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396B Posssibility of Asteroid Hitting Earth (2)
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136. Thanks for the "Math Refresher" in SpanishI'm 40 years old and passed my junior high and High school and I never understood Algebra, the substitution and replacement rules for the "X" or "Y's"
When I accessed your page I thought it was just another informative science page, but I'm so surprised by the way how you explain the formulas Off course this came to me about 35 or so many years late but still enjoyable, I really appreciate this small gift that you gave me on this night.
I will definitely introduce this page to my daughters I'm sure they will really appreciate it too.
Please keep up your excellent job
By the way I read the Spanish version
(English spelling and wording were corrected)
...and more:I'm 32 years-old Mexican living in Mexico.
Please receive this quick note to thank you for setting up a web page on the Pythagoras' Theorema, but specially for your infinite kindness of providing a Spanish translation.
I just stumble with the fact that I don't remember my high school math. And I needed it to deduct the measurement for 2 hypothenuses that are part of a sculpture (a triangular prism) of which I'm writing a paper on. After some web-surfing I got directed to your page. And... problem solved, I was able to get the paper written with its right measurements.
Again, thanks sharing your knowledge.
...still more (Spanish, though not math):  I have just visited your website and am so excited that I found something in Spanish! I teach Integrated Physics and Chemistry at Pampa High School, Pampa, Texas. Every year I have students who are ESL and have very little or no understanding of English, I have been trying to find some resources on Lab Safety and metric measurements in Spanish. I like what you have on your site, but it is not appropriate for these students. Do you know of any resources on the internet in Spanish that I could use as a supplement to my lessons. I would appreciate any information that you could give me.
Thank you ... Diana
...and still more (math, from Africa):Good afternoon,
I accessed the site containing (M-1)Algebra--the basic ideas and it appears to be the stuff that I need to familiarize myself with. I want to write the GMAT exam for MBA, for the 2nd time. Last was in May 2001 and my score was rather dismal. My math knowledge is minimal, and I realize that math in fact seems to be fun. If the basic idea is to isolate all that one does not need in the process of solving a problem and remain with what one needs on both sides, geez, I do wish I had known this ages back. I just hope I will be able to figure out all that you explain.
Best regards ... Rosemary
137. The Pressure of SunlightAfter visiting NASA solar sailing page I saw your name as a reference.
Do you know who validated solar pressure? Did anybody actually measure the pressure a (hefty) laser puts on a mirror? Michelson measured the speed of light but did anybody actually measure light's pressure?
ReplyLight pressure was predicted by Maxwell's theory of electromagnetic waves. It was verified in 1910 by the Russian physicist Pyort Nikolaevich Lebedev. See
Google has additional references under his name, which today is honored by the Lebedev Institute for Physics. Incidentally, the radiometer--vanes in an evacuated light bulb, which rotate when light shines on them--does NOT demonstrate light pressure, but an effect associated with the rarefied gas left in the bulb. See
I was not aware of the reference in the NASA page solar sailing page. It probably was to my page on solar sails
or to its mirror site. If you go there again you will find at the beginning and end of that page links to the index page of the "Stargazer" collection, where I am sure you'll find other interesting stuff.
I do not put much faith in Russian science. Not necessarily because of some bias (which is there), but because Lebedeev did not use laser directly (laser was not invented yet).
My question -- whether a laser was used on mirror to measure light's pressure directly -- is answered in the negative.
I appreciate all kinds of conjectures from great scientists about light pressure but, if there are so many of them then a simple laser-on-mirror experiment should be done to confirm that.
Incidentally, I am putting my money on the fact that light's momentum is virtual at reflection and a laser will not be able to impart pressure on mirror. After all, light bouncing between two parallel mirrors would make a perpetual motion machine. (So much for the Russian science.)
138. How is the instant the seasons change determined?Hello Professor Stern,
I came along to your site regarding seasons while researching a question - How do scientists approximate the time (down to the minute) that seasons change? Can you answer this question or point me in the right direction to find the answer.
During the year, the Sun seems to follow a large circle around the sky, among the "fixed" stars which define the constellations and whose positions are essentially fixed, because they are so far. That circle is known as the ecliptic, and is described on the web site.
Because of the Earth's rotation, the sphere of stars also seems to rotate. It has two "poles" around which the rotation takes place, one in the northern half of the sky and the other in the southern half. The first is visible only north of the equator, the second only south of it. The line halfway between the poles is knows as the equator of the celestial sphere, or the celestial equator. This, too, is described in the first sections of "Stargazers".
