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396B Posssibility of Asteroid Hitting Earth (2)
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170. Spacecraft AttitudeWell, it started when I wanted to know how can a spacecraft calculate its self-position in space. I started by assuming it'd take reference to stars surrounding it, but I didn't know if the movement of the stars were negligible so it would need to compensate for it, and if so -- how would the measurements take place, and with what kind of resolution.. and so on...
ReplyConcerning the "self-position" of a satellite, you probably mean its orientation in space or "attitude." It is a big subject and much depends on how accurate do you want it to be determined. If you are happy with half a degree, a "sun sensor" and an optical "horizon sensor" may give you enough information--for other uses, star cameras exist and have been used.
A related question is, how do you rotate a satellite from one orientation to another? The usual way is to have gyroscopes, flywheels in constant rotation. Forcing the rotation axis into a new position causes the entire satellite to rotate in a way that preserves its angular momentum, then when you think it has rotated far enough, you return the axis of the gyroscope to its original direction and the satellite stops rotating. You can mount the gyroscope on gimbals and point it in different directions, or have independent gyroscopes for the (x,y,z) axes and spin them up and down. The Hubble telescope, whose attitude (orientation) must be constantly adjusted, has several gyroscopes, and one of the reason for visiting it from time to time is to replace gyroscopes, because their bearings etc. do wear out.
Still another question is how to you express the direction of the satellite in space. You need two angles in spherical coordinates, which express the right ascension and declination of some axis on it, and you also need an angle to characterize rotation around that axis. With those you express (x,y,z) coordinates of points on the satellite. Rotating the satellite in space to a new (x',y',z') requires a calculation with matrices. A simple example is in problem 8 of
which continues at
171 What makes the Earth rotate?Hi
Dont mind a question, do you? I was just doing a little competition, and I came across this: What makes the Earth rotate?
ReplyWe believe that the entire solar system started as a cloud of gas and dust which gradually pulled itself together by its own gravity. In such a process, a measure of rotation, known as angular momentum, is preserved. Angular momentum of a collection of matter is proportional to its mass, the square of the average distance R of the mass from the rotation axis and to the frequency F of rotations per second (or per year, or century... whatever units you choose).
Any time a rotating cloud of matter gets reduced in size--as it did when Earth originated--R gets smaller. Therefore, to keep the angular momentum the same, F has to increase. It follows that even if the original nebula rotated very slowly, by the time it formed the Earth, it must have speeded up quite a bit.
An analogy can be taken from the tornado, forming during times of severe thunderstorm activity. Usually thunderstorms are associated with vertical hot air rising and giving up energy (see section S-1A in "Stargazers") but there also exist horizontal flows of warm humid air into the storm (low down) and cold dry air out of it (high up).
Usually everything is pretty symmetric, but sometimes a row of thunderstorms is formed ("squall line") and the flow of air in neighboring storms can set one of them slowly rotating. That is all it takes! As humid warm air is sucked into the rotating thunderstorm, its rotation gets faster and faster, and a tornado can result. Unlike hurricanes,though, tornadoes are too small for their direction of rotation to be related to the Earth's rotation.
172. Energy from the Earth's Rotation?... just personally, I have had an interest of tapping other sources of energy besides oil, etc. for the earth's energy needs. I have had an idea about gathering the 'rotational kinetic energy' produced by the spin of the earth? Its just something on the side that is an interest.
ReplyUsing the energy of the Earth's rotation is an interesting idea, and it is feasible, but the way it's done is probably not what you had in mind. The problem is, to affect the rotation you need some other object which brakes it. It must be outside the earth--nothing on Earth will do it.
Such an object exists: our Moon. The interaction between the Earth's rotation and the motion of the Moon raises tides in the oceans, and schemes to extract energy from tides therefore takes energy from the rotation of the earth and from the orbital energy of the Moon. Such schemes exist, but they are hard to implement (tides are low, salt water is corrosive) and give a low yield.
See also next question, below.
173. How were planets created?Dear Dr. Stern,
In your reply to Q 78 you write - "...It was established not when the planets cooled but before that, a relic of the swirling of the cloud of gas and dust from which the solar system (and the sun) formed..."
My question is: From where did the "swirling cloud of gas and dust come? What is the SCIENTIFIC explanation????
ReplyFrom the evidence we have, all matter in the universe appeared--very hot, very dense--in the "big bang", about 13.8 billion years ago. I hope you do not expect me to recount all that evidence--expanding universe, microwave radiation etc.
Applying to this event what we know about atomic nuclei tells us that when that matter cooled enough to form atoms, these were mainly hydrogen, some helium and a little lithium. The materials from which planets are made (also, you and me)--elements such as carbon, nitrogen, oxygen and the rest-- --are more complicated, and must have appeared later. Without those elements, no dust would exiss--and no Earth, either, since our planet is largely made up of them.
