wide-body jetliner has been adapted for such launches). It is a solid-fuel rocket, whose first stage is equipped with small wings, and it has launched some moderate-size satellites into low Earth orbit.
It is an inexpensive option, but carries some risk. The rocket is dropped from the carrying aircraft at about 40,000' above the ocean, and at a safe distance, it is ignited. It usually fires OK, but I vaguely recall cases when it did not, or when its upper stage failed to ignite, especially in the early days. If you ever visit the US, you can see a "Pegasus" at the Udvar-Hazy Museum of the Smithsonian at Dulles airport, outside Washington, DC.
(2) The speed of the payload can be 3 times faster than that of the exhaust jet, but the mass of that payload is much, much smaller than that of the fuel consumed. It is that inequality that makes spaceflight at all possible, and at the same time, underscores the "but just barely" disclaimer.
This is discussed in some simple terms in the last part ("Rocket Motion") of
Again, any trick which can improve performance is valuable, such as ion rockets (section #33 of "Stargazers") and planetary gravity-assist maneuvers (section #35).
Best wishes to our friends down under
271. The Equation of ContinuityI came across your website on google and was very interested by the questions-and-answers page. I hope my question makes sense.
It is as follows:
If you turn on a water tap, and set it so that only enough water is running to keep a constant small flow of water at the top of the tap, why is it that if you hold your hand near the drain 10 inches (20cm) below, you will feel the water dripping rather than running in a stream? Closer observation shows that the water stream narrows the further it is down from the tap, and then about 6 inches down from the tap will start dripping rather than flowing continuously? How come?
ReplyExcellent question, excellent observation. The brief answer is that what you see is caused by the acceleration of the water, as it falls due to gravity. It is all tied to a fundamental property of fluid flow, known as the "equation of continuity."
Imagine you have a stream of fluid--say in a pipe of varying cross-section, and say it is a fluid which cannot be compressed. Water is very nearly such a fluid--air and gases are not (they may get compressed as they flow) and for them the equation gets a little more complicated. But we are talking now of water, which is simpler.
As you go down the pipe, its cross sectional area A (say in square centimeters) may vary, and the velocity V of the fluid (say, in centimeters per second) may vary too. The equation of continuity says that these two variations are connected: where the pipe in narrow (A small) the flow is fast, and where it is wide (big A) the flow is slow.
How much water passes each second some selected cross section A? You may imagine the water stacked behind A, in a column of cross section A, waiting to get across. All the water molecules within distance V centimeters will make it in that one second, so the total volume crossing each second is
But by the nature of the pipe, the flow V crossing any cross section is the same in every cross section. So we get, along the pipe
That is the equation of continuity in its simplest form: where the pipe narrows down (small A) the velocity v increases by the same proportion. You may already know this from experience: if you use a watering hose in the garden and narrow down the exit (with your thumb or a nozzle), the emerging jet moves faster. (No, you don't gain new energy, because narrowing the exit slows the flow in the rest of the hose and decreases energy loss by friction there.)
After these preliminaries, back to water running from your faucet. Like everything falling freely due to gravity, the emerging water also speeds up (accelerates). Which means, its velocity V increases, and by the equation of continuity
its cross section decreases. Hence you get a narrowing stream, You need no pipe to confine the stream in this case.
But the stream is not moving like a single object, either: the lower part moves faster than the upper part, and the difference tends to tear the stream apart. Up to a certain distance (6 inches in your case) the difference can be accommodated by a gradual narrowing down of the stream. But sooner or later this is not enough, and that is where the stream breaks up into droplets.
I tried to be as clear as I can: please tell me if I succeeded and you understood!
ResponseThank you very much. I entirely understood your explanation. You're awesome at explaining things like this!
I like how you first explained about the pipe and garden hose. I recall from a visit to a university engineering camp that water flows the fastest in the centre of a pipe and the slowest at the edges, but that anywhere in the pipe, even through various sizes of pipe, the water is flowing at the same rate of ml/second, though the velocity of the moving water may change depending on pipe size. (Next the professor showed us how dye moves in a glass pipe, etc.)
I can see how your equation applies to the faucet question because of the way that water "needs" less space when it is moving fast enough, just like it needs to move faster in a smaller pipe.
Thank you very much. I've been wondering about this for something of the order of 2 years now, and I could never find anyone to ask who would know.
