How we can find the radius of the earth sphere?
437010
its a "sphere" so to speak...so it does have a radius..\n\nthe problem is that since it spins, its a bit wider at the equator..so you need to specify the radius at that location or the polar radius...\n\nmost ppl care about the equatorial radius.\n\nthe Earth's equatorial radius, or semi-major axis, is the distance from its centre to the equator and equals 6,378.135 km (≈3,963.189 mi; ≈3,443.917 nmi).
430210
Volume of a shphere  = 4/3 * pi * R^3\n\nwhere pi = 3.142 (approx)\nand R is the radius
544990
Well as far as a formula i can't help you there. i just happen to own a program that allows me to draw perfect shapes of any size and diminsion them accordingly (auto CADD). and at the top of your 300 ft house, or where ever it is that your measuring from the horizon at sea level is a tad over 21 miles away (21.20389 to be analy exact). first you must know the the earths circumfrance (i know i can't spell) which is 20,890,637 ft, next you have to know the distance you are away from the earth (which you do, 300 ft). now you have to know the formula for finding a line that intersects your hieght and crosses tangentally across the earths surface. know you have your triangle and apply the Pythagorean theorem. you have an answer.
1093481
The Earth rotates 360 degrees (2 pi radians) in 23 hours, 56 minutes, and 4.09054 seconds.  In other words, its angular velocity is (2 pi)/(86,164.09054 sec).  You could google the answer this one, since the Earth's angular velocity is a well known constant (7.292115 x 10^-5 rad/sec).\n\nThe equatorial radius is 6,378.137 km.  Your distance from the Earth's rotational pole can be found by multiplying the equatorial radius by the cosine of your latitude.  For example, if you're located at 38 degrees N latitude, your distance from the rotational axis is 6,091.59 km.\n\nYour linear speed is found by multiplying your distance from the Earth's rotational axis by the angular velocity.  For example, at 38 deg N latitude, your velocity would be 444.2 meters/sec.\n\nYou have to know the latitude a person is located at in order to calculate their velocity.  You can find the approximate latitude on any globe.  An online map or a Gazeteer could give you more exact coordinates.  Googling the coordinates for the local airport would also get you a very accurate latitude for your city.
136253
You're already on a website.  What is it with people these days?  Look in your book.  \nInteresting though: the derivative of the area of a circle is equal to the circumference of it.  Works the same for a sphere too.
78436
The diameter of the earth at the equator is 7,926.41 miles
152159
At the Equator, Earth's radius is --\n6,378 km\n3,963 miles\n6,378,000 meters
1059357
Thats why they asked theoretically.\n\nsame radius as the solar system\n1267650600228229401496703205376 kms\n\nOf which you allready knew, otherwise u would not be asking
329806
The distance from the center to a point on the sphere is the radius.  Find that by the distance formula, then double it to get the diameter.\n\n(The distance formula says to subtract the two x coordinates and square this difference.  Then do the same for the two y coordinates and z-coordinates.  Add these 3 answers and find the square root of the sum.  That's the distance.)
905767
You are making it needlessly hard on yourself.  The planet has the same mass as the earth, and someone on it weight the same as on earth, therefore the surface of the planet is the same distance from its center as the earth radius, i.e. this planet has the same dimension as earth.\nSo, the radius of the planet is 6,372.797 km, same as earth's.\nYou don't need to compute anything.  The orbital velocity of the planet has nothing to do with it.\nIf the weight was different, you only need to scale it remembering that the weight would vary as an inverse of the square of the radius.
101953
There's a simple way to go about this. The answer is reasonably accurate too.\n\nNow, we know that the formula for the volume of a sphere is 4/3 times pi times the cube of earth's radius.\n\nGiven the value of pi as 3.14 and the earth's radius as 6400 km, and also assuming that the earth is a perfect sphere, we get the volume as 1.09806622 * 10^12 cubic kilometers.
47332
Find the radius from the diameter by dividing it by 2.  So r = 44000 miles\n\nSurface area of any sphere = 4 * pi * r * r\n\npi = 22/7\n \nPlug in the values and you will get the surface area of Jupiter.
931645
first answer was close to right.\n\nthere is one qualification that seems need.\n(do this all in meters for easyness of calculation)\n\nR = radius of earth = 6400 km = 6.4 * 10^6 meters\nd = distance to horizon in meters\nh = height in meters\n\nSimilar triangles:\n\nR / d  = d / 2h\n\nd^2 = 2Rh\n\nd = sqrt (2Rh)\n\n....  works pretty well for small "h"\n\neg 2 meters height =>\n\nd = sqrt (2 * 6.4 * 10^6 * 2 ) = sqrt (25.6 * 10^6) \n=5.06 km\n\n[[ online calculator gives 5.5 km ]\n\nmine's a LOT easier to use ...\n\nI use "2h" instead of "h" for similar triangles\nbecause of discrepancy of right angle at the\nradius from eearth-center to observer and\nr-a from earth-center to distant object.\n[intersection happens "h" below earth surface]\n\nthat sounds confusing but I can't draw a picture\n\nmy eqn is easiest possible tho ..
350355
Earth is an oblate spheroid, resembling a sphere flattened at the poles.\n\nThe volume of an oblate spheroid is\n\n    V = 4/3 X 3.14 X a^2 X b\n\nwhere π is the mathematical constant pi, and a and b are the equatorial and polar radii.\n\nSubstituting a = 6,378,135 m and b = 6,356,750 m\n\nThe Huge EarthThe mean radius of the earth is approximately 6.4 million meters (exact = 6.37 x 106 m). Its volume is then:\n\n(4/3) x 3.14 x 64000003 m3\n\nThis comes to 1,097,509,500,000,000,000,000 cubic meters.
716401
In classical times, educated people pretty much understood that the Earth is a sphere.\n\n"Eratosthenes (276 BC - 194 BC) estimated Earth's circumference around 240 BC. He had heard about a place in Egypt where the Sun was directly overhead at the summer solstice and used geometry to come up with a circumference of 250,000 stades. This estimate astonishes some modern writers, as it is within 2% of the modern value of the equatorial circumference, 40,075 kilometres. However, the length of a 'stade' is not precisely known; Eratosthenes' figure falls short if we do not use a fairly generous estimate for this length." (http://en.wikipedia.org/wiki/Spherical_Earth)
809750
Ratio of what?\nSurface area?\nVolume?\n\nI will go with volume.\n\nassume radius of sphere is = 1\n\nsphere = 4/3pi\ncylinder = 2pi\ncone = 2/3 pi\n\neat the pi's (get rid of them) and multiply what's left by 1.5\n\nSphere:Cylinder:Cone = 2:3:1\n\nI hope that helps
820928
I am guessing that you want to know how big one of the sections are, so that's what I will try to determine.\n\nThe first step is to find the surface area of the hemisphere with a radius of 18 meters:\nThe surface area of a sphere, with r as the radius, is: A = (4)(π)(r²). Since this is a hemisphere, we can divide the expression equal to the surface area by 2, leaving us: A = (2)(π)(r²).\nA = 2(π)(18²) = 2(π)(324) = 648(π)\n\nSince we know that the surface area of the hemisphere is 648π m², we can divide this by 12 to find out the surface area of one of the sections:\n648π / 12 = 54π meters² ≈ 169.65 meters²