It is generally held--and again, evidence exists, as well as theory, in which the late Hans Bethe, who just passed away at 98, had a big part--that heavier elements are "cooked," in part in the processes which even now power the Sun, but in addition (and especially the heaviest ones) in the sudden collapse of a supernova, which preceded the solar system. That includes such elements as uranium, which decay radioactively. From the radioactive content of moon rocks, they were dated about 4.7 billion years ago, and it is believed the Earth formed around that time, too.
The material from which the solar system formed must have been the cloud of dust and gas left from the supernova (or maybe from more than one), gradually pulled together by gravity. And it must have been swirling even then, because the amount of swirl--angular momentum--is preserved in mechanical systems. Furthermore, astronomers have observed (using the Hubble telescope, among others), disks of swirling dust which presumably mark the birth of other planetary systems, or perhaps systems of binary stars.
That, in a nutshell, is the "scientific explanation." We weren't there when it happened (the way the Almighty was, perhaps--see Job 38, v. 4), but we have plenty of evidence. You may also look up http://seds.lpl.arizona.edu/nineplanets/nineplanets/origin.html.
174. Does Precession of the Equinoxes shift our Seasons?A very good old friend of mine a few years ago told me that the reason our winters are starting to arrive later and later is because in a few thousand years, the axis of the earth will eventually change and our winter season will become summer season and vice versa. I'm no longer in touch with him but I always knew him to be a very knowledgeable person in the fields of cosmology and meteorology. But I have yet to read anywhere or meet anyone who could verify his claim. Is there any truth to his claim? I have noticed the change in the past twenty years so I'd really to get to the bottom of it and understand why this is happening.
ReplyYour friend had in mind a real phenomenon of nature, but it actually happens differently, and the results are not the ones your friend credited to it.
The Earth orbits the Sun in a large flat plane, known as the ecliptic. The reasons we have winter, summer etc., is that the axis around which the Earth turns is not perpendicular to the ecliptic, but makes an angle of about 23.5 degrees to that perpendicular.
Thus in the summer, the northern hemisphere is tilted towards the Sun, the north pole gets 24-hour sunshine and countries north of the equator also get longer days and more concentrated sunlight. Six months later, the axis still faces the same way in space, but now the Sun is on the other side, the north pole is in the shade all the time, the northern hemisphere gets longer nights and sunlight falls there at a more shallow angle, reducing its power to heat the land.
All this is described in "Seasons of the year" at http://www.phy6.org/stargaze/Sseason.htm which also shows how seasons south of the equator are in opposite parts of the year.
Now what your friend is aware of is that the direction of the Earth axis changes. It always makes an angle of about 23.5 degrees with the line perpendicular to the ecliptic plane--but while keeping this angle, it wanders around a cone, whose axis is perpendicular to the ecliptic. It takes about 26,000 years to go completely around that cone. The technical name for this phenomenon is "Precession of the Equinoxes" and it is described in
Precession, however, does not make the seasons different from what they are now. The picture given in "Seasons of the Year," of the Earth tilted by 23.5 degrees while orbiting the Sun, still describes the situation, except that over the thousands of years, it is slowly rotated in space, relative to the rest of the universe. (If you made a flat paper model of it and put it on a table top, what happens is that the model gets slowly rotated on top of the table). The orbital plane remains the same, only the Earth rotation axis goes around a cone.
Since the picture is the same, the seasons are the same. Our calendars are all adjusted to include this phenomenon, to keep (for instance) the date of the shortest day of the year around December 21.
What does change are the background stars. Since the pattern rotates relative to them, the seasonal presentation of the stars slowly rotates, too. Nowadays we see the constellation Orion in the sky after sunset in winter, and the constellation of Scorpio in mid summer, with its bright portion of the Milky Way. In 13,000 years, the rotation will make Orion shine in the evening in midsummer, and Scorpio in mid-winter. The pole star, too, will shift, and the evening constellations may also shift closer or further from the horizon. (Stars have motions of their own, too, and in 13,000 years, constellations may somewhat change.).
All that, however, deals with the stars at night. Seasons are tied strictly to the relative positions of Earth and Sun, and that will not change.
I have not heard about winters arriving later, but it may seem that way, because winters are also getting warmer, part of the "global warming" trend. There is little doubt that global warming is happening. Some people still wonder whether it is part of a slow fluctuation which will ultimately reverse itself. More and more however accept that it is caused by the increase of molecular gases (carbon dioxide, methane) due to human activities, which makes it harder for the atmosphere to radiate (shine) its heat back to space.
