What is an equatorial radius?
954725
Using the earth as an example, it would be the distance from the earth's axis of rotation (the line between the north and south poles).... to the equator at the surface.
1035386
Astronomers generally use either the equatorial or ecliptical reference frames. The standard reference plane for the equatorial reference frame is the earth's equator. The standard reference plane for the ecliptical reference frame is the plane of the earth's orbit around the Sun. The intersection of these two planes is a line that points in one direction towards the vernal equinox, and in the other direction towards the autumal equinox.\n\nIn both systems, the zero point of longitude (for ecliptic) or Right Ascension (for equatorial) is defined as the direction of the vernal equinox, also called the First point of Aries.
61553
well, the place to start is the periodic table.  uranium's configuration number is [Rn] 5f3 6d1 7s2, which means the outermost electrons are in the n = 7 shell.\nto calculate the radius of 7s electrons requires some seriously ugly quantum mechanics, so i'll just present you with some results:\nhttp://web.mit.edu/3.091/www/pt/pert1.html\nwhich states that the atomic radius is 1.38 angstroms.  other places on the web list 1.45 angstroms and 1.7 angstroms, so about 1.5 seems like a pretty good average.  (an angstrom is 10^-10 m, or 0.1 nanometer)\n\nas for the nuclear radius, this has to be measured from proton-scattering experiments.  typically they are around 1 femtometer, which is 10^-15m or 0.00001 of an angstrom.  (see the wikipedia page for more detail)\n\ncheers!
141519
It is the length divided by the radius
106990
Period is proportional to R^3/2.\nThis is an answer I posted to bombay2pac2000 under Astronomy & Space 8 hours previously, and it applies here as well.\nKepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2 \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*(R^3/2)/SQRT(GM)
559917
First, the speed of the rotation of the earth relative to where you are on the earth.  At the equator, the circumference of the Earth is roughly 24000 miles.  This is convenient, because the Earth rotates about its axis once every 24 hours.  Therefore an hour = 1000 miles of equatorial rotation.  Understand the fastest moving part of the Earth is the equator, since an object on the equator travels 24000 around the earth's Axis every day.  If you are closer to the poles, you are on a "thinner" slice of the earth, where the circumferene may only be 20000miles.  In that case you are moving at 833 miles per hour.  The fact that the Earth moves faster at the Equator than at the poles is responsible for the Coriolus Effect which is really important to people who study meteorology, or who wonder why their toilets get all swirly when flushed.  Why we don't "feel" the rotation is, simply put, because we are rotating exactly the same way as the earth.  In physics, this is called a "frame of reference".  For most intents and purposes, the earth is at "rest" relative to us.  As a human, you make up part of the mass of the earth, and you spin with it in a constant rotation.  It's only when we look to things that are not directly connected to the earth that we really see the effects of it's rotation (ie Tides of water, moments of clouds and weather, etc. . .)  Hope that helps.
3444
The diameter of the earth at the equator is 7,926.41 miles (12,756.32 kilometers).\nBut, if you measure the earth through the poles the diameter is a bit shorter - 7,901 miles (12,715.43 km). This the earth is a tad wider (25 miles / 41 km) than it is tall, giving it a slight bulge at the equator. This shape is known as an ellipsoid or more properly, geoid (earth-like).
665724
The diameter of the earth at the equator in kilometers is approximately 12,756 kilometers.
954337
In the earth:\nthe distance between the object and the poles (or equator)\nwhat hemisphere? The rotation changes (clockwise => N / opposite in the North)
929196
I feel the question should be what is the circumference of earth -\nIt is -\nThe average radius of the Earth is 3,959 miles (6,374 kilometers).\nThe equitorial diameter of the Earth (distance from one side of the Earth to the other at the equator) is about 7,926 miles.\n\nThe ratio of the circumference to the diameter of a circle (circumference/diameter) \nis written as the symbol pi. \nPi is approximately 3.141592.\n3.14159265\n3.1415926535\n\nTherefore, to determine the circumference from the diameter given above:\nequitorial diameter x 3.141592 = equitorial circumference\n| |\n7,926 x 3.141592 = 24,900\n| |\nThe earth has a circumference of approximately 24,900 miles.\n\n\nMore precisely the circumference of the earth (or distance around earth)\nat the equator is 24,902 mi / 40,076 km. \n\nso the answer is 24,902 miles or 40,076 kms.
233745
The mean radius is 6371.0 km.  The radius at any latitude is equal to that times the cosine of the latitude.  The Earth turns once on it's axis in 23 hr 56 min 4.0989 sec, or 23.93447 hrs.\n\nIt takes that long to travel 2(Pi)(6371)(cos(lat)).  Divide that distance by the time above and you're in the ball park (neglecting the difference in polar and equatorial radii).\n\nNote: The figure given for the Earth's rotation is correct according to Astrophysical Data: Planets and Stars.  The 24-hour day is measured from noon to noon - in relation to the Sun.  The Earth kind of rolls around its orbit, so after it has turned thru one rotation it has also moved along its orbit.  It has to turn for another 53.9 sec past the full rotation for the Sun to be back to due south again.
919628
The volume of a spherical shell that has inner radius r and outer radius R is given by:\n\nV = 4/3 * pi * (R^3 - r^3)\n\nThe heights of the boundaries between the various layers of the atmosphere are not constant -- for instance, the top of troposphere is at about 7 km above the Earth's surface at the poles, and ~17 km at the equator.  For the sake of this calculation, I'll assume the following average values for the heights of the various layers (see source), as measured from the surface of the Earth:\n\nTroposphere: 0 - 12 km\nStratosphere: 12km - 50 km\nMesosphere: 50 km - 85 km\nThermosphere: 85 km - 640 km\nExosphere: 640 km - 10,000 km\n\nTo get the radii corresponding to these heights, we need to add the average radius of the Earth (6378 km) to each of these values.  Then we can plug those radii into the equation above to calculate the volume of each layer.  If you do this, you should get:\n\nV(troposphere) = 6.15*10^9 km^3\nV(stratosphere) = 1.96*10^10 km^3\nV(mesosphere) = 1.83*10^10 km^3\nV(thermosphere) = 3.17*10^11 km^3\nV(exosphere) = 1.70*10^13 km^3
502090
Centre is 1.5,2 and the Radius is 3
1061210
If you're not taking earth rotation into account, the value of GMe is 9.8 * (6.38E6)^2, so the square of the radius required is that value divided by .660, the root of which is 2.4584521E7.  Subtracting the earth radius from that gives\n\n1.8204521E7, or about 18 million km.\n\nIf you're considering a point that stays directly above the earth as it rotates, the nswer depends on the latitude and is the solution of a cubic equation.   I'm not going there for you!
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 ..
14152
The circumference of the earth at the equator is 24,901.55 miles (40,075.16 kilometers).\n\nBut, if you measure the earth through the poles the circumference is a bit shorter - 24,859.82 miles (40,008 km). This the earth is a tad wider than it is tall, giving it a slight bulge at the equator. This shape is known as an ellipsoid or more properly, geoid (earth-like).
883944
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.