Get a Straight Answer

    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

    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).

    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.

    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.

    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?

Reply

    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 for your questions, you may find the required information at

    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?

Reply

    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.

    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?"

Reply

    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.

    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.

Reply

    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

    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?

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    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!

    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

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    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.

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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.

    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"?

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    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?

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    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

Solar time = standard time + 4 (T-L-E) minutes[1]

Where

Standard time = local time minus one hour in summer.
T = longitude of time zone (in degrees).[2]
L = longitude of site (in degrees).[3]
E = ephemeris (in degrees). See table.[4]

Ephemeris tableM

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.

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    The 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.

[Written about the planned first shuttle launch on 13 July 2005, after years of delay caused by the Columbia disaster. That launch was later delayed]

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?

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    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.