175. "Zenial Days" on HawaiiI live in Kaunakakai, Hawaii, on the island of Molokai. The busiest intersection on the island is more or less at 21.0893 N, 157.0227 W. We'll use that for Kaunakakai.
As I understand it, the longest day of the year in our tropical town is not the summer solstice. We should have two longest days of the year, one shortly before the summer solstice, one shortly afterward. The sun should be directly overhead once as it travels north to the Tropic of Cancer, and once as it heads back south for its rendezvous with the equator and the equinox.
How can I find out exactly when those dates are? The sunrise and sunset times on our local tide calendar are too imprecise for me to figure it out that way. Is there a celestial almanac that will list the latitude of the sun for each date? Is there a formula a 7th grade English teacher can figure out?
ReplyYou indeed seem to have a clear understanding of the motions of the Sun.
Every day the Sun rises somewhere near east and sets somewhere near west, and at noon (in the continental US) it passes to the south and reaches its largest elevation angle above the horizon (largest for the day). In those states, that angle is always less then 90°, so the Sun never passes overhead--"at zenith." Also (because of the tilt of the Earth's axis), that angle depends on the date. In summer it grows larger as the date approaches midsummer day on 21 June, then it decreases again (see "From Stargazers to Starships").
Hawaii is close to the equator, so the Sun's noontime elevation there in midsummer actually exceeds 90°, making it pass north of the zenith. The noon elevations before and after that day get smaller, and you are asking, when are they 90° , so that the noontime Sun passes exactly overhead.
I would guess those dates are May 26 and July 18. (These are "zenial days" and were apparently noted by Maya astronomers).
But you seem to want more than numbers--you also want understanding. I therefore ask you to draw a picture on a sheet of paper, a simplified north-south cross-section of the celestial sphere at Kaunakakai. Then go through the reasoning below, very slowly, making sure you understand every step. You might want to draw additional sketches, and also to consult suitable parts of "From Stargazers to Starships."
First draw a horizontal line (x-axis), which represents the north-south axis on the ground (north at right). Add to it (above it) a perpendicular line (y-axis) which points up to zenith, straight overhead. The origin is where you stand.
On the right, add a straight line northwards from the origin, making an angle 21.09° with the x-axis (you need no accuracy, this is just a sketch). It represents the direction to the celestial pole (or the pole star, very nearly) from Kaunakakai. And finally, a line from the origin, making a 90° angle with the preceding one--i.e. tilted southwards from the zenith, by an angle 21.09 degrees. It represents the direction to the celestial equator.
Over the year the Sun (in its apparent motion relative to the stars) traverses the ecliptic, a circle on the celestial sphere making an angle 23.5° with the celestial equator. At equinox it is on the intersection of equator and ecliptic, during the summer it is north of the equator, during winter south of it.
On any day of the year, it will have on that sphere a certain "distance" (actually, an angle) north or south of the equator. Suppose it is summer. Imagine a line on the celestial sphere, from the celestial pole to where the Sun is, and you continue it until it hits the equator. The size (in degrees) of that last section is the Sun's "declination angle" at that time, its elevation angle above the equator. On midsummer day it is 23.5°, at equinox, zero, and you can extend the definition to find its value on midwinter day as (–23.5°).
Now go back to your drawing, of the north-south sky at Kaunakakai. On midsummer day at noon, the direction to the Sun makes an angle of 23.5° with the direction to the celestial equator. You can add that direction to your sketch as a broken line, if you want, and because 23.5° is larger than the latitude of Kaunakakai, it is tilted past zenith, northward of the y-axis. Not the highest possible elevation! You want the Sun to be at zenith, which means the declination--the elevation (above the equator) of the point of the ecliptic where the Sun is located--has to be 21.09°, or about 21 degrees, 5.5 minutes.
The web has many sites which calculate or tabulate the position of the Sun, and the US Naval Observatory publishes a yearly astronomical almanac, but (as of 2016) the site answering your questions, http://www.astrosoftware.net/data/2005/sun.html , is no longer posted on the web . It had a tabulation of many variables, some specific to a location at latitude 45 north--but all you need is the declination of the Sun, which was tabulated day by day. Look on what days it comes closest to 21 degrees, 5.5 minutes, and you have your answer.
Please let me know how clear the above explanation has been. I could probably do a better job in a short session with you in front of a blackboard or a sheet of paper, but I'm retired and no one will pay me to fly to Hawaii to do so. (Curses!).
176. Sun's Temperature and Energy Density of SunlightI cannot seem to find the temperature of the Sun at the surface. Memory tells me is of the order of 10 to the 28th degrees centigrade. Please tell me where I am wrong.
Secondly, is it possible to determine the Joules per cubic centimeter (cc) at the surface or is there another way of discussing solar energy? I'm not talking about solar panels. I'm trying to get to the real meaning or power of a cubic centimeter of solar energy. It's going to be in Joules per cc--equal to what?