These two circles--the ecliptic and the celestial equator--intersect (as any two circles on a sphere must do, if each divides the sphere in two equal halves). The seasons are defined somewhat arbitrarily by this geometry. The two points of intersection are the spring and fall equinox, and when the Sun is there, night and day are equal (hence the name). The first is viewed as the start of spring,: see story on the Persian New Year in the section on the calendar, in the yellow box there. The second is the start of fall, around September 23.
Summer starts when the Sun is as far north of the ecliptic as it can get--that is the summer solstice, around June 21, the longest day. And winter starts when the Sun is as far south as it can get, and that is when nights are longest, around December 21.
Of course, we can't observe the Sun's position among the stars. But we can infer it, say by noting the times when on two consecutive days it passes exactly to the south. Using an accurate clock and observing the stars exactly halfway between those times allows astronomers to deduce the position of the Sun in the sky at that time, to the minute.
139. Operation of Ion RocketsI am really confused on how the accelerator grids function [in ion rockets]. I know that the negative grid accelerates the ions out of the ionization chamber but what exactly does the positive grid do? Does it act as a decelerator or as a repulsive force for the positive xenon ions? Are both grids on at the same time? How do the positive xenon ions get past the positive grid? Does one grid have a higher potential than the other?
I have been to many websites and I have read that the ions are accelerated as the result of "the potential difference between the grids"... what does this mean?
Any clarification will be greatly, greatly appreciated.
Reply"Potential" is a mathematical term also known colloquially as "voltage" because its value is measured in units called "volts." It is somewhat like height in a gravity field: a stone falls from a high location to a low one, and gains energy in proportion to the height difference it crosses. A proton or a positive xenon ion "falls" from high potential to low potential, and gains energy.
One difference: there also exist negative particles, like electrons, for whom "up" and "down" are reversed. Electrons gain energy moving from low potential (very negative) to high potential (very positive). Except for this reversal, everything is similar.
Still with me?
So we have two parallel grids, positive (say) on top and negative on bottom. The space in between is where electric forces can be observed, and a region like that is known as "electric field." Electrons released in this space move up (TO the high voltage), positive ions like those of xenon move down (AWAY FROM high voltage).
When positive ions cross the grid and move out of the in-between space (where a strong electric field exists), forces change, and in particular, electric forces quickly get very weak. The negative grid still attracts the ion (now pulling it back), but the positive one still repels it, and as you get away from the two parallel grids, those two forces quickly tend to cancel. So the ion keeps coasting with whatever energy it got in the electric field. That is what we wanted it to do, wasn't it?
140. Physical Librations of the MoonDear Dr. Stern, I truly enjoyed your web page on librations of the moon.
"In addition to the preceding modes, there also exist "physical" librations, actual pendulum-like nodding and wobbling of the Moon around its equilibrium position, like the spring-attached head of one of those "bobblehead dolls" popular as souvenirs. The main mode is pole-to-pole nodding and amounts to about a degree and a half. The longitudinal side-to-side oscillation, on the other hand, has an amplitude of only a quarter of a minute of arc, much too small to be measured by ordinary telescope-based methods. It can however be quite easily checked by laser signals bounced back from the sets of reflecting prisms left by Apollo 11, 14 and 15 astronauts on the Moon, part of the ALSEP or "Apollo Lunar Surface Experiment Package," or by the set placed there by the Soviet Luna 21 robot lander. The results are in agreement with the theory of such motions, which was extended to cover motions previously neglected as unobservable."
I would be very interested to learn the following:
1. What are the periods of these two modes of oscillations?
2. Are these oscillations damped? If so, what is the damping time constant? Or the data collected cover to short an observation period to make any conclusions about the damping?
I think my claim about physical librations came from the book by Cook, cited at the end of my web page on librations. I am away from the library now, so instead I made a quick check on the web. Here is what I found at
As for the period, I found the following at http://www.astro.amu.edu.pl/html/IAU_Coll/Abstracts/i21.htm
"The Lunar Physical Librations"
Newhall X X, Williams J.G.
Jet Proplusion Laboratory, California Institute of Technology Pasadena, CA 91109-8099
A Fourier analysis of the integrated libration angles produced results for both the forced and free librations. The forced terms arise from external torques and are predictable. There are three free libration modes, each representing a departure from an equilibrium condition. They have known periods, but their amplitudes and phases are not predictable and must be measured.
