How did they measure the distance fomr the north pole to the south pole? (In order to find the perfect middle which is marked the equator?) I got smart with my teacher and Iasked, she said I don't know and I said she HAD to know! So she said I HAD to find out how...I can't find it on google or anything! Please help =D
137289
Surveys that are conducted over areas so large that the curvature of the earth gets in the way use a method called Geoditic Surveying or Geodesy.  It's a bit complicated to explain and was something I studied for many months so rather than explain it here you can read up on it from this webpage... http://www.pobonline.com/CDA/Archives/50d9817cac0f6010VgnVCM100000f932a8c0____\n\nNowadays of course, we use satelites which can measure distances to within a metre.
670870
You are right, that because of centripetal acceleration you will weigh a tiny amount less at the equator than at the poles. Try not to think of centripetal acceleration as a force though; what's really going on is that objects which are in motion like to go in a straight line and so it takes some force to make them go round in a circle. So some of the force of gravity is being used to make you go round in a circle at the equator (instead of flying off into space) while at the pole this is not needed.\n\nThis means that the centripetal acceletation at the equator is about 0.03 m/s2 (metres per seconds squared). Compare this to the acceleration due to gravity which is about 10 m/s2 and you can see how tiny an effect this is - you would weigh about 0.3% less at the equator than at the poles!
119943
Why can't we see the flag on the moon?  \nA.    Lets assume that: \nThe height of the flag (a) pole in ft. = 6 \nThe average distance to the Moon (approx) in miles = 221,000 \nWhich means that the distance to the Moon (approx) in feet (b) = 1,166,880,000 \nB.    Given the above: \nThe tangent (t) of the above in radians [tan(a/b)] = 0.0000000051, in radians \nConverted to Degrees [180*(t/pi)] = 0.000000295 degrees \nThis means that a telescope would have to be able to resolve an angle of 2.95 x 10-07 degrees to see the flag left on the moon.\nC.    That means:\nSince one Arc Degree (1/360 of a circle) = 1.000000000 Deg. \nAnd one Arc Minute (1/60 of a degree) = 0.016666667 Deg. \nAnd one Arc Second (1/60 of an arc-minute) = 0.000277778 Deg. \nWe can calculate that:  \nThe Angle to be resolved is: 0.000000295 Deg. \nAnd the Angle to be resolved in Arc Seconds is: 0.001060596 \n\nSo, to see the flag, a telescope must be able to resolve an angle less than an 1/1000th of an arc second.\n\n"So how big of a telescope would that require?", you ask.\nThe Dawes limit calculates how close two objects can be and still be resolved by a telescope. \nThe formula is:  Resolution (in arc seconds) = 4.56/Diameter. \nSo, to determine the required diameter, the formula is:  D = 4.56/Resolution(arc seconds) \n\nDoing the math, that is 4.56/.001060596 arc seconds, to see the flag a telescope would require a mirror 4,299.47 inches wide! That is 358.29 Feet!\nA super-telescope THAT big would have to be in orbit, to eliminate the turbulence of the earth's atmosphere.
339707
The distance to the center of the earth and most planets is greater at the equator than at the poles. This is due to the spinning of the object on its axis.
164803
The first measurement of the equator was done by Eratosthenes of Samos--centuries ago.  He computed it as being roughly 25,000 miles, which is really quite accurate.\n\nEratosthenes was a hotshot at geometry.  He knew that on the spring equinox, the sun cast no shadow at noon in (as I recall it) Syene, Egypt (? on the spelling); in Samos, the shadow at the same time cast a shadow that made an angle of about 6 degrees with the top of the stick.  Using the known distance between the two places, the already accepted fact that the earth was spherical, and a little geometry, he figured out the equitoral diameter, and did a very fine job of it as well.\n\nHe later recomputed the figure, by the way, getting 18,000 miles.  Columbus repeated the calculation, long afterward, and got 19,400.  The Portuguese (the premier mapmakers of Columbus' day) had computed it as about 24,800 or so... They knew the world was round--and bigger than ol' Christopher Columbus thought...  An aside, but a fun one...