Can you help me with this? I was watching a movie in which a man invented a laser which had a zillion times more power per cc than the Sun. Or, is that an incorrect way of thinking about the Sun's energy?
ReplyThe radiation from the visible surface of the Sun approximates what is given out by a hot object at temperature 5800° absolute, give or take a few degrees (in Sun1lite.htm I wrote 5780 deg.)..
The number is not important. Students and teachers should not memorize too much--at most, remember where the answer can be found. The important question is, "how to be know"? There exists a general law by which colors are distributed in the emissions of hot dense objects ("black bodies" which favor no color--the way emissions from gases then to favor colors characteristic of their atoms). The hotter the object, the shorter the wavelength, that is the further the peak emission migrates from infra red to dull red to bright red to orange to bright white (all colors). And maybe beyond. Colors of stars tell how hot they are.
That law is also important because it gave us the first inklings of quantum theory. See more in http://www.phy6.org/stargaze/Q4.htm That file has not yet released, pending more work, but you may like it.
Asking for the volume density of sunlight is the wrong question. The mantra for electromagnetic radiation (e.g. light) is "spreads like a wave, deposits article like a particle, like a photon." A wave in theory fills all space, though not equally (brighter in the light than in shade). Considering 1 cc of this volume does not tell much. A photon is a little point, an atom absorbing radiation--also not useful in terms of volume.
What is important it the flux--the rate of energy flow through a unit area (think of rain falling: how many gallons per minute per acre [or liters per second per hectare]?). Surround the Sun with a sphere at the Earth's orbit: through every meter squared flow 1380 watts of power, 1380 joule each second. That is the "Solar Constant"--see first paragraph of section S-7.
We are about 200 solar radii from the Sun, and light weakens as the inverse of the distance squared, so 200 squared is 40,000, and the energy flow at the "surface" of the Sun is about 40,000 times larger. The energy density there is about (3/2)nkT, where T=5800 deg., n is the density of the gas (which you will have to look up) and k is "Boltzmann's constant" which you will also have to look up. A laser is a non-thermal source of radiation--as far from a "black body" as you can get--so one really cannot assign to it a temperature. The energy flux of a laser beam, or course, can be high--laser beams can melt holes in thin metal, and sunlight cannot, so it is much more than 1380 joule/meter squared. How much higher, depends on the laser.
177. Teaching about energy in 8th grade(from a message by an 8th grade teacher)
My 8th grade students are having problems with physical science -- they are a Title 1 school that is not meeting "proficiency levels" in math and in language arts. I spent a large amount of time, starting with measurements and the metric system. But nothing seems to be connecting in their long term memory banks. They conduct experiments and write down notes, but they do not understand the concepts and they don't appear curious. I am not reaching them. ..
From my reply: ...I would recommend that in the remaining month you stress energy and work--it's a concept one encounters more and more.
From the response: "In what way will their exposure to work and energy be a more significant learning experience for them?"
ReplyYou asked "In what way will their exposure to work and energy be a more significant learning experience for them?" (for your students). In several ways.
You wrote that you were teaching 8th grade. Your students are probably getting their first exposure to physics in a systematic ways, and therefore it is important to convince them that understanding it (not always easy!) is useful and provides knowledge which they can apply.
Energy is fundamental in this respect. They surely have heard about the cost of energy, about conserving it and about its various forms: moving cars and trains, heating homes and lighting them, running computers and TVs--all these involve energy. Food gives them something called "calories"--that is a unit of energy too. Therefore, as they learn about energy, they encounter examples in which physics enters their lives (and that of the community around them) in many ways.
Energy is also a good framework for unifying physics. Physics has many branches--mechanics (study of motions), heat, electricity, optics, atomic and nuclear, for starters (also, the physics which forms the foundation of chemistry and biology!)--all these come together in energy, as the table in the energy unit of "stargazers" shows. And there exists a good analogy, described there: energy is like money, paying for various processes. You can even hint about the second law of thermodynamics--tell them that heat is a 'soft currency", what you invest in heat can never be fully recovered but spreads around, more and more diffuse.
Work is the basic example of energy: overcoming a force over distance--such as lifting a brick against gravity, or climbing stairs--also overcoming friction, which creates heat (e.g. in the braking of a car, or armor piercing shells, which melt their way through a tank's armor, as their kinetic energy becomes heat) Work has many familiar examples makes the concept real.
The simplest example of kinetic and potential energy is of course a pendulum--or a kid on a bike, coasting down into a valley and losing speed as one gets up the next hill. Or a roller coaster. Burt Rutan's "Spaceship One" last year fired its rocket to accelerate to 3.5 times the speed of sound, then coasted up to 62 miles, then down again, braking its fall in the atmosphere by swiveling parts of its wings to create an obstacle.