272. The Four Corners MonumentDr. Stern,
I think you are just the man to help me if you have the time. I have recently done some research to find the latitude and longitude of the Four Corners Monument [where Utah, Cororado, New Mexico and Arizona touch] and have found several different measurements. Where would be the best place to get the most accurate information, and if asked to give the point of intersection down to the nearest hundreths decimal point exactly how would you write this? It is very important to my research that I get this exactly the way it was asked of me. The coordinates that could possibly be correct for The Four Corners is 36° 59' 56.31532" Lat and 109° 02' 42.62019". Thank you very much for your time.
ReplyI guess the most accurate source is the National Geodetic Survey, which gives the same values as the ones you cited, on
But what does it mean? What kind of research are you conducting, that demands such accuracy?
One second of arc is abut 100' along a line of longitude (less along most lines of latitude). So carrying the position to two decimals gives position within about one foot.
Where I am not sure is in the system of coordinates; the Earth is an ellipsoid, and I don't know how this is taken into account. NGS coordinates are "adjusted" to fit its own system, which makes its own choices. Bottom line: if you really need all that accuracy, you know much more geodesy than I do, and should be able to write a better explanation than I can.
273. Which among the brightest stars is closest to N Pole?(by a student of astronomy)
I know that the star closest to the celestial north pole is Polaris, the pole star. However, of the 20 brightest stars (Sirius, then Canopus, then Alpha Centauri, etc.), which is closest to the celestial north pole? Do you use both the declination and the R.A. to get this answer or do you only use the declination? Which star is this?
ReplyYou use only declination--see
Declination gives distance from the pole, right ascension gives direction out of the 360° around the pole. Of course, some stars (Alpha Centauri for instance) are closer to the southern pole of the sky--and their distance from that pole depends on (180°–declination).
Follow up:Thank you. Your information that I only use declination was a big help. It appears that of the brightest star, the closest to the celestial north pole (declination +46) is Capella. Also, the brightest star closest to the celestial south pole (declination –60) is Alpha Centauri.
Finally, the brightest star closest to the celestial equator (declination +5) is Procyon.
Here is a link to the list of brightest stars:
I need to find out which of these 3 stars can be seen from California. Would you have any idea how I would go about finding this out?
Follow up Reply:Look up chapter on "Navigation".
If your location is north of the equator (as California is), all stars north of the equator should be visible from there. So Capella should be visible, unless it is in the sky during the daytime.
Suppose you are at northern geographic latitude L. Then the pole star P (or more accurately, the northern celestial pole) is north of you at elevation L degrees above the horizon.
Let point Q be on the horizon, exactly south of you. The angle between the direction to Q and a point on the northern horizon is exactly 180 degrees, so the angle between that direction and the direction to P is (180°–L) degrees.
The southern pole of the sky is 180° from the northern pole (add it to the drawing!), so it will be at an angle L degrees below the southern horizon. The only stars you will never see are those whose distance from the southern pole of the heavens is less than L. These will stay below the horizon at all times.
If you are at 30 degrees latitude north, Alpha Centauri will be skimming the southern horizon for a brief period once in 24 hours. You just might glimpse it, if it happens at night and if smog does not interfere (refraction of starlight by the atmosphere extends your view by an extra degree or two). Further north, no chance. Procyon at –5° should pose no problem, since it is 85° from the southern pole of the heavens.
To see a star, the season must be right, too. Consult a sky chart to see when (date and hour) any star passes the north-south line. That is the best time to see it: if it does so in the middle of the day, of course, you may well miss it.
274. Gravity AssistI discovered your personal page while doing a Google search and thought I'd take my chances in submitting a question.
I'm curious about the idea of using a "solar assist" as a slingshot effect to send a body to another star. Specifically, I'm thinking about the asteroid Eros. (The details regarding mass and whatever else is pertinent can be found on the Wikipedia entry for Eros.)
Is such a thing conceivable? Could a gravity boost be capable of sending such a body out of the solar system and to another star, even if it were to take centuries to arrive? What would be the basic details of such a maneuver?
The idea is a premise I have for a science fiction story and I would like it to be as accurate and didactic as possible. Unfortunately, my abilities in advanced mathematical formulas and thinking are terribly wanting. I hope this question intrigues you and you will have time for a response.