The results for the forced terms are in general agreement with theory, except that a long-period term was found that cannot be identified with a single, constant period. For the free libration in longitude, theory predicts a period of 1056 days, a value almost identical to that of one of the forced terms. A term having precisely that period was found; its amplitude (a combination of the free and forced librations) is 1.8''.
A latitude libration at 74.59 years was found, with an elliptical amplitude of 3.3'' by 8.2''. Because free librations will normally damp out in a comparatively short time, the presence of these terms suggests some sort of stimulation mechanism."
141. The De-Laval NozzleA dumb question from a retired engineer (ex NERVA program)
"There aint anything for the gas to push against!"
The expanding gas delivers--I suppose-a hydrostatic pressure in all directions, and a component of the force on the nozzle helps push the rocket.
I guess this is a non-problem based on naive physics. None of the propellant has anything to "push against" after all.
ReplyYour question ain't dumb (and didn't NERVA have a De-Laval nozzle?).
You can look at the situation in one of two ways. One, the "bell" of the nozzle looks like a paraboloid, and the exit from the narrow section is near its focus. Hot molecules coming from it can be viewed like rays of light emitted from a flashlight bulb, hitting the flashlight reflector and bouncing off, all into directions parallel to its axis.
OK, there exist many collisions on the way.. so look in another way at The hot gas expands outwards, cools by expansion and loses heat energy. The heat energy goes into expansion velocity, which after hitting the "bell" is all rerouted parallel to the axis. The thrust is exerted against the bell. Is that what you meant?
Of course, rigorous calculations also exist. I'm just trying to make the result plausible.
Response:Yes, the NERVA rocket did have a De Laval nozzle. We even did a thrust measurement with a down-firing reactor, in Nevada air.
You have a very nice analogies and explanations. R P Feynman would applaud you.
142. Why does the space shuttle rotate at take-off?David,
I have been perplexed for years on why the Space Shuttle rotates during liftoff at initial flight stage. I can not find the answer anywhere I have searched or read on this subject. I have asked many people and no one has a clue why this maneuver takes place. I have sifted through reams of documents on space flight and this fact is never discussed. It is probably some simple reason.
I am talking about shortly after liftoff the shuttle rotates. Why?
I am a computer specialist with a large Computer company. I have a pilots license and have been flying for over thirty years. I am knowledgeable about flying and I a an avid flying enthusiast. I would be forever grateful if you could answer this elusive question.
I have not paid that much attention to shuttle lift-offs, and don't know the answer. But I looked on the web, and an answer DOES exist there. See:
143. Cold FusionI am most interested in Cold Fusion. Do there exist any in-depth websites you can recommend for this subject?
ReplyCold fusion is not my area of expertise, but I can recommend a "Wikipedia" article at
which seems to be written by someone knowledgeable. Please look up also
about muon-catalyzed fusion, the only case of cold fusion accepted by all scientists. Muons are unstable nuclear fragments produced in high-energy collisions (almost all cosmic ray particles at sea level are secondary muons from collisions high in the atmosphere) and they behave like extra-heavy electrons. They are equally likely to have a positive or negative electric charge.
Injected into liquid or compressed deuterium--heavy hydrogen, with an extra neutron, combining like hydrogen into 2-atom molecules--negative muons, once they slow down, can replace electrons in deuterium molecules. This draws their two atoms close together, to where the quantum wave-functions of their nuclei overlap (on the quantum level, particles are somewhat spread out), and once this happens, these atoms fuse very rapidly. The fact muons need high-energy collisions to be created, that they decay radioactively after only two microseconds, and that even before that they are likely to be lost to a competing process, all these are reasons the process is "interesting but not useful." Perhaps Pons and Fleischmann (who started the "cold fusion" controversy) believed a similar process occurred in their set-up, catalyzed not by muons but by the palladium in their electrodes
144. What if a neutron star hit our sun?Hello Dr. Stern
I'm dong a bit of research for a Science Fiction novel I'm writing. While I realize that one can take great liberties with such things, I prefer to stay (in some situations) as close to "fact" as I can.
To this end, I was wondering if you could outline, in a very broad sense obviously, what effect a collision between a neutron star and our sun would have.