Then, if you want something extra--you can tell (section 18d) about work against electric attraction, creating high voltage--in a Van de Graaf generator (The Boston Museum of Science has some giant ones, demonstrated to visitors), in lightning and in prying apart sheets coming out of a xerox copier, stuck together electrically. That however is extra.
Not much math is needed. You mix 5 liters of water at 60 degrees C with 2 liters at 10 degrees--what is the final temperature? (Energy is conserved). How many joules needed for a 60-kg kid to climb to the next floor, 3 meters up? How much is that in calories? If the kid ate a 6-gram chunk of chocolate (about 1/4 of a bar), providing 9 calories per gram (that's for fat--the sugar lowers it a bit), and converts the calories with 10% efficiency--how high can that kid climb, on this energy? And so on. Only arithmetic is used, no algebra (except maybe in the formula for kinetic energy, which can be explained separately)..
178. About the jetstreamOne question was bothering me since long, thanks again in advance for the answer
I saw some time back on science TV channel a documentary which says that within the oceans are rivers which flow in certain directions and the flow within is relatively faster. The documentary further says that sometimes submarines get themselves into these rivers to travel faster or to just to save some fuel.
My question is does the earth atmosphere also has such channels? & if so, can they be utilized by commercial aircrafts the same way as a sub does?
Regards & keep up the good work.
ReplyYou asked if "the earth atmosphere also has such channels?" Good question, and the answer is, yes it does.
The predominant flow at middle latitudes consists of the "Westerlies", a flow around the Earth from west to east, subject to large excursion north and south, the "Rossby Waves". This is discussed in
The flow is fastest at higher altitudes, where it is known as the "jet stream" and yes, airliners do take advantage of it. You may look up airline schedules for some direct flights between, say, New York and San Francisco. I would not be surprised if the estimated time of flight TO San Francisco is longer than the one FROM San Francisco (do not neglect the 3-hour difference in local time!). At one time airlines actually selected routes according to where the jet stream was strongest, but I do not think they do it any more.
179. What would a breach in a space station do?I'm 37-year-old librarian and a bit of a science fiction fan, and I found your site while trying to determine how accurate certain scenes in some movies and television shows are. While watching a Dr. Who episode, I became annoyed at something I had previously noticed during several Star Trek episodes (ALL series). The scene is always this: a hole in a spaceship or space station opens up, and one character gets sucked out into space, while the rest hold on for dear life until the breach is sealed or they are rescued.
I realize the scene is already inaccurate because the characters in the ship or station are still breathing, while in reality, the air would almost instantaneously leave the room. I wonder, however, how likely it is that any of them would not be immediately sucked out into space if this really happened. At first, I assumed they would be dragged out with the air. I then wondered if the ship's or space station's gravity would keep them inside if they were not directly in front of the opening. What would really happen if this occurred?
ReplyI like science fiction, too, especially the older stuff which has some science in it. Modern writing gravitates towards fantasy, drifting from the highly improbable to the downright impossible.
Also, your are probably right, a large hole of a spaceship would result in very rapid loss of all air, in the process sucking out all sorts of other loose objects. Even passengers, though I would suspect light objects would go first, and passengers may be too big to go through the hole. The ship's gravity is no match.
If I were designing a space station the design might include in it a number of hollow balls, like the rubber balls kids play with, but made of real tough plastic, maybe Kevlar, with an outer layer of soft rubber. If the breach is not too, big, sooner or later one of the balls will be sucked into it and seal it. Ah, here is a nice story for you to write!
A question like yours was asked before. See question #152 (listed in error as #150) "Sudden decompression, 5 miles up".
180. Gravity at the Earth's centerWhat is the gravitational field at the mass centroid of the earth?
ReplyThe answer is zero. No gravity at the center!
Newton proved this for a sphere--see 3rd paragraph in the reply to
Look at this another way: if gravity existed there, in what direction would it pull? There is no preferred direction in the middle of a spherical distribution of mass.
181. Freak waves on the oceanI have attached a text document copied from the BBC which talks about waves tracked by the ESA satellites on the worlds oceans in the recent past. The article mentions their association with calm seas and mild weather.
I actually researched into this some years ago out of personal interest and found that on many occasions mariners have told stories of the sea rising up from a calm surface to a great height often even baring the sea floor. After closely assessing the data from the ESA I concluded that many of these waves are not physically possible based only on the wind and ocean currents and that there was no seismic activity in the region that would warrant such a wave.
The waves are typically very steep, travel very quickly and are generally short lived. they remind me very much of the flares often seen on the surface of the sun. (much smaller of course).