ReplyQuick reply: yes, it is possible, though the time scale is likely to be many thousands of years, because stars are even more distant than you think. See
Eros is not a good candidate, and neither is the Sun (getting too close to it causes too much heating). But if you can catch hold of a comet far enough on the fringes of the solar system (where a moderate push makes a great difference in the trajectory), and have it swing by Jupiter or even Earth--yeah, you could make it reach some star, maybe in 100,000 years or more.
275. Does the Centrifugal Force Exist?(Question by a student studying to become a science teacher)
I am working on one that has been told to me by college professors. The oblate shape of the Earth is a result of centrifugal force. Since this is an assignment for one of my graduate courses, our professor has a list of web sites listed for these. I will email her and ask if I may share them.
But is it appropriate to invoke the centrifugal force?
ReplyDoes a centrifugal force exist? Or is it a myth, a "fictitious force," as too many physics students are told?
For my own stand, see
as well as sections close to the above. Mainly I agree with the teacher who argued that centrifugal force was no myth, nothing fictitious, but rather, it is an inertial force, a concept deserving its place in the standard curriculum.
Classical physics always obeys Newton's laws, but the choice of frame of reference is optional. You can view an airplane wing as moving relative to the air, or do the same calculation with air flowing across an immobile wing. The forces remain the same, because all that's being added is a constant velocity of fixed magnitude and direction.
Accelerated frames of reference add inertial forces, and adding something one could do without may needlessly complicate simple situations. But not always. Two cases exist where such addition may be justified. One case involves the dynamics of the atmosphere and oceans. There you'd rather handle all motions in the frame of the rotating Earth.
And in high school, you may want to engage the student's intuition. The student knows from experience that on a rotating carousel, or in a car rounding a corner, a body is thrown outwards. The term centrifugal force is used in common speech, because it is intuitive. This is where a wise teacher acknowledges that in a rotating frames the laws are modified, accepts the intuitive as legitimate, and explains the modification. For static equilibrium in a rotating frame, you just add the centrifugal force. If motion occurs in a rotating frame, you also have to deal with the Coriolis force, but high school students are usually not expected to go so far.
You may teach any way you wish. If your students end up confused--or spend more effort than is absolutely necessary to understan simple ideas, that's your problem, and theirs.
A happy new year y'all, those who celebrate it. Happy equinox to the rest.
276. How do we know Earth is bigger than the Moon?While teaching about gravity, I asked the class which is bigger the earth or the moon. Everyone said that the moon was smaller than the earth. I wanted to get them thinking so I asked them, "How do you know?" Some said "Because NASA tells us", others said "because the moon orbits the earth". I told them not to just swallow everything that NASA (or anyone else) says blindly. Then I made a "mistake": I said that the moon traces an arc across the sky, but so does the sun. Why would we say that the moon orbits us, but we orbit the sun??
ReplyThe Earth is larger than the Moon. There exist total eclipses of the Moon, demonstrating that the shadow of the Earth is larger than the Moon.
The above argument was used by Aristarchus around 270 BC to estimate the distance of the Moon (quite accurately at 60 Earth radii) and also that of the Sun (quite inaccurately, but it made no difference--it led him to the first heliocentric theory). Aristarchus thought the Moon was about half the width of the Earth; actually it is less than 1/3.
To be sure, the shadow of an object such as the Earth can widen with distance, if the light source is very compact. In the case of a lunar eclipse, however, the source is the Sun, occupying a width of about half a degree of the sky. That fact makes the shadow ultimately shrink, since there is no shadow at distances from which the Earth appears smaller than the Sun.
The calculation can be found on
The "no shadow zone" starts at about 4 times the distance of the Moon, not far from the L2 point where the Webb Space Telescope is to be placed.
277. Sound waves on the Sun?I have a question about our sun. The sun has an atmosphere, which ought to transmit sound. What would it sound like to be on the sun? Have there been any probes to the surface to listen to the sun?
ReplyYour simple question leads to many complications, most of them going far beyond my limited knowledge, or even that of experts.
Sound waves certainly exist on the Sun, and are likely to be exceedingly LOUD. No instruments have ever landed on the Sun, nor are any likely to do so (they would boil away instantly) but all we can observe, especially the constant churning of the solar surface, suggests a high density of sound-like waves.
Furthermore, the visible "surface" of the Sun, at about 6000 degrees K, has above it a rarefied corona of about 1,000,000 deg. K. Assuming the required energy comes from the Sun below, how does it do so, without heating the surface layers? Sound waves have been suggested--and to create a million degrees, they need to be pretty intense.