If this fictional neutron star were moving at a good clip, could it pass right through, ripping our poor sol's guts out? (So to speak!) Would the sun implode? Explode? Would the added mass convert the Neutron star into a black hole? How long would these processes take? Would a near miss pull mass/gas from the sun causing it to cool?
What I need for this to work out (for me, not the poor people in the book) is for the sun to "cool" or even be extinguished, but relatively gradually over several months at least, though a year would be better, to a point that would render the planet very, very cold and uninhabitable.
Thanks for any reply/time you can lend to this project.
Luckily, I do not have to think too hard about your questions, because someone has already done it for me. Try to get hold of the "Scientific American" of November 2002, and look up the article "When Stars Collide." which is also featured (with a snazzy illustration) on the cover. Yes, the collision with a neutron star could lead to a black hole or just to a more massive neutron star. The results to humanity would be devastating and abrupt.
You look for a scenario that would be less abrupt, and I am not really an astrophysicist. On the other hand, science fiction is less constraining than science (and I exclude here fantasy, where anything goes). Namely, in science you have to prove that your claim is probably correct, while in science fiction it is enough to make it hard to prove it wrong (and being just improbable is not disqualifying).
So maybe you could have a star of some sort (may be a white dwarf, which allows an interesting transition) pass close enough--maybe strike the Sun peripherally--to tear off a good fraction of its mass, say 10%-20%, and carry it away. The remaining Sun would be smaller and therefore dimmer. I leave it to you to figure out details.
There exists an old Sci-Fi book "The Black Cloud" by Fred Hoyle. Hoyle's literary merits are less impressive than his scientific ones (at least, some of those), but "Black Cloud" is probably his best (and first). A dark cloud approaches the Sun and envelops it, dimming it temporarily (though ultimately sunlight must get through). But the cloud stops at the Sun and does not go flying on, because (as it turns out), it's intelligent and can control its own motion. I write this to point out that a dark cloud may not be the solution you look for, but read Hoyle anyway, he has some interesting views on the ups and downs of climate in such circumstances, e.g. the role of humidity.
ReplyI can't thank you enough for pointing me in the right direction. Unfortunately a white dwarf is a little too visible for my purposes, so I guess I will have to take some liberties somewhere. Not too much as I grew up on such authors as Robert Heinlein, Ben Bova and Robert L. Forward. (Who wrote a truly remarkable book about life evolving on a Neutron star, "Dragon's Egg") All of whom tried their best to stick to known science where ever possible.
Again, thanks for the help!
145. Why did the Moon appear red?My 3 year old wants to know why the moon looked red on Aug 1. Can you help me answer her question?
Hard to answer without more information! It's like asking a doctor over the telephone why does my back hurt. There can be many answers. The doctor needs to know more.
My guess, though, is that the moon was near the horizon. It's the air, and dust in the air, and water droplets, which change the color, and moonlight coming from near the horizon goes through the greatest thickness of air. The sun also looks red near the horizon--more on some days, when the air has more matter floating in it. Such as smog. little droplets collected by exhaust from cars and machinery.
One thing, for sure: whatever colors the moon is in the air. If you go up into space (or even in a jet plane, which goes above the dense and dirty part of the atmosphere) the moon is always sort of white. And by the way, in ancient Hebrew, the language of the Bible, the name of the moon ("L'vanah") meant "The White One".
146. Centrifuge for Whirling AstronautsDr. Stern
I am involved in planning a human centrifuge for the International Space Station or for exploration missions.
Several folks have proposed using a short-arm human centrifuge. Correct me if I am wrong, but doesn't the g-force get less as you move toward the center of the radius? In other words, if I have an astronaut seat in a chair in a 1.6 meter radius centrifuge, and he gets one g at his feet, but wouldn't his head get something like .5 g?
It would appear to me, based on equations and mathematical models, that the larger the centrifuge, the less dynamic change across someone standing on the centrifuge. If I have a large 10 meter centrifuge, the distance from the center of the radius to my head is much greater, and the change in g from my feet to my head less.
Now here is where things get a little complicated. What is I put two centrifuges together but in opposite directions. For instance lets say I have a centrifuge that is 6 meters in radius turning in a clockwise direction, and then just inside that centrifuge (perhaps on a layer of ball-bearings) I have another centrifuge that is say 5.95 meters in radius and turning in the opposite counterclockwise direction. Would the G force be additive? If they are at the same speed, would the net movement of someone on the dual centrifuge be zero?