Do you think there is any foundation to the idea that these waves and flares could be caused by natural 'eruptions' of electro-magnetic fields from the cores of the sun and Earth
ReplyFreak waves have been known for a long time. The "Queen Mary" was almost sunk by one during WW II--see
An Italian liner (the Michelangelo?) had part of its bridge swept away in the North Atlantic, and only recently a serious wave hit a cruise liner:
The cause is not completely known, but it is certainly not electromagnetic energy--there is not enough such energy in electromagnetic disturbances arriving from space (nothing arrives from below) and its much too spread out.
Different waves can combine randomly, and sometimes reinforce each other. That may happen, but solitary high waves may be different. There exist mathematical models of "solitons", solitary waves which do not diminish with distance from the source, as ordinary waves do. They may play a role; maybe steady winds, too. The trouble with freak waves, like some other phenomena, is that they happen unexpectedly at unpredictable locations, and do not last long, making it hard to collect reliable measurements about them.
182. Citation on "Bad Greenhouse" web pageI have serious questions about your terminology. In your web page
you make atmospheric analogies with greenhouses and blankets. You also, later, refer to water being 'forced out' of the air.
I refer you to
which wrote: " But don't ever teach nonsense by claiming that the radiation is trapped, or that the atmosphere reradiates, or that the atmosphere behaves as a greenhouse (or parked car), or that greenhouse gases behave as a blanket." Also
according to which: "The idea that it is the air which determines the amount of water vapor which can be present through some sort of holding capacity is an eighteenth century idea which was shown to be false both empirically and theoretically about two hundred years ago! The fact that it is still taught in our schools and defended by teachers and (gulp) professors, is a testimony to the mindless persistence of myth."
I looked up Dr. Fraser's web page and don't care either for his style or for his claims. His style is opinionated, and the passages you cite are a fair example of that. It reminds one of the joke of the cleaning woman who tidied up the church after services, and found the pastor's notes for his sermon, with a notation in one place "argument is weak here--raise voice."
I try to be careful not to inject opinions, but instead cite the reasons--observations, interpretations--why a claim is made. Dr. Fraser's arguments have a different style.
The atmosphere absorbs heat, yet its temperature does not rise without limit: it must re-radiate. It does so in the infra-red, because it consists of molecules, and these absorb and emit infra-red--their energy levels are in the IR range, whereas in the visible (bigger photons) they are transparent. This is the "long wave" radiation in
Of course, whatever the atmosphere emits is often reabsorbed by other layers in the atmosphere, or even by the ground. But that is the essence of the greenhouse effect--it does not prevent the flow of heat to the high atmosphere (where it is radiated to space), it just impedes it. Fraser is right on one thing--without greenhouse gases, Earth would be mighty cold.
Water vapor has a complex role. It too, is a greenhouse gas, an important one. In addition, it contains energy. It takes 100 calories to bring a gram of liquid water from freezing to boiling, but 540 calories to evaporate it, and in that sense, humid air is a store of energy of two types--in its temperature and in its content of water vapor. It can release energy either by cooling down (e.g. radiating to space) or by getting rid of the water (as in a thunderstorm). Air which got rid of its water is then warmer and rises, and can give up its heat in the high levels.
In short: I stand by what I wrote
183. How can radio waves carry sound?Firstly could I thank you for producing such an educational and informative website. I am now beginning to understand the wonders of magnetism and electromagnetic waves, and also be fascinated by the fact this "energy " is completely natural. Maybe God does exist!!
Anyway I digress. The question that I have relates to radio communications. I have read that radio comms use a carrier wave to "transport" sound waves. Could you tell me how that is actually achieved. How is the sound wave of somebody`s voice, ie a radio DJ physically attached to the carrier wave. I would appreciate it very much if you could answer this question and I will continue to use your website to learn more about other related topics.
ReplyTo give a full answer to your question is not easy. You are of course aware that sound waves and radio waves are quite differentt--radio involves electric and magnetic phenomena, while sound does not, it depends on the exchange of pressure and motion.
The energy of a piece of music transmitted to you over the radio has undergone several changes:
(1) The sound vibrations are converted to electric vibrations, using an instrument known as a microphone.
(2) Another electric vibration is created, of much higher frequency, and the electric vibrations corresponding to the music are encoded as variations in it.
(3) The high-frequency electric vibration (with its encoded signal) is channeled to an antenna, an electric conductor open to space, which radiates it away as radio waves.
(4) Another antenna, in your radio receiver, captures some of the radio wave. Of course, only a tiny part of the energy reaches this antenna--the rest if broadcast in all other possible directions. As a result, the signal received is very weak, and needs to be amplified: the energy for this is largely derived from the battery or electric power supply of your radio.