However... sound is not the only wave mode. Sound in our atmosphere (or in solids) is the periodic interchange of motion (kinetic energy) and compression (one form of potential energy). The surface of the Sun his hot enough to conduct electricity, and that allows modes where magnetic fields get periodically compressed (Alfvén waves), also electric oscillations, and to make things really complex, all sorts of hybrid modes between various forms of energy. Your "sound" turns into something very complicated. Many modes get reflected back from the low densities of the corona, which is why to the best of my knowledge, the heating of the corona is still unsolved.
If you want to get a taste of what's involved, ask Google about (say) "photospheric sonic waves."
278. What if we had to give up use of Satellites?What would be the effect on life on earth if every satellite now in orbit suddenly crashed? i.e. how would it affect communications, and which facilities and services would be hit?
I'm a retired journalist in the UK, currently writing a science fiction novel, and have googled myself blind trying to find an overall answer to this question - with only limited success.
ReplyIt is hard to imagine anything causing such a grand crash! Synchronous orbits are probably good for billions of years, and long before that synchronous satellites will be dead garbage, to be brought down to earth again or be blasted away from us (which may be easier).
However I will allow that a prolonged blast of radiation from the Sun (unprecedented, unlikely--but in SF you are allowed almost anything that does not blatantly contradict physics) might disable them by burning out the solar panels and solid state chips. It needs to be prolonged--say a day--so that satellites in the Earth's shadow are not spared. Humanity will be shielded well enough by the atmosphere, except perhaps some high-flying jets above the pole.
The most profound effect would be the loss of all navigational statellites, especially the Global Positioning System (GPS) of the USA, the Russian parallel GLONASS and the system Europe is now trying to assemble. These systems can tell your position within decimeters or meters and are relied upon by drivers, pilots, ships, the military and of course guided missiles of all sorts. Loss of those will force people to go back to earlier systems--reading maps, using radio signals (if these work--maybe sextants and precise clocks if not), weapons guided by the eye of a nearby controller, etc. Back to 1970 or so.
Global communications will also be disrupted, but I am not sure by how much, since fiber optic cables now also exist, even under the ocean. They can take a lot of the traffic. The fluffy stuff on the internet--internet blogs, spam, e-bay, legit advertising, maybe my own educational files, all these may take a vacation. Space science, of course, would be interrupted even more severely--except solar physics, which would be desperately trying to figure out, how the disaster happened.
Beyond that--you are author and commander. Go to it!
279. Rotation of VenusDoes Venus spin in the retrograde direction, or is it just tilted at an extreme axis, like Uranus, that it just seems to be spinning in the retrograde direction? Is there any proof , either from Hubble or any other satellites suggesting the retrograde direction of Venus? If there is proof, then what are some of the reasons scientists believe this retrograde direction exists?
ReplyI quote from
Thus the rotation is truly retrograde. "Hubble" cannot tell about the rotation, all it can see are clouds high in the atmosphere. The first data have instead come from the Arecibo radio telescope, built in a deep valley on Puerto Rico. I remember how frustrated radio astronomers were by being able to observe only one side of the planet, because every time Venus came close to Earth, it presented the same face. "Magellan" of course has mapped it all out.
We do not know the reason planets rotate the way they do,. Presumably , when the solar system first collected itself, material accreted in large "planetisimals" which later collided to form our planets. Our own large moon may be left over from that stage. In any case, in the last few collisions which contributed most of the mass of the planet, the encounters among the "planetisimals" may have been somewhat random, causing the final rotation to vary.
280. "Proper Motion" of starsCan you say if sufficient data is available for us to show the relative motion of the nearest stars? Perhaps someone has attempted to illustrate this. Our human timescale is too small to observe any appreciable motion, perhaps we need to stretch observations to cover 100 (or 1000) years!
ReplyUsing the magnification of modern telescopes and the mechanical precision of their mounting, astronomers do not have to wait centuries to see the change is positions of stars--though by now they do have about two centuries of data to work with. The "proper motion of stars" is described on many web sites, for instance
The table at the end of that web site lists some stars with large proper motions. Since we know the distances of the nearest stars from parallax measurements, this can usually be converted to relative velocities in km/sec.