3856
I'm not sure about the distance to the moon, but I can explain the rest...\n\nLots of questions here... \n\nFirst, they found out about the curvature of the earth mainly by observing the earth's shadow on the moon during a lunar eclipse. You can easily see the Earth's shadow as curved and not straight.\n\nThe experiment with the sticks, performed by Eratosthenes, was to determine the circumference of the earth, and thus, how big it is.\n\nYou're absolutely right with the argument that the shade produced by a stick on a flat surface would be different than that on a curved surface.\n\nEratosthenes went with the [correct] assumption that the sun was so far away, that its light rays striking the earth would be parallel to each other.\n\nIn truth, this only occurs for an object that is infinitely far away, but the sun is SO far away, that this difference, from two different places on Earth, is negligable.\n\nSo if you have a flat surface, then no matter where you are on the surface, the shadow cast by a stick should be the same. If its noon, and the light rights are parallel, and the Earth is flat, then there should be no shadow no matter where you are on the surface.\n\nNow, Eratosthenes saw that at noon, in two different cities, that in one city, there was no shadow cast by the stick, meaning the sun was directly overhead, and at the same time in the other city, there was a small shadow, and measured the angle the stick made with the shadow. \n\nHe noticed the angle was about 7 degrees, which means its 1/50th of a full circle (360 degrees). Therefore, the two cities were 1/50th the way around the globe. Knowing how far the cities were, he multiplied their distance by 50 and got the distance around the globe. c = 2*pi*r, so he could then calculate the radius of the earth as well!\n\nThat's pretty much it, hoped you understood it. Its really amazing how they could get the size of the earth simply by using two old sticks and a little thought! :)
145023
The ball will oscillate between the North Pole and the South Pole just like a pendulum bob with a time period of about 84 mins, i.e, it will travel to south pole and back to north pole in approx 84 mins.\n\n(It will be executing Simple Harmonic Motion)
431996
That work was much more time consuming and tedious for early explorers than it is for us today.  These sites should help you .... and here are some answers to your questions ...\n\n1.  In the third century B.C., the Greek astronomer and mathematician Eratosthenes estimated the Earth’s size by comparing the shadows cast at two locations at noon. He used his knowledge of triangles, circles, and angles to calculate the Earth’s circumference to within one percent of the actual measurement.\n\nThroughout history, mapmakers have been able to provide detailed maps of small areas, but it wasn't until the 17th century that mapmakers could make accurate measurements over larger areas. At that time, a new technique of measuring distance called triangulation revolutionized mapmaking. To determine the distance to a far-off point, mapmakers first measured the distance between two relatively close points and then measured the angle between each of these two points and the distant point using a transit or a theodolite. Once these angles and the distance between the two baseline points were known, the distance to the far-off point could be calculated using geometry. \n\nOne of the first official uses of triangulation for mapmaking was in 1669 when King Louis XIV of France commissioned Jean Picard and Jean Dominique Cassini to map the coast of France. This first accurate measurement of France showed it to be significantly smaller than earlier maps did. The King is said to have declared, "Your work has cost me a large part of my State!" \n\n2.  In the northern hemisphere, latitude may be derived by measuring the angle made between the North Star and the horizon. Crude instruments to obtain this angle have existed for thousands of years and instruments like the sextant and octant, only slightly less precise than those available to us, were in use.\n\nLatitude may also be calculated by measuring the altitude of the sun, moon, or certain planets and stars above the horizon on known days and reading latitude from tables designed for that purpose. It was not much more difficult for the captains and early explorers to measure these altitudes with a sextant or octant and to calculate latitude using one of the three ephemerae or astronomical almanacs they carried. These contained tables showing the daily position of celestial bodies such as the sun, the moon, and key stars. \n\n3.  Longitudinal observations were prone to error because chronometers did not have quartz batteries to keep them running.  When it ran down, it had to be re-calibrated using local time: using observation of the sun's zenith or local "noon" and then setting the chronometer by estimating the current longitudinal position. Over the course of a trip, the small incremental errors produced by this became larger ones.\n\nThere were other methods of calculating longitude available to early explorers, using astronomical observation. But many astronomical readings had to be acquired over the course of a night in order to obtain sufficiently precise data to determine longitude. It was asking a lot for men exhausted by the rigors of their daily work, to spend three or four hours at night taking sightings of the moon and stars, recording observations, and making calculations by candlelight or firelight. It was only natural that errors would exist in data obtained in this manner. Even the most skilled astronomer or surveyor would have been hard pressed to make highly accurate observations under such circumstances.
648074
Yeah where's angle 1 and 2.\nBut remember the angle opposite the longest side has the greastest measurement.
328890
You have to align the mount with the celstial equator.  Essentially, point the axis of the mount up at the North celestial pole.  rather than try that rather difficult stunt directly, find out your latitude.  The North celestial pole will be due north and have an altitude of your latitude above the northern horizon.  If the mount has dec and RA markings, there ought to be some sort of angle measuring device so you can measure the altitude of the axis of the equitorial mount when the mount is level.\n\nYou will also need built in levels or a plumb bob to make sure your telescope is not tilted when you do this.  Finally, you might want to devise some trick to make sure the north south orientation is correct, like lay out a long string that is aligned properly.  You can't just use a compass to do this either, since the north magnetic pole is not at the geographic north pole. The proper GPS device can help with this. Look up "magnetic declination" for more information on finding true north with a magnetic compass.\n\nBest of luck!  Finally, one word of warning:  a good refractor costs a fortune!  You probably should stick to fairly bright objects, or you might be dissappointed.  Also, look for a local astronomy club or local star parties in the newspaper.  Those guys can help you with learning telescope observing a great deal.