ReplyCentrifuges are not my area, and I am not sure to what uses they are put, What you say about centrifuges is correct, but of course persons whirled around in a centrifuge do not keep their legs and heads at different radial distances. Rather, they tend to keep their spine parallel to the axis of rotation--like those riders inside a whirling drum in amusement park rides. On such persons the force is more or less uniform--like the force on an astronaut lying flat on a couch in an accelerating spacecraft during launch.
There may, though, exist Coriolis forces on the flow inside blood vessels. I do not know how important that is, but it may get more pronounced in a short-arm centrifuge. That is your field, not mine.
What you propose, counter-rotating centrifuges inside each other, will not work--the motions will subtract, because they must be considered in the frame of the non-rotating universe. (Handling a counter-rotating circular motion inside a rotating frame takes more math, but presumably leads to the same result). The problem is discussed on my web site
in the last section there "Jet planes above the equator." A jet plane flying along the equator moves in a great circle and therefore reduced the effective gravity inside its cabin. Does it make any difference whether it flies east to west or west to east? Surprisingly, it does. The rotation of the eastward jet is added to that of the Earth, that of the westward jet is subtracted.
Follow-up ResponseThanks so much! That was very helpful. As you are aware, we are looking at both short-arm and long-arm centrifuges for exploration. The Coriolis effect on the neurovestibular system will indeed be a problem with the short-arm centrifuge, but the vehicle torques (to say nothing of the expense, upmass, and volume) would be a problem with the long-arm centrifuge.
I appreciate the insight. I was not sure how to calculate a counter-rotating centrifuge, or whether the forces would be additive or subtracting, or would simply negate one another. I had the picture of Stanley Kubrick's "2001: A Space Odyssey" centrifuge in my mind
147. What Holds Galaxies Together?Dear Sir,
What force holds galaxies together? I mean, Pluto alone is barely clinging on to the edge of our motley bunch of planets... the nearest star is over 4 light years away and I'd bet its not feeling a great deal of attraction to us gravitationally at least, and yet stars are clustered together in galaxies. Is there another force at work here? Something that we can't yet measure? Or can we measure it, and know about it, and the only reason it is not well known is that it hasn't been publicized as much?
Thanks for your time,
ReplyAbout galaxies... ours has a big black hole at its center, and astronomers guess that other galaxies do, too. See "The Black Hole at the Center of our Galaxy" at
However, that might not be the whole story. If a central body held the galaxy, you would expect its stars to rotate like the planets in the solar system--near ones fast, distant ones slowly. Vera Rubin showed that in distant galaxies, the fringes did move more slowly, but much of the galaxy rotated almost like a rigid body. It's a puzzlement, perhaps clue to some "dark mass" we don't see. Stay tuned.
148. View of Earth and Moon from MarsI'm a retired mathematician/computer scientist, now mentoring some students on Science Fair projects at Presentation High School, San Jose CA. I've just been reading your interesting article: "Can an Astronaut on Mars distinguish Earth from its Moon?" at Pumas web site
According to the article, the astronaut could easily see two separate objects, but it does not mention the question: whether the Astronaut could tell which is Earth and which is the Moon.
I guess the main question would be, could he tell which is larger and by approximately what factor?
ReplyThe difference in brightness would be quite evident. The Earth has a radius about 3 times that of the Moon, so if the reflectivity of both bodies is the same, the Earth would be 9 times as bright. (Actually, I think, the Earth's reflectivity is higher.) Since the dimmer object will be seen moving at fixed intervals from one side of the brighter object to the other, it would be clear that here is an orbiting satellite.
149. Appearance of the Moon (1)Hi! I am a Spanish woman and I like to observe the moon. I have always had a doubt about the vision of the moon shape from different places on the earth. I think that when I was in Mexico I saw the moon many times in a horizontal position, when it is not full or new, like a boat in the sea. And I think that from Spain I watch the moon in a different manner, a vertical manner. But I can't remember it well and I'm not sure.
I would like to know if the view of the moon shape is the same from each different point of the earth in the same moment or if it changes depending on the different geographic places from we are observing the moon from the earth. I have studied History of Art and haven't read too much about astronomy; also I haven't found information about this issue.
ReplyYou ask a simple question, but the answer is complicated. Yes, the Moon does appear differently from different locations--both the divisions between light and dark on the Moon (as in "half moon") and the markings on the surface, when the Moon is "full". What determines the difference is more complicated, and depends on an astronomical concept, the ecliptic, discussed in
http://www.phy6.org/stargaze/Secliptc.htm (Meclipt.htm in Spanish).
The exact definition of the ecliptic is the plane in which the Earth orbits the Sun. The Moon and planets also all orbit very close to that plane, so you can imagine all the solar system (including our Moon) to be flat like a sheet of paper, and any object in it is always somewhere on the sheet.
Earth is on the sheet too, and from our viewpoint, we view that sheet sideways, from the edge. It cuts the sphere of the sky into two halves, along a line which circles the sky (and which is also called "ecliptic"--the circle of the ecliptic). Wherever the Sun, Moon or planets may be, they are on that line, or near it. By the way, the star groupings on it are known as the zodiac, and astrologers pay great attention to them.
The motion of the Sun during one day depends mainly on the rotation of the Earth, and always makes an angle of 23.5 degree with the ecliptic. That is not a big angle, so you could imagine the ecliptic is always close to the line followed by the Sun during the day. The Moon will also be somewhere near that line, and the division between dark and light--a straight line with "half moon", or the line connecting the tips of a crescent moon--is always perpendicular to it.
Now the Sun moves differently across the sky, as seen from different places: in Alaska it may just roll around the horizon, never getting far from it, while on the equator it rises from the horizon straight up and goes across the zenith, or nearly so. The ecliptic behaves in a similar way. It follows that in the arctic, the dark-light division is close to perpendicular to the horizon, while near the equator, it is close to parallel.
It is always perpendicular to the direction where the Sun is in the sky, which has to be somewhere else on the ecliptic. If you saw the crescent Moon shaped like a boat above the horizon, you were probably close to the equator (where the Sun comes straight up), and saw it just before sunrise, with the Sun coming up soon after it
The markings on the Moon also change with location. The full Moon always has the same markings, in the same directions, but when you are in Australia, your "up" is in a different direction than in Spain
150. Appearance of the Moon (2): Does it "roll around"?I enjoy your web site very much and appreciate the time and effort you have made. If you have a moment perhaps you could answer a question that is perplexing me greatly.
In the last few months, I have noticed that each night, after the Moon passes its greatest distance from the horizon, it appears to rotate clockwise approx. 90 degrees. That is a LOT of rotation!
In other words, when the Moon is highest above the horizon, the crater Tycho, (for example), is at the 5 o'clock position on the moon's face, (as is normally expected and pictured) but within the space of just several hours it rotates to nearly the 8 O'clock position. No new or additional surface of the moon is revealed, just a rotation of what was already visible. This seemingly odd occurrence is easily visible to anyone who cares to look up.
What is up with such a thing? I don't recall that ever occurring before. Am I just not getting enough sleep?
Initial ReplyDear Cliff I have written about librations of the Moon at
but what you claim to have seen is much, much bigger. Are you sure your eye isn't playing tricks on you, that what you see is a change in Tycho's position relative to the edge of the shadow area on the Moon? That would be my guess.
Follow-up ResponseI am including some photos taken from the morning of July 3 from 0200 to 0630 hours. In them, you can see the rotation that I am speaking of. Note also that it almost appears as though the moon is laying over on it's side as the post apogee transit progresses.
(I hasten to repeat that this effect could be caused by a variety of Earth-based causes and does not mean that I think the moon is "rolling over")
Reply to follow-upYour pictures are convincing! You are certainly right that the Moon is not "rolling over" but that it is something on Earth. My guess on that follows below.
Let me start by noting that the Moon looks different at different latitudes--different in Argentina, say, than in the US. The different orientation is not the Moon's but the observer's--the "up" direction in Argentina is quite different from what it is in the State of Washington, say. We see "the man in the Moon," others may not (in the Orient it's "the rabbit in the Moon").
If the axis of Moon and Earth were both perpendicular to the plane of the Moon's orbit (which is close to the plane of the ecliptic), only latitude would matter, and as the rotation of the Earth carries an observer around, the angle between the vertical and the Moon's axis would (I think) stay the same.
Actually, however, the Earth axis has an inclination of 23.5 degrees, which means that, relative to the Earth-Moon line, we travel up and down in "effective latitude" every day. It is my guess that this is what you observed--that the Moon seems to rotate relative to the "up" direction, although you keep seeing exactly the same area on the Moon. It is just you who is oriented differently.
Does this make sense to you?
ResponseYes, it does make sense. Thank you. It just seemed so odd to me that this apparent radical rotation of features occurred only in the second half of the nightly transit and with such regularity. It may be that I had never noticed such a thing in the past and so when I did finally see it...It was surprising that they could appear to move nearly 90 degrees in a few hours.
151. Altitude of the tip of the Big DipperDear Sir,
I have a question to ask you! I am 16 yrs old and for my astronomy class, I was given an assignment to measure the altitude of the tip of the handle of the Big Dipper. But I am having a little problem with trying to get an answer. Any ideas, suggestions or help from you will be appreciated.
It is a strange assignment, to say the least, because the altitude of the tail star of the Big Dipper changes all the time. As you should know, the sky appears to rotate around a fixed spot, the pole of the sky, which now happens to be very close to the star Polaris.
(If you don't realize that, stop here and read the first 3 sections for "From Stargazers to Starships.")
The tail star of the Big Dipper thus follows a circle around the pole star, taking about 24 hours for a full circuit. Its altitude--its distance from the horizon--will depend on the position in that circle where you find it.
Let me instead suggest you measure two other angles: the altitude of the pole star (which should equal your geographical latitude--see those sections of "Stargazers")--and the angle between the tail star of the Big Dipper and the pole star (see the flag of Alaska in section #1-c, which has both those stars near the top of the flag.
That measurement tells you the radius of the circle in which the tail star moves, while the first measurement tells you the altitude of the center of the circle. Add them to get the highest altitude of the tail star, subtract to get the lowest (if negative, it's then below the horizon), and other information can also be obtained.
You can use a telescope, and from the declination of the tail star you can find the radius, in degrees. A telescope can also tell you the altitude at any time, although you need some calculations. Or you can build a simple cross-staff like the one in section #5-B, except you can use simpler sights, because you will perform only two measurements.
I hope you can do the rest
152. Sudden decompression, 5 miles upDear Dr David Stern
I am a therapist at a home for fourteen abused teenaged boys. As part of their therapy they are given problems for them, as a group, to solve. A recent problem, which has generated a lot of discussion goes as follows -
A rocket ship is returning to Earth after a flight to the moon. Various things happen on the journey forcing the boys to make difficult decisions: like putting a crew member out of the airlock into space. This particular problem ended with the crew, as they reached 5 miles high running out of breathable air. What would happen if the crew opened the air lock at this point?
Would the cabin fill with air, and the ship continue to descend without problems?
Would the ship become uncontrollable and crash?
Would the crew roast in the heat from the engines?
Or would something totally different happen? If you have any enlightening thoughts on the problem, we wait with bated breath to hear them.
What a question! At 5 miles, the air is very cold and has about 1/3 of the seal-level density. You could breathe it, but might get unconscious after a while from lack of oxygen. The question is really, how bad is the air you are breathing at this point. It probably has much more oxygen, but build-up of carbon dioxide may befuddle your mind. I don't know what to recommend, but waiting a little longer seems best.
Will the ship crash? At that point, it is probably no longer guided by its passengers. It may be descending by parachute (like moon-return capsule) or gliding down on stubby wings like the space shuttle. Opening a door will make no difference. Engines are no longer active--fuel is so precious in rocket flight that all of it is used going up.
There exists a similar question, more practical. You are flying in an airliner at 5 miles (a reasonable altitude) and suddenly a poorly secured hatch flies open. Or else, a bomb was carried aboard, and it blows a hole in the side of the cabin. What happens next?
Such things have taken place, and by all accounts they are terrifying. The air whooshes out, and if you are (say) a stewardess standing in the aisle, you may be flung to your death, though passengers strapped into their seats are likely to stay there. A moderately small hole will just cause a very rapid drop in pressure, the air will expand and cool, and since cabin air is fairly humid, the cabin will immediately fill with fog. As emergency instructions tell you, oxygen masks will automatically drop down from bins in the ceiling, helping you breathe more easily, while the pilots frantically descend to denser air.
A real big rupture could damage the integrity of the airplane. If the passenger cabin decompresses much faster than the cargo space below, or vice versa, the floor may buckle, for instance. A Turkish airliner crashed near Paris, many years ago, because a poorly secured cargo bay door came off in flight. None of the 300-odd passengers survived.
Maybe the next problem to your group will be of a happier sort!
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