A related question is the motion of the solar system relative to the average positions of stars in the galaxy. The Sun moves towards a "solar apex" near the present position of the star Vega--see for instance
The site suggests we are orbiting the center of our galaxy at 200-250 km/sec.
281. Doppler shift from the Big BangHello, I am Mark at a Pennsylvania University; and I am conducting a project on earths' precession. I want to know how to measure a Doppler effect to determine which direction our galaxy is moving; resulting from the Big Bang...
ReplyWhoa there, Mark!
The Earth's precession is a very local effect, connected with the Moon's gravity pulling the equatorial bulge of the Earth. The Big Bang involves the entire universe: read about it in
Our galaxy might have a small motion of its own, but in the main, it moves away from ALL other galaxies in the universe: the result of the Big Bang is that the average spacing between galaxies just keeps growing, as the space occupied by the universe increases. There seems to be no preferred direction in this.
There exists a Doppler effect, of course, but it has no preferred direction. It is evidenced by the cosmic microwave background (see section stargaze/Q4.htm#h54 ) which started as the radiation of a very hot gas at short wavelengths, when the universe was very young. The Doppler effect has shifted it into the microwave range, and the shift seems very close to identical in all directions.
282 Science Fiction Flight to MarsConcerning solar flares:
If I send a huge disk Intrepid [A science fiction spaceship] loaded with cargo and passengers (miners and tourists) to Mars and the trip takes only 25 days (Cascade Generator, Heisenberg Anomaly Compensator and Steinmetz Accelerator [devices thought up by the author of the story]), would solar flares be a problem? Why or why not?
Once on the surface, assuming only minimal terraforming has been accomplished near the "cities", what sort of protective clothing (armor?), if any, would be required to prevent death or injury while searching for artifacts on the open desert?
Okay, it's a SciFi novel but I want it to make sense.
Stay well and watch the skies.
ReplyI have never heard of a Cascade Generator, Heisenberg Anomaly Compensator or Steinmetz Accelerator, but flying to Mars in 25 days sounds too far out--you not only need build up a huge velocity, but also kill most of it when approaching your destination.
Presumably, you would write of a time when Mars is already developed, and it may be believable that humans and/or robots residing there could manufacture rocket fuel.
So why not have an efficient flight of not 25 but 250 days, with a cylindrical fuel tank, having a shelter cavity in its center, where passengers can hide. Solar flares with enough radiation to harm humans are quite rare, their radiation does not last too long, and it usually can be stopped by shields--or in bad cases, enough of it can. Some risk remains, but spaceflight will always be risky.
Then on Mars you refill the tank and get shielding from the fuel, again. The fuel carried is for the final descent, on both legs of the trip (other tanks and rockets create the interplanetary velocity).
On the Mars surface, if the Sun erupts, you will probably get some warning by radio from sentinel satellites close to the Sun (Earth may be in a poor position to observe the eruption). For explorers on the surface, that may give some time (say, 10 minutes) to scoop up dirt and pour it into hollow shield-tanks.
Personally I believe that exploration of Mars, as well as any mining and manufacture, would in the end fall to intelligent robots, which do not need the elaborate life support which you and I would require. Indeed, I wonder what reason would bring people to Mars, other than the public mystique of exploration. It is a cold, hostile place, with a thin unbreathable atmosphere and no food resources. Machines will go there and explore it long before our descendants do
But of course, in science fiction, anything is possible! Good luck
283 Motion of the MoonI saw your web site, and I am looking for either a good book or message board to ask questions about the Moon. I find what is available to not be enough for me to understand the motion of the Moon, Sun, and Earth. Some of my questions would be like, "What is the exact longitude and latitude that the Moon takes in revolving around Earth?", as in the longitude and latitude on Earth the Moon's orbital plane takes on Earth.
Also questions about how to apply and use the additional concepts of the Earth's tilt and seasons to the Moon's irregular orbiting patterns, since we see the Sun and the Moon taking different courses across the sky. A lot of the public's lack of understanding may arise from the way pictures are drawn in books and web sites. This project might be for publication purposes, so I thought I would e-mail to see if there is a book, web site, or chat line that is credible and appropriate. It's just not the kind of information I have been able to find, but I will keep looking.
ReplyThe Moon and Sun each move in their orbital plane. The one of the Sun is known as "The Ecliptic"--see
The difference between the courses of Sun and Moon across the sky arises mainly from the fact the two planes are inclined to each other by about 5 degrees (one reason we do not have eclipses every month).