478715
You need radius and distance\n\nEarth radius: 6,372 km\n\nMoon:\nDiameter: 3,472 km\nRadius: 1,736 km\nMean distance: 384,401 km (less radius of Earth = 378029 km)\n\nSun:\n\nDiameter: 1.392×10^6 km\nRadius: 6.96×10^5 km\nMean distance:149.6×10^6 km\n\nBy trigonometry, the angle is twice the inverse tangent of radius divided by the distance.\n\nMoon: 0.009032 radians (0.5175 degrees)\nMoon corrected for radius of Earth:\n0.009184 radians (0.5262 degrees)\nSun: 0.009305 radians (0.5331 degrees)\n\nNote These are mean values.  I calculated from the center of the Earth and the surface by subtracting the radius of Earth.  There are several numbers you can plug in from my sources.  I could not draw a diagram to explain my calculation due to the limits of Y!A.
7412
It depends a lot on how high above the water your eyes are and how tall the distant object is.\n\nThe radius of the Earth at the equator is 6378135 meters.  Assume your eyes are 2 meters above sea-level, and the distant object is at sea-level.\n\nYou have a right triangle, where the longer leg is the radius of the Earth, and the hypotenuse is 6378315 + 2.  The shorter leg is the distance to the horizon.\n\nThe distance to the horizon = √(6378317^2 - 6378315^2) = 5051 meters, or about 3 miles.\n\n\nIf the distant object is 2 meters above sea-level, then it'll disappear when it's another 3 miles past the horizon, or 6 miles from you.
660500
pole\nmagnetic pole, the two nearly opposite ends of the planet where the earth's magnetic intensity is the greatest, as the north and south magnetic poles. For the magnetic north, it is the direction from any point on the earth's surface linking the horizontal component of the magnetic lines of force with the observer and north magnetic pole; it is similar for magnetic south. The north magnetic pole, first located (1831) by English explorer Sir James C. Ross, is now about 78°N and 104°W in the Queen Elizabeth Islands of northern Canada. The south magnetic pole, reached (1909) by English geologists Sir T. W. E. David and Sir Douglas Mawson, is now about 66°S and 139°E on the Adélie Coast of Antarctica. The magnetic poles are not fixed but follow circular paths with diameters of about 100 miles (160 km). Studies of paleomagnetism also indicate that the earth's magnetic field has reversed its polarity many times in the geologic past. The best hypothesis to date for the origin of terrestrial magnetism is the self-exciting dynamo theory, where the earth's magnetic field is generated by the interaction of motion and electrical currents in the earth's liquid outer core.\n\n///\nMagnetic fields are produced by the motion of electrical charges. For example, the magnetic field of a bar magnet results from the motion of negatively charged electrons in the magnet. The origin of the Earth's magnetic field is not completely understood, but is thought to be associated with electrical currents produced by the coupling of convective effects and rotation in the spinning liquid metallic outer core of iron and nickel. This mechanism is termed the dynamo effect. \nRocks that are formed from the molten state contain indicators of the magnetic field at the time of their solidification. The study of such "magnetic fossils" indicates that the Earth's magnetic field reverses itself every million years or so (the north and south magnetic poles switch). This is but one detail of the magnetic field that is not well understood\n\n....\nIn order to explain the fact that ferromagnetic materials with spontaneous magnetisation could exist in the demagnetised state Weiss proposed the concept of magnetic domains. Weiss built on earlier work carried out by Ampère, Weber and Ewing suggesting their existence. The findings of this work revealed that within a domain large numbers of atomic moments are aligned  typically 1012-1018, over a much larger volume than was previously suspected. The magnetisation within the domain is saturated and will always lie in the easy direction of magnetisation when there is no externally applied field. The direction of the domain alignment across a large volume of material is more or less random and hence the magnetisation of a specimen can be zero.\n\nMagnetic domains exist in order to reduce the energy of the system. A uniformly magnetised specimen as shown in figure 5(a) has a large magnetostatic energy associated with it. This is the result of the presence of magnetic free poles at the surface of the specimen generating a demagnetising field, Hd. From the convention adopted for the definition of the magnetic moment for a magnetic dipole the magnetisation within the specimen points from the south pole to the north pole, while the direction of the magnetic field points from north to south. Therefore, the demagnetising field is in opposition to the magnetisation of the specimen. The magnitude of Hd is dependent on the geometry and magnetisation of the specimen. In general if the sample has a high length to diameter ratio (and is magnetised in the long axis) then the demagnetising field and the magnetostatic energy will be low.\n\nThe break up of the magnetisation into two domains as illustrated in figure 5(b) reduces the magnetostatic energy by half. In fact if the magnet breaks down into N domains then the magnetostatic energy is reduced by a factor of 1/N, hence figure 5(c) has a quarter
361922
What are the co-ordinates of the other point? just use the midppint formula for a segment. All the best!\n\nLife's Lovely! Love & Live Life!
602660
Do you have a small scale?  Put a magnet on the scale with the north pole up.  Mount another magnet above that magnet with the north pole down.  Move the top magnet to different heights and records the height and the "weight" on the scale.  Mount the second  magnet with the south pole down and repeat.  Draw a graph comparing height and weight with the north pole  up and the north pole down.  What shapes are the graphs?
59168
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
68322
Polar Diameter: 7900 mi (12720 KM)\nPolar Circumference:  40008 km\n\nSo from the north pole to south pole it is about 20004 KM or about 12430 miles