(5) The original electric vibration, corresponding to the sound of the music and encoded in the radio signal, is extracted from the high-frequency "carrier wave."
(6) The electric signal is re-converted to sound, using an instrument known as loudspeaker.
In principle, you could feed the electric vibration from (1) into an antenna, recover it at your radio (step 4), amplify it and turn it back to sound using a loudspeaker. In practice, it won't work, because radio waves propagate much more efficiently when the frequency is high. It therefore pays to create a high frequency "carrier wave" and let it carry the signal, encoded as a variation of the strength of the carrier wave ("amplitude variation" or AM), or (a later development, more challenging but with certain advantages) as a variation of the frequency of the carrier wave ("frequency modulation" or FM).
There exist various designs of microphones. A simple one has a little box filled with carbon grains, and one side of the box can move in and out, moved by a round membrane. The sound of music hits the membrane, makes it oscillate, and this compresses the carbon grains and changes the electric resistance. An electric current led through the box therefore varies with the sound. Other, newer methods also exist.
The loudspeaker also has a membrane. Attached to its middle is a coil, and inside that is a magnet. The electric current with the signal flows through the coil, and as the current gets stronger or weaker, the coil senses a magnetic force which is stronger or weaker, and it pushes and pulls the membrane, creating sound which imitates the electric signal.
The high-frequency signal and amplification are generated by special electronic amplifiers--transistors nowadays, electronic vacuum tubes before transistors were invented. There is more, much more. But I hope you get the idea.
184. Do Cosmic Rays produce lightning?We are a French tv production company that specializes in documentaries and are currently working on a program about weather reports on the solar system.
We recently came across the electrical discharge theory that claims the origins of lighting on earth is actually from a constant shower of cosmic gamma rays from the sun. We would like to use the theory as a subject for our program so we've contacted you to ask the almighty question, "is the theory really true"?
ReplyLightning is clearly generated by thunderstorms (and sometimes volcanic eruptions--a different process). Cosmic rays are found all over Earth, but lightning occurs only in thunderstorms, and clearly gets its energy from their violent air motion--a combination of hot warm air rising and raindrops (or hail) falling down. See
Processes driven by that combination create high voltages. The problem is that air is an insulator, and for lightning to discharge the voltage, the insulation must somehow break down. It probably does so somewhat like a wedge splits wood--the "sharp tip" of concentrated voltage gradually extends the strong electric field, until it reaches the ground.
Cosmic rays are not gamma rays and do not come from the Sun. They are fast protons and other nuclei (mainly protons), filling our galaxy and probably originating in supernova explosions: see
As they collide with atmospheric atoms, they tear off electrons ("ionize the atoms") and by this allow the air to conduct electricity (for a very short moment, until those electrons find their way back). Cosmic rays also undergo nuclear collisions which create other fragments, and these too add to the ionization. The idea (I believe) is that cosmic ray ionization helps the high voltage break down the insulation of the air. In addition, the lightning process concentrates the high voltage, and it can then accelerate electrons (torn off by cosmic rays, say) to high energies, to create more ionization which also helps get the lightning started.
That is the little I know or guess. Better get some expert to help you with the program, before you start serious work.
185. Star positions shifted by the atmosphereMy name is Siddhartha and I live in India. I am 12 years old and am in the 8th grade.
Recently while studying about 'The Universe', our Physics teacher told us that from the earth, the stars do not appear at their original position due to refraction of light coming from the stars as it passes through the 4 layers of the atmosphere of the earth. That means, even stars like Alpha Centauri, the star nearest to the earth after sun, when viewed from earth, do not appear at their original position. That even means that the Sun also, when viewed from earth, does not appear to be in its original position.
My question is that how can the distance between the original position and the refracted position of the stars, that are in space, be measured? That means that how can the original position of the star be determined?
ReplyTwelve years old and studying physics in the 8th grade! Wow, I admire your ability and hope you will realize its full extent as you get older.
Yes, the positions of the stars and Sun are shifted by refraction in the atmosphere, but the shift is small and occurs mainly near the horizon, where sunlight and starlight slices diagonally and has a relatively long path through the atmosphere.
This was discovered by Tycho Brahe even before the telescope was introduced, but the effect is easiest to measure using a modern telescope. Telescopes (all but the simplest) have a clockwork or motor which slowly turns them to compensate for the rotation of the Earth--see picture of a telescope mounting in
Without this, the rotations of the Earth (which is greatly magnified by the telescope) causes any star to quickly drift to the edge of the field of view and be lost. If your telescope has crosshairs in the middle of the field of view and you place a star there, those crosshairs track where the star SHOULD be, taking into account only the rotation of the Earth.
Place in the crosshair a star near the horizon--for instance, the evening star Venus just before it sets. You will see that it drifts more and more away from where it is supposed to be, because as it gets closer to the horizon, refraction by the atmosphere shifts its image. You may even measure the refraction using this trick.
Astronomers of course know all this and take it into account. Before the "Hubble" telescope was sent into orbit, a huge computerized catalog of star positions was assembled for it, and since the telescope is outside the atmosphere, these all are true positions and no correction is necessary.
The Sun's position is also shifted--when you watch the Sun setting, it is really already below the horizon, and only the bending of light makes it visible. We therefore get (at any location) a little extra sunlight, our day is made a few minutes longer. Some time ago I got asked about this and you might like to read the correspondence here.
Give my greetings to your teacher!
186. The equation of timeI have enjoyed your article "A Millennium of Geomagnetism" and your website on sundials. I have a question regarding ephemeris. My research project involves magnetostratigraphy and we use a solar compass to orient our rock cores. We use the following equation to calculate solar time:
Solar time = standard time + 4 (T-L-E) minutes
Standard time = local time minus one hour in summer.
T = longitude of time zone (in degrees).
L = longitude of site (in degrees).
E = ephemeris (in degrees). See table.
Dec. 30 --Jan. 7: + 1 deg
8 --Jan. 19: + 2 deg
Jan. 20 --Mar. 15: + 3 deg
Mar. 16 --Mar. 25: + 2 deg
Mar. 26 --Apr. 5: + 1 deg
Apr. 6 --Apr. 26: 0
Apr. 27 --June 1: - 1 deg
June 2 --June 26: 0 (summer solstice)
June 27 --Aug. 21: +1 deg
Aug. 22 --Sept. 8: 0
Sept. 9 --Sept. 20: - 1 deg
Sept. 21 --Oct. 1: - 2 deg
Oct. 2 --Oct. 16: - 3 deg
Oct. 17 --Nov. 18: - 4 deg
Nov. 19 --Dec. 3: + 3 deg
Dec. 4 --Dec. 12: - 2 deg
Dec. 13 --Dec. 20: - 1 deg
Dec. 21 --Dec. 29: 0 (winter solstice)
I would like to know more about the role of ephemeris in the solar time equation and solar compasses. I would greatly appreciate any suggestions you could offer.
ReplyThe dictionary defines "Ephemeris" as a tabulation of astronomical positions. While the quantity you denote E is something like that, it is usually called "The Equation of Time." It is a historical name--it's not what we call an equation, but probably "something that makes (time) equal."
A SOLAR day is the time from noon to noon, and noon is when the Sun passes exactly to the south, which is also the noon indicated by a sundial.
You would think that since the Earth rotates at a fixed rate, all solar days are equally long, but they are not. The vary because (1) the Earth moves around the Sun in a slightly eccentric path, and its velocity varies (Kepler's 2nd law), and (2) the angle between the Earth's orbital plane ("ecliptic") and the Earth's equatorial plane (i.e. the tilt angle of the Earth) also creates a small variation.
Our clocks are based on the "mean solar day" averaged over the year, and the times of "noon by the clock" are always spaced apart the same 24 hours. The south passage of the Sun can vary from this time by up to 15 minutes or so, and the difference is your E, the equation of time. Look up
The equation of time is mentioned briefly on the web page about the sundial, at the end of the section on "Accuracy." Want more? See an earlier question by a user, What is the Analemma? linked at the end there, which also cited yet another link.
187. Launch window of the Space Shuttle
I was wondering if there was a reason that the space shuttle...or other space launches have to take off at a certain time? For instance, today's launch should have been at 3:51, why not 3:50 or 4:00?...is there a reason physically or is it just NASA following a timeline?
ReplyI can only guess, like you. Shuttles have been launched at all times, even at night, but this time NASA declared it wants to launch in daylight, so that cameras can track the flight in detail and check for debris, etc.
In addition, the shuttle is meant to reach the space station, which has its own orbit. To match speeds with the station, the shuttle orbit should be in the same plane as the orbit of the space station. The direction of launch is also prescribed: it must be eastwards (to take advantage of the Earth's rotation--a small extra margin, but every bit helps) and at an angle to match the inclination of the orbit of the space station, I think it's 57 degrees.
So imagine the orbital plane of the space station: it passes through the center of the Earth and cuts the equatorial plane of the Earth at 57 degrees. As the Earth rotates, that plane stays fixed in space, but the point where it intersects the line of latitude of Cape Canaveral (28.5 deg north) marches around that line, going all around it each rotation of the Earth. Only twice in each 24 hours is Cape Canaveral located at that intersection, and a shuttle launched at such a time will match the orbital plane of the space station. However, only one of those times is suitable--the other one would require launch in the opposite direction.
So the result is a very narrow "launch window," as you noted.
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