The motion of the Sun and Moon across the sky is approximately uniform. To calculate it precisely takes a fair amount of math, though it has been done and results are published in astronomical almanacs. You may find one in the library of any department of astronomy, or any observatory (including one of a serious amateur), and some parts are on the web. My own web files may help, too, though they are aimed at a low level. Look up their Q&A section, too, e.g.
Other than that, you will just have to keep looking.
Further commentThanks for the information and the web sites! They have greatly helped me plan a book. I know a lot has been published, but I have some ideas, too, and would like to publish something a little different. I just needed to know more about the moon's elliptical motion around the Earth and what longitude and latitude coordinates it might involve. I don't think such as system has been developed yet, and probably star charts don't include the moon or the moon's center or edges. Trying to plot or observe coordinates from Earth - if that kind of stuff does exist, I can't find it. Libraries don't seem to have answers to my question, but ironically, I found an astronomy book that went into depth about the moon, given away for free by our library. What you provided helps me in trying to understand that book. Thanks again!
Further ReplyI do not know how much mathematics, physics and astronomy you have learned, but the motion of the Moon is not a simple job at any level.
All the web material I refer to below is in "From Stargazers to Starships." Suppose first that the Moon moves exactly according to Kepler's laws (sections 10-12 of "Stargazers"), and you know the size of the orbit (semi-major axis a), its eccentricity e, the orbital period T and the time it passes perigee (date, hour etc.). At that time the mean anomaly M (section 12b) is zero.
To trace the moon's position at some time t after passing perigee, first determine M at that time (first equation in section 12b--that's easy). Then from this derive the eccentric anomaly E, by solving "Kepler's equation", also in that section: that is harder. Finally, from E derive the true anomaly f, which is the polar angle covered in the orbital plane. It is a complicated trigonometric relation, not given on my site.
If the orbital plane were the same as the Earth's equator, f would measure the longitude, or rather the difference in longitude from the point under the perigee point of the Moon.
But of course, the Moon's plane is inclined to the equator, by the inclination angle I (see the part of section 12b titled "the orbit in space"; here we refer to the equator, astronomers may perhaps use the ecliptic as reference), and you also need two more angles, the argument of perigee and the longitude of the ascending node, to properly place perigee in the orbital plane and find the right intersection between that plane and the equator.
What you have now are two systems of coordinates: one in the orbital plane of the Moon, and one in the equatorial plane of the Earth. The orbital motion gives the angle f of the Moon's motion in the first, while longitude and latitude are angles in the second system. To transform angles from one system to another you need something called a rotation matrix.
Rotations in the two dimensions are hard enough--this is handled in problem 8, section M-12 of "Stargazers." To handle rotations in 3 dimensions, you better be familiar with 3-by-3 matrices and vectors in 3 dimensions.
Hey, wake up! Because the motion of the Moon is even more complicated.
First, of course, the Moon does not orbit the center of the Earth, but the combined center of gravity of the two--inside the Earth, below the Moon, about 1/4 the way to the Earth's center. Second, the motion is disturbed by the Sun and the bulge of the Earth (also by Jupiter etc., to a smaller extent). The result is very complicated.
The Astronomical Almanac by the US Naval Observatory has "low accuracy formulas" for the position of the Moon, and these have about 40 terms, "Astronomical Algorithms" by Jan Meeus has longer, more accurate formulas.
Do you really want to go into this?
In the 1800s, a French astronomer named Delauney spent 20 years calculating a formula for the motion of the Moon. His result filled a book, a fairly thick one. All his calculations were done using paper and pen, in terms of sines, cosines etc.; a modern computer repeated his work in 5 minutes and found, amazingly, only a few (three?) small errors.
As for books about astronomy: the mother lode for this sort of literature (at very reasonable prices, too) is the Willman-Bell publishing house in Richmond, Virginia http://www.willbell.com/
From them you can buy, for instance "Epic Moon: A history of lunar exploration in the age of the telescope," by William P. Sheehan and Thomas A. Dobbins, Hardbound, 6 by 9 inches, 364 pages, 186 Illustrations, 2 Lbs. 7 Ozs. ship wt. $29.95. Perhaps it's a better idea to read this book first and then think whether you want to write one of your own.
Live long and